More practice questions

4. Q4. For PCl₅(g) ⇌ PCl₃(g) + Cl₂(g) at equilibrium, argon is added (a) at constant volume, (b) at constant total pressure. At constant temperature and assuming ideal gases, what happens to the degree of dissociation of PCl₅ in each case?

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   (a) No change: the partial pressures of PCl₅, PCl₃ and Cl₂ are unchanged. (b) The volume increases, the partial pressures of the reacting gases fall, and the equilibrium shifts towards more moles of gas: more PCl₅ dissociates.

5. Q5. Using NCERT Table 6.5 (page 178), is N₂O₄(g) ⇌ 2NO₂(g) endothermic or exothermic? Is that consistent with NCERT’s ΔH = −57.2 kJ mol⁻¹ for 2NO₂(g) ⇌ N₂O₄(g)?

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   K_p rises from 47.9 at 400 K to 1700 at 500 K, so the dissociation is endothermic. That fits: the reverse reaction, dimerisation, has ΔH = −57.2 kJ mol⁻¹, so the dissociation has ΔH = +57.2 kJ mol⁻¹.

6. Q6. In a simple one-peak energy-profile model, a reaction has E_a(forward) = 120 kJ mol⁻¹ and ΔH = −40 kJ mol⁻¹. A catalyst lowers E_a(forward) to 100 kJ mol⁻¹. Find E_a(reverse) before and after, and say what happens to K.

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   E_a(reverse) = E_a(forward) − ΔH = 120 + 40 = 160 kJ mol⁻¹ before and 100 + 40 = 140 kJ mol⁻¹ after: both are lowered by the same 20 kJ mol⁻¹, because both directions pass through the same transition state. ΔH, ΔG^⊖ and so K are unchanged; equilibrium is only reached faster.

7. Q7. Why is ammonia made at around 500 °C and 200 atm with an iron catalyst, rather than at room temperature, where K_p is 6.8 × 10⁵?

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   At room temperature K is large but the reaction is far too slow. A higher temperature gives a useful rate, at the cost of a smaller K (NCERT Table 6.5 already gives 3.6 × 10⁻² at 500 K; at 500 °C, 773 K, it is far smaller); the iron catalyst speeds it up further; the high pressure pushes the equilibrium towards ammonia (4 mol of gas become 2); and removing the ammonia as it forms keeps the reaction going forward.

8. Q8. In the CO₂(g) + C(s) ⇌ 2CO(g) mixture of Example 4, more graphite is added at constant volume and temperature. What happens?

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   The equilibrium gas composition is unchanged if graphite was already present and gas volume is unchanged. Graphite is a pure solid, so it does not appear in K_p = (p_CO)²/p_CO2, and adding it does not change Q_p.