Question

Two moles of ammonia was introduced in an evacuated vessel of $1\, L$ capacity. At high temperature, the gas undergoes partial dissociation according to the equation:
$2 NH _{3}( g ) \rightleftharpoons N _{2}( g )+3 H _{2}( g )$
At equilibrium the concentration of ammonia was found to be $1\, mol$. What is the value of $K$ ?

• A. $3 / 4=0.75\, mol ^{2}\, L ^{-2}$
• B. $3 / 2=1.5\, mol ^{2}\, L ^{-2}$
• C. $27 / 16=1.7\, mol ^{2}\, L ^{-2}$
• D. $27 / 64=0.42\, mol ^{2}\, L ^{-2}$

Answer 1

Answer: (C)

Let $\alpha$ be the degree of dissociation:

Hence we have :



At equilibrium : $\left[ NH _{3}\right]=2-2 \alpha=1$ or $\alpha=1 / 2$

$\left[ N _{2}\right]=\frac{1}{2},\left[ H _{2}\right]=\frac{3}{2},\left[ NH _{3}\right]=1$

$\therefore K=\frac{\left[ N _{2}\right]\left[ H _{2}\right]^{3}}{\left[ NH _{3}\right]^{2}}=\frac{\frac{1}{2} \times \frac{3}{2} \times \frac{3}{2} \times \frac{3}{2}}{1 \times 1}=\frac{27}{16}=1.7\, mol\,{}^{2}\, L ^{-2}$