Chemical Equilibrium Question 2

Question 2 - 29 January - Shift 1

Water decomposes at $2300 K$

$H_2 O(g) \to H_2(g)+\frac{1}{2} O_2(g)$

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 $2300 K$

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Answer: (2)

Solution:

Formula: Equilibrium constatnt in terms of $K_p$

$H_2 O(g) \rightarrow H_2(g)+\frac{1}{2} O_2(g)$

$P_0[1-\alpha] \quad P_0 \alpha \quad \frac{P_0 \alpha}{2} \quad$ partial pr. at eq.

$P_0[1+\frac{\alpha}{2}]=1$

$K_p=\frac{(P _{H_2})(P _{O_2})^{1 / 2}}{P _{H_2 O}}$

$\frac{(P_0 \alpha)(\frac{P_0 \alpha}{2})^{1 / 2}}{P_0[1-\alpha]}=2 \times 10^{-3}$

since $\alpha$ is negligible w.r.t 1 so $P_0=1$ and $1-\alpha \approx 1$

$\frac{\alpha \sqrt{\alpha}}{\sqrt{2}}=2 \times 10^{-3}$

$\alpha^{3 / 2}=2^{3 / 2} \times 10^{-3}$

$\alpha=2^{3 / 2 \times 2 / 3} \times 10^{-3 \times 2 / 3}$

$\alpha=2 \times 10^{-2} \quad % \alpha=2 %$