Question

Calculate $K_p$ for the equilibrium,
$N{H}_{4}H{S}_{(s)}\rightleftharpoons N{H}_{3(g)}+H_2{S}_{(g)}$
if the total pressure inside the reaction vessel is $1.12atm$ 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 $=x+x=2x$

We know $K_p=p_{N{H}_3}\times p_{H_{2}S}$

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

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

IInd method : Both $N{H}_3$ and $H_{2}S$ have same number of moles at equilibrium so, have same mole fraction and thus equal partial pressures.

i.e., $p_{N{H}_3}=p_{H_{2}S}=\frac{1.12}{2}$

$K_p=p_{N{H}_3}\times p_{H_{2}S}=\frac{1.12}{2}\times \frac{1.12}{2}=0.3136$