Question

Some solid $N{H}_{4}HS$ is paced in flask containing $0.5$ atm of $N{H}_3$. What would be the pressures of $N{H}_3$ and $H_{2}S$ when equilibrium is reached?

$N{H}_{4}HS\left(s\right)⟺N{H}_3\left(g\right)+H_{2}S\left(g\right);K_p=0.11$

• A. $1.32$ atm and$0.33$ atm
• B. $6.65$ atm and $1.65$atm
• C. $2.5$ atm and $0.65$ atm
• D. $0.67$ atm and $0.17$atm

Answer 1

Answer: (D)

$0.67$ atm and $0.17$atm

The expression for the equilibrium constant is $K_p=\left[N{H}_3\right]\left[H_{2}S\right]$.

Substitute values in the above expression.

$0.11=(0.5+x)x$ or $x=0.17atm$

$0.5+x=0.67atm$.

Thus the pressures of ammonia and hydrogen sulfide are 0.67 atm & 0.17atm respectively.