Δ Gf⁰ and Δ Hf⁰ for Fe(g) are 370.7 kJ mol⁻¹ and 416.3 kJ mol⁻¹ at 298 K, respectively. Assuming Δ Hf⁰ is… · ABC26PH0616

Subject: Physical Chemistry · Chapter: Thermodynamics · Topic: Thermodynamics – General · Exam: GATE 2005-2021 · Marks: 2 · Difficulty: Medium

$\Delta \mathrm{G}_{\mathrm{f}}^{0}$ and $\Delta \mathrm{H}_{\mathrm{f}}^{0}$ for $\mathrm{Fe}(\mathrm{g})$ are $370.7 \mathrm{kJ} \mathrm{mol}^{-1}$ and $416.3 \mathrm{kJ} \mathrm{mol}^{-1}$ at 298 K , respectively. Assuming $\Delta \mathrm{H}_{\mathrm{f}}^{0}$ is constant in the interval 250 K to 375 K, $\Delta \mathrm{G}_{\mathrm{f}}^{0}$ (rounded off to the nearest integer) for $\mathrm{Fe}(\mathrm{g})$ at 375 K is:
(a)$338 \mathrm{kJ} \mathrm{mol}^{-1}$
(b)$310 \mathrm{kJ} \mathrm{mol}^{-1}$
(c)$325 \mathrm{kJ} \mathrm{mol}^{-1}$
(d)$359 \mathrm{kJ} \mathrm{mol}^{-1}$
Answer
Answer: D ✓ checked by 4AB · confidence high

The source book printed no answer; this one was worked out and checked — see the explanation.

Explanation
ΔS_f = (ΔH_f − ΔG_f)/T = (416.3 − 370.7)/298 = 0.1530 kJ K⁻¹ mol⁻¹. At 375 K: ΔG_f = 416.3 − 375 × 0.1530 = 358.9 ≈ 359 kJ mol⁻¹.

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