10 mole of `NH_(3)` is heated at 15 atm from `27^(@C` to `347^(@)C` assuming volume constant. The pressure at equilibrium is found to be 50 atm. The equilibrium constant for dissociation of `NH_(3)` :
`2NH_(3) hArr N_(2) + 3H_(2), Delta H = 91.94 kJ` can be written as `K_(p) = (p_(N_(2)) xx (p_(H_(2)))^(3))/((P_(NH_(3))^(2)))("atm")^(2)`
The equilibrium constant `K_(p)` for the reaction is :

Options

A. `7.08 xx 10^(2)`
B. `3.06 xx 10^(2)`
C. `7.6 xx 10^(2)`
D. `1.53 xx 10^(3)`

Answer

D

Explanation

`K_(p) = (n_(N_(2)) xx (n_(H_(2))))/((n_(NH_(3)))) xx [(P)/(Sigma n)]^(2) = ((6.13)/(2) xx [(6.13 xx 3)/(2)]^(3))/([10-6.13]^(2)) xx [(50)/(10+6.13)]^(2)= 1.528 xx 10^(3) " atm"^(2)`