Question

The equilibrium constant for decomposition of $ H_2O $ (g) $ H_2O(g) \rightleftharpoons H_2(g) + \frac{1}{2} O_2(g) \quad (\Delta G^\circ = 92.34 \, \text{kJ mol}^{-1}) $ is $ 8.0 \times 10^{-3} $ at 2300 K and total pressure at equilibrium is 1 bar. Under this condition, the degree of dissociation ($ \alpha $) of water is _____ $\times 10^{-2}$ (nearest integer value). [Assume $ \alpha $ is negligible with respect to 1]

Hint:

- For dissociation problems, express equilibrium composition in terms of \( \alpha \) - When \( \alpha \ll 1 \), approximations simplify calculations - Remember to account for stoichiometric coefficients in \( K_p \) expression - \( K_p \) has pressure units that must balance the reaction equation

Answer 1

Correct Answer: 5

Step 1: Write the equilibrium expression For the reaction:
H₂O(g) → H₂(g) + ½ O₂(g)
The equilibrium constant K_p is given by:
K_p = (P_H2 • P_O2^1/2) / P_H2O

Step 2: Express partial pressures in terms of α
Let initial moles of H₂O = 1
At equilibrium:
Moles of H₂O = 1 - α
Moles of H₂ = α
Moles of O₂ = α/2
Total moles = 1 + α/2 ≈ 1 (since α ≪ 1)
Partial pressures (total pressure = 1 bar):
P_H2O = (1 - α) ≈ 1
P_H2 = α
P_O2 = α/2

Step 3: Substitute into K_p expression
8.0 • 10^-3 = (α • (α/2)^1/2) / 1
8.0 • 10^-3 = α^3/2 / √2

Step 4: Solve for α
α^3/2 = 8.0 • 10^-3 • √2 = 1.131 • 10^-2
α = (1.131 • 10^-2)^2/3 = 0.049 ≈ 0.05
Expressed as   • 10^-2:
α = 5 • 10^-2