Chemical Equilibrium

Derivation of the relationship between $K_p$ and $K_c$

Consider a general reversible reaction in the gaseous state:

$aA + bB \rightleftharpoons cC + dD$

Law of mass action in terms of concentration:

$K_c = \frac{[C]^c [D]^d}{[A]^a [B]^b}$

Law of mass action in terms of partial pressure:

$K_p = \frac{p_C^c \, p_D^d}{p_A^a \, p_B^b}$

Step 1: Use the ideal gas equation for each gaseous species.

For an ideal gas, $pV = nRT$, so the partial pressure of a gas is

$p = \frac{n}{V}RT = CR\,T$

where $C = n/V$ is the molar concentration of that gas.

Step 2: Write the partial pressure of each species in terms of its concentration.

$p_A = [A]RT, \quad p_B = [B]RT, \quad p_C = [C]RT, \quad p_D = [D]RT$

Step 3: Substitute these into the expression for $K_p$.

$K_p = \frac{([C]RT)^c \, ([D]RT)^d}{([A]RT)^a \, ([B]RT)^b}$

Step 4: Separate the concentration terms from the $(RT)$ terms.

$K_p = \frac{[C]^c [D]^d}{[A]^a [B]^b} \times \frac{(RT)^{c+d}}{(RT)^{a+b}}$

$K_p = K_c \,(RT)^{(c+d)-(a+b)}$

Step 5: Introduce $\Delta n_g$.

Let $\Delta n_g$ = (number of moles of gaseous products) − (number of moles of gaseous reactants)

$\Delta n_g = (c + d) - (a + b)$

Therefore,

$\boxed{K_p = K_c \,(RT)^{\Delta n_g}}$

where $R$ is the universal gas constant ($0.0821\ \text{L atm K}^{-1}\text{mol}^{-1}$ when pressure is in atm) and $T$ is the absolute temperature in kelvin.

Special cases:

• If $\Delta n_g = 0$, then $(RT)^0 = 1$, so $K_p = K_c$.
• If $\Delta n_g > 0$, then $K_p > K_c$.
• If $\Delta n_g < 0$, then $K_p < K_c$.

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Example where $K_p > K_c$

For $K_p$ to be greater than $K_c$, we need $\Delta n_g > 0$, i.e. the number of moles of gaseous products must exceed the number of moles of gaseous reactants.

Consider the dissociation of phosphorus pentachloride:

$\text{PCl}_5(g) \rightleftharpoons \text{PCl}_3(g) + \text{Cl}_2(g)$

Here, $\Delta n_g = 2 - 1 = 1$.

$K_p = K_c (RT)^1 = K_c \,RT$

Since $RT > 1$ at ordinary temperatures, $K_p > K_c$ for this reaction.

The general relation is $K_p = K_c(RT)^{\Delta n}$, where $\Delta n$ = moles of gaseous products $-$ moles of gaseous reactants.

(i) $\mathrm{PCl_5 \rightleftharpoons PCl_3 + Cl_2}$: $\Delta n = 2-1 = +1$, so

$K_p = K_c\,(RT)$

(ii) $\mathrm{N_2 + 3H_2 \rightleftharpoons 2NH_3}$: $\Delta n = 2-4 = -2$, so

$K_p = K_c\,(RT)^{-2} = \frac{K_c}{(RT)^2}$

By Le Chatelier's principle:

Increase of pressure: The system shifts to the side with fewer moles of gas. Reactants have 3 moles and the product has 2 moles, so the equilibrium shifts forward and the yield of $\mathrm{SO_3}$ increases.

Decrease of temperature: The forward reaction is exothermic (it releases heat), so lowering the temperature favours the forward reaction, and the yield of $\mathrm{SO_3}$ increases.

Basicity of an acid is the number of replaceable hydrogen ions ($\mathrm{H^+}$) present in one molecule of the acid.

In $\mathrm{NaOH + H_2SO_4 \rightarrow NaHSO_4 + H_2O}$ only one H⁺ of H₂SO₄ is replaced by Na⁺, so the basicity of H₂SO₄ in this reaction is 1 (although it is a dibasic acid in complete neutralisation).

Equilibrium constant ($K_c$) is the ratio of the product of equilibrium concentrations of the products to that of the reactants, each raised to the power of its stoichiometric coefficient, at a given temperature.

For $\mathrm{3Fe(s) + 4H_2O(g) \rightleftharpoons Fe_3O_4(s) + 4H_2(g)}$, pure solids are left out of the expression: $K_c=\frac{[\mathrm{H_2}]^4}{[\mathrm{H_2O}]^4}$

Le Chatelier's principle: if a system at equilibrium is disturbed by changing concentration, pressure or temperature, the equilibrium shifts in the direction that reduces (opposes) the change.

1. Concentration. Adding a reactant or removing a product shifts the equilibrium forward. In $\mathrm{N_2 + 3H_2 \rightleftharpoons 2NH_3}$, removing $\mathrm{NH_3}$ as it forms (by liquefying it) keeps driving the reaction to the right.

2. Pressure (only for gases with a change in the number of moles). Increasing pressure favours the side with fewer gas moles. In the Haber process ($4$ mol $\rightarrow$ $2$ mol) and the Contact process $\mathrm{2SO_2 + O_2 \rightleftharpoons 2SO_3}$ ($3 \rightarrow 2$), high pressure raises the yield. For $\mathrm{H_2 + I_2 \rightleftharpoons 2HI}$ ($\Delta n = 0$) pressure has no effect.

3. Temperature. Raising the temperature favours the endothermic direction; lowering it favours the exothermic direction. Haber and Contact reactions are exothermic, so low temperature gives more product, but a moderate temperature ($450$ to $500^\circ\mathrm{C}$) is used so that the rate is reasonable. In $\mathrm{N_2O_4 \rightleftharpoons 2NO_2}$ (endothermic) heating deepens the brown colour.

4. Catalyst and inert gas. A catalyst speeds up both directions equally, so it only helps equilibrium to be reached sooner, with no change in yield.

Other applications: physical equilibria such as melting of ice under pressure ($\mathrm{ice \rightleftharpoons water}$, volume decreases), and the solubility of gases in liquids (soda water bottle: opening it lowers pressure and $\mathrm{CO_2}$ escapes).

Law of mass action (Guldberg and Waage): at constant temperature, the rate of a chemical reaction is directly proportional to the product of the active masses (molar concentrations) of the reactants, each raised to the power of its stoichiometric coefficient in the balanced equation.

For $aA + bB \rightarrow \text{products}$:

$\text{rate} \propto [A]^a[B]^b \quad\Rightarrow\quad \text{rate} = k[A]^a[B]^b$

where $k$ is the rate constant. For equilibrium $aA + bB \rightleftharpoons cC + dD$ it gives $K_c = \dfrac{[C]^c[D]^d}{[A]^a[B]^b}$.

Law of mass action

Guldberg and Waage (1864) stated: at constant temperature, the rate of a chemical reaction is directly proportional to the product of the active masses (molar concentrations) of the reactants, each raised to the power of its stoichiometric coefficient.

For a reaction $aA + bB \rightarrow \text{products}$:

$\text{Rate} \propto [A]^a[B]^b \Rightarrow \text{Rate} = k[A]^a[B]^b$

where $k$ is the rate constant.

Equilibrium constant

For a reversible reaction

$aA + bB \rightleftharpoons cC + dD$

the forward rate is $r_f = k_f[A]^a[B]^b$ and the backward rate is $r_b = k_b[C]^c[D]^d$. At equilibrium, $r_f = r_b$:

$k_f[A]^a[B]^b = k_b[C]^c[D]^d$

$K_c = \frac{k_f}{k_b} = \frac{[C]^c[D]^d}{[A]^a[B]^b}$

$K_c$ is the equilibrium constant (concentrations at equilibrium). For gaseous reactions, in terms of partial pressures,

$K_p = \frac{p_C^c\,p_D^d}{p_A^a\,p_B^b}, \qquad K_p = K_c(RT)^{\Delta n_g}$

Important points

• $K$ depends only on temperature; it is not changed by concentration, pressure or a catalyst.
• A large $K$ means the products are favoured at equilibrium (the reaction goes nearly to completion); a small $K$ means the reactants are favoured.
• Pure solids and liquids are omitted from the expression.
• If the reaction is reversed, $K' = 1/K$.

Example: For $\mathrm{N_2 + 3H_2 \rightleftharpoons 2NH_3}$,

$K_c = \frac{[NH_3]^2}{[N_2][H_2]^3}$

(c) Low pressure and high temperature

The forward reaction is exothermic ($\Delta H = -196\ \text{kJmol}^{-1}$) and involves a decrease in the number of moles of gas (3 mol → 2 mol). So the reverse reaction is favoured by raising the temperature (endothermic direction) and by lowering the pressure (direction of increasing volume).