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 $2300 K$
Show Answer
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 $1-\alpha \approx 1$
$\frac{\alpha \sqrt{\alpha}}{\sqrt{2}}=2 \times 10^{-3}$
$\alpha^{3 / 2}=2^{3 / 2} \times 10^{-3}$
$\alpha=2^{3 / 2 \times 2 / 3} \times 10^{-3 \times 2 / 3}$
$\alpha=2 \times 10^{-2} \quad % \alpha=2 %$
