Question

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

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

$\text{The given reaction is}$

$\begin{array}{c}{H}_{2}O\leftrightharpoons H_2+\frac{1}{2}{O}_2\\ 100\\ 1-\alpha \alpha \frac{1}{2}\alpha \end{array}$

$\begin{array}{c}\text{Total mole}=1+\frac{\alpha }{2}\\ \text{At equilibrium,}{P}_{H_{2}O}=\frac{1-\alpha }{1+\frac{\alpha }{2}}P=\frac{2(1-\alpha )}{(2+\alpha )}P\end{array}$

$\begin{array}{cc}{P}_{H_2}& =\frac{\alpha }{1+\frac{\alpha }{2}}P=\frac{2\alpha }{2+\alpha }P\\ \text{and,}{P}_{O_2}& =\frac{\alpha /2}{1+\frac{\alpha }{2}}P=\frac{\alpha }{2+\alpha }P\\ \text{Now,}Kp& =\frac{\left(P_{H_2}\right)\times {\left(P_{O_2}\right)}^{1/2}}{P_{H_{2}O}}.\end{array}$

$\begin{array}{cc}& =\frac{\left(\frac{2\alpha P}{2+\alpha }\right)\times {\left(\frac{\alpha P}{2+\alpha }\right)}^{1/2}}{\frac{2(1-\alpha )P}{(2+\alpha )}}\\ =& \frac{{\alpha }^{3/2}{P}^{1/2}}{(1-\alpha )(2+\alpha {)}^{1/2}}\end{array}$