Question

The equilibrium constant $\left(K_p\right)$ for the decomposition of a gaseous water:
$H_2{O}_{(}g)\rightleftharpoons H_{2\left(g\right)}+O_{2\left(g\right)}$
Is related to degree of dissociation at total pressure ‘p’ is given by

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

Answer 1

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

$H_2{O}_{(}g)\rightleftharpoons H_{2\left(g\right)}+O_{2\left(g\right)}$
$\begin{array}{cccc}Initialcom& 1& 0& 0\\ Atequation& 1-\alpha & \alpha & \frac{\alpha }{2}\\ Totalnoofmolesate{q}^b=& 1-\alpha +& \alpha +& \frac{\alpha }{2}\\ & =1+\frac{\alpha }{2}& & \end{array}$

$P\left(H_{2}O\right)=\frac{1-\alpha }{1+\frac{\alpha }{2}}P$ $P\left(O_2\right)=\frac{\frac{\alpha }{2}}{1+\frac{\alpha }{2}}P$
$P\left(H_2\right)=\frac{\alpha }{1+\frac{\alpha }{2}}P$
$K_p=\frac{P\left(H_2\right)(P{O}_2{)}^{\frac{1}{2}}}{\left(P{H}_{2}O\right)}=\frac{{\alpha }^{\frac{3}{2}{p}^{\frac{1}{2}}}}{(1-\alpha )(2+\alpha {)}^{\frac{1}{2}}}$