15 chemical equilibrium15 chemical equilibrium visualizing concepts 15.1 (a) k f > k r. according...

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15 Chemical Equilibrium

Visualizing Concepts

15.1 (a) k f > k r . According to the Arrhenius equation [14.19], As the magnitude of E a increases, k decreases. On the energy profile, E a is the difference in energy between the starting point and the energy at the top of the barrier. Clearly this difference is smaller for the forward reaction, so k f > k r .

(b) From the Equation [15.5], the equilibrium constant = k f /k r . Since k f > k r , the equilibrium constant for the process shown in the energy profile is greater than 1.

15.4 Analyze/Plan. Given that element A = red and element B = blue, evaluate the species in the reactant and product boxes, and write the reaction. Answer the remaining questions based on the balanced equation. Solve. (a) reactants: 4A 2 + 4B; products: 4A 2 B

balanced equation: A 2 + B → A 2 B

(b)

(c) Δn = Σn(prod) – Σn(react) = 1 – 2 = –1. (d) , Equation [15.14].

If you have a balanced equation, calculate Δn. Use Equation [15.14] to calculate K p from K c , or vice versa.

15.6 Analyze. Given box diagrams, reaction type, and value of K c , determine whether each reaction mixture is at equilibrium.

Plan. Analyze the contents of each box, express them as concentrations (see Solution 5.3). Write the equilibrium expression, calculate Q for each mixture, and compare it to K c . If Q = K, the mixture is at equilibrium. If Q < K, the reaction shifts right (more product). If Q > K, the reaction shifts left (more reactant).

Solve.

For this reaction, Δn = 0, so the volume terms cancel in the equilibrium expression. In this case, the number of each kind of particle can be used as a representation of moles (see Solution 5.3) and molarity.

(a) Mixture (i): 1A 2 , 1B 2 , 6AB;

Q > K c , the mixture is not at equilibrium.

15 Chemical Equilibrium Solutions to Exercises

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Mixture (ii): 3A 2 , 2B 2 , 3AB;

Q = K c , the mixture is at equilibrium.

Mixture (iii): 3A 2 , 3B 2 , 2AB;

Q < K c , the mixture is not at equilibrium.

(b) Mixture (i) proceeds toward reactants.

Mixture (iii) proceeds toward products. 15.9 If temperature increases, K of an endothermic reaction increases and K of an exothermic

reaction decreases. Calculate the value of K for the two temperatures and compare. For this reaction, Δn = 0 and K p = K c . We can ignore volume and use number of particles as a measure of moles and molarity. K c = [A][AB]/[A 2 ][B] (1) 300 K, 3A, 5AB, 1A 2 , 1B; K c = (3)(5)/(1)(1) = 15 (2) 500 K, 1A, 3AB, 3A 2 , 3B; K c = (1)(3)/(3)(3) = 0.33 K c decreases as T increases, so the reaction is exothermic.

Equilibrium; The Equilibrium Constant

15.11 Analyze/Plan. Given the forward and reverse rate constants, calculate the equilibrium constant using Equation [15.5]. At equilibrium, the rates of the forward and reverse reactions are equal. Write the rate laws for the forward and reverse reactions and use their equality to answer part (b). Solve.

(a) Equation [15.5];

For this reaction, K p = K c = 0.12.

(b) rate f = rate r ; k f [A] = k r [B]

Since k f < k r , in order for the two rates to be equal, [A] must be greater than [B].

15.13 Analyze/Plan. Follow the logic in Sample Exercises 15.1 and 15.6. Solve.

(a) (b)

(c) (d)

(e)

homogeneous: (a), (b), (d); heterogeneous: (c), (e)

15.15 Analyze. Given the value of K c , predict the contents of the equilibrium mixture.

Plan. If K c >> 1, products dominate; if K c << 1, reactants dominate. Solve.


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(a) mostly reactants (K c << 1)

(b) mostly products (K c >> 1)

15.17 Analyze/Plan. Follow the logic in Sample Exercise 15.2. Solve. PCl3(g) + Cl2(g) ⇌ PCl5(g), Kc = 0.042.

= 0.042(RT) – 1 = 0.042/RT

15.19 Analyze. Given K c for a chemical reaction, calculate K c for the reverse reaction.

Plan. The equilibrium expressions for the reaction and its reverse are the reciprocals of each other, and the values of K c are also reciprocal. Evaluate which species are favored by examining the magnitude of K c . Solve.

(a)

(b) K c < 1 when NOBr is the product, and K c > 1 when NOBr is the reactant. At this temperature, the equilibrium favors NO and Br 2 .

15.21 Analyze. Given K p for a reaction, calculate K p for a related reaction.

Plan. The algebraic relationship between the K p values is the same as the algebraic relationship between equilibrium expressions. Solve.

(a)

(b)

(c) ; Δn = 2 – 3 = –1; T = 1000 K

K p = K c (RT) – 1 = K c /RT; K c = K p (RT)

K c = 3.4225(0.08206)(1000) = 280.85 = 281

15.23 Analyze/Plan. Follow the logic in Sample Exercise 15.5. Solve.

CoO(s) + H2(g) ⇌ Co(s) + H2O(g) K1 = 67

Co(s) + CO2(g) ⇌ CoO(s) + CO(g) K2 = 1/490

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CoO(s) + H2(g) + Co(s) + CO2(g) ⇌ Co(s) + H2O (g) + CoO(s) + Co(g)

H2(g) + CO2(g) ⇌ H2O(g)+ CO(g)


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(a)

(b) The molar concentration, the ratio of moles of a substance to volume occupied by the substance, is a constant for pure solids and liquids.

Calculating Equilibrium Constants

15.27 Analyze/Plan. Follow the logic in Sample Exercise 15.8 using concentrations rather than pressures. Solve.


2NO(g) + Cl2(g) ⇌2NOCl(g)

15.31 Analyze/Plan. Follow the logic in Sample Exercise 15.9. Since the container volume is 1.0 L, mol = M. Solve.

(a) First calculate the change in [NO], 0.10 – 0.062 = 0.038 = 0.04 M. From the stoichiometry of the reaction, calculate the changes in the other pressures. Finally, calculate the equilibrium pressures.

2NO(g) + 2H2(g) ⇌ N2(g) + 2H2O(g)

initial 0.10 M 0.050 M 0 M 0.10 M

change –0.038 M –0.038 M +0.019 M +0.038 M

equil. 0.062 M 0.012 M 0.019 M 0.138 M

Strictly speaking, the change in [NO] has one decimal place and thus one sig fig. This limits equilibrium pressures to one decimal place for all but H2O, and K c to one sig fig. We compute the extra figures and then round.

(b)

15.33 Analyze/Plan. Follow the logic in Sample Exercise 15.9, using partial pressures, rather than concentrations. Solve.

(a)


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(b) The change in is 3.51 – 3.28 = 0.2276 = 0.23 atm. From the reaction stoichiometry, calculate the change in the other pressures and the equilibrium pressures.

CO2(g) + H2(g) ⇌ CO(g) + H2O(g)

initial 4.10 atm 2.05 atm 0 atm 3.28 atm

change –0.23 atm –0.23 atm +0.23 +0.23 atm

equil 3.87 atm 1.82 atm 0.23 atm 3.51 atm

(c)

Without intermediate rounding, equilibrium pressures are and in good agree-

ment with the value above.

Applications of Equilibrium Constants

15.35 (a) A reaction quotient is the result of the law of mass action for a general set of concentrations, whereas the equilibrium constant requires equilibrium concentrations.

(b) In the direction of more products, to the right.

(c) If Q c = K c , the system is at equilibrium; the concentrations used to calculate Q must be equilibrium concentrations.

15.37 Analyze/Plan. Follow the logic in Sample Exercise 15.10. We are given molarities, so we calculate Q directly and decide on the direction to equilibrium. Solve.

(a)

The reaction will proceed left to attain equilibrium.

(b)

The reaction will proceed right to attain equilibrium.

(c)

The reaction is at equilibrium.


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15.39 Analyze/Plan. Follow the logic in Sample Exercise 15.11. We are given concentrations, so write the K c expression and solve for [Cl 2 ]. Change molarity to partial pressure using the ideal gas equation and the definition of molarity. Solve.

Check. Our values are self-consistent.

15.41 Analyze/Plan. Follow the logic in Sample Exercise 15.11. In each case, change given masses to molarities solve for the equilibrium molarity of the desired component, and calculate mass of that substance present at equilibrium. Solve.

(a)

[Br] = (K c [Br 2 ]) 1 /2 = [(1.04 × 10 – 3)(0.007666)] 1 /2 = 0.002824 = 0.00282 M

Check. K c = (0.002824) 2 /(0.007666) = 1.04 × 10 – 3

(b)

[HI] = [(55.3) (0.01389) (0.008589)] 1 /2 = 0.08122 = 0.081 M

Check.


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15.43 Analyze/Plan. Follow the logic in Sample Exercise 15.12. Since molarity of NO is given directly, we can construct the equilibrium table straight away. Solve.

2NO(g) ⇌ N2(g) + O2(g)

initial 0.200 M 0 0

change –2x +x +x

equil. 0.200 – 2x +x +x

x = (2.4 × 10 3 ) 1 /2 (0.200 – 2x); x = 9.798 – 97.98x; 98.98x = 9.798, x = 0.09899 = 0.099 M

[N 2 ] = [O 2 ] = 0.099 M; [NO] = 0.200 – 2(0.09899) = 0.00202 = 0.002 M

Check. K c = (0.09899) 2 /(0.00202) 2 = 2.4 × 10 3

15.45 Analyze/Plan. Write the K p expression, substitute the stated pressure relationship, and solve for Solve.

When P NOBr = P NO, these terms cancel and This is true for all cases where P NOBr = P N O.

15.47 (a) CaSO4(s) ⇌ Ca2+(aq) + SO42–(aq)

At equilibrium, [Ca 2 +] = [SO 42 –] = x

K c = 2.4 × 10 – 5 = x 2 ; x = 4.9 × 10 – 3 M Ca 2 + and SO 42 –

(b) A saturated solution of CaSO 4 (aq) is 4.9 × 10 – 3 M. 3.0 L of this solution contain:

A bit more than 2.0 g CaSO 4 is needed in order to have some undissolved CaSO 4 (s) in equilibrium with 3.0 L of saturated solution.

15.49 Analyze/Plan. Follow the approach in Solution 15.43. Calculate [IBr] from mol IBr and construct the equilibrium table.

Solve. [IBr] = 0.500 mol/1.00 L = 0.500 M

Since no I 2 or Br 2 was present initially, the amounts present at equilibrium are produced by the reverse reaction and stoichiometrically equal. Let these amounts equal x. The amount of HBr that reacts is then 2x. Substitute the equilibrium molarities (in terms of x) into the equilibrium expression and solve for x.


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I2 + Br2 ⇌ 2IBr

initial 0 M 0 M 0.500 M

change +x M +x M –2x M

equil. x M x M (0.500 – 2x) M

x = 0.02669 = 0.0267 M; [I 2 ] = [Br 2 ] = 0.0267 M

[IBr] = 0.500 – 2(0.02669) = 0.4466 = 0.447 M

LeChâtelier’s Principle


(a) Shift equilibrium to the right; more SO 3 (g) is formed, the amount of SO 2 (g) decreases.

(b) Heating an exothermic reaction decreases the value of K. More SO 2 and O 2 will form, the amount of SO 3 will decrease.

(c) Since, Δn = –1, a change in volume will affect the equilibrium position and favor the side with more moles of gas. The amounts of SO 2 and O 2 increase and the amount of SO 3 decreases; equilibrium shifts to the left.

(d) No effect. Speeds up the forward and reverse reactions equally.

(e) No effect. Does not appear in the equilibrium expression.

(f) Shift equilibrium to the right; amounts of SO 2 and O 2 decrease.

15.53 Analyze/Plan. Given certain changes to a reaction system, determine the effect on K p , if any. Only changes in temperature cause changes to the value of K p . Solve.

(a) no effect (b) no effect (c) increase equilibrium constant (d) no effect

15.55 Analyze/Plan. Use Hess’s Law, reactants, to calculate ΔH o . According to the sign of ΔH° , describe the effect of temperature on the value of K. According to the value of Δn, describe the effect of changes to container volume. Solve.

(a)

ΔH° = 33.84 kJ + 81.6 kJ – 3(90.37 kJ) = –155.7 kJ

(b) The reaction is exothermic because it has a negative value of ΔH° . The


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equilibrium constant will decrease with increasing temperature.

(c) Δn does not equal zero, so a change in volume at constant temperature will affect the fraction of products in the equilibrium mixture. An increase in container volume would favor reactants, while a decrease in volume would favor products.

Additional Exercises 15.57 (a) Since both the forward and reverse processes are elementary steps, we can write

the rate laws directly from the chemical equation.

rate f = k f [CO][Cl 2 ] = rate r = k r [COCl][Cl]

For a homogeneous equilibrium in the gas phase, we usually write K in terms of partial pressures. In this exercise, concentrations are more convenient because the rate constants are expressed in terms of molarity. For this reaction, the value of K is the same regardless of how it is expressed, because there is no change in the moles of gas in going from reactants to products.

(b) Since the K is quite small, reactants are much more plentiful than products at equilibrium.

15.60

The change in [SOCl 2 ] = 0.56(1.00 M) = 0.56 M SO2Cl2(g) ⇌ SO2(g) + Cl2(g)

initial 1.00 M 0 0

change –0.56 M +0.56 M +0.56 M

equil. 0.44 M +0.56 M +0.56 M

15.63 (a) A(g) ⇌ 2B(g) initial 0.75 atm 0

change –0.39 atm +0.78 atm

equil. 0.36 atm 0.78 atm

P t = P A + P B = 0.36 atm + 0.78 atm = 1.14 atm

(b)


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15.66 (a) K p = 0.052; ; Δn = 2 – 0 = 2; K c = K p /(RT) 2

K c = 0.052/[0.08206)(333)] 2 = 6.964 × 10 – 5 = 7.0 × 10 – 5

(b) PH 3 BCl 3 is a solid and its concentration is taken as a constant, C.

[BCl 3 ] = 0.0128 mol/0.500 L = 0.0256 M

PH3BCl3 ⇌ PH3 + BCl3

initial C 0 M 0.0256 M

change +x M +x M

equil. C +x M (0.0256 + x) M

Check. The numerator in the quadratic has 1 sig fig, which leads to [PH3] with 1 sig fig. Kc = (2 × 10−3)(0.0256 + 2 × 10−3) = 6 × 10−5. Using 2 or 3 figures for [PH3] leads to closer agreement.

15.69 In general, the reaction quotient is of the form

(a)

Q > K p . Therefore, the reaction will shift toward reactants, to the left, in moving toward equilibrium.

(b)

Q > K p . Therefore, the reaction will shift toward reactants, to the left, in moving toward equilibrium.

(c)

Q < K p . Therefore, the reaction mixture will shift in the direction of more product, to the right, in moving toward equilibrium.

15.73

If P C O is 150 torr, can never exceed 760 – 150 = 610 torr. Then Q = 610/150 = 4.1. Since this is far less than K, the reaction will shift in the direction of more product. Reduction will therefore occur.

15.76 Analyze/Plan. Equilibrium pressures of H 2 , I 2 , HI → K p → equilibrium table → new equilibrium pressures. Solve.


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H2(g) + I2(g) ⇌ 2HI(g);

H2(g) + I2(g) ⇌ 2HI(g)

initial 1.34 atm 1.34 atm 9.30 atm + 1.20 atm

change +x atm +x atm –2x atm

equil. (1.34+x) atm (1.34+x) atm (10.50–2x) atm

Check.

15.79 The patent claim is false. A catalyst does not alter the position of equilibrium in a system, only the rate of approach to the equilibrium condition.

Integrative Exercises

15.80 Calculate the initial [IO4−], and then construct an equilibrium table to determine

[H4IO6−] at equilibrium.

IO 4− (aq) + 2H 2 O(l) ⇌ H 4 IO 6

− (aq)

initial change equil.

0.0724 M −x

0.0724 − x

0 +x +x

Because Kc is relatively large and [IO 4− ] is relatively small, we cannot assume x is small

relative to 0.0724.


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0.035(0.0724 – x) = x; 0.002534 – 0.035x = x ; 0.002534 = 1.035x

x = 0.002534/1.035 = 0.002448 = 0.0024 M H 4 IO 6− at equilibrium

15.83 (a) At equilibrium, the forward and reverse reactions occur at equal rates.

(b) One expects the reactants to be favored at equilibrium since they are lower in energy.

(c) A catalyst lowers the activation energy for both the forward and reverse reactions; the “hill” would be lower.

(d) Since the activation energy is lowered for both processes, the new rates would be equal and the ratio of the rate constants, k f /k r , would remain unchanged.

(e) Since the reaction is endothermic (the energy of the reactants is lower than that of the products, ΔE is positive), the value of K should increase with increasing temperature.

15.86 Mole % = pressure %. Since the total pressure is 1 atm, mol %/100 = mol fraction = partial pressure.

Temp (K) (atm) P C O (atm) K P

1123 0.0623 0.9377 14.1

1223 0.0132 0.9868 73.8

1323 0.0037 0.9963 2.7 × 10 2

1473 0.0006 0.9994 1.7 × 10 3 (2 × 10 3 )

Because K grows larger with increasing temperature, the reaction must be endothermic in the forward direction.

15 chemical equilibrium15 chemical equilibrium visualizing concepts 15.1 (a) k f > k r. according...

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