Question

The equilibrium pressure of $N{H}_{4}CN\left(s\right)\rightleftharpoons N{H}_3\left(g\right)+HCN\left(g\right)$ is $0.3\text{atm}$. Calculate $K_p$ If $N{H}_{4}CN\left(S\right)$ is allowed to decompose in presence of $N{H}_3$ at $0.25\text{atm}$. Calculate partial pressure of $HCN$ at equilibrium.
$(\text{Given}:\sqrt{0.1525}\approx 0.39)$

• A. $K_p=0.0225{\text{atm}}^2\text{and}0.07\text{atm}$
• B. $K_p=0.0225{\text{atm}}^2\text{and}0.7\text{atm}$
• C. $K_p=0.0112{\text{atm}}^2\text{and}0.06\text{atm}$
• D. $K_p=0.0112{\text{atm}}^2\text{and}0.6\text{atm}$

Answer 1

The correct option is A $K_p=0.0225{\text{atm}}^2\text{and}0.07\text{atm}$

$N{H}_{4}CN\left(s\right)\rightleftharpoons \underset{P}{N{H}_3}\left(g\right)+\underset{P}{HCN\left(g\right)}$
At eqbm.
Total pressure $=2P=0.3$ atm
$p=0.15$ atm
Also $K_p=P_{N{H}_3}^1\times P_{HCN}^1=0.15\times 0.15=0.0225at{m}^2$
Now $P_{N{H}_3}=0.25$ atm

$$\begin{array}{ccccc}\left(N{H}_4\right(CN\left)\right(S)& \rightleftharpoons & N{H}_3\left(g\right)& +& HCN\left(g\right)\\ \text{Initial}& & 0.25& & 0\\ \text{at eq}& & 0.25+p^{\prime }& & p^{\prime }\end{array}$$
$K_p=p^1(0.25+p^1)\Rightarrow (p^{\prime }{)}^2+0.25p^{\prime }-0.0225=0p^{\prime }=0.07\text{atm}$