For the reaction , the degree of dissociation (where ) is directly proportional to . If the initial — Chemical Equilibrium Chemistry Question
Question
For the reaction $N_2O_4(g) \rightleftharpoons 2NO_2(g)$, the degree of dissociation $\alpha$ (where $\alpha \ll 1$) is directly proportional to $\sqrt{V}$. If the initial volume of the container is expanded by a factor of $64$ at constant temperature, by what exact integer factor does the degree of dissociation increase?
💡 Solution & Explanation
The relation is $\alpha \propto \sqrt{V}$. Let the initial volume be $V$ and initial dissociation be $\alpha_1$. When the volume becomes $64V$, the new dissociation $\alpha_2 \propto \sqrt{64V} = 8\sqrt{V}$. Therefore, $\alpha_2 = 8 \alpha_1$, meaning it increases by a factor of 8.