Question

The equilibrium,
$N{H}_{4}HS\left(s\right)\rightleftharpoons N{H}_3\left(g\right)+H_{2}S\left(g\right)$, is followed to set-up at $127{}^{\circ }\text{C}$ in a closed vessel. The total pressure at equilibrium was $20\text{atm}$. The $K_c$ for the reaction is:
(Given $R=0.08{\text{L atm K}}^{-1}{\text{mol}}^{-1}$)

• A. $0.097{\text{M}}^2$
• B. $0.085{\text{M}}^2$
• C. $3.045{\text{M}}^2$
• D. $1.04{\text{M}}^2$

Answer 1

The correct option is A $0.097{\text{M}}^2$

Since $N{H}_{4}HS$ is a solid $K_p$ is expressed as:
$K_p=pN{H}_3\times p{H}_{2}S$
Also,
$pN{H}_3=p{H}_{2}S=\frac{20}{2}\text{atm}$
$\therefore K_p=pN{H}_3\times p{H}_{2}S=100{\text{atm}}^2$
$\Rightarrow K_p=K_c(RT{)}^{\Delta n}$
$\Rightarrow 100=K_c\times (0.08\times 400{)}^2$
$\Rightarrow K_c=0.097{\text{M}}^2$