Solid ammonium hydrogen sulphide decomposes according to the equation $N{H}_{4}H{S}_{\left(s\right)}\rightleftharpoons N{H}_{3\left(g\right)}+H_2{S}_{\left(g\right)}\text{ at }25°\mathrm{c}.$  If the total pressure at equilibrium is 0.5 atm, the equilibrium constant $\left(K_p\right)$ is

The given reaction belongs to heterogeneous equilibrium, in which gaseous products are always formed in equal number of moles. So, partial pressures of gases  $P_{N{H}_3}=0.25\text{ atm and }{P}_{H_{2}S}=0.25\text{ atm }$

$\therefore K_p=\left(P_{N{H}_3}\right)\left(P_{H_{2}S}\right)$

$=0.0625{ \mathrm{atm}}^2$