Homogeneous Equilibria

Homogeneous Equilibria

A homogeneous equilibrium is one in which all reactants and products (and any catalysts, if applicable) are present in the same phase. By this definition, homogeneous equilibria take place in solutions. These solutions are most commonly either liquid or gaseous phases, as shown by the examples below:

$C_2H_2 \; (aq) + 2\, Br_2\, (aq)$ ⇌ $C_2H_2Br_4\, (aq)$ $K_c=\frac{[C_2H_2Br_4]}{[C_2H_2][Br_2]^2}$
$I_2\, (aq) + I^-\, (aq)$ $\rightleftharpoons $ $I_3^-\, (aq)$ $K_c=\frac{[I_3^-]}{[I_2][I^-]}$
$HF\, (aq) + H_2O\, (l)$ $\rightleftharpoons $ $H_3O^+\, (aq) +F^-\, (aq)$ $K_c=\frac{[H_3O^+][F^-]}{[HF]}$
$NH_3\, (aq) + H_2O\, (l)$ $\rightleftharpoons $ $NH_4^+\, (aq) + OH^-\, (aq)$ $K_c=\frac{[NH_4^+][OH^-]}{[NH_3]}$

These examples all involve aqueous solutions, those in which water functions as the solvent. In the last two examples, water also functions as a reactant, but its concentration is not included in the reaction quotient. The reason for this omission is related to the more rigorous form of the Q (or K) expression mentioned previously in this chapter, in which chemical activities (approximated by concentrations) for liquids and solids are equal to 1 and needn’t be included. Consequently, reaction quotients include concentration or pressure terms only for gaseous and solute species.

In order for this conversion to work (and because of the “more rigorous” form referenced above), all units of concentration in $Q_c$ and $K_c$ must be given in M, and all units of pressure in $Q_P$ and $K_P$ in bar.

The equilibria below all involve gas-phase solutions:

$C_2H_6\, (g)$ $\rightleftharpoons $ $C_2H_4\, (g) + H_2\, (g)$ $K_c=\frac{[C_2H_4][H_2]}{[C_2H_6]}$
$3\, O_2\, (g)$ $\rightleftharpoons $ $2\, O_3\, (g)$ $K_c=\frac{[O_3]^2}{[O_2]^3}$
$N_2\, (g) + 3\, H_2\, (g)$ $\rightleftharpoons $ $2\, NH_3\, (g)$ $K_c=\frac{[NH_3]^2}{[N_2][H_2]^3}$
$C_3H_8\, (g) + 5\, O_2\, (g) $ $\rightleftharpoons $ $ 3\, CO_2\, (g) + 4\, H_2O\, (g)$ $K_c=\frac{[CO_2]^3[H_2O]^4}{[C_3H_8][O_2]^5}$

For gas-phase solutions, the equilibrium constant may be expressed in terms of either the molar concentrations ($\,{K_c}$) or partial pressures ($\,{K_p}$) of the reactants and products. A relation between these two $\,{K}$ values may be simply derived from the ideal gas equation and the definition of molarity:

For the gas-phase reaction: $mA+nB⇌xC+yD$:$$K_P=\frac{(P_C)^x(P_D)^y}{(P_A)^m(P_B)^n}$$

where $\Delta \, n$ is the difference in the molar amounts of product and reactant gases, in this case:$$\Delta n=(x+y)−(m+n)$$

Calculation of $\, K_P$

Write the equations relating $\, K_c$ to $\, K_P$ for each of the following reactions:

(a)$C_2H_6\, (g) ⇌ C_2H_4\, (g) + H_2\, (g)$

(b)$CO\, (g) + H_2O\, (g) ⇌ CO_2\, (g) +H_2\, (g)$

(c)$N_2\, (g) + 3H_2\, (g) ⇌ 2NH_3\, (g)$

(d) $\, K_c$ is equal to 0.28 for the following reaction at 900 °C:

$$CS_2\, (g) + 4H_2\, (g) ⇌ CH_4\, (g) + 2H_2S\, (g)$$

What is $\, {K_P}$ at this temperature?

Solution
(a) Δ$\, n = (2) − (1) = 1$

$\, {K_P} = \, {K_c}(\, {RT})^{Δ\, n} = \, {K_c}(\, {RT})^1 \\ \mathbf{K_P = {K_c}(\, {RT})}$

(b) $Δ\, n = (2) − (2) = 0$

$\, {K_P} = \, {K_c}(\, {RT})^{Δ\, n} = \, {K_c}(\, {RT})^0 \\ \mathbf{K_P = K_c}$

(c) $Δ\, n = (2) − (1 + 3) = −2$

$\, {K_P} = \, {K_c}(\, {RT})^{Δ\, n} = \, {K_c}(\, {RT})^{−2} \\ \mathbf{K_P = \frac{\, {K_c}}{(\, {RT})^2}}$

(d) $\, {K_P} = \, {K_c}(\, {RT})^{Δ\, n} = (0.28)[(0.0821)(1173)]^{−2} = \mathbf{3.0×10^{−5}}$

Check Your Learning

Write the equations relating $\, {K_c}$ to $\, {K_P}$ for each of the following reactions:

(a)$\quad 2SO_2\, (g) +O_2\, (g) ⇌ 2SO_3\, (g)$

(b)$\quad N_2O_4\, (g) ⇌ 2NO_2\, (g)$

(c)$\quad C_3H_8\, (g) + 5O_2\, (g) ⇌ 3CO_2\, (g) + 4H_2O\, (g)$

(d) At 227 °C, the following reaction has $\, {K_c} = 0.0952$:

$$CH_3OH\, (g) ⇌ CO\, (g) +2H_2\, (g)$$

What would be the value of $\, {K_P}$ at this temperature?

Answer

(a) $\quad {K_P}=\, {K_c}(\, {RT})^{−1}$
(b) $\quad {K_P}=\, {K_c}(\, {RT})$
(c) $\quad {K_P}=\, {K_c}(\, {RT})$
(d)$\quad 160\ \text{or} \ 1.6×10^2$