Question

Calculate the $K_p$ for the equilibrium,
$N{H}_{4}H{S}_{(}s)\rightleftharpoons N{H}_{3\left(g\right)}+H_2{S}_{(}g)$
if the total pressure inside the reaction vessel is $1.12$ atm at ${105}^0$C.

• A. $0.56$
• B. $1.25$
• C. $0.31$
• D. $0.63$

Answer 1

The correct option is C $0.31$

Given reaction is :-

${NH}_{4}HS\left(s\right)\rightleftharpoons {NH}_3\left(g\right)+H_{2}S\left(g\right)$

Now, $Kp=\frac{P{NH}_3.{PH}_{2}S}{P{NH}_{4}HS}$

Now, ${NH}_{4}HS$ being solid, its pressure is considered to be $1$. Since, volume and temperature are constant, so no. of moles $\alpha$ partial pressure

$\Rightarrow Kp=P{NH}_3.{PH}_{2}S$ (from the stoichionmetry of reaction)

$\Rightarrow Kp={\left(P{NH}_3\right)}^2$

Given, total pressure at eqm $=1.12$

$\Rightarrow {PNH}_3=\frac{1.12}{2}=0.56atm$

$\Rightarrow Kp=0.56\times 0.56=0.31$