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Q. If in the reaction $N_{2}O_{4} = 2NO_{2}$, $\alpha$ is that part of $N_{2}O_{4}$ which dissociates, then the number of moles at equilibrium will be

Equilibrium

Solution:

$\underset{\overset{1}{(1-\alpha) }}{N_{2}O_{4}} \rightleftharpoons \underset{\overset{0}{2\alpha }}{2NO_{2}}$
Total mole at equilibrium $= (1- \alpha) + 2\alpha =1+\alpha$