At 525 K, (g) is 80% dissociated at 1 atm. Sufficient inert gas is added at constant pressure to pro — Chemical Equilibrium Chemistry Question

💡 Solution & Explanation

Step 1 - Write the equilibrium and Kp expression For \(\ce{PCl5(g) <=> PCl3(g) + Cl2(g)}\), starting with 1 mol, degree of dissociation \(\alpha\): Total equilibrium moles = \(1 + \alpha\) \[K_p = \frac{\alpha^2 P_{\text{reactive}}}{1 - \alpha^2}\] Step 2 - Calculate Kp from initial conditions Initial: \(\alpha = 0.80\), \(P_{\text{reactive}} = 1\) atm (no inert gas): \[K_p = \frac{(0.80)^2 \times 1}{1 - (0.80)^2} = \frac{0.64}{0.36} = \frac{16}{9} \approx 1.778 \text{ atm}\] Step 3 - Find new reactive pressure after inert gas addition Total pressure = 1 atm (constant). Inert gas partial pressure = 0.9 atm. \[P_{\text{reactive}} = 1.0 - 0.9 = 0.1 \text{ atm}\] Step 4 - Solve for new degree of dissociation Temperature unchanged, so \(K_p = 1.778\) atm: \[1.778 = \frac{\alpha'^2 \times 0.1}{1 - \alpha'^2}\] \[1.778 - 1.778\alpha'^2 = 0.1\alpha'^2\] \[1.778 = 1.878\alpha'^2\] \[\alpha'^2 = \frac{1.778}{1.878} \approx 0.9467\] \[\alpha' = \sqrt{0.9467} \approx 0.973\] \[\text{Percentage dissociation} = \boxed{97.3\%}\] Step 5 - Explain all options * **Option (A) 97.3%**: Correct. Adding inert gas at constant pressure reduces the effective partial pressure of reacting gases from 1 atm to 0.1 atm, shifting equilibrium forward — dissociation increases from 80% to 97.3%. * **Option (B) 80%**: Incorrect. This would be the result only if inert gas were added at constant volume (no pressure dilution). At constant pressure, dissociation increases. * **Option (C) 65.6%**: Incorrect. Dissociation cannot decrease when the effective pressure on reactive species drops. * **Option (D) 4.7%**: Incorrect. Mathematical error — possibly from incorrect formula or wrong Kp calculation.