The observed molar mass () of a gaseous equilibrium mixture undergoing dissociation is related to th — Chemical Equilibrium Chemistry Question
Question
The observed molar mass ($M_{mix}$) of a gaseous equilibrium mixture undergoing dissociation is related to the theoretical molar mass ($M_{th}$) by the expression $M_{mix} = \frac{M_{th}}{1 + \alpha(n - 1)}$. For which of the following reversible gaseous reactions is it strictly true that $M_{mix} = \frac{M_{th}}{1 + \alpha}$?
💡 Solution & Explanation
The expression $M_{mix} = \frac{M_{th}}{1 + \alpha}$ matches the general formula when $n - 1 = 1$, which means $n = 2$. This implies that $1\text{ mole}$ of reactant must yield $2\text{ moles}$ of products. $N_2O_4 \rightleftharpoons 2NO_2$ ($1 \to 2$). $PCl_5 \rightleftharpoons PCl_3 + Cl_2$ ($1 \to 2$). $COCl_2 \rightleftharpoons CO + Cl_2$ ($1 \to 2$). However, $2NH_3 \rightleftharpoons N_2 + 3H_2$ scales to $1\text{ NH}_3 \rightleftharpoons 0.5\text{ N}_2 + 1.5\text{ H}_2$ ($1 \to 2$), so $n=2$ works there too if defined per mole of $NH_3$, BUT standard derivation defines $n$ based on the balanced equation. If written as $2NH_3$, total moles $= 2 + 2\alpha$. $M_{mix} = \frac{2M_{th}}{2+2\alpha} = \frac{M_{th}}{1+\alpha}$. Thus all A, B, and D apply perfectly for $1\text{ mole}$ stoichiometry.