The equilibrium constant `(K_(p))` for the decomposition of gaseous `H_(2)O`
`H_(2)O(g)hArr H_(2)(g)+(1)/(2)O_(2)(g)`
is related to the degree of dissociation `alpha` at a total pressure P by

Options

A. (A) `K=(alpha_(2)P^(1//2))/((1+alpha)(2-alpha)^(1//2))`
B. (B)`K=(alpha^(3//2)P^(1//2))/((1+alpha)(2+alpha)^(1//2))`
C. (C) `K=(alpha^(3)P^(1//2))/(sqrt2)`
D. (D) `K=(alpha^(3)P^(3//2))/((1-alpha)(2+alpha)^(1//2))`

Answer

B

Explanation

`H_(2)O(g)toH_(2)(g)+(1)/(2)O_(2)(g)`
`{:(1,0,0),(1-alpha,alpha,alpha//2):}`
Total moles at equilibrium`=1-alpha+alpha+alpha//2=1+alpha//2`
Let the total pressure at equilibrium be`=P`
So `P_(H_(2^o))=(1-alpha)/(1+alpha//2)xxP`
`P_(H_(2))=(alpha)/(1+alpha//2)xxP`
`P_(O_(2))=(alpha//2)/(1+alpha//2)xxP`
So `K_(P)=((Po_(2))^(1//2)(PH_(2)))/(PH_(2)O)`