Q. The equilibrium constant $(K_{p})$ for the decomposition of gaseous $H_{2}O$ $H_{2}O(g) \rightleftharpoons H_{2}(g) + \frac{1}{2}O_{2}(g)$ is related to degree of dissociation $(\alpha)$ at a total pressure $p$ is given by
A) $K_{p} = \frac{\alpha^{3}p^{\frac{1}{2}}}{(1 + \alpha)(2 + \alpha)^{\frac{1}{2}}}$
B) $K_{p} = \frac{\alpha^{3}p^{\frac{3}{2}}}{(1 - \alpha)(2 + \alpha)^{\frac{1}{2}}}$
C) $K_{p} = \frac{\alpha^{\frac{3}{2}}p^{2}}{(1 - \alpha)(2 + \alpha)^{\frac{1}{2}}}$
D) $K_{p} = \frac{\alpha^{\frac{3}{2}}p^{\frac{1}{2}}}{(1 - \alpha)(2 + \alpha)^{\frac{1}{2}}}$
Explanation
(d)
$H_{2}O(g) \rightleftharpoons H_{2}(g) + \frac{1}{2}O_{2}(g)$
1 0 0 Initial
$(1 - \alpha)\ \ \ \ \ \ \ \alpha\ \ \ \ \ \ \ \ \ \ \ \frac{\alpha}{2}$ At equilibrium
Total moles = $(1 - \alpha) + \alpha + \frac{\alpha}{2} = 1 + \frac{\alpha}{2}$
Partial pressure of $H_{2}O = p\left( \frac{1 - \alpha}{1 + \frac{\alpha}{2}} \right)$
Partial pressure of $H_{2} = p\left( \frac{\alpha}{1 + \frac{\alpha}{2}} \right)$
Partial pressure of $O_{2} = p\left( \frac{\alpha}{2 + \alpha} \right)$
$K_{p} = \frac{pH_{2}.pO_{2}^{\frac{1}{2}}}{pH_{2}O}$
$K_{p} = \frac{\alpha^{\frac{3}{2}}p^{\frac{1}{2}}}{(1 - \alpha)(2 + \alpha)^{\frac{1}{2}}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.