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Hegelian Dialectic: What It Is & How Does It Apply To You?

no problem reaction solution

Hegelian dialectic is ‘PROBLEM-REACTION-SOLUTION’ and how it works is like this:

  • The government creates or exploits a problem in which attributes blame to others.
  • The people react by asking the govt for protecton and help (safety and security) to help solve the problem.
  • Then, the government offers the solution that was planned by them long before the crisis occured.

What’s the outcome? The outcome of all of this is: the rights and liberties are exch an ged for the illusion of protection and help. 

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This was so informative, I’m shocked this isn’t taught in schools, I knew the government had a reason And a method, but never heard of this, this is astounding , and I’m to understand this is as old as the early 1800’S, with the 2nd central bank until Andrew Jackson was against it, which created the federal reserve, someone tell me if this is correct, I love American history, and I want know facts

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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

no problem reaction solution

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.

no problem reaction solution

Check Your Learning 4.3.1 – Determining Relative Changes in Concentration 

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

no problem reaction solution

          (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.

no problem reaction solution

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:

no problem reaction solution

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

no problem reaction solution

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:

no problem reaction solution

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:

no problem reaction solution

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.

no problem reaction solution

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.

no problem reaction solution

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

no problem reaction solution

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.

no problem reaction solution

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

no problem reaction solution

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

no problem reaction solution

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

no problem reaction solution

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 :

no problem reaction solution

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,

no problem reaction solution

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

no problem reaction solution

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

no problem reaction solution

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

no problem reaction solution

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

no problem reaction solution

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.

no problem reaction solution

Retain a few extra significant figures to minimize rounding problems.

no problem reaction solution

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

no problem reaction solution

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).

image

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

no problem reaction solution

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

no problem reaction solution

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

no problem reaction solution

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

no problem reaction solution

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.

no problem reaction solution

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?

no problem reaction solution

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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problem-reaction-solution

English [ edit ], noun [ edit ].

problem - reaction - solution ( uncountable )

  • A theory postulated by David Icke in which the government (or another higher power) manipulates the population by introducing a problem and then using their own means to solve that problem.

no problem reaction solution

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7 Best Replies To “No Problem” In A Text Or In Person

  • January 8, 2024

Last Updated on January 8, 2024 by Ketan

There is always someone who says ‘No problem’ habitually to whatever you say .  

It could be their ‘cool’ nature.

Which keeps them so relaxed toward everything happens. 

Mostly, they act like they’ve no issue at all. 

Well, you can’t be sure if they really mean it or just say it. 

And, that you can figure out based on their tone. 

If you get ‘no problem’ over text or in person, you need to be sure why they say so, before replying to them.

Some people say it to make you feel relaxed. While some show they don’t mind it at all. 

How To Respond To “No Problem”?

In most cases, people use ‘Okay, no problem’ ‘or ‘Not a problem’ in response when you thank them after getting their help. 

This way they actually say, ‘Don’t worry, I like to help you out.’

Whereas, some say it when you ask them to do something for you. 

Here saying ‘No worries’ means, ‘I have no issue helping you, So relax.’

So, when someone says ‘No worries’ to you, you better choose what to reply based on the context of this conversation. 

If you want to say something after ‘No problem’ here are the best replies you can share. 

How To Respond To No Problem

1. “Are you sure, Really?”

Your friend might say ‘No problem ‘ as part of their habit. 

You asked them for some favor that sounds challenging to you.  

But they still tell you to not worry about it.

With this reply, you just confirm that they are okay with it.

2. “No, I truly appreciate your help.” 

Some people just love to help others. 

And when you thank them for some reason, they say ‘No problemo’ instead of saying ‘welcome’. 

This proves that this person doesn’t want you to panic about anything. 

So, you express your thankfulness to them sincerely.  

READ NEXT:  Flirty Replies To ‘How Was Your Day?’ Text

3. “I can’t thank you enough. And, I mean it.” 

You are seriously saying that what they do for you is a great help.

As they say ‘No problem’ this makes you feel more relaxed as they’re with you. 

With this reply, you just want to acknowledge their support. 

4. “Surprising, how can you have a solution for everything?” 

No matter what you ask for them, they always fix it for you. 

It truly surprises you, how they can be so good at everything.

So, this one is a respectful and funny response to ‘No problem’ at the same time. 

Funny Response To No Problem

5. “Of course, when you’re with me I’ve nothing to worry about.” 

This person is trustworthy and you rely on them so much. 

So far they’ve proved that they can do anything for you. 

They say ‘no problem’ every time and actually mean it.

And your response to it, shows you value their presence.  

READ NEXT:  Clever Comebacks To Rude Remarks

6. “How can you be so chill, dude?”

Sometimes it just irritates you when your friend always texts with ‘No worries’ to anything you say to them. 

You find being so calm and collected is so difficult. 

But this friend has no impact or worries about anything at all.

So this is how you reply like you’re asking about their secret of being so chill. 

7. “Take it seriously, else It will be a big trouble.”

This person always acts so chill no matter what the situation is.

But now, you want them to be serious.

Again they say ‘No worries’, tell them that this is a different case.

What Does “No Problem” Mean?

When someone says ‘No problem’ or ‘No worries’ they mean ‘Relax, you’re not bothering me’ Or ‘I’m fine’. 

This phrase is more personal and relaxing when being used in place of a ‘Welcome’ response after someone apologizes or thanks for something. 

Because this way you show that ‘I’m happy to help.

Please don’t think much about it.’. Or  ‘Please don’t mind, I’ve no issue’. 

People often use ‘no worries’ habitually over text or in person. 

What Does No Problem Mean

Saying ‘No problem’ isn’t always perfect. 

Because different people say it with their unique purposes. 

Either in a relaxed tone or sarcastic. 

So you better check if the person says ‘No problem’, Are they telling you to calm down and worry about nothing? 

Or they just taunt you like ‘ Never mind , I don’t care about you much’. 

Based on the context and your connection, you can choose a funny or casual response to ‘No problem’ from someone. 

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no problem reaction solution

Specialized in marketing, with 'communication' as a favorite subject, Ketan P. is a head writer at 'Better Responses'. He loves to share his unique perspectives and ways to make everyday conversations a bit 'lively'.

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no problem reaction solution

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Totality of Evidence

Problem – Reaction – Solution

Mass-scale mind manipulation technique, problem > reaction > solution.

Once you know this, you’ll see it everywhere!

David Ike explains the mind-manipulation technique which he calls Problem > Reaction > Solution

“This “ Problem-Reaction-Solution ” technique is not just used to create and control wars, it can also be applied to any situation where you want to produce a particular outcome . …It can also be used to introduce new laws that restrict our civil liberties.

So, whenever you see some big news story – whether it be an alleged terrorist attack, a run on a currency, the accidental loss of people’s data, or an allegation that some poor country has developed nuclear weapons, [or the emergence of a novel virus] – the first question you need to ask yourself is this:

“ Who benefits from me believing this information? “ David Ike

The Hegelian Dialectic

Georg Hegel (1770-1831) a German Philosopher “came up with the basic system for achieving the goals of the Illuminati called The Hegalian Dialectic ” – WATCH

A system based on 3 basic steps:

  • Create the Problem
  • Control the Reaction
  • Offer the Solution

Dialectic : The art or practice of arriving at the truth by the exchange of logical arguments – DICTIONARY . “The dialectic refers to a movement of the mind in search for truth.”- REF

Hegelian Dialectic - Problem - Reaction -Solution

David Ike did not discover the manipulation formula, Problem-Reaction-Soultion, he merely put Georg Hegel’s founding systematic philosophy into layperson’s speak.

David Ike explains how it works today:

… it works like this : Let’s say you want to centralize power into fewer and fewer hands through the UN or NATO. If you did that openly and said, “This is what we want to do”, there’d be a reaction against that. People would say, “Hey, this is a fascist state you want to create. Were not having this.” But through this technique of Problem-Reaction-Solution , you can actually manipulate people to demand you do what you want to do anyway. So it works like this : First of all you create the problem , but you get someone else to be blamed for it. You then report that problem through the media in the way you want it reported – because the media is owned by the same people who own the banks, etc. You get the public to react to your problem by saying, “Something must be done; this can’t go on; what are THEY going to do about it?”. [FEAR] And at that point, THEY, who have covertly created the problem and blamed someone else, who gleaned that reaction of “Do Something”, then offer the solution to the problems they have created . So if you take the world wars for example – after the first world war, which the financial centers of London and Wall Street, etc, funded all sides, power on this planet was in fewer hands than ever before (via the League of Nations). After the second world war, it was even fewer hands on the wheels of power. And as a result of the second world war, we had the creation of the United Nations [think WORLD HEALTH ORGANISATION ] and NATO ; and we had this great centralization of global power . David Ike May 2009

Centralisation of Power > Dilution of Freedoms – WATCH

David Ike explains the powerful mass manipulation technique of Problem-Reaction-Solution

People don’t look beyond the “news” These are random events no longer, when you know where we are being led

“ The fundamental reason for a war is to change the nature of post-war society ” Not all wars are kinetic! Think the war on coronavirus – WATCH

Summary of the Mind-manipulation technique called Problem-Reaction-Solution (date unknown) – EXCERPT

David Ike – YouTube Channel ARCHIVE for you to explore Explore to find more on David Ike – BITCHUTE

David Ike & Others has been trying to “Red Pill” the world for years!

Interesting links for you to explore.

January 10, 2024 – The Aussie Wire: Big Pharma’s Dirty Secret: Rebekah Barnett Shines a Light on Covert Pharma Advertising! – WATCH , How Big Pharma harnesses our tax money and news media to market their drugs – READ

  • 1. Pharma pays for market research
  • 2. Pharma crafts results into press release urging take up of the target product, forwards to news syndicator, not their own site
  • 3. News syndicator lightly edits press release, publishes as news
  • 4. Local news outlets republish syndicated article as local news
  • Voila! A plethora of advertorials disguised as ‘local news’ for the small price of conducting a public poll (as an example).

August 26, 2023 – Phillip Altman Substack: HERE WE GO AGAIN !- The global control playbook works – READ

Problem Reaction Solution Playbook 2023

January 22, 2022 – Science ABC: What Is Dialectics? What Is The Triad Thesis? “Dialectics as a process is essential for all human beings to understand because in actuality, most of us think or carry out arguments through this process. We make use of dialectics all the time, but it happens subconsciously .” – READ

  • “While it’s true that Hegel wrote about dialectics as a process, he never actually mentioned the words “thesis, antithesis and synthesis” in his writings… Hegel instead made use of terms like “concrete, absolute and universal”.”

The Triad Thesis:

  • Thesis : the moment of understanding (Problem)
  • Antithesis : Dialectical moment (Opposition, the negative, Reaction)
  • Synthesis : Speculative moment (Rational, Solution)
  • “The synthesis is a statement that arises from the thesis and the antithesis, and continues the discussion.”

May 2021 – Oracle Films: DAVID ICKE Interview – WATCH

August 20, 2020 – COMPILATION – Problem Reaction Solution (1hr) – WATCH

March 11, 2020 – Coronavirus – Problem Reaction Solution – David Icke – WATCH , ARCHIVE

August 2016 – Agenda 21, The Plan To Kill You – David Icke – ARCHIVE

June 2016 – The REAL EU agenda : The David Icke Videocast/Podcast Trailer – ARCHIVE

May 25, 2016 – The Christian Observer: What is The HEGELIAN DIALECTIC? And why is it important to understand it? – READ

September 18, 2014 – David Icke Awesome NEW Interview DNA, Royals and Iluminati History (1h39m) – ARCHIVE

May 29, 2014 – The David Icke Videocast: Mind Controlled Killers In Service To The State – ARCHIVE

May 23, 2014 – Live at Oxford Debating Society: Mind Control and the New World Order – David Icke (2hrs) – ARCHIVE

May 23, 2014 – David Icke The Reptilians the Schism Obama and the New world Order (2hrs) – ARCHIVE

March 29, 2014 – David Icke’s Best Presentation Ever – The Elephant in the Living Room – ARCHIVE “Most conspiracy researchers won’t touch this subject” says Icke. While David has not been shy telling the world about reptilians, he waited 20 years to finally share this info!

August 22, 2012 – David Icke Problem, Reaction, Solution – explaining the take over of the middle east – WATCH

2010 – Problem – Reaction – Solution – David Icke Talking In 2010 – WATCH , SOURCE

A description of mass manipulation techniques using the media based on the Hegelian dialectic as explained through David Icke. (3min summary) – WATCH

January 30, 2009 – DAVID ICKE – Manipulation Techniques – Problem Reaction Solution – WATCH , ARCHIVE

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14.E: Oxidation-Reduction Reaction (Exercises)

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14.2: Oxidation-Reduction Reactions

  • Is this reaction a redox reaction? Explain your answer. 2K(s) + Br 2 (ℓ) → 2KBr(s)
  • Is this reaction a redox reaction? Explain your answer. 2NaCl(aq) + Pb(NO 3 ) 2 (aq) → 2NaNO 3 (aq) + PbCl 2 (s)
  • Which substance loses electrons and which substance gains electrons in this reaction? 2Mg(s) + O 2 (g) → 2MgO
  • Which substance loses electrons and which substance gains electrons in this reaction? 16Fe(s) + 3S 8 (s) → 8Fe 2 S 3 (s)
  • Which substance is oxidized and which substance is reduced in this reaction? 2Li(s) + O 2 (g) → Li 2 O 2 (s)
  • Which substance is oxidized and which substance is reduced in this reaction? 2Fe(s) + 3I 2 (s) → 2FeI 3 (s)
  • What are two different definitions of oxidation?
  • What are two different definitions of reduction?
  • SO 3 2 −
  • Ca 3 (PO 3 ) 2
  • (NH 4 ) 2 Se
  • NO 2 −
  • Zn(C 2 H 3 O 2 ) 2
  • Identify what is being oxidized and reduced in this redox reaction by assigning oxidation numbers to the atoms. 2NO + Cl 2 → 2NOCl
  • Identify what is being oxidized and reduced in this redox reaction by assigning oxidation numbers to the atoms. Sr + SO 3 → SrSO 3
  • Identify what is being oxidized and reduced in this redox reaction by assigning oxidation numbers to the atoms. 2KrF 2 + 2H 2 O → 2Kr + 4HF + O 2
  • Identify what is being oxidized and reduced in this redox reaction by assigning oxidation numbers to the atoms. SO 3 + SCl 2 → SOCl 2 + SO 2
  • Identify what is being oxidized and reduced in this redox reaction by assigning oxidation numbers to the atoms. 2Rb + MgCl 2 → 2RbCl + Mg
  • Identify what is being oxidized and reduced in this redox reaction by assigning oxidation numbers to the atoms. 2C 8 H 18 + 25O 2 → 16CO 2 + 18H 2 O
  • yes because oxidation numbers are changing

lose: Mg; gain: O

oxidized: Li; reduced: O

  • increase in oxidation number; loss of electrons
  • S: +6; O: −2
  • S: +4; O: −2
  • Ca: +2; P: +3; O: −2
  • N: +2; O: −2
  • N: +4; O: −2
  • Cr: +2; Cl: −1
  • Cr: +3; Cl: −1
  • C: 0; H: +1; O: −2
  • N: −3; H: +1
  • Rb: +1; S: +6; O: −2
  • Zn: +2; C: 0; H: +1; O: −2
  • oxidized: N; reduced: Cl
  • oxidized: O; reduced: Kr
  • oxidized: Rb; reduced: Mg

14.3: Balancing Redox Reactions

  • Na + F 2 → NaF
  • Al 2 O 3 + H 2 → Al + H 2 O
  • Fe 2 S 3 + O 2 → Fe 2 O 3 + S
  • Cu 2 O + H 2 → Cu + H 2 O
  • CH 4 + O 2 → CO 2 + H 2 O
  • P 2 O 5 + Cl 2 → PCl 3 + O 2
  • PbCl 2 + FeCl 3 → PbCl 4 + FeCl 2
  • SO 2 + F 2 → SF 4 + OF 2
  • Ca + H + → Ca 2 + + H 2
  • Sn 2 + → Sn + Sn 4 + (Hint: both half reactions will start with the same reactant.)
  • Fe 3 + + Sn 2 + → Fe + Sn 4 +
  • Pb 2 + → Pb + Pb 4 + (Hint: both half reactions will start with the same reactant.)
  • Na + Hg 2 Cl 2 → NaCl + Hg
  • Al 2 O 3 + C → Al + CO 2
  • Br − + I 2 → I − + Br 2
  • CrCl 3 + F 2 → CrF 3 + Cl 2
  • Cu + NO 3 − → Cu 2 + + NO 2
  • Fe + MnO 4 − → Fe 3 + + Mn
  • CrO 3 + Ni 2 + → Cr 3 + + Ni 3 +
  • OsO 4 + C 2 H 4 → Os + CO 2
  • ClO − + Ti 2 + → Ti 4 + + Cl −
  • BrO 3 − + Ag → Ag + + BrO 2
  • H 2 O 2 + NO → N 2 O 3 + H 2 O
  • VO 2 + + NO → V 3+ + NO 2
  • Explain why this chemical equation is not balanced and balance it if it can be balanced: Cr 2 + + Cl 2 → Cr 3 + + 2Cl −
  • Explain why this equation is not balanced and balance it if it can be balanced: O 2 + 2H 2 O + Br 2 → 4OH − + 2Br −
  • 2Na + F 2 → 2NaF
  • Al 2 O 3 + 3H 2 → 2Al + 3H 2 O

CH 4 + 2O 2 → CO 2 + 2H 2 O

  • 2P 2 O 5 + 6Cl 2 → 4PCl 3 + 5O 2
  • Ca + 2H + → Ca 2 + + H 2
  • 2Sn 2 + → Sn + Sn 4 +
  • 2Na + Hg 2 Cl 2 → 2NaCl + 2Hg
  • 2Al 2 O 3 + 3C → 4Al + 3CO 2
  • 4H + + Cu + 2NO 3 − → Cu 2 + + 2NO 2 + 2H 2 O in acidic solution; 2H 2 O + Cu + 2NO 3 − → Cu 2 + + 2NO 2 + 4OH − in basic solution
  • 24H + + 3MnO 4 − + 7Fe → 7Fe 3 + + 3Mn + 12H 2 O in acidic solution; 12H 2 O + 3MnO 4 − + 7Fe → 7Fe 3 + + 3Mn + 24OH − in basic solution
  • 2H + + ClO − + Ti 2 + → Cl − + H 2 O + Ti 4 + in acidic solution; H 2 O + ClO − + Ti 2 + → Cl − + Ti 4 + + 2OH − in basic solution
  • 2H + + BrO 3 − + Ag → BrO 2 + H 2 O + Ag + in acidic solution; H 2 O + BrO 3 − + Ag → BrO 2 + Ag + + 2OH − in basic solution
  • The charges are not properly balanced. The correct balanced equation is 2Cr 2 + + Cl 2 → 2Cr 3 + + 2Cl − .

14.4: Applications of Redox Reactions - Voltaic Cells

  • Draw the voltaic cell represented by this reaction and label the cathode, the anode, the salt bridge, the oxidation half cell, the reduction half cell, the positive electrode, and the negative electrode. Use Fig. 14.4.1 as a guide. Zn + 2Ag + → Zn 2 + + 2Ag
  • Draw the voltaic cell represented by this reaction and label the cathode, the anode, the salt bridge, the oxidation half cell, the reduction half cell, the positive electrode, and the negative electrode. Use Fig. 14.4.1 as a guide. 3Mg + 2Cr 3 + → 3Mg 2 + + 2Cr
  • What is the voltage of this half reaction? 2F − → F 2 + 2e −
  • What is the voltage of this half reaction? Na → Na + + e −
  • What is the voltage of the voltaic cell in Exercise 1? Consult Table 14.4.1.
  • What is the voltage of the voltaic cell in Exercise 2? Consult Table 14.4.1.
  • Balance this redox reaction and determine its voltage. Is it spontaneous? Li + + Al → Li + Al 3 +
  • Balance this redox reaction and determine its voltage. Is it spontaneous? Pb 2 + + Ni → Pb + Ni 2 +
  • Balance this redox reaction and determine its voltage. Is it spontaneous? Cu 2 + + Ag + Cl − → Cu + AgCl
  • Balance this redox reaction and determine its voltage. Is it spontaneous? Mn 2 + + Br 2 → MnO 4 − + Br −
  • Which reaction represents the cathode reaction in Exercise 7? The anode reaction?
  • Which reaction represents the cathode reaction in Exercise 8? The anode reaction?
  • Which reaction represents the cathode reaction in Exercise 9? The anode reaction?
  • Which reaction represents the cathode reaction in Exercise 10? The anode reaction?
  • A voltaic cell is based on this reaction: Ni + 2Au + → Ni 2 + + 2Au; If the voltage of the cell is 0.33 V, what is the standard reduction potential of the Au + + e − → Au half reaction?
  • A voltaic cell is based on this reaction: 3Pb + 2V 3 + → 3Pb 2 + + 2V; If the voltage of the cell is −0.72 V, what is the standard reduction potential of the V 3+ + 3e − → V half reaction?
  • What species is being oxidized and what species is being reduced in a dry cell?
  • What species is being oxidized and what species is being reduced in an alkaline battery?
  • What species is being oxidized and what species is being reduced in a silver oxide button battery?
  • What species is being oxidized and what species is being reduced in a lead storage battery?
  • Based on the data in Table 14.4.1, what is the highest voltage battery you can construct?
  • Based on the data in Table 14.4.1, what is the lowest voltage battery you can construct? (This may be more challenging to answer than Exercise 21.)

The anode/oxidation half cell/negative electrode has a rod with Zn & cathode/reduction half cell/positive electrode has a rod with Ag.

  • −2.87 V

3Li + + Al → 3Li + Al 3 + ; −1.39 V; not spontaneous

Cu 2 + + 2Ag + 2Cl − → Cu + 2AgCl; 0.12 V; spontaneous

cathode reaction: Li + + e − → Li; anode reaction: Al → Al 3 + + 3e −

cathode reaction: Cu 2 + + 2e − → Cu; anode reaction: Ag + Cl − → AgCl + e −

oxidized: Zn; reduced: Mn

oxidized: Zn; reduced: Ag

5.92 V from the reaction of F 2 and Li

14.5: Electrolysis

  • Define electrolytic cell .
  • How does the operation of an electrolytic cell differ from a voltaic cell?
  • List at least three elements that are produced by electrolysis.
  • Write the half reactions for the electrolysis of the elements listed in Exercise 3.
  • Based on Table 14.4.1, what voltage must be applied to an electrolytic cell to electroplate copper from Cu 2 + ?
  • Based on Table 14.4.1, what voltage must be applied to an electrolytic cell to electroplate aluminum from Al 3 + ?
  • an electrochemical cell in which charge is forced through and a nonspontaneous reaction occurs

any three of the following: Al, K, Li, Na, Cl 2 , or Mg

Additional Exercises

  • Oxidation was once defined as chemically adding oxygen to a substance. Use this reaction to argue that this definition is consistent with the modern definition of oxidation: 2Mg + O 2 → 2MgO
  • Reduction was once defined as chemically adding hydrogen to a substance. Use this reaction to argue that this definition is consistent with the modern definition of reduction: C 2 H 2 + 2H 2 → C 2 H 6
  • Kr (krypton)
  • krypton tetrafluoride (KrF 4 )
  • dioxygen difluoride (O 2 F 2 )
  • lithium hydride (LiH)
  • potassium peroxide (K 2 O 2 )
  • potassium fluoride (KF)
  • Na 2 Cr 2 O 7
  • Balance this redox reaction by inspection: S 8 + O 2 → SO 2
  • Balance this redox reaction by inspection: C 18 H 38 + O 2 → CO 2 + H 2 O
  • Balance this redox reaction by the half reaction method by assuming an acidic solution: Cr 2 O 7 2 − + Fe → Cr 3 + + Fe 3 +
  • Balance the redox reaction in Exercise 9 by the half reaction method by assuming a basic solution.
  • The uranyl ion (UO 2 2 + ) is a fairly stable ion of uranium that requires strong reducers to reduce the oxidation number of uranium further. Balance this redox reaction using the half reaction method by assuming an acidic solution. UO 2 2+ + HN 3 → U + N 2
  • Balance the redox reaction in Exercise 11 by the half reaction method by assuming a basic solution.
  • Zinc metal can be dissolved by acid, which contains H + ions. Demonstrate that this is consistent with the fact that this reaction has a spontaneous voltage: Zn + 2H + → Zn 2 + + H 2
  • Copper metal cannot be dissolved by acid, which contains H + ions. Demonstrate that this is consistent with the fact that this reaction has a nonspontaneous voltage: Cu + 2H + → Cu 2 + + H 2
  • A disproportionation reaction occurs when a single reactant is both oxidized and reduced. Balance and determine the voltage of this disproportionation reaction. Use the data in Table 14.4.1 - Standard Reduction Potentials of Half Reactions : Cr 2 + → Cr + Cr 3 +
  • A disproportionation reaction occurs when a single reactant is both oxidized and reduced. Balance and determine the voltage of this disproportionation reaction. Use the data in Table 14.4.1 - Standard Reduction Potentials of Half Reactions : Fe 2 + → Fe + Fe 3 +
  • What would be overall reaction for a fuel cell that uses CH 4 as the fuel?
  • What would be overall reaction for a fuel cell that uses gasoline (general formula C 8 H 18 ) as the fuel?
  • When NaCl undergoes electrolysis, sodium appears at the cathode. Is the definition of cathode the same for an electrolytic cell as it is for a voltaic cell?
  • When NaCl undergoes electrolysis, chlorine appears at the anode. Is the definition of anode the same for an electrolytic cell as it is for a voltaic cell?
  • An award is being plated with pure gold before it is presented to a recipient. If the area of the award is 55.0 cm 2 and will be plated with 3.00 µm of Au, what mass of Au will be plated on the award? The density of Au is 19.3 g/cm 3 .

As oxygen is added to magnesium, it is being oxidized. In modern terms, the Mg atoms are losing electrons and being oxidized, while the electrons are going to the O atoms.

  • Kr: +4; F: −1
  • O: +1; F: −1

14H + + Cr 2 O 7 2 − + 2Fe → 2Cr 3 + + 7H 2 O + 2Fe 3 +

6HN 3 + UO 2 2 + → U + 2H 2 O + 9N 2 + 2H +

The voltage of the reaction is +0.76 V, which implies a spontaneous reaction.

3Cr 2 + → Cr + 2Cr 3 + ; −0.50 V

yes because reduction occurs at the cathode

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    Question 4. Fluorine reacts with ice and results in the change: H20 (S) + F2 (g) ——-> HF (g) + HOF (g) Justify that this reaction is a redox reaction. Answer: Writing the O.N. of each atom above its symbol, we have, Here, the O.N. of F decreases from 0 in F 2 to -1 in HF and increases from 0 in F 2 to +1 in HOF.