Question

In a container equilibrium, $N_2{O}_4\left(g\right)\rightleftharpoons 2N{O}_2\left(g\right)$ is attained at ${25}^{\circ }C$. The total equilibrium pressure in container is $380\text{torr}$. If equilibrium constant of above equilibrium is $0.667\text{atm}$, then degree of dissociation of $N_2{O}_4$ at this temperature will be:

• A. $\frac{1}{3}$
• B. $\frac{1}{2}$
• C. $\frac{2}{3}$
• D. $\frac{1}{4}$

Answer 1

The correct option is B $\frac{1}{2}$

$$\begin{array}{cccc}& N_2{O}_4\left(g\right)& \rightleftharpoons & 2N{O}_2\left(g\right)\\ \text{Initial}:& 1& & \\ \text{Final}:& 1-\alpha & & 2\alpha \end{array}$$
$P_{eq}=380\text{torr}=\frac{1}{2}\text{atm}$
$K_p=0.667\text{atm}=\frac{2}{3}\text{atm}{K}_p=\frac{{\left[P_{N{O}_2}\right]}^2}{P_{N_2{O}_4}}=\frac{{\left(\frac{2\alpha }{1+\alpha }{P}_{eq}\right)}^2}{\left(\frac{1-\alpha }{1+\alpha }\right)P_{eq}}\Rightarrow \frac{2}{3}=\frac{(2\alpha {)}^2\times P_{eq}}{(1-\alpha )(1+\alpha )}[\text{Given,}{P}_{eq}=\frac{1}{2}\text{atm}]\Rightarrow \alpha =\frac{1}{2}$