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 equilibrium constant expression, K, for the reaction can be represented as:

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

1. Start by defining the initial moles and changes at equilibrium:

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

2. Define changes during the reaction:

   • Let x be the moles of CH₄ that react. Then,
     • At equilibrium:
       • CH₄: (25.0 - x)
       • H₂O: (25.0 - 2x)
       • CO₂: (0.0 + x)
       • H₂: (0.0 + 4x)

3. Given that at equilibrium the container contains 6.12 moles of H₂: [ 4x = 6.12 \implies x = 1.53 \text{ mol} ]

4. Now substitute the value of x into the equilibrium concentrations:

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

5. Calculate the equilibrium concentrations (in Molarity) for a 1000 L container:

   • [ [CH_4] = \frac{23.47}{1000} = 0.02347 \text{ M} ]
   • [ [H_2O] = \frac{21.94}{1000} = 0.02194 \text{ M} ]
   • [ [CO_2] = \frac{1.53}{1000} = 0.00153 \text{ M} ]
   • [ [H_2] = \frac{6.12}{1000} = 0.00612 \text{ M} ]

6. Finally, plug these values into the equilibrium constant expression: [ K = \frac{(0.00612)^4 (0.00153)}{(0.02347)(0.02194)^2} \approx 1.90 \times 10^{-3} ]