One mole of a gaseous mixture containing , , and possesses a mean molar mass of . Upon intense heati — Mole Concept and Some Basic Concepts of Chemistry Chemistry Question
Question
One mole of a gaseous mixture containing $N_2$, $NO_2$, and $N_2O_4$ possesses a mean molar mass of $55.4\text{ g/mol}$. Upon intense heating to a high temperature where all the $N_2O_4$ completely dissociates into $NO_2$ ($N_2O_4 \rightarrow 2NO_2$), the mean molar mass drops to a lower value of $39.57\text{ g/mol}$. What was the exact initial mole ratio of $N_2 : NO_2 : N_2O_4$ in the mixture?
💡 Solution & Explanation
Let the initial moles of $N_2, NO_2, N_2O_4$ be $x, y, z$. Given: $x + y + z = 1$. Initial mean molar mass: $\frac{28x + 46y + 92z}{1} = 55.4 \Rightarrow 28x + 46y + 92z = 55.4$ (Eq 1). After heating, $z$ moles of $N_2O_4$ dissociate into $2z$ moles of $NO_2$. Total moles after heating = $x + y + 2z$. New mean molar mass = $\frac{55.4}{x + y + 2z} = 39.57 \Rightarrow x + y + 2z = \frac{55.4}{39.57} \approx 1.4$. Since $x + y + z = 1$, substituting this gives: $1 + z = 1.4 \Rightarrow z = 0.4$. Substitute $z = 0.4$ into Eq 1: $28x + 46y + 92(0.4) = 55.4 \Rightarrow 28x + 46y = 18.6$ (Eq 2). We know $x + y + 0.4 = 1 \Rightarrow x + y = 0.6 \Rightarrow 28x + 28y = 16.8$ (Eq 3). Subtract (3) from (2): $18y = 1.8 \Rightarrow y = 0.1$. Then $x = 0.6 - 0.1 = 0.5$. Ratio $x : y : z = 0.5 : 0.1 : 0.4$.