Chapter 15

Chemical Equilibrium

15.9

a.
b.	K = 7.2 ´ 10^2

15.10

[SO₂] = 0.0124 M

[O₂] = 0.031 M

[SO₃] = ?

Solving for [SO₃] gives

[SO₃]² = K [SO₂]²[O₂] = (2.8 ´ 10²)( 0.0124)²(0.031) = 1.33 ´ 10^-3

[SO₃] = = 0.037 M

15.11

The problem states that the system is at equilibrium, so we simply substitute the equilibrium concentrations into the equilibrium expression to calculate K_c.

Step 1: Calculate the concentrations of the components in units of mol/L. The molarities can be calculated by simply dividing the number of moles by the volume of the flask.

Step 2: Once the molarities are known, K_c can be found by substituting the molarities into the equilibrium expression.

Think About It:

If you forget to convert moles to moles/liter, will you get a different answer? Under what circumstances will the two answers be the same?

15.12

Using the mole amounts and volume given in the problem, we calculate the concentrations of H₂ and CH₃OH.

[H₂] = 0.0021 M

[CH₃OH] = 0.00062 M

The equilibrium expression is

Solving for [CO] gives

[CO] = = 8.8 ´ 10³ M

15.21

a.

With Kc = 10, products are favored at equilibrium. Because the coefficients for both A and B are one, we expect the concentration of B to be 10 times that of A at equilibrium. Diagram (a) is the best choice with 10 B molecules and 1 A molecule.

b.

With Kc = 0.10, reactants are favored at equilibrium. Because the coefficients for both A and B are one, we expect the concentration of A to be 10 times that of B at equilibrium. Diagram (d) is the best choice with 10 A molecules and 1 B molecule. You can calculate Kc in each case without knowing the volume of the container because the mole ratio between A and B is the same. Volume will cancel from the Kc expression. Only moles of each component are needed to calculate Kc.

15.22

Note that we are comparing similar reactions at equilibrium – two reactants producing one product, all with coefficients of one in the balanced equation.

a.	The reaction, A + C AC has the largest equilibrium constant. Of the three diagrams, there is the most product present at equilibrium.
b.	The reaction, A + D AD has the smallest equilibrium constant. Of the three diagrams, there is the least amount of product present at equilibrium.

15.23

When the equation for a reversible reaction is written in the opposite direction, the equilibrium constant becomes the reciprocal of the original equilibrium constant.

15.24

Using Equation 15.4 of the text: K_P = K_c(0.0821 T)^Dn

where, Dn = 2 - 3 = -1

and T = (1273 + 273) K = 1546 K

K_P = (2.24 ´ 10²²)(0.0821 ´ 1546)^-1 = 1.76 ´ 10²⁰

15.25

Strategy:	The relationship between Kc and KP is given by Equation 15.4 of the text. What is the change in the number of moles of gases from reactant to product? Recall that Dn = moles of gaseous products - moles of gaseous reactants What unit of temperature should we use?
Solution:	The relationship between Kc and KP is given by Equation 15.4 of the text. KP = Kc(0.0821 T)^Dn Rearrange the equation relating KP and Kc, solving for Kc. Because T = 623 K and Dn = 3 - 2 = 1, we have:

15.26

We can write the equilibrium expression from the balanced equation and substitute in the pressures.

Think About It:

Do we need to know the temperature?

15.27

The equilibrium expressions are:

a.

Substituting the given equilibrium concentration gives:

or 8.2 ´ 10^-2

b.

Substituting the given equilibrium concentration gives:

Think About It:

Is there a relationship between the Kc values from parts (a) and (b)?

15.28

The equilibrium expression for the two forms of the equation are:

The relationship between the two equilibrium constants is

K_P can be found as shown below.

K_P = K_c'(0.0821 T)^Dn = (2.6 ´ 10⁴)(0.0821 ´ 1000)^-1 = 3.2 ´ 10²

15.29

Because pure solids do not enter into an equilibrium expression, we can calculate K_P directly from the pressure that is due solely to CO₂(g).

Now, we can convert K_P to K_c using the following equation.

K_P = K_c(0.0821 T)^Dn

15.30

We substitute the given pressures into the reaction quotient expression.

The calculated value of Q_P is less than K_P for this system. The system will change in a way to increase Q_P until it is equal to K_P. To achieve this, the pressures of PCl₃ and Cl₂ must increase, and the pressure of PCl₅ must decrease.

Think About It:

Could you actually determine the final pressure of each gas?

15.31

Strategy:

Because they are constant quantities, the concentrations of solids and liquids do not appear in the equilibrium expressions for heterogeneous systems. The total pressure at equilibrium that is given is due to both NH3 and CO2. Note that for every 1 atm of CO2 produced, 2 atm of NH3 will be produced due to the stoichiometry of the balanced equation. Using this ratio, we can calculate the partial pressures of NH3 and CO2 at equilibrium.

Solution:

The equilibrium expression for the reaction is The total pressure in the flask (0.363 atm) is a sum of the partial pressures of NH3 and CO2. Let the partial pressure of CO2 = x. From the stoichiometry of the balanced equation, you should find that Therefore, the partial pressure of NH3 = 2x. Substituting into the equation for total pressure gives: 3x = 0.363 atm Substitute the equilibrium pressures into the equilibrium expression to solve for KP.

15.32

If the CO pressure at equilibrium is 0.497 atm, the balanced equation requires the chlorine pressure to have the same value. The initial pressure of phosgene gas can be found from the ideal gas equation.

Since there is a 1:1 mole ratio between phosgene and CO, the partial pressure of CO formed (0.497 atm) equals the partial pressure of phosgene reacted. The phosgene pressure at equilibrium is:

CO(g) + Cl₂(g) COCl₂(g)

Initial (atm): 0 0 1.31

Change (atm): +0.497 +0.497 -0.497

Equilibrium (atm): 0.497 0.497 0.81

The value of K_P is then found by substitution.

15.33

Let x be the initial pressure of NOBr. Using the balanced equation, we can write expressions for the partial pressures at equilibrium.

P_NOBr = (1 - 0.34)x = 0.66x

P_NO = 0.34x

The sum of these is the total pressure.

0.66x + 0.34x + 0.17x = 1.17x = 0.25 atm

x = 0.214 atm

The equilibrium pressures are then

P_NOBr = 0.66(0.214) = 0.141 atm

P_NO = 0.34(0.214) = 0.073 atm

P= 0.17(0.214) = 0.036 atm

We find K_P by substitution.

K_P =9.6 ´ 10^-3

The relationship between K_P and K_c is given by

K_P = K_c(RT)^Dn

We find K_c (for this system Dn = +1)

K_c =3.9 ´ 10^-4

15.34

The target equation is the sum of the first two.

H₂S ⇄ H⁺ + HS

HS ⇄ H⁺ + S^2-

H₂S ⇄ 2H⁺ + S^2-

Since this is the case, the equilibrium constant for the combined reaction is the product of the constants for the component reactions (Section 15.3 of the text). The equilibrium constant is therefore:

K_c = K_c'K_c" = 9.5 ´ 10^-27

Think About It:

What happens in the special case when the two component reactions are the same? Can you generalize this relationship to adding more than two reactions? What happens if one takes the difference between two reactions?

15.35

K = (6.5 ´ 10^-2)(6.1 ´ 10^-5)

K = 4.0 ´ 10^-6

15.36

Given:

For the overall reaction:

15.37

To obtain 2SO₂ as a reactant in the final equation, we must reverse the first equation and multiply by two. For the equilibrium, 2SO₂(g) 2S(s) + 2O₂(g)

Now we can add the above equation to the second equation to obtain the final equation. Since we add the two equations, the equilibrium constant is the product of the equilibrium constants for the two reactions.

2SO₂(g) 2S(s) + 2O₂(g)

2S(s) + 3O₂(g) 2SO₃(g)

2SO₂(g) + O₂(g) 2SO₃(g)

15.39

Strategy:	Use the law of mass action to write the equilibrium expression. The equilibrium expression must yield Kc = 2 for the reaction to be at equilibrium.
Setup:	The equilibrium expression has the form of concentrations of products over concentrations of reactants, each raised to the appropriate power. Plug in the concentrations of all three species to evaluate Kc.
Solution:	Diagram (a): Diagram (b): Diagram (c): Diagram (d): Since Kc = 2, diagrams (a), (b), and (c) represent and equilibrium mixture of A2, B2, and AB.

15.40

Strategy:	Use the law of mass action to write the equilibrium expression. The mixtures at equilibrium will have the same Kc value.
Setup:	The equilibrium expression has the form of concentrations of products over concentrations of reactants, each raised to the appropriate power. Plug in the concentrations of all three species to evaluate Kc.
Solution:	Diagram (a): Diagram (b): Diagram (c): Diagram (d): For diagrams (a), (c), and (d), Kc = 3. Therefore, diagram (b) is not at equilibrium.

15.41

Strategy:

The equilibrium constant Kc is given, and we start with a mixture of CO and H2O. From the stoichiometry of the reaction, we can determine the concentration of H2 at equilibrium. Knowing the partial pressure of H2 and the volume of the container, we can calculate the number of moles of H2.

Solution:

The balanced equation shows that one mole of water will combine with one mole of carbon monoxide to form one mole of hydrogen and one mole of carbon dioxide. (This is the reverse reaction.) Let x be the depletion in the concentration of either CO or H2O at equilibrium (why can x serve to represent either quantity?). The equilibrium concentration of hydrogen must then also be equal to x. The changes are summarized as shown in the table. H2 + CO2 H2O + CO Initial (M): 0 0 0.0300 0.0300 Change (M): +x +x -x -x Equilibrium (M): x x (0.0300 - x) (0.0300 - x) The equilibrium constant is: Substituting, Taking the square root of both sides, we obtain: x = 0.0173 M The number of moles of H2 formed is: 0.0173 mol/L ´ 10.0 L = 0.173 mol H2

15.42

Since the reaction started with only pure NO₂, the equilibrium concentration of NO must be twice the equilibrium concentration of O₂, due to the 2:1 mole ratio of the balanced equation. Therefore, the equilibrium partial pressure of NO is (2 ´ 0.25 atm) = 0.50 atm.

We can find the equilibrium NO₂ pressure by rearranging the equilibrium expression, then substituting in the known values.

15.43

Notice that the balanced equation requires that for every two moles of HBr consumed, one mole of H₂ and one mole of Br₂ must be formed. Let 2x be the depletion in the concentration of HBr at equilibrium. The equilibrium concentrations of H₂ and Br₂ must therefore each be x. The changes are shown in the table.

H₂ + Br₂ 2HBr

Initial (M): 0 0 0.267

Change (M): +x +x -2x

Equilibrium (M): x x (0.267 - 2x)

The equilibrium constant relationship is given by:

Substitution of the equilibrium concentration expressions gives

Taking the square root of both sides we obtain:

x = 1.80 ´ 10^-4

The equilibrium concentrations are:

[H₂] = [Br₂] = 1.80 ´ 10^-4 M

[HBr] = 0.267 - 2(1.80 ´ 10^-4) = 0.267 M

Think About It:

If the depletion in the concentration of HBr at equilibrium were defined as x, rather than 2x, what would be the appropriate expressions for the equilibrium concentrations of H2 and Br2? Should the final answers be different in this case?

15.44

We follow the procedure outlined in Section 15.4 of the text to calculate the equilibrium concentrations.

Step 1: The initial concentration of I₂ is 0.0456 mol/2.30 L = 0.0198 M. The stoichiometry of the problem shows 1 mole of I₂ dissociating to 2 moles of I atoms. Let x be the amount (in mol/L) of I₂ dissociated. It follows that the equilibrium concentration of I atoms must be 2x. We summarize the changes in concentrations as follows:

I₂(g) 2I(g)

Initial (M): 0.0198 0.000

Change (M): -x +2x

Equilibrium (M): (0.0198 - x) 2x

Step 2: Write the equilibrium expression in terms of the equilibrium concentrations. Knowing the value of the equilibrium constant, solve for x.

4x² + (3.80 ´ 10^-5)x - (7.52 ´ 10^-7) = 0

The above equation is a quadratic equation of the form ax² + bx + c = 0. The solution for a quadratic equation is

Here, we have a = 4, b = 3.80 ´ 10^-5, and c = -7.52 ´ 10^-7. Substituting into the above equation,

x = 4.29 ´ 10^-4 M or x = -4.39 ´ 10^-4 M

The second solution is physically impossible because you cannot have a negative concentration. The first solution is the correct answer.

Step 3: Having solved for x, calculate the equilibrium concentrations of all species.

[I] = 2x = (2)(4.29 ´ 10^-4 M) = 8.58 ´ 10^-4 M

[I₂] = (0.0198 - x) = [0.0198 - (4.29 ´ 10^-4)] M = 0.0194 M

Think About It:

We could have simplified this problem by assuming that x was small compared to 0.0198. We could then assume that 0.0198 - x » 0.0198. By making this assumption, we could have avoided solving a quadratic equation.

15.45

Since equilibrium pressures are desired, we calculate K_P.

K_P = K_c(0.0821 T)^Dn = (4.63 ´ 10^-3)(0.0821 ´ 800)¹ = 0.304

COCl₂(g) CO(g) + Cl₂(g)

Initial (atm): 0.760 0.000 0.000

Change (atm): -x +x +x

Equilibrium (atm): (0.760 - x) x x

x² + 0.304x - 0.231 = 0

x = 0.352 atm

At equilibrium:

P_CO =

15.46

a.

The equilibrium constant, Kc, can be found by simple substitution.

b.

The magnitude of the reaction quotient Qc for the system after the concentration of CO2 becomes 0.50 mol/L, but before equilibrium is reestablished, is: The value of Qc is smaller than Kc; therefore, the system will shift to the right, increasing the concentrations of CO and H2O and decreasing the concentrations of CO2 and H2. Let x be the depletion in the concentration of CO2 at equilibrium. The stoichiometry of the balanced equation then requires that the decrease in the concentration of H2 must also be x, and that the concentration increases of CO and H2O be equal to x as well. The changes in the original concentrations are shown in the table. CO2 + H2 CO + H2O Initial (M): 0.50 0.045 0.050 0.040 Change (M): -x -x +x +x Equilibrium (M): (0.50 - x) (0.045 - x) (0.050 + x) (0.040 + x) The equilibrium expression is: 0.52(x^2 - 0.545x + 0.0225) = x^2 + 0.090x + 0.0020 0.48x^2 + 0.373x - (9.7 ´ 10^-3) = 0 The positive root of the equation is x = 0.025. The equilibrium concentrations are: [CO2] = (0.50 - 0.025) M = 0.48 M [H2] = (0.045 - 0.025) M = 0.020 M [CO] = (0.050 + 0.025) M = 0.075 M [H2O] = (0.040 + 0.025) M = 0.065 M

15.47

The equilibrium expression for the system is:

The total pressure can be expressed as:

If we let the partial pressure of CO be x, then the partial pressure of CO₂ is:

Substitution gives the equation:

This can be rearranged to the quadratic:

x² + 1.52x - 6.84 = 0

The solutions are x = 1.96 and x = -3.48; only the positive result has physical significance (why?). The equilibrium pressures are

P_CO = x = 1.96 atm

15.48

The initial concentrations are [H₂] = 0.80 mol/5.0 L = 0.16 M and [CO₂] = 0.80 mol/5.0 L = 0.16 M.

H₂(g) + CO₂(g) H₂O(g) + CO(g)

Initial (M): 0.16 0.16 0.00 0.00

Change (M): -x -x +x +x

Equilibrium (M): 0.16 - x 0.16 - x x x

Taking the square root of both sides, we obtain:

x = 0.11 M

The equilibrium concentrations are:

[H₂] = [CO₂] = (0.16 - 0.11) M = 0.05 M

[H₂O] = [CO] = 0.11 M

15.49

Strategy:	We are given the concentrations of all species in a biochemical reaction and must determine whether the reaction will proceed in the forward direction or the reverse direction in order to establish equilibrium. This requires calculating the value of Q and comparing it to the value of K, which is given in the problem.
Solution:	The reaction quotient is Q = Because K is 1.11, Q is greater than K. Thus, the reaction will proceed to the left (the reverse reaction will occur) in order to establish equilibrium. The forward reaction will not occur.

15.54

Strategy:	According to Le Châtelier's principle, increasing the temperature of an equilibrium mixture causes a shift toward the product side (to the right) for an endothermic reaction, and a shift toward the reactant side (to the left) for an exothermic reaction.
Setup:	Reactions where ∆H is negative are exothermic.
Solution:	Equilibria (a), (d), and (e) are exothermic and will shift to the left when the temperature is increased.

15.55

Strategy:	According to Le Châtelier's principle, increasing the volume causes a shift toward the side with the larger number of moles of gas. Decreasing the volume of an equilibrium mixture causes a shift toward the side of the equation with the smaller number of moles of gas. For an equilibrium with equal numbers of gaseous moles on both sides, a change in volume does not cause the equilibrium to shift in either direction.
Setup:	Determine the numbers of moles of gas in both the reactants and products for each reaction.
Solution:	There are an equal number of moles of gas in the reactants and products for reaction (b). A change in volume will not affect the position of its equilibrium.

15.56

Strategy:	According to Le Châtelier's principle, adding a reactant to an equilibrium mixture shifts the equilibrium toward the product side (to the right) of the equation.
Setup:	Only equilibria containing H2 as a reactant will shift to the right when H2 is added.
Solution:	Equilibria (a) and (c) contain H2 as a reactant and will shift to the right when more H2 is added.

15.57

Strategy:

Use Le Châtelier's principle to determine which of the following scenarios will cause the reaction equilibrium to shift to the right.

Solution:

a.	Decreasing the volume of an equilibrium mixture causes a shift toward the side of the equation with the smaller number of moles of gas. The equilibrium will shift to the left.
b.	Increasing the volume of an equilibrium mixture causes a shift toward the side with the larger number of moles of gas. The equilibrium will shift to the right.
c.	Altering the amount of a solid does not change the position of the equilibrium because doing so does not change the value of Q.
d.	The addition of more reactant will shift the equilibrium to the right.
e.	The removal of a product will shift the equilibrium to the right.

(b), (d), and (e) will cause the equilibrium to shift to the right.

15.58

a.	Addition of more Cl2(g) (a reactant): The equilibrium would shift to the right.
b.	Removal of SO2Cl2(g) (a product): The equilibrium would shift to the right.
c.	Removal of SO2(g) (a reactant): The equilibrium would shift to the left.

15.59

a.	Removal of CO2(g) from the system: The equilibrium would shift to the right.
b.	Addition of more solid Na2CO3: The equilibrium would be unaffected. [Na2CO3] does not appear in the equilibrium expression.
c.	Removal of some of the solid NaHCO3: The equilibrium would be unaffected. Same reason as (b).

15.60

a.	This reaction is endothermic. (Why?) According to Section 15.5, an increase in temperature favors an endothermic reaction, so the equilibrium constant should become larger.
b.	This reaction is exothermic. Such reactions are favored by decreases in temperature. The magnitude of Kc should decrease.
c.	In this system heat is neither absorbed nor released. A change in temperature should have no effect on the magnitude of the equilibrium constant.

15.61

Strategy:

A change in pressure can affect only the volume of a gas, but not that of a solid or liquid because solids and liquids are much less compressible. The stress applied is an increase in pressure. According to Le Châtelier's principle, the system will adjust to partially offset this stress. In other words, the system will adjust to decrease the pressure. This can be achieved by shifting to the side of the equation that has fewer moles of gas. Recall that pressure is directly proportional to moles of gas: PV = nRT so P µ n.

Solution:

a.	Changes in pressure ordinarily do not affect the concentrations of reacting species in condensed phases because liquids and solids are virtually incompressible. Pressure change should have no effect on this system.
b.	No effect. Same situation as (a).
c.	Only the product is in the gas phase. An increase in pressure should favor the reaction that decreases the total number of moles of gas. The equilibrium should shift to the left, that is, the amount of B should decrease and that of A should increase
d.	In this equation there are equal moles of gaseous reactants and products. A shift in either direction will have no effect on the total number of moles of gas present. There will be no effect on the position of the equilibrium when the pressure is increased.
e.	A shift in the direction of the reverse reaction (shift to the left) will have the result of decreasing the total number of moles of gas present.

15.62

a.	A pressure increase will favor the reaction (forward or reverse?) that decreases the total number of moles of gas. The equilibrium should shift to the right, i.e., more I2 will be produced at the expense of I.
b.	If the concentration of I2 is suddenly altered, the system is no longer at equilibrium. Evaluating the magnitude of the reaction quotient Qc allows us to predict the direction of the resulting equilibrium shift. The reaction quotient for this system is: Increasing the concentration of I2 will increase Qc. The equilibrium will be reestablished in such a way that Qc is again equal to the equilibrium constant. More I will form. The system shifts to the left to establish equilibrium.
c.	The forward reaction is exothermic. A decease in temperature will shift the system to the right to reestablish equilibrium.

15.63

Strategy:

(a) What does the sign of DH° indicate about the heat change (endothermic or exothermic) for the forward reaction? (b) The stress is the addition of Cl₂ gas. How will the system adjust to partially offset the stress? (c) The stress is the removal of PCl₃ gas. How will the system

adjust to partially offset the stress? (d) The stress is an increase in pressure. The system will adjust to decrease the pressure. Remember, pressure is directly proportional to moles of gas.

(e) What is the function of a catalyst? How does it affect a reacting system not at

equilibrium? at equilibrium?

Solution:

a.	The stress applied is the heat added to the system. Note that the reaction is endothermic (DH° > 0). Endothermic reactions absorb heat from the surroundings; therefore, we can think of heat as a reactant. heat + PCl5(g) ⇄ PCl3(g) + Cl2(g) The system will adjust to remove some of the added heat by undergoing a decomposition reaction (Shift to the right)
b.	The stress is the addition of Cl2 gas. The system will shift in the direction to remove some of the added Cl2. The system shifts to the left until equilibrium is reestablished.
c.	The stress is the removal of PCl3 gas. The system will shift to replace some of the PCl3 that was removed. The system shifts to the right until equilibrium is reestablished.
d.	The stress applied is an increase in pressure. The system will adjust to remove the stress by decreasing the pressure. Recall that pressure is directly proportional to the number of moles of gas. In the balanced equation we see 1 mole of gas on the reactants side and 2 moles of gas on the products side. The pressure can be decreased by shifting to the side with the fewer moles of gas. The system will shift to the left to reestablish equilibrium.
e.	The function of a catalyst is to increase the rate of a reaction. If a catalyst is added to the reacting system not at equilibrium, the system will reach equilibrium faster than if left undisturbed. If a system is already at equilibrium, as in this case, the addition of a catalyst will not affect either the concentrations of reactant and product, or the equilibrium constant. A catalyst has no effect on equilibrium position.

15.64

a.	Increasing the temperature favors the endothermic reaction so that the concentrations of SO2 and O2 will increase while that of SO3 will decrease.
b.	Increasing the pressure favors the reaction that decreases the number of moles of gas. The concentration of SO3 will increase.
c.	Increasing the concentration of SO2 will lead to an increase in the concentration of SO3 and a decrease in the concentration of O2.
d.	A catalyst has no effect on the position of equilibrium.
e.	Adding an inert gas at constant volume has no effect on the position of equilibrium.

15.65

No change. A catalyst has no effect on the position of the equilibrium.

15.66

a.	If helium gas is added to the system without changing the pressure or the temperature, the volume of the container must necessarily be increased. This will decrease the partial pressures of all the reactants and products. A pressure decrease will favor the reaction that increases the number of moles of gas. The position of equilibrium will shift to the left.
b.	If the volume remains unchanged, the partial pressures of all the reactants and products will remain the same. The reaction quotient Qc will still equal the equilibrium constant, and there will be no change in the position of equilibrium.

15.67

For this system,

This means that to remain at equilibrium, the pressure of carbon dioxide must stay at a fixed value as long as the temperature remains the same.

a.	If the volume is increased, the pressure of CO2 will drop (Boyle's law, pressure and volume are inversely proportional). Some CaCO3 will break down to form more CO2 and CaO. (Shift to right)
b.	Assuming that the amount of added solid CaO is not so large that the volume of the system is altered significantly, there should be no effect. If a huge amount of CaO were added, this would have the effect of reducing the volume of the container. What would happen then?
c.	Assuming that the amount of CaCO3 removed doesn't alter the container volume significantly, there should be no effect. Removing a huge amount of CaCO3 will have the effect of increasing the container volume. The result in that case will be the same as in part (a).
d.	The pressure of CO2 will be greater and will exceed the value of KP. Some CO2 will combine with CaO to form more CaCO3. (Shift to left)
e.	Carbon dioxide combines with aqueous NaOH according to the equation CO2(g) + NaOH(aq) ® NaHCO3(aq) This will have the effect of reducing the CO2 pressure and causing more CaCO3 to break down to CO2 and CaO. (Shift to the right)
f.	Carbon dioxide does not react with hydrochloric acid, but CaCO3 does. CaCO3(s) + 2HCl(aq) ® CaCl2(aq) + CO2(g) + H2O(l) The CO2 produced by the action of the acid will combine with CaO as discussed in (d) above. (Shift to the left)
g.	This is a decomposition reaction. Decomposition reactions are endothermic. Increasing the temperature will favor this reaction and produce more CO2 and CaO. (Shift to the right)

15.68

Exothermic.

15.69

a.	2O3(g) 3O2(g) DH= -284.4 kJ/mol.
b.	Equilibrium would shift to the left. The number of O3 molecules would increase and the number of O2 molecules would decrease.

15.70

Strategy:

Use Le Châtelier's principle to predict the direction of shift for each case. A shift to the left will cause a decrease in the amount of HbO₂ and a shift to the right will cause an increase in the amount of HbO₂.

Solution:

a.	Decreasing the temperature of an equilibrium mixture causes a shift toward the product side for an exothermic reaction (∆H < 0). The equilibrium will shift to the right and increase the amount of HbO2.
b.	According to Henry’s Law (Equation 13.3), the solubility of a gas in a liquid is proportional to the pressure of the gas over the solution. Increasing the pressure of O2 would increase its concentration. The equilibrium would shift to the right and increase the amount of HbO2.
c.	The removal of a reactant will shift the equilibrium to the left. The amount of HbO2 will decrease.

15.71

Strategy:	According to Le Châtelier's principle, adding a product to an equilibrium mixture shifts the equilibrium toward the reactant side (to the left) of the equation.
Solution:	The addition of acid (product) will shift the equilibrium to the left. As a result, the concentration of oxyhemoglobin will decrease as acidosis occurs.

15.72

The relevant relationships are:

and

K_P = K_c(0.0821 T)^Dn = K_c(0.0821 T) Dn = +1

We set up a table for the calculated values of K_c and K_P.

T (°C) K_c K_P

200 56.9(0.0821 ´ 473) = 2.21 ´ 10³

300 3.41(0.0821 ´ 573) = 1.60 ´ 10²

400 2.10(0.0821 ´ 673) = 116

Since K_c (and K_P) decrease with temperature, the reaction is exothermic.

15.73

a.	The equation that relates KP and Kc is: KP = Kc(0.0821 T)^Dn For this reaction, Dn = 3 - 2 = 1
b.	A mixture of H2 and O2 can be kept at room temperature because of a very large activation energy. The reaction of hydrogen with oxygen is infinitely slow without a catalyst or an initiator. The action of a single spark on a mixture of these gases results in the explosive formation of water.

15.74

Using data from Appendix 2 we calculate the enthalpy change for the reaction.

The enthalpy change is negative, so the reaction is exothermic. The formation of NOCl will be favored by low temperature.

A pressure increase favors the reaction forming fewer moles of gas. The formation of NOCl will be favored by high pressure.

15.75

a.

Calculate the value of KP by substituting the equilibrium partial pressures into the equilibrium expression.

b.

The total pressure is the sum of the partial pressures for the two gaseous components, A and B. We can write: PA + PB = 1.5 atm and PB = 1.5 - PA Substituting into the expression for KP gives: Solving the quadratic equation, we obtain: PA = 0.69 atm and by difference, PB = 0.81 atm Check that substituting these equilibrium concentrations into the equilibrium expression gives the equilibrium constant calculated in part (a).

15.76

a.	The balanced equation shows that equal amounts of ammonia and hydrogen sulfide are formed in this decomposition. The partial pressures of these gases must just be half the total pressure, i.e., 0.355 atm. The value of KP is
b.	We find the number of moles of ammonia (or hydrogen sulfide) and ammonium hydrogen sulfide. From the balanced equation the percent decomposed is
c.	If the temperature does not change, KP has the same value. The total pressure will still be 0.709 atm at equilibrium. In other words the amounts of ammonia and hydrogen sulfide will be twice as great, and the amount of solid ammonium hydrogen sulfide will be: [0.1205 - 2(0.0582)]mol = 0.0041 mol NH4HS

15.77

Total number of moles of gas is:

0.020 + 0.040 + 0.96 = 1.02 mol of gas

You can calculate the partial pressure of each gaseous component from the mole fraction and the total pressure.

Calculate K_P by substituting the partial pressures into the equilibrium expression.

15.78

Since the reactant is a solid, we can write:

The total pressure is the sum of the ammonia and carbon dioxide pressures.

From the stoichiometry,

Therefore:

Substituting into the equilibrium expression:

K_P = (0.212)²(0.106) = 4.76 ´ 10^-3

15.79

Set up a table that contains the initial concentrations, the change in concentrations, and the equilibrium concentration. Assume that the vessel has a volume of 1 L.

H₂ + Cl₂ 2HCl

Initial (M): 0.47 0 3.59

Change (M): +x +x –2x

Equilibrium (M): (0.47 + x) x (3.59 - 2x)

Substitute the equilibrium concentrations into the equilibrium expression, then solve for x. Since
Dn = 0, K_c = K_P.

Solving the quadratic equation,

x = 0.103

Having solved for x, calculate the equilibrium concentrations of all species.

[H₂] = 0.573 M [Cl₂] = 0.103 M [HCl] = 3.384 M

Since we assumed that the vessel had a volume of 1 L, the above molarities also correspond to the number of moles of each component.

From the mole fraction of each component and the total pressure, we can calculate the partial pressure of each component.

Total number of moles = 0.573 + 0.103 + 3.384 = 4.06 mol

P=0.28 atm

P=0.051 atm

P=1.67 atm

15.80

Set up a table that contains the initial concentrations, the change in concentrations, and the equilibrium concentrations. The initial concentration of I₂(g) is 0.054 mol/0.48 L = 0.1125 M. The amount of I₂ that dissociates is (0.0252)(0.1125 M) = 0.002835 M. We carry extra significant figures throughout this calculation to minimize rounding errors.

I₂ 2I

Initial (M): 0.1125 0

Change (M): -0.002835 +(2)(0.002835)

Equilibrium (M): 0.1097 0.005670

Substitute the equilibrium concentrations into the equilibrium expression to solve for K_c.

K_P = K_c(0.0821 T)^Dn

K_P = (2.93 × 10^-4)(0.0821 × 860)¹ = 0.021

15.81

This is a difficult problem. Express the equilibrium number of moles in terms of the initial moles and the change in number of moles (x). Next, calculate the mole fraction of each component. Using the mole fraction, you should come up with a relationship between partial pressure and total pressure for each component. Substitute the partial pressures into the equilibrium expression to solve for the total pressure, P_T.

The reaction is:

N₂ + 3 H₂ 2 NH₃

Initial (mol): 1 3 0

Change (mol): -x -3x 2x

Equilibrium (mol): (1 - x) (3 - 3x) 2x

x = 0.35 mol

Substituting x into the following mole fraction equations, the mole fractions of N₂ and H₂ can be calculated.

The partial pressures of each component are equal to the mole fraction multiplied by the total pressure.

Substitute the partial pressures above (in terms of P_T) into the equilibrium expression, and solve for P_T.

P_T = 5.0 ´ 10¹ atm

15.82

For the balanced equation:

15.83

We carry an additional significant figure throughout this calculation to minimize rounding errors. The initial molarity of SO₂Cl₂ is:

The concentration of SO₂ at equilibrium is:

Since there is a 1:1 mole ratio between SO₂ and SO₂Cl₂, the concentration of SO₂ at equilibrium (0.01725 M) equals the concentration of SO₂Cl₂ reacted. The concentrations of SO₂Cl₂ and Cl₂ at equilibrium are:

SO₂Cl₂(g) SO₂(g) + Cl₂(g)

Initial (M): 0.02500 0 0

Change (M): -0.01725 +0.01725 +0.01725

Equilibrium (M): 0.00775 0.01725 0.01725

Substitute the equilibrium concentrations into the equilibrium expression to calculate K_c.

or 0.038

15.84

a.	The reaction is endothermic because A—B bonds are broken and no new bonds form. Therefore, when temperature is decreased, the equilibrium will shift to the left causing more AB to form.
b.	Equilibrium will shift to the right (toward the larger number of gaseous moles).
c.	Adding He will have no effect on the position of the equilibrium.
d.	Adding a catalyst will have no effect on the position of the equilibrium.

15.85

For a 100% yield, 2.00 moles of SO₃ would be formed (why?). An 80% yield means 2.00 moles ´ (0.80)  1.60 moles SO₃ is formed.

The amount of SO₂ remaining at equilibrium = (2.00 - 1.60)mol = 0.40 mol

The amount of O₂ reacted = ´ (amount of SO₂ reacted) = (´ 1.60)mol = 0.80 mol

The amount of O₂ remaining at equilibrium = (2.00 - 0.80)mol = 1.20 mol

Total moles at equilibrium = moles SO₂ + moles O₂ + moles SO₃ = (0.40 + 1.20 + 1.60)mol = 3.20 moles

P_total = 3.3×10² atm

15.86

We carry an additional significant figure throughout this calculation to minimize rounding errors.

If there were no dissociation, then the pressure would be:

We construct a table to determine how much I₂ has dissociated.

I₂(g) 2I(g)

Initial P (atm): 0.953 0

Change in P (atm): -x +2x

Equilibrium P (atm): 0.953 - x 2x

Knowing the total pressure, we can write

0.953 - x + 2x = 1.51 atm

and solve for x:

x = 0.557

Therefore, the equilibrium pressures are P= 0.396 atm and P= 1.114 atm. Using these pressures in the equilibrium expression gives

K_P = 3.1

15.87

Of the original 1.05 moles of Br₂, 1.20% has dissociated. The amount of Br₂ dissociated in molar concentration is:

Setting up a table:

Br₂(g) 2Br(g)

Initial (M): 0

Change (M): -0.0129 +2(0.0129)

Equilibrium (M): 1.06 0.0258

15.88

Panting decreases the concentration of CO₂ because CO₂ is exhaled during respiration. This decreases the concentration of carbonate ions, shifting the equilibrium to the left. Less CaCO₃ is produced. Two possible solutions would be either to cool the chickens' environment or to feed them carbonated water.

15.89

According to the ideal gas law, pressure is directly proportional to the concentration of a gas in mol/L if the reaction is at constant volume and temperature. Therefore, pressure may be used as a concentration unit.

The reaction is:

N₂ + 3H₂ 2NH₃

Initial (atm): 0.862 0.373 0

Change (atm): -x -3x +2x

Equilibrium (atm): (0.862 - x) (0.373 - 3x) 2x

At this point, we need to make two assumptions that 3x is very small compared to 0.373 and that x is very small compared to 0.862. Hence,

0.373 - 3x » 0.373

and

0.862 - x » 0.862

Solving for x.

x = 2.20 ´ 10^-3 atm

The equilibrium pressures are:

Think About It:

Was the assumption valid that we made above? Typically, the assumption is considered valid if x is less than 5 percent of the number that we said it was very small compared to. Is this the case?

15.90

a.	The sum of the mole fractions must equal one. and According to the hint, the average molar mass is the sum of the products of the mole fraction of each gas and its molar mass. (XCO ´ 28.01 g) + [(1 - XCO) ´ 44.01 g] = 35 g Solving, XCO = 0.56 and
b.	Solving for the pressures PCO = XCOPtotal = (0.56)(11 atm) = 6.2 atm

15.91

a.	The equation is: fructose glucose Initial (M): 0.244 0 Change (M): -0.131 +0.131 Equilibrium (M): 0.113 0.131 Calculating the equilibrium constant,
b.

15.92

If you started with radioactive iodine in the solid phase, then you should fine radioactive iodine in the vapor phase at equilibrium. Conversely, if you started with radioactive iodine in the vapor phase, you should find radioactive iodine in the solid phase. Both of these observations indicate a dynamic equilibrium between solid and vapor phase.

15.93

a.	There is only one gas phase component, O2. The equilibrium constant is simply
b.	From the ideal gas equation, we can calculate the moles of O2 produced by the decomposition of CuO. From the balanced equation, (23%)
c.	If a 1.0 mol sample were used, the pressure of oxygen would still be the same (0.49 atm) and it would be due to the same quantity of O2. Remember, a pure solid does not affect the equilibrium position. The moles of CuO lost would still be 3.7 ´ 10^-2 mol. Thus the fraction decomposed would be: (3.7%)
d.	If the number of moles of CuO were less than 0.037 mol, the equilibrium could not be established because the pressure of O2 would be less than 0.49 atm. Therefore, the smallest number of moles of CuO needed to establish equilibrium must be slightly greater than 0.037 mol.

15.94

If there were 0.88 mole of CO₂ initially and at equilibrium there were 0.11 moles, then (0.88 - 0.11) moles = 0.77 moles reacted.

NO + CO₂ NO₂ + CO

Initial (mol): 3.9 0.88 0 0

Change (mol): -0.77 -0.77 +.077 +0.77

Equilibrium (mol): (3.9 - 0.77) 0.11 0.77 0.77

Solving for the equilibrium constant:

In the balanced equation there are equal number of moles of products and reactants; therefore, the volume of the container will not affect the calculation of K_c. We can solve for the equilibrium constant in terms of moles.

15.95

We first must find the initial concentrations of all the species in the system.

Calculate the reaction quotient by substituting the initial concentrations into the appropriate equation.

We find that Q_c is less than K_c. The equilibrium will shift to the right, decreasing the concentrations of H₂ and I₂ and increasing the concentration of HI.

We set up the usual table. Let x be the decrease in concentration of H₂ and I₂.

H₂ + I₂ 2 HI

Initial (M): 0.298 0.410 0.369

Change (M): -x -x +2x

Equilibrium (M): (0.298 - x) (0.410 - x) (0.369 + 2x)

The equilibrium expression is:

This becomes the quadratic equation

50.3x² - 39.9x + 6.49 = 0

The smaller root is x = 0.228 M. (The larger root is physically impossible.)

Having solved for x, calculate the equilibrium concentrations.

[H₂] = (0.298 - 0.228) M = 0.07 M

[I₂] = (0.410 - 0.228) M = 0.18 M

[HI] = [0.36 + 2(0.228)] M = 0.83 M

15.96

Since we started with pure A, then any A that is lost forms equal amounts of B and C. Since the total pressure is P, the pressure of B + C = P - 0.14 P = 0.86 P. The pressure of B = C = 0.43 P.

15.97

The gas cannot be (a) because the color became lighter with heating. Heating (a) to 150°C would produce some HBr, which is colorless and would lighten rather than darken the gas.

The gas cannot be (b) because Br₂ doesn't dissociate into Br atoms at 150°C, so the color shouldn't change.

The gas must be (c). N₂O₄(colorless) ® 2NO₂(brown) is consistent with the observations. The reaction is endothermic so heating darkens the color. Above 150°C, the NO₂ breaks up into colorless NO and O₂:

2NO₂(g) ® 2NO(g) + O₂(g)

An increase in pressure shifts the equilibrium back to the left, restoring the color by producing NO₂.

15.98

As ionic size decreases, charge density increases. Thus, Ba²⁺ has the lowest charge density and Mg²⁺ has the highest charge density. The higher the charge density, the greater the Coulombic attraction between cation and anion.

15.99

Given the following:

a.	Temperature must have units of Kelvin. KP = Kc(0.0821 T)^Dn KP = (1.2)(0.0821 ´ 648)^(2^-4) = 4.2 ´ 10^-4
b.	Recalling that, Therefore,
c.	Since the equation N2(g) + H2(g) NH3(g) is equivalent to [N2(g) + 3H2(g) 2NH3(g)] then, for the reaction: N2(g) + H2(g) NH3(g) equals for the reaction: N2(g) + 3H2(g) 2NH3(g) Thus,
d.	For KP in part (b): KP = (0.83)(0.0821 ´ 648)^+2 = 2.3 ´ 10^3 and for KP in part (c): KP = (1.1)(0.0821 ´ 648)^-1 = 2.1 ´ 10^-2

15.100

a.	PNO = 1.0 ´ 10^-6 atm
b.	PNO = 2.6 ´ 10^-16 atm
c.	Since KP increases with temperature, it is endothermic.
d.	Lightening. The electrical energy promotes the endothermic reaction.

15.101

a.	Color deepens.

b.	Increases.

c.	Decreases.

d.	Increases.

e.	Unchanged.

15.102

The vapor pressure of water is equivalent to saying the partial pressure of H₂O(g).

15.103

Potassium is more volatile than sodium. Therefore, its removal shifts the equilibrium from left to right.

15.104

We can calculate the average molar mass of the gaseous mixture from the density.

Let be the average molar mass of NO₂ and N₂O₄. The above equation becomes:

= 50.4 g/mol

The average molar mass is equal to the sum of the molar masses of each component times the respective mole fractions. Setting this up, we can calculate the mole fraction of each component.

We can now calculate the partial pressure of NO₂ from the mole fraction and the total pressure.

We can calculate the partial pressure of N₂O₄ by difference.

Finally, we can calculate K_P for the dissociation of N₂O₄.

15.105

In this problem, you are asked to calculate K_c.

Step 1: Calculate the initial concentration of NOCl. We carry an extra significant figure throughout this calculation to minimize rounding errors.

Step 2: Let's represent the change in concentration of NOCl as -2x. Setting up a table:

2NOCl(g) 2NO(g) + Cl₂(g)

Initial (M): 1.667 0 0

Change (M): -2x +2x +x

Equilibrium (M): 1.667 - 2x 2x x

If 28.0 percent of the NOCl has dissociated at equilibrium, the amount consumed is:

(0.280)(1.667 M) = 0.4668 M

In the table above, we have represented the amount of NOCl that reacts as 2x. Therefore,

2x = 0.4668 M

x = 0.2334 M

The equilibrium concentrations of NOCl, NO, and Cl₂ are:

[NOCl] = (1.667 - 2x)M = (1.667 - 0.4668)M = 1.200 M

[NO] = 2x = 0.4668 M

[Cl₂] = x = 0.2334 M

Step 3: The equilibrium constant K_c can be calculated by substituting the above concentrations into the equilibrium expression.

or 3.5 ´ 10^-2

15.106

a.

Since both reactions are endothermic (DH° is positive), according to Le Châtelier’s principle the products would be favored at high temperatures. Indeed, the steam-reforming process is carried out at very high temperatures (between 800°C and 1000°C). It is interesting to note that in a plant that uses natural gas (methane) for both hydrogen generation and heating, about one-third of the gas is burned to maintain the high temperatures. In each reaction there are more moles of products than reactants; therefore, we expect products to be favored at low pressures. In reality, the reactions are carried out at high pressures. The reason is that when the hydrogen gas produced is used captively (usually in the synthesis of ammonia), high pressure leads to higher yields of ammonia.

b.

(i) The relation between Kc and KP is given by Equation 15.4 of the text: KP = Kc(0.0821 T)^Dn Since Dn = 4 - 2 = 2, we write: KP = (18)(0.0821 ´ 1073)^2 = 1.4 ´ 10^5 (ii) Let x be the amount of CH4 and H2O (in atm) reacted. We write: CH4 + H2O CO + 3H2 Initial (atm): 15 15 0 0 Change (atm): -x -x +x +3x Equilibrium (atm): 15 - x 15 - x x 3x The equilibrium constant is given by: Taking the square root of both sides, we obtain: which can be expressed as 5.2x^2 + (3.7 ´ 10^2x) - (5.6 ´ 10^3) = 0 Solving the quadratic equation, we obtain x = 13 atm (The other solution for x is negative and is physically impossible.) At equilibrium, the pressures are: PCO = 13 atm

15.107

a.	Shifts to right.
b.	Shifts to right.
c.	No change.
d.	No change.
e.	No change.
f.	Shifts to left.

15.108

K_P = (1.1)(1.1) = 1.2

15.109

a.

Assuming the self-ionization of water occurs by a single elementary step mechanism, the equilibrium constant is just the ratio of the forward and reverse rate constants.

b.

The product can be written as: [H^+][OH^-] = K[H2O] What is [H2O]? It is the concentration of pure water. One liter of water has a mass of 1000 g (density = 1.00 g/mL). The number of moles of H2O is: The concentration of water is 55.5 mol/1.00 L or 55.5 M. The product is: [H^+][OH^-] = (1.8 ´ 10^-16)(55.5) = 1.0 ´ 10^-14 We assume the concentration of hydrogen ion and hydroxide ion are equal. [H^+][OH^-] = (1.0 ´ 10^-14)^1/2 = 1.0 ´ 10^-7 M

15.110

At equilibrium, the value of K_c is equal to the ratio of the forward rate constant to the rate constant for the reverse reaction.

k_f = (12.6)(5.1 ´ 10^-2) = 0.64

The forward reaction is third order, so the units of k_f must be:

rate = k_f[A]²[B]

k_f = 0.64 /M²×s

15.111

The equilibrium is: N₂O₄(g) 2NO₂(g)

Volume is doubled so pressure is halved. Let’s calculate Q_P and compare it to K_P.

Equilibrium will shift to the right. Some N₂O₄ will react, and some NO₂ will be formed. Let
x = amount of N₂O₄ reacted.

N₂O₄(g) 2NO₂(g)

Initial (atm): 0.10 0.075

Change (atm): -x +2x

Equilibrium (atm): 0.10 - x 0.075 + 2x

Substitute into the K_P expression to solve for x.

4x² + 0.413x - 5.68 ´ 10^-3 = 0

x = 0.0123

At equilibrium:

Think About It:

; close enough to 0.113

15.112

a.	Combine Ni with CO above 50°C. Pump away the Ni(CO)4 vapor (shift equilibrium to right), leaving the solid impurities behind.
b.	Consider the reverse reaction: Ni(CO)4(g) ® Ni(s) + 4CO(g) DH° = (4)(-110.5 kJ/mol) - (1)(-602.9 kJ/mol) = 160.9 kJ/mol The decomposition is endothermic, which is favored at high temperatures. Heat Ni(CO)4 above 200°C to convert it back to Ni.

15.113

a.	Molar mass of PCl5 = 208.2 g/mol
b.	PCl5 PCl3 + Cl2 Initial (atm) 1.03 0 0 Change (atm) -x +x +x Equilibrium (atm) 1.03 - x x x x^2 + 1.05x - 1.08 = 0 x = 0.639 At equilibrium:
c.	PT = (1.03 - x) + x + x = 1.03 + 0.639 = 1.67 atm
d.	(62.0%)

15.114

a.
b.
c.

15.115

a.	KP = PHg = 0.0020 mmHg = 2.6 ´ 10^-6 atm = 2.6 ´ 10^-6 (equil. constants are expressed without units)
b.	Volume of lab = (6.1 m)(5.3 m)(3.1 m) = 100 m^3 [Hg] = Kc The concentration of mercury vapor in the room is: Yes. This concentration exceeds the safety limit of 0.05 mg/m^3.

15.116

Initially, at equilibrium: [NO₂] = 0.0475 M and [N₂O₄] = 0.487 M. At the instant the volume is halved, the concentrations double.

[NO₂] = 2(0.0475 M) = 0.0950 M and [N₂O₄] = 2(0.487 M) = 0.974 M. The system is no longer at equilibrium. The system will shift to the left to offset the increase in pressure when the volume is halved. When a new equilibrium position is established, we write:

N₂O₄ 2 NO₂

0.974 M + x 0.0950 M – 2x

4x² – 0.3846x + 4.52 ´ 10^-3 = 0

Solving

x = 0.0824 M (impossible) and x = 0.0137 M

At the new equilibrium,

[N₂O₄] = 0.974 + 0.0137 = 0.988 M

[NO₂] = 0.0950 – (2 ´ 0.0137) = 0.0676 M

As we can see, the new equilibrium concentration of NO₂ is greater than the initial equilibrium concentration (0.0475 M). Therefore, the gases should look darker!

15.117

There is a temporary dynamic equilibrium between the melting ice cubes and the freezing of water between the ice cubes.

15.118

a.	A catalyst speeds up the rates of the forward and reverse reactions to the same extent.
b.	A catalyst would not change the energies of the reactant and product.
c.	The first reaction is exothermic. Raising the temperature would favor the reverse reaction, increasing the amount of reactant and decreasing the amount of product at equilibrium. The equilibrium constant, K, would decrease. The second reaction is endothermic. Raising the temperature would favor the forward reaction, increasing the amount of product and decreasing the amount of reactant at equilibrium. The equilibrium constant, K, would increase.
d.	A catalyst lowers the activation energy for the forward and reverse reactions to the same extent. Adding a catalyst to a reaction mixture will simply cause the mixture to reach equilibrium sooner. The same equilibrium mixture could be obtained without the catalyst, but we might have to wait longer for equilibrium to be reached. If the same equilibrium position is reached, with or without a catalyst, then the equilibrium constant is the same.

15.119

First, let's calculate the initial concentration of ammonia.

Let's set up a table to represent the equilibrium concentrations. We represent the amount of NH₃ that reacts as 2x.

2NH₃(g) N₂(g) + 3H₂(g)

Initial (M): 0.214 0 0

Change (M): -2x +x +3x

Equilibrium (M): 0.214 - 2x x 3x

Substitute into the equilibrium expression to solve for x.

Taking the square root of both sides of the equation gives:

Rearranging,

5.20x² + 1.82x - 0.195 = 0

Solving the quadratic equation gives the solutions:

x = 0.086 M and x = -0.44 M

The positive root is the correct answer. The equilibrium concentrations are:

[NH₃] = 0.214 - 2(0.086) = 0.042 M

[N₂] = 0.086 M

[H₂] = 3(0.086) = 0.26 M

15.120

To determine DH°, we need to plot ln K_P versus 1/T (y vs. x).

ln KP	1/T
4.93	0.00167
1.63	0.00143
–0.83	0.00125
–2.77	0.00111
–4.34	0.00100

The slope of the plot equals -DH°/R.

DH° = -1.15 × 10⁵ J/mol = -115 kJ/mol

15.121

a.	From the balanced equation N2O4 2NO2 Initial (mol): 1 0 Change (mol): –x +2x Equilibrium (mol): (1 - x) 2x The total moles in the system = (moles N2O4 + moles NO2) = [(1 - x) + 2x] = 1 + x. If the total pressure in the system is P, then: and KP =
b.	Rearranging the KP expression: 4x^2P = KP - x^2KP x^2(4P + KP) = KP KP is a constant (at constant temperature). If P increases, the fraction (and therefore x) must decrease. Equilibrium shifts to the left to produce less NO2 and more N2O4 as predicted.

15.122

a.

We start by writing the van’t Hoff equation at two different temperatures. Assuming an endothermic reaction, DH° > 0 and T2 > T1. Then, < 0, meaning that < 0 or K1 < K2. A larger K2 indicates that there are more products at equilibrium as the temperature is raised. This agrees with LeChatelier’s principle that an increase in temperature favors the forward endothermic reaction. The opposite of the above discussion holds for an exothermic reaction.

b.

Treating H2O(l) H2O(g) DHvap = ? as a heterogeneous equilibrium, . We substitute into the equation derived in part (a) to solve for DHvap. -1.067 = DH°(-2.466 × 10^-5) DH° = 4.34 × 10^4 J/mol = 434 kJ/mol

15.123

Initially, the pressure of SO₂Cl₂ is 9.00 atm. The pressure is held constant, so after the reaction reaches equilibrium, . The amount (pressure) of SO₂Cl₂ reacted must equal the pressure of SO₂ and Cl₂ produced for the pressure to remain constant. If we let , then the pressure of SO₂Cl₂ reacted must be 2x. We set up a table showing the initial pressures, the change in pressures, and the equilibrium pressures.

SO₂Cl₂(g) SO₂(g) + Cl₂(g)

Initial (atm): 9.00 0 0

Change (atm): -2x +x +x

Equilibrium (atm): 9.00 - 2x x x

Again, note that the change in pressure for SO₂Cl₂ (-2x) does not match the stoichiometry of the reaction, because we are expressing changes in pressure. The total pressure is kept at 9.00 atm throughout.

x² + 4.10x - 18.45 = 0

Solving the quadratic equation, x = 2.71 atm. At equilibrium,

15.124

Using Equation 14.8 of the text, we can calculate k_-1.

Then, we can calculate k₁ using the expression

(see Section 15.2 of the text)

k_-1 = 6.5 × 10⁴ s^-1

k₁ = 6.4 × 10⁸ s^-1

15.125

We start with a table.

A₂ + B₂ 2AB

Initial (mol): 1 3 0

Change (mol): +x

Equilibrium (mol): x

After the addition of 2 moles of A,

A₂ + B₂ 2AB

Initial (mol): x

Change (mol): +x

Equilibrium (mol): 3 - x 3 - x 2x

We write two different equilibrium constants expressions for the two tables.

We equate the equilibrium expressions and solve for x.

-6x + 9 = -8x + 12

x = 1.5

We substitute x back into one of the equilibrium expressions to solve for K.

Substitute x into the other equilibrium expression to see if you obtain the same value for K. Note that we used moles rather than molarity for the concentrations, because the volume, V, cancels in the equilibrium expressions.

15.126

a.	First, we calculate the moles of I2. Let x be the number of moles of I2 that dissolves in CCl4, so (1.26 × 10^-4 - x)mol remains dissolved in water. We set up expressions for the concentrations of I2 in CCl4 and H2O. Next, we substitute these concentrations into the equilibrium expression and solve for x. 83(1.26 × 10^-4 - x) = 6.67x x = 1.166 × 10^-4 The fraction of I2 remaining in the aqueous phase is:
b.	The first extraction leaves only 7% I2 in the water. The next extraction with 0.030 L of CCl4 will leave only (0.07)(0.07) = 5 × 10^-3. This is the fraction remaining after the second extraction which is only 0.5%.
c.	A single extraction using 0.060 L of CCl4 gives: The concentration of I2 is the same as in part (a). Substituting these concentrations into the equilibrium expression and solving for x gives: 83(1.26 × 10^-4 - x) = 3.33x x = 1.211 × 10^-4 The fraction of I2 remaining in the aqueous phase is:

15.127

Strategy:	The concentrations of solids and pure liquids do not appear in KC. So, even though the given equilibrium is heterogeneous, Equation 15.4 still applies.
Solution:	There are 3 moles of gas molecules on the product side and 6 moles on the reactant side. So, Dn = 3 – 6 = –3.

15.128

According to Le Châtelier’s principle, an exothermic equilibrium can be shifted to the right by lowering the temperature. Also, raising the pressure will favor the production of product since the equilibrium will respond to this stress by minimizing the number of gas particles. From the standpoint of optimizing the yield, temperature should be low and pressure should be high. To optimize the rate of production, both the forward and reverse reactions should be very rapid to ensure that equilibrium is established quickly. From a kinetic standpoint, temperature and pressure (concentration) should both be high.

15.129

Strategy:

According to Equation 15.4, K_P and K_C are the same when Dn = 0:

So, look for reactions that show no change in the number of gas molecules as the reaction occurs. Also, note that Equation 15.4 does not apply to heterogeneous equilibria if liquid phases are solutions.

Solution:

a.	KP ¹ KC (Dn ¹ 0)
b.	KP ¹ KC (Dn ¹ 0)
c.	Equation 15.4 is not applicable (some species are aqueous).
d.	KP = KC (Dn = 0)
e.	KP ¹ KC (Dn ¹ 0)
f.	Equation 15.4 is not applicable (some species are aqueous).
g.	KP = KC (Dn = 0)
h.	KP ¹ KC (Dn ¹ 0)

15.130

a.

Strategy:	Use the law of mass action to write the equilibrium expression.
Setup:	The equilibrium expression has the form of the concentrations of products over the concentration of reactants, each raised to a power equal to its stoichiometric coefficient in the balanced chemical equation.
Solution:

b.

Strategy:	To evaluate Kc, plug in the equilibrium concentrations of all three species into the equilibrium expression derived in part ‘a’.
Solution:

c.

Strategy:	Begin by writing the equilibrium expression for the precipitation reaction. Take the sum of the two reactions and write the net reaction. The overall equilibrium constant is the product of the individual constants.
Setup:	The equilibrium expression for the aqueous reaction was calculated in part (a): The equilibrium expression for the precipitation reaction is Taking the sum of the two reactions gives: OD1A(aq) + OF2A(aq) OD17X(aq) OD17X(aq) + A771A(aq) OD17X-A77(s) OD1A(aq) + OF2A(aq) + A771A(aq) OD17X-A11(s)
Solution:

d.

Strategy:	Use the concentrations provided to calculate Qc, and then compare Qc with Kc.
Setup:
Solution:	The calculated value of Qc is greater than Kc. Therefore, the reaction is not at equilibrium and must proceed to the left to establish equilibrium.

e.

Strategy:	Construct an equilibrium table to determine the equilibrium concentration of each species in terms of an unknown (x); solve for x, and use it to calculate the equilibrium molar concentrations.
Setup:	Insert the starting concentrations that we know into the equilibrium table: OD1A + OF2A OD17X Initial (M): 1.0 1.0 0 Change (M): Equilibrium (M):
Solution:	We define the change in concentration of one of the reactants as x. Because there is no product at the start of the reaction, the reactant concentration must decrease; that is, this reaction must proceed in the forward direction to reach equilibrium. According to the stoichiometry of the chemical reaction, the reactant concentrations will both decrease by the same amount (x), and the product concentration will increase by that amount (x). Combining the initial concentration and the change in concentration for each species, we get expressions (in terms of x) for the equilibrium concentrations: OD1A + OF2A OD17X Initial (M): 1.0 1.0 0 Change (M): -x -x +x Equilibrium (M): (1.0 - x) (1.0 - x) x Next, we insert these expressions for the equilibrium concentrations into the equilibrium expression and solve for x. Collecting terms we get This is a quadratic equation of the form ax^2 + bx + c = 0. The solution for the quadratic equation (see Appendix 1) is Here we have a = 3.8 × 10^2, b = –7.61 × 10^2, c = 3.8 × 10^2, so x = 1.05 or x = 0.95 Only the second of these values, 0.95, makes sense because concentration cannot be a negative number (1 – 1.05 = –0.5). Using the calculated values of x, we can determine the equilibrium concentration of each species as follows: [OD17X] = 0.95 M [OD1A] = [OF2A] = (1 – 0.95) = 0.05 M