Question

Ammonia gas at $15$ atm is introduced in a rigid vessel at $300$K. At equilibrium, the total pressure of the vessel is found to be $40.11$ atm at ${300}^0$C. The degree of dissociation of $N{H}_3$ will be :

• A. $0.6$
• B. $0.4$
• C. Unpredictable
• D. None of these

Answer 1

The correct option is B $0.4$

$\begin{array}{c}{P}_1=15,9+m\\ T_1=300K\\ P_1=?\\ T_2=300+273=573K\end{array}$

$\begin{array}{c}ByPV=nRT\\ \Rightarrow \frac{PV}{nT}=R+\text{const.}\\ \Rightarrow \frac{P_1{V}_1}{n_1{T}_1}=\frac{P_2{V}_2}{n_2{T}_2}\end{array}$

$\begin{array}{cc}\Rightarrow \frac{P_1}{T_1}& =\frac{P_2}{T_2}\\ \Rightarrow \frac{15}{500}& =\frac{P_2}{573}\\ P_2& =28.65\text{atm}\end{array}$

$\begin{array}{c}2{NH}_3\rightleftharpoons N_2+3H_2\\ \left(28.65\right)\\ 28.65-2pp3p\end{array}$

$\begin{array}{cc}\text{(a)}\epsilon q^n\to 28.65-2p+p+3p& =28.65+2p\\ 28\cdot 65+2p=40.11& \end{array}$

$p=5.73$

$\begin{array}{c}\alpha =\frac{\text{Diss. Pressure}}{\text{Total pressure}}=\frac{2\times 5.73}{28.65}=0.4\\ \therefore \alpha =0.4\end{array}$