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4.3 – Solving Equilibrium Problems

We know that at equilibrium, the value of the reaction quotient of any reaction is equal to its equilibrium constant. Thus, we can use the mathematical expression for Q to determine a number of quantities associated with a reaction at equilibrium or approaching equilibrium. While we have learned to identify in which direction a reaction will shift to reach equilibrium, we want to extend that understanding to quantitative calculations. We do so by evaluating the ways that the concentrations of products and reactants change as a reaction approaches equilibrium, keeping in mind the stoichiometric ratios of the reaction. This algebraic approach to equilibrium calculations will be explored in this section.

Relative Changes in Concentration

Changes in concentrations or pressures of reactants and products occur as a reaction system approaches equilibrium. In this section we will see that we can relate these changes to each other using the coefficients in the balanced chemical equation describing the system. We use the decomposition of ammonia as an example.

On heating, ammonia reversibly decomposes into nitrogen and hydrogen according to this equation:

2 NH 3 ( g) ⇌ N 2 (g) + 3 H 2 (g)

If a sample of ammonia decomposes in a closed system and the concentration of N 2 increases by 0.11 mol/L, the change in the N 2 concentration, Δ[N 2 ] = [N 2 ] f – [N 2 ] i , is 0.11 M. The change is positive because the concentration of N 2 increases .

The change in the H 2 concentration, Δ[H 2 ], is also positive—the concentration of H 2 increases as ammonia decomposes. The chemical equation tells us that the change in the concentration of H 2 is three times the change in the concentration of N 2 because for each mole of N 2 produced, 3 moles of H 2 are produced.

[H 2 ] = 3 × [N 2 ]

= 3 × 0.11 mol/L = 0.33 mol/L

The change in concentration of NH 3 , Δ[NH 3 ], is twice that of Δ[N 2 ]; the equation indicates that 2 moles of NH 3 must decompose for each mole of N 2 formed. However, the change in the NH 3 concentration is negative because the concentration of ammonia decreases as it decomposes.

∆[NH 3 ] = – 2 × ∆[N 2 ] = – 2 × 0.11 mol/L = – 0.22 mol/L

We can relate these relationships directly to the coefficients in the equation

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Note that all the changes on one side of the arrows are of the same sign and that all the changes on the other side of the arrows are of the opposite sign.

If we did not know the magnitude of the change in the concentration of N 2 , we could represent it by the symbol + x .

The changes in the other concentrations would then be represented as:

∆[H 2 ] = 3 × ∆[N 2 ] = + 3x

∆[NH 3 ] = – 2 × ∆[N 2 ] = – 2x

The coefficients in the Δ terms are identical to those in the balanced equation for the reaction.

The simplest way for us to find the coefficients for the concentration changes in any reaction is to use the coefficients in the balanced chemical equation. The sign of the coefficient is positive when the concentration increases; it is negative when the concentration decreases.

Example 4.3.1 – Determining Relative Changes in Concentration 

Complete the changes in concentrations for each of the following reactions.

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Check Your Learning 4.3.1 – Determining Relative Changes in Concentration 

Complete the changes in concentrations for each of the following reactions:

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          (a) + 2 x , + x , − 2 x; (b) + x , − 2 x; (c) + 4 x , + 7 x , − 4 x , − 6 x or − 4 x , − 7 x , + 4 x , + 6 x

Calculations Involving Equilibrium Concentrations or Pressures

 Because the value of the reaction quotient of any reaction at equilibrium is equal to its equilibrium constant, we can use the mathematical expression for Q to determine a number of quantities associated with a reaction at equilibrium. It may help if we keep in mind that Q = K (at equilibrium) in all of these situations and that there are only two basic types of equilibrium problems:

  • Calculation of an equilibrium constant. If concentrations/partial pressures of reactants and products at equilibrium are known, the value of the equilibrium constant for the reaction can be calculated.
  • C alculation of  equilibrium concentrations/partial pressures. If the value of the equilibrium constant and all of the equilibrium concentrations/pressures, except one, are known, the remaining unknown can be calculated. In addition, if the value of the equilibrium constant and a set of concentrations or pressures of reactants and products that are not at equilibrium are known, the quantity at equilibrium can be calculated.

In the following discussion, we will examine examples of equilibrium calculations involving solutes and values of K in concentration units ( K C ). However, please note that the problem-solving procedures equally hold for reactions involving gases and values of K in pressure units ( K P ).

Calculation of an Equilibrium Constant

In order to calculate an equilibrium constant, enough information must be available to determine the equilibrium concentrations of all reactants and products. Armed with the concentrations, we can solve the equation for K , as it will be the only unknown.

In the previous section, we learned how to determine the equilibrium constant of a reaction if we know the concentrations of reactants and products at equilibrium. The following example shows how to use the stoichiometry of the reaction and a combination of initial concentrations and equilibrium concentrations to determine an equilibrium constant. This technique, commonly called an ICE chart – for Initial, Change, and Equilibrium – will be helpful in solving many equilibrium problems. A chart is generated beginning with the equilibrium reaction in question. The initial concentrations of the reactants and products are provided in the first row of the ICE table (these essentially time-zero concentrations that assume no reaction has taken place). The next row of the table contains the changes in concentrations that result when the reaction proceeds toward equilibrium (don’t forget to account for the reaction stoichiometry). The last row contains the concentrations once equilibrium has been reached.

Example 4.3.2 – Calculation of an Equilibrium Constant – 1

Iodine molecules react reversibly with iodide ions to produce triiodide ions.

I 2 ( aq ) + I – ( aq ) ⇌ I 3 – ( aq )

If a solution with the concentrations of I 2 and I − both equal to 1.000 × 10 −3 mol/L before reaction gives an equilibrium concentration of I 2 of 6.61 × 10 −4 mol/L, what is the equilibrium constant for the reaction?

We will begin this problem by calculating the changes in concentration as the system goes to equilibrium. Then we determine the equilibrium concentrations and, finally, the equilibrium constant. First, we set up a table with the initial concentrations, the changes in concentrations, and the equilibrium concentrations using − x as the change in concentration of I 2 .

I 2 +I − ⇌ I 3 −

Since the equilibrium concentration of I 2 is given, we can solve for x . At equilibrium the concentration of I 2 is  6.61 × 10 −4 M, therefore:

1.000 × 10 -3 – x = 6.61 × 10 -4

x = 1.000 × 10 -3 – 6.61 × 10 -4

x = 3.39 × 10 -4 mol/L

Now we can fill in the table with the concentrations at equilibrium.

We now calculate the value of the equilibrium constant.

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This value for K makes sense – it is close to 1, indicating that, at equilibrium, the system will contain comparable amounts of reactants and products. This is true when we look at the equilibrium concentration in the ICE table, or even visualize the close proximity of concentration curves (in a graph) of species when equilibrium is reached (Figure 4.3.1).

image

Figure 4.3.1. With a value of K relatively close to 1, the concentrations of reactants and products approach each other when the system moves towards equilibrium.

Check Your Learning 4.3.2 – Calculation of an Equilibrium Constant – 1

Ethanol and acetic acid react to form water and ethyl acetate, the solvent responsible for the odor of some nail polish removers:

C 2 H 5 OH + CH 3 CO 2 H  ⇌  CH 3 CO 2 C 2 H 5 + H 2 O

When 1.00 mol each of C 2 H 5 OH and CH 3 CO 2 H are allowed to react in 1 L of the solvent dioxane, equilibrium is established when 0.13 mol of each of the reactants remains. Calculate the equilibrium constant for the reaction. (Note: Water is not a solvent in this reaction.)

Example 4.3.3 – Calculation of an Equilibrium Constant – 2

A 1.00 mol sample of NOCl was placed in a 2.00 L reactor and heated to 227°C until the system reached equilibrium. The contents of the reactor were then analyzed and found to contain 0.056 mol of Cl 2 . Calculate K C at this temperature. The equation for the decomposition of NOCl to NO and Cl 2 is as follows:

2 NOCl ( g) ⇌ 2 NO (g) + Cl 2 (g)

The first step in any such problem is to balance the chemical equation for the reaction (if it is not already balanced) and use it to derive the equilibrium constant expression. In this case, the equation is already balanced, and the equilibrium constant expression is as follows:

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Initially, the system contains 1.00 mol of NOCl in a 2.00 L container. Thus [NOCl] i = 1.00 mol/2.00 L = 0.500 mol/L. The initial concentrations of NO and Cl 2 are 0 mol/L because initially no products are present. Moreover, we are told that at equilibrium the system contains 0.056 mol of Cl 2 in a 2.00 L container, so [Cl 2 ] f = 0.056 mol/2.00 L = 0.028 mol/L. We insert these values into the following table:

We use the stoichiometric relationships given in the balanced chemical equation to find the change in the concentration of Cl 2 , the substance for which initial and final concentrations are known:

Δ[Cl 2 ] = [0.028 mol/L (final) − 0.00 mol/L (initial)] = + 0.028 mol/L

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Similarly, 2 mol of NOCl are consumed for every 1 mol of Cl 2 produced, so the change in the NOCl concentration is as follows:

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We sum the numbers in the [NOCl] and [NO] columns to obtain the final concentrations of NO and NOCl:

[NO] f = 0.000 M + 0.056 M = 0.056 M

[NOCl] f = 0.500 M + (−0.056 M) = 0.444 M

We can now complete the table:

We can now calculate the equilibrium constant for the reaction:

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Check Your Learning 4.3.3 – Calculation of an Equilibrium Constant – 2

The German chemist Fritz Haber (1868–1934; Nobel Prize in Chemistry 1918) was able to synthesize ammonia (NH 3 ) by reacting 0.1248 M H 2 and 0.0416 M N 2 at about 500°C (Figure 4.3.2). At equilibrium, the mixture contained 0.00272 M NH 3 . What is K C for the reaction N 2 + 3H 2 ⇌ 2NH 3 at this temperature? What is the value of K P ?

image

Figure 4.3.2. The original laboratory apparatus designed by Fritz Haber and Robert Le Rossignol in 1908 for synthesizing ammonia from its elements. A metal catalyst bed, where ammonia was produced, is in the large cylinder at the left. The Haber-Bosch process used for the industrial production of ammonia uses essentially the same process and components but on a much larger scale. Unfortunately, Haber’s process enabled Germany to prolong World War I when German supplies of nitrogen compounds, which were used for explosives, had been exhausted in 1914.

K C = 0.105; K P = 2.61 × 10 −5

Calculation of Equilibrium Concentration(s)

In these types of equilibrium problems, if we know the equilibrium constant for a reaction and know the concentrations at equilibrium of all reactants and products except one, we can calculate the missing concentration.

Example 4.3.4 – Calculation of a Missing Equilibrium Concentration

Nitrogen oxides are air pollutants produced by the reaction of nitrogen and oxygen at high temperatures. At 2000°C, the value of the equilibrium constant K C for the reaction, N 2 (g) + O 2 ( g) ⇌ 2 NO (g) , is 4.1 × 10 −4 . Calculate the equilibrium concentration of NO (g) in air at 1.00 atm pressure and 2000°C. The equilibrium concentrations of N 2 and O 2 at this pressure and temperature are 0.036 M and 0.0089 M, respectively.

We are given all of the equilibrium concentrations except that of NO. Thus, we can solve for the missing equilibrium concentration by rearranging the equation for the equilibrium constant.

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Thus [NO] is 3.6 × 10 −4 mol/L at equilibrium under these conditions.

We can check our answer by substituting all equilibrium concentrations into the expression for the reaction quotient, Q C , to see whether it is equal to the equilibrium constant, and thus confirm that the system is indeed at equilibrium.

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The answer checks; our calculated value gives the equilibrium constant within the error associated with the significant figures in the problem.

Check Your Learning 4.3.4 – Calculation of a Missing Equilibrium Concentration

The equilibrium constant for the reaction of nitrogen and hydrogen to produce ammonia at a certain temperature is 6.00 × 10 −2 . Calculate the equilibrium concentration of ammonia if the equilibrium concentrations of nitrogen and hydrogen are 4.26 M and 2.09 M, respectively.

In another scenario, if we know the equilibrium constant for a reaction and a set of concentrations of reactants and products that are not at equilibrium , we can calculate the changes in concentrations as the system comes to equilibrium, as well as the new concentrations at equilibrium. The typical procedure can be summarized in four steps.

   1. Determine the direction the reaction proceeds to come to equilibrium.

         a. Write a balanced chemical equation for the reaction.

         b. If the direction in which the reaction must proceed to reach equilibrium is

not readily obvious, calculate Q from the initial values and compare it to K to

determine the direction of change.

   2. Determine the relative changes needed to reach equilibrium, then write the

equilibrium concentrations in terms of these changes.

         a. Define the changes in the initial concentrations that are needed for the

reaction to reach equilibrium. Generally, we represent the smallest change

with the symbol x and express the other changes in terms of the smallest

         b. Define missing equilibrium concentrations in terms of the initial

concentrations and the changes in concentration determined in (a).

   3. Solve for the change and the equilibrium concentrations.

         a. Substitute the equilibrium concentrations into the expression for the

equilibrium constant, solve for x , and check any assumptions used to find x .

         b. Calculate the equilibrium concentrations.

   4. Check the arithmetic.

   5. Check the calculated equilibrium concentrations by substituting them into the

equilibrium expression and determining whether they give the equilibrium

Sometimes a particular step may differ from problem to problem—it may be more complex in some problems and less complex in others. However, every calculation of equilibrium concentrations from a set of initial concentrations will involve these steps.

In solving equilibrium problems that involve changes in concentration, it is again very convenient to set up an ICE table.

Example 4.3.5 – Calculation of Concentration Changes as a Reaction Goes to Equilibrium

Under certain conditions, the equilibrium constant K C for the decomposition of PCl 5 (g) into PCl 3 (g) and Cl 2 (g) is 0.0211. What are the equilibrium concentrations of PCl 5 , PCl 3 , and Cl 2 if the initial concentration of PCl 5 was 1.00 M?

Use the stepwise process described earlier.

1. Determine the direction the reaction proceeds.

The balanced equation for the decomposition of PCl 5 is

PCl 5 ( g) ⇌ PCl 3 (g) + Cl 2 (g)

Because we have no products initially, Q = 0 and the reaction must proceed to the right (towards products).

2. Determine the relative changes needed to reach equilibrium, then write the equilibrium concentrations in terms of these changes.

Let us represent the increase in concentration of PCl 3 by the symbol x . The other changes may be written in terms of x by considering the coefficients in the chemical equation.

– x              + x             + x

The changes in concentration and the expressions for the equilibrium concentrations are:

PCl 5 ⇌ PCl 3 + Cl 2

3. Solve for the change and the equilibrium concentrations.

Substituting the equilibrium concentrations into the equilibrium constant equation gives

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This equation contains only one variable, x , the change in concentration. We can write the equation as a quadratic equation and solve for x using the quadratic formula.

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Appendix C   shows us an equation of the form ax 2 + bx + c = 0 can be rearranged to solve for x :

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In this case, a = 1, b = 0.0211, and c = −0.0211. Substituting the appropriate values for a , b , and c yields:

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In this case, the second solution (−0.156) is physically impossible because we know the change must be a positive number (otherwise we would end up with negative values for concentrations of the products). Thus, x = 0.135 M.

The equilibrium concentrations are

[PCl 5 ] = 1.00 – 0.135 = 0.87 M

[PCl 3 ] = x = 0.135 M

[Cl 2 ] = x = 0.135 M

4. Check the arithmetic.

Substitution into the expression for K c (to check the calculation) gives

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The equilibrium constant calculated from the equilibrium concentrations is equal to the value of K c given in the problem (when rounded to the proper number of significant figures). Thus, the calculated equilibrium concentrations are confirmed.

Check Your Learning 4.3.5 – Calculation of Concentration Changes as a Reaction Goes to Equilibrium

Acetic acid, CH 3 CO 2 H, reacts with ethanol, C 2 H 5 OH, to form water and ethyl acetate, CH 3 CO 2 C 2 H 5 .

CH 3 CO 2 H + C 2 H 5 OH ⇌ CH 3 CO 2 C 2 H 5 + H 2 O

The equilibrium constant for this reaction at a certain temperature, using dioxane as a solvent, is 4.0. What are the equilibrium concentrations when 0.15 mol CH 3 CO 2 H, 0.15 mol C 2 H 5 OH, 0.40 mol CH 3 CO 2 C 2 H 5 , and 0.40 mol H 2 O are mixed in enough dioxane solvent to make 1.0 L of solution?

[CH 3 CO 2 H] = 0.36 M, [C 2 H 5 OH] = 0.36 M, [CH 3 CO 2 C 2 H 5 ] = 0.17 M, [H 2 O] = 0.17 M

Check Your Learning 4.3.6 – Calculation of Concentration Changes as a Reaction Goes to Equilibrium

A 1.00-L flask is filled with 1.00 moles of H 2 and 2.00 moles of I 2 . The value of the equilibrium constant K C for the reaction of hydrogen and iodine reacting to form hydrogen iodide is 50.5 under the given conditions. What are the equilibrium concentrations of H 2 , I 2 , and HI in moles/L?

H 2 (g) + I 2 ( g) ⇌ 2 HI (g)

[H 2 ] = 0.06 M, [I 2 ] = 1.06 M, [HI] = 1.88 M

Now let’s consider another example where we can utilize a square root shortcut method to facilitate problem-solving. If we find that the fractional term consisting of reactants (denominator) and products (numerator) has perfect squares, we can take the square root of both sides when solving for x .

Example 4.3.6 – Concentration Changes as a Reaction Goes to Equilibrium – Square Root Shortcut

The water–gas shift reaction is important in several chemical processes, such as the production of H 2 for fuel cells. This reaction can be written as follows:

H 2 (g) + CO 2 ( g) ⇌ H 2 O (g) + CO (g)

K C = 0.106 at 700 K. If a mixture of gases that initially contains 0.0150 M H 2 and 0.0150 M CO 2 is allowed to equilibrate at 700 K, what are the final concentrations of all substances present?

The initial concentrations of the reactants are [H 2 ] i = [CO 2 ] i = 0.0150 M. Just as before, we will focus on the change in the concentrations of the various substances between the initial and final states. If we define the change in the concentration of H 2 O as x , then Δ[H 2 O] = + x . We can use the stoichiometry of the reaction to express the changes in the concentrations of the other substances in terms of x . For example, 1 mol of CO is produced for every 1 mol of H 2 O, so the change in the CO concentration can be expressed as Δ[CO] = + x . Similarly, for every 1 mol of H 2 O produced, 1 mol each of H 2 and CO 2 are consumed, so the change in the concentration of the reactants is Δ[H 2 ] = Δ[CO 2 ] = − x . We enter the values in the following table and calculate the final concentrations.

We can now use the equilibrium equation and the given K to solve for x :

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We could solve this equation with the quadratic formula, but it is far easier to solve for x by recognizing that the left side of the equation is a perfect square; that is,

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Taking the square root of both sides of this equation yields,

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The final concentrations of all species in the reaction mixture are as follows:

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We can check our work by inserting the calculated values back into the equilibrium constant expression:

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To two significant figures, this K C is the same as the value given in the problem, so our answer is confirmed.

Check Your Learning 4.3.7 – Concentration Changes as a Reaction Goes to Equilibrium – Square Root Shortcut

Hydrogen gas reacts with iodine vapor to give hydrogen iodide according to the following chemical equation:

K C = 54.0 at 425°C. If 0.172 M H 2 and I 2 are injected into a reactor and maintained at 425°C until the system equilibrates, what is the final concentration of each substance in the reaction mixture?

[HI] f = 0.270 M; [H 2 ] f = [I 2 ] f = 0.037 M

Sometimes it is possible to use chemical insight to find solutions to equilibrium problems without actually solving a quadratic (or more complicated) equation. First, however, it is useful to verify that equilibrium can be obtained starting from two extremes: all (or mostly) reactants and all (or mostly) products .

Consider the ionization of 0.150 M HA, a weak acid.

HA ( aq ) + H 2 O ( l ) ⇌ H 3 O + ( aq ) + A – ( aq )          K C =6.80×10 -4

The most obvious way to determine the equilibrium concentrations would be to start in a system containing only reactants. This could be called the “all reactant” starting point. Using x for the amount of acid ionized at equilibrium, this is the ICE table and solution.

HA ( aq ) + H 2 O ( l ) ⇌ H 3 O + ( aq ) + A – ( aq )

Setting up and solving the quadratic equation gives

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Using the positive (physical) root, the equilibrium concentrations are

[HA] = 0.150 – x = 0.140 M

[H 3 O + ] = [A – ] = x = 0.00977 M

A less obvious way to solve the problem would be to assume all the HA ionizes first, then the system comes to equilibrium. This could be called the “all product” starting point. Assuming all of the HA ionizes gives

[HA] = 0.150 – 0.150 = 0 M

[H 3 O + ] = 0 + 0.150 = 0.150 M

[A – ] = 0 + 0.150 = 0.150 M

Using these as initial concentrations and “ y ” to represent the concentration of HA at equilibrium, this is the ICE table for this starting point.

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Retain a few extra significant figures to minimize rounding problems.

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Rounding each solution to three significant figures gives

y = 0.160 M        or        y = 0.140 M

Using the physically significant root (0.140 M) gives the equilibrium concentrations as

[HA] = y = 0.140 M

[H 3 O + ] = 0.150 – y = 0.010 M

[A – ] = 0.150 – y = 0.010 M

Thus, the two approaches give the same results (to three decimal places ), and show that both starting points lead to the same equilibrium conditions (Figure 4.3.3). The “all reactant” starting point resulted in a relatively small change ( x ) because the system was close to equilibrium, while the “all product” starting point had a relatively large change ( y ) that was nearly the size of the initial concentrations. Notice that the graph in part (a) of Figure 4.3.3 experiences little change in concentration; hence, it can be said that a system that starts “close” to equilibrium will require only a ”small” change in conditions ( x ) to reach equilibrium.

image

Figure 4.3.3. Regardless of wherever you start, whether it be with 100% reactants in (a) or 100% products in (b), we end up at the same equilibrium point regardless. (a) The change in the concentrations of reactants and products is depicted as the HA ( aq ) ⇌ H + ( aq ) + A – ( aq ) reaction approaches equilibrium, when going from an “all reactant” starting point. (b) The change in concentrations of reactants and products is depicted as the reaction HA ( aq ) ⇌ H + ( aq ) + A – ( aq ) approaches equilibrium, when going from an “all product” starting point.

Recall that a small value of K means that very little of the reactants form products and a large K means that most of the reactants form products. If the system can be arranged so it starts “close” to equilibrium, then if the change ( x ) is small compared to any initial concentrations, it can be neglected. The following two examples demonstrate this.

Example 4.3.7 – Approximate Solution Starting Close to Equilibrium

What are the concentrations at equilibrium of a 0.15 M solution of HCN?

HCN ( aq ) + H 2 O (l) ⇌ H 3 O + ( aq ) + CN – ( aq )               K = 4.9 x 10 -10

Using “ x ” to represent the concentration of each product at equilibrium gives this ICE table.

HCN ( aq ) + H 2 O (l) ⇌ H 3 O + ( aq ) + CN – ( aq )

The exact solution may be obtained using the quadratic formula with

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Thus [H 3 O + ] = [CN – ] = x = 8.6 × 10 –6 M and [HCN] = 0.15 – x = 0.15 M.

In this case, chemical intuition can provide a simpler solution. From the equilibrium constant and the initial conditions, x must be small compared to 0.15 M. More formally, if x ≪ 0.15, then 0.15 – x ≈ 0.15 (Figure 4.3.4 visually demonstrates this).

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Figure 4.3.4. Concentrations of the reactant and products are shown initially and at equilibrium for the following reaction: HCN ( aq ) +H 2 O (l ) ⇌ H 3 O + ( aq ) + CN – ( aq ) . The reaction starts with only HCN ( aq ) , but even at equilibrium, you can tell that the relative amounts of H + and CN – are so little that they are negligible – there is practically still only reagent which demonstrates the extremely small K value of the reaction and confirms the validity of the assumption that 0.15 – x ≈ 0.15.

If this assumption is true, then it simplifies obtaining x

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In this example, solving the exact (quadratic) equation and using approximations gave the same result to two significant figures. While most of the time the approximation is a bit different from the exact solution, as long as the error is less than 5%, the approximate solution is considered valid. In this problem, the 5% applies to (0.15 – x ) ≈ 0.15 M, so if

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is less than 5%, as it is in this case, the assumption is valid. The approximate solution is thus a valid solution.

Check Your Learning 4.3.8 – Approximate Solution Starting Close to Equilibrium

What are the equilibrium concentrations in a 0.25 M NH 3 solution?

NH 3 ( aq ) + H 2 O (l ) ⇌ NH 4 + ( aq ) + OH – ( aq )                           K = 1.8 x 10 -5

Assume that x is much less than 0.25 M and calculate the error in your assumption.

[OH − ] = [NH 4 + ] = 0.0021 M; [NH 3 ] = 0.25 M, error = 0.84%

The second example requires that the original information be processed a bit, but it still can be solved using a small x approximation.

Example 4.3.8 – Approximate Solution After Shifting Starting Concentration

Copper(II) ions form a complex ion in the presence of ammonia

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If 0.010 mol Cu 2+ is added to 1.00 L of a solution that is 1.00 M NH 3 what are the concentrations when the system comes to equilibrium?

The initial concentration of copper(II) is 0.010 M. The equilibrium constant is very large so it would be better to start with as much product as possible because “all products” is much closer to equilibrium than “all reactants” (Figure 4.3.5). Therefore, to simplify our calculations, let us assume that the reaction goes 100% to completion.  Note that Cu 2+ is the limiting reactant; if all 0.010 M of it reacts to form product the concentrations would

[Cu 2 + ]= 0.010 – 0.010 = 0 M

[Cu(NH 3 ) 4 2+ ] = 0.010 M

[NH 3 ] = 1.00 – 4 x 0.010 = 0.96 M

image

Figure 4.3.5. The K value is very large for the equilibrium reaction Cu 2+ ( aq ) + 4NH 3 ( aq ) ⇌ Cu(NH 3 ) 4 2+ ( aq ) , so products are very heavily favoured. The change in concentration of both products and reactants is almost minimal since as it stands, the relative concentration of all species initially almost already corresponds to the relative amounts at equilibrium, where the product is close to its maximum possible concentration.

Using these “shifted” values as initial concentrations with x as the free copper(II) ion concentration at equilibrium gives this ICE table.

Cu 2+ ( aq ) +4NH 3 ( aq ) ⇌Cu(NH 3 ) 4 2+ ( aq )

Since we are starting close to equilibrium, x should be small so that

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Select the smallest concentration for the 5% rule – dividing a value by the smallest value possible will yield the largest possible error to really put the 5% rule to the test.

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This is much less than 5%, so the assumptions are valid. The concentrations at equilibrium are

[Cu 2+ ] = x = 2.4 x 10 -16 M

[NH 3 ] = 0.96 – 4x = 0.96 M

[Cu(NH 3 ) 4 2+ ] = 0.010 – x = 0.010 M

If we subtract x from 0.010 M, for example, we end up with 0.00999999… M which, when counting significant figures, rounds back up to 0.010 M anyway.

Overall, we started with a much higher concentration of reactant compared to product – note that we use up Cu 2+ so initially we have the highest possible concentration of products. But since K is very large, the change in concentration is kept minimal since this reaction mixture almost practically corresponds to what is expected at equilibrium.

By starting with the maximum amount of product, this system was near equilibrium and the change ( x ) was very small – this very small change was particularly driven by the complete absence of Cu 2+ initially (we discuss this in the context of Le Châtelier’s Principle in the next section). With only a small change required to get to equilibrium, the equation for x was greatly simplified and gave a valid result well within the 5% error maximum.

Check Your Learning 4.3.9 – Approximate Solution After Shifting Starting Concentration

What are the equilibrium concentrations when 0.25 mol Ni 2+ is added to 1.00 L of 2.00 M NH 3 solution?

Ni 2+ ( aq ) + 6 NH 3 ( aq ) ⇌ Ni(NH 3 ) 6 2+ ( aq )               K C = 5.5 x 10 8

With such a large equilibrium constant, first form as much product as possible, then assume that only a small amount ( x ) of the product shifts left. Calculate the error in your assumption.

[ Ni( NH 3 ) 6 2+ ] = 0.25 M, [NH 3 ] = 0.50 M, [Ni 2+ ] = 2.9 × 10 –8 M, error = 1.2 × 10 –5 %

★ Questions

   1. In the equilibrium reaction A + B ⇌ C, what happens to K if the concentrations       of the reactants are doubled? tripled? Can the same be said about the                           equilibrium reaction A ⇌ B + C?

   2. The following table shows the reported values of the equilibrium P{O 2 } at three     temperatures for the reaction Ag 2 O ( s) ⇌ 2 Ag (s) + 1/2 O 2 (g) for which Δ H ° = 31       kJ/mol. Are these data consistent with what you would expect to occur? Why or       why not?

3. Given the equilibrium system N 2 O 4 ( g) ⇌ 2 NO 2 (g), what happens to K P if the       initial pressure of N 2 O 4 is doubled? If K P is 1.7 × 10 −1 at 2300°C, and the system         initially contains 100% N 2 O 4 at a pressure of 2.6 × 10 2 bar, what is the equilibrium     pressure of each component?

4. At 430°C, 4.20 mol of HI in a 9.60 L reaction vessel reaches equilibrium                   according to the following equation: H 2 (g) + I 2 ( g) ⇌ 2 HI (g). At equilibrium, [H 2 ]       = 0.047 M and [HI] = 0.345 M. What are K and K P for this reaction?

5. Methanol, a liquid used as an automobile fuel additive, is commercially                     produced from carbon monoxide and hydrogen at 300°C according to the                   following reaction: CO (g) + 2 H 2 (g) ⇌ CH 3 OH (g) and K P = 1.3 × 10 −4 . If 56.0 g of       CO is mixed with excess hydrogen in a 250 mL flask at this temperature, and the       hydrogen pressure is continuously maintained at 100 bar, what would be the             maximum percent yield of methanol? What pressure of hydrogen would be               required to obtain a minimum yield of methanol of 95% under these conditions?

★★ Questions

   6. Starting with pure A, if the total equilibrium pressure is 0.969 atm for the               reaction A ( s) ⇌ 2 B (g) + C (g), what is K P (hint: must use the unit “bar” when           working with K P )?

7. The decomposition of ammonium carbamate to NH 3 and CO 2 at 40°C is written     as NH 4 CO 2 NH 2 (s) ⇌ 2 NH 3 (g) + CO 2 If the partial pressure of NH 3 at equilibrium       is 0.242 atm, what is the equilibrium partial pressure (in atm) of CO 2 ? What is the     total gas pressure of the system (in atm)? What is K P (hint: must use the unit             “bar” when working with K p )?

8. At 375 K, K P for the reaction SO 2 Cl 2 ( g) ⇌ SO 2 (g) + Cl 2 (g) is 2.4, with pressures     expressed in atmospheres. At 303 K, K P is 2.9 × 10 −2 .

a. What is K for the reaction at each temperature?

b. If a sample at 375 K has 0.100 M Cl 2 and 0.200 M SO 2 at equilibrium, what is           the concentration of SO 2 Cl 2 ?

c. If the sample given in part b is cooled to 303 K, what is the pressure inside              the bulb (in atm)?

9. Experimental data on the system Br 2 (l ) ⇌ Br 2 ( aq ) are given in the following         table. Graph [Br 2 ] versus moles of Br 2 (l) present; then write the equilibrium               constant expression and determine K .

10. Data accumulated for the reaction n-butane ( g) ⇌ isobutane (g) at equilibrium     are shown in the following table. What is the equilibrium constant for this                 conversion? If 1 mol of n -butane is allowed to equilibrate under the same reaction     conditions, what is the final number of moles of n -butane and isobutane?

solving equilibrium problems

11. Solid ammonium carbamate (NH 4 CO 2 NH 2 ) dissociates completely to ammonia      and carbon dioxide when it vaporizes:

NH 4 CO 2 NH 2 ( s) ⇌ 2 NH 3 (g) + CO 2 (g)

At 25°C, the total pressure of the gases in equilibrium with the solid is 0.116 atm. What is the equilibrium partial pressure of each gas (in atm)? What is K p (hint: must use the unit “bar” when working with K P )? If the concentration of CO 2 is doubled and then equilibrates to its initial equilibrium partial pressure + x atm, what change in the NH 3 concentration is necessary for the system to restore equilibrium?

   12. The equilibrium constant for the reaction COCl 2 ( g) ⇌ CO (g) + Cl 2 (g) is K P = 2.2     × 10 −10 at 100°C. If the initial concentration of COCl 2 is 3.05 × 10 −3 M, what is the       partial pressure of each gas at equilibrium at 100°C (in bar)? What assumption           can be made to simplify your calculations?

13. Aqueous dilution of IO4 − results in the following reaction:

IO 4 – ( aq ) + 2 H 2 O (l ) ⇌ H 4 IO 6 – ( aq )

and K = 3.5 × 10 −2 . If you begin with 50 mL of a 0.896 M solution of IO 4 − that is diluted to 250 mL with water, how many moles of H 4 IO 6 − are formed at equilibrium?

★★★ Questions

   14. Iodine and bromine react to form IBr, which then sublimes. At 184.4°C, the            overall reaction proceeds according to the following equation:

I 2 (g) + Br 2 ( g)  ⇌ 2 IBr (g)

K P = 1.2 × 10 2 (Hint: assumed to have used “bar” as a unit while calculating K P ). If you begin the reaction with 7.4 g of I 2 vapor and 6.3 g of Br 2 vapor in a 1.00 L container, what is the concentration of IBr (g) at equilibrium (gmol-1)? What is the partial pressure of each gas at equilibrium (in bar)? What is the total pressure of the system (in bar)?

   15. For the reaction

C (s) + 12 N 2 (g) + 5/2 H 2 ( g) ⇌ CH 3 NH 2 (g)

K = 1.8 × 10 −6 . If you begin the reaction with 1.0 mol of N 2 , 2.0 mol of H 2 , and sufficient C(s) in a 2.00 L container, what are the concentrations of N 2 and CH 3 NH 2 at equilibrium (gmol-1)? What happens to K if the concentration of H 2 is doubled?

   1. The K value is now raised to the respected factor. When it is doubled K is now        K 2 and when it is tripled, K is K 3

   2. These results are not expected, as with an increase in temperature an increase       in pressure should occur. ∆H० is a positive value in this case (31 kJ/mol) indicating       that it is an endothermic reaction. This being said, with an increase in                           temperature the reaction will shift forward meaning that more product will be           produced. If more product is produced, more oxygen gas is present, therefore             increasing its pressure.

   3. Kp would remain the same, P N2O2 = 2.3 x 10 2 bar, P NO2 = 6.6 bar

   4. K = 53.88; Kp = 53.88

   5. 215 bar MeOH; 383 bar H 2

   6. Kp = 0.140

   7. Partial Pressure of CO 2 = 0.121 atm; total gas pressure of the system = 0.363             atm; Kp = 7.37 x 10 -3

   8. (a) K = 7.8 x 10 -2 at 375K, K = 1.2 x 10 -3 at 303K ; (b) 0.256 M; (c) 14.13 atm

image

   10. K = 2.5; the final moles will be 0.3 mol of n-butane and 0.7 mol of isobutane.

   11. P NH3 = 0.0773 atm, P CO2 = 0.0387 atm; K p = 2.411 x 10 -4 ; the concentration of           NH 3 will drop in order for the equilibrium to restore.

   12. P COCl2 = 0.042 bar, P CO = 4.59 x 10 -6 bar, P Cl2 = 4.59 x 10 -6 bar; An assumption      to be made is that the total volume is 1L.

   13. H 4 IO 6 – = 9.09 x 10 -3 mol

   14. P I2 = 7.7 bar, P Br2 = 47 bar, P IBr = 2.1 x10 2 bar = 0.054 M, P T = 2.6 x 10 2 bar

   15. [N 2 ] ≈ 0.99 M, [H 2 ] ≈ 2 M, [CH 3 NH 2 ] = 1.02 x 10 -5 M; if the [H 2 ] doubles, K               remains unchanged (only temperature can alter a K value)

General Chemistry for Gee-Gees Copyright © by Kevin Roy; Mahdi Zeghal; Jessica M. Thomas; and Kathy-Sarah Focsaneanu is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License , except where otherwise noted.

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  • If Possible, Take the Square Root of Both Sides
  • Using the Quadratic Equation
  • Solving Equations Containing x 3 , x 4 , etc.
  • The Method of Successive Approximations
  • Assume That the Change is Small
  • Rearrange to the form:  ax 2 + bx + c = 0.
  • Substitute the coefficients into the quadratic equation and solve for x.
  • Using logarithms
  • Finding the n th root
  • Using either method, the first step is to simplify the equation.
  • Take the log of both sides of the equation.
  • Factor out the exponent (log x n = n log x) and find the log of the number.
  • Solve for "log x".
  • Solve for x (if log x = y, then x = 10 y ).
  • Finding the n th root.
  • Most calculators will have a button labeled:
  • Enter in the number for which you want the n th root.
  • Press the button for finding the n th root followed by the exponent, n, and the "=" or "enter" button to obtain the result.
  • assume an approximate value for the variable that will simplify the equation
  • solve for the variable
  • use the answer as the second apporximate value and solve the equation again
  • repeat this process until a constant value for the variable is obtained
  • Approximate a value for the variable that will simplify the equation.
  • Simpify the equation and solve for the variable.
  • Using the result, make a second approximation.
  • Using the second approximation, simplify the equation and solve for the variable
  • Repeat the process until a constant value is obtained.
  • K and Q Are Very Close in Size
  • K and Q Lie on Opposite Sides of One (K>>Q or K<<Q)
  • Calculate Q , the reaction quotient and compare to K.
  • Make an ICE chart .
  • Subsitute concentrations into the equilibrium expression. Assume that [A] - x = [A], simplify the equation, and solve for the change.
  • Check to see if the change is less than 5% of the starting quantity, or within the limits set by your instructor.
  • Calculate the equilibrium amounts if asked to do so.
  • Check your work.
  • Calculate Q and compare to K.
  • Substitute into the equilibrium expression.  Assume that 0.500 - x ~ 0.500.  Simplify equation and solve for the change.
  • Check answer to see if it is within limits set by your instructor. (Here we use 5%.)
  • Determine the equilibrium concentrations of each species
  • Check work.
  • When K>>Q and K > 1, assume 100% conversion into products, followed by the back reaction to establish equilibrium.  When K<<Q and K < 1, assume 100% conversion into reactants, followed by the forward reaction to establish equilibrium.
  • Make an ICE chart to determine change and equilibrium quantities starting with those resulting from the 100% conversion.
  • Substitute quantities into the equilibrium expression.
  • Assume the change is near zero such that "[A] - x" is equal to "[A]."
  • Solve for the variable.
  • Check to see if the change is less than 5% of the maximum amount, or within the limits set by your instructor. If not, use the method of approximations , a programmable calculator, or other method to solve.
  • Solve for the equilibrium concentrations if asked to do so.
  • Initially the [HI] = 0, so K >>Q  and K is > 1. The change in the concentration of each species will be large so we calculate the quantity of product formed assuming 100% conversion.
  • Make an ICE chart starting with the concentrations after the 100% conversion.
  • Substitute equilibrium amounts into the equilibrium expression.
  • Assume the change in the concentration of the product is 0.  Substitute into the equation and solve for "x."
  • Check to see if the change is within the limits set by your instructor. (Here we use 5%.)
  • Calculate the equilibrium concentrations.

4.5 Solving Equilibrium Problems

solving equilibrium problems

An “equilibrium problem” in general chemistry typically involves one to solve for the equilibrium concentrations of the reaction given some starting information. This section introduces a variety of equilibrium type problems, each one illustrating the method in solving for the equilibrium concentrations.

4.5.1 ICE Table

One tool we use to solve equilibrium problems is the ICE table . ICE is an acronym that stands for I nitial pressure/concentration, C hange in pressure/concentration, and E quilibrium pressure/concentration. The table is then populated with values that have units of either Molarity or pressures. The purpose of the ICE table is to simply keep our data organized. That’s it. By adhering to the table format, we can conveniently keep track of what we are doing. It is a very useful tool! Here’s how it works.

Take the following reversible reaction.

\[\mathrm{N_2O_4}(g) \rightleftharpoons 2\mathrm{NO}_2 (g)\]

We would write our ICE table as follows

4.5.2 Example 1

The following reaction begins with [N 2 O 4 ] = 1.00 M .

\[\mathrm{N_2O_4}(g)\rightleftharpoons 2\mathrm{NO}_2 (g) \qquad K = 4.46\times 10^{-3}\] Determine the equilibrium concentrations of the reaction.

Write out ICE Table

Here I populate the initial concentration row using the information given to me by the problem.

Determine the direction of reaction

Since the reaction is beginning with all reactant, the reaction must move right. The change in concentration of the reactants is therefore going to be a negative quantity (e.g.  –x ) because reactants will be consumed. We don’t know the magnitude of x but we know it will be negative. The “1” in front of the x is due to the stoichiometric coefficient from the balanced chemical equation.

The change in concentration of the products must therefore be +2x . Since the reaction will proceed right, products will be produced and its change in concentration is positive.

Stoichiometrically, for every 1 mole of reactant that is consumed, 2 moles of product is produced. Our ICE table now looks like the following

Fill in the Equilibrium Row

The equilibrium row of the table is simply the first two rows (the I and C rows) added together. What this tells us is this. The equilibrium concentration of NO 2 is the initial concentration minus how far the reaction proceeds to reach equilibrium ( –x ). Since the reaction is moving right, the final concentration of N 2 O 4 must be less than the initial concentration.

On the other hand, initially there is zero NO 2 product. The reaction is going to proceed right and product will be produced. Therefore, the initial concentration of the product is 0. However, its concentration will be the magnitude of 2x at equilibrium.

Write out the Equilibrium Expression

Next, write out the equilibrium expression.

\[\dfrac{[\mathrm{NO_2}]^2}{[\mathrm{N_2O_4}]} = K\]

Substitute in Equilibrium Concentration Terms

Substitute in the equilibrium concentrations from the table as well as the equilibrium constant into the equilibrium expression.

\[\dfrac{(2x)^2}{(1.00 - x)} = 4.46\times 10^{-3}\] The next step is to solve for x . There can be multiple ways to do this, some of which are more straightforward than others. I will introduce these methods below.

4.5.2.1 Quadratic Method

Sometimes we can solve for x by using the quadratic formula if our equilibrium expression gives us a quadratic equation. A quadratic equation takes on the form

\[\color{red}{a}x^2 + \color{blue}{b}x + \color{green}{c} = 0\] Notice how there is an x 2 and x term in the same equation. Here, x cannot be isolated and solved for. We must use the quadratic equation (below) to solve for x .

\[x = \dfrac{-\color{blue}{b}\pm\sqrt{\color{blue}{b}^2-4\color{red}{a}\color{green}{c}}}{2\color{red}{a}}\]

Let us rearrange our equilibrium expression into the form of a quadratic.

\[\begin{align*} \dfrac{(2x)^2}{(1.000 - x)} &= 4.46\times 10^{-3} \\[1.5ex] 4x^2 &= 4.46\times 10^{-3}(1.000 -x) \\[1ex] &= 4.46\times 10^{-3} - 4.46\times 10^{-3}x \\[1ex] \color{red}{4}x^2 &+ \color{blue}{4.46\times 10^{-3}}x \color{green}{-4.46\times 10^{-3}} = 0 \end{align*}\]

Next, solve the equation using the quadratic equation.

\[\begin{align*} x &= \dfrac{-\color{blue}{b}\pm\sqrt{\color{blue}{b}^2-4\color{red}{a}\color{green}{c}}}{2\color{red}{a}}\\[2ex] &= \dfrac{-\color{blue}{4.46\times 10^{-3}} \pm\sqrt{(\color{blue}{4.46\times 10^{-3}})^2 - 4(\color{red}{4})(\color{green}{-4.46\times 10^{-3}})}}{2(\color{red}{4})}\\[2ex] &\approx 0.03284~~\mathrm{or}~~-0.03395 \end{align*}\]

Here, we see two solutions (two roots) from the ± operation. We will always disregard the negative root. Therefore,

\[x = 0.03284\]

Determine equilibrium concentrations

Now that we know what x is, plug it back into the equilibrium row of the ICE table.

\[\begin{align*} [\mathrm{N_2O_4}] &= 1.000 - x \\[1.5ex] &= 1.000- 0.03284 \\[1.5ex] &= 0.9672~M \\[1.5ex] [\mathrm{NO_2}] &= 2x \\[1.5ex] &= 2(0.03284~M) \\[1.5ex] &= 0.06568 \end{align*}\]

4.5.2.2 Small x Approximation Method

Another method we can use to solve for x is the small x approximation. Remember our “reaction line” exercises presented earlier? Our reaction is starting near equilibrium ( Q is located at all reactants and K is small therefore Q is starting near K ). The magnitude of x , in this case, is small . It might be useful to invoke the small x approximation because of this.

For this method, we revisit our equilibrium expression.

\[\dfrac{(2x)^2}{(1.00 \color{red}{- x})} = 4.46\times 10^{-3}\]

Notice the x in the denominator of the fraction. If x were a very small number, the denominator will essentially not change (it will essentially be equal to 0.100 if x is indeed “small enough”). Therefore, we can throw out the “ –x term in the denominator to give

\[\dfrac{(2x)^2}{(1.00)} = 4.46\times 10^{-3}\] This allows us to directly solve for x as we will not have a quadratic equation to solve for.

\[\begin{align*} \dfrac{(2x)^2}{(1.000)} &= 4.46\times 10^{-3} \\[1.5ex] (2x)^2 &= 4.46\times 10^{-3} \\[1.5ex] 2x &= \sqrt{4.46\times 10^{-3}} \\[1.5ex] x &= \dfrac{\sqrt{4.46\times 10^{-3}}}{2}\\[1.5ex] &= 0.0334 \end{align*}\]

We assumed that x was small but was it? We must test x to ensure that the small x approximation was a good approximation.

If x is within 5% of the number it was being subtracted from (or added to), the small x approximation is good enough.

Note: The “5%” threshold is somewhat arbitrary. However, I use the “5%” rule for the purpose of this class (as many other classes/textbooks do). However, in practice , this threshold may not be small enough!

To test x , take the value of x and divide by the number it was being subtracted from (or added to). Multiply by 100% and see what the result is!

\[\begin{align*} \dfrac{0.0334}{1.000} \times 100\% = 3.34\% \end{align*}\]

Since x was smaller than 5% of “1.000”, we move accept that the approximation was reasonable and use this value for x to determine the equilibrium concentrations.

Note: If x was 5% or larger, we would have to STOP , backtrack, and solve the equilibrium expression using the quadratic method.

\[\begin{align*} [\mathrm{N_2O_4}] &= 1.000~M - x \\[1.5ex] &= 1.000~M - 0.0334~M \\[1.5ex] &= 0.9666~M \\[1.5ex] [\mathrm{NO_2}] &= 2x \\[1.5ex] &= 2(0.0334~M) \\[1.5ex] &= 0.0668~M \end{align*}\]

Compare the equilibrium concentrations from the small x approximation to those determined with the quadratic. They are very similar!

4.5.2.3 Practice

Try to solve the equilibrium concentrations for the same reaction with different starting amounts!

Don’t peek before you’ve tried it!

No product. Shift right. \[\begin{align*} \dfrac{[\mathrm{NO_2}]^2}{[\mathrm{N_2O_4}]} &= K \\[1ex] \dfrac{(2x)^2}{0.670-x} &= 4.65\times 10^{-3} \end{align*}\] Quadratic approach: \[\begin{align*} 4x^2 &= 4.65\times 10^{-3} (0.670-x) \\ 4x^2 &= 3.12\times 10^{-3} - 4.65\times 10^{-3}x \\ 4x^2 + 4.65\times 10^{-3}x - 3.12\times 10^{-3} &= 0 \\[2ex] x &= \dfrac{-b\pm \sqrt{b^2-4ac}}{2a} \\ &= \dfrac{-4.65\times 10^{-3} \pm \sqrt{(4.65\times 10^{-3})^2 - 4(4)(- 3.12\times 10^{-3})}}{2(4)} \\ &= 0.0274 \quad \mathrm{and} \quad -0.0285 \end{align*}\] First root: \[\begin{align*} [\mathrm{N_2O_4}] &= 0.670 - x = 0.670 - 0.0274 = 0.643~M \\ [\mathrm{NO_2}] &= 2x = 2(0.0274) = 0.0548~M \end{align*}\]

Small \(x\) approach (less than 5%): \[\begin{align*} \dfrac{(2x)^2}{0.670} &= 4.65\times 10^{-3} \\ 4x^2 &= (4.65\times 10^{-3})(0.670) \\ x &= \sqrt{\dfrac{(4.65\times 10^{-3})(0.670)}{4}} \\ &= 0.0279 \\[2ex] \dfrac{x}{[\mathrm{N_2O_4}]_{\mathrm{i}}} \times 100\% &= \dfrac{0.0279}{0.670}\times 100\% = 4.16\%\\[2ex] [\mathrm{N_2O_4}] &= 0.670 - x = 0.670 - 0.0279 = 0.642~M \\ [\mathrm{NO_2}] &= 2x = 2(0.0279) = 0.0558~M \end{align*}\]

\[\begin{align*} \dfrac{[\mathrm{NO_2}]^2}{[\mathrm{N_2O_4}]} &= Q \\[1ex] \dfrac{(0.0500)^2}{0.446} &= \\ 5.61\times 10^{-3} &= \end{align*}\] \(Q > K\) ; \(5.61\times 10^{-3} > 4.66\times 10^{-3}\) – Shift left \[\begin{align*} \dfrac{[\mathrm{NO_2}]^2}{[\mathrm{N_2O_4}]} &= K \\[1ex] \dfrac{(0.0500-2x)^2}{0.446+x} &= 4.66\times 10^{-3} \end{align*}\] Quadratic approach: \[\begin{align*} 2.5\times 10^{-3} - 0.1x - 0.1x + 4x^2 &= 4.66\times 10^{-3} (0.446 + x) \\ 2.5\times 10^{-3} - 0.2x + 4x^2 &= 2.08\times 10^{-3} + 4.66\times 10^{-3}x \\ 4x^2 - 0.205x + 4.2\times 10^{-4} &= 0 \end{align*}\] \[\begin{align*} x = \dfrac{-b\pm \sqrt{b^2-4ac}}{2a} &= \dfrac{-(-0.205) \pm \sqrt{(-0.205)^2 - 4(4)(4.2\times 10^{-4})}}{2(4)} \\ &= \dfrac{0.205 \pm \sqrt{3.53\times 10^{-2}}}{8} \\ &= 4.91 \times 10^{-2} \quad \mathrm{and} \quad 2.14\times 10^{-3} \end{align*}\] First root: \[\begin{align*} [\mathrm{N_2O_4}] &= 0.446 + x = 0.446 + 4.91\times 10^{-2} = 0.495 \\ [\mathrm{NO_2}] &= 0.0500-2x = 0.0500 - 2(4.91\times 10^{-2}) = -0.0482 \quad \mathbf{UNPHYSICAL} \end{align*}\] Second root: \[\begin{align*} [\mathrm{N_2O_4}] &= 0.446 + x = 0.446 + 2.14\times 10^{-3} = 0.448 \\ [\mathrm{NO_2}] &= 0.0500-2x = 0.0500 - 2(2.14\times 10^{-3}) = 0.0457\\[2ex] \dfrac{(0.0457)^2}{0.448} &= 4.66\times 10^{-3} \quad \mathbf{RIGHT} \end{align*}\]

Small \(x\) approach (less than 5%): \[\begin{align*} \dfrac{(0.0500-2x)^2}{0.446} &= 4.66\times 10^{-3} \\ (0.0500-2x)^2 &= 2.08\times 10^{-3} \\ 0.0500-2x &= 4.56\times 10^{-2} \\ -2x &= -4.41\times 10^{-3} \\ x &= 2.205\times 10^{-3} \\ \dfrac{x}{[\mathrm{NO_2}]_{\mathrm{i}}} \times 100\% &= \dfrac{2.205\times 10^{-3}}{0.446}\times 100\% = 0.49\%\\[2ex] [\mathrm{N_2O_4}] &= 0.446 + x = 0.446 + 2.205\times 10^{-3} = 0.448~M \\ [\mathrm{NO_2}] &= 0.0500-2x = 0.0500 - 2(2.205\times 10^{-3}) = 0.046~M \end{align*}\]

\[\begin{align*} \dfrac{[\mathrm{NO_2}]^2}{[\mathrm{N_2O_4}]} &= Q \\[1ex] \dfrac{(0.0300)^2}{0.500} &= \\ 1.8\times 10^{-3} &= \end{align*}\] \(Q < K\) ; \(1.81\times 10^{-3} < 4.60\times 10^{-3}\) – Shift right \[\begin{align*} \dfrac{[\mathrm{NO_2}]^2}{[\mathrm{N_2O_4}]} &= K \\[1ex] \dfrac{(0.0300+2x)^2}{0.500-x} &= 4.60\times 10^{-3} \end{align*}\] Quadratic approach: \[\begin{align*} 9.0\times 10^{-4} + 0.06x + 0.06x + 4x^2 &= 4.60\times 10^{-3} (0.500 - x) \\ 9.0\times 10^{-4} + 0.12x + 4x^2 &= 2.3\times 10^{-3} - 4.60\times 10^{-3}x \\ 4x^2 + 0.1246x - 1.4\times 10^{-3} &= 0 \end{align*}\] \[\begin{align*} x = \dfrac{-b\pm \sqrt{b^2-4ac}}{2a} &= \dfrac{-0.1246 \pm \sqrt{(0.1246)^2 - 4(4)(-1.4\times 10^{-3})}}{2(4)} \\ &= \dfrac{-0.1246 \pm \sqrt{3.79\times 10^{-2}}}{8} \\ &= 8.77 \times 10^{-3} \quad \mathrm{and} \quad -3.99\times 10^{-2} \end{align*}\] First root: \[\begin{align*} [\mathrm{N_2O_4}] &= 0.500 - x = 0.500 - 8.77\times 10^{-3} = 0.491 \\ [\mathrm{NO_2}] &= 0.0300+2x = 0.0300 + 2(8.77\times 10^{-3}) = 0.0475 \end{align*}\] Small \(x\) approach (less than 5%): \[\begin{align*} \dfrac{(0.0300-2x)^2}{0.500} &= 4.60\times 10^{-3} \\ (0.0300+2x)^2 &= 2.3\times 10^{-3} \\ 0.0300+2x &= 4.80\times 10^{-2} \\ 2x &= 1.80\times 10^{-2} \\ x &= 8.98\times 10^{-3} \\ \dfrac{x}{[\mathrm{NO_2}]_{\mathrm{i}}} \times 100\% &= \dfrac{8.98\times 10^{-3}}{0.500}\times 100\% = 1.8\% \\[2ex] [\mathrm{N_2O_4}] &= 0.500 - x = 0.500 - 8.89\times 10^{-3} = 0.491~M \\ [\mathrm{NO_2}] &= 0.0300+2x = 0.0300 + 2(8.89\times 10^{-3}) = 0.0478~M \end{align*}\]

\[\begin{align*} \dfrac{[\mathrm{NO_2}]^2}{[\mathrm{N_2O_4}]} &= Q \\[1ex] \dfrac{(0.0400)^2}{0.600} &= \\ 2.67\times 10^{-3} &= \end{align*}\] \(Q < K\) ; \(2.67\times 10^{-3} < 4.60\times 10^{-3}\) – Shift right \[\begin{align*} \dfrac{[\mathrm{NO_2}]^2}{[\mathrm{N_2O_4}]} &= K \\[1ex] \dfrac{(0.0400+2x)^2}{0.600-x} &= 4.60\times 10^{-3} \end{align*}\] Quadratic approach: \[\begin{align*} 1.6\times 10^{-3} + 0.08x + 0.08x + 4x^2 &= 4.60\times 10^{-3} (0.600 - x) \\ 9.0\times 10^{-4} + 0.16x + 4x^2 &= 2.76\times 10^{-3} - 4.60\times 10^{-3}x \\ 4x^2 + 0.1646x - 1.16\times 10^{-3} &= 0 \end{align*}\] \[\begin{align*} x = \dfrac{-b\pm \sqrt{b^2-4ac}}{2a} &= \dfrac{-0.1646 \pm \sqrt{(0.1646)^2 - 4(4)(-1.16\times 10^{-3})}}{2(4)} \\ &= \dfrac{-0.1646 \pm \sqrt{4.565\times 10^{-2}}}{8} \\ &= 6.13 \times 10^{-3} \quad \mathrm{and} \quad -4.73\times 10^{-2} \end{align*}\] First root: \[\begin{align*} [\mathrm{N_2O_4}] &= 0.600 - x = 0.600 - 6.13\times 10^{-3} = 0.594 \\ [\mathrm{NO_2}] &= 0.0400+2x = 0.0400 + 2(6.13\times 10^{-3}) = 0.0523 \end{align*}\] Small \(x\) approach (less than 5%): \[\begin{align*} \dfrac{(0.0400-2x)^2}{0.600} &= 4.60\times 10^{-3} \\ (0.0400+2x)^2 &= 2.76\times 10^{-3} \\ 0.0400+2x &= 5.25\times 10^{-2} \\ 2x &= 1.25\times 10^{-2} \\ x &= 6.27\times 10^{-3} \\ \dfrac{x}{[\mathrm{NO_2}]_{\mathrm{i}}} \times 100\% &= \dfrac{6.27\times 10^{-3}}{0.600}\times 100\% = 1.12\%\\[2ex] [\mathrm{N_2O_4}] &= 0.600 - x = 0.600 - 6.27\times 10^{-3} = 0.594~M \\ [\mathrm{NO_2}] &= 0.0400+2x = 0.0400 + 2(6.27\times 10^{-3}) = 0.0525~M \end{align*}\]

No reactant. Shift left. \[\begin{align*} \dfrac{[\mathrm{NO_2}]^2}{[\mathrm{N_2O_4}]} &= K \\[1ex] \dfrac{(0.200-2x)^2}{x} &= 4.63\times 10^{-3} \end{align*}\] Quadratic approach: \[\begin{align*} 4.0\times 10^{-2} - 0.4x - 0.4x + 4x^2 &= 4.63\times 10^{-3}x \\ 4.0\times 10^{-2} - 0.8x + 4x^2 &= 4.63\times 10^{-3}x \\ 4x^2 - 0.8046x + 4.0\times 10^{-2} &= 0 \end{align*}\] \[\begin{align*} x = \dfrac{-b\pm \sqrt{b^2-4ac}}{2a} &= \dfrac{-(-0.8046) \pm \sqrt{(-0.8046)^2 - 4(4)(4.0\times 10^{-2})}}{2(4)} \\ &= \dfrac{0.8046 \pm \sqrt{7.381\times 10^{-3}}}{8} \\ &= 1.11 \times 10^{-1} \quad \mathrm{and} \quad 8.98\times 10^{-2} \end{align*}\] First root: \[\begin{align*} [\mathrm{N_2O_4}] &= x = 0.111 \\ [\mathrm{NO_2}] &= 0.200-2x = 0.200 - 2(1.11\times 10^{-1}) = -0.022 \quad \mathbf{UNPHYSICAL} \end{align*}\] Second root: \[\begin{align*} [\mathrm{N_2O_4}] &= x = 0.0898 \\ [\mathrm{NO_2}] &= 0.200-2x = 0.200 - 2(8.98\times 10^{-2}) = 0.0204 \end{align*}\] Small \(x\) approach (less than 5%): \[\begin{align*} \dfrac{(0.200-2x)^2}{x} &= 4.63\times 10^{-3} \\ \dfrac{(0.200)^2}{x} &= 4.63\times 10^{-3} \\ x &= 8.64 \\ \dfrac{x}{[\mathrm{NO_2}]_{\mathrm{i}}} \times 100\% &= \dfrac{8.64}{0.200}\times 100\% = 4.32\times 10^3\% \quad \mathbf{TOO~LARGE} \end{align*}\]

4.5.3 Example 2

Sometimes you do not end up with a quadratic and/or have to use the small x approximation!

Take this problem for example. For the following reaction

\[\begin{align*} \mathrm{H_2}(g) + \mathrm{I_2}(g) \rightleftharpoons \mathrm{2HI}(g) \qquad K = 54.3 \end{align*}\]

the initial concentrations are

\[\begin{align*} [\mathrm{H_2}]_{\mathrm{i}} = 0.500~M \quad [\mathrm{I_2}]_{\mathrm{i}} = 0.500~M \end{align*}\]

Solve for the equilibrium concentrations.

Fill out the ICE table.

Write out the equilibrium expression and solve for x .

\[\begin{align*} \dfrac{[\mathrm{HI}]^2}{[\mathrm{H_2}][\mathrm{I_2}]} &= K \\[1.5ex] \dfrac{(2x)^2}{(0.500-x)(0.500-x)} &= 54.3 \\[1.5ex] \dfrac{(2x)^2}{(0.500-x)^2} &= 54.3 \\[1.5ex] \sqrt{\dfrac{(2x)^2}{(0.500-x)^2}} &= \sqrt{54.3} \\[1.5ex] \dfrac{2x}{0.500-x} &= 7.369 \\[1.5ex] 2x &= 7.369 (0.500 - x) \\ 2x &= 3.6845 - 7.369x \\ 9.369x &= 3.6845 \\ x &= 0.3933 \end{align*}\]

Plug in x into the equilibrium terms and solve for the equilibrium expressions.

\[\begin{align*} [\mathrm{H_2}] &= 0.500 - x = 0.500 - 0.3933 = 0.1067~M \\ [\mathrm{I_2}] &= 0.500 - x = 0.500 - 0.3933 = 0.1067~M \\ [\mathrm{HI}] &= 2x = 2(0.3933) = 0.7866~M \end{align*}\]

4.5.4 Example 3

We can solve equilibrium problems by starting with some equilibrium concentrations instead of only initial concentrations!

For the following reaction

\[\begin{align*} \mathrm{N_2}(g) + \mathrm{O_2}(g) \rightleftharpoons \mathrm{2NO}(g) \qquad K = 4.1\times 10^{-4} \end{align*}\]

determine the equilibrium concentration of NO given

\[\begin{align*} [\mathrm{N_2}]_{\mathrm{eq}} &= 0.05~M \\[1.5ex] [\mathrm{O_2}]_{\mathrm{eq}} &= 0.002~M \end{align*}\]

Fill out an ICE table.

We have two equilibrium concentrations to fill in. We don’t know any of the other information.

Write out the equilibrium expression and solve for the NO concentration!

\[\begin{align*} \dfrac{[\mathrm{NO}]^2}{[\mathrm{N_2}][\mathrm{O_2}]} &= K \\[1.5ex] \dfrac{[\mathrm{NO}]^2}{(0.05)(0.002)} &= 4.1\times 10^{-4} \\[1.5ex] [\mathrm{NO}]^2 &= 4.1\times 10^{-4} \times (0.05)(0.002) \\[1.5ex] \sqrt{[\mathrm{NO}]^2} &= \sqrt{4.1\times 10^{-4} \times (0.05)(0.002)} \\[1.5ex] [\mathrm{NO}] &= 2.02\times 10^{-4}~M \end{align*}\]

4.5.5 Example 4

Sometimes we can start an equilibrium problem with both initial and equilibrium concentrations and solve for the equilibrium constant!

For the reaction

\[\begin{align*} \mathrm{I_2}(aq) + \mathrm{I^-}(aq) \rightleftharpoons \mathrm{I_3^-}(aq) \end{align*}\]

determine the equilibrium constant, K , given

\[\begin{align*} [\mathrm{I_2}]_{\mathrm{i}} &= 0.1~M \\ [\mathrm{I^-}]_{\mathrm{i}} &= 0.08~M \\ [\mathrm{I_2}]_{\mathrm{eq}} &= 0.05~M \end{align*}\]

Recognize that we have an initial concentration and an equilibrium concentration for I 2 . We can directly solve for x .

\[\begin{align*} [\mathrm{I_2}]_{\mathrm{i}} - x &= [\mathrm{I_2}]_{\mathrm{eq}} \\[1.5ex] x &= [\mathrm{I_2}]_{\mathrm{i}} - [\mathrm{I_2}]_{\mathrm{eq}} \\ &= 0.1~M - 0.05~M \\ &= 0.05~M \end{align*}\]

Write out the equilibrium expression, substitute in the equilibrium terms, and then plug in x . Solve for K .

\[\begin{align*} K &= \dfrac{[\mathrm{I_3^-}]}{[\mathrm{I_2}][\mathrm{I^-}]} \\[1.5ex] &= \dfrac{x}{(0.05)(0.08-x)} \\[1.5ex] &= \dfrac{0.05}{(0.05)(0.08-0.05)} \\[1.5ex] &= 33.33 \end{align*}\]

  • Nahkleh Group
  • Robinson Group
  • Weaver Group
  • Bodner Group

Acid-Base Equilibria

Water, Acids, and Bases

pH, pOH, and Ka

Equilibrium Problems Involving Strong Acids

Weak Acids and Equilibrium

Equilibrium Problems Involving Bases

Mixtures and Buffers

Diprotic and Triprotic Acids and Bases

Compounds that could be either Acids or Bases

Solving Equilibrium Problems Involving Bases

With minor modifications, the techniques applied to equilibrium calculations for acids are valid for solutions of bases in water.

Consider the calculation of the pH of an 0.10 M NH 3 solution. We can start by writing an equation for the reaction between ammonia and water.

Strict adherence to the rules for writing equilibrium constant expressions leads to the following equation for this reaction.

But, taking a lesson from our experience with acid-dissociation equilibria, we can build the [H 2 O] term into the value of the equilibrium constant. Reactions between a base and water are therefore described in terms of a base-ionization equilibrium constant , K b .

For NH 3 , K b is 1.8 x 10 -5 .

We can organize what we know about this equilibrium with the format we used for equilibria involving acids.

Substituting this information into the equilibrium constant expression gives the following equation.

K b for ammonia is small enough to allow us to consider the assumption that C is small compared with the initial concentration of the base.

Solving this approximate equation gives the following result.

pOH = - log (1.3 x 10 -3 ) = 2.89

Which, in turn, can be used to calculate the pH of the solution.

pH = 14 - pOH = 11.11

Equilibrium problems involving bases are relatively easy to solve if the value of K b for the base is known. The first step in many base equilibrium calculations involves determining the value of K b for the reaction from the value of K a for the conjugate acid.

As an example, let's calculate the pH of a 0.030 M solution of sodium benzoate (C 6 H 5 CO 2 Na) in water from the value of K a for benzoic acid (C 6 H 5 CO 2 H): K a = 6.3 x 10 -5 .

Benzoic acid and sodium benzoate are members of a family of food additives whose ability to retard the rate at which food spoils has helped produce a 10-fold decrease in the incidence of stomach cancer. To save time and space, we'll abbreviate benzoic acid as HOBz and sodium benzoate as NaOBz. Benzoic acid, as its name implies, is an acid. Sodium benzoate is a salt of the conjugate base, the OBz - or benzoate ion.

Whenever sodium benzoate dissolves in water, it dissociates into its ions.

The benzoate ion then acts as a base toward water, picking up a proton to form the conjugate acid and a hydroxide ion.

OBz - ( aq ) + H 2 O( l ) HOBz( aq ) + OH - ( aq )

The base-ionization equilibrium constant expression for this reaction is therefore written as follows.

The next step in solving the problem involves calculating the value of K b for the OBz - ion from the value of K a for HOBz.

The K a and K b expressions for benzoic acid and its conjugate base both contain the ratio of the equilibrium concentrations of the acid and its conjugate base. K a is proportional to [OBz - ] divided by [HOBz], and K b is proportional to [HOBz] divided by [OBz - ].

Two changes have to made to derive the K b expression from the K a expression: We need to remove the [H 3 O + ] term and introduce an [OH - ] term. We can do this by multiplying the top and bottom of the K a expression by the OH - ion concentration.

Rearranging this equation gives the following result.

The two terms on the right side of this equation should look familiar. The first is the inverse of the K b expression, the second is the expression for K w .

K a x K b = K w

According to this equation, the value of K b for the reaction between the benzoate ion and water can be calculated from K a for benzoic acid.

Now that we know K b for the benzoate ion, we can calculate the pH of an 0.030 M NaOBz solution with the techniques used to handle weak-acid equilibria. We start, once again, by building a representation for the problem.

We then substitute this information into the K b expression.

We then solve the approximate equation for the value of C .

pOH = - log [2.2 x 10 -6 ] = 5.66

The problem asked for the pH of the solution, however, so we use the relationship between pH and pOH to calculate the pH.

pH = 14 - pOH = 8.34

Two assumptions were made in this calculation.

  • We assumed that all of the OH - ion at equilibrium came from the reaction between the benzoate ion and water. (In other words, we ignored the contribution to the OH - ion concentration from the dissociation of water.)

We have already confirmed the validity of the first assumption. What about the second? The OH - ion concentration obtained from this calculation is 2.1 x 10 -6 M , which is 21 times the OH - ion concentration in pure water. According to LeChatelier's principle, however, the addition of a base suppresses the dissociation of water. This means that the dissociation of water makes a contribution of significantly less than 5% to the total OH - ion concentration in this solution. It can therefore be legitimately ignored.

Two factors affect the OH - ion concentration in aqueous solutions of bases: K b and C b . We can ignore the dissociation of water when K b C b for a weak base is larger than 1.0 x 10 -13 . When K b C b is smaller than 1.0 x 10 -13 , we have to include the dissociation of water in our calculations.

  • 12.2 Examples of Static Equilibrium
  • Introduction
  • 1.1 The Scope and Scale of Physics
  • 1.2 Units and Standards
  • 1.3 Unit Conversion
  • 1.4 Dimensional Analysis
  • 1.5 Estimates and Fermi Calculations
  • 1.6 Significant Figures
  • 1.7 Solving Problems in Physics
  • Key Equations
  • Conceptual Questions
  • Additional Problems
  • Challenge Problems
  • 2.1 Scalars and Vectors
  • 2.2 Coordinate Systems and Components of a Vector
  • 2.3 Algebra of Vectors
  • 2.4 Products of Vectors
  • 3.1 Position, Displacement, and Average Velocity
  • 3.2 Instantaneous Velocity and Speed
  • 3.3 Average and Instantaneous Acceleration
  • 3.4 Motion with Constant Acceleration
  • 3.5 Free Fall
  • 3.6 Finding Velocity and Displacement from Acceleration
  • 4.1 Displacement and Velocity Vectors
  • 4.2 Acceleration Vector
  • 4.3 Projectile Motion
  • 4.4 Uniform Circular Motion
  • 4.5 Relative Motion in One and Two Dimensions
  • 5.2 Newton's First Law
  • 5.3 Newton's Second Law
  • 5.4 Mass and Weight
  • 5.5 Newton’s Third Law
  • 5.6 Common Forces
  • 5.7 Drawing Free-Body Diagrams
  • 6.1 Solving Problems with Newton’s Laws
  • 6.2 Friction
  • 6.3 Centripetal Force
  • 6.4 Drag Force and Terminal Speed
  • 7.2 Kinetic Energy
  • 7.3 Work-Energy Theorem
  • 8.1 Potential Energy of a System
  • 8.2 Conservative and Non-Conservative Forces
  • 8.3 Conservation of Energy
  • 8.4 Potential Energy Diagrams and Stability
  • 8.5 Sources of Energy
  • 9.1 Linear Momentum
  • 9.2 Impulse and Collisions
  • 9.3 Conservation of Linear Momentum
  • 9.4 Types of Collisions
  • 9.5 Collisions in Multiple Dimensions
  • 9.6 Center of Mass
  • 9.7 Rocket Propulsion
  • 10.1 Rotational Variables
  • 10.2 Rotation with Constant Angular Acceleration
  • 10.3 Relating Angular and Translational Quantities
  • 10.4 Moment of Inertia and Rotational Kinetic Energy
  • 10.5 Calculating Moments of Inertia
  • 10.6 Torque
  • 10.7 Newton’s Second Law for Rotation
  • 10.8 Work and Power for Rotational Motion
  • 11.1 Rolling Motion
  • 11.2 Angular Momentum
  • 11.3 Conservation of Angular Momentum
  • 11.4 Precession of a Gyroscope
  • 12.1 Conditions for Static Equilibrium
  • 12.3 Stress, Strain, and Elastic Modulus
  • 12.4 Elasticity and Plasticity
  • 13.1 Newton's Law of Universal Gravitation
  • 13.2 Gravitation Near Earth's Surface
  • 13.3 Gravitational Potential Energy and Total Energy
  • 13.4 Satellite Orbits and Energy
  • 13.5 Kepler's Laws of Planetary Motion
  • 13.6 Tidal Forces
  • 13.7 Einstein's Theory of Gravity
  • 14.1 Fluids, Density, and Pressure
  • 14.2 Measuring Pressure
  • 14.3 Pascal's Principle and Hydraulics
  • 14.4 Archimedes’ Principle and Buoyancy
  • 14.5 Fluid Dynamics
  • 14.6 Bernoulli’s Equation
  • 14.7 Viscosity and Turbulence
  • 15.1 Simple Harmonic Motion
  • 15.2 Energy in Simple Harmonic Motion
  • 15.3 Comparing Simple Harmonic Motion and Circular Motion
  • 15.4 Pendulums
  • 15.5 Damped Oscillations
  • 15.6 Forced Oscillations
  • 16.1 Traveling Waves
  • 16.2 Mathematics of Waves
  • 16.3 Wave Speed on a Stretched String
  • 16.4 Energy and Power of a Wave
  • 16.5 Interference of Waves
  • 16.6 Standing Waves and Resonance
  • 17.1 Sound Waves
  • 17.2 Speed of Sound
  • 17.3 Sound Intensity
  • 17.4 Normal Modes of a Standing Sound Wave
  • 17.5 Sources of Musical Sound
  • 17.7 The Doppler Effect
  • 17.8 Shock Waves
  • B | Conversion Factors
  • C | Fundamental Constants
  • D | Astronomical Data
  • E | Mathematical Formulas
  • F | Chemistry
  • G | The Greek Alphabet

Learning Objectives

By the end of this section, you will be able to:

  • Identify and analyze static equilibrium situations
  • Set up a free-body diagram for an extended object in static equilibrium
  • Set up and solve static equilibrium conditions for objects in equilibrium in various physical situations

All examples in this chapter are planar problems. Accordingly, we use equilibrium conditions in the component form of Equation 12.7 to Equation 12.9 . We introduced a problem-solving strategy in Example 12.1 to illustrate the physical meaning of the equilibrium conditions. Now we generalize this strategy in a list of steps to follow when solving static equilibrium problems for extended rigid bodies. We proceed in five practical steps.

Problem-Solving Strategy

Static equilibrium.

  • Identify the object to be analyzed. For some systems in equilibrium, it may be necessary to consider more than one object. Identify all forces acting on the object. Identify the questions you need to answer. Identify the information given in the problem. In realistic problems, some key information may be implicit in the situation rather than provided explicitly.
  • Set up a free-body diagram for the object. (a) Choose the xy -reference frame for the problem. Draw a free-body diagram for the object, including only the forces that act on it. When suitable, represent the forces in terms of their components in the chosen reference frame. As you do this for each force, cross out the original force so that you do not erroneously include the same force twice in equations. Label all forces—you will need this for correct computations of net forces in the x - and y -directions. For an unknown force, the direction must be assigned arbitrarily; think of it as a ‘working direction’ or ‘suspected direction.’ The correct direction is determined by the sign that you obtain in the final solution. A plus sign ( + ) ( + ) means that the working direction is the actual direction. A minus sign ( − ) ( − ) means that the actual direction is opposite to the assumed working direction. (b) Choose the location of the rotation axis; in other words, choose the pivot point with respect to which you will compute torques of acting forces. On the free-body diagram, indicate the location of the pivot and the lever arms of acting forces—you will need this for correct computations of torques. In the selection of the pivot, keep in mind that the pivot can be placed anywhere you wish, but the guiding principle is that the best choice will simplify as much as possible the calculation of the net torque along the rotation axis.
  • Set up the equations of equilibrium for the object. (a) Use the free-body diagram to write a correct equilibrium condition Equation 12.7 for force components in the x -direction. (b) Use the free-body diagram to write a correct equilibrium condition Equation 12.11 for force components in the y -direction. (c) Use the free-body diagram to write a correct equilibrium condition Equation 12.9 for torques along the axis of rotation. Use Equation 12.10 to evaluate torque magnitudes and senses.
  • Simplify and solve the system of equations for equilibrium to obtain unknown quantities. At this point, your work involves algebra only. Keep in mind that the number of equations must be the same as the number of unknowns. If the number of unknowns is larger than the number of equations, the problem cannot be solved.
  • Evaluate the expressions for the unknown quantities that you obtained in your solution. Your final answers should have correct numerical values and correct physical units. If they do not, then use the previous steps to track back a mistake to its origin and correct it. Also, you may independently check for your numerical answers by shifting the pivot to a different location and solving the problem again, which is what we did in Example 12.1 .

Note that setting up a free-body diagram for a rigid-body equilibrium problem is the most important component in the solution process. Without the correct setup and a correct diagram, you will not be able to write down correct conditions for equilibrium. Also note that a free-body diagram for an extended rigid body that may undergo rotational motion is different from a free-body diagram for a body that experiences only translational motion (as you saw in the chapters on Newton’s laws of motion). In translational dynamics, a body is represented as its CM, where all forces on the body are attached and no torques appear. This does not hold true in rotational dynamics, where an extended rigid body cannot be represented by one point alone. The reason for this is that in analyzing rotation, we must identify torques acting on the body, and torque depends both on the acting force and on its lever arm. Here, the free-body diagram for an extended rigid body helps us identify external torques.

Example 12.3

The torque balance.

w 1 = m 1 g w 1 = m 1 g is the weight of mass m 1 ; m 1 ; w 2 = m 2 g w 2 = m 2 g is the weight of mass m 2 ; m 2 ;

w = m g w = m g is the weight of the entire meter stick; w 3 = m 3 g w 3 = m 3 g is the weight of unknown mass m 3 ; m 3 ;

F S F S is the normal reaction force at the support point S .

We choose a frame of reference where the direction of the y -axis is the direction of gravity, the direction of the x -axis is along the meter stick, and the axis of rotation (the z -axis) is perpendicular to the x -axis and passes through the support point S . In other words, we choose the pivot at the point where the meter stick touches the support. This is a natural choice for the pivot because this point does not move as the stick rotates. Now we are ready to set up the free-body diagram for the meter stick. We indicate the pivot and attach five vectors representing the five forces along the line representing the meter stick, locating the forces with respect to the pivot Figure 12.10 . At this stage, we can identify the lever arms of the five forces given the information provided in the problem. For the three hanging masses, the problem is explicit about their locations along the stick, but the information about the location of the weight w is given implicitly. The key word here is “uniform.” We know from our previous studies that the CM of a uniform stick is located at its midpoint, so this is where we attach the weight w , at the 50-cm mark.

Now we can find the five torques with respect to the chosen pivot:

The second equilibrium condition (equation for the torques) for the meter stick is

When substituting torque values into this equation, we can omit the torques giving zero contributions. In this way the second equilibrium condition is

Selecting the + y + y -direction to be parallel to F → S , F → S , the first equilibrium condition for the stick is

Substituting the forces, the first equilibrium condition becomes

We solve these equations simultaneously for the unknown values m 3 m 3 and F S . F S . In Equation 12.17 , we cancel the g factor and rearrange the terms to obtain

To obtain m 3 m 3 we divide both sides by r 3 , r 3 , so we have

To find the normal reaction force, we rearrange the terms in Equation 12.18 , converting grams to kilograms:

Significance

Check your understanding 12.3.

Repeat Example 12.3 using the left end of the meter stick to calculate the torques; that is, by placing the pivot at the left end of the meter stick.

In the next example, we show how to use the first equilibrium condition (equation for forces) in the vector form given by Equation 12.7 and Equation 12.8 . We present this solution to illustrate the importance of a suitable choice of reference frame. Although all inertial reference frames are equivalent and numerical solutions obtained in one frame are the same as in any other, an unsuitable choice of reference frame can make the solution quite lengthy and convoluted, whereas a wise choice of reference frame makes the solution straightforward. We show this in the equivalent solution to the same problem. This particular example illustrates an application of static equilibrium to biomechanics.

Example 12.4

Forces in the forearm.

Notice that in our frame of reference, contributions to the second equilibrium condition (for torques) come only from the y -components of the forces because the x -components of the forces are all parallel to their lever arms, so that for any of them we have sin θ = 0 sin θ = 0 in Equation 12.10 . For the y -components we have θ = ± 90 ° θ = ± 90 ° in Equation 12.10 . Also notice that the torque of the force at the elbow is zero because this force is attached at the pivot. So the contribution to the net torque comes only from the torques of T y T y and of w y . w y .

and the y -component of the net force satisfies

Equation 12.21 and Equation 12.22 are two equations of the first equilibrium condition (for forces). Next, we read from the free-body diagram that the net torque along the axis of rotation is

Equation 12.23 is the second equilibrium condition (for torques) for the forearm. The free-body diagram shows that the lever arms are r T = 1.5 in . r T = 1.5 in . and r w = 13.0 in . r w = 13.0 in . At this point, we do not need to convert inches into SI units, because as long as these units are consistent in Equation 12.23 , they cancel out. Using the free-body diagram again, we find the magnitudes of the component forces:

We substitute these magnitudes into Equation 12.21 , Equation 12.22 , and Equation 12.23 to obtain, respectively,

When we simplify these equations, we see that we are left with only two independent equations for the two unknown force magnitudes, F and T , because Equation 12.21 for the x -component is equivalent to Equation 12.22 for the y -component. In this way, we obtain the first equilibrium condition for forces

and the second equilibrium condition for torques

The magnitude of tension in the muscle is obtained by solving Equation 12.25 :

The force at the elbow is obtained by solving Equation 12.24 :

The negative sign in the equation tells us that the actual force at the elbow is antiparallel to the working direction adopted for drawing the free-body diagram. In the final answer, we convert the forces into SI units of force. The answer is

The second equilibrium condition, τ T + τ w = 0 , τ T + τ w = 0 , can be now written as

From the free-body diagram, the first equilibrium condition (for forces) is

Equation 12.26 is identical to Equation 12.25 and gives the result T = 433.3 lb . T = 433.3 lb . Equation 12.27 gives

We see that these answers are identical to our previous answers, but the second choice for the frame of reference leads to an equivalent solution that is simpler and quicker because it does not require that the forces be resolved into their rectangular components.

Check Your Understanding 12.4

Repeat Example 12.4 assuming that the forearm is an object of uniform density that weighs 8.896 N.

Example 12.5

A ladder resting against a wall.

the net force in the y -direction is

and the net torque along the rotation axis at the pivot point is

where τ w τ w is the torque of the weight w and τ F τ F is the torque of the reaction F . From the free-body diagram, we identify that the lever arm of the reaction at the wall is r F = L = 5.0 m r F = L = 5.0 m and the lever arm of the weight is r w = L / 2 = 2.5 m . r w = L / 2 = 2.5 m . With the help of the free-body diagram, we identify the angles to be used in Equation 12.10 for torques: θ F = 180 ° − β θ F = 180 ° − β for the torque from the reaction force with the wall, and θ w = 180 ° + ( 90 ° − β ) θ w = 180 ° + ( 90 ° − β ) for the torque due to the weight. Now we are ready to use Equation 12.10 to compute torques:

We substitute the torques into Equation 12.30 and solve for F : F :

We obtain the normal reaction force with the floor by solving Equation 12.29 : N = w = 400.0 N . N = w = 400.0 N . The magnitude of friction is obtained by solving Equation 12.28 : f = F = 150.7 N . f = F = 150.7 N . The coefficient of static friction is μ s = f / N = 150.7 / 400.0 = 0.377 . μ s = f / N = 150.7 / 400.0 = 0.377 .

The net force on the ladder at the contact point with the floor is the vector sum of the normal reaction from the floor and the static friction forces:

Its magnitude is

and its direction is

We should emphasize here two general observations of practical use. First, notice that when we choose a pivot point, there is no expectation that the system will actually pivot around the chosen point. The ladder in this example is not rotating at all but firmly stands on the floor; nonetheless, its contact point with the floor is a good choice for the pivot. Second, notice when we use Equation 12.10 for the computation of individual torques, we do not need to resolve the forces into their normal and parallel components with respect to the direction of the lever arm, and we do not need to consider a sense of the torque. As long as the angle in Equation 12.10 is correctly identified—with the help of a free-body diagram—as the angle measured counterclockwise from the direction of the lever arm to the direction of the force vector, Equation 12.10 gives both the magnitude and the sense of the torque. This is because torque is the vector product of the lever-arm vector crossed with the force vector, and Equation 12.10 expresses the rectangular component of this vector product along the axis of rotation.

Check Your Understanding 12.5

For the situation described in Example 12.5 , determine the values of the coefficient μ s μ s of static friction for which the ladder starts slipping, given that β β is the angle that the ladder makes with the floor.

Example 12.6

Forces on door hinges.

We select the pivot at point P (upper hinge, per the free-body diagram) and write the second equilibrium condition for torques in rotation about point P :

We use the free-body diagram to find all the terms in this equation:

In evaluating sin β , sin β , we use the geometry of the triangle shown in part (a) of the figure. Now we substitute these torques into Equation 12.32 and compute B x : B x :

Therefore the magnitudes of the horizontal component forces are A x = B x = 100.0 N . A x = B x = 100.0 N . The forces on the door are

The forces on the hinges are found from Newton’s third law as

Check Your Understanding 12.6

Solve the problem in Example 12.6 by taking the pivot position at the center of mass.

Check Your Understanding 12.7

A 50-kg person stands 1.5 m away from one end of a uniform 6.0-m-long scaffold of mass 70.0 kg. Find the tensions in the two vertical ropes supporting the scaffold.

Check Your Understanding 12.8

A 400.0-N sign hangs from the end of a uniform strut. The strut is 4.0 m long and weighs 600.0 N. The strut is supported by a hinge at the wall and by a cable whose other end is tied to the wall at a point 3.0 m above the left end of the strut. Find the tension in the supporting cable and the force of the hinge on the strut.

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Solving for Equilibrium

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  • Onkar Gulati

When a force is applied on a body, it may change the state of motion of the body. Motions created by the applied force can be translational or rotational, but not all forces cause translational or rotational motion. You may like understanding the concept of equilibrium .

Understanding balanced and unbalanced forces

Equilibrium of a particle, gravitational force or weight.

Balanced Force: Two or more forces are said to be balanced if the net force of the all the forces is equal to zero . Balanced forces do not cause any motion in the object they are acting on. \(_\square\)
Unbalanced Force: Two or more forces are said to be unbalanced if the net force of the all the forces is equal to not equal to zero . Unbalanced forces always cause motion in the object they are acting on. \(_\square\)

A rigid body is one that does not deform under the application of an external force. For a rigid body to be in a state of equilibrium, the net force on the object must be zero. That is, if an object remains motionless, then Newton's second law, \(F=ma\), tells us that the total external force acting on the object is zero. The total external force on an object is also zero, if the body moves with a constant nonzero velocity.

The whole goal when solving for equilibrium is to find out what the various forces have to do so that there is zero net force on each body \(\left(\sum { F=0 }\right). \) Since force is a vector, we have to take into consideration the direction in which the force acts.

Consider an object that is on top of a frictionless table. From the figure below what single force should be acted so that the object is in equilibrium? What will be the direction of the force? For convention let the right direction be positive, and left direction negative. Also, let the force required be \(X.\) Now balance all the forces: \[\begin{align} \sum {F} &=0 \\ X+4\text{ N}+3.5\text{ N}-3\text{ N}-1.5\text{ N}&=0\\ \Rightarrow X&=4.5\text{ N}-7.5\text{ N}\\ &=-3\text{ N}. \end{align}\] Since the negative sign means left direction, a force of \(3N\) must act to the left so that the object is in equilibrium. \(_\square\)

A common force when solving for equilibrium is weight. Weight is the direct application of Newton's law of gravitation and for an object near the surface of the earth it is given by

where \(m\) is the mass of the body, and \(g\) is the gravitational acceleration and is equal to approximately \(9.8\text{ m/s}^2\). Note that weight always acts downwards toward the center of the earth.

Another common force when solving for equilibrium is tension . Tension is the general name for the pulling force of a rope, stick, cable, etc. Tension has a unit of force and can be measured in newtons.

If a block of mass \(3\text{ kg}\) is hanging from a ceiling by a massless rope, what will be the tension on the rope? Take \(g\) to be \(9.8\text{ m/s}^{2}\). fkjashf \[\] We know that the block is acted upon by a gravitational force equal to \(mg\), and since the object is motionless there must be another force to balance it. That force is tension, which we can get as follows: \[\begin{align} \sum F&=0 \\ T-mg&=0\\ \Rightarrow T&=(3\text{ kg})\left(9.8\text{ m/s}^{2}\right)\\ &=29.4\text{ N}. \ _\square \end{align}\]
A stick-man who has a mere mass of \(3\text{ kg}\) is sitting on a swing and is at rest, as shown in the figure below. If the platform he is sitting upon has a mass of \(5\text{ kg},\) what will be the tension in each individual rope that holds the swing? Assume that the weight is equally divided by the two ropes. ![dkf;l](https://i.imgur.com/YUzIjov.png](https://i.imgur.com/YUzIjov.png){: .center} Consider the system of the stick-man and the platform. Since the mass is equally distributed, let the tension in each rope be \(T.\) dslkf Then since the system is in equilibrium, we have \[\begin{align} \sum F&=0 \\ T+T-(5\text{ kg})(g)-(3\text{ kg})(g)&=0\\ 2T&=8g\cdot\text{ kg}\\ \Rightarrow T&=4g\cdot \text{ kg}\\ &\approx4\times 9.8 \text{ m/s}^2 \cdot \text{ kg}\\ &= 39.2\text{ N}. \ _\square \end{align}\]

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Physics LibreTexts

8.4: Solving Statics Problems

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learning objectives

  • Formulate and apply six steps to solve static problems

Statics is the study of forces in equilibrium. Recall that Newton’s second law states:

\[\mathrm{∑F=ma}\]

Therefore, for all objects moving at constant velocity (including a velocity of 0 — stationary objects), the net external force is zero. There are forces acting, but they are balanced — that is to say, they are “in equilibrium. ”

When solving equilibrium problems, it might help to use the following steps:

  • First, ensure that the problem you’re solving is in fact a static problem—i.e., that no acceleration (including angular acceleration) is involved Remember:\(\mathrm{∑F=ma=0}\) for these situations. If rotational motion is involved, the condition \(\mathrm{∑τ=Iα=0}\) must also be satisfied, where is torque, is the moment of inertia, and is the angular acceleration.
  • Choose a pivot point. Often this is obvious because the problem involves a hinge or a fixed point. If the choice is not obvious, pick the pivot point as the location at which you have the most unknowns. This simplifies things because forces at the pivot point create no torque because of the cross product:\(\mathrm{τ=rF}\)
  • Write an equation for the sum of torques, and then write equations for the sums of forces in the x and y directions. Set these sums equal to 0. Be careful with your signs.
  • Solve for your unknowns.
  • Insert numbers to find the final answer.
  • Check if the solution is reasonable by examining the magnitude, direction, and units of the answer. The importance of this last step cannot be overstated, although in unfamiliar applications, it can be more difficult to judge reasonableness. However, these judgments become progressively easier with experience.
  • First, ensure that the problem you’re solving is in fact a static problem—i.e., that no acceleration (including angular acceleration ) is involved.
  • Choose a pivot point — use the location at which you have the most unknowns.
  • Write equations for the sums of torques and forces in the x and y directions.
  • Solve the equations for your unknowns algebraically, and insert numbers to find final answers.
  • torque : A rotational or twisting effect of a force; (SI unit newton-meter or Nm; imperial unit foot-pound or ft-lb)
  • moment of inertia : A measure of a body’s resistance to a change in its angular rotation velocity

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  • Curation and Revision. Provided by : Boundless.com. License : CC BY-SA: Attribution-ShareAlike

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  • OpenStax College, College Physics. September 17, 2013. Provided by : OpenStax CNX. Located at : http://cnx.org/content/m42173/latest/?collection=col11406/1.7 . License : CC BY: Attribution
  • OpenStax College, College Physics. September 17, 2013. Provided by : OpenStax CNX. Located at : http://cnx.org/content/m42167/latest/?collection=col11406/1.7 . License : CC BY: Attribution
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Engineering LibreTexts

3.4: Particles in Two Dimensions

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  • Daniel W. Baker and William Haynes
  • Colorado State University via Engineeringstatics

Introduction

In this section we will study situations where everything of importance occurs in a 2-dimensional plane and the third dimension is not involved. Studying two-dimensional problems is worthwhile, because they illustrate all the important principles of engineering statics while being easier to visualize and less mathematically complex.

We will normally work in the “plane of the page,” that is, a two-dimensional Cartesian plane with a horizontal \(x\) axis and a vertical \(y\) axis discussed in Section 2.3 previously. This coordinate system can represent either the front, side, or top view of a system as appropriate. In some problems it may be worthwhile to rotate the coordinate system, that is, to establish a coordinate system where the \(x\) and \(y\) axes are not horizontal and vertical. This is usually done to simplify the mathematics by avoiding simultaneous equations.

General Procedure

The general procedure for solving equilibrium of a particle problems in two dimensions is to:

  • Identify the particle. The particle will be the object or point where the lines of action of all the forces intersect.
  • Establish a coordinate system. Normally this will be a system with the origin at the particle and a horizontal \(x\) axis and a vertical \(y\) axis, though it may be advantageous to align one axis with an unknown.
  • Draw a free-body diagram. The FBD shows the object and all the forces acting on it, and defines the symbols we will use. Every force should be labeled with a roman letter to represent its magnitude and, unless it aligns with a coordinate axis, a greek letter or degree measure for its direction.
  • State any given values and identify the unknown values.
  • Find trivial angles. Some angles may be easily found from the geometry of the problem. If that is the case, draw a simple, labeled triangle and use trigonometry to determine the measure of the angle.
  • Count knowns and unknowns. At this point you should have no more than two unknowns remaining. If you don't, reread the problem and look for overlooked information. When solving mechanics problems, it is always helpful to know what you know and what you are looking for and this information changes as you work through your solution.
  • Formulate equilibrium equations. Based on the free-body diagram, and using the symbols you have selected, formulate an equilibrium equation using one of the methods described in this section. The choice of method is up to you, and as you gain experience you will be able to identify the ‘best’ approach.
  • Simplify. Use algebra to simplify the equilibrium equations. Get them into a form where the unknown values are alone on the left of the equals sign. Work symbolically as long as you can and avoid the temptation to insert numeric values prematurely, because this tends to lead to errors and obscures the relationships between the forces and angles.
  • Substitute values for symbols. When your equilibrium equations have been fully simplified in symbolic form, pull out your calculator and substitute the known values and calculate the unknowns. Indicate the units of your results, and underline or box your answers.
  • Check your work. Have you made any algebra or trig mistakes? If you add the forces graphically do they appear to add to zero? Do the results seem reasonable given the situation? Have you included appropriate units? If you have time, work the problem using another approach and compare answers.

Force Triangle Method

The force triangle method is applicable to situations where there are (exactly) three forces acting on a particle, and no more than two unknown magnitudes or directions.

If such a particle is in equilibrium then the three forces must add to zero. Graphically, if you arrange the force vectors tip-to-tail, they will form a closed, three-sided polygon, i.e. a triangle. This is illustrated in Figure 3.4.1.

Move the slider to rearrange the forces acting on the particle into the corresponding force triangle.

Question 3.4.2 .

Why do the forces always form a closed polygon?

Because their resultant is zero.

The force triangle is a graphical representation of the vector equilibrium equation (3.1.1). It can be used to solve for the unknown values in a number of different ways, which will be illustrated in the next two examples. In Example 3.4.3 We will use a graphical approach to find the forces causing equilibrium, and in Example 3.4.4 we will use trigonometry to to solve for the unknown forces mathematically.

In the next example we will use technology to draw a scaled diagram of the force triangle representing the equilibrium situation. We are using Geogebra to make the drawing, but you could use CAD, another drawing program, or even a ruler and protractor as you prefer. Since the diagram is accurately drawn, the lengths and angles represent the the magnitudes and directions of the forces which hold the particle in equilibrium.

Example 3.4.3 . Frictionless Incline.

A force \(P\) is being applied to a \(\lb{100}\) block resting on a frictionless incline as shown. Determine the magnitude and direction of force \(P\) and of the contact force on the bottom of the block.

\begin{align*} P \amp = \lb{43.8 }\text{ at }10° \measuredangle\\ N \amp = \lb{102}\text{ at }115° \measuredangle \end{align*}

1. Assumptions.

We must assume that the block is in equilibrium, that is, either motionless or moving at a constant velocity in order to use the equilibrium equations. We will represent the block’s weight and the force between the incline and the block as concentrated forces. The force of the inclined surface on the block must act in a direction which is normal to the surface since it is frictionless and can't prevent motion along the surface.

The knowns here are the weight of the block, the direction of the applied force, and the slope of the incline. The slope of the incline provides the direction of the normal force.

The unknown values are the magnitudes of forces \(P\) and \(N\text{.}\)

We define three symbols, \(W\text{,}\) \(N\text{,}\) and \(P\text{,}\) representing the weight, normal force, and the applied force respectively. The angles could be given symbols too, but since we know their values it isn't necessary.

3. Free Body Diagram.

Notice that the force \(N\) is represented as acting 25° from the \(y\) axis, which is 90° away from the direction of the surface.

4. Force Triangle.

Use the known information to carefully and accurately construct the force triangle.

  • Draw force \(\vec{N}\) from point \(C\) back to point \(A\text{.}\)

5. Results.

In steps 6 and 7, Geogebra tells us that p = (0.438;10.0°) which means force \(P\) is 0.438 units long with a direction of 10°, similarly n = (1.02;115°) means \(N\) is 1.02 units long at 115°. These angles are measured counterclockwise from the positive \(x\) axis.

These are not the answers we are looking for, but we’re close. Remember that for this diagram, our scale is

\[ 1 \text{ unit} = 100 \text{ lbs}\text{,} \nonumber \]

so scaling the lengths of p and n by this factor gives

\begin{align*} P \amp = (\unit{0.438}) (\lb{100}/\unit{})\\ \amp = \lb{43.8} \text{ at } 10° \measuredangle\\ N \amp = ( \unit{1.02}) ({\lb{100}}/\unit{})\\ \amp = \lb{102} \text{ at } 115° \measuredangle\text{.} \end{align*}

If you use technology such as Geogebra, as we did here, or CAD software to draw the force triangle, it will accurately produce the solution.

If technology isn't available to you, such as during an exam, you can still use a ruler and protractor draw the force triangle, but your results will only be as accurate as your diagram. In the best case, using a sharp pencil and carefully measuring lengths and angles, you can only expect about two significant digits of accuracy from an hand drawn triangle. Nevertheless, even a roughly drawn triangle can give you an idea of the correct answers and be used to check your work after you use another method to solve the problem.

Trigonometric Method

The general approach for solving particle equilibrium problems using the trigonometric method is to:

  • Draw and label a free-body diagram.
  • Rearrange the forces into a force triangle and label it.
  • Identify the knowns and unknowns.
  • Use trigonometry to find the unknown sides or angles of the triangle.

There must be no more than two unknowns to use this method, which may be either magnitudes or directions. During the problem setup you will probably need to use the geometry of the situation to find one or more angles.

If the force triangle has a right angle you can use Section B.2 to find the unknown values, but in most cases the triangle will be oblique and you will need to use either or both of the Law of Sines or the Law of Cosines to find the sides or angles.

Example 3.4.4 . Cargo Boom.

A \(\kN{24}\) crate is being lowered into the cargo hold of a ship. Boom \(AB\) is \(\m{20}\) long and acts at a 40° angle from kingpost \(AC\text{.}\) The boom is held in this position by topping lift \(BC\) which has a 1:4 slope.

Determine the forces in the boom and in the topping lift.

\begin{align*} T\amp= \kN{17.16 } \amp C \amp= \kN{25.9} \end{align*}

1. Draw diagrams.

Start by identifying the particle and drawing a free-body diagram. The particle in this case is point \(B\) at the end of the boom because it is the point where all three forces intersect. Let \(T\) be the tension of the topping lift, \(C\) be the force in the boom, and \(W\) be the weight of the load. Let \(\alpha\) and \(\beta\) be the angles that forces \(T\) and \(C\) make with the horizontal.

Rearrange the forces acting on point \(B\) to form a force triangle as was done in the previous example.

Angle \(\alpha\) can be found from the slope of the topping lift.

\[ \alpha = \tan^{-1}\left(\frac{1}{4}\right) = 14.0°\text{.} \nonumber \]

Angle \(\beta\) is the complement of the 40° angle the boom makes with the vertical kingpost.

\[ \beta = 90° - 40° = 50° \nonumber \]

Use these values to find the three angles in the force triangle.

\begin{align*} \theta_1 \amp = \alpha + \beta = 64.0°\\ \theta_2 \amp = 90° - \alpha = 76.0°\\ \theta_3 \amp = 90° - \beta = 40.0° \end{align*}

With the angles and one side of the force triangle known, apply the Law of Sines to find the two unknown sides.

\[ \frac{\sin \theta_1}{W} = \frac{\sin \theta_2}{C} = \frac{\sin \theta_3}{T} \nonumber \]

\begin{align*} T \amp = W \left(\frac{\sin \theta_3}{\sin \theta_1}\right) \amp C \amp = W \left(\frac{\sin \theta_2}{\sin \theta_1}\right)\\ T \amp= \kN{24} \left(\frac{\sin 40.0°}{\sin 64.0°}\right) \amp C \amp =\kN{24} \left(\frac{\sin 76.0°}{\sin 64.0°}\right)\\ T\amp= \kN{17.16}\amp C \amp= \kN{25.9} \end{align*}

The general statement of equilibrium of forces, (3.1.1), can be expressed as the sum of forces in the \(\ihat\text{,}\) \(\jhat\) and \(\khat\) directions

\begin{equation} \Sigma\vec{F} = \Sigma F_x\ \ihat + \Sigma F_y\ \jhat + \Sigma F_z\ \khat = 0\text{.}\tag{3.4.1} \end{equation}

This statement will can only be true if all three coefficients of the unit vectors are themselves equal to zero, leading to this scalar interpretation of the equilibrium equation

\begin{equation} \Sigma\vec{F} = 0 \implies \begin{cases}\Sigma F_x = 0\\\Sigma F_y = 0\\ \Sigma F_z = 0 \end{cases} \qquad (\textrm{three dimensions})\text{.}\label{scalar-equilibrium-3d}\tag{3.4.2} \end{equation}

In other words the single vector equilibrium equation is equivalent to three independent scalar equations, one for each coordinate direction.

In two-dimensional situations, no forces act in the \(\khat\) direction leaving just these two equilibrium equations to be satisfied

\begin{equation} \Sigma\vec{F} = 0 \implies \begin{cases}\Sigma F_x = 0\\\Sigma F_y = 0 \end{cases} \qquad (\textrm{two-dimensions}) \text{.}\label{equations-2d-particle}\tag{3.4.3} \end{equation}

We will use this equation as the basis for solving two-dimensional particle equilibrium problems in this section and equation (3.4.2) for three-dimensional problems in Section 3.5.

You are undoubtedly familiar with utility poles which carry electric, cable and telephone lines, but have you ever noticed as you drive down a winding road that the poles will switch from one side of the road to the other and back again? Why is this?

If you consider the forces acting on the top of a pole beside a curving section of road you'll observe that the tensions of the cables produce a net force towards the road. This force is typically opposed by a “guy wire” pulling in the opposite direction which prevents the pole from tipping over due to unbalanced forces. The power company tries to keep poles beside road segments with convex curvature. If they didn't switch sides, the guy wire for poles at concave curves would extend into the road... which is a poor design.

Example 3.4.5 . Utility Pole.

Consider the utility pole next to the road shown below. A top view is shown in the right hand diagram. If each of the six cables pulls with a force of \(\kN{10.0}\text{,}\) determine the magnitude of the tension in the guy wire.

pole.png

\begin{gather*} G = \kN{14.5} \end{gather*}

A utility pole isn't two-dimensional, but we will consider the top view and forces in the horizontal plane only.

It also isn't a concurrent force problem because the lines of action of the forces don't all intersect at a single point. However, we can make it into one by replacing the forces of the three cables in each direction with a single force three times larger. This is an example of an equivalent transformation , a trick engineers use frequently to turn complex situations into simpler ones. It works here because all the tensions are equal, and the outside wires are equidistant from the center wire. You must be careful to to justify all equivalent transformations, because they will lead to errors if they are not applied correctly. Equivalent transformations will be discussed in greater detail in Section 4.6 later.

\(T = \kN{10.0}\)

3. Procedure.

Begin by drawing a neat, labeled, free-body diagram of the pole, establishing a coordinate system and indicating the directions of the forces. Although it is not necessary, it simplifies this problem considerably to note the symmetry and establish the \(x\) axis along the axis of symmetry. Let \(T\) be the tension in one wire, and \(G\) be the tension of the guy wire.

To solve apply the equations of equilibrium. The symmetry of this problem means that the \(\Sigma F_x\) equation is sufficient.

\begin{align*} \Sigma F_x \amp = 0\\ G - 6\,T_x \amp= 0\\ G \amp= 6\,(T \cos 76°)\\ \amp= \kN{14.5} \end{align*}

This problem could have also been solved using the force triangle method. See Subsection 3.4.3.

In the next example we look at the conditions of equilibrium by considering the load and the constraints, rather than taking a global equilibrium approach which considers both the load and reaction forces.

Example 3.4.6 . Slider.

Three forces act on a machine part which is free to slide along a vertical, frictionless rod. Forces \(A\) and \(B\) have a magnitude of \(\N{20}\) and force \(C\) has a magnitude of \(\N{30}\text{.}\) Force \(B\) acts \(\alpha\) degrees from the horizontal, and force \(C\) acts at the same angle from the vertical.

Determine the angle \(\alpha\) required for equilibrium, and the magnitude and direction of the reaction force acting on the slider.

The question asks for the reaction force. The reaction force \(\vec{R'}\) is equal and opposite to force \(\vec{R}\text{.}\)

\begin{align*} \vec{R'} \amp = - \vec{R} = \N{26.15} \leftarrow \\ \amp = \langle \N{-26.15}, 0 \rangle \end{align*}

We are given magnitudes of forces \(A= \N{20}\text{,}\) \(B = \N{20}\text{,}\) and \(C = \N{30}\text{.}\) The unknowns are angle \(\alpha\) and resultant force \(R\text{.}\)

This interactive diagram for Example 3.4.6 will help you visualize the relationship between angle \(\alpha\) and the resultant load force \(R\text{.}\) When in equilibrium, the rod will supply a reaction force equal and opposite to the resultant load.

2. Procedure.

Since the rod is frictionless, it cannot prevent the slider from moving vertically. Consequently the slider will only be in equilibrium is if the resultant of the three load forces is horizontal. Since a horizontal force has no \(y\) component, we can establish this equilibrium condition:

\[ R_y = \Sigma F_y = A_y + B_y + C_y = 0 \nonumber \]

Inserting the known values into the equilibrium relation and simplifying gives an equation in terms of unknown angle \(\alpha\text{.}\)

\begin{align*} R_y = A_y + B_y + C_y \amp = 0\\ A + B \sin \alpha - C \cos \alpha \amp = 0\\ 20 + 20 \sin \alpha - 30 \cos \alpha \amp = 0\\ 2 + 2 \sin \alpha - 3 \cos \alpha \amp= 0 \end{align*}

This is a single equation with a single unknown, although is is not particularly easy to solve with algebra. One approach is described here . An alternate approach is to use technology to graph the function \(y(x) = 2 + 2 \sin x - 3 \cos x\text{.}\) The roots of this equation correspond to values of \(\alpha\) which satisfy the equilibrium condition above. The root occurring closest to \(x=0\) will be the answer corresponding to our problem, in this case \(\alpha = 22.62°\) which you can verify by plugging it back into the equilibrium equation. Note that -90° also satisfies this equation, but it is not the solution we are looking for.

Once \(\alpha\) is known, we can find the reaction force by adding the \(x\) components of \(A\text{,}\) \(B\text{,}\) and \(C\text{.}\)

\begin{align*} R_x \amp = A_x + B_x + C_x\\ \amp = A + B \cos \alpha + C \sin \alpha\\ \amp = 0 + 10 \cos(22.62°) + 20 \sin(22.62°)\\ \amp = \N{26.15} \end{align*}

By inspection the sum of \(A\text{,}\) \(B\text{,}\) and \(C\) acts to the right, and the resultant force \(\vec{R}\) is the vector sum of \(R_x\) and \(R_y\text{.}\) In this case, because \(R_y\) = 0,

\begin{align*} \vec{R} \amp = R_x \rightarrow \\ \amp = \langle \N{26.15}, 0 \rangle \end{align*}

The next example demonstrates how rotating the coordinate system can simplify the solution. In the first solution, the standard orientation of the \(x\) and \(y\) axes is chosen, and in the second the coordinate system is rotated to align with one of the unknowns, which enables the solution to be found without solving simultaneous equations.

Example 3.4.7 . Roller.

A lawn roller which weighs \(\lb{160}\) is being pulled up a \(\ang{10}\) slope at a constant velocity.

Determine the required pulling force \(P\text{.}\)

\[ P = \lb{32.1} \nonumber \]

  • Select a coordinate system, in this case horizontal and vertical.
  • Draw a free-body digram
  • Solve the equations of equilibrium using the scalar approach.

Solving simultaneously for \(P\)

\begin{align*} 0.643 P + 0.985 ( 4.40 P) \amp = \lb{160}\\ 4.98 P \amp = \lb{160}\\ P \amp = \lb{32.1} \end{align*}

1. Strategy.

  • Rotate the standard coordinate system 10∘ clockwise to align the new y′ axis with with force .N.
  • Draw a free-body digram and calculate the angles between the forces and the rotated coordinate system.
  • Solve for force P directly.

\begin{align*} \Sigma F_{x'} \amp = 0 \\ -P_{x'} + W_{x'} \amp = 0 \\ P \cos \ang{30} \amp = W \sin \ang{10} \\ P \amp = \lb{160} \left( \frac{0.1736}{0.866} \right)\\ P \amp = \lb{32.1} \end{align*}

Multi-Particle Equilibrium

When two or more particles interact with each other there will always be common forces between them as a result of Newton’s Third Law, the action-reaction principle.

Consider the two boxes with weights \(W_1\) and \(W_2\) connected to each other and the ceiling shown in the interactive diagram. Position one shows the physical arrangement of the objects, position two shows their free-body diagrams, and position three shows simplified FBDs where the objects are represented by points. The boxes were freed by replacing the cables with tension forces \(T_A\) and \(T_B\text{.}\)

From the free-body digrams you can see that cable \(B\) only supports the weight of the bottom box, while cable \(A\) and the ceiling support the combined weight. The tension \(T_B\) is common to both diagrams. Recognizing the common force is the key to solving multi-particle equilibrium problems.

Example 3.4.8 . Two hanging weights.

A \(\N{100}\) weight \(W\) is supported by cable \(ABCD\text{.}\) There is a frictionless pulley at \(B\) and the hook is firmly attached to the cable at point \(C\text{.}\)

What is the magnitude and direction of force \(\vec{P}\) required to hold the system in the position shown?

The particles are points \(B\) and \(C\text{.}\) The common force is the tension in rope segment \(BC\text{.}\)

\begin{align*} P\amp = \N{84.3} \amp \theta \amp = \ang{54.7} \text{ CCW from } -x \text{ axis.}\\ \vec{P} \amp= \langle \N{-48.7}, \N{-68.8} \rangle \end{align*

Following the General Procedure we identify the particles as points A and B, and draw free-body digrams of each. We label the rope tensions \(A\text{,}\) \(C\text{,}\) and \(D\) for the endpoints of the rope segments, and label the angles of the forces \(\alpha\text{,}\) \(\beta\text{,}\) and \(\phi\text{.}\) We will use the standard cartesian coordinate system and use the scalar components method 3.4.5.

Weight \(W\) was given, and we can easily find angles \(\alpha\text{,}\) \(\beta\text{,}\) and \(\phi\) so the knowns are:

\begin{align*} W \amp = \N{100}\\ \alpha \amp = \tan^{-1} \left( \frac{40}{20} \right) = 63.4°\\ \beta \amp = \tan^{-1} \left( \frac{10}{80} \right) = 7.13°\\ \phi \amp = \tan^{-1} \left( \frac{50}{50} \right) = 45° \end{align*}

Counting unknowns we find that there are two on the free-body diagram of particle \(C\) (\(C\) and \(D)\text{,}\) but four on particle \(B\text{,}\) (\(A\) \(C\text{,}\) \(P\) and \(\theta\)).

Two unknowns on particle \(C\) means it is solvable since there are two equilibrium equations (3.4.3) available, so we begin there.

2. Solve Particle C.

\begin{align*} \Sigma F_x \amp = 0 \amp \Sigma F_y \amp = 0\\ - C_x + D_x \amp = 0 \amp C_y + D_y - W \amp = 0\\ C \cos \beta \amp = D \cos \phi \amp C \sin \beta + D \sin \phi \amp= W\\ C \amp = D \left(\frac{\cos \ang{45}}{\cos \ang{7.13}}\right) \amp C \sin \ang{7.13} + D \sin \ang{45} \amp = \N{100}\\ C \amp = 0.713 D \amp 0.124 C + 0.707 D \amp = \N{100} \end{align*}

Solving these two equations simultaneously gives

\begin{align*} C \amp =\N{89.5} \amp D \amp = \N{125.7}. \end{align*}

With particle \(C\) solved, we can use the results to solve particle \(B\text{.}\) There are three unknowns remaining, tension \(A\text{,}\) magnitude \(P\text{,}\) and direction \(\theta\text{.}\) Unfortunately, we still only have two available equilibrium equations. When you find yourself in this situation with more unknowns than equations, it generally means that you are missing something. In this case it is the pulley. When a cable wraps around a frictionless pulley the tension doesn't change. The missing information is that \(A = C\text{.}\) Knowing this, the magnitude and direction of force \(\vec{P}\) can be determined.

3. Solve Particle B.

Referring to the FBD for particle \(B\) we can write these equations.

\begin{align*} \Sigma F_x \amp = 0 \amp \Sigma F_y \amp = 0\\ - A_x - P_x + C_x \amp = 0 \amp A_y - P_y - C_y \amp = 0\\ P \cos \theta \amp = C \cos\beta - A \cos\alpha \amp P \sin \theta \amp = A \sin\alpha -C \sin\beta \end{align*}

Since \(A = C = \N{89.5}\text{,}\) substituting and solving simultaneously gives

\begin{align*} P \cos \theta \amp = \N{48.7} \amp P \sin \theta \amp = \N{68.9}\\ P\amp = \N{84.3} \amp \theta \amp = \ang{54.7}. \end{align*}

These are the magnitude and direction of vector \(\vec{P}\text{.}\) If you wish, you can express\(\vec{P}\) in terms of its scalar components. The negative signs on the components have been applied by hand since \(\vec{P}\) points down and to the left.

\begin{align*} \vec{P} \amp= \langle - P \cos\theta, - P \sin\theta \rangle\\ \amp = \langle \N{-48.7}, \N{-68.8} \rangle \end{align*}

solving equilibrium problems

15.3 Solving Equilibrium Problems

Learning objective.

  • To solve quantitative problems involving chemical equilibriums.

There are two fundamental kinds of equilibrium problems: (1) those in which we are given the concentrations of the reactants and the products at equilibrium (or, more often, information that allows us to calculate these concentrations), and we are asked to calculate the equilibrium constant for the reaction; and (2) those in which we are given the equilibrium constant and the initial concentrations of reactants, and we are asked to calculate the concentration of one or more substances at equilibrium. In this section, we describe methods for solving both kinds of problems.

Calculating an Equilibrium Constant from Equilibrium Concentrations

We saw in the exercise in Example 6 in Section 15.2 "The Equilibrium Constant" that the equilibrium constant for the decomposition of CaCO 3 (s) to CaO(s) and CO 2 (g) is K = [CO 2 ]. At 800°C, the concentration of CO 2 in equilibrium with solid CaCO 3 and CaO is 2.5 × 10 −3 M. Thus K at 800°C is 2.5 × 10 −3 . (Remember that equilibrium constants are unitless.)

A more complex example of this type of problem is the conversion of n -butane, an additive used to increase the volatility of gasoline, to isobutane (2-methylpropane). This reaction can be written as follows:

Equation 15.26

and the equilibrium constant K = [isobutane]/[ n -butane]. At equilibrium, a mixture of n -butane and isobutane at room temperature was found to contain 0.041 M isobutane and 0.016 M n -butane. Substituting these concentrations into the equilibrium constant expression,

Equation 15.27

Thus the equilibrium constant for the reaction as written is 2.6.

solving equilibrium problems

The reaction between gaseous sulfur dioxide and oxygen is a key step in the industrial synthesis of sulfuric acid:

A mixture of SO 2 and O 2 was maintained at 800 K until the system reached equilibrium. The equilibrium mixture contained 5.0 × 10 −2 M SO 3 , 3.5 × 10 −3 M O 2 , and 3.0 × 10 −3 M SO 2 . Calculate K and K p at this temperature.

Given: balanced equilibrium equation and composition of equilibrium mixture

Asked for: equilibrium constant

Write the equilibrium constant expression for the reaction. Then substitute the appropriate equilibrium concentrations into this equation to obtain K .

Substituting the appropriate equilibrium concentrations into the equilibrium constant expression,

To solve for K p , we use Equation 16.18 , where Δ n = 2 − 3 = −1:

Hydrogen gas and iodine react to form hydrogen iodide via the reaction

A mixture of H 2 and I 2 was maintained at 740 K until the system reached equilibrium. The equilibrium mixture contained 1.37 × 10 −2 M HI, 6.47 × 10 −3 M H 2 , and 5.94 × 10 −4 M I 2 . Calculate K and K p for this reaction.

Answer: K = 48.8; K p = 48.8

Chemists are not often given the concentrations of all the substances, and they are not likely to measure the equilibrium concentrations of all the relevant substances for a particular system. In such cases, we can obtain the equilibrium concentrations from the initial concentrations of the reactants and the balanced chemical equation for the reaction, as long as the equilibrium concentration of one of the substances is known. Example 9 shows one way to do this.

A 1.00 mol sample of NOCl was placed in a 2.00 L reactor and heated to 227°C until the system reached equilibrium. The contents of the reactor were then analyzed and found to contain 0.056 mol of Cl 2 . Calculate K at this temperature. The equation for the decomposition of NOCl to NO and Cl 2 is as follows:

Given: balanced equilibrium equation, amount of reactant, volume, and amount of one product at equilibrium

Asked for: K

A Write the equilibrium constant expression for the reaction. Construct a table showing the initial concentrations, the changes in concentrations, and the final concentrations (as initial concentrations plus changes in concentrations).

B Calculate all possible initial concentrations from the data given and insert them in the table.

C Use the coefficients in the balanced chemical equation to obtain the changes in concentration of all other substances in the reaction. Insert those concentration changes in the table.

D Obtain the final concentrations by summing the columns. Calculate the equilibrium constant for the reaction.

A The first step in any such problem is to balance the chemical equation for the reaction (if it is not already balanced) and use it to derive the equilibrium constant expression. In this case, the equation is already balanced, and the equilibrium constant expression is as follows:

To obtain the concentrations of NOCl, NO, and Cl 2 at equilibrium, we construct a table showing what is known and what needs to be calculated. We begin by writing the balanced chemical equation at the top of the table, followed by three lines corresponding to the initial concentrations, the changes in concentrations required to get from the initial to the final state, and the final concentrations.

B Initially, the system contains 1.00 mol of NOCl in a 2.00 L container. Thus [NOCl] i = 1.00 mol/2.00 L = 0.500 M. The initial concentrations of NO and Cl 2 are 0 M because initially no products are present. Moreover, we are told that at equilibrium the system contains 0.056 mol of Cl 2 in a 2.00 L container, so [Cl 2 ] f = 0.056 mol/2.00 L = 0.028 M. We insert these values into the following table:

C We use the stoichiometric relationships given in the balanced chemical equation to find the change in the concentration of Cl 2 , the substance for which initial and final concentrations are known:

According to the coefficients in the balanced chemical equation, 2 mol of NO are produced for every 1 mol of Cl 2 , so the change in the NO concentration is as follows:

Similarly, 2 mol of NOCl are consumed for every 1 mol of Cl 2 produced, so the change in the NOCl concentration is as follows:

We insert these values into our table:

D We sum the numbers in the [NOCl] and [NO] columns to obtain the final concentrations of NO and NOCl:

We can now complete the table:

We can now calculate the equilibrium constant for the reaction:

The German chemist Fritz Haber (1868–1934; Nobel Prize in Chemistry 1918) was able to synthesize ammonia (NH 3 ) by reacting 0.1248 M H 2 and 0.0416 M N 2 at about 500°C. At equilibrium, the mixture contained 0.00272 M NH 3 . What is K for the reaction N 2 + 3 H 2 ⇌ 2 NH 3 at this temperature? What is K p ?

Answer: K = 0.105; K p = 2.61 × 10 −5

Calculating Equilibrium Concentrations from the Equilibrium Constant

To describe how to calculate equilibrium concentrations from an equilibrium constant, we first consider a system that contains only a single product and a single reactant, the conversion of n -butane to isobutane ( Equation 15.26 ), for which K = 2.6 at 25°C. If we begin with a 1.00 M sample of n -butane, we can determine the concentration of n -butane and isobutane at equilibrium by constructing a table showing what is known and what needs to be calculated, just as we did in Example 9.

solving equilibrium problems

The original laboratory apparatus designed by Fritz Haber and Robert Le Rossignol in 1908 for synthesizing ammonia from its elements. A metal catalyst bed, where ammonia was produced, is in the large cylinder at the left. The Haber-Bosch process used for the industrial production of ammonia uses essentially the same process and components but on a much larger scale. Unfortunately, Haber’s process enabled Germany to prolong World War I when German supplies of nitrogen compounds, which were used for explosives, had been exhausted in 1914.

The initial concentrations of the reactant and product are both known: [ n -butane] i = 1.00 M and [isobutane] i = 0 M. We need to calculate the equilibrium concentrations of both n -butane and isobutane. Because it is generally difficult to calculate final concentrations directly, we focus on the change in the concentrations of the substances between the initial and the final (equilibrium) conditions. If, for example, we define the change in the concentration of isobutane (Δ[isobutane]) as + x , then the change in the concentration of n -butane is Δ[ n -butane] = − x . This is because the balanced chemical equation for the reaction tells us that 1 mol of n -butane is consumed for every 1 mol of isobutane produced. We can then express the final concentrations in terms of the initial concentrations and the changes they have undergone.

Substituting the expressions for the final concentrations of n -butane and isobutane from the table into the equilibrium equation,

Rearranging and solving for x ,

We obtain the final concentrations by substituting this x value into the expressions for the final concentrations of n -butane and isobutane listed in the table:

We can check the results by substituting them back into the equilibrium constant expression to see whether they give the same K that we used in the calculation:

This is the same K we were given, so we can be confident of our results.

Example 10 illustrates a common type of equilibrium problem that you are likely to encounter.

The water–gas shift reaction is important in several chemical processes, such as the production of H 2 for fuel cells. This reaction can be written as follows:

K = 0.106 at 700 K. If a mixture of gases that initially contains 0.0150 M H 2 and 0.0150 M CO 2 is allowed to equilibrate at 700 K, what are the final concentrations of all substances present?

Given: balanced equilibrium equation, K , and initial concentrations

Asked for: final concentrations

A Construct a table showing what is known and what needs to be calculated. Define x as the change in the concentration of one substance. Then use the reaction stoichiometry to express the changes in the concentrations of the other substances in terms of x . From the values in the table, calculate the final concentrations.

B Write the equilibrium equation for the reaction. Substitute appropriate values from the table to obtain x .

C Calculate the final concentrations of all species present. Check your answers by substituting these values into the equilibrium constant expression to obtain K .

A The initial concentrations of the reactants are [H 2 ] i = [CO 2 ] i = 0.0150 M. Just as before, we will focus on the change in the concentrations of the various substances between the initial and final states. If we define the change in the concentration of H 2 O as x , then Δ[H 2 O] = + x . We can use the stoichiometry of the reaction to express the changes in the concentrations of the other substances in terms of x . For example, 1 mol of CO is produced for every 1 mol of H 2 O, so the change in the CO concentration can be expressed as Δ[CO] = + x . Similarly, for every 1 mol of H 2 O produced, 1 mol each of H 2 and CO 2 are consumed, so the change in the concentration of the reactants is Δ[H 2 ] = Δ[CO 2 ] = − x . We enter the values in the following table and calculate the final concentrations.

B We can now use the equilibrium equation and the given K to solve for x :

We could solve this equation with the quadratic formula, but it is far easier to solve for x by recognizing that the left side of the equation is a perfect square; that is,

(The quadratic formula is presented in Essential Skills 7 in Section 15.7 "Essential Skills" .) Taking the square root of the middle and right terms,

C The final concentrations of all species in the reaction mixture are as follows:

We can check our work by inserting the calculated values back into the equilibrium constant expression:

To two significant figures, this K is the same as the value given in the problem, so our answer is confirmed.

Hydrogen gas reacts with iodine vapor to give hydrogen iodide according to the following chemical equation:

K = 54 at 425°C. If 0.172 M H 2 and I 2 are injected into a reactor and maintained at 425°C until the system equilibrates, what is the final concentration of each substance in the reaction mixture?

Answer: [HI] f = 0.270 M; [H 2 ] f = [I 2 ] f = 0.037 M

In Example 10, the initial concentrations of the reactants were the same, which gave us an equation that was a perfect square and simplified our calculations. Often, however, the initial concentrations of the reactants are not the same, and/or one or more of the products may be present when the reaction starts. Under these conditions, there is usually no way to simplify the problem, and we must determine the equilibrium concentrations with other means. Such a case is described in Example 11.

In the water–gas shift reaction shown in Example 10, a sample containing 0.632 M CO 2 and 0.570 M H 2 is allowed to equilibrate at 700 K. At this temperature, K = 0.106. What is the composition of the reaction mixture at equilibrium?

Given: balanced equilibrium equation, concentrations of reactants, and K

Asked for: composition of reaction mixture at equilibrium

A Write the equilibrium equation. Construct a table showing the initial concentrations of all substances in the mixture. Complete the table showing the changes in the concentrations ( x ) and the final concentrations.

B Write the equilibrium constant expression for the reaction. Substitute the known K value and the final concentrations to solve for x .

C Calculate the final concentration of each substance in the reaction mixture. Check your answers by substituting these values into the equilibrium constant expression to obtain K .

A [CO 2 ] i = 0.632 M and [H 2 ] i = 0.570 M. Again, x is defined as the change in the concentration of H 2 O: Δ[H 2 O] = + x . Because 1 mol of CO is produced for every 1 mol of H 2 O, the change in the concentration of CO is the same as the change in the concentration of H 2 O, so Δ[CO] = + x . Similarly, because 1 mol each of H 2 and CO 2 are consumed for every 1 mol of H 2 O produced, Δ[H 2 ] = Δ[CO 2 ] = − x . The final concentrations are the sums of the initial concentrations and the changes in concentrations at equilibrium.

B We can now use the equilibrium equation and the known K value to solve for x :

In contrast to Example 10, however, there is no obvious way to simplify this expression. Thus we must expand the expression and multiply both sides by the denominator:

Collecting terms on one side of the equation,

This equation can be solved using the quadratic formula:

Only the answer with the positive value has any physical significance, so Δ[H 2 O] = Δ[CO] = +0.148 M, and Δ[H 2 ] = Δ[CO 2 ] = −0.148 M.

We can check our work by substituting these values into the equilibrium constant expression:

Because K is essentially the same as the value given in the problem, our calculations are confirmed.

The exercise in Example 8 showed the reaction of hydrogen and iodine vapor to form hydrogen iodide, for which K = 54 at 425°C. If a sample containing 0.200 M H 2 and 0.0450 M I 2 is allowed to equilibrate at 425°C, what is the final concentration of each substance in the reaction mixture?

Answer: [HI] f = 0.0882 M; [H 2 ] f = 0.156 M; [I 2 ] f = 9.2 × 10 −4 M

In many situations it is not necessary to solve a quadratic (or higher-order) equation. Most of these cases involve reactions for which the equilibrium constant is either very small ( K ≤ 10 −3 ) or very large ( K ≥ 10 3 ), which means that the change in the concentration (defined as x ) is essentially negligible compared with the initial concentration of a substance. Knowing this simplifies the calculations dramatically, as illustrated in Example 12.

Atmospheric nitrogen and oxygen react to form nitric oxide:

K p = 2.0 × 10 −31 at 25°C. What is the partial pressure of NO in equilibrium with N 2 and O 2 in the atmosphere (at 1 atm, P N 2 = 0.78 atm and P O 2 = 0.21 atm)?

Given: balanced equilibrium equation and values of K p , P O 2 , and P N 2

Asked for: partial pressure of NO

A Construct a table and enter the initial partial pressures, the changes in the partial pressures that occur during the course of the reaction, and the final partial pressures of all substances.

B Write the equilibrium equation for the reaction. Then substitute values from the table to solve for the change in concentration ( x ).

C Calculate the partial pressure of NO. Check your answer by substituting values into the equilibrium equation and solving for K .

A Because we are given K p and partial pressures are reported in atmospheres, we will use partial pressures. The initial partial pressure of O 2 is 0.21 atm and that of N 2 is 0.78 atm. If we define the change in the partial pressure of NO as 2 x , then the change in the partial pressure of O 2 and of N 2 is − x because 1 mol each of N 2 and of O 2 is consumed for every 2 mol of NO produced. Each substance has a final partial pressure equal to the sum of the initial pressure and the change in that pressure at equilibrium.

B Substituting these values into the equation for the equilibrium constant,

In principle, we could multiply out the terms in the denominator, rearrange, and solve the resulting quadratic equation. In practice, it is far easier to recognize that an equilibrium constant of this magnitude means that the extent of the reaction will be very small; therefore, the x value will be negligible compared with the initial concentrations. If this assumption is correct, then to two significant figures, (0.78 − x ) = 0.78 and (0.21 − x ) = 0.21. Substituting these expressions into our original equation,

C Substituting this value of x into our expressions for the final partial pressures of the substances,

From these calculations, we see that our initial assumption regarding x was correct: given two significant figures, 2.0 × 10 −16 is certainly negligible compared with 0.78 and 0.21. When can we make such an assumption? As a general rule, if x is less than about 5% of the total, or 10 −3 > K > 10 3 , then the assumption is justified. Otherwise, we must use the quadratic formula or some other approach. The results we have obtained agree with the general observation that toxic NO, an ingredient of smog, does not form from atmospheric concentrations of N 2 and O 2 to a substantial degree at 25°C. We can verify our results by substituting them into the original equilibrium equation:

The final K p agrees with the value given at the beginning of this example.

Under certain conditions, oxygen will react to form ozone, as shown in the following equation:

K p = 2.5 × 10 −59 at 25°C. What ozone partial pressure is in equilibrium with oxygen in the atmosphere ( P O 2 = 0.21  atm ) ?

Answer: 4.8 × 10 −31 atm

Another type of problem that can be simplified by assuming that changes in concentration are negligible is one in which the equilibrium constant is very large ( K ≥ 10 3 ). A large equilibrium constant implies that the reactants are converted almost entirely to products, so we can assume that the reaction proceeds 100% to completion. When we solve this type of problem, we view the system as equilibrating from the products side of the reaction rather than the reactants side. This approach is illustrated in Example 13.

The chemical equation for the reaction of hydrogen with ethylene (C 2 H 4 ) to give ethane (C 2 H 6 ) is as follows:

K = 9.6 × 10 18 at 25°C. If a mixture of 0.200 M H 2 and 0.155 M C 2 H 4 is maintained at 25°C in the presence of a powdered nickel catalyst, what is the equilibrium concentration of each substance in the mixture?

Given: balanced chemical equation, K , and initial concentrations of reactants

Asked for: equilibrium concentrations

A Construct a table showing initial concentrations, concentrations that would be present if the reaction were to go to completion, changes in concentrations, and final concentrations.

B Write the equilibrium constant expression for the reaction. Then substitute values from the table into the expression to solve for x (the change in concentration).

C Calculate the equilibrium concentrations. Check your answers by substituting these values into the equilibrium equation.

A From the magnitude of the equilibrium constant, we see that the reaction goes essentially to completion. Because the initial concentration of ethylene (0.155 M) is less than the concentration of hydrogen (0.200 M), ethylene is the limiting reactant; that is, no more than 0.155 M ethane can be formed from 0.155 M ethylene. If the reaction were to go to completion, the concentration of ethane would be 0.155 M and the concentration of ethylene would be 0 M. Because the concentration of hydrogen is greater than what is needed for complete reaction, the concentration of unreacted hydrogen in the reaction mixture would be 0.200 M − 0.155 M = 0.045 M. The equilibrium constant for the forward reaction is very large, so the equilibrium constant for the reverse reaction must be very small. The problem then is identical to that in Example 12. If we define − x as the change in the ethane concentration for the reverse reaction, then the change in the ethylene and hydrogen concentrations is + x . The final equilibrium concentrations are the sums of the concentrations for the forward and reverse reactions.

B Substituting values into the equilibrium constant expression,

Once again, the magnitude of the equilibrium constant tells us that the equilibrium will lie far to the right as written, so the reverse reaction is negligible. Thus x is likely to be very small compared with either 0.155 M or 0.045 M, and the equation can be simplified [(0.045 + x ) = 0.045 and (0.155 − x ) = 0.155] as follows:

C The small x value indicates that our assumption concerning the reverse reaction is correct, and we can therefore calculate the final concentrations by evaluating the expressions from the last line of the table:

We can verify our calculations by substituting the final concentrations into the equilibrium constant expression:

This K value agrees with our initial value at the beginning of the example.

Hydrogen reacts with chlorine gas to form hydrogen chloride:

K p = 4.0 × 10 31 at 47°C. If a mixture of 0.257 M H 2 and 0.392 M Cl 2 is allowed to equilibrate at 47°C, what is the equilibrium composition of the mixture?

Answer: [H 2 ] f = 4.8 × 10 −32 M; [Cl 2 ] f = 0.135 M; [HCl] f = 0.514 M

When an equilibrium constant is calculated from equilibrium concentrations, molar concentrations or partial pressures are substituted into the equilibrium constant expression for the reaction. Equilibrium constants can be used to calculate the equilibrium concentrations of reactants and products by using the quantities or concentrations of the reactants, the stoichiometry of the balanced chemical equation for the reaction, and a tabular format to obtain the final concentrations of all species at equilibrium.

Key Takeaway

  • Various methods can be used to solve the two fundamental types of equilibrium problems: (1) those in which we calculate the concentrations of reactants and products at equilibrium and (2) those in which we use the equilibrium constant and the initial concentrations of reactants to determine the composition of the equilibrium mixture.

Conceptual Problems

Describe how to determine the magnitude of the equilibrium constant for a reaction when not all concentrations of the substances are known.

Calculations involving systems with very small or very large equilibrium constants can be dramatically simplified by making certain assumptions about the concentrations of products and reactants. What are these assumptions when K is (a) very large and (b) very small? Illustrate this technique using the system A + 2B ⇌ C for which you are to calculate the concentration of the product at equilibrium starting with only A and B. Under what circumstances should simplifying assumptions not be used?

Numerical Problems

Please be sure you are familiar with the topics discussed in Essential Skills 7 ( Section 15.7 "Essential Skills" ) before proceeding to the Numerical Problems.

In the equilibrium reaction A + B ⇌ C, what happens to K if the concentrations of the reactants are doubled? tripled? Can the same be said about the equilibrium reaction 2 A ⇌ B + C?

The following table shows the reported values of the equilibrium P O 2 at three temperatures for the reaction Ag 2 O(s) ⇌ 2 Ag(s) + 1 2 O 2 (g) , for which Δ H ° = 31 kJ/mol. Are these data consistent with what you would expect to occur? Why or why not?

Given the equilibrium system N 2 O 4 (g) ⇌ 2 NO 2 (g), what happens to K p if the initial pressure of N 2 O 4 is doubled? If K p is 1.7 × 10 −1 at 2300°C, and the system initially contains 100% N 2 O 4 at a pressure of 2.6 × 10 2 atm, what is the equilibrium pressure of each component?

At 430°C, 4.20 mol of HI in a 9.60 L reaction vessel reaches equilibrium according to the following equation: H 2 (g) + I 2 (g) ⇌ 2 HI(g) . At equilibrium, [H 2 ] = 0.047 M and [HI] = 0.345 M. What are K and K p for this reaction?

Methanol, a liquid used as an automobile fuel additive, is commercially produced from carbon monoxide and hydrogen at 300°C according to the following reaction: CO ( g ) + 2H 2 ( g ) ⇌ CH 3 OH ( g ) , and K p = 1.3 × 10 −4 . If 56.0 g of CO is mixed with excess hydrogen in a 250 mL flask at this temperature, and the hydrogen pressure is continuously maintained at 100 atm, what would be the maximum percent yield of methanol? What pressure of hydrogen would be required to obtain a minimum yield of methanol of 95% under these conditions?

Starting with pure A, if the total equilibrium pressure is 0.969 atm for the reaction A(s) ⇌ 2 B(g) + C(g), what is K p ?

The decomposition of ammonium carbamate to NH 3 and CO 2 at 40°C is written as NH 4 CO 2 NH 2 (s) ⇌ 2 NH 3 (g) + CO 2 (g) . If the partial pressure of NH 3 at equilibrium is 0.242 atm, what is the equilibrium partial pressure of CO 2 ? What is the total gas pressure of the system? What is K p ?

At 375 K, K p for the reaction SO 2 Cl 2 (g) ⇌ SO 2 (g) + Cl 2 (g) is 2.4, with pressures expressed in atmospheres. At 303 K, K p is 2.9 × 10 −2 .

  • What is K for the reaction at each temperature?
  • If a sample at 375 K has 0.100 M Cl 2 and 0.200 M SO 2 at equilibrium, what is the concentration of SO 2 Cl 2 ?
  • If the sample given in part b is cooled to 303 K, what is the pressure inside the bulb?

For the gas-phase reaction a A ⇌ b B, show that K p = K ( RT ) Δ n assuming ideal gas behavior.

For the gas-phase reaction I 2 ⇌ 2 I, show that the total pressure is related to the equilibrium pressure by the following equation:

Experimental data on the system Br 2 (l) ⇌ Br 2 (aq) are given in the following table. Graph [Br 2 ] versus moles of Br 2 (l) present; then write the equilibrium constant expression and determine K .

Data accumulated for the reaction n -butane(g) ⇌ isobutane(g) at equilibrium are shown in the following table. What is the equilibrium constant for this conversion? If 1 mol of n -butane is allowed to equilibrate under the same reaction conditions, what is the final number of moles of n -butane and isobutane?

Solid ammonium carbamate (NH 4 CO 2 NH 2 ) dissociates completely to ammonia and carbon dioxide when it vaporizes:

At 25°C, the total pressure of the gases in equilibrium with the solid is 0.116 atm. What is the equilibrium partial pressure of each gas? What is K p ? If the concentration of CO 2 is doubled and then equilibrates to its initial equilibrium partial pressure + x atm, what change in the NH 3 concentration is necessary for the system to restore equilibrium?

The equilibrium constant for the reaction COCl 2 (g) ⇌ CO(g) + Cl 2 (g) is K p = 2.2 × 10 −10 at 100°C. If the initial concentration of COCl 2 is 3.05 × 10 −3 M, what is the partial pressure of each gas at equilibrium at 100°C? What assumption can be made to simplify your calculations?

Aqueous dilution of IO 4 − results in the following reaction: IO 4 − (aq) + 2 H 2 O(l) ⇌ H 4 IO 6 − (aq), and K = 3.5 × 10 −2 . If you begin with 50 mL of a 0.896 M solution of IO 4 − that is diluted to 250 mL with water, how many moles of H 4 IO 6 − are formed at equilibrium?

Iodine and bromine react to form IBr, which then sublimes. At 184.4°C, the overall reaction proceeds according to the following equation:

K p = 1.2 × 10 2 . If you begin the reaction with 7.4 g of I 2 vapor and 6.3 g of Br 2 vapor in a 1.00 L container, what is the concentration of IBr(g) at equilibrium? What is the partial pressure of each gas at equilibrium? What is the total pressure of the system?

For the reaction C(s)  +  1 2 N 2 (g) + 5 2 H 2 (g) ⇌ CH 3 NH 2 (g), K = 1.8 × 10 −6 . If you begin the reaction with 1.0 mol of N 2 , 2.0 mol of H 2 , and sufficient C(s) in a 2.00 L container, what are the concentrations of N 2 and CH 3 NH 2 at equilibrium? What happens to K if the concentration of H 2 is doubled?

Tasmanian election to focus on cost of living, health and housing, with minority government and stadium lurking in background

A premier speaks at a press conference on the balcony of a hotel, with the Hobart waterfront in the background.

The Liberals, Labor and the Greens have started the Tasmanian election campaign in furious agreement — the state has many problems that need solving.

Tasmania's public health system continues to struggle to respond to community demand, the social housing waiting list is growing longer, and cost of living — including rent and housing — is causing significant hardship.

All three major parties listed cost of living, health and housing as their priorities as soon as the election was called — aligning with a recent Tasmanian community sentiment survey.

Premier Jeremy Rockliff released a video advertisement to coincide with his visit to the governor, highlighting some of his government's economic and infrastructure credentials, but conceding many Tasmanians were struggling.

It featured phrases like "our health system needs to be better", "we need to do more to support those struggling with the cost of living", and "for many Tasmanians, their first home lies beyond their reach and the rental prices are still too high".

The Liberals have been in power in Tasmania since 2014, but Mr Rockliff said they had a plan to address these issues by 2030.

"This election is about who is best able to restore stability and certainty Tasmania needs, so we can take action on the issues affecting you right now," he said.

"We do have a strong plan which is all about addressing those important issues, but the parliament has become unworkable."

Rebecca White has shoulder-length blonde hair and is speaking into a radio microphone.

Labor, too, was eager to focus on cost of living.

Leader Rebecca White announced a policy to cap power prices, and lower them by 22.5 per cent on July 1 — a move that Mr Rockliff described as "a gimmick".

Ms White said that the state's problems had only been exacerbated by the government.

"If they haven't fixed the problems of cost of living, or health, or housing by now, they never will," she said.

Both parties will release more details of their plans for health and housing as the campaign continues.

Rockliff open to negotiating, White rules out party deals

Both parties face an uphill battle to form majority government.

Even the strong personal popularity of former premier Peter Gutwein could only grant the Liberals a one-seat majority in 2021.

An expanded lower house — from 25 to 35 seats — makes the path to majority even narrower under the Hare-Clark multi-member electorate system.

The questions about minority government started straight away.

A premier arrives at a press conference on the balcony of a hotel.

Mr Rockliff said the Liberals had the best chance of reaching majority, but left the door open to negotiations.

"I will respect the outcome of the election and the will of the voters, and of course have the maturity to govern sensibly in the best interests of all Tasmanians," he said.

"When it comes to everyone else, some key principles will apply.

"I will not agree to anything that constrains me or my government. I will not be trading ministerial positions for policies."

Mr Rockliff was referring to the most recent period of Labor government — when the party governed in coalition with the Greens.

The next crossbench is likely to be even more complicated, with the possibility of the Jacqui Lambie Network picking up a seat, among more independents.

Former Labor leader David O'Byrne will run in Franklin , while former Liberal attorney-general Elise Archer will run in Clark — both as independents, both with grievances against their former parties, and both with strong local electoral history.

David O'Byrne has a chuckle.

Ms White said she would not form government through "deals" with minor parties.

"I can categorically rule out doing a deal with the Greens or any other party to form government," she said.

"If the people of Tasmania vote to return a parliament where it's not clear if any party has won a majority, we'd be asking the parliament to support our plan for Tasmania."

Election 'not a referendum' on stadium, but AFL deal looms large

The final breakdown in talks between Mr Rockliff and independents John Tucker and Lara Alexander appeared to centre on their disagreements over the proposed Macquarie Point stadium and AFL training centre.

But the stadium did not feature in Mr Rockliff's election announcement, nor in his first advert.

It was also the bottom of the list of priorities for Tasmanians in a recent 3P Advisory survey.

Mr Rockliff said the election would not be a "referendum" on the stadium, but argued he had achieved a "good deal" with the AFL.

Premier Jeremy Rockliff and AFL CEO Gillon McLachlan sit at a table and each sign a document.

The deal includes a requirement for a roofed stadium at Macquarie Point, which Labor argues is "not a priority", and questions the legitimacy of the $715 million price tag.

But Mr Rockliff has this morning promised a re-elected majority Liberal government would cap government spending on the stadium at $375 million, seeking private sector investment for any shortfall.

It's likely to feature heavily as the election draws closer, however, with the Tasmania AFL team colours and name to be announced one week out from the March 23 poll.

Ms White said she would renegotiate the deal with the AFL.

"A Labor government will sit down like adults with the AFL and negotiate a better deal for Tasmanians," she said.

"Rather than have the AFL push their deal on this state which is what Peter Gutwein and Jeremy Rockliff both accepted."

Greens want balance of power to exert environmental influence

Labor and the Liberals disagree on many things in Tasmania, but they also have quite a bit in common — their support for the salmon and forestry sectors.

Both have expressed concern at a Commonwealth review of salmon farms in Macquarie Harbour due to their impact on the endangered maugean skate .

They also reject any moves to phase out native forest logging, particularly as other states move away from the industry.

A woman with blonde hair sits at a desk and speaks

Greens leader Rosalie Woodruff said her party was aiming for balance of power, to influence the major parties on environmental matters.

"What we've seen from the Liberal government and also Labor politicians is all too much a focus on boosting profits for corporate mates," she said.

"We will work collaboratively.

"Most importantly, we want to win back the seats of Bass and Lyons.

"We want to hold the balance of power at the next election so that we can hold the government to account and make real change on the issues Tasmanians care about."

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Chemistry LibreTexts

2.E: Chemical Equilibrium (Practice Problems with Answers)

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These are homework exercises to accompany the Textmap created for "Chemistry: The Central Science" by Brown et al. Complementary General Chemistry question banks can be found for other Textmaps and can be accessed here . In addition to these publicly available questions, access to private problems bank for use in exams and homework is available to faculty only on an individual basis; please contact Delmar Larsen for an account with access permission.

15.1: The Concept of Equilibrium

Conceptual problems.

  • What is meant when a reaction is described as “having reached equilibrium”? What does this statement mean regarding the forward and reverse reaction rates? What does this statement mean regarding the amounts or concentrations of the reactants and the products?
  • Is it correct to say that the reaction has “stopped” when it has reached equilibrium? Explain your answer and support it with a specific example.
  • Why is chemical equilibrium described as a dynamic process? Describe this process in the context of a saturated solution of \(NaCl\) in water. What is occurring on a microscopic level? What is happening on a macroscopic level?
  • oxygen and hemoglobin in the human circulatory system
  • iodine crystals in an open beaker
  • the combustion of wood
  • the amount of \(\ce{^{14}C}\) in a decomposing organism

Conceptual Answer

1. When a reaction is described as "having reached equilibrium" this means that the forward reaction rate is now equal to the reverse reaction rate. In regards to the amounts or concentrations of the reactants and the products, there is no change due to the forward reaction rate being equal to the reverse reaction rate.

2. It is not correct to say that the reaction has "stopped" when it has reached equilibrium because it is not necessarily a static process where it can be assumed that the reaction rates cancel each other out to equal zero or be "stopped" but rather a dynamic process in which reactants are converted to products at the same rate products are converted to reactants. For example, a soda has carbon dioxide dissolved in the liquid and carbon dioxide between the liquid and the cap that is constantly being exchanged with each other. The system is in equilibrium and the reaction taking place is: \(CO_{2}\,(g)+2\,H_{2}O\,(l)\rightleftharpoons H_{2}CO_3\,(aq)\).

3. Chemical equilibrium is described as a dynamic process because there is a movement in which the forward and reverse reactions occur at the same rate to reach a point where the amounts or concentrations of the reactants and products are unchanging with time. Chemical equilibrium can be described in a saturated solution of \(NaCl\) as on the microscopic level \(Na^+\) and \(Cl^−\) ions continuously leave the surface of an \(NaCl\) crystal to enter the solution, while at the same time \(Na^+\) and \(Cl^−\) ions in solution precipitate on the surface of the crystal. At the macroscopic level, the salt can be seen to dissolve or not dissolve depending whether chemical equilibrium was established.

a. Exists in a state of equilibrium as the chemical reaction that occurs in the body is: \(Hb\,(aq)+4\,H_{2}O\,(l)\rightleftharpoons Hb(O_{2})_{4}\,(aq)\).

b. Exists in a state of equilibrium as the chemical reaction occurs is: \(H_{2}\,(g)+I_{2}\,(g) \rightleftharpoons 2\,HI\,(g)\)

c. Does not exist in a state of equilibrium as it is not a reversible process. The chemical reaction that takes place is: \(6\,C_{10}H_{15}O_{7}\,(s)+heat\rightarrow C_{50}H_{10}O\,(s)+10\,CH_{2}O\,(g)\).

d. Does not exist in a state of chemical equilibrium as it is not a reversible process. The chemical reaction that takes place is: \(CH_{2}O+O_{2}\rightarrow H_{2}O\,(l)+CO_{2}\,(g)+nutrients\).

15.2: The Equilibrium Constant

  • For an equilibrium reaction, what effect does reversing the reactants and products have on the value of the equilibrium constant?
  • \(2\,HF\,(g)\rightleftharpoons H_{2}\,(g)+F_{2}\,(g)\)
  • \(C\,(s) + 2\,H_{2}\,(g) \rightleftharpoons CH_{4}\,(g)\)
  • \(H_{2}C=CH_{2}\,(g) + H_{2}\,(g) \rightleftharpoons C_{2}H_{6}\,(g)\)
  • \(2\,Hg\,(l) + O_{2}\,(g) \rightleftharpoons 2\,HgO\,(s)\)
  • \(NH_{4}CO_{2}NH_{2}\,(s) \rightleftharpoons 2NH_{3}\,(g)+CO_{2}\,(g)\)
  • \(C\,(s) + O_{2}\,(g) \rightleftharpoons CO_{2}\,(g)\)
  • \(2\,Mg\,(s) + O_{2}\,(g) \rightleftharpoons 2\,MgO\,(s)\)
  • \(AgCl\,(s) \rightleftharpoons Ag^+\,(aq)+Cl^−\,(aq)\)
  • If an equilibrium reaction is endothermic, what happens to the equilibrium constant if the temperature of the reaction is increased? if the temperature is decreased?
  • Industrial production of \(NO\) by the reaction \(N_{2}\,(g)+O_{2}\,(g) \rightleftharpoons 2\,NO_\,(g)\) is carried out at elevated temperatures to drive the reaction toward the formation of the product. After sufficient product has formed, the reaction mixture is quickly cooled. Why?
  • How would you differentiate between a system that has reached chemical equilibrium and one that is reacting so slowly that changes in concentration are difficult to observe?
  • What is the relationship between the equilibrium constant, the concentration of each component of the system, and the rate constants for the forward and reverse reactions?
  • \(CO\,(g) + H_2O\,(g) \rightleftharpoons CO_{2}\,(g)+H_{2}\,(g)\)
  • \(PCl_{3}\,(g)+Cl_{2}\,(g) \rightleftharpoons PCl_{5}\,(g)\)
  • \(2\,O_{3}\,(g) \rightleftharpoons 3\,O_{2}\,(g)\)

\(2\,NO\,(g)+O_{2}\,(g) \rightleftharpoons 2\,NO_{2}\,(g)\)

\(\dfrac{1}{2}\,H_2\,(g)+12\,I_{2}\,(g) \rightleftharpoons HI\,(g)\)

\(cis-stilbene\,(soln) \rightleftharpoons trans-silbene\,(soln)\)

  • Why is it incorrect to state that pure liquids, pure solids, and solvents are not part of an equilibrium constant expression?
  • \(2\,S\,(s)+3\,O_{2}\,(g) \rightleftharpoons 2\,SO_{3}\,(g)\)
  • \(C\,(s) + CO_{2}\,(g) \rightleftharpoons 2\,CO\,(g)\)
  • \(2\,ZnS\,(s)+3\,O_{2}\,(g) \rightleftharpoons 2\,ZnO\,(s)+2\,SO_{2}\,(g)\)
  • \(2\,HgO\,(s) \rightleftharpoons 2\,Hg\,(l)+O_{2}\,(g)\)
  • \(H_{2}\,(g)+I_{2}\,(s) \rightleftharpoons 2\,HI\,(g)\)
  • \(NH_4CO_2NH_{2}\,(s) \rightleftharpoons 2\,NH_{3}\,(g)+CO_{2}\,(g)\)
  • At room temperature, the equilibrium constant for the reaction \(2\,A\,(g) \rightleftharpoons B\,(g)\) is 1. What does this indicate about the concentrations of \(A\) and \(B\) at equilibrium? Would you expect \(K\) and \(K_p\) to vary significantly from each other? If so, how would their difference be affected by temperature?
  • For a certain series of reactions, if \(\dfrac{[OH^−][HCO_3^−]}{[CO_3^{2−}]} = K_1\) and \(\dfrac{[OH^−][H_2CO_3]}{[HCO_3^−]} = K_2\), what is the equilibrium constant expression for the overall reaction? Write the overall equilibrium equation.
  • In the equation for an enzymatic reaction, \(ES\) represents the complex formed between the substrate \(S\) and the enzyme protein \(E\). In the final step of the following oxidation reaction, the product \(P\) dissociates from the \(ESO_2\) complex, which regenerates the active enzyme:

Give the overall reaction equation and show that \(K = K_1 \times K_2 \times K_3\).

Conceptual Answers

1. By reversing the reactants and products for an equilibrium reaction, the equilibrium constant becomes: \(K′ = \dfrac{1}{K}\).

a. This equilibrium is homogenous as all substances are in the same state.

b. This equilibrium is heterogeneous as not all substances are in the same state.

c. This equilibrium is homogeneous as all substances are in the same state.

d. This equilibrium is heterogeneous as not all substances are in the same state.

a. This equilibrium is heterogeneous as not all substances are in the same state.

c. This equilibrium is heterogeneous as not all substances are in the same state.

4. According to Le Chatelier’s principle, equilibrium will shift in the direction to counteract the effect of a constraint (such as concentration of a reactant, pressure, and temperature). Thus, in an endothermic reaction, the equilibrium shifts to the right-hand side when the temperature is increased which increases the equilibrium constant and the equilibrium shifts to the left-hand side when the temperature is decreased which decreases the equilibrium constant.

5. After sufficient industrial production of \(NO\) by the reaction of \(N_{2}\,(g)+O_{2}\,(g) \rightleftharpoons 2\,NO_\,(g)\) at elevated temperatures to drive the reaction toward the formation of the product, the reaction mixture is cooled quickly because it quenches the reaction and prevents the system from reverting to the low-temperature equilibrium composition that favors the reactants.

6. To differentiate between a system that has reached equilibrium and one that is reacting slowly that changes in concentrations are difficult to observe we can use Le Chatelier’s principle to observe any shifts in the reaction upon addition of a constraint (such as concentration, pressure, or temperature).

7. The relationship between the equilibrium constant, the concentration of each component of a system, and the rate constants for the forward and reverse reactions considering a reaction of a general form: \(a\,A+b\,B \rightleftharpoons c\,C+d\,D\) is \(K=\dfrac{[C]^c[D]^d}{[A]^a[B]^b}=\dfrac{k_f}{k_r}\)

\(K=\dfrac{[CO_2][H_2]}{[CO][H_2O]}\)

\(K_p=\dfrac{(P_{CO_2})(P_{H_2})}{(P_{CO})(P_{H_2O})}\)

\(K=\dfrac{[PCl_5]}{[PCl_3][Cl_2]}\)

\(K_p=\dfrac{(P_{PCl_5})}{(P_{Cl_3})(P_{Cl_2})}\)

\(K=\dfrac{[O_2]^3}{[O_3]^2}\)

\(K_p=\dfrac{(P_{O_2})^3}{(P_{O_3})^2}\)

  • \(K=\dfrac{[NO_2]^2}{[NO]^2[O_2]}\)

\(K_p=\dfrac{(P_{NO_{2}})^2}{(P_{NO})^2(P_{O_2})}\)

\(K=\dfrac{[HI]}{[H_2]^{\dfrac{1}{2}}[O_2]}\)

\(K_p=\dfrac{(P_{HI})}{(P_{H_{2}})^{\frac{1}{2}}(P_{O_{2}})}\)

\(K=\dfrac{trans-stilbene}{cis-stilbene}\)

\(K_p=\dfrac{(P_{trans-stilbene})}{(P_{cis-stilbene})}\)

10. It is incorrect to state that pure liquids, pure solids, and solvents are not part of an equilibrium constant expression because they are not reactive enough or cause a change in the concentrations of the ions or the species that exist in the gas phase.

\(K=\dfrac{[SO_3]^2}{[O_2]^3}\)

\(K_p=\dfrac{(P_{SO_3})^2}{(P_{O_2})^3}\)

\(K=\dfrac{[CO]^2}{[CO_2]}\)

\(K_p=\dfrac{(P_{CO})^2}{(P_{CO_2})}\)

\(K=\dfrac{[SO_2]^2}{[O_2]^3}\)

\(K_p=\dfrac{(P_{SO_2})^2}{(P_{O_2})^3}\)

\(K=[O_2]\)

\(K_p=(P_{O_{2}})\)

\(K=\dfrac{[HI]^2}{[H_2]}\)

\(K_p=\dfrac{(P_{HI})^2}{(P_{H_2})}\)

\(K=[NH_3]^2[CO_2]\)

\(K_p=(P_{NH_3})^2(P_{CO_2})\)

\(K=\dfrac{[B]}{[A]^2} \rightarrow 1=\frac{[B]}{[A]^2} \rightarrow [A]^2=[B] \rightarrow [A]=\sqrt{B}\)

\(K\) and \(K_p\) vary by \(RT\), but it largely depends on \(T\) as \(R\) is constant. A raise or decrease in temperature would cause a difference.

\(K_p=K(RT)^{Δn}=K(RT)^{-1}=\dfrac{K}{RT}\)

\(Δn=(total\,moles\,of\,gas\,on\,the\,product\,side)-(total\,of\,moles\,on\,the\,reactant\,side)=1-2=−1\)

\(K=\dfrac{K_1}{K_2}={\dfrac{\dfrac{[OH^−][HCO_3^−]}{[CO_3^{2−}]}}{\dfrac{[OH^−][H_2CO_3]}{[HCO_3^−]}}=\dfrac{[HCO_3^−]^2}{[CO_3^{2-}][H_2CO_3]}}\)

\(CO_3^{2-}\,(g)+H_2CO_{3}\,(g) \rightleftharpoons 2\,HCO_3^{-}\,(g)\)

\(K = K_1 \times K_2 \times K_3=\frac{[ES]}{[E][S]}\times\frac{[ESO_2]}{[ES][O_2]}\times\frac{[E][P]}{[ESO_2]}=\frac{[P]}{[S][O_2]}\)

\(S+O_2 \rightleftharpoons P\)

Numerical Problems

  • Explain what each of the following values for \(K\) tells you about the relative concentrations of the reactants versus the products in a given equilibrium reaction: \(K = 0.892\); \(K = 3.25 \times 10^8\); \(K = 5.26 \times 10^{−11}\). Are products or reactants favored at equilibrium?
  • \(N_2O_{4}\,(g) \rightleftharpoons 2\,NO_{2}\,(g)\)
  • \(\frac{1}{2}\,N_2O_{4}\,(g) \rightleftharpoons NO_{2}\,(g)\)
  • \(\frac{1}{2}N_{2}\,(g)+\frac{3}{2}H_{2}\,(g) \rightleftharpoons NH_{3}\,(g)\)
  • \(\frac{1}{3}N_{2}\,(g)+H_{2}\,(g) \rightleftharpoons \frac{2}{3}NH_{3}\,(g)\)

How are these two expressions mathematically related to the equilibrium constant expression for

\[N_{2}\,(g)+3\,H_{2}\,(g) \rightleftharpoons 2\,NH_{3}\,(g) ?\]

  • \(C\,(s) + 2\,H_2O\,(g) \rightleftharpoons CO_{2}\,(g)+2\,H_{2}\,(g)\)
  • \(SbCl_{3}\,(g)+Cl_{2}\,(g) \rightleftharpoons SbCl_{5}\,(g)\)
  • Give an equilibrium constant expression for each reaction.

a. \(2\,NO\,(g) + O_{2}\,(g) \rightleftharpoons 2\,NO_{2}\,(g)\)

b. \(\frac{1}{2}H_{2}\,(g)+\frac{1}{2}I_{2}\,(g) \rightleftharpoons HI\,(g)\)

c. \(CaCO_{3}\,(s) + 2\,HOCl\,(aq) \rightleftharpoons Ca^{2+}\,(aq) + 2\,OCl^−\,(aq) + H_2O\,(l) + CO_{2}\,(g)\)

6. Calculate \(K\) and \(K_p\) for each reaction.

  • \(2\,NOBr\,(g) \rightleftharpoons 2\,NO\,(g)+Br_2\,(g)\): at 727°C, the equilibrium concentration of \(NO\) is 1.29 M, \(Br_2\) is 10.52 M, and \(NOBr\) is 0.423 M.
  • \(C\,(s) + CO_{2}\,(g) \rightleftharpoons 2\,CO\,(g)\): at 1,200 K, a 2.00 L vessel at equilibrium has partial pressures of 93.5 atm \(CO_2\) and 76.8 atm \(CO\), and the vessel contains 3.55 g of carbon.

7. Calculate \(K\) and \(K_p\) for each reaction.

  • \(N_2O_4\,(g) \rightleftharpoons 2\,NO_{2}\,(g)\): at the equilibrium temperature of −40°C, a 0.150 M sample of \(N_2O_4\) undergoes a decomposition of 0.456%.
  • \(CO\,(g)+2\,H_{2}\,(g) \rightleftharpoons CH_3OH\,(g)\): an equilibrium is reached at 227°C in a 15.5 L reaction vessel with a total pressure of \(6.71 \times 10^2\) atm. It is found to contain 37.8 g of hydrogen gas, 457.7 g of carbon monoxide, and 7,193 g of methanol.

8. Determine \(K\) and \(K_p\) (where applicable) for each reaction.

  • \(2\,H_2S\,(g) \rightleftharpoons 2\,H_{2}\,(g)+S_{2}\,(g)\): at 1065°C, an equilibrium mixture consists of \(1.00 \times 10^{−3}\) M \(H_2\), \(1.20 \times 10^{−3}\) M \(S_2\), and \(3.32 \times 10^{−3}\) M \(H_2S\).
  • \(Ba(OH)_{2}\,(s) \rightleftharpoons 2\,OH^−\,(aq)+Ba^{2+}\,(aq)\): at 25°C, a 250 mL beaker contains 0.330 mol of barium hydroxide in equilibrium with 0.0267 mol of barium ions and 0.0534 mol of hydroxide ions.

9. Determine \(K\) and \(K_p\) for each reaction.

  • \(2\,NOCl\,(g) \rightleftharpoons 2\,NO\,(g)+Cl_{2}\,(g)\): at 500 K, a 24.3 mM sample of \(NOCl\) has decomposed, leaving an equilibrium mixture that contains 72.7% of the original amount of \(NOCl\).
  • \(Cl_{2}\,(g)+PCl_{3}\,(g) \rightleftharpoons PCl_{5}\,(g)\): at 250°C, a 500 mL reaction vessel contains 16.9 g of \(Cl_2\) gas, 0.500 g of \(PCl_3\), and 10.2 g of \(PCl_5\) at equilibrium.

10. The equilibrium constant expression for a reaction is \(\dfrac{[CO_2]^2}{[SO_2]^2[O_2]}\). What is the balanced chemical equation for the overall reaction if one of the reactants is \(Na_2CO_{3}\,(s)\)?

11. The equilibrium constant expression for a reaction is \(\dfrac{[NO][H_{2}O]^{\dfrac{3}{2}}}{[NH_3][O_2]^{\dfrac{5}{4}}}\). What is the balanced chemical equation for the overall reaction?

12. Given \(K =\dfrac{k_f}{k_r}\), what happens to the magnitude of the equilibrium constant if the reaction rate of the forward reaction is doubled? What happens if the reaction rate of the reverse reaction for the overall reaction is decreased by a factor of 3?

13. The value of the equilibrium constant for \[2\,H_{2}\,(g)+S_{2}\,(g) \rightleftharpoons 2\,H_2S\,(g)\] is \(1.08 \times 10^7\) at 700°C. What is the value of the equilibrium constant for the following related reactions

  • \(H_{2}\,(g)+12\,S_{2}\,(g) \rightleftharpoons H_2S\,(g)\)
  • \(4\,H_{2}\,(g)+2\,S_{2}\,(g) \rightleftharpoons 4\,H_2S\,(g)\)
  • \(H_2S\,(g) \rightleftharpoons H_{2}\,(g)+12\,S_{2}\,(g)\)

Numerical Answers

1. In the given equilibrium reaction where \(K = 0.892\approx1\) has a concentration of the reactants that is approximately equal to the concentration of the products so neither formation of the reactants or products is favored. In the given equilibrium reaction where \(K = 3.25 \times 10^8>1\) has a concentration of the products that is relatively small compared to the concentration of the reactants so the formation of the products is favored. In the given equilibrium reaction where \(K = 5.26 \times 10^{−11}<1\) has a concentration of products that is relatively large compared to that of the concentration of the reactants so the formation of the reactants is favored.

a. \(K=\dfrac{[NO_2]^{2}}{[N_2O_4]}\)

b. \(K=\dfrac{[NO_2]}{[N_2O_4]^{\dfrac{1}{2}}}\)

Although the equilibrium constant expressions have a 2:1 ratio of concentration for the products to the concentration of the reactants for the same species involved to get the \(K\) value for a. We would need to square it to get the \(K\) value for b.

\(K’=\dfrac{[NH_3]}{[N_2]^{\dfrac{1}{2}}[H_2]^{\dfrac{3}{2}}}\)

\(K’’=\dfrac{[NH_3]^{\dfrac{2}{3}}}{[N_2]^{\dfrac{1}{2}}[H_2]}\)

\(K=\dfrac{[NH_3]^2}{[N_2][H_2]^3}\)

\(K'=K^{\dfrac{1}{2}}\)

\(K''=K^{\dfrac{1}{3}}\)

a. \(K=\dfrac{[H_2]^2[CO_2]}{[H_{2}O]^2}\)

b. \(K=\dfrac{[SbCl_5]}{[SbCl_3][Cl_2]}\)

c. \(K=\dfrac{[O_2]^3}{[O_3]^2}\)

  • \(K=\dfrac{[HI]}{[H_2]^{\dfrac{1}{2}}[I_2]^{\dfrac{1}{2}}}\)
  • \(K=\dfrac{[Ca^{2+}][OCl^−]^2[CO_2]}{[HOCl]^2}\)

\(K=\dfrac{[NO]^2[Br]}{[NOBr]^2}=\frac{[1.29\,M]^2[10.52\,M]}{[0.423\,M]^2}=97.8\)

\(K_p=K(RT)^{Δn}=(97.8)((0.08206\frac{L\cdot atm}{mol\cdot K})((727+273.15)K))^{3-2}=8.03x10^{4}\)

\(K_p=K(RT)^{Δn}=K(RT)^{2-1}=K(RT) \rightarrow K=\dfrac{K_p}{RT}=\frac{63.1}{(0.08206\frac{L\cdot atm}{mol\cdot K})(1,200\,K)}=6.41\)

\(K=\dfrac{(P_{CO})^2}{P_{CO_2}}=\frac{(76.8\,atm)^2}{93.5\,atm}=63.1\)

\(K=\dfrac{[NO_{2}]^2}{[N_{2}O_{4}]}=\frac{[0.001368\,M]}{[0.149316\,M]}=1.25x10^{-5}\)

\([NO_2]=(2)(0.150\,M)(0.00456)=0.001368\,M\)

\([N_2O_4]=(0.150\,M)(1-0.00456)=0.149316\,M\)

\(K_p=K(RT)^{Δn}=(1.25x10^{-5})((0.08206\frac{L\cdot atm}{mol\cdot K})((-40+273.15)K))^{2-1}=2.39x10^{-4}\)

\(K=\dfrac{[CH_{3}OH]}{[CO][H_{2}]^2}=\frac{[14.5\,M]}{[1.05\,M][1.21\,M]^2}=9.47\)

\([CH_{3}OH]=7,193\,g\,CH_{3}OH \times \frac{1\,mol\,CH_{3}OH}{32.04\,g\,CH_{3}OH} \times \frac{1}{15.5\,L}=14.5\,M\)

\([CO]=457.7\,g\,CO \times \frac{1\,mol\,CO}{28.01\,g\,CO} \times \frac{1}{15.5\,L}=1.05\,M\)

\([H_{2}]=37.8\,g\,H_{2} \times \frac{1\,mol\,H_{2}}{2.02\,g\,H_{2}} \times \frac{1}{15.5\,L}=1.21\,M\)

\(K_p=K(RT)^{Δn}=(9.47)((0.08206\frac{L\cdot atm}{mol\cdot K})((227+273.15)K))^{1-3}=5.62x10^{-2}\)

\(K=\dfrac{[H_{2}]^2[S_{2}]}{[H_{2}S]^2}=\frac{[1.00 \times 10^{-3}\,M]^2[1.20 \times 10^{-3}\,M]}{[3.32 \times\ 10^{-3}\,M]^2}=1.09 \times 10^{-4}\)

\(K_p=K(RT)^{Δn}=(1.09 \times 10^{-4})((0.08206\frac{L\cdot atm}{mol\cdot K})((1065+273.15)K))^{3-2}=1.20 \times 10^{-2}\)

\(K=[OH^{-}]^2[Ba^{2+}]=[0.2136\,M]^2[0.1068\,M]=4.87 \times 10^{-3}\)

\([OH^{-}]=0.0534\,mol\,OH^{-} \times \frac{1}{0.25\,L}=0.2136\,M\)

\([Ba^{2+}]=0.0267\,mol\,Ba^{2+} \times \frac{1}{0.25\,L}=0.1068\,M\)

\(K_p=K(RT)^{Δn}=(4.87 \times 10^{-3})((0.08206\frac{L\cdot atm}{mol\cdot K})((25+273.15)K))^{3-1}=2.92\)

\(K=\dfrac{[NO]^2[Cl_{2}]}{[NOCl]^2}=4.59 \times 10^{-4}\)

\(K_p=K(RT)^{Δn}=(4.59 \times 10^{-4})((0.08206\frac{L\cdot atm}{mol\cdot K})(500K))^{3-2}=1.88 \times 10^{-2}\)

\(K=\dfrac{[PCl_5]}{[Cl_{2}][PCl_{3}]}=\frac{[9.80 \times 10^{-2}\,M]}{[4.77 \times 10^{-1}\,M][7.28 \times 10^{-3}\,M]}=28.2\)

\([PCl_{5}]=10.2\,g\,PCl_{5} \times \frac{1\,mol\,PCl_{5}}{208.2388\,g\,PCl_{5}} \times \frac{1}{0.5\,L}=9.80 \times 10^{-2}\,M\)

\([Cl_{2}]=16.9\,g\,Cl_{2} \times \frac{1\,mol\,Cl_{2}}{70.9\,g\,Cl_{2}} \times \frac{1}{0.5\,L}=4.77 \times 10^{-1}\,M\)

\([PCl_{3}]=0.500\,g\,PCl_{3} \times \frac{1\,mol\,PCl_{3}}{137.33\,g\,PCl_{3}} \times \frac{1}{0.5\,L}=7.28 \times 10^{-3}\,M\)

\(K_p=K(RT)^{Δn}=(28.2)((0.08206\frac{L\cdot atm}{mol\cdot K})(250+273.15)K)^{1-2}=6.57 \times 10^{-1}\)

10. \(2\,SO_{2}\,(g)+O_{2}\,(g)+2\,Na_{2}CO_{3}\,(s) \rightleftharpoons 2\,CO_{2}\,(g)+2\,Na_{2}SO_{4}\,(s)\)

11. \(NH_{3}\,(g) + \frac{5}{4}\,O_{2}\,(g)⇌NO \,(g)+\frac{3}{2}\,H_{2}O\,(g)\)

a. \(K= \dfrac{[H_{2}S]}{[H_{2}][S_{2}]^\frac{1}{2}}=K’^{\frac{1}{2}}=(1.08 \times 10^{7})^{\frac{1}{2}}=3.29 \times 10^{3}\) b. \(K= \dfrac{[H_{2}S]^4}{[H_{2}]^4[S_{2}]^2}=K’^{2}=(1.08 \times 10^{7})^{2}=1.17 \times 10^{14}\) c. \(K= \dfrac{[H_{2}][S_2]^\frac{1}{2}}{[H_{2}S]}=K’^{-\frac{1}{2}}= (1.08 \times 10^{7})^{-\frac{1}{2}}=3.04 \times 10^{-4}\)

15.3: Interpreting & Working with Equilibrium Constants

  • Describe how to determine the magnitude of the equilibrium constant for a reaction when not all concentrations of the substances are known.
  • Calculations involving systems with very small or very large equilibrium constants can be dramatically simplified by making certain assumptions about the concentrations of products and reactants. What are these assumptions when \(K\) is (a) very large and (b) very small? Illustrate this technique using the system \(A+2B \rightleftharpoons C\) for which you are to calculate the concentration of the product at equilibrium starting with only A and B. Under what circumstances should simplifying assumptions not be used?

1. The magnitude of the equilibrium constant for a reaction depends on the form in which the chemical reaction is written. For example, writing a chemical reaction in different but chemically equivalent forms causes the magnitude of the equilibrium constant to be different but can be related by comparing their respective magnitudes.

a. When \(K\) is very large the reactants are converted almost entirely to products, so we can assume that the reaction proceeds 100% to completion.

\(K= \dfrac{[C]}{[A][B]^2}=\frac{[C]}{very\,small}=\frac{1}{0}= \infty \rightarrow [C]= \infty\)

b. When \(K\) is very small the reactants do not tend to form products readily, and the equilibrium lies to the left as written, favoring the formation of the reactants.

\(K=\dfrac{[C]}{[A][B]^2}=\frac{very\,small}{[A][B]^2}=\frac{0}{1}=0 \rightarrow [C]=0\)

Simplifying assumptions should not be used if the equilibrium constant is not known to be very large or very small.

Please be sure you are familiar with the quadratic formula before proceeding to the Numerical Problems.

  • In the equilibrium reaction \(A+B \rightleftharpoons C\), what happens to \(K\) if the concentrations of the reactants are doubled? tripled? Can the same be said about the equilibrium reaction \(2\,A \rightleftharpoons B+C\)?
  • The following table shows the reported values of the equilibrium \(P_{O_2}\) at three temperatures for the reaction \(Ag_{2}O\,(s) \rightleftharpoons 2\,Ag\,(s)+ \frac{1}{2}\,O_{2}\,(g)\), for which ΔH° = 31 kJ/mol. Are these data consistent with what you would expect to occur? Why or why not?
  • Given the equilibrium system \(N_2O_{4}\,(g) \rightleftharpoons 2\,NO_{2}\,(g)\), what happens to \(K_p\) if the initial pressure of \(N_2O_4\) is doubled? If \(K_p\) is \(1.7 \times 10^{−1}\) at 2300°C, and the system initially contains 100% \(N_2O_4\) at a pressure of \(2.6 \times 10^2\) atm, what is the equilibrium pressure of each component?
  • At 430°C, 4.20 mol of \(HI\) in a 9.60 L reaction vessel reaches equilibrium according to the following equation: \[H_{2}\,(g)+I_{2}\,(g) \rightleftharpoons 2\,HI\,(g)\] At equilibrium, \([H_2] = 0.047\;M\) and \([HI] = 0.345\;M\) What are \(K\) and \(K_p\) for this reaction?
  • Methanol, a liquid used as an automobile fuel additive, is commercially produced from carbon monoxide and hydrogen at 300°C according to the following reaction: \[CO\,(g)+2\,H_{2}\,(g) \rightleftharpoons CH_3OH\,(g)\] with \(K_p = 1.3 \times 10^{−4}\). If 56.0 g of \(CO\) is mixed with excess hydrogen in a 250 mL flask at this temperature, and the hydrogen pressure is continuously maintained at 100 atm, what would be the maximum percent yield of methanol? What pressure of hydrogen would be required to obtain a minimum yield of methanol of 95% under these conditions?
  • Starting with pure A, if the total equilibrium pressure is 0.969 atm for the reaction \(A\,(s) \rightleftharpoons 2\,B\,(g)+C\,(g)\), what is \(K_p\)?
  • The decomposition of ammonium carbamate to \(NH_3\) and \(CO_2\) at 40°C is written as \(NH_4CO_2NH_{2}\,(s) \rightleftharpoons 2\,NH_{3}\,(g)+CO_{2}\,(g)\). If the partial pressure of \(NH_3\) at equilibrium is 0.242 atm, what is the equilibrium partial pressure of \(CO_2\)? What is the total gas pressure of the system? What is \(K_p\)?
  • What is \(K\) for the reaction at each temperature?
  • If a sample at 375 K has 0.100 M \(Cl_2\) and 0.200 M \(SO_2\) at equilibrium, what is the concentration of \(SO_2Cl_2\)?
  • If the sample given in part b is cooled to 303 K, what is the pressure inside the bulb?
  • For the gas-phase reaction \(a\,A \rightleftharpoons b\,B\), show that \(K_p = K(RT)^{Δn}\) assuming ideal gas behavior.
  • For the gas-phase reaction \(I_2 \rightleftharpoons 2\,I\), show that the total pressure is related to the equilibrium pressure by the following equation: \[P_T=\sqrt{K_{p}P_{I_{2}}} + P_{I_{2}}\]
  • Experimental data on the system \(Br_{2}\,(l) \rightleftharpoons Br_{2}\,(aq)\) are given in the following table. Graph \(Br_{2}\,(aq)\) versus moles of \(Br_{2}\,(l)\) present; then write the equilibrium constant expression and determine K.
  • Data accumulated for the reaction (\n-butane(g) \rightleftharpoons isobutane(g)\) at equilibrium are shown in the following table. What is the equilibrium constant for this conversion? If 1 mol of n-butane is allowed to equilibrate under the same reaction conditions, what is the final number of moles of n-butane and isobutane?
  • Solid ammonium carbamate (\(NH_{4}CO_{2}NH_{2}\)) dissociates completely to ammonia and carbon dioxide when it vaporizes: \(NH_{4}CO_{2}NH_{2}\,(s) \rightleftharpoons 2\,NH_{3}\,(g)+CO_{2}\,(g)\) At 25°C, the total pressure of the gases in equilibrium with the solid is 0.116 atm. What is the equilibrium partial pressure of each gas? What is \(K_p\)? If the concentration of \(CO_2\) is doubled and then equilibrates to its initial equilibrium partial pressure +x atm, what change in the \(NH_{3}\) concentration is necessary for the system to restore equilibrium?
  • The equilibrium constant for the reaction \(COCl_{2}\,(g) \rightleftharpoons CO\,(g)+Cl_{2}\,(g)\) is \(K_p = 2.2 \times 10^{−10}\) at 100°C. If the initial concentration of \(COCl_{2}\) is \(3.05 \times 10^{−3}\; M\), what is the partial pressure of each gas at equilibrium at 100°C? What assumption can be made to simplify your calculations?
  • Aqueous dilution of \(IO_{4}^{−}\) results in the following reaction: \[IO^−_{4}\,(aq)+2\,H_{2}O_(l)\, \rightleftharpoons H_4IO^−_{6}\,(aq)\] with \(K = 3.5 \times 10^{−2}\). If you begin with 50 mL of a 0.896 M solution of \(IO_4^−\) that is diluted to 250 mL with water, how many moles of \(H_4IO_6^−\) are formed at equilibrium?
  • Iodine and bromine react to form \(IBr\), which then sublimes. At 184.4°C, the overall reaction proceeds according to the following equation: \[I_{2}\,(g)+Br_{2}\,(g) \rightleftharpoons 2\,IBr\,(g)\] with \(K_p = 1.2 \times 10^2\). If you begin the reaction with 7.4 g of \(I_2\) vapor and 6.3 g of \(Br_2\) vapor in a 1.00 L container, what is the concentration of \(IBr\,(g)\) at equilibrium? What is the partial pressure of each gas at equilibrium? What is the total pressure of the system?
  • For the reaction \[2\,C\,(s) + \,N_{2}\,(g)+5\,H_{2}\, \rightleftharpoons 2\,CH_{3}NH_{2}\,(g)\] with \(K = 1.8 \times 10^{−6}\). If you begin the reaction with 1.0 mol of \(N_2\), 2.0 mol of \(H_2\), and sufficient \(C\,(s)\) in a 2.00 L container, what are the concentrations of \(N_2\) and \(CH_3NH_2\) at equilibrium? What happens to \(K\) if the concentration of \(H_2\) is doubled?

1. In both cases, the equilibrium constant will remain the same as it does not depend on the concentrations.

2. No, the data is not consistent with what I would expect to occur because enthalpy is positive indicating that the reaction is endothermic thus heat is on the left side of the reaction. As the temperature is raised \(P_{O_2}\) would be expected to increase to counteract the constraint.

If the initial pressure of \(N_2O_4\) was doubled then \(K_p\) is one half of the original value.

\(K_p=\dfrac{(P_{NO_{2}})^2}{(P_{N_{2}O_{4}})} \rightarrow 1.7 \times 10^{-1} = \frac{(2x)^2}{2.6 \times 10^2 -x} \rightarrow 44.2-0.17x=4x^2 \rightarrow x \approx 3.303\,atm\)

\(P_{N_{2}O_{4}}=2.6 \times 10^2-x=260-3.303=2.6 \times 10^2\,atm\)

\(P_{NO_{2}}=2x=(2)(3.303)=6.6\,atm\)

\(K=\frac{[HI]^2}{[H_2][I_2]}=\frac{0.345\,M}{(0.047\,M)(0.047\,M)}=157\)

\(K_p=K(RT)^{Δn}=(157)((0.08206\frac{L\cdot atm}{mol\cdot K})(430+273.15)K)^{2-2}=157\)

\(Maximum\;Percent\;Yield=\frac{Actual}{Theoretical} \times 100\%=\frac{212.593}{376.127} \times 100\%=56.52\% \approx 57\%\)

\(PV=nRT \rightarrow P=\frac{(1.999\;mol)(0.08206\frac{L\cdot atm}{mol\cdot K})(300+273.15)K}{0.250\;L}=376.127\;atm\)

\([CO]=56.0\;g\;CO \times \frac{1\;mol\;CO}{28.01\,g\,CO}=1.999\,mol\,CO\)

\(K_p=\frac{P_{CH_{3}OH}}{(P_{CO})(P_{H_2})^{2}} \rightarrow 1.3 \times 10^{-4}= \frac{x}{(376.02-x)(100^2)} \rightarrow 1.3= \frac{x}{376.07-x} \rightarrow 488.965-1.3x=x \rightarrow 488.965=2.3x \rightarrow x=212.593\,atm\)

\(K_p=\frac{(P_{CH_{3}OH})}{(P_{CO})(P_{H_2})^2} \rightarrow 1.3 \times 10^{-4}= \frac{357.320}{(376.127-357.320)(P_{H_{2}})^2} \rightarrow 1.3 \times 10^{-4}= \frac{357.320}{(18.80635)(P_{H_{2}})^2} \rightarrow 0.002444(P_{H_{2}})^2=357.320 \rightarrow (P_{H_{2}})=382.300 \approx 3.8 \times 10^2\,atm\)

\(Minimum\,Percent\,Yield=\frac{Actual}{Theoretical} \times 100 \% \rightarrow 95\%=\frac{x}{376.127} \times 100\% \rightarrow x=357.320\,atm\)

6. \(K_p=\frac{(P_B)^2(P_C)}{P_A}=\frac{[2x]^2[x]}{[0.969-x]}=\frac{4x^3}{0.969-x}\)

\(P_{CO_{2}}=P_{tot}-P_{NH_{3}}=P_{tot}-0.242\,atm\)

\(P_{tot}=P_{NH_3}+P_{CO_2}=0.242\,atm+P_{CO_2}\)

\(K_p=(P_{NH_3})^2(P_{CO_2})=(0.242\,atm)^2(P_{CO_2})\)

\(At\,375\,K:K_p=K(RT)^{Δn} \rightarrow K=\frac{K_p}{(RT)^{Δn}}=\frac{2.9 \times 10^{-2}}{((0.08206\frac{L\cdot atm}{mol\cdot K})(375\,K))^{2-1}}=7.80 \times 10^{-2}\)

\(At\,303\,K:K_p=K(RT)^{Δn} \rightarrow K=\frac{K_p}{(RT)^{Δn}}=\frac{2.9 \times 10^{-2}}{((0.08206\frac{L\cdot atm}{mol\cdot K})(303\,K))^{2-1}}=1.17 \times 10^{-3}\)

b. \(K=\frac{[SO_{2}][Cl_{2}]}{[SO_{2}Cl_{2}]} \rightarrow [SO_{2}Cl_{2}]=\frac{[0.200\,M][0.100\,M]}{7.80 \times 10^{-2}}=2.56 \times 10^{-1}\,M\)

c. If the sample given in part b is cooled to 303 \(K\), the pressure inside the bulb would decrease.

\(K_p=\frac{(P_B)^b}{(P_A)^a}=\frac{((\frac{n_B}{V})(RT))^b}{((\frac{n_A}{V})(RT))^a}=\frac{[B]^{b}(RT)^b}{[A]^a(RT)^a}=K(RT)^{b-a}=K(RT)^{Δn}\)

\(PV=nRT \rightarrow P=\frac{n}{V}RT\)

\(K=\frac{[B]^{b}}{[A]^{a}}\)

\(Δn=b-a\)

\(P_T=P_I+P_{I_2}=\sqrt{K_pP_{I_2}}+P_{I_2}\)

\(K_p=\frac{(P_I)^2}{P_{I_2}} \rightarrow (P_I)^2=K_p(P_{I_2}) \rightarrow P_I=\sqrt{K_pP_{I_2}}\)

The graph should be a positive linear correlation.

\([Br_2\,(l)]=1.0\,g\,Br_2 \times \frac{1\,mol\,Br_2}{159.808\,g\,Br_2} \times \frac{1}{0.1\,L}=6.26 \times 10^{-2}\)

\(K=\frac{[Br_2\,(aq)]}{[Br_2\,(l)]}=\frac{[Br_2\,(aq)]}{1}=[Br_2\,(aq)]\)

12. \(K=\frac{[isobutane]}{[n-butane]}=\frac{x}{1-x}\)

\(P_{NH_3}=2x=2(0.0387)=7.73 \times 10^{-2}\,atm\)

\(P_{CO_2}=x=3.87 \times 10^{-2}\,atm\)

\(K_p=(P_{NH_3})^2(P_{CO_2})=(2x)^2(x)=4x^3=4(0.0387)^3=2.32 \times 10^{-4}\)

\(P_{tot}=P_{NH_3}+P_{CO_2} \rightarrow 0.116=2x+x \rightarrow 0.116=3x \rightarrow x=0.0387\)

If the concentration of \(CO_{2}\) is doubled and then equilibrates to its initial equilibrium partial

pressure +x atm, the concentration of \(NH_{3}\) should also be doubled for the system to restore

equilibrium.

\(P_{COCl_{2}}=9.34 \times 10^{-2}-x=9.34 \times 10^{-2}-9.34 \times 10^{-22}=9.34 \times 10^{-2}\,atm\)

\(P_{CO}=x=9.34 \times 10^{-22}\,atm\)

\(P_{Cl_{2}}=x=9.34 \times 10^{-22}\,atm\)

Assume that the equilibrium mainly lies on the reactants side because the \(K_p\) value is less than 1.

\(K_p=\frac{(P_{CO})(P_{Cl_{2}})}{(P_{COCl_{2}})} \rightarrow 2.2 \times 10^{-10} =\frac{x^{2}}{9.34 \times 10^{-2}-x} \rightarrow 2.0548 \times 10^{-11}-2.2 \times 10^{-10}x=x^{2} \rightarrow x^{2}+2.2 \times 10^{-10}x-2.0548 \times 10^{-11}=0 \rightarrow x=9.34 \times 10^{-22}\)

\(PV=nRT \rightarrow P=\frac{nRT}{V}=(3.05 \times 10^{-3})(0.08206\frac{L\cdot atm}{mol\cdot K})(100+273.15)K=9.34 \times 10^{-2}\)

15. \([H_{4}IO_{6}^{-}]=x=1.6 \times 10^{-3}\,mol\)

\(K=\frac{[H_{4}IO_{6}^{-}]}{[IO_{4}^{-}]} \rightarrow 3.5 \times 10^{-2} =\frac{x}{(0.0448-x)} \rightarrow x=1.568 \times 10^{-3}\)

\(IO_{4}^{-}:50\,mL\,IO_{4}^{-} \times \frac{1\,L\,IO_{4}^{-}}{1,000\,mL\,IO_{4}^{-}} \times \frac{0.896\,mol\,IO_{4}^{-}}{1\,L\,IO_{4}^{-}}=0.0448\,mol\)

16. \(PV=nRT \rightarrow \frac{P}{RT}=\frac{n}{V} \rightarrow \frac{12.9468}{(0.08206\frac{L\cdot atm}{mol\cdot K})(184.4+273.15)K}=3.5 \times 10^{-1}\,M\)

\(K_p=\frac{(P_{IBr})^2}{(P_{I_{2}})(P_{Br_{2}})} \rightarrow 1.2 \times 10^{-2} = \frac{2x}{(1.096-x)(1.479-x)}=\frac{2x}{x^2-2.575x+1.62098} \rightarrow 0.012x^2-0.0309x+0.0194518=2x \rightarrow 0.012x^2-2.0309+0.0194518=0 \rightarrow x=12.9468\)

\(PV=nRT \rightarrow P=\frac{nRT}{V}=(2.92 \times 10^{-2}\,M)(0.08206\frac{L\cdot atm}{mol\cdot K})(184.4+273.15)K=1.096\,atm\)

\([I_{2}]=7.4\,g\,I_{2} \times \frac{1\,mol\,I_{2}}{253\,g\,I_{2}} \times \frac{1}{1.00\,L}=2.92 \times 10^{-2}\,M\)

\(PV=nRT \rightarrow P={nRT}{V}=(3.94 \times 10^{-2}\,M)(0.08206\frac{L\cdot atm}{mol\cdot K})(184.4+273.15)K=1.479\,atm\)

\([Br_{2}]=6.3\,g\,Br_{2} \times \frac{1\,mol\,Br_{2}}{159.808\,g\,Br_{2}}=3.94 \times 10^{-2}\,M\)

\([N_{2}]=0.5-12x=0.5-12(0.000471330)=0.494\,M\)

\([CH_{3}NH_{2}]=2x=2(0.0004713300=9.43 \times 10^{-4}\,M\)

\(If\,the\,concentration\,of\,H_{2}\,is\,doubled\,,then\,K=\frac{[CH_{3}NH_{2}]^{2}}{[N_2][H_2]^5}=\frac{(9.43 \times 10^{-4})^{2}}{(0.494)(1.998)^5}=5.65 \times 10^{-8}\)

\(2 \times [H_{2}]=2(1.0-2.5x)=2(1.0-2.5(0.000471330))=1.998\,M\)

\(K=\frac{[CH_{3}NH_{2}]^2}{[N_2][H_2]^5} \rightarrow 1.8 \times 10^{-6}=\frac{(2x)^2}{(0.5-x)(1.0-5x)^5} \rightarrow x=0.000471330\,M\)

\([N_2]=1.00\,mol\,N_2 \times \frac{1}{2.00\,L}=0.5\,M\)

\([H_2]=2.00\,mol\,H_2 \times \frac{1}{2.00\,L}=1.0\,M\)

15.4: Heterogeneous Equilibria

15.5: calculating equilibrium constants, 15.6: applications of equilibrium constants.

1. During a set of experiments, graphs were drawn of [reactants] versus [products] at equilibrium. Using Figure 15.8 and Figure 15.9 as your guides, sketch the shape of each graph using appropriate labels.

  • \(H_2O\,(l) \rightleftharpoons H_2O\,(g)\)
  • \(2\,MgO\,(s) \rightleftharpoons 2\,Mg\,(s)+O_{2}\,(g)\)
  • \(2\,PbS\,(g)+3\,O_{2}\,(g) \rightleftharpoons 2\,PbO\,(s)+2\,SO_{2}\,(g)\)

2. Write an equilibrium constant expression for each reaction system. Given the indicated changes, how must the concentration of the species in bold change if the system is to maintain equilibrium?

  • \(2\,NaHCO_{3}\,(s) \rightleftharpoons Na_2CO_{3}\,(s) + CO_{2}\,(g)+ H_2O\,(g)\): \([CO_2]\) is doubled.
  • \(N_2F_{4}\,(g) \rightleftharpoons 2\,NF_{2}\,(g)\): \([NF_{2}]\) is decreased by a factor of 2.
  • \(H_{2}\,(g) + I_{2}\,(g) \rightleftharpoons 2\,HI\,(g)\): \([I_2]\) is doubled.

3. Write an equilibrium constant expression for each reaction system. Given the indicated changes, how must the concentration of the species in bold change if the system is to maintain equilibrium?

  • \(CS_{2}\,(g) + 4\,H_{2}\,(g) \rightleftharpoons CH_{4}\,(g) + 2\,H_2S\,(g)\): \([CS_2]\) is doubled.
  • \(PCl_{5}\,(g) \rightleftharpoons PCl_{3}\,(g) + Cl_{2}\,(g)\): \([Cl_2]\) is decreased by a factor of 2.
  • \(4\,NH_{3}\,(g) + 5\,O_{2}\,(g) \rightleftharpoons 4\,NO\,(g) + 6\,H_2O\,(g)\): \([NO]\) is doubled.

a. According to Figure 15.8, we could obtain a graph with the x-axis labeled \([H_2O]\,(l)\,(M)\) and y-axis labeled \([H_2O]\,(g)\,(M)\). The graph should have a positive linear correlation. For any equilibrium concentration of \(H_2O\,(g)\), there is only one equilibrium \(H_2O\;(l)\). Because the magnitudes of the two concentrations are directly proportional, a large \([H_2O]\,(g)\) at equilibrium requires a large \([H_2O]\,(l)\) and vice versa. In this case, the slope of the line is equal to \(K\).

b. According to Figure 15.9, we could obtain a graph with the x-axis labeled \([O2]\,(M)\) and y-axis labeled \([MgO]\,(M)\). Because \(O_2\,(g)\) is the only one in gaseous form, the graph would depend on the concentration of \(O_2\).

c. According to Figure 15.8, we could obtain a graph with the x-axis labeled \([O_3]\,(M)\) and y-axis labeled \([O2]\,(M)\). The graph should have a positive linear correlation. For every \(3\,O_2\,(g)\) there is \(2\,O_3\,(g)\). Because the magnitudes of the two concentrations are directly proportional, a large \([O3]\,(g)\) at equilibrium requires a large \([O_2]\,(g)\) and vice versa. In this case, the slope of the line is equal to \(K\).

d. According to figure 15.8, we could obtain a graph with the x-axis labeled \([O2]\,(M)\) and y-axis labeled \([SO2]\,(M)\). The graph should have a positive linear correlation. For every \(3\,O_{2}\,(g)\) there is \(2\,SO_2\,(g)\). Because the magnitudes of the two concentrations are directly proportional, a large \(O_2\,(g)\) at equilibrium requires a large \(SO_2\,(g)\) and vice versa. In this case, the slope of the line is equal to \(K\).

\(K=[Na_{2}CO_{3}][CO_2][H_{2}O]\)

If \([CO_2]\) is doubled, \([H_2O]\) should be halved if the system is to maintain equilibrium.

\(K=\frac{[NF_2]^2}{[N_{2}F_{4}]}\)

If \([NF_2]\) is decreased by a factor of 2, then \([N_{2}F_{4}]\) must also be decreased by a factor of 2 if the system is to maintain equilibrium.

\(K=\frac{[HI]^2}{[H_2][I_{2}]}\)

If \([I_{2}]\) is doubled then \([HI]\) must also be doubled if the system is to maintain equilibrium.

\(K=\dfrac{[CH_4][H_2S]^2}{[CS_2][H_2]^4}\)

If \([CS_2]\) is doubled then \([H_2]\) must be decreased by a factor of 2√4≅ 1.189 if the system is to maintain equilibrium.

\(K=\dfrac{[PCl_3]}{[Cl_2][PCl_5]}\)

If \([Cl_2]\) is halved then \([PCl_5]\) must also be halved if the system is to maintain equilibrium.

\(K=\dfrac{[NO]^4[H_2O]^6}{[NH_3][O_2]^5}\)

If \([NO]\) is doubled then \([H_2O]\) must also be multiplied by 22/3≅1.587 if the system is to maintain equilibrium.

  • The data in the following table were collected at 450°C for the reaction \(N_{2}\,(g)+3\,H_{2}\,(g) \rightleftharpoons 2\,NH_{3}\,(g)\):

The reaction equilibrates at a pressure of 30 atm . The pressure on the system is first increased to 100 atm and then to 600 atm . Is the system at equilibrium at each of these higher pressures? If not, in which direction will the reaction proceed to reach equilibrium?

  • For the reaction \(A \rightleftharpoons B+C\), \(K\) at 200°C is 2.0. A 6.00 L flask was used to carry out the reaction at this temperature. Given the experimental data in the following table, all at 200°C, when the data for each experiment were collected, was the reaction at equilibrium? If it was not at equilibrium, in which direction will the reaction proceed?
  • The following two reactions are carried out at 823 K:

\[CoO\,(s)+H_{2}\,(g) \rightleftharpoons Co\,(s)+H_2O\,(g) \text{ with } K=67\]

\[CoO\,(s)+CO\,(g) \rightleftharpoons Co\,(s)+CO_{2}\,(g) \text{ with } K=490\]

  • Write the equilibrium expression for each reaction.
  • Calculate the partial pressure of both gaseous components at equilibrium in each reaction if a 1.00 L reaction vessel initially contains 0.316 mol of \(H_2\) or \(CO\) plus 0.500 mol \(CoO\).
  • Using the information provided, calculate Kp for the following reaction: \[H_{2}\,(g)+CO_{2}\,(g) \rightleftharpoons CO\,(g)+H_2O\,(g)\]
  • Describe the shape of the graphs of [reactants] versus [products] as the amount of \(CoO\) changes.
  • Hydrogen iodide (HI) is synthesized via \(H_{2}\,(g)+I_{2}\,(g) \rightleftharpoons 2\,HI\,(g)\), for which \(K_p = 54.5\) at 425°C. Given a 2.0 L vessel containing \(1.12 \times 10^{−2}\,mol\) of \(H_2\) and \(1.8 \times 10^{−3}\,mol\) of \(I_2\) at equilibrium, what is the concentration of \(HI\)? Excess hydrogen is added to the vessel so that the vessel now contains \(3.64 \times 10^{−1}\,mol\) of \(H_2\). Calculate \(Q\) and then predict the direction in which the reaction will proceed. What are the new equilibrium concentrations?

The system is not at equilibrium at each of these higher pressures. To reach equilibrium, the reaction will proceed to the right to decrease the pressure because the equilibrium partial pressure is less than the total pressure.

\(K_p=\frac{[NH_{3}]^2}{[N_{2}][H_{2}]^3}=\frac{[15.20]^2}{[19.17][65.13]^3}=4.4 \times 10^{-5}\)

\(K_p=\frac{[NH_{3}]^{2}}{[N_{2}][H_{2}]^3}=\frac{[321.6]^2}{[56.74][220.8]^3}= 1.7 \times 10^{-4}\)

1. \(K=\frac{[B][C]}{[A]}=\frac{[2.50][2.50]}{[2.50]}=2.50\)

2. \(K_p=K(RT)^{Δn} \rightarrow K=\frac{K_p}{(RT)^{Δn}}=\frac{19.0}{((0.08206\frac{L\cdot atm}{mol\cdot K})(200+273.15)K)^{2-1})}=0.49\)

\(K_p=\frac{(P_B)(P_C)}{(P_A)}=\frac{(1.75)(14.15)}{1.30}=19.0\)

3. \(K=\frac{[B][C]}{[A]^{2}}=\frac{(18.72)(6.51)}{12.61}=9.7\)

Experiment 1 is about the same as the given \(K\) value and thus considered to be about equilibrium. The second experiment has a \(K\) value that is about 1 so neither the formation of the reactants or products is favored. The third experiment has a \(K\) value that is larger than 1 so the formation of the products is favored.

\(K=\frac{[H_{2}O]}{[H_{2}]}\)

\(K=\frac{[CO_{2}]}{[Co]}\)

\([H_2]=[CO]=0.316\,mol\,H_{2} \times \frac{1}{1.00\,L}=0.316\,M\)

\([CoO]=0.5\,mol\,CoO \times \frac{1}{1.00\,L}=0.5\,M\)

Reaction 1:

\(PV=nRT \rightarrow P=\frac{nRT}{V}=(4.65 \times 10^{-3})(0.08206\frac{L\cdot atm}{mol\cdot K})(823\,K)=0.314\)

\([H_{2}]=0.316-x=0.316-0.311=4.65x10^{-3}\)

\(PV=nRT \rightarrow P=\frac{nRT}{V}=(0.311)(0.08206\frac{L\cdot atm}{mol\cdot K})(823\,K)=21.0\)

\([H_{2}O]=x=0.311\)

\(K=\frac{[H_{2}O]}{[H_{2}]} \rightarrow 67=\frac{x}{0.316-x} \rightarrow x=0.311\)

Reaction 2:

\(PV=nRT \rightarrow P=\frac{nRT}{V}=(0.001)(0.08206\frac{L\cdot atm}{mol\cdot K})(823\,K)=6.75 \times 10^{-2}\,atm\)

\([CO]=0.316-x=0.316-0.315=0.001\,M\)

\(PV=nRT \rightarrow P=\frac{nRT}{V}=(0.315)(0.08206\frac{L\cdot atm}{mol\cdot K})(823\,K)=21.3\,atm\)

\([CO_{2}]=x=0.315\,M\)

\(K=\frac{[CO_{2}]}{[Co]} \rightarrow 490=\frac{x}{0.316-x} \rightarrow x=0.315\)

\(H_{2}\,(g)+CO_{2}\,(g) \rightleftharpoons CO\,(g)+H_{2}O\,(g)\)

\(K_p=\frac{(P_{CO})(P_{H_{2}O})}{(P_{H_{2}})(P_{CO_{2}})}=\frac{(6.75 \times 10^{-2})(21)}{(0.314)(21.3)}=0.21\)

d. The shape of the graphs [reactants] versus [products] does not change as the amount of \(CoO\) changes because it is a solid.

\(PV=nRT \rightarrow \frac{n}{V}=\frac{P}{RT} \rightarrow \frac{n}{V}=\frac{0.101798}{(0.08206\frac{L\cdot atm}{mol\cdot K})((425+273.15)\,K)}=1.5 \times 10^{-4}\,M\,HI\)

\([HI]=2x=2(0.050899)=0.101798\,atm\)

\(K_p=\frac{(P_{HI})^2}{(P_{H_{2}})(P_{I_{2}})} \rightarrow 54.5=\frac{(2x)^2}{(3.2 \times 10^{-1}-x)(5.16 \times 10^{-2}-x)} \rightarrow x=0.050899\,atm\)

\(PV=nRT \rightarrow P=\frac{nRT}{V}=(5.6 \times 10^{-3})(0.08206\frac{L\cdot atm}{mol\cdot K})((425+273.15)\,K)=3.2 \times 10^{-1}\,atm\)

\([H_{2}]=1.12 \times 10^{-2}\,mol\,H_{2} \times \frac{1}{2.0\,L}=5.6 \times 10^{-3}\,M\)

\(PV=nRT \rightarrow P=\frac{nRT}{V}=(9.0 \times 10^{-4})(0.08206\frac{L\cdot atm}{mol\cdot K})((425+273.15)\,K)=5.16 \times 10^{-2}\,atm\)

\([I_2]= 1.8 \times 10^{-3}\,mol\,I_{2} \times \frac{1}{2.0\,L}=9.0 \times 10^{-4}\,M\)

For excess hydrogen:

\(Q=\frac{[HI]}{[H_{2}][I_{2}]}=\frac{1.8 \times 10^{-3}}{(594.410)(1.09 \times 10^{-3})}=2.8 \times 10^{-3}\)

The reaction will proceed to the right to reach equilibrium.

\(PV=nRT \rightarrow \frac{n}{V}=\frac{P}{RT} \rightarrow \frac{0.103162}{(0.08206\frac{L\cdot atm}{mol\cdot K})((425+273.15)\,K)}=1.8 \times 10^{-3}\,M\)

\([HI]=2x=2(0.051581)=0.103162\,atm\)

\(PV=nRT \rightarrow \frac{n}{V}=\frac{P}{RT} \rightarrow \frac{10.375}{(0.08206\frac{L\cdot atm}{mol\cdot K})((425+273.15)\,K)}=594.410\,M\)

\([H_{2}]=10.427-x=10.427-0.051581=10.375\,atm\)

\(PV=nRT \rightarrow \frac{n}{V}=\frac{P}{RT} \rightarrow \frac{1.9 \times 10^{-5}}{(0.08206\frac{L\cdot atm}{mol\cdot K})((425+273.15)\,K)}=1.09 \times 10^{-3}\,M\)

\([I_{2}]=5.16 \times 10^{-2}-x=5.16 \times 10^{-2}-0.051581=1.9 \times 10^{-5}\,atm\)

\(K_p=\frac{(P_{HI})^2}{(P_{H_{2}})(P_{I_{2}})} \rightarrow 54.5=\frac{(2x)^2}{(10.427-x)(5.16 \times 10^{-2}-x)} \rightarrow x=0.051581\,atm\)

\(PV=nRT \rightarrow P=\frac{nRT}{V}=(0.182\,M)(0.08206\frac{L\cdot atm}{mol\cdot K})((425+273.15)\,K)=10.427\,atm\)

\([H_{2}]=3.64 \times 10^{-1}\,mol\,H_{2} \times \frac{1}{2.0\,L}=0.182\,M\)

15.7: Le Chatelier's Principle

COMMENTS

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  21. 6.7: Solving Equilibrium Problems

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