Question

Ammonium hydrogen sulphide dissociates according to the equation $N{H}_{4}HS\left(s\right)\rightleftharpoons N{H}_3\left(g\right)+H_{2}S\left(g\right).$ If the observed pressure of the mixture is $1.12atm$ at ${106}^{o}C,$ what is the equilibrium constant $K_p$ of the reaction?

• A. $0.314at{m}^2$
• B. $2.000at{m}^2$
• C. $1.000at{m}^2$
• D. $0.480at{m}^2$

Answer 1

The correct option is A $0.314at{m}^2$

$\begin{array}{ccccc}N{H}_{4}HS\left(s\right)& \rightleftharpoons & N{H}_3\left(g\right)& +& H_{2}S\left(g\right)\\ t=0& & 0& & 0\\ t=t_{eq}& & xatm& & xatm\end{array}$
since solid has no role in pressure

so, total pressure , $P_T=2x=1.12$
$x=0.56$
So
$K_p=P_{N{H}_3}\times P_{H_{2}S}$
$K_p=x\times x=x^2$

putting values,
$K_P=0.314at{m}^2$


Theory:

Writing $K_p$:

Example 1:
$N_2\left(g\right)+3H_2\left(g\right)\rightleftharpoons 2N{H}_3\left(g\right)$
$K_p=\frac{(P_{N{H}_3}{)}_{eq}^2}{\left(P_{N_2}{)}_{eq}\right(P_{H_2}{)}_{eq}^3}$


Example 2:

$\frac{1}{2}{H}_2\left(g\right)+\frac{1}{2}{I}_2\rightleftharpoons HI\left(g\right)$

$K_p=\frac{(P_{HI}{)}_{eq}}{\left(P_{H_2}{)}_{eq}^{\frac{1}{2}}\right(P_{I_2}{)}_{eq}^{\frac{1}{2}}}$