Question

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

Answer 1

$2N{H}_3\rightleftharpoons N_2+3H_{2}O$

$T_1=300K15atm=P_1{T}_2=573K\frac{15}{300}\times \left(573\right)atm=P_2$ $\because$ $P_2=\left(\frac{P_1}{T_1}\right)T_2$

$\because$ Degree of dissociation $\alpha$ is given by,

$\left(\frac{15}{300}\right)\times 573\times (1+\alpha )=40.111+\alpha =1.4\alpha =0.4$