Question

Ammonium hydrogen sulphide dissociates according to the equation $N{H}_{4}HS\left(s\right)\rightleftharpoons N{H}_3\left(g\right)+H_{2}S\left(s\right)$.
The total pressure at equilibrium at $400K$ is found to be $1atm$
The equilibrium constant $K_p$ of the reaction is:

• A. $1at{m}^2$
• B. $0.25at{m}^2$
• C. $0.5atm$
• D. $0.25atm$

Answer 1

The correct option is B $0.25at{m}^2$

Answer is option (B)

GIVEN:

Total pressure of ammonia + hydrogen sulfide = $1atm$

Therefore, pressure of ammonia=pressure of hydrogen sulphide= $=\frac{1}{2}=0.5atm$

The values of $K_P$ can be calculated as,

$K_P=\left({NH}_{3}atm\right)\left(H_{2}Satm\right)$

$K_P=0.25{atm}^2$