Chapter 13

The reaction can proceed in both the forward and reverse directions.

When a system has reached equilibrium, no further changes in the reactant and product concentrations occur; the reactions continue to occur, but at equivalent rates.

The concept of equilibrium does not imply equal concentrations, though it is possible.

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.

(a) K_c = [Ag⁺][Cl⁻] < 1. AgCl is insoluble; thus, the concentrations of ions are much less than 1 M; (b) $K_c=\frac{1}{[{\text{Pb}}^{\mathrm{2+}}]{\left[{\text{Cl}}^{\text{−}}\right]}^2}$ > 1 because PbCl₂ is insoluble and formation of the solid will reduce the concentration of ions to a low level (<1 M).

Since $K_c=\frac{[{\text{C}}_6{\text{H}}_6]}{{[{\text{C}}_2{\text{H}}_2]}^3},$ a value of K_c ≈ 10 means that C₆H₆ predominates over C₂H₂. In such a case, the reaction would be commercially feasible if the rate to equilibrium is suitable.

K_c > 1

(a) $Q_c=\frac{[{\text{CH}}_3\text{Cl}][\text{HCl}]}{[{\text{CH}}_4][{\text{Cl}}_2]};$ (b) $Q_c=\frac{{[\text{NO}]}^2}{[{\text{N}}_2][{\text{O}}_2]};$ (c) $Q_c=\frac{{[{\text{SO}}_3]}^2}{{[{\text{SO}}_2]}^2[{\text{O}}_2]};$ (d) Q_c = [SO₂]; (e) $Q_c=\frac{1}{[{\text{P}}_4]{[{\text{O}}_2]}^5};$ (f) $Q_c=\frac{{[\text{Br}]}^2}{[{\text{Br}}_2]};$ (g) $Q_c=\frac{[{\text{CO}}_2]}{[{\text{CH}}_4]{[{\text{O}}_2]}^2};$ (h) Q_c = [H₂O]⁵

(a) Q_c 25 proceeds left; (b) Q_P 0.22 proceeds right; (c) Q_c undefined proceeds left; (d) Q_P 1.00 proceeds right; (e) Q_P 0 proceeds right; (f) Q_c 4 proceeds left

The system will shift toward the reactants to reach equilibrium.

(a) homogenous; (b) homogenous; (c) homogenous; (d) heterogeneous; (e) heterogeneous; (f) homogenous; (g) heterogeneous; (h) heterogeneous

This situation occurs in (a) and (b).

(a) K_P = 1.6 $\times$ 10⁻⁴; (b) K_P = 50.2; (c) K_c = 5.31 $\times$ 10⁻³⁹; (d) K_c = 4.60 $\times$ 10⁻³

$K_P=P_{{\text{H}}_2\text{O}}=0.042.$

$Q_c=\frac{[{\text{NH}}_4{}^{\text{+}}][{\text{OH}}^{\text{−}}]}{[{\text{NH}}_3]}$

The amount of CaCO₃ must be so small that $P_{{\text{CO}}_2}$ is less than K_P when the CaCO₃ has completely decomposed. In other words, the starting amount of CaCO₃ cannot completely generate the full $P_{{\text{CO}}_2}$ required for equilibrium.

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.

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.

Add N₂; add H₂; decrease the container volume; heat the mixture.

(a) ΔT increase = shift right, ΔP increase = shift left; (b) ΔT increase = shift right, ΔP increase = no effect; (c) ΔT increase = shift left, ΔP increase = shift left; (d) ΔT increase = shift left, ΔP increase = shift right.

(a) $K_c=\frac{[{\text{CH}}_3\text{OH}]}{{[{\text{H}}_2]}^2[\text{CO}]};$ (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.

(a) $K_c=\frac{[\text{CO}][{\text{H}}_2]}{[{\text{H}}_2\text{O}]};$ (b) [H₂O] no change, [CO] no change, [H₂] no change; (c) [H₂O] decreases, [CO] decreases, [H₂] decreases; (d) [H₂O] increases, [CO] increases, [H₂] decreases; (f) [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.

Only (b)

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).

(a)

$K_c=\frac{{[\text{C}]}^2}{\left[\text{A}\right]{\left[\text{B}\right]}^2}.$ [A] = 0.1 M, [B] = 0.1 M, [C] = 1 M; and [A] = 0.01, [B] = 0.250, [C] = 0.791.

K_c = 6.00 $\times$ 10⁻²

K_c = 0.50

K_P = 1.9 $\times$ 10³

K_P = 3.06

(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 $\times$ 10⁻⁴ M; (f) x, 0.1 atm.

Activities of pure crystalline solids equal 1 and are constant; however, the mass of Ni does change.

[NH₃] = 9.1 $\times$ 10⁻² M

P_BrCl = 4.9 $\times$ 10⁻² atm

[CO] = 2.04 $\times$ 10⁻⁴ M

$P_{{\text{H}}_{\text{2}}}{}_{\text{O}}=3.64\times {10}^{\text{−3}}\text{atm}$

Calculate Q based on the calculated concentrations and see if it is equal to K_c. Because Q does equal 4.32, the system must be at equilibrium.

(a) [NO₂] = 1.17 $\times$ 10⁻³ M
[N₂O₄] = 0.128 M
(b) Percent error $=\frac{5.87\times {10}^{\mathrm{-4}}}{0.129}\times 100\%=0.455\%.$ The change in concentration of N₂O₄ is far less than the 5% maximum allowed.

(a) [H₂S] = 0.810 atm
[H₂] = 0.014 atm
[S₂] = 0.0072 atm
(b) The 2x is dropped from the equilibrium calculation because 0.014 is negligible when subtracted from 0.824. The percent error associated with ignoring 2x is $\frac{0.014}{0.824}\times 100\%=1.7\%,$ which is less than allowed by the “5% test.” The error is, indeed, negligible.

[PCl₅] = 1.80 M; [Cl₂] = 0.195 M; [PCl₃] = 0.195 M.

[NO₂] = 0.19 M
[NO] = 0.0070 M
[O₂] = 0.0035 M

$P_{{\text{O}}_3}=4.9\times {10}^{\mathrm{-26}}\text{atm}$

507 g

33 g

(a) Both gases must increase in pressure.
(b)$P_{{\text{N}}_{\text{2}}}{}_{{\text{O}}_{\text{4}}}=8.0\text{atm and}{P}_{{\text{NO}}_{\text{2}}}=1.0\text{atm}$

(a) 0.33 mol.
(b) [CO₂] = 0.50 M Added H₂ forms some water to compensate for the removal of water vapor and as a result of a shift to the left after H₂ is added.

$P_{{\text{H}}_{\text{2}}}=8.64\times {10}^{\text{−11}}\text{atm}$
$P_{{\text{O}}_{\text{2}}}=0.250\text{atm}$
$P_{{\text{H}}_{\text{2}}}{}_{\text{O}}=0.500\text{atm}$

(a) $K_c=\frac{{[{\text{NH}}_3]}^4{[{\text{O}}_2]}^7}{{[{\text{NO}}_2]}^4{[{\text{H}}_2\text{O}]}^6}.$ (b) [NH₃] must increase for Q_c to reach K_c. (c) That decrease in pressure would decrease [NO₂]. (d) $P_{{\text{O}}_{\text{2}}}=49\text{torr}$

[fructose] = 0.15 M

$P_{{\text{N}}_{\text{2}}}{}_{{\text{O}}_{\text{3}}}=1.90\text{atm and}{P}_{\text{NO}}=P_{{\text{NO}}_{\text{2}}}=1.90\text{atm}$

(a) HB ionizes to a greater degree and has the larger K_c.
(b) K_c(HA) = 5 $\times$ 10⁻⁴
K_c(HB) = 3 $\times$ 10⁻³