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 given reaction is: 

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

At \( t = 0 \), the amount of water vapor is 1 mole.

At equilibrium, \( t = t_{\text{eq}} \), and the fraction of dissociation is denoted as \( \alpha \), where:

\( nT = 1 + \frac{\alpha}{2} \approx 1 \) (since \( \alpha \ll 1 \))

The equilibrium constant \( k_p \) is given by:

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

Given that \( P = 1 \), we can calculate:

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

Simplifying the equation:

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

Now solving for \( \alpha \):

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

Taking the cube root:

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

Answer 2

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