Question

Calculate $K_{p}$ for the equilibrium,
$NH _{4} HS _{(s)} \rightleftharpoons NH _{3( g )}+ H _{2} S _{(g)}$
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 1

Answer: (C)

Total gaseous moles at equilibrium $=x+x=2 x$

We know $K_{p}=p_{ NH _{3}} \times p_{ H _{2} S }$

but partial pressure $(p)=$ mole fraction $\times$ total pressure $(P)$

$K_{p}=\left(\frac{x}{2 x} \times P\right)\left(\frac{x}{2 x} \times P\right)=\left(\frac{P}{2}\right)^{2}=\left(\frac{1.12}{2}\right)^{2}=0.3136$

IInd method : Both $NH _{3}$ and $H _{2} S$ have same number of moles at equilibrium so, have same mole fraction and thus equal partial pressures.

i.e., $p_{ NH _{3}}=p_{ H _{2} S }=\frac{1.12}{2}$

$K_{p}=p_{ NH _{3}} \times p_{ H _{2} S }=\frac{1.12}{2} \times \frac{1.12}{2}=0.3136$