Question

Calculate $K_p$ for the equilibrium,
$NH_4HS_{(s)} \rightleftharpoons NH_{3(g)} + H_2S_{(g)}$
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 $= x + x = 2x$
We know, $K_p = p_{NH_{3}}\times p_{H_{2}S}$
But, partial pressure $(p) =$ mole fraction $\times$ total pressure $(P)$
$K_{p}=\left(\frac{x}{2x}\times P\right)\left(\frac{x}{2x}\times P\right)$
$=\left(\frac{P}{2}\right)^{2}=\left(\frac{1.12}{2}\right)^{2}$
$=0.3136$