The reaction involved is NH₄HS(g) $$\rightleftharpoons$$ NH₃(g) + H₂S(g).

Mass of ammonium hydrogen sulphide decomposed $$ = {{3.06 \times 30} \over {100}} = 0.918$$ g

Moles of NH₄HS decomposed $$ = {{0.918} \over {51}} = 0.018$$

Moles of ammonia formed = 0.018 ; Moles of hydrogen sulphide formed = 0.018

Thus, $$[N{H_3}] = {{0.018} \over 2} = 0.009$$ M ; $$[{H_2}S] = {{0.018} \over 2} = 0.009$$ M

Applying the law of chemical equilibrium on the reaction involved.

$${K_c} = {{[N{H_3}(g)][{H_2}s(g)]} \over {[N{H_4}HS]}} = {{0.009 \times 0.009} \over 1}$$ (Molar conc. of solid is equal to one)

$$ = 8.1 \times {10^{ - 5}}$$

$$\therefore$$ $${K_p} = {K_c} \times {(RT)^{\Delta n}}$$

Here R = 0.081 lit atm K^$$-$$1 mol^$$-$$1 , T = 27$$^\circ$$C = 300 K ; $$\Delta$$ⁿ = 2

Substituting the values, we get $${K_c} = 8.1 \times {10^{ - 5}} \times {(0.081 \times 300)^2} = 4.78 \times {10^{ - 2}}$$