More practice questions

4. Q4. For N₂(g) + 3H₂(g) ⇌ 2NH₃(g) at 298 K, Δ_rH^⊖ = −92.2 kJ mol⁻¹ and Δ_rS^⊖ = −198.7 J K⁻¹ mol⁻¹. Find Δ_rG^⊖ and K.

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   Δ_rG^⊖ = −92.2 − 298 × (−0.1987) = −92.2 + 59.2 = −33.0 kJ mol⁻¹. log K = 33 000/5706, about 5.78, so K ≈ 6 × 10⁵.

5. Q5. Using the same values, estimate Δ_rG^⊖ and K for the ammonia synthesis at 500 K. Why is the reaction still run hot in industry?

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   Δ_rG^⊖ ≈ −92.2 + 500 × 0.1987 = +7.15 kJ mol⁻¹; log K = −7150/(2.303 × 8.314 × 500), about −0.75, so K ≈ 0.18. K falls as the temperature rises (exothermic, ΔS negative), but at room temperature the reaction is far too slow; a higher temperature (with a catalyst and high pressure) gives a useful rate.

6. Q6. The ionic product of water is 1.0 × 10⁻¹⁴ at 298 K. Find Δ_rG^⊖ for H₂O(l) ⇌ H⁺(aq) + OH⁻(aq).

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   Δ_rG^⊖ = −2.303 × 8.314 × 298 × log(1.0 × 10⁻¹⁴) = −5706 × (−14), about +79.9 kJ mol⁻¹.

7. Q7. At equilibrium Δ_rG = 0. Does that mean Δ_rG^⊖ = 0 too? Explain with NCERT Problem 5.12.

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   No. Δ_rG refers to the actual mixture, which at equilibrium sits at the minimum; Δ_rG^⊖ refers to all substances in their standard states and equals −RT ln K. For oxygen and ozone at equilibrium Δ_rG = 0, while Δ_rG^⊖ = +163 kJ mol⁻¹, because K is only 2.47 × 10⁻²⁹.

8. Q8. A student writes Δ_rG^⊖ = −2.303 RT ln K for K = 100 at 298 K. What answer does she get, and what is the right one?

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   She gets −2.303 × 8.314 × 298 × ln 100 = −2.303 × 2478 × 4.605, about −26.3 kJ mol⁻¹. The 2.303 belongs only with log: −RT ln 100 = −2.303 RT log 100 = −2478 × 4.605, about −11.4 kJ mol⁻¹.