(molecular mass = M) dissociates: (g) ⇌ (g) + (g). If total pressure is P and density is d at temper — Chemical Equilibrium Chemistry Question

💡 Solution & Explanation

Step 1 - Set up the ICE table For \(\ce{PCl5(g) <=> PCl3(g) + Cl2(g)}\), starting with 1 mol, let \(\alpha\) = degree of dissociation: | | PCl5 | PCl3 | Cl2 | |--|--|--|--| | Initial | 1 | 0 | 0 | | Change | -α | +α | +α | | Equilibrium | 1-α | α | α | Total moles at equilibrium: \(n_{\text{total}} = 1 + \alpha\) Step 2 - Relate observed molar mass to degree of dissociation Total mass is conserved = M (initial molar mass of PCl5). So: \[M_{\text{obs}} = \frac{M}{1 + \alpha} \implies \alpha = \frac{M}{M_{\text{obs}}} - 1\] Step 3 - Express observed molar mass in terms of density From the ideal gas equation in density form (\(P = \frac{dRT}{M_{\text{obs}}}\)): \[M_{\text{obs}} = \frac{dRT}{P}\] Step 4 - Solve for α \[\alpha = \frac{M}{M_{\text{obs}}} - 1 = \frac{M}{\frac{dRT}{P}} - 1 = \boxed{\frac{PM}{dRT} - 1}\] Step 5 - Explain each option * **Option (A) PM/dRT**: Incorrect. This equals \(M/M_{\text{obs}} = 1+\alpha\), not \(\alpha\) itself. * **Option (B) PM/dRT - 1**: Correct. Equals \(M/M_{\text{obs}} - 1 = \alpha\). * **Option (C) dRT/PM - 1**: Incorrect. This equals \(M_{\text{obs}}/M - 1 = 1/(1+\alpha) - 1\), which is negative. * **Option (D) dRT/PM**: Incorrect. This equals \(M_{\text{obs}}/M = 1/(1+\alpha)\).