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]
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]
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Correct Answer: 5
Approach Solution - 1
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:
\( \alpha = \sqrt[3]{128 \times 10^{-6}} = 5.03 \times 10^{-2} \)
Approach Solution -2
Step 1: Write the equilibrium expression For the reaction:
H2O(g) → H2(g) + ½ O2(g)
The equilibrium constant Kp is given by:
Kp = (PH2 • PO21/2) / PH2O
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):
PH2O = (1 - α) ≈ 1
PH2 = α
PO2 = α/2
Step 3: Substitute into Kp expression
8.0 • 10-3 = (α • (α/2)1/2) / 1
8.0 • 10-3 = α3/2 / √2
Step 4: Solve for α
α3/2 = 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
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