Question

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

Answer 1

$N_2{O}_4\rightleftharpoons 2N{O}_2$

$1$

$1-\alpha$ $2\alpha \beta$

Original volume $=1$

Existing volume $=(1+\alpha )$

$\therefore 1=\frac{80}{100}\times (1+\alpha )$

$\Rightarrow 5=4+4\alpha$

$\Rightarrow \alpha =\frac{1}{4}$