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)))^(2))/((P_(NH_(3))^(2)))("atm")^(2)`
The increase in pressure and temperature on the reaction in equilibrium favours :

Options

A. forward reaction in both cases
B. less dissociation of `NH_(3)`
C. backward reaction and forward reaction respectively
D. more formation of `N_(2)` but less formation of `H_(2)`

Answer

C

Explanation

Increase in pressure favours the reaction showing decrease in volume, i.e., backward reaction and increase in temperature favours endothermic reaction, i.e., forward one.