Question

1 mol of XY(g) and 0.2 mol of Y(g) are mixed in 1 L vessel. At equilibrium, 0.6 mol of Y(g) is present. The value of K for the reaction $XY\left(g\right)\rightleftharpoons X\left(g\right)+Y\left(g\right)$ is :

• A. $0.04mol{L}^{-1}$
• B. $0.06mol{L}^{-1}$
• C. $0.36mol{L}^{-1}$
• D. $0.40mol{L}^{-1}$

Answer 1

The correct option is D $0.40mol{L}^{-1}$

$\underset{\underset{1-x}{1}}{XY\left(g\right)}\rightleftharpoons \underset{\underset{x}{0}}{X\left(g\right)}+\underset{\underset{0.2+x}{0.2}}{Y\left(g\right)}$
At eq. $\left[X\right]=0.2+x=0.6$
$\therefore x=0.4mol=\frac{0.4}{1}=0.4M$
$\left[XY\right]=1-0.4mol=0.6mol=\frac{0.6}{1}=0.6M$
$\left[X\right]=x=0.4=\frac{0.4}{1}=0.4M$
$K=\frac{\left[X\right]\left[Y\right]}{\left[XY\right]}=\frac{0.4\times 0.6}{0.6}=0.4mol{L}^{-1}$