Question

The equilibrium constant for the decomposition of water, $H_{2}O\left(g\right)\rightleftharpoons H_2\left(g\right)+\frac{1}{2}{O}_2\left(g\right)$ is given by:

($\alpha =$degree of decomposition, $P=$ Total pressure at equilibrium)

• A. $K_p=\frac{{\alpha }^3{P}^{1/2}}{(1-\alpha ){(2-\alpha )}^{1/2}}$
• B. $K_p=\frac{{\alpha }^{3/2}{P}^{1/2}}{(1-\alpha ){(2+\alpha )}^{1/2}}$
• C. $K_p=\frac{{\alpha }^{3/2}{P}^{1/2}}{(1-\alpha )\sqrt{2}}$
• D. $K_p=\frac{{\alpha }^3{P}^{3/2}}{(1-\alpha ){(2+\alpha )}^{1/2}}$

Answer 1

Answer: (C)

$K_p=\frac{{\alpha }^{3/2}{P}^{1/2}}{(1-\alpha )\sqrt{2}}$

The given reaction is :-

$H_{2}O\left(g\right)\rightleftharpoons H_2\left(g\right)+1/2O_2\left(g\right)$

initial pressure : $Y$ $0$ $0$ (let)

At eqm : $Y(1-\alpha )$ $Y\alpha$ $\frac{Y\alpha }{2}$ (given)

Now, Total pressure at eqm = P

$\Rightarrow Y+\frac{Y\alpha }{2}=P\Rightarrow \frac{Y(1+\alpha )}{2}=P$

$\Rightarrow Y=\left(\frac{2P}{2+\alpha }\right)\to \left(I\right)$

Now,

$K_P=\frac{p_{H_2}.{\left(p_{O_2}\right)}^{1/2}}{p_{H_{2}O}}$

$=\frac{Y\alpha .{\left(\frac{Y\alpha }{2}\right)}^{1/2}}{Y(1-\alpha )}$

$K_P=\frac{Y^{1/2}{\alpha }^{3/2}}{\sqrt{2}(1-\alpha )}=\frac{{\alpha }^{3/2}{P}^{1/2}}{(1-\alpha )\sqrt{2+\alpha }}$ (from (I))