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For the process A(l) → 2B(g), the internal energy change is ΔU = 2.1 kcal and the entropy change is ΔS = 20 cal K⁻¹ at 300 K. The Gibbs free energy change, in kcal, is:

1. −2.70
2. 2.70
3. −3.30
4. 3.30

Correct answer: −2.70

Solution

dH = dU + (delta n)RT = 2.1 + 2(1.987e-3)(300) = 2.1 + 1.19 = 3.29 kcal. dG = dH - T dS = 3.29 - 300(0.020) = 3.29 - 6.0 = -2.70 kcal.

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