Question

The equilibrium constant ($K_p$) for the decomposition of gaseous $H_{2}O$ and the equilibrium equation, $H_{2}O\left(g\right)\rightleftharpoons H_2\left(g\right)+\frac{1}{2}{O}_2\left(g\right)$, is related to degree of dissociation ($\alpha$) at a total pressure $P$ is given by:

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

Answer 1

The correct option is D $K_p=\frac{{\alpha }^{3/2}{P}^{1/2}}{(1-\alpha ){(2+\alpha )}^{1/2}}$

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,

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

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

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