OpenStax Chemistry 2e 13.1: Chemical Equilibria

OpenStax Chemistry 2e 13.1: Chemical Equilibria

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Chemistry 2e 13: Fundamental Equilibrium Concepts

13.1: Chemical Equilibria 1. What does it mean to describe a reaction as “reversible”? Solution The reaction can proceed in both the forward and reverse directions. 2. When writing an equation, how is a reversible reaction distinguished from a nonreversible reaction? Solution The two types of reactions are distinguished by the arrow used in the equation: is used for equilibria reactions and is used for “one-way” reactions. 3. If a reaction is reversible, when can it be said to have reached equilibrium? Solution When a system has reached equilibrium, no further changes in the reactant and product concentrations occur; the forward and reverse reactions continue to proceed, but at equal rates. 4. Is a system at equilibrium if the rate constants of the forward and reverse reactions are equal? Solution No, the rate constants having the same value does not mean that the rates of the reactions are equal, which defines a system being at equilibrium. 5. If the concentrations of products and reactants are equal, is the system at equilibrium? Solution Not necessarily. A system at equilibrium is characterized by constant reactant and product concentrations, but the values of the reactant and product concentrations themselves need not be equal.


13.2: Equilibrium Constants 6. Explain why there may be an infinite number of values for the reaction quotient of a reaction at a given temperature but there can be only one value for the equilibrium constant at that temperature. Solution For a general reaction of the type A B C Dm n x y � �� , the reaction quotient Qc is written

as [C] [D] [A] [B]

x y

c m nQ . When the reaction reaches equilibrium, the forward and reverse reactions

are proceeding at the same rate. Each component’s net concentration does not change with time; accordingly, the reaction quotient will take on one value, which is called the equilibrium constant. Prior to equilibrium being reached, the rates of the forward and reverse reactions are not equal and the concentration of each component will be different at different times prior to equilibrium being achieved. Because it is possible to have an infinite number of concentrations before equilibrium, it is possible to calculate an infinite number of values of Qc. 7. Explain why an equilibrium between Br2(l) and Br2(g) would not be established if the container were not a closed vessel shown in Figure 13.4. Solution


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Equilibrium cannot be established between the liquid and the gas phase if the top is removed from the bottle because the system is not closed; one of the components of the equilibrium, the Br2 vapor, would escape from the bottle until all liquid disappeared. Thus, more liquid would evaporate than can condense back from the gas phase to the liquid phase. 8. If you observe the following reaction at equilibrium, is it possible to tell whether the reaction started with pure NO2 or with pure N2O4?

2 2 42NO ( ) N O ( )g g� �� Solution For the reaction 2 2 42NO ( ) N O ( )g g� �� , when equilibrium is achieved, the following

relationship will be satisfied: 2 42

2

[N O ] [NO ]cK . Thus, at equilibrium, both reactants and products

will be present. Because the same position of equilibrium will be achieved, no matter from which direction (either combination of NO2 or decomposition of N2O4), it is not possible to tell in which direction the reaction took place to achieve the equilibrium position. Prior to achieving equilibrium, it is possible to tell in which direction the reaction is proceeding. For example, in this system, if the color of the reaction mixture were a deep reddish brown that became a lighter intensity with time, then it would be possible to recognize that equilibrium is being approached as a result of the brown NO2 molecules combining to form the colorless N2O4 molecules. 9. Among the solubility rules previously discussed is the statement: All chlorides are soluble except Hg2Cl2, AgCl, PbCl2, and CuCl. (a) Write the expression for the equilibrium constant for the reaction represented by the equation

–AgCl( ) Ag ( ) Cl ( )s aq aq � �� . Is Kc > 1, < 1, or ≈ 1? Explain your answer. (b) Write the expression for the equilibrium constant for the reaction represented by the equation

2 –2Pb ( ) 2Cl ( ) PbCl ( )aq aq s � �� . Is Kc > 1, < 1, or ≈ 1? Explain your answer.

Solution (a) Kc = [Ag+][Cl–] < 1. AgCl is insoluble; thus, the concentrations of ions are much less than 1 M;

(b) 22

1 Pb Cl

cK

> 1 because PbCl2 is insoluble and formation of the solid will reduce

the concentration of ions to a low level (< 1 M). 10. Among the solubility rules previously discussed is the statement: Carbonates, phosphates, borates, and arsenates—except those of the ammonium ion and the alkali metals—are insoluble. (a) Write the expression for the equilibrium constant for the reaction represented by the equation

2 23 3CaCO ( ) Ca ( ) CO ( )s aq aq � �� . Is Kc > 1, < 1, or ≈ 1? Explain your answer.

(b) Write the expression for the equilibrium constant for the reaction represented by the equation 2 3–

4 3 4 23Ba ( ) 2PO ( ) Ba PO ( )aq aq s � �� . Is Kc > 1, < 1, or ≈ 1? Explain your answer. Solution (a) 2 2

3 Ca COcK < 1. CaCO3 is insoluble in water, and the concentrations of the ions

are much less than 1 M. (b) 3 22 34

1 Ba PO

cK

> 1. Ba3(PO4)2 is insoluble in water, and


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the concentrations of the ions are much less than 1 M. Division of small concentrations into 1 gives a number larger than 1. 11. Benzene is one of the compounds used as octane enhancers in unleaded gasoline. It is manufactured by the catalytic conversion of acetylene to benzene: 2 2 6 63C H ( ) C H ( )g g� �� . Which value of Kc would make this reaction most useful commercially? Kc ≈ 0.01, Kc ≈ 1, or Kc ≈ 10. Explain your answer. Solution

Since 6 63

2 2

[C H ] = [C H ]cK , a value of Kc ≈ 10 means that C6H6 predominates over C2H2. In such a

case, the reaction would be commercially feasible if the rate to equilibrium is suitable. 12. Show that the complete chemical equation, the total ionic equation, and the net ionic equation for the reaction represented by the equation 2 3KI( ) I ( ) KI ( )aq aq aq � �� give the same expression for the reaction quotient. KI3 is composed of the ions K+ and –

3I . Solution

The reaction quotient for the complete form of the equation is 3

2

[KI ] = [KI][I ]cQ .

The total ionic equation would be + – + –2 3K ( ) + I ( ) + I ( ) K ( ) + I ( )aq aq aq aq aq� �� . This would

lead to 2

3

[K ][I ][I ] = [K ][I ]cQ

, which simplifies to 2

3

[I ][I ] = [I ]cQ

when the potassium ion

concentration is cancelled out. The net ionic equation would be – –2 3I ( ) + I ( ) I ( )aq aq aq� ��

(since potassium is a spectator ion), for which 2

3

[I ][I ] = [I ]cQ

. Since 3 3[KI ] = [I ] and [KI] = [I–],

all three expressions are equivalent. 13. For a titration to be effective, the reaction must be rapid and the yield of the reaction must essentially be 100%. Is Kc > 1, < 1, or ≈ 1 for a titration reaction? Solution Kc > 1; the product must be formed in overwhelmingly large proportions. 14. For a precipitation reaction to be useful in a gravimetric analysis, the product of the reaction must be insoluble. Is Kc > 1, < 1, or ≈ 1 for a useful precipitation reaction? Solution Kc is the product of all reaction products divided by the product of the reactants, with all concentrations of reactants and products raised to their respective stoichiometric powers. As the precipitate is formed, the amount of its corresponding ions in solution declines to lower levels. Because the value of the numerator is 1 and the concentrations in the denominator are very small, the value of Kc will be much greater than 1 for a useful precipitation reaction. 15. Write the mathematical expression for the reaction quotient, Qc, for each of the following reactions: (a) 4 2 3CH ( ) Cl ( ) CH Cl( ) HCl( )g g g g � ��

(b) 2 2N ( ) O ( ) 2NO( )g g g � ��

(c) 2 2 32SO ( ) O ( ) 2SO ( )g g g � ��


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(d) 3 2BaSO ( ) BaO( ) SO ( )s s g� �� (e) 4 2 4 10P ( ) 5O ( ) P O ( )g g s � �� (f) 2Br ( ) 2Br( )g g� �� (g) 4 2 2 2CH ( ) 2O ( ) CO ( ) 2H O( )g g g l � �� (h) 4 2 4 2CuSO 5H O( ) CuSO ( ) 5H O( )s s g� �� Solution

(a)

3

4 2

CH Cl HCl =

CH ClcQ ; (b)

2

2 2

NO =

N OcQ ; (c)

23

22 2

SO =

SO OcQ ; (d) Qc= [SO2]; (e)

54 2

1 = P O

cQ ; (f)

2

2

Br =

BrcQ ; (g)

2

24 2

CO =

CH OcQ ; (h) Qc= [H2O]5

16. Write the mathematical expression for the reaction quotient, Qc, for each of the following reactions: (a) 2 2 3N ( ) 3H ( ) 2NH ( )g g g � �� (b) 3 2 24NH ( ) 5O ( ) 4NO( ) 6H O( )g g g g � �� (c) 2 4 2N O ( ) 2NO ( )g g� �� (d) 2 2 2CO ( ) H ( ) CO( ) H O( )g g g g � �� (e) 4 3NH Cl( ) NH ( ) HCl( )s g g� �� (f) 3 2 22

2Pb NO ( ) 2PbO( ) 4NO ( ) O ( )s s g g � ��

(g) 2 2 22H ( ) O ( ) 2H O( )g g l � �� (h) 8S ( ) 8S( )g g� �� Solution

Answer: (a)

23

32 2

NHH N

cQ ; (b)

4 62

4 53 2

NO H ONH O

cQ ; (c)

22

2 4

NON OcQ ; (d)

2

2 2

CO H OCO HcQ ;

(e) Qc= [NH3][HCl], (f) Qc= [NO2]4[O2]; (g) 2

2 2

1H O

cQ ; (h)

8

8

SScQ

17. The initial concentrations or pressures of reactants and products are given for each of the following systems. Calculate the reaction quotient and determine the direction in which each system will proceed to reach equilibrium. (a) 3 2 22NH ( ) N ( ) 3H ( ) 17cg g g K � �� ; [NH3] = 0.20 M, [N2] = 1.00 M, [H2] = 1.00 M (b) 4

3 2 22NH ( ) N ( ) 3H ( ) 6.8 10Pg g g K � �� ; NH3 = 3.0 atm, N2 = 2.0 atm, H2 = 1.0 atm (c) 3 2 22SO ( ) 2SO ( ) O 0.230cg g g K � �� ; [SO3] = 0.00 M, [SO2] = 1.00 M, [O2] = 1.00 M (d) 3 2 22SO ( ) 2SO ( ) O ( ) = 16.5Pg g g K� �� ; SO3 = 1.00 atm, SO2 = 1.00 atm, O2 =


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1.00 atm (e) 4

22NO( ) Cl ( ) 2NOCl( ) 4.6 10cg g g K � �� ; [NO] = 1.00 M, [Cl2] = 1.00 M, [NOCl] = 0 M (f) 2 2N ( ) O ( ) 2NO( ) 0.050Pg g g K � �� ; NO = 10.0 atm, N2 = O2 = 5 atm Solution

(a) 3 3

2 22 2

3

[N ][H ] (1.00)(1.00) 25[NH ] (0.20)cQ

Qc > Kc, proceeds left;

(b) 2 2

3

3 3N H

2 2NH

( ) (2.0)(1.0) 0.22( ) (3.0)P

P PQ

P

QP < KP, proceeds right;

(c) 2 2

2 22

3

[SO ] [O ] (1.00) (1.00) undefined[SO ] (0)cQ

Qc > Kc, proceeds left;

(d) 2 2

3

2 2SO O

2 2SO

( ) (1.00) (1.00) 1.00( ) (1.00)P

P PQ

P

QP< KP, proceeds right;

(e) 2

2 2NOCl

2 2NO Cl

( ) (0) 0( ) (1.00) (1.00)P

PQP P

QP < KP, proceeds right;

(f) 2 2

2 2

[NO] (10.0) 4[N ][O ] (5.00)(5.00)cQ

Qc>Kc, proceeds left 18. The initial concentrations or pressures of reactants and products are given for each of the following systems. Calculate the reaction quotient and determine the direction in which each system will proceed to reach equilibrium. (a) 3 2 22NH ( ) N ( ) 3H ( ) 17cg g g K � �� ; [NH3] = 0.50 M, [N2] = 0.15 M, [H2] = 0.12 M (b) 4

3 2 22NH ( ) N ( ) 3H ( ) 6.8 10Pg g g K � �� ; NH3 = 2.00 atm, N2 = 10.00 atm, H2 = 10.00 atm

(c) 3 2 22SO ( ) 2SO ( ) O ( ) 0.230cg g g K � �� ; [SO3] = 2.00 M, [SO2] = 2.00 M, [O2] = 2.00 M (d) 3 2 22SO ( ) 2SO ( ) O ( ) 6.5 atmPg g g K � �� ; SO2 = 1.00 atm, O2 = 1.130 atm, SO3 = 0 atm (e) 3

22NO( ) Cl ( ) 2NOCl( ) 2.5 10Pg g g K � �� ; NO = 1.00 atm, Cl2 = 1.00 atm, NOCl = 0 atm (f) 2 2N ( ) O ( ) 2NO( ) 0.050cg g g K � �� ; [N2] = 0.100 M, [O2] = 0.200 M, [NO] = 1.00 M Solution


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(a) 3 3

32 22 2

3

[N ][H ] (0.15)(0.12) 1.0 10[NH ] (0.50)cQ ; Qc < Kc, proceeds right;

(b) 2 2

3

3 3N H 3

2 2NH

( ) (10.00)(10.00) 2.50 10( ) (2.00)P

P PQ

P ; QP < KP, proceeds right;

(c) 2 2

2 22 2

3

[SO ] [O ] (2.00) (2.00) 2.00[SO ] (2.00)cQ ; Qc > Kc, proceeds left;

(d) 2 2

3

2 2SO O

2 2SO

( ) (1.00) (1.130) undefined but very large( ) (0)P

P PQ

P ; QP > KP, proceeds left;

(e) 2

2NOCl

2 2NO Cl

(0) 0( ) (1.00) (1.00)P

PQP P

;QP < KP, proceeds right;

(f) 2 2

2 2

[NO] (1.00) 50[N ][O ] (0.100)(0.200)cQ ; Qc > Kc, proceeds left

19. The following reaction has KP = 4.50 10–5 at 720 K.

2 2 3N ( ) + 3H ( ) 2NH ( )g g g� �� If a reaction vessel is filled with each gas to the partial pressures listed, in which direction will it shift to reach equilibrium? P(NH3) = 93 atm, P(N2) = 48 atm, and P(H2) = 52 atm Solution

The reaction quotient expression for this problem is 3

2 2

2NH

3N H

( ) =

( )( )p

PQ

P P. Plugging in the given

values of partial pressures gives

2

3

(93)(48) (52)

, so Qp = 1.3 10–3. Since this value is larger

than KP (4.50 10–5), the system will shift toward the reactants to reach equilibrium. 20. Determine if the following system is at equilibrium. If not, in which direction will the system need to shift to reach equilibrium?

2 2 2 2SO Cl ( ) SO ( ) + Cl ( ) g g g� �� [SO2Cl2] = 0.12 M, [Cl2] = 0.16 M and [SO2] = 0.050 M. Kc for the reaction is 0.078. Solution

The reaction quotient expression is 2 2

2 2

[SO ][Cl ] = [SO Cl ]cQ . Plugging in the values provided gives

(0.050)(0.16) = = 0.067(0.12)cQ . The value of Qc is lower than the value of Kc, so this reaction will

shift toward the products. 21. Which of the systems described in Exercise 15 are homogeneous equilibria? Which are heterogeneous equilibria? Solution (a) homogenous; (b) homogenous; (c) homogenous; (d) heterogeneous; (e) heterogeneous; (f) homogenous; (g) heterogeneous; (h) heterogeneous


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22. Which of the systems described in Exercise 16 are homogeneous equilibria? Which are heterogeneous equilibria? Solution (a) homogenous; (b) homogenous; (c) homogenous; (d) homogenous; (e) heterogeneous; (f) heterogeneous; (g) heterogeneous; (h) homogenous 23. For which of the reactions in Exercise 15 does Kc (calculated using concentrations) equal KP (calculated using pressures)? Solution When the number of gaseous components are the same on both sides of the equilibrium expression, Kc will equal KP. This situation occurs in (a) and (b). 24. For which of the reactions in Exercise 16 does Kc (calculated using concentrations) equal KP (calculated using pressures)? Solution Only (d) has the same number of gaseous reactants as gaseous products, which makes Δg = 0 in the expression ( ) n

p cK K RT 25. Convert the values of Kc to values of KP or the values of KP to values of Kc. (a) 2 2 3N ( ) 3H ( ) 2NH ( ) 0.50 at 400 Ccg g g K � ��

(b) 2 2H I 2HI 50.2 at 448 CcK � �� (c) –25

2 4 2 2 4 2Na SO •10H O( ) Na SO ( ) 10H O( ) 4.08 10 at 25 CPs s g K � �� (d) 2 2H O( ) H O( ) 0.122 at 50 CPl g K � �� Solution

( ) nP cK K RT , where Δn is the sum of gaseous products minus the sum of gaseous reactants.

(a) Δn = (2) – (1 + 3) = –2, KP = 0.50[0.08206 673.15]–2 = 1.6 10–4; (b) Δn = (2) – (1 + 1) = 0, KP = Kc(RT)0 = Kc= 50.2; (c) Δn = (10) – (0) = 10, Kc = KP(RT)–Δn, Kc = 4.08 10–25[0.08206 298.15]–10 = 5.34 10–39; (d) Δn = (1) – (0) = 1, Kc = 0.122(0.08206 323.15)–1 = 4.60 10–

3 26. Convert the values of Kc to values of KP or the values of KP to values of Kc. (a) –2

2 2Cl ( ) Br ( ) 2BrCl( ) 4.7 10 at 25 Ccg g g K � �� (b) 2 2 32SO ( ) O ( ) 2SO ( ) = 48.2 at 500 CPg g g K � �� (c) –44

2 2 2 2CaCl • 6H O( ) CaCl ( ) 6H O( ) 5.09 10 at 25 CPs s g K � �� (d) 2 2H O( ) H O( ) 0.196 at 60 CPl g K � �� Solution KP = Kc(RT)Δn, where Δn is the sum of gaseous products minus the sum of gaseous reactants. (a) Δn = (2) – (1 + 1) = 0, KP = Kc(RT)0 = Kc = 4.7 10–2; (b) Δn = (2) – (2 + 1) = –1, Kc = KP(RT)–

(–1) = 48.2(0.08206)(773.15) = 3.06 103 (c) Δn = (6) – (0 + 0) = 6, Kc = KP(RT)–Δn = 5.09 10–

44(0.08206 298.15)–6 = 2.37 10–52; (d) Δn = (1) – (0) – 1, Kc = KP(RT)–1 = 0.196(0.08206 3.3315)–1 = 7.17 10–3 27. What is the value of the equilibrium constant expression for the change

2 2H O( ) H O( )l g� �� at 30 C? (see Appendix E) Solution


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The equilibrium expression for this transformation is 2H O PK P . The vapor pressure of H2O at

30 ºC is 31.8 torr. Converting to atmospheres gives 31.8 torr 1 atm 760 torr

0.042 atm .

Therefore, 2H O 0.042PK P .

28. Write the expression of the reaction quotient for the ionization of HOCN in water. Solution The reaction is

2 3H O( ) HOCN( ) H O ( ) OCN ( )l aq aq aq � �� Because H2O is in the condensed phase, its concentration is constant at a given temperature;

thus, we would write the reaction quotient for this reaction as 3[H O ][OCN ] [HOCN]cQ

29. Write the reaction quotient expression for the ionization of NH3 in water. Solution

+3 2 4NH ( ) H O( ) NH ( ) OH ( )aq l aq aq � �� . Because the concentration of water is a

constant, the term [H2O] is normally incorporated into the reaction quotient as well as the final

equilibrium constant. 4

3

[NH ][OH ] [NH ]cQ

30. What is the approximate value of the equilibrium constant KP for the change 2 5 2 5 2 5 2 5C H OC H ( ) C H OC H ( ) l g� �� at 25 °C. (The equilibrium vapor pressure for this

substance is 570 torr at 25 °C.) Solution At 25 °C, the equilibrium may be expressed most easily in terms of pressure. There is no term for the liquid ether, C2H5OC2H5, because the activity of the pure liquid is 1.The vapor pressure of the ether is about 570 torr = 0.75 atm, so

2 5 2 5C H OC H 0.75 atmPK P


13.3: Shifting Equilibria: Le Châtelier’s Principle 31. The following equation represents a reversible decomposition:

3 2CaCO ( ) CaO( ) CO ( )s s g� �� Under what conditions will decomposition in a closed container proceed to completion so that no CaCO3 remains? Solution The amount of CaCO3 must be so small that

2COP is less than KP when the CaCO3 has completely decomposed. In other words, the starting amount of CaCO3 cannot completely generate the full

2COP required for equilibrium. 32. Explain how to recognize the conditions under which changes in volume will affect gas-phase systems at equilibrium. Solution An effect arises only when the number of moles of gaseous reactants in the equation that describes the equilibrium reaction differs from the number of moles of gaseous products. As the


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volume of a gaseous system decreases, the gases become compressed and the total number of molecules per unit volume increases, producing greater molar concentrations. This change in concentration represents a stress on the system, and the system will act to remove the stress, if possible. The pressure change will have an effect on the equilibrium only if the number of moles of gaseous reactants differs from the number of moles of gaseous products, shifting the condition of equilibrium to the side with the fewer number of moles. 33. What property of a reaction can we use to predict the effect of a change in temperature on the value of an equilibrium constant? Solution The change in enthalpy may be used. If the reaction is exothermic, the heat produced can be thought of as a product. If the reaction is endothermic the heat added can be thought of as a reactant. Additional heat would shift an exothermic reaction back to the reactants but would shift an endothermic reaction to the products. Cooling an exothermic reaction causes the reaction to shift toward the product side; cooling an endothermic reaction would cause it to shift to the reactants’ side. 34. The following reaction occurs when a burner on a gas stove is lit:

4 2 2 2CH ( ) 2O ( ) CO ( ) 2H O( )g g g g � �� Is an equilibrium among CH4, O2, CO2, and H2O established under these conditions? Explain your answer. Solution No, it is not at equilibrium. Because the system is not confined, products continuously escape from the region of the flame; reactants are also added continuously from the burner and surrounding atmosphere. 35. A necessary step in the manufacture of sulfuric acid is the formation of sulfur trioxide, SO3, from sulfur dioxide, SO2, and oxygen, O2, shown here. At high temperatures, the rate of formation of SO3 is higher, but the equilibrium amount (concentration or partial pressure) of SO3 is lower than it would be at lower temperatures.

2 2 32SO ( ) O ( ) 2SO ( )g g g (a) Does the equilibrium constant for the reaction increase, decrease, or remain about the same as the temperature increases? (b) Is the reaction endothermic or exothermic? Solution (a) For the yield to decrease from that obtained at a lower temperature, the value of K must decrease with an increase in temperature. (b) The observation that K decreases with an increase in temperature indicates the reaction is exothermic. 36. Suggest four ways in which the concentration of hydrazine, N2H4, could be increased in an equilibrium described by the following equation:

2 2 2 4N ( ) 2H ( ) N H ( ) 95 kJg g g H � �� Solution Add N2; add H2; decrease the container volume; heat the mixture. 37. Suggest four ways in which the concentration of PH3 could be increased in an equilibrium described by the following equation:

4 2 3P ( ) 6H ( ) 4PH ( ) 110.5 kJg g g H � �� Solution Add P4; add H2; decrease the container volume; heat the mixture.


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38. How will an increase in temperature affect each of the following equilibria? How will a decrease in the volume of the reaction vessel affect each? (a) 3 2 22NH ( ) N ( ) 3H ( ) 92 kJg g g H � �� (b) 2 2N ( ) O ( ) 2NO( ) 181 kJg g g H � �� (c) 3 22O ( ) 3O ( ) – 285 kJg g H � �� (d) 2 3CaO( ) CO ( ) CaCO ( ) –176 kJs g s H � �� Solution (a) T increase = shift right, V decrease = shift left; (b) T increase = shift right, V decrease = no effect; (c) T increase = shift left, V decrease = shift left; (d) T increase = shift left, V decrease = shift right. 39. How will an increase in temperature affect each of the following equilibria? How will a decrease in the volume of the reaction vessel affect each? (a) 2 2 22H O( ) 2H ( ) O ( ) = 484 kJg g g H � �� (b) 2 2 3N ( ) 3H ( ) 2NH ( ) = – 92.2 kJg g g H � �� (c) 22Br( ) Br ( ) – 224 kJg g H � �� (d) 2 2H ( ) I ( ) 2HI( ) 53 kJg s g H � �� Solution (a) T increase = shift right, V decrease = shift left; (b) T increase = shift left, V decrease = shift right; (c) T increase = shift left, V decrease = shift right; (d) T increase = shift right, V decrease = shift left. 40. Methanol can be prepared from carbon monoxide and hydrogen at high temperature and pressure in the presence of a suitable catalyst. (a) Write the expression for the equilibrium constant (Kc) for the reversible reaction

2 32H ( ) CO( ) CH OH( ) – 90.2 kJg g g H � �� (b) What will happen to the concentrations of H2, CO, and CH3OH at equilibrium if more H2 is added? (c) What will happen to the concentrations of H2, CO, and CH3OH at equilibrium if CO is removed? (d) What will happen to the concentrations of H2, CO, and CH3OH at equilibrium if CH3OH is added? (e) What will happen to the concentrations of H2, CO, and CH3OH at equilibrium if the temperature of the system is increased? (f) What will happen to the concentrations of H2, CO, and CH3OH at equilibrium if more catalyst is added? Solution

(a) 32

2

[CH OH] [H ] [CO]cK ; (b) [H2] increases, [CO] decreases, [CH3OH] increases; (c) [H2]

increases, [CO] decreases, [CH3OH] decreases; (d) [H2] increases, [CO] increases, [CH3OH] increases; (e) [H2] increases, [CO] increases, [CH3OH] decreases; (f) no changes. 41. Nitrogen and oxygen react at high temperatures. (a) Write the expression for the equilibrium constant (Kc) for the reversible reaction

2 2N ( ) O ( ) 2NO( ) 181 kJg g g H � ��


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(b) What will happen to the concentrations of N2, O2, and NO at equilibrium if more O2 is added? (c) What will happen to the concentrations of N2, O2, and NO at equilibrium if N2 is removed? (d) What will happen to the concentrations of N2, O2, and NO at equilibrium if NO is added? (e) What will happen to the concentrations of N2, O2, and NO at equilibrium if the volume of the reaction vessel is decreased? (f) What will happen to the concentrations of N2, O2, and NO at equilibrium if the temperature of the system is increased? (g) What will happen to the concentrations of N2, O2, and NO at equilibrium if a catalyst is added? Solution

(a) 2

2 2

[NO] [N ][O ]cK ; (b) [N2] decreases, [O2] increases, [NO] increases; (c) [N2] decreases, [O2]

increases, [NO] decreases; (d) [N2] increases, [O2] increases, [NO] increases; (e) [N2] increases, [O2] increases, [NO] increases(same moles in decreased volume); (f) [N2] decreases, [O2] decreases, [NO] increases; (g) no changes occur. 42. Water gas, a mixture of H2 and CO, is an important industrial fuel produced by the reaction of steam with red hot coke, essentially pure carbon. (a) Write the expression for the equilibrium constant for the reversible reaction

2 2C( ) H O( ) CO( ) H ( ) 131.30 kJs g g g H � �� (b) What will happen to the concentration of each reactant and product at equilibrium if more C is added? (c) What will happen to the concentration of each reactant and product at equilibrium if H2O is removed? (d) What will happen to the concentration of each reactant and product at equilibrium if CO is added? (e) What will happen to the concentration of each reactant and product at equilibrium if the temperature of the system is increased? Solution

(a) 2

2

[CO][H ] [H O]cK ; (b) [H2O] no change, [CO] no change, [H2] no change; (c) [H2O]

decreases, [CO] decreases, [H2] decreases; (d) [H2O] increases, [CO] increases, [H2] decreases; (e) [H2O] decreases, [CO] increases, [H2] increases. In (b), (c), (d), and (e), the mass of carbon will change, but its concentration (activity) will not change. 43. Pure iron metal can be produced by the reduction of iron(III) oxide with hydrogen gas. (a) Write the expression for the equilibrium constant (Kc) for the reversible reaction

2 3 2 2Fe O ( ) 3H ( ) 2Fe( ) 3H O( ) 98.7 kJs g s g H � �� (b) What will happen to the concentration of each reactant and product at equilibrium if more Fe is added? (c) What will happen to the concentration of each reactant and product at equilibrium if H2O is removed? (d) What will happen to the concentration of each reactant and product at equilibrium if H2 is added? (e) What will happen to the concentration of each reactant and product at equilibrium if the volume of the reaction vessel is decreased?


Page 12 of 31

(f) What will happen to the concentration of each reactant and product at equilibrium if the temperature of the system is increased? Solution

(a) 3

23

2

[H O] [H ]cK ; (b) no changes occur; (c) [H2] decreases, [H2O] decreases; (d) [H2] increases,

[H2O] increases; (e) [H2] increases, [H2O] increases (concentrations rise due to decreased volume, but there is no shift in the equilibrium); (f) [H2] decreases, [H2O] increases. In (b), (c), (d), (e), and (f), the mass of Fe will change, but its concentration (activity) will not change. 44. Ammonia is a weak base that reacts with water according to this equation:

+ –3 2 4NH ( ) H O( ) NH ( ) OH ( )aq l aq aq � ��

Will any of the following increase the percent of ammonia that is converted to the ammonium ion in water? (a) Addition of NaOH (b) Addition of HCl (c) Addition of NH4Cl Solution Only (b). In (a), addition of a strong base forces the equilibrium toward forming more NH3(aq). In (b), the addition of HCl consumes OH–, shifting the equilibrium right and increasing the conversion of NH3 to NH4

+. In (c), 4NH ion causes the equilibrium to shift to the left, forming more NH3(aq). 45. Acetic acid is a weak acid that reacts with water according to this equation:

3 2 2 3 3 2CH CO H( ) H O( ) H O ( ) CH CO ( )aq aq aq aq � �� Will any of the following increase the percent of acetic acid that reacts and produces 3 2CH CO ion? (a) Addition of HCl (b) Addition of NaOH (c) Addition of NaCH3CO2 Solution Only (b). In (a), the addition of HCl increase the H3O+ concentration and shifts the equilibrium to the left, decreasing the acetate ion concentration and increasing the acetic acid concentration. In (b), the addition of a strong base consumes H3O+ and shifts the equilibrium to the right, forming acetate ion. In (c), addition of the acetate ion shifts the equilibrium left, increasing the concentration of acetic acid. 46. Suggest two ways in which the equilibrium concentration of Ag+ can be reduced in a solution of Na+, Cl–, Ag+, and 3NO , in contact with solid AgCl.

–3 3Na ( ) Cl ( ) Ag ( ) NO ( ) AgCl( ) Na ( ) NO ( )aq aq aq aq s aq aq � ��

– 65.9 kJH Solution Add NaCl or some other salt that produces Cl– to the solution. Cooling the solution forces the equilibrium to the right, precipitating more AgCl(s). 47. How can the pressure of water vapor be increased in the following equilibrium?

2 2H O( ) H O( ) 41 kJl g H � �� Solution


Page 13 of 31

Add heat or reduce the pressure. 48. A solution is saturated with silver sulfate and contains excess solid silver sulfate: Ag2SO4(s) ⇌ 2Ag+(aq) + SO4

2−(aq) A small amount of solid silver sulfate containing a radioactive isotope of silver is added to this solution. Within a few minutes, a portion of the solution phase is sampled and tests positive for radioactive Ag+ ions. Explain this observation. Solution (a) Though the solution is saturated, the dynamic nature of the solubility equilibrium means the opposing processes of solid dissolution and precipitation continue to occur (just at equal rates, meaning the dissolved ion concentrations and the amount of undissolved solid remain constant). The radioactive Ag+ ions detected in the solution phase come from dissolution of the added solid, and their presence is countered by precipitation of nonradioactive Ag+. 49. The amino acid alanine has two isomers, α-alanine and β-alanine. When equal masses of these two compounds are dissolved in equal amounts of a solvent, the solution of α-alanine freezes at the lowest temperature. Which form, α-alanine or β-alanine, has the larger equilibrium constant for ionization –(HX H X ) � �� ? Solution The freezing point depression is proportional to the number of particles produced in a solvent. For a weak electrolyte, T = ikfm, where i is the number of ions produced from a solute. Because both isomers have identical molecular masses and are dissolved in the same amount of solvent, kf and m are constants. Any difference in the reduction of the freezing point, therefore, must reflect a difference in the degree of ionization, i, of the two forms of alanine into fragments—namely, a hydrogen ion and the anion. A greater number of ions will be produced by the form with the larger equilibrium constant, which results in a lower freezing point for that species. Since α-alanine has a lower freezing point and, consequently, the larger freezing point depression, it must have the larger number of ions in solution and has the larger value of Kc.


13.4: Equilibrium Calculations 50. A reaction is represented by this equation:

3A( ) 2B( ) 2C( ) 1 10caq aq aq K � �� (a) Write the mathematical expression for the equilibrium constant. (b) Using concentrations ≤ 1 M, identify two sets of concentrations that describe a mixture of A, B, and C at equilibrium. Solution

2

2

[C] A BcK . There are many different sets of equilibrium concentrations; two are [A] = 0.1

M, [B] = 0.1 M, [C] = 1 M; and [A] = 0.01, [B] = 0.250, [C] = 0.791. 51. A reaction is represented by this equation:

–42W( ) X( ) 2Y( ) 5 10caq aq aq K � �� (a) Write the mathematical expression for the equilibrium constant. (b) Using concentrations of ≤ 1 M, identify two sets of concentrations that describe a mixture of W, X, and Y at equilibrium. Solution


Page 14 of 31

(a) 2

2

[X][Y] [W]cK . (b) Two of the many possible sets of concentrations are [X] = 0.05 M, [Y] =

0.015 M, [W] = 0.15 M; and [X] = 0.050 M, [Y] = 0.01 M, [W] = 0.10 M 52. What is the value of the equilibrium constant at 500 C for the formation of NH3 according to the following equation?

2 2 3N ( ) 3H ( ) 2NH ( )g g g � �� An equilibrium mixture of NH3(g), H2(g), and N2(g) at 500 C was found to contain 1.35 M H2, 1.15 M N2, and 4.12 10–1M NH3. Solution The reaction may be written as

2 2 3N ( ) 3H ( ) 2NH ( )g g g � �� The equilibrium constant for the reaction is

2 1 2 23

3 3 32 2

[NH ] (4.12 10 ) (0.170) [N ][H ] (1.15)(1.35) (1.15)(2.46)cK

= 0.0600 = 6.00 10–2 53. Hydrogen is prepared commercially by the reaction of methane and water vapor at elevated temperatures.

4 2 2CH ( ) H O( ) 3H ( ) CO( )g g g g � �� What is the equilibrium constant for the reaction if a mixture at equilibrium contains gases with the following concentrations: CH4, 0.126 M; H2O, 0.242 M; CO, 0.126 M; H2 1.15 M, at a temperature of 760 C? Solution

32

4 2

[H ] [CO] [1.15] [0.126] 6.28[CH ][H O] [0.126][0.242]cK

54. A 0.72-mol sample of PCl5 is put into a 1.00-L vessel and heated. At equilibrium, the vessel contains 0.40 mol of PCl3(g) and 0.40 mol of Cl2(g). Calculate the value of the equilibrium constant for the decomposition of PCl5 to PCl3 and Cl2 at this temperature. Solution The decomposition of PCl5 to PCl3 and Cl2 is given as

5 3 2

3 2

5

PCl ( ) PCl ( ) Cl ( )[PCl ][Cl ]

[PCl ]c

g g g

K

� ��

Let x = change in [PCl5].

(0.40)(0.40) 0.50

(0.32)cK


Page 15 of 31

55. At 1 atm and 25 C, NO2 with an initial concentration of 1.00 M is 0.0033% decomposed into NO and O2. Calculate the value of the equilibrium constant for the reaction.

2 22NO ( ) 2NO( ) O ( )g g g� �� Solution The decomposition forms 3.3 10−5 1.00 M = 3.3 10−5 M NO and 1.65 10−5 M O2. The amount of NO2 remaining is 1.00 M – 3.3 10−5 M ≈ 1.00 M

5 2 5142

2 22

[NO][O ] (3.3 10 ) (1.65 10 ) 1.8 10[NO ] (1.00 )c

M MKM

56. Calculate the value of the equilibrium constant KP for the reaction

22NO( ) Cl ( ) 2NOCl( )g g g � �� from these equilibrium pressures: NO, 0.050 atm; Cl2, 0.30 atm; NOCl, 1.2 atm. Solution

2 23

2 2 32

[NOCl] (1.2) 1.44 1.9 10[NO] [Cl ] (0.050) (0.30) (2.5 10 )(0.30)PK

57. When heated, iodine vapor dissociates according to this equation:

2I ( ) 2I( )g g� �� At 1274 K, a sample exhibits a partial pressure of I2 of 0.1122 atm and a partial pressure due to I atoms of 0.1378 atm. Determine the value of the equilibrium constant, KP, for the decomposition at 1274 K. Solution

2

2 2I

I

( ) (0.1378) 0.16920.1122P

PKP

58. A sample of ammonium chloride was heated in a closed container.

4 3NH Cl( ) NH ( ) HCl( )s g g� �� At equilibrium, the pressure of NH3(g) was found to be 1.75 atm. What is the value of the equilibrium constant KP for the decomposition at this temperature? Solution Because the decomposition must generate the same pressure of HCl as NH3, 1.75 atm of HCl must be present.

3p NH HCl (1.75 atm)(1.75 atm) 3.06K P P 59. At a temperature of 60 C, the vapor pressure of water is 0.196 atm. What is the value of the equilibrium constant KP for the vaporization equilibrium at 60 C?

2 2H O( ) H O( )l g� �� Solution In heterogeneous equilibrium systems such as this one, the concentration of the condensed phase is constant at a given temperature. Thus, the equilibrium expression for the transformation

2 2H O( ) H O( )l g� �� is written as 2H OPK P . At 60 ºC, the pressure of water vapor at

equilibrium with liquid water is 0.196 atm; therefore, the equilibrium constant is KP = 0.196. 60. Complete the changes in concentrations (or pressure, if requested) for each of the following reactions. (a)


Page 16 of 31

3 2 22SO ( ) 2SO ( ) + O ( )

___ ___ +___ ___ 0.125

g g g

xM

� ��

(b)

3 2 2 24NH ( ) + 3O ( ) 2N ( ) + 6H O( )

___ 3 ___ ______ 0.24 ___ ___

g g g g

xM

� ��

(c) Change in pressure:

4 2 2 22CH ( ) C H ( ) + 3H ( )

___ ______ 25 torr ___

g g g

x

� ��

(d) Change in pressure:

4 2 2CH ( ) + H O( ) CO(g) + 3H ( )

___ ___ ______ 5 atm ___ ___

g g g

x

� ��

(e)

4 3

4

NH Cl( ) NH ( ) + HCl( )

___ 1.03 10 ___

s g g

xM

� ��

(f) change in pressure:

4Ni( ) + 4CO( ) Ni(CO) ( )

4 ___ 0.40 atm ___

s g g

x

� ��

Solution (a) –2x, 2x, –0.250 M, 0.250 M; (b) 4x, –2x, –6x, 0.32 M, –0.16 M, –0.48 M; (c) –2x, 3x, –50 torr, 75 torr; (d) x, – x, –3x, 5 atm, –5 atm, –15 atm; (e) x, 1.03 10–4M; (f)x, 0.1 atm. 61. Complete the changes in concentrations (or pressure, if requested) for each of the following reactions. (a)

2 2 22H ( ) + O ( ) 2H O( )

___ ___ +2___ ___ 1.50

g g g

xM

� ��

(b)

2 2 4 2CS ( ) + 4H ( ) CH ( ) + 2H S( )

___ ___ ___0.020 ___ ___ ___

g g g g

xM

� ��


Page 17 of 31

(c) Change in pressure:

2 2H ( ) + Cl ( ) 2HCl( )

___ ___1.50 atm ___ ___

g g g

x

� ��

(d) Change in pressure:

3 2 2 22NH ( ) + 2O ( ) N O( ) + 3H O( )

___ ___ ___ ___ ___ ___ 60.6 torr

g g g g

x

� ��

(e)

4 3 2

6

NH HS( ) NH ( ) + H S( )

___ 9.8 10 ___

s g g

xM

� ��

(f) Change in pressure:

5Fe( ) + 5CO( ) Fe(CO) ( )

___ ___ 0.012 atm

s g g

x

� ��

Solution (a) –2x, –x, –1.50 M, –0.75 M; (b) 4x, –x, –2x, 0.080 M, –0.020 M, –0.040 M; (c) x, –2x, 1.50

atm, –3.00 atm; (d) 2 3

x , 23

x , 13

x , –40.4 torr, –40.4 torr, 20.2 torr; (e) x, 9.8 10–6M; (f) –

5x, –0.060 atm. 62. Why are there no changes specified for Ni in Exercise 60, part (f)? What property of Ni does change? Solution Activities of pure crystalline solids equal 1 and are constant; however, the mass of Ni does change. 63. Why are there no changes specified for NH4HS in Exercise 61, part (e)? What property of NH4HS does change? Solution Activities of pure crystalline solids equal 1 and are constant; however, the mass of NH4HS does change. 64. Analysis of the gases in a sealed reaction vessel containing NH3, N2, and H2 at equilibrium at 400C established the concentration of N2 to be 1.2 M and the concentration of H2 to be 0.24 M.

2 2 3N ( ) 3H ( ) 2NH ( ) 0.50 at 400 °Ccg g g K � �� Calculate the equilibrium molar concentration of NH3. Solution Write the equilibrium constant expression and solve for [NH3].

2 23 3

3 32 2

[NH ] [NH ] 0.50[N ][H ] [1.2][0.24]cK

[NH3]2 = 1.2 (0.24)3 0.50 = 0.0083


Page 18 of 31

[NH3] = 9.1 10–2 M 65. Calculate the number of moles of HI that are at equilibrium with 1.25 mol of H2 and 1.25 mol of I2 in a 5.00–L flask at 448 C.

2 2H I 2HI 50.2 at 448 °CcK � �� Solution Write the equilibrium constant expression, and then solve for [HI]. From [HI], determine the moles of HI present.

2 2

2 2

[HI] [HI] 50.21.25 1.25[H ][I ]5.00 5.00

cK

[HI]2 = (0.250)2 50.2 = 3.14 [HI] = 1.77 M 5 L 1.77 M = 8.85 mol HI 66. What is the pressure of BrCl in an equilibrium mixture of Cl2, Br2, and BrCl if the pressure of Cl2 in the mixture is 0.115 atm and the pressure of Br2 in the mixture is 0.450 atm?

–22 2Cl ( ) Br ( ) 2BrCl( ) 4.7 10Pg g g K � ��

Solution Write the equilibrium constant expression and solve for PBrCl.

2 2

2 22BrCl BrCl

Cl Br

2 2 3BrCl

( ) ( ) 4.7 10(0.115)(0.450)

( ) 0.115 0.450 4.7 10 2.43 10 atm

PP PK

P P

P

PBrCl = 4.9 10–2atm 67. What is the pressure of CO2 in a mixture at equilibrium that contains 0.50 atm H2, 2.0 atm of H2O, and 1.0 atm of CO at 990 C?

2 2 2H ( ) CO ( ) H O( ) CO( ) 1.6 at 990 °CPg g g g K � �� Solution Write the equilibrium constant expression and solve for

2COP .

2

2 2 2

2

H O CO

H CO CO

CO

(2.0)(1.0) 1.6(0.50)

2.0 1.0 2.5 atm0.50 1.6

P

P PK

P P P

P

68. Cobalt metal can be prepared by reducing cobalt(II) oxide with carbon monoxide. 2

2CoO( ) CO( ) Co( ) CO ( ) 4.90 10 at 550 Ccs g s g K � �� What concentration of CO remains in an equilibrium mixture with [CO2] = 0.100 M? Solution Write the equilibrium constant expression and solve for [CO].

22

42

[CO ] 0.100 4.90 10[CO] [CO]

0.100[CO] = 2.04 10 4.90 10

cK

M


Page 19 of 31

69. Carbon reacts with water vapor at elevated temperatures.

2 2C( ) H O( ) CO( ) H ( ) 0.2 at 1000 Ccs g g g K � �� Assuming a reaction mixture initially contains only reactants, what is the concentration of CO in an equilibrium mixture with [H2O] = 0.500 M at 1000 C? Solution

From the equilibrium constant expression, 2 2

2

[CO][H ] [CO][H ] 0.2[H O] 0.500cK . Since [CO]

must equal [H2], let the concentrations of both CO and H2 be x. x2 = 0.500 0.2 = 0.01; x = [CO] = [H2] = 0.3 M. 70. Sodium sulfate 10–hydrate, Na2SO4•10H2O, dehydrates according to the equation

–252 4 2 2 4 2Na SO •10H O( ) Na SO ( ) 10H O( ) 4.08 10 at 25 CPs s g K � ��

What is the pressure of water vapor at equilibrium with a mixture of Na2SO4•10H2O and NaSO4? Solution Because two of the substances involved in the equilibrium are solids, their activities are 1, and their pressures are constant and do not appear in the equilibrium expression.

2

2

25 10H O

10 25 3H O

4.08 10 ( )

4.08 10 3.64 10 atm

PK P

P

71. Calcium chloride 6–hydrate, CaCl2•6H2O, dehydrates according to the equation –44

2 2 2 2CaCl • 6H O( ) CaCl ( ) 6H O( ) 5.09 10 at 25 CPs s g K � �� What is the pressure of water vapor at equilibrium with a mixture of CaCl2•6H2O and at 25 C? Solution Two of the components of this system are solids and have activities of 1. Their pressures are constant and do not enter into the equilibrium expression.

2

2

6 44H O

6 44 8H O

( ) 5.09 10

5.09 10 6.09 10 atm

PK P

P

72. A student solved the following problem and found the equilibrium concentrations to be [SO2] = 0.590 M, [O2] = 0.0450 M, and [SO3] = 0.260 M. How could this student check the work without reworking the problem? The problem was: For the following reaction at 600 °C:

2 2 3 c2SO ( ) O ( ) 2SO ( ) 4.32g g g K � �� What are the equilibrium concentrations of all species in a mixture that was prepared with [SO3] = 0.500 M, [SO2] = 0 M, and [O2] = 0.350 M? Solution Calculate Q based on the calculated concentrations and see if it is equal to Kc. Because Q does equal 4.32, the system must be at equilibrium. 73. A student solved the following problem and found [N2O4] = 0.16 M at equilibrium. How could this student recognize that the answer was wrong without reworking the problem? The problem was: What is the equilibrium concentration of N2O4 in a mixture formed from a sample of NO2 with a concentration of 0.10 M?

2 2 42NO ( ) N O ( ) 160cg g K � �� Solution

2CaCl


Page 20 of 31

The stoichiometry of the reaction between 2NO2 and N2O4 forces the concentration of N2O4 to be no larger than one-half of that of NO2. Therefore, no concentration value for N2O4 > 0.05 M is possible. 74. Assume that the change in concentration of N2O4 is small enough to be neglected in the following problem. (a) Calculate the equilibrium concentration of both species in 1.00 L of a solution prepared from 0.129 mol of N2O4 with chloroform as the solvent.

–52 4 2N O ( ) 2NO ( ) 1.07 10cg g K � �� in chloroform

(b) Confirm that the change is small enough to be neglected. Solution (a) Write the starting conditions, change, and equilibrium constant in tabular form.

Since K is very small, ignore x in comparison with 0.129 M. The equilibrium expression is

2 2 25 2

2 45

2 7

[NO ] (2 ) (2 ) 1.07 10 [N O ] (0.129 ) 0.129

0.129 1.07 10 3.45 104

cx xK

x

x

x = 5.87 10–4 The concentrations are: [NO2] = 2x = (5.87 10–4)(2) = 1.17 10–3 M [N2O4] = 0.129 – x = 0.129 – 5.87 10–4 = 0.128 M: (b) The assumption that x is negligibly small compared to 0.129 is confirmed by comparing the initial concentration of the N2O4 to its concentration at equilibrium (they differ by just 1 in the least significant digit’s place). 75. Assume that the change in concentration of COCl2 is small enough to be neglected in the following problem. (a) Calculate the equilibrium concentration of all species in an equilibrium mixture that results from the decomposition of COCl2 with an initial concentration of 0.3166 M.

–102 2COCl ( ) CO( ) Cl ( ) 2.2 10cg g g K � ��

(b) Confirm that the change is small enough to be neglected. Solution (a) Write the starting conditions, change, and equilibrium conditions in tabular form.


Page 21 of 31

Since Kc is very small, ignore x in comparison with 0.3166 M. The equilibrium constant expression is

2102

2

[CO][Cl ] 2.2 10[COCl ] 0.3166c

xK

x2 = 6.965 10–11 x = 8.3 10–6 = [CO] = [Cl2] [COCl2] = 0.3166 – 8.3 10–6 = 0.3166 (b) The assumption that x is negligibly small compared to 0.3166 is confirmed by comparing the initial concentration of the COCl2 to its concentration at equilibrium (they are identical when recorded to the proper number of significant digits). 76. Assume that the change in pressure of H2S is small enough to be neglected in the following problem. (a) Calculate the equilibrium pressures of all species in an equilibrium mixture that results from the decomposition of H2S with an initial pressure of 0.824 atm.

62 2 22H S( ) 2H ( ) S ( ) 2.2 10Pg g g K � ��

(b) Confirm that the change is small enough to be neglected. Solution (a) Write the balanced equation, and then set up a table with initial pressures and the changed pressures using x as the change in pressure. The simplest way to find the coefficients for the x values is to use the coefficient in the balanced equation.

2 2

2

2 2S H6

2 2H S

( )( ) [ ][2 ] 2.2 10 ( ) [0.824 2 ]P

P P x xKP x

Neglecting the change in reactant concentration (2x), as instructed, yields: 2

62

36

[ ][2 ]2.2 10 [0.824]

42.2 10 [0.679]

x x

x

1.494 10–6 = 4x3

3.73 10–7 = x3 7.20 10–3 = x Final equilibrium pressures: [H2S] = 0.824 – 2x = 0.824 – 2(7.20 10–3) = 0.824 – 0.0144 = 0.810 atm [H2] = 2x = 2(7.2 10–3) = 0.014 atm [S2] = [x] = 0.0072 atm


Page 22 of 31

(b) The assumption that 2x is negligibly small compared to 0.824 is confirmed by comparing the initial concentration of the H2S to its concentration at equilibrium (0.824 atm versus 0.810 atm, a difference of less than 2%). 77. What are all concentrations after a mixture that contains [H2O] = 1.00 M and [Cl2O] = 1.00 M comes to equilibrium at 25 C?

2 2H O( ) Cl O( ) 2HOCl( ) 0.0900cg g g K � �� Solution As all species are given in molar concentration, a simple Kc equilibrium can be solved using the balanced equation.

2 2

2 2

[HOCl] [2 ] 0.0900 [H O][Cl O] [1.00 2 ][1.00 ]c

xKx x

2

2

40.0900 1.00 2.00

xx x

0.0900 – 0.180x + 0.0900x2 = 4x2 Arrange the terms in the form of the quadratic equation: ax2 + bx + c = 0 3.91x2 + 0.180x – 0.0900 = 0 Then solve using the quadratic formula:

22 0.180 (0.180) 4(3.91)( 0.0900) 4 2 2(3.91)

0.180 (0.0324) 1.4076 0.182 1.200 7.82 7.82

0.130 or 0.176

b b acxa

M M

Since concentrations must be positive, only the positive root (0.130) is physically significant. Final equilibrium concentrations: [H2O] = [Cl2O] = 1.00 – x = 1.00 – 0.130 = 0.870 M [HOCl] = 2x = 2(0.130 M) = 0.260 M 78. What are the concentrations of PCl5, PCl3, and Cl2 in an equilibrium mixture produced by the decomposition of a sample of pure PCl5 with [PCl5] = 2.00 M?

5 3 2PCl ( ) PCl ( ) Cl ( ) 0.0211cg g g K � �� Solution As all species are gases and are in M concentration units, a simple Kc equilibrium can be solved using the balanced equation:


Page 23 of 31

3 2

5

[PCl ][Cl ] [ ][ ] 0.0211 [PCl ] [2.00 ]c

x xKx

220.0211 0.0422 0.0211

[2.00 ]x x x

x

Begin by arranging the terms in the form of the quadratic equation: ax2 + bx + c = 0 x2 + 0.0211x – 0.0422 = 0 Next, solve for x using the quadratic formula.

22 0.0211 (0.0211) 4(1)( 0.0422) 4 2 2(1)

b b acxa

0.0211 0.0004452 0.1688 0.0211 0.4114 2 2

= 0.195 M or –0.216 M Since concentrations must be positive, only the positive root (0.195) is physically significant. The final equilibrium concentrations are: [PCl5] = 2.00 – x = 2.00 – 0.195 = 1.80 M; [PCl3] = [Cl2] = x = 0.195 M. 79. Calculate the number of grams of HI that are at equilibrium with 1.25 mol of H2 and 63.5 g of iodine at 448 C.

2 2H I 2HI 50.2 at 448 CcK � �� Solution The number of moles of I2 is

21

2

2 2

63.5 gmol 0.250 mol I253.809 g mol[HI]

[H ][I ]cK

The unit for each concentration term is moles per liter. If the volume were known for this exercise, the number of moles in each term should be divided by this volume. However, there are two terms in the numerator and two terms in the denominator, so these volumes cancel one another. Consequently, for any expression with the same number of numerator terms as denominator terms, the number for moles can be used in place of moles per liter. In this exercise, the volume is not needed even though it is given. (mol HI)2 = K mol H2 mol I2 = 50.2 1.25 mol 0.250 mol = 15.7 mol2 mol HI = 215.7 mol 3.96 mol


Page 24 of 31

Mass(HI) = 3.96 mol 127.9124 g/mol = 507 g 80. Butane exists as two isomers, n–butane and isobutane.

KP= 2.5 at 25 C What is the pressure of isobutane in a container of the two isomers at equilibrium with a total pressure of 1.22 atm? Solution

isobutane

-butane

2.5Pn

PKP

Let x be the partial pressure of n-butane. 1.22 atm 2.5x

x

2.5x + x = 1.22 3.5x = 1.22 x = 0.35 atm = Pn-butane Pisobutane = 1.22 – x = 0.87 atm 81. What is the minimum mass of CaCO3 required to establish equilibrium at a certain temperature in a 6.50-L container if the equilibrium constant (Kc) is 0.50 for the decomposition reaction of CaCO3 at that temperature?

3 2CaCO ( ) CaO( ) CO ( )s s g� �� Solution At equilibrium the concentration of CO2 = Kc = 0.50 M. The number of moles CO2 in the system is then mol CO2 = 6.5 L 0.50 mol/L = 3.3 mol. The minimum amount of CaCO2 must then exceed:

32 3

2

1 mol CaCO3.3 mol CO = 3.3 mol CaCO1 mol CO

32

3

100.1 g CaCO3.3 mol CO = 330 g1 mol CaCO

82. The equilibrium constant (Kc) for this reaction is 1.60 at 990 C:

2 2 2H ( ) CO ( ) H O( ) CO( )g g g g � �� Calculate the number of moles of each component in the final equilibrium mixture obtained from adding 1.00 mol of H2, 2.00 mol of CO2, 0.750 mol of H2O, and 1.00 mol of CO to a 5.00-L container at 990 C. Solution For the reaction the equilibrium expression is

2

2 2

[H O][CO] 1.60[H ][CO ]cK

Initially, the concentrations of the components are

21.00 mol[H ] 0.200 5.00 L

M


Page 25 of 31

22.00 mol[CO ] 0.400 5.00 L

M

20.75 mol[H O] 0.150 5.00 L

M

1.0 mol[CO] 0.200 5.00 L

M

To discern the direction in which the reaction will proceed, compute Qc:

푄 =(0.150)(0.200)(0.200)(0.400) = 0.375

Since Qc < Kc (1.60), the reaction will proceed in the forward direction to establish equilibrium. An ICE table may now be developed:

2

2

(0.150 )(0.200 + ) (0.300 0.350 ) 1.6 = (0.200 )(0.400 ) (0.080 0.600 )c

x x x xKx x x x

Rearranging the expression gives 0.60x2 – 1.31x + 0.098 = 0 Using the quadratic formula to solve for x,

22 1.31 ( 1.31) (4)(0.60)(0.098) 4 2 (2)(0.60)

1.31 (1.2169) 0.0776 or 2.111.2

b b acxa

The x = 2.11 value does not make physical sense because the maximum value that x could achieve is 0.200 (e.g., if all the 0.200 M H2 reacted). Therefore, use x = 0.0776 M. The concentrations present are [H2] = 0.200 – 0.0776 = 0.122 M [CO2] = 0.400 – 0.0776 = 0.322 M [H2O] = 0.150 + 0.0776 = 0.228M [CO] = 0.200 + 0.0776 = 0.278M The total number of moles of each component present is [H2] = 0.122 M 5.00 L = 0.610 mol [CO2] = 0.322 M 5.00 L = 1.61 mol [H2O] = 0.228M 5.00 L = 1.14 mol [CO] = 0.278M 5.00 L = 1.39 mol 83. In a 3.0-L vessel, the following equilibrium partial pressures are measured: N2, 190 torr; H2, 317 torr; NH3, 1.00 103 torr.

2 2 3N ( ) 3H ( ) 2NH ( )g g g � ��


Page 26 of 31

(a) How will the partial pressures of H2, N2, and NH3 change if H2 is removed from the system? Will they increase, decrease, or remain the same? (b) Hydrogen is removed from the vessel until the partial pressure of nitrogen, at equilibrium, is 250 torr. Calculate the partial pressures of the other substances under the new conditions. Solution (a) According to Le Châtelier’s principle, if some of a reactant is removed, the system will shift left. Removing some of a gas lowers its pressure. The removal of H2 results in decrease in the partial pressure of H2, an increase in the partial pressure of N2, and a decrease in the partial pressure of NH3. (b) The reaction is

2 2 3N ( ) 3H ( ) 2NH ( )g g g � �� Converting to atm (760 torr = 1 atm)

3

2

2

3NH

H

N

1 atm = 1.00 10 torr = 1.32 atm760 atm

1 atm = 317 torr = 0.417 atm760 torr

1 atm = 190 torr = 0.250 atm760 torr

1250 torr = 0.329 atm760 torr

P

P

P

so the equilibrium constant is

3

2 2

2 2NH

3 3N H

( ) (1.32) = = = 96.1( ) (0.250)(0.417)P

PK

P P From the balanced equation, 1 mol N2 produces 2 mol NH3. The pressure change, which is equivalent to the increased pressure of N2, increase 0.329 atm – 0.250 atm = 0.079 atm. Therefore, the pressure of NH3 must decrease by 2(0.079); its pressure is 1.32 – 0.158 = 1.16 atm. To maintain equilibrium, hydrogen must be removed from the reaction mixture. Let x = pressure of H2 in the final mixture. Then,

2

3

23

(1.16) = = 96.1(0.329)

(1.16) = (0.329)(96.1)

PKx

x

x = 0.349 atm 2H = = 0.349 atm = 265 torrP x

84. The equilibrium constant (Kc) for this reaction is 5.0 at a given temperature.

2 2 2CO( ) H O( ) CO ( ) H ( )g g g g � �� (a) On analysis, an equilibrium mixture of the substances present at the given temperature was found to contain 0.20 mol of CO, 0.30 mol of water vapor, and 0.90 mol of H2 in a liter. How many moles of CO2 were there in the equilibrium mixture? (b) Maintaining the same temperature, additional H2 was added to the system, and some water vapor was removed by drying. A new equilibrium mixture was thereby established containing 0.40 mol of CO, 0.30 mol of water vapor, and 1.2 mol of H2 in a liter. How many moles of CO2


Page 27 of 31

were in the new equilibrium mixture? Compare this with the quantity in part (a), and discuss whether the second value is reasonable. Explain how it is possible for the water vapor concentration to be the same in the two equilibrium solutions even though some vapor was removed before the second equilibrium was established. Solution

(a) For this reaction, 2 2

2

[CO ][H ] 5.0[CO][H O]cK . The concentrations at equilibrium are 0.20M

CO, 0.30 M H2O, and 0.90 M H2. Substitution gives 2[CO ][0.90] 5.0 [0.20][0.30]

K ;

25.0(0.20)(0.30)[CO ] 0.33

0.90M ; Amount of CO2 = 0.33 mol 1 = 0.33 mol.

(b) At the particular temperature of reaction, Kc remains constant at 5.0. The new concentrations are 0.40 M CO, 0.30 M H2O, and 1.2 M H2.

2[CO ] 1.2 5.0

0.40 0.30

2[CO ] 0.50 M Amount of CO2 = 0.50 mol 1 = 0.50 mol. Added H2 forms some water as a result of a shift to the left after H2 is added. 85. Antimony pentachloride decomposes according to this equation:

5 3 2SbCl ( ) SbCl ( ) Cl ( )g g g� �� An equilibrium mixture in a 5.00-L flask at 448 C contains 3.85 g of SbCl5, 9.14 g of SbCl3, and 2.84 g of Cl2.How many grams of each will be found if the mixture is transferred into a 2.00-L flask at the same temperature? Solution

The equilibrium expression 3 2

5

[SbCl ][Cl ] [SbCl ]cK shows us that if the volume of the reaction

vessel is changed, the system’s equilibrium is disrupted and shifts the reaction in such a way as to reestablish the equilibrium position. To calculate the equilibrium constant, first calculate the equilibrium concentrations in molar terms for the components present.

55

33

22

1 mol SbCl 1(SbCl ) (3.85 g) 0.00258 299.0135 g 5 L

1 mol SbCl 1(SbCl ) (9.14 g) 0.00801 228.1081 g 5 L

1 mol Cl 1(Cl ) (2.84 g) 0.00801 70.90542 g 5 L

M M

M M

M M

From these equilibrium concentrations, calculate the equilibrium constant for this reaction. 3 2

5

[SbCl ][Cl ] (0.00801)(0.00801) 0.0249[SbCl ] 0.00258cK

Changing the mixture to a 2.00-L flask would change the concentrations as follows:


Page 28 of 31

55

33

22

1 mol SbCl 1(SbCl ) (3.85 g) 0.00644 299.0135 g 2 L

1 mol SbCl 1(SbCl ) (9.14 g) 0.0200 228.1081 g 2 L

1 mol Cl 1(Cl ) (2.84 g) 0.0200 70.90542 g 2 L

M M

M M

M M

The reaction quotient becomes 3 2

5

[SbCl ][Cl ] (0.0200)(0.0200) 0.0621[SbCl ] 0.00644

Q

The reduction in volume of the reaction vessel from 5 L to 2 L has disrupted the equilibrium of the system. Because Qc (0.0621) > Kc(0.02493), the reaction is going to shift to the left (i.e., toward the reactants) to restore equilibrium. Calculate the new concentrations to which the system will shift to reestablish equilibrium as a result of the change of volume from 5 L to 2 L.

5 3 2SbCl ( ) SbCl ( ) Cl ( )g g g� ��

Putting the new equilibrium concentrations into the equilibrium expression gives

2(0.0200 )(0.0200 ) (0.0200 ) = 0.0249 (0.00644 ) (0.00644 )c

x x xKx x

Rearrangement gives x2 – 0.0649x + 0.00024 = 0 Using the quadratic formula and solving for x gives x = 0.00394 Calculate the new equilibrium concentrations as SbCl5 = 0.00644 + 0.00394 = 0.0104 M SbCl3 = 0.0200 – 0.00394 = 0.0161 M Cl2 = 0.0200 – 0.00395 = 0.0161 M The component masses after equilibrium is re-established are Mass SbCl5 = (2.00 L)(0.0104 mol/L)(299.0135 g/mol) = 6.22 g Mass SbCl3 = (2.00 L)(0.0161 mol/L)(228.1081 g/mol) = 7.35 g Mass Cl2 = (2.00 L)(0.0161 mol/L)(70.90542 g/mol) = 2.28 g 86. Consider the equilibrium

2 2 3 24NO ( ) 6H O( ) 4NH ( ) 7O ( )g g g g � �� (a) What is the expression for the equilibrium constant (Kc) of the reaction? (b) How must the concentration of NH3 change to reach equilibrium if the reaction quotient is less than the equilibrium constant?


Page 29 of 31

(c) If the reaction were at equilibrium, how would an increase in the volume of the reaction vessel affect the pressure of NO2? (d) If the change in the pressure of NO2 is 28 torr as a mixture of the four gases reaches equilibrium, how much will the pressure of O2 change? Solution

(a) 4 7

3 24 6

2 2

[NH ] [O ] [NO ] [H O]cK .

(b) Because [NH3] is in the numerator of Kc, [NH3] must increase for Qc to reach Kc. (c) The increase in system volume would lower the partial pressures of all reactants (including NO2). (d) The relative changes in pressures are related by the stoichiometry of the reaction.

2 2O NO7 7 (28 torr) 49 torr4 4

P P

87. The binding of oxygen by hemoglobin (Hb), giving oxyhemoglobin (HbO2), is partially regulated by the concentration of H3O+ and dissolved CO2 in the blood. Although the equilibrium is complicated, it can be summarized as

2 3 2 2 2 2HbO ( ) H O ( ) CO ( ) CO Hb H O ( ) H O( )aq aq g g l � �� (a) Write the equilibrium constant expression for this reaction. (b) Explain why the production of lactic acid and CO2 in a muscle during exertion stimulates release of O2 from the oxyhemoglobin in the blood passing through the muscle. Solution

(a)

2 2

2 3 2

CO Hb H O

HbO H O COcK

.

(b) Lactic acid releases H3O+. This, along with the increased concentration of CO2, shifts the equilibrium to the right, releasing O2. 88. Liquid N2O3 is dark blue at low temperatures, but the color fades and becomes greenish at higher temperatures as the compound decomposes to NO and NO2. At 25 °C, a value of KP = 1.91 has been established for this decomposition. If 0.236 moles of N2O3 are placed in a 1.52-L vessel at 25 °C, calculate the equilibrium partial pressures of N2O3(g), NO2(g), and NO(g). Solution Write the balanced equilibrium expression. With all of the species as gases, it is a straightforward KP problem to solve. However, all species must be converted to pressures, from other units related to concentration. For N2O3, with 0.236 mol in 1.52 L at 25 ºC: PV = nRT

0.236 mol

nP RTV

1.52 L(0.08206 L atm)(298.15 K )

mol K 3.80 atm

Write the balanced equation and the equilibrium changes:


Page 30 of 31

2

2 3

NO NO

N O

( )( )

( )P

P PK

P

2

1.91 (3.80 )

xx

7.258 – 1.91x = x2 0 = x2 +1.91x – 7.258 0 = ax2 + bx + c

2

2

4

21.91 (1.91) 4(1)( 7.258)

2(1)

1.91 3.648 29.032 2

1.91 32.680 2

1.91 5.717 2

b b acx

a

= 1.90 atm or –3.81 atm As negative pressure is physically relevant and so the positive root is used. The final pressures are:

2 3N OP = 3.80 – x = 3.80 – 1.90 = 1.90 atm and 2NO NO 1.90 atmP P x .

89. A 1.00-L vessel at 400 C contains the following equilibrium concentrations: N2, 1.00 M; H2, 0.50 M; and NH3, 0.25 M. How many moles of hydrogen must be removed from the vessel to increase the concentration of nitrogen to 1.1 M? The equilibrium reaction is N2(g) + 3H2(g) = 2NH3(g) Solution The reaction is

2 2 3N ( ) 3H ( ) 2NH ( )g g g � �� The equilibrium constant is calculated from the equilibrium data.

2 23

3 32 2

[NH ] (0.25) 0.50[N ][H ] (1.00)(0.50)cK

Now the concentration of H2 must be changed to increase the concentration of nitrogen to 1.1 M. The increase in nitrogen from 1.00 M to 1.1 M (an increase by 0.1 mol in the 1.00-L vessel) requires a corresponding decrease in NH3 of 0.2 mol (i.e., the stoichiometry is 2NH3 for every


Page 31 of 31

1N2). The decrease in 0.2 mol of NH3 leaves 0.25 – 0.20 = 0.05 mol of NH3 at the new position of equilibrium. Solve for the [H2] at equilibrium:

2

32

32

(0.05) 0.50(1.1)[H ]

0.00250.50[H ] 0.0022731.1

cK

[H2]3 = 0.004545 [H2] = 0.17 mol/L This is the new equilibrium amount of H2. The amount of H2 that would have to be removed to reach this level is

3

2

from original hydrogen: 0.50 mol 0.17 mol 0.33 molfrom decomposition of NH 0.30 moltotal moles of H that must be removed: 0.63 mol

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