The equilibrium constant for the decomposition of water H2O(g)⇌H2(g)+12O2(g)is given by : 2 (α = degree of dissociation of H2O(g); p = Total equilibrium pressure)

K p = α 2 p 1 / 2 ( 1 + α ) ( 2 − α ) 1 / 2

K p = α 2 p 1 / 2 ( 1 + α ) ( 2 − α ) 1 / 2

K p = α 3 p 3 / 2 ( 1 − α ) ( 2 + α ) 1 / 2

K p = α 3 / 2 p 1 / 2 ( 1 − α ) ( 2 + α ) 1 / 2

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The equilibrium constant for the decomposition of water $H_{2} O (g) ⇌ H_{2} (g) + \frac{1}{2} O_{2} (g)$ is given by : 2 ( $α$ = degree of dissociation of $H_{2} O (g)$ ; p = Total equilibrium pressure)

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$K_{p} = \frac{α^{3} p^{1 / 2}}{\sqrt{2}}$

$K_{p} = \frac{α^{2} p^{1 / 2}}{(1 + α) (2 - α)^{1 / 2}}$

$K_{p} = \frac{α^{3} p^{3 / 2}}{(1 - α) (2 + α)^{1 / 2}}$

$K_{p} = \frac{α^{3 / 2} p^{1 / 2}}{(1 - α) (2 + α)^{1 / 2}}$

answer is B.

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Here,

$\begin{array}{l} H_{2} O (g) ⇌ H_{2} (g) + \frac{1}{2} O_{2} (g) \\ a (1 - α) aα \frac{aα}{2} \end{array}$

$x = \frac{1 - α}{1 + \frac{α}{2}} = \frac{α}{1 + \frac{α}{2}} = \frac{\frac{α}{2}}{1 + \frac{α}{2}}$

$K_{P} = \frac{αP \times {(\frac{α}{2} P)}^{1 / 2}}{(1 - α) P \times {(1 + \frac{α}{2})}^{1 / 2}}$

$K_{p} = \frac{α^{3 / 2} \times P^{1 / 2}}{(1 - α) \times (2 + α)^{1 / 2}}$

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