Question

8 mol of a gas $A{B}_3$ are introduced into a 1.0 $d{m}^3$ vessel. It dissociates as $2A{B}_3\left(g\right)\rightleftharpoons A_2\left(g\right)+3B_2\left(g\right)$
At equilibrium, 2 mol of $A_2$ is found to be present. The equilibrium constant for the reaction is :

• A. $2mo{l}^2{L}^{-2}$
• B. $3mo{l}^2{L}^{-2}$
• C. $27mo{l}^2{L}^{-2}$
• D. $36mo{l}^2{L}^{-2}$

Answer 1

Answer: (C)

$27mo{l}^2{L}^{-2}$

$\underset{\underset{\frac{8-2x}{1}}{8}}{2A{B}_3\left(g\right)}\rightleftharpoons \underset{\underset{\frac{x}{1}}{0}}{A_2\left(g\right)}+\underset{\underset{\frac{3x}{1}}{0}}{3B_2\left(g\right)}$

Since volume $=1d{m}^3=1L$

At eq. $\left[A_2\right]=2mol=x$

$\therefore \left[A{B}_3\right]=8-2\times 2=4M$

$\left[A_2\right]=2M$

$\left[B_2\right]=3\times 2=6M$

$K=\frac{\left[B_2{]}^3\right[A_2]}{[A{B}_3{]}^2}=\frac{6\times 6\times 6\times 2}{4\times 4}=27mo{l}^{-2}{L}^{-2}$