Question

A small amount of $N{H}_{4}HS$ solid is placed in a flask containing ammonia at 0.5atm pressure. The $N{H}_{4}HS$ will decompose to give $N{H}_3$ and $H_{2}S$. The reaction attains equilibrium and the total equilibrium pressure is 0.84atm. The equilibrium constant for the decomposition of $N{H}_{4}HS$ is

• A. 0.11
• B. 0.17
• C. 0.18
• D. 0.30

Answer 1

The correct option is A 0.11

$N{H}_{4}HS\rightleftharpoons N{H}_3+H_{2}S$
$\begin{array}{cccc}& & & \\ Initial& -& 0.5& 0\\ Equilibrum& -& 0.5+p& p\end{array}$
0.5+p+p=0.84
p = 0.17
$K_p=pN{H}_3\times p{H}_{2}S=0.67\times 0.17=0.114at{m}^2$