Question

The equilibrium constant $\left( K _{ p }\right)$ for the decomposition of water $H _{2} O ( g ) \rightarrow H _{2}( g )+\frac{1}{2} O _{2}( g )$ Is given by
Where $\alpha$ is the degree of dissociation and $p$ is the pressure of the reaction mixture at equilibrium.

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

Answer 1

Answer: (D)

$K _{ p }=\frac{(\alpha)(\alpha / 2)^{1 / 2}}{(1-\alpha)}\left(\frac{ p }{1+\frac{\alpha}{2}}\right)^{1 / 2}$
$=\frac{\alpha^{3 / 2} P ^{1 / 2} \sqrt{2}}{\sqrt{2}(1-\alpha)(2+\alpha)^{1 / 2}}$
$=\frac{\alpha^{3 / 2} P ^{1 / 2}}{(1-\alpha)(2+\alpha)^{1 / 2}}$