Question
Calculate $${K_p}$$ for the equilibrium, $$N{H_4}H{S_{\left( s \right)}} \rightleftharpoons N{H_{3\left( g \right)}} + {H_2}{S_{\left( g \right)}}$$ if the total pressure inside the reaction vessel is $$1.12\,atm$$ at $$105{\,^ \circ }C.$$
A.
0.56
B.
1.25
C.
0.31
D.
0.63
Answer :
0.31
Solution :
\[\begin{matrix}
{} & N{{H}_{4}}H{{S}_{\left( s \right)}} & \rightleftharpoons & N{{H}_{3\left( g \right)}} \\
\text{Initial}\,\,\text{moles} & 1 & {} & 0 \\
\text{At}\,\,\text{equilibrium} & \left( 1-x \right) & {} & x \\
{} & {} & {} & {} \\
\end{matrix}\begin{matrix}
+ \\
{} \\
{} \\
{} \\
\end{matrix}\begin{matrix}
{{H}_{2}}{{S}_{\left( g \right)}} \\
0 \\
x \\
{} \\
\end{matrix}\]
Total gaseous moles at equilibrium $$ = x + x = 2x$$
We know,$${\text{ }}{K_p} = {p_{N{H_3}}} \times {p_{{H_2}S}}$$
But, partial pressure $$(p)$$ = mole fraction × total pressure $$(P)$$
$$\eqalign{
& {K_p} = \left( {\frac{x}{{2x}} \times P} \right)\left( {\frac{x}{{2x}} \times P} \right) \cr
& \,\,\,\,\,\,\,\,\, = {\left( {\frac{P}{2}} \right)^2} \cr
& \,\,\,\,\,\,\,\,\, = {\left( {\frac{{1.12}}{2}} \right)^2} \cr
& \,\,\,\,\,\,\,\,\, = 0.3136 \cr} $$