Question

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

What is the $K_p$ of the reaction if the total pressure at equilibrium is 30 atm?

• A. $15at{m}^2$
• B. $225at{m}^2$
• C. $30at{m}^2$
• D. $15atm$

Answer 1

The correct option is B $225at{m}^2$

The equilibrium reaction is $N{H}_{4}HS\left(s\right)\rightleftharpoons N{H}_3\left(g\right)+H_{2}S\left(g\right)$.

The expression for the equilibrium constant is $K_p=\frac{P_{N{H}_3}\times P_{H_{2}S}}{P_{N{H}_{4}HS}}=P_{N{H}_3}\times P_{H_{2}S}$ as $P_{N{H}_{4}HS}=1$.

The mole fractions of $N{H}_3\left(g\right)$ and $H_{2}S\left(g\right)$ are 0.5 each.

The partial pressures of $N{H}_3\left(g\right)$ and $H_{2}S\left(g\right)$ are $0.5\times 30=15$ each.

Substitute the values in the expression for the equilibrium constant.

$K_P=15\times 15=225at{m}^2$.

Hence, the correct option is $\text{B}$