Question

For ${NH}_{4}HS\left(s\right)\rightleftharpoons {NH}_3\left(g\right)+H_{2}S\left(g\right)$ the observed,pressure for reaction mixture in equilibrium is 1.12 atm at ${106}^{0}C$.

What is the value of $K_P$ for the reaction?

• A. $K_P=0.04{atm}^2$
• B. $K_P=0.3136{atm}^2$
• C. $K_P=3.199,{atm}^2$
• D. $K_P=2.76{atm}^2$

Answer 1

The correct option is B $K_P=0.3136{atm}^2$

${NH}_{4}HS\left(s\right)\leftrightharpoons {NH}_3\left(g\right)+H_{2}S\left(g\right)$

Total pressure is 1.12 atm one half of this is equal to partial pressure of ammonia. It is also equal to partial pressure of hydrogen sulphide.

$\begin{array}{cc}{P}_{N{H}_3}& =P_{H_{2}S}=\frac{1}{2}\times 1.12=0.56\text{at}\\ K_P& =P_{N{H}_3}\times P_{H_{2}S}\\ & =0.56\times 0.56\\ & =0.314\end{array}$

Thus, the value of $K_p$ for the reaction is 0.3136 .