Question

Water decomposes at 2300 K:
\[2\text{H}_2\text{O}(g) \rightleftharpoons 2\text{H}_2(g) + \text{O}_2(g).\]
The percent of water decomposing at 2300 K and 1 bar is — (Nearest integer).
Equilibrium constant for the reaction is \(2 \times 10^{-3}\) at 2300 K.

Hint:

For equilibrium decomposition problems:
• Write the equilibrium constant expression carefully using stoichiometric coefficients.
• Assume small x if K is very small to simplify calculations.
• Use the percentage decomposition formula x × 100 to find the final result.

Answer 1

Correct Answer: 2

\[\text{H}_2\text{O}(g) \rightleftharpoons \text{H}_2(g) + \frac{1}{2} \text{O}_2(g)\]
At equilibrium, the partial pressures are:
\[P_0[1 - \alpha] \quad \text{for H}_2\text{O}, \quad P_0\alpha \quad \text{for H}_2, \quad \frac{P_0\alpha}{2} \quad \text{for O}_2.\]
The total pressure at equilibrium is:
\[P_0 \left[ 1 + \frac{\alpha}{2} \right] = 1 \quad \dots \text{(i)}.\]
The equilibrium constant \(K_p\) is given by:
\[K_p = \frac{(P_0\alpha)\left(\frac{P_0\alpha}{2}\right)^{1/2}}{P_0[1 - \alpha]}.\]
Substituting values:
\[\frac{(P_0\alpha)\left(\frac{P_0\alpha}{2}\right)^{1/2}}{P_0[1 - \alpha]} = 2 \times 10^{-3}.\]
Simplify:
\[\frac{\alpha \sqrt{\frac{\alpha}{2}}}{1 - \alpha} = 2 \times 10^{-3}.\]
Since \(\alpha\) is negligible with respect to 1, we approximate \(P_0 = 1\) and \(1 - \alpha \approx 1\):
\[\alpha \sqrt{\frac{\alpha}{2}} = 2 \times 10^{-3}.\]
Further simplification:
\[\alpha \cdot \sqrt{\frac{\alpha}{2}} = 2 \times 10^{-3},\]
\[\alpha^{3/2} \cdot \frac{1}{\sqrt{2}} = 2 \times 10^{-3},\]
\[\alpha^{3/2} = 2\sqrt{2} \times 10^{-3}.\]
Raise to the power \(2/3\):
\[\alpha = \left( 2\sqrt{2} \times 10^{-3} \right)^{2/3}.\]
Evaluate:
\[\alpha = 2 \times 10^{-2}.\]
Percentage decomposition:
\[\%\alpha = 2\%.\]
Final Answer: \(2\%\).