Hydrogen gas, H2(g), can be produced by reacting methane, CH4(g), with steam, H2O(g), at 300 °C in the presence of a suitable catalyst - VCE - SSCE Chemistry - Question 5 - 2023 - Paper 1

The expression for the equilibrium constant, K, at equilibrium is given by:

$K = \frac{[H_2]^4 [CO_2]}{[CH_4][H_2O]^2}$

To find the equilibrium constant, we first need to establish the equilibrium concentrations of each substance.

Initial Concentrations:

• CH4: 25.0 mol
• H2O: 25.0 mol
• CO2: 0.00 mol
• H2: 0.00 mol

Change in Concentrations:

Assuming x mol of CH4 reacts:

• CH4: 25.0 - x
• H2O: 25.0 - 2x
• CO2: x
• H2: 4x

Given that after equilibrium the amount of H2 is 6.12 mol, we have:

$4x = 6.12 \Rightarrow x = \frac{6.12}{4} = 1.53 \, \text{mol}$

Now we can calculate the equilibrium concentrations:

• CH4: 25.0 - 1.53 = 23.47 mol
• H2O: 25.0 - 2(1.53) = 21.94 mol
• CO2: 1.53 mol
• H2: 6.12 mol

Equilibrium Concentrations (in Molarity):

Given volume is 100.0 L:

• [CH4] = \frac{23.47}{100.0} = 0.2347 , M
• [H2O] = \frac{21.94}{100.0} = 0.2194 , M
• [CO2] = \frac{1.53}{100.0} = 0.0153 , M
• [H2] = \frac{6.12}{100.0} = 0.0612 , M

Final Calculation for K:

Now substituting these values back into the equilibrium expression:

$K = \frac{(0.0612)^4 (0.0153)}{(0.2347)(0.2194)^2}$

Calculating this gives: $K \approx 1.90 \times 10^{-3} \text{ M}^{-1}$.