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) for the reaction can be written as:

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

To calculate the equilibrium constant, we first need the equilibrium concentrations of all species:

1. Initial Concentrations:

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

2. Change in Concentrations:

   • Let the change in concentration of CH4 be -x, then:
   • Change in H2O will be -2x,
   • Change in CO2 will be +x,
   • Change in H2 will be +4x.

3. Equilibrium Concentrations:

   • CH4 = 25.0 - x
   • H2O = 25.0 - 2x
   • CO2 = 0 + x = x
   • H2 = 0 + 4x = 4x

Given that after equilibrium the concentration of H2 = 6.12 mol,

• Thus, 4x = 6.12 → x = 1.53 mol.

4. Substituting Values:

   • 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

5. Equilibrium Concentrations in Molarity (for a 100 L container):

   • [CH4] = 0.2347 M
   • [H2O] = 0.2194 M
   • [CO2] = 0.0153 M
   • [H2] = 0.0612 M

6. Final Calculation of K:

   • Thus,

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

By calculating this, we find:

$K \approx 1.90 \times 10^{-3} M^2$