Question

In a closed container, $N{O}_2$ was taken which dimerises to give $N_2{O}_4$. After some time, mole fraction of $N_2{O}_4$ was found to be $\frac{2}{3}$ in the container. Calculate the $\%$ dimerisation of $N{O}_2$?

• A. $80\%$
• B. $33.3\%$
• C. $66.6\%$
• D. $50\%$

Answer 1

The correct option is A $80\%$

Given that $N{O}_2$ dimerises to give $N_2{O}_4$ when taken in a closed container.
The reaction is as follows:

$2N{O}_2\left(g\right)\rightleftharpoons N_2{O}_4\left(g\right)$

After some time, the mole fractions of $2N{O}_2$ and $N_2{O}_4$ are $\frac{1}{3}$ and $\frac{2}{3}$ respectively.

Thus, if total number of moles is $3$, the number of moles of $2N{O}_2$ and $N_2{O}_4$ are $1$ and $2$ respectively.

$2$ moles of $N_2{O}_4$ comes from $4$ moles of $2N{O}_2$.

Hence, total number of moles of $2N{O}_2$ present initially is $1+4=5$ .

The percentage dimerization of $2N{O}_2$ is $\frac{4}{5}\times 100=80\%$.