For dissociation of a gas as: (g) ⇌ 2(g) + 1/2 (g). The reaction is performed at constant temperatur — Chemical Equilibrium Chemistry Question
Question
For dissociation of a gas $N_2O_5$ as: $N_2O_5$(g) ⇌ 2$NO_2$(g) + 1/2 $O_2$(g). The reaction is performed at constant temperature and volume. If D is the vapour density of equilibrium mixture, P_o is initial pressure of $N_2O_5$(g) and M is molecular mass of $N_2O_5$, then the correct information(s) at the equilibrium is/are:
💡 Solution & Explanation
Step 1 - Formulate the Equilibrium Table (ICE Table) The dissociation of nitrogen pentoxide gas: \[\ce{N2O5(g) <=> 2NO2(g) + 1/2 O2(g)}\] Start with 1 mol \ce{N2O5}, degree of dissociation = $x$: \[\begin{array}{lccccc} \text{Species} & \ce{N2O5(g)} & \ce{<=>} & \ce{2NO2(g)} & + & \ce{1/2 O2(g)} \ \hline \text{Initial moles} & 1 & & 0 & & 0 \ \text{Change} & -x & & +2x & & +0.5x \ \text{Equilibrium moles} & 1-x & & 2x & & 0.5x \ \end{array}\] Total moles at equilibrium: $n_{\text{total}} = (1-x) + 2x + 0.5x = 1 + 1.5x$ Step 2 - Relate Vapour Density to Degree of Dissociation By mass conservation, total mass = $M$ (molecular mass of \ce{N2O5}). Average molar mass of mixture: $M_{\text{obs}} = \dfrac{M}{1 + 1.5x}$ Since vapour density $D = \dfrac{M_{\text{obs}}}{2}$: \[D = \frac{M}{2(1 + 1.5x)} \implies 1 + 1.5x = \frac{M}{2D}\] Step 3 - Verify Option (A): Total Pressure at Equilibrium At constant $T$ and $V$: pressure is proportional to moles. \[\frac{P_{\text{total}}}{P_o} = \frac{n_{\text{total}}}{n_{\text{initial}}} = \frac{1 + 1.5x}{1} = \frac{M}{2D}\] \[P_{\text{total}} = \frac{P_o M}{2D}\] **Option (A) is correct.** Step 4 - Verify Option (B): Degree of Dissociation \[1.5x = \frac{M}{2D} - 1 = \frac{M - 2D}{2D} \implies x = \frac{M - 2D}{3D}\] **Option (B) is correct.** Step 5 - Verify Option (C): Partial Pressure of \ce{N2O5(g)} \[P_{\ce{N2O5}} = \frac{1-x}{1+1.5x} \times P_{\text{total}} = (1-x) P_o\] \[1 - x = 1 - \frac{M-2D}{3D} = \frac{3D - M + 2D}{3D} = \frac{5D - M}{3D}\] \[P_{\ce{N2O5}} = \frac{(5D - M) P_o}{3D}\] **Option (C) is correct.** Step 6 - Verify Option (D): Partial Pressure of \ce{O2(g)} \[P_{\ce{O2}} = \frac{0.5x}{1+1.5x} \times P_{\text{total}} = \frac{x}{2} P_o = \frac{1}{2} \cdot \frac{M-2D}{3D} \cdot P_o = \frac{(M-2D) P_o}{6D}\] Option (D) states the denominator is $2D$, but the correct denominator is $6D$. **Option (D) is incorrect.** The correct options are **A, B, and C**.