Question

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

• A. $K_p=\frac{{\alpha }^{3/2}{P}^{1/2}}{(1+\alpha )(2+\alpha {)}^{1/2}}$
• 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. $K_p=\frac{{\alpha }^{3/2}{P}^{1/2}}{(1-\alpha )(2+\alpha {)}^{1/2}}$
• E. $K_0=\frac{{\alpha }^{2}P}{\sqrt{2}}$

Answer 1

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

$\begin{array}{c}\text{Total mole}=1+\frac{\alpha }{2}\\ \text{At eqlbrium}\\ P_{H_{20}}=\frac{1-\alpha }{1+\frac{\alpha }{2}}P=\frac{2(1-\alpha )}{2+\alpha }P\end{array}$

$\begin{array}{cc}{P}_{H_2}& =\frac{\alpha }{1+\frac{\alpha }{2}}\cdot P=\frac{2\alpha }{2+\alpha }\cdot P\\ P_{O_2}& =\frac{\alpha /2}{1+\frac{\alpha }{2}}P=\frac{\alpha }{2+\alpha }P\\ K_P& =\frac{P_{H_2}\times {\left(P_{O_2}\right)}^{1/2}}{P_{H_{2}O}}\end{array}$

$=\frac{\frac{2\alpha P}{2+\alpha }\cdot {\left(\frac{\alpha P}{2+\alpha }\right)}^{1/2}}{\frac{2(1-\alpha )P}{2+\alpha }}$

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