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.

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 Br_{2} 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) ${K}_{c}=\phantom{\rule{0.2em}{0ex}}\frac{1}{[{\text{Pb}}^{\mathrm{2+}}]{\left[{\text{Cl}}^{\text{\u2212}}\right]}^{2}}$ > 1 because PbCl

_{2}is insoluble and formation of the solid will reduce the concentration of ions to a low level (<1

*M*).

Since ${K}_{c}=\phantom{\rule{0.2em}{0ex}}\frac{[{\text{C}}_{6}{\text{H}}_{6}]}{{[{\text{C}}_{2}{\text{H}}_{2}]}^{3}},$ a value of *K _{c}* ≈ 10 means that C

_{6}H

_{6}predominates over C

_{2}H

_{2}. In such a case, the reaction would be commercially feasible if the rate to equilibrium is suitable.

(a) ${Q}_{c}=\phantom{\rule{0.2em}{0ex}}\frac{[{\text{CH}}_{3}\text{Cl}][\text{HCl}]}{[{\text{CH}}_{4}][{\text{Cl}}_{2}]};$ (b) ${Q}_{c}=\phantom{\rule{0.2em}{0ex}}\frac{{[\text{NO}]}^{2}}{[{\text{N}}_{2}]\phantom{\rule{0.2em}{0ex}}[{\text{O}}_{2}]};$ (c) ${Q}_{c}=\phantom{\rule{0.2em}{0ex}}\frac{{[{\text{SO}}_{3}]}^{2}}{{[{\text{SO}}_{2}]}^{2}[{\text{O}}_{2}]};$ (d) *Q _{c}* = [SO

_{2}]; (e) ${Q}_{c}=\phantom{\rule{0.2em}{0ex}}\frac{1}{[{\text{P}}_{4}]{[{\text{O}}_{2}]}^{5}};$ (f) ${Q}_{c}=\phantom{\rule{0.2em}{0ex}}\frac{{[\text{Br}]}^{2}}{[{\text{Br}}_{2}]};$ (g) ${Q}_{c}=\phantom{\rule{0.2em}{0ex}}\frac{[{\text{CO}}_{2}]}{[{\text{CH}}_{4}]{[{\text{O}}_{2}]}^{2}};$ (h)

*Q*= [H

_{c}_{2}O]

^{5}

(a) *Q*_{c} 25 proceeds left; (b) *Q _{P}* 0.22 proceeds right; (c)

*Q*undefined proceeds left; (d)

_{c}*Q*1.00 proceeds right; (e)

_{P}*Q*0 proceeds right; (f)

_{P}*Q*4 proceeds left

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

(a) *K _{P}* = 1.6 $\times $ 10

^{−4}; (b)

*K*= 50.2; (c)

_{P}*K*= 5.31 $\times $ 10

_{c}^{−39}; (d)

*K*= 4.60 $\times $ 10

_{c}^{−3}

${Q}_{c}=\phantom{\rule{0.2em}{0ex}}\frac{[{\text{NH}}_{4}{}^{\text{+}}][{\text{OH}}^{\text{\u2212}}]}{[{\text{HN}}_{3}]}$

The amount of CaCO_{3} must be so small that ${P}_{{\text{CO}}_{2}}$ is less than *K _{P}* when the CaCO

_{3}has completely decomposed. In other words, the starting amount of CaCO

_{3}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.

(a) Δ*T* increase = shift right, Δ*P* increase = shift left; (b) Δ*T* increase = shift right, Δ*P* increase = no effect; (c) Δ*T* increase = shift left, Δ*P* increase = shift left; (d) Δ*T* increase = shift left, Δ*P* increase = shift right.

(a) ${K}_{c}=\phantom{\rule{0.2em}{0ex}}\frac{[{\text{CH}}_{3}\text{OH}]}{{[{\text{H}}_{2}]}^{2}[\text{CO}]};$ (b) [H_{2}] increases, [CO] decreases, [CH_{3}OH] increases; (c), [H_{2}] increases, [CO] decreases, [CH_{3}OH] decreases; (d), [H_{2}] increases, [CO] increases, [CH_{3}OH] increases; (e), [H_{2}] increases, [CO] increases, [CH_{3}OH] decreases

(a) ${K}_{c}=\phantom{\rule{0.2em}{0ex}}\frac{[\text{CO}][{\text{H}}_{2}]}{[{\text{H}}_{2}\text{O}]};$ (b) [H_{2}O] no change, [CO] no change, [H_{2}] no change; (c) [H_{2}O] decreases, [CO] decreases, [H_{2}] decreases; (d) [H_{2}O] increases, [CO] increases, [H_{2}] decreases; (f) [H_{2}O] decreases, [CO] increases, [H_{2}] increases. In (b), (c), (d), and (e), the mass of carbon will change, but its concentration (activity) will not change.

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*).

${K}_{c}=\phantom{\rule{0.2em}{0ex}}\frac{{[\text{C}]}^{2}}{\left[\text{A}\right]{\left[\text{B}\right]}^{2}}.$ [A] = 0.1 *M*, [B] = 0.1 *M*, [C] = 1 *M*; and [A] = 0.01, [B] = 0.250, [C] = 0.791.

(a) −2*x*, 2*x*, −0.250 *M*, 0.250 *M*; (b) 4*x*, −2*x*, −6*x*, 0.32 *M*, −0.16 *M*, −0.48 *M*; (c) −2*x*, 3*x*, −50 torr, 75 torr; (d) *x*, − *x*, −3*x*, 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.

${P}_{{\text{H}}_{\text{2}}}{}_{\text{O}}=3.64\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{\text{\u22123}}\phantom{\rule{0.2em}{0ex}}\text{atm}$

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) [NO_{2}] = 1.17 $\times $ 10^{−3} *M*

[N_{2}O_{4}] = 0.128 *M*

(b) Percent error $=\phantom{\rule{0.2em}{0ex}}\frac{5.87\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{\mathrm{-4}}}{0.129}\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}100\%=0.455\%.$ The change in concentration of N_{2}O_{4} is far less than the 5% maximum allowed.

(a) [H_{2}S] = 0.810 atm

[H_{2}] = 0.014 atm

[S_{2}] = 0.0072 atm

(b) The 2*x* is dropped from the equilibrium calculation because 0.014 is negligible when subtracted from 0.824. The percent error associated with ignoring 2*x* is $\frac{0.014}{0.824}\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}100\%=1.7\%,$ which is less than allowed by the “5% test.” The error is, indeed, negligible.

${P}_{{\text{O}}_{3}}=4.9\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{\mathrm{-26}}\phantom{\rule{0.2em}{0ex}}\text{atm}$

(a) Both gases must increase in pressure.

(b)${P}_{{\text{N}}_{\text{2}}}{}_{{\text{O}}_{\text{4}}}=8.0\phantom{\rule{0.2em}{0ex}}\text{atm and}\phantom{\rule{0.2em}{0ex}}{P}_{{\text{NO}}_{\text{2}}}=1.0\phantom{\rule{0.2em}{0ex}}\text{atm}$

(a) 0.33 mol.

(b) [CO]_{2} = 0.50 *M* Added H_{2} forms some water to compensate for the removal of water vapor and as a result of a shift to the left after H_{2} is added.

${P}_{{\text{H}}_{\text{2}}}=8.64\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{\text{\u221211}}\phantom{\rule{0.2em}{0ex}}\text{atm}$

${P}_{{\text{O}}_{\text{2}}}=0.250\phantom{\rule{0.2em}{0ex}}\text{atm}$

${P}_{{\text{H}}_{\text{2}}}{}_{\text{O}}=0.500\phantom{\rule{0.2em}{0ex}}\text{atm}$

(a) ${K}_{c}=\phantom{\rule{0.2em}{0ex}}\frac{{[{\text{NH}}_{3}]}^{4}{[{\text{O}}_{2}]}^{7}}{{[{\text{NO}}_{2}]}^{4}{[{\text{H}}_{2}\text{O}]}^{6}}.$ (b) [NH_{3}] must increase for *Q _{c}* to reach

*K*. (c) That decrease in pressure would decrease [NO

_{c}_{2}]. (d) ${P}_{{\text{O}}_{\text{2}}}=49\phantom{\rule{0.2em}{0ex}}\text{torr}$

${P}_{{\text{N}}_{\text{2}}}{}_{{\text{O}}_{\text{3}}}=1.90\phantom{\rule{0.2em}{0ex}}\text{atm and}\phantom{\rule{0.2em}{0ex}}{P}_{\text{NO}}={P}_{{\text{NO}}_{\text{2}}}=1.90\phantom{\rule{0.2em}{0ex}}\text{atm}$

(a) HB ionizes to a greater degree and has the larger *K _{c}*.

(b)

*K*(HA) = 5 $\times $ 10

_{c}^{−4}

*K*(HB) = 3 $\times $ 10

_{c}^{−3}

In each of the following, the value of Δ*G* is not given at the temperature of the reaction. Therefore, we must calculate Δ*G* from the values Δ*H*° and Δ*S* and then calculate Δ*G* from the relation Δ*G* = Δ*H*° − *T*Δ*S*°.

(a) *K* = 1.29;

(b) *K* = 2.51 $\times $ 10^{−3};

(c) *K* = 4.83 $\times $ 10^{3};

(d) *K* = 0.219;

(e) *K* = 16.1

The standard free energy change is $\text{\Delta}{G}_{298}^{\xb0}=\text{\u2212}RT\phantom{\rule{0.2em}{0ex}}\text{ln}\phantom{\rule{0.2em}{0ex}}K=\text{4.84 kJ/mol}.$ When reactants and products are in their standard states (1 bar or 1 atm), *Q* = 1. As the reaction proceeds toward equilibrium, the reaction shifts left (the amount of products drops while the amount of reactants increases): *Q* < 1, and $\text{\Delta}{G}_{298}$ becomes less positive as it approaches zero. At equilibrium, *Q* = *K*, and Δ*G* = 0.

1.0 $\times $ 10^{−8} atm. This is the maximum pressure of the gases under the stated conditions.

$x=1.29\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{-5}\phantom{\rule{0.2em}{0ex}}\text{atm}={P}_{{\text{O}}_{2}}$