CHAPTER 14
Chemical Equilibrium
■ SOLUTIONS TO EXERCISES
Note on significant figures: If the final answer to a solution needs to be rounded off, it is given first with
one nonsignificant figure, and the last significant figure is underlined. The final answer is then rounded to
the correct number of significant figures. In multistep problems, intermediate answers are given with at
least one nonsignificant figure; however, only the final answer has been rounded off.
14.1. Use the table approach, and give the starting numbers, the change, and the equilibrium number of
moles of each.
Amt. (mol) CO(g) + H2O(g) CO2(g) + H2(g)
Starting 1.00 1.00 0 0
Change −x −x +x +x
Equilibrium (1.00 − x) (1.00 − x) x x = 0.43
Because we are given that x = 0.43 in the statement of the problem, we can use that to calculate
the equilibrium amounts of the reactants and products:
Equilibrium amount CO = 1.00 − 0.43 = 0.57 mol
Equilibrium amount H2O = 1.00 − 0.43 = 0.57 mol
Equilibrium amount CO2 = x = 0.43 mol
Equilibrium amount H2 = x = 0.43 mol
14.2. For the equation 2NO2 + 7H2 2NH3 + 4H2O, the expression for the equilibrium constant, Kc, is
[ NH3 ]2 [ H 2 O ]4
Kc =
[ NO2 ]2 [ H 2 ]7
Notice that each concentration term is raised to a power equal to its coefficient in the chemical
equation.
For the equation NO2 + 7/2H2 NH3 + 2H2O, the expression for the equilibrium constant, Kc, is
[ NH3 ][ H 2 O ]2
Kc =
[ NO2 ][ H 2 ]7/ 2
Note the correspondence between the power and coefficient for each molecule.
© 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
, Chapter 14: Chemical Equilibrium 495
14.3. The chemical equation for the reaction is
CO(g) + H2O(g) H2(g) + CO2(g)
The expression for the equilibrium constant for this reaction is
[ CO2 ][H 2 ]
Kc =
[ CO ][ H 2 O ]
We obtain the concentration of each substance by dividing the moles of substance by its volume.
The equilibrium concentrations are as follows: [CO] = 0.057 M, [H2O] = 0.057 M, [CO2] = 0.043
M, and [H2] = 0.043 M. Substituting these values into the equation for the equilibrium constant
gives
(0.043)(0.043)
Kc = = 0.569 = 0.57
(0.057)(0.057)
14.4. Use the table approach, and give the starting concentrations, the change, and the equilibrium
concentration of each by dividing moles by volume in liters.
Conc. (M) 2H2S(g) 2H2(g) + S2(g)
Starting 0.0100 0 0
Change −2x +2x (= 0.00285) +x
Equilibrium 0.0100 − 2x 2x x
Because the problem states that 0.00285 M H2 was formed, we can use the 0.00285 M to calculate
the other concentrations. The S2 molarity should be one-half that, or 0.001425 M, and the H2S
molarity should be 0.0100 − 0.00285, or 0.00715 M. Substituting into the equilibrium expression
gives
[ H 2 ]2 [S2 ] ( 0.00285) 2 ( 0.001425)
Kc = 2
= = 2.26 10−4 = 2.3 10−4
[ H 2S] ( 0.00715) 2
14.5. Use the expression that relates Kc to Kp:
Kp = Kc(RT)n
The n term is the sum of the coefficients of the gaseous products minus the sum of the
coefficients of the gaseous reactants. In this case, n = (2 − 1) = 1, and Kp is
Kp = (3.26 10−2) (0.0821 464)1 = 1.241 = 1.24
14.6. For a heterogeneous equilibrium, the concentration terms for liquids and solids are omitted
because such concentrations are constant at a given temperature and are incorporated into the
measured value of Kc. For this case, Kc is defined as
[Ni(CO)4 ]
Kc =
[CO]4
14.7. Because the equilibrium constant is very large (> 104), the equilibrium mixture will contain
mostly products. Rearrange the Kc expression to solve for [NO2].
[NO2] = K c [O2 ][NO]2 = (4.0 1013 )(2.0 106 )(2.0 106 ) 2 = 1.78 10−2 = 1.8 10−2 = 0.018 M
© 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
, 496 Chapter 14: Chemical Equilibrium
14.8. First, divide moles by volume in liters to convert to molar concentrations, giving 0.00015 M CO2
and 0.010 M CO. Substitute these values into the reaction quotient and calculate Q.
[CO]2 (0.010) 2
Q= = = 0.666 = 0.67
[CO 2 ] (0.00015)
Because Q = 0.67 and is less than Kc, the reaction will go to the right, forming more CO.
14.9. Rearrange the Kc expression, and substitute for Kc (= 0.0415) and the given moles per 1.00 L to
solve for moles per 1.00 L of PCl5.
[ PCl3 ][ Cl2 ] ( 0.020) ( 0.020)
[PCl5] = = = 9.63 10−3 = 9.6 10−3 M
Kc 0.0415
Because the volume is 1.00 L, the moles of PCl5 = 0.0096 mol.
14.10. Use the table approach, and give the starting number, the change, and the equilibrium number of
moles of each.
Amt. (mol) H2(g) + I2(g) 2HI(g)
Starting 0.500 0.500 0
Change −x −x +2x
Equilibrium 0.500 − x 0.500 − x 2x
Substitute the equilibrium concentrations into the expression for Kc (= 49.7).
[HI]2 (2 x) 2 (2 x) 2
Kc = ; 49.7 = =
[H 2 ][I 2 ] (0.500 x)(0.500 x) (0.500 x) 2
Taking the square root of both sides of the right-hand equation and solving for x gives
2x
±7.05 = , or ±7.05(0.500 − x) = 2x
(0.500 x)
Using the positive root, x = 0.390.
Using the negative root, x = 0.698 (this must be rejected because 0.698 is greater than the 0.500
starting number of moles).
Substituting x = 0.390 mol into the last line of the table to solve for equilibrium concentrations
gives these amounts: 0.11 mol H2, 0.11 mol I2, and 0.78 mol HI.
14.11. Use the table approach for starting, change, and equilibrium concentrations of each species.
Conc. (M) PCl5(g) PCl3(g) + Cl2(g)
Starting 1.00 0 0
Change −x +x +x
Equilibrium 1.00 − x x x
Substitute the equilibrium-concentration expressions from the table into the equilibrium equation,
and solve for x using the quadratic formula.
[PCl3][Cl2] = Kc [PCl5] = 0.0211(1.00 − x) = x2
x2 + 0.0211x − 0.0211 = 0
© 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
, Chapter 14: Chemical Equilibrium 497
0.0211 (0.0211) 2 4(0.0211) 0.0211 0.2913
x= =
2 2
x = −0.1562 (impossible; reject), or x = 0.13509 = 0.135 M (logical)
Solve for the equilibrium concentrations using x = 0.135 M: [PCl5] = 0.86 M, [Cl2] = 0.135 M,
and [PCl3] = 0.135 M.
14.12. a. Increasing the pressure will cause a net reaction to occur from right to left, and more
CaCO3 will form.
b. Increasing the concentration of hydrogen will cause a net reaction to occur from right to
left, forming more Fe and H2O.
14.13. a. Because there are equal numbers of moles of gas on both sides of the equation, increasing
the pressure will not increase the amount of product.
b. Because the reaction increases the number of moles of gas, increasing the pressure will
decrease the amount of product.
c. Because the reaction decreases the number of moles of gas, increasing the pressure will
increase the amount of product.
14.14. Because this is an endothermic reaction and absorbs heat, high temperatures will be more
favorable to the production of carbon monoxide.
14.15. Because this is an endothermic reaction and absorbs heat, high temperatures will give the best
yield of carbon monoxide. Because the reaction increases the number of moles of gas, decreasing
the pressure will also increase the yield.
■ ANSWERS TO CONCEPT CHECKS
14.1. The statement that when reactant A decreases by an amount x, product C increases by amount x
implies that A and C have the same coefficients. The statement that when reactant B decreases by
an amount x, product C increases by amount 2x implies that the coefficient of C is twice that of B.
Therefore, the coefficient of A is twice that of B. The simplest equation satisfying these
conditions is 2A + B 2C.
14.2. To answer this question, find the relationship between the two species present, using the
equilibrium-constant expression and its value. For the first reaction, A(g) B(g), with K = 2,
this becomes
[B]
K=2=
[A]
This reduces to [B] = 2[A]. This corresponds to the container that has twice as many balls of one
color than of the other color, namely container IV. Here, the blue molecules are B (eight of them),
and the red molecules are A (four of them).
© 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Chemical Equilibrium
■ SOLUTIONS TO EXERCISES
Note on significant figures: If the final answer to a solution needs to be rounded off, it is given first with
one nonsignificant figure, and the last significant figure is underlined. The final answer is then rounded to
the correct number of significant figures. In multistep problems, intermediate answers are given with at
least one nonsignificant figure; however, only the final answer has been rounded off.
14.1. Use the table approach, and give the starting numbers, the change, and the equilibrium number of
moles of each.
Amt. (mol) CO(g) + H2O(g) CO2(g) + H2(g)
Starting 1.00 1.00 0 0
Change −x −x +x +x
Equilibrium (1.00 − x) (1.00 − x) x x = 0.43
Because we are given that x = 0.43 in the statement of the problem, we can use that to calculate
the equilibrium amounts of the reactants and products:
Equilibrium amount CO = 1.00 − 0.43 = 0.57 mol
Equilibrium amount H2O = 1.00 − 0.43 = 0.57 mol
Equilibrium amount CO2 = x = 0.43 mol
Equilibrium amount H2 = x = 0.43 mol
14.2. For the equation 2NO2 + 7H2 2NH3 + 4H2O, the expression for the equilibrium constant, Kc, is
[ NH3 ]2 [ H 2 O ]4
Kc =
[ NO2 ]2 [ H 2 ]7
Notice that each concentration term is raised to a power equal to its coefficient in the chemical
equation.
For the equation NO2 + 7/2H2 NH3 + 2H2O, the expression for the equilibrium constant, Kc, is
[ NH3 ][ H 2 O ]2
Kc =
[ NO2 ][ H 2 ]7/ 2
Note the correspondence between the power and coefficient for each molecule.
© 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
, Chapter 14: Chemical Equilibrium 495
14.3. The chemical equation for the reaction is
CO(g) + H2O(g) H2(g) + CO2(g)
The expression for the equilibrium constant for this reaction is
[ CO2 ][H 2 ]
Kc =
[ CO ][ H 2 O ]
We obtain the concentration of each substance by dividing the moles of substance by its volume.
The equilibrium concentrations are as follows: [CO] = 0.057 M, [H2O] = 0.057 M, [CO2] = 0.043
M, and [H2] = 0.043 M. Substituting these values into the equation for the equilibrium constant
gives
(0.043)(0.043)
Kc = = 0.569 = 0.57
(0.057)(0.057)
14.4. Use the table approach, and give the starting concentrations, the change, and the equilibrium
concentration of each by dividing moles by volume in liters.
Conc. (M) 2H2S(g) 2H2(g) + S2(g)
Starting 0.0100 0 0
Change −2x +2x (= 0.00285) +x
Equilibrium 0.0100 − 2x 2x x
Because the problem states that 0.00285 M H2 was formed, we can use the 0.00285 M to calculate
the other concentrations. The S2 molarity should be one-half that, or 0.001425 M, and the H2S
molarity should be 0.0100 − 0.00285, or 0.00715 M. Substituting into the equilibrium expression
gives
[ H 2 ]2 [S2 ] ( 0.00285) 2 ( 0.001425)
Kc = 2
= = 2.26 10−4 = 2.3 10−4
[ H 2S] ( 0.00715) 2
14.5. Use the expression that relates Kc to Kp:
Kp = Kc(RT)n
The n term is the sum of the coefficients of the gaseous products minus the sum of the
coefficients of the gaseous reactants. In this case, n = (2 − 1) = 1, and Kp is
Kp = (3.26 10−2) (0.0821 464)1 = 1.241 = 1.24
14.6. For a heterogeneous equilibrium, the concentration terms for liquids and solids are omitted
because such concentrations are constant at a given temperature and are incorporated into the
measured value of Kc. For this case, Kc is defined as
[Ni(CO)4 ]
Kc =
[CO]4
14.7. Because the equilibrium constant is very large (> 104), the equilibrium mixture will contain
mostly products. Rearrange the Kc expression to solve for [NO2].
[NO2] = K c [O2 ][NO]2 = (4.0 1013 )(2.0 106 )(2.0 106 ) 2 = 1.78 10−2 = 1.8 10−2 = 0.018 M
© 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
, 496 Chapter 14: Chemical Equilibrium
14.8. First, divide moles by volume in liters to convert to molar concentrations, giving 0.00015 M CO2
and 0.010 M CO. Substitute these values into the reaction quotient and calculate Q.
[CO]2 (0.010) 2
Q= = = 0.666 = 0.67
[CO 2 ] (0.00015)
Because Q = 0.67 and is less than Kc, the reaction will go to the right, forming more CO.
14.9. Rearrange the Kc expression, and substitute for Kc (= 0.0415) and the given moles per 1.00 L to
solve for moles per 1.00 L of PCl5.
[ PCl3 ][ Cl2 ] ( 0.020) ( 0.020)
[PCl5] = = = 9.63 10−3 = 9.6 10−3 M
Kc 0.0415
Because the volume is 1.00 L, the moles of PCl5 = 0.0096 mol.
14.10. Use the table approach, and give the starting number, the change, and the equilibrium number of
moles of each.
Amt. (mol) H2(g) + I2(g) 2HI(g)
Starting 0.500 0.500 0
Change −x −x +2x
Equilibrium 0.500 − x 0.500 − x 2x
Substitute the equilibrium concentrations into the expression for Kc (= 49.7).
[HI]2 (2 x) 2 (2 x) 2
Kc = ; 49.7 = =
[H 2 ][I 2 ] (0.500 x)(0.500 x) (0.500 x) 2
Taking the square root of both sides of the right-hand equation and solving for x gives
2x
±7.05 = , or ±7.05(0.500 − x) = 2x
(0.500 x)
Using the positive root, x = 0.390.
Using the negative root, x = 0.698 (this must be rejected because 0.698 is greater than the 0.500
starting number of moles).
Substituting x = 0.390 mol into the last line of the table to solve for equilibrium concentrations
gives these amounts: 0.11 mol H2, 0.11 mol I2, and 0.78 mol HI.
14.11. Use the table approach for starting, change, and equilibrium concentrations of each species.
Conc. (M) PCl5(g) PCl3(g) + Cl2(g)
Starting 1.00 0 0
Change −x +x +x
Equilibrium 1.00 − x x x
Substitute the equilibrium-concentration expressions from the table into the equilibrium equation,
and solve for x using the quadratic formula.
[PCl3][Cl2] = Kc [PCl5] = 0.0211(1.00 − x) = x2
x2 + 0.0211x − 0.0211 = 0
© 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
, Chapter 14: Chemical Equilibrium 497
0.0211 (0.0211) 2 4(0.0211) 0.0211 0.2913
x= =
2 2
x = −0.1562 (impossible; reject), or x = 0.13509 = 0.135 M (logical)
Solve for the equilibrium concentrations using x = 0.135 M: [PCl5] = 0.86 M, [Cl2] = 0.135 M,
and [PCl3] = 0.135 M.
14.12. a. Increasing the pressure will cause a net reaction to occur from right to left, and more
CaCO3 will form.
b. Increasing the concentration of hydrogen will cause a net reaction to occur from right to
left, forming more Fe and H2O.
14.13. a. Because there are equal numbers of moles of gas on both sides of the equation, increasing
the pressure will not increase the amount of product.
b. Because the reaction increases the number of moles of gas, increasing the pressure will
decrease the amount of product.
c. Because the reaction decreases the number of moles of gas, increasing the pressure will
increase the amount of product.
14.14. Because this is an endothermic reaction and absorbs heat, high temperatures will be more
favorable to the production of carbon monoxide.
14.15. Because this is an endothermic reaction and absorbs heat, high temperatures will give the best
yield of carbon monoxide. Because the reaction increases the number of moles of gas, decreasing
the pressure will also increase the yield.
■ ANSWERS TO CONCEPT CHECKS
14.1. The statement that when reactant A decreases by an amount x, product C increases by amount x
implies that A and C have the same coefficients. The statement that when reactant B decreases by
an amount x, product C increases by amount 2x implies that the coefficient of C is twice that of B.
Therefore, the coefficient of A is twice that of B. The simplest equation satisfying these
conditions is 2A + B 2C.
14.2. To answer this question, find the relationship between the two species present, using the
equilibrium-constant expression and its value. For the first reaction, A(g) B(g), with K = 2,
this becomes
[B]
K=2=
[A]
This reduces to [B] = 2[A]. This corresponds to the container that has twice as many balls of one
color than of the other color, namely container IV. Here, the blue molecules are B (eight of them),
and the red molecules are A (four of them).
© 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
