Question

At 273 K and 1 atm, 10 litre of $N_2{O}_4$ decomposes to $N{O}_2$ according to equation :
$N_2{O}_4\rightleftharpoons 2N{O}_2\left(g\right)$
What is degree of dissociation (a) when the original volume is 25% less that of existing volume?

• A. 0.25
• B. 0.33
• C. 0.66
• D. 0.5

Answer 1

The correct option is B 0.33

Let the initial volume of $N_2{O}_4$ be $x$ and initial volume of $N{O}_2$ is $0$

If the degree of dissociation is $a$, then final volume of $N_2{O}_4$ is $x(1-a)$ and $N{O}_2$ is $2ax$.

$\begin{array}{cccc}& N_2{O}_4& ⟶& 2N{O}_2\\ \text{Initial}& x& & 0\\ \text{It equlibrium}& x(1-a)& & 2ax\end{array}$

Total initial volume $=x+0=x$

Final volume $=x(1-a)+2ax=x+ax=x(1+a)$

It is given that the initial volume is $25\%$ less than the final volume

$x=0.75\times (1+a)$

$1+a=1.33$

$a=0.33$

$\therefore$ option B is correct