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

The reaction is:

\( \text{H}_2\text{O}(g) \rightleftharpoons \text{H}_2(g) + \frac{1}{2}\text{O}_2(g) \)

Initial condition: At \( t = 0 \), moles of water vapor = 1.

At equilibrium (\( t = t_{\text{eq}} \)), dissociation fraction is \( \alpha \). Total moles \( nT = 1 + \frac{\alpha}{2} \approx 1 \) (assuming \( \alpha \ll 1 \)).

The equilibrium constant \( k_p \) is defined as:

\( k_p = \frac{P_{\text{H}_2} P_{\text{O}_2}^{1/2}}{P_{\text{H}_2O}} = \frac{(\alpha P) (\frac{\alpha}{2} P)^{1/2}}{(1 - \alpha) P} \)

Given \( P = 1 \):

\( 8 \times 10^{-3} = \frac{\alpha^{3/2}}{\sqrt{2}} \)

This simplifies to:

\( \alpha^{3/2} = 8\sqrt{2} \times 10^{-3} \)

Solving for \( \alpha \):

\( \alpha^3 = 128 \times 10^{-6} \)

Taking the cube root yields:

\( \alpha = \sqrt[3]{128 \times 10^{-6}} = 5.03 \times 10^{-2} \)