Question

The pressure of $N{H}_{4}HS\left(s\right)$ and $N{H}_3\left(g\right)$ mixture is $0.2$ atm. On heating this mixture, the pressure becomes $0.6$ atm at equilibrium.

The $K_p$ for the reaction is:

• A. $0.04at{m}^2$
• B. $0.008at{m}^2$
• C. $0.08at{m}^2$
• D. $0.004at{m}^2$

Answer 1

Answer: (C)

$0.08at{m}^2$

The reaction is as follows: $N{H}_{4}HS\left(s\right)\to N{H}_3+H_{2}S$

Initially: - ($0.2$) -
At equilibrium: - ($0.2+x$) ($x$)

Given that total pressure is $0.6$ atm.

Then, $0.2+2x=0.6$

$2x=0.4$

or, $x=0.2$.

Therefore, $K_p=P_{N{H}_3}\times P_{H_{2}S}=0.4\times 0.2=0.08at{m}^2$.

So, the correct option is $C$