Question

A definite amount of solid ${NH}_{4}HS$ is placed in a flask already containing gas at a certain temperature and $0.50$ atm pressure. ${NH}_{4}HS$ decomposes to give ${NH}_3$ and $H_{2}S$ and at equilibrium total pressure in flask is $0.84atm$. The equilibrium constant for the reaction is:

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

Answer 1

The correct option is D $0.11$

The reactions takes place as:

${NH}_{4}HS\to N{H}_{3\left(g\right)}+H_2{S}_{\left(g\right)}$

${NH}_{4}HS\to N{H}_{3\left(g\right)}+H_2{S}_{\left(g\right)}startAt0.50equili.0.5+xatm+xatmthen0.5+x+x=2x+0.5=0.84\left(given\right)\Rightarrow 0.17atm{P}_{{NH}_3}=0.5+0.17=0.67atm,P_{H_{2}S}=0.17atmthereforeK=P_{{NH}_3}\times P_{H_{2}S}$

$0.67\times 0.17at{m}^2$

$=0.11$ hence the proof.