Chemistry - Chemical Equilibrium Question with Solution | TestHub

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

$t=0$: $n$

$t=equilibrium$: $n(1-\alpha )$ $2n\alpha$

Let vapour density at $t=0$ is $V{D}_1$ and at equilibrium be $V{D}_2$.

$V{D}_1=\frac{M_{}}{2}=\frac{28+64}{2}=46$.

$V{D}_2=\frac{\left(1-\alpha \right)92+\alpha \left(46\right)2}{2(1+\alpha )}=30$.

$2\left(1+\alpha \right)=\frac{92}{30}=3.0667$

$\alpha =0.5333$

% dissociation =$53.33\%$

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