Calculating the equilibrium constant (K)

• An equilibrium concentration can be calculated using the K value and the equilibrium concentrations of the other species (worked example 1).
• Conversely, the equilibrium constant (K) can be determined using the initial concentrations of reactants and the equilibrium concentrations of products (worked example 2).
• Additionally, the initial concentrations can be derived from the equilibrium concentrations and the K value (worked example 3).

Worked example 1 - Calculating equilibrium concentration using K

When propanoic acid was allowed to reach equilibrium with ethanol at 30°C, it was found that the equilibrium mixture contained 1.8 mol dm^-3^ propanoic acid and 3.0 mol dm^-3^ ethanol.

CH_3_CH_2_COOH_(aq)_ + C_2_H_5_OH_(aq)_ ⇌ CH_3_CH_2_COOC_2_H_5(aq)_ + H_2_O_(l)_

If the K value is 0.35 at 30°C, determine the concentration of ethyl propanoate (CH_3_CH_2_COOC_2_H_5_) at equilibrium. Give your answer to 2 significant figures.

Step 1: Write the equilibrium constant (K) expression

$K =\frac{ [CH_3CH_2COOC_2H_5]}{ [CH_3CH_2COOH][C_2H_5OH]}$

Step 2: Rearrange expression

$[CH_3CH_2COOC_2H_5]=K\times[CH_3CH_2COOH][C_2H_5OH]$

Step 3: Substitution and correct evaluation

$[CH_3CH_2COOC_2H_5]=0.35\times1.8\times3.0=1.9\text{ mol dm}^{-3}$

Worked example 2 - Calculating K using initial concentrations

PCl_5_ decomposes at 650 K in a sealed vessel according to the equation:

PCl_5(g)_ ⇌ PCl_3(g)_ + Cl_2(g)_

The initial concentration of PCl_5_ is 0.0625 mol dm^-3. At equilibrium, the concentration of Cl_2_ is 0.025 mol dm^-3.

Calculate the equilibrium constant, K, for the reaction. Give your answer to 2 significant figures.

Step 1: Determine equilibrium concentrations

$\text{[PCl}_5\text{] }=0.0625-0.025=0.0375\text{ mol dm}^\text{-3}$

$\text{[PCl}_3\text{] }=\text{ [Cl}_2\text{] }=0.025\text{ mol dm}^\text{-3}$

Step 2: Write the equilibrium constant (K) expression

$K= \frac{[PCl_3][Cl_2]}{[PCl_5]}$

Step 3: Substitution and correct evaluation

$K= \frac{(0.025\times0.025)}{0.0375} = 0.017$

Worked example 3 - Calculating initial concentration using K

For the equilibrium 3H_2(g)_ + N_2(g)_ ⇌ 2NH_3(g)_ at 500 K, the equilibrium constant K = 0.59, and the equilibrium concentrations of N_2_ and NH_3_ are 0.30 mol dm^-3^ and 0.20 mol dm^-3^, respectively.

Calculate the equilibrum concentration, in mol dm^-3^, of H_2_. Give your answers to 2 significant figures.

Step 1: Construct initial and equilibrium concentration table

Step 2: Equation

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

Step 3: Rearrange equation:

$[H_2]=\sqrt[3]{\frac{[NH_3]^2}{K\times[N_2]}}$

Step 4: Substitution and correct evaluation

$[H_2]_\text{equil }= x =\sqrt[3]{\frac{0.20^2}{0.59\times0.30}}=0.61\text{ mol dm}^{-3}$

Step 5: Calculate [H₂]_initial and [N₂]_initial

$[H_2]_{inital }= y = 0.61 - (-0.30) = 0.91\text{ mol dm}^{-3}$

$[N_2]_{inital }= z = 0.30 - (-0.10) = 0.41\text{ mol dm}^{-3}$

Reaction quotient indicates position relative to equilibrium

The reaction quotient (Q) has the same mathematical form as the equilibrium constant expression but uses the actual concentrations of reactants and products at any point in time, not necessarily at equilibrium.

For the general reaction:

aA + bB ⇌ dD + eE

The expression for Q is:

$\text{Q }=\frac{[D]^d[E]^e}{[A]^a[B]^b}$

By comparing the values of Q and K, we can determine which direction the reaction will proceed to reach equilibrium:

• If Q < K - The forward reaction is favoured to reach equilibrium.
• If Q > K - The reverse reaction is favoured to reach equilibrium.
• If Q = K - The reaction is already at equilibrium.

Worked example 4 - Calculating Q

Given the reaction:

N_2(g) + 3H_2(g) ⇌ 2NH_3(g)

At 500 K, the equilibrium constant (K) is 0.45. In a mixture containing nitrogen, hydrogen, and ammonia, each species has a concentration of 0.60 mol dm^-3^.

Calculate the reaction quotient (Q) to 2 significant figures and determine the direction in which the reaction will proceed to reach equilibrium.

Step 1: Equation

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

Step 2: Substitution and correct evaluation

$\text{Q }=\frac{(0.60)^2}{(0.60)\times(0.60)^3}=2.8$

Step 3: Comparison with equilibrium constant (K)

Since Q > K, the reverse reaction is favoured to reach equilibrium.

Simplifying calculations for weak acids and bases

For weak acids (HA) and bases (B) that ionise to a small extent in water:

HA_(aq) ⇌ H⁺_(aq) + A^-_(aq)

B_(aq) + H₂O_(l) ⇌ BH⁺_(aq) + OH^-_(aq)

The equilibrium constants (K) for weak acids/bases are usually very small, indicating that the equilibrium lies far to the left. This means that the equilibrium concentrations of HA and B are only slightly lower than their initial concentrations.

To simplify calculations, we can assume that the equilibrium concentrations of HA and B are approximately equal to their initial concentrations:

[HA]_eq ≈ [HA]_initial

[B]_eq ≈ [B]_initial

Additionally, since the acid or base dissociates to form equal amounts of H⁺ (or BH⁺) and A^- (or OH^-), we can assume that their equilibrium concentrations are approximately equal:

[H⁺] ≈ [A^-]

[BH⁺] ≈ [OH^-]

These approximations allow us to simplify the equilibrium constant expression and calculate the equilibrium concentrations of ions, as demonstrated in the following worked example.

Worked example 5 - Calculating equilibrium concentrations

For the ionisation of methanoic acid (HCOOH) in water:

HCOOH_(aq) ⇌ HCOO^-_(aq) + H⁺_(aq) K = 1.77 × 10^-4 at 298 K

Calculate [H⁺] at equilibrium in a 0.150 mol dm^-3 HCOOH solution. Give your answer to 3 significant figures.

Step 1: Assume [HCOOH]_eq ≈ [HCOOH]_initial

[HCOOH]_eq ≈ [HCOOH]_initial = 0.150 mol dm^-3

Step 2: Equilibrium expression

Assume [HCOO^-] ≈ [H⁺]:

$K = \frac{[HCOO^{-}][H^{+}]}{[HCOOH]}=\frac{[H^{+}]^2}{[HCOOH]}$

Step 3: Rearrange equation

$[H^{+}]=\sqrt{K\times[HCOOH]}$

Step 4: Substitution and correct evaluation

$[H^{+}]=\sqrt{(1.77\times10^{-4})\times0.150}=5.15\times10^{-3}\text{ mol dm}^{-3}$

Gibbs energy and equilibrium

The standard Gibbs energy change (ΔG^⦵) indicates the spontaneity of a reaction under standard conditions. It is related to K by the equation:

$\Delta{G}^{\ominus}=-RTln{K}$

Where:

ΔG^⦵ = standard Gibbs energy change (kJ mol^-1)

R = gas constant (8.31 J K^-1^ mol^-1^)

T = temperature (K)

K = equilibrium constant

The table below summarises the relationship between the equilibrium constant (K), the standard Gibbs energy change (ΔG^⦵).

Worked example 6 - Calculating ΔG^⦵ from K

For the reaction: SO_2(g) + ½O_2(g) ⇌ SO_3(g) at 300 K, K = 500

Calculate ΔG^⦵, in kJ mol^-1, for the forward reaction. Give your answer to 3 significant figures.

The gas constant, R = 8.31 J K^-1 mol^-1.

Step 1: Equation

$\Delta{G}^{\ominus}=-RTln{K}$

Step 2: Substitution and correct evaluation

$\Delta{G}^{\ominus}=-8.31\times300\times\ln{500}=-15,500\text{ J mol}^{-1}$

Step 3: Conversion of J mol^-1 into kJ mol^-1

To convert from J mol^-1 into kJ mol^-1, divide by 1,000

$-$15,500 J mol^-1 = $-$15.5 kJ mol^-1