Problem 2 - Entrance Test

For a reaction, ΔG° = −20 kJ/mol at 300 K. If ΔS° = 50 J/mol·K, what is ΔH°?

A. −35 kJ/molB. −40 kJ/molC. −50 kJ/molD. −60 kJ/mol

Correct: D

Using ΔG° = ΔH° − TΔS°, −20 = ΔH° − 300×0.050 → ΔH° = −20 + 15 = −5 kJ/mol. Wait, no. Correct approach: Rearranged formula: ΔH° = ΔG° + TΔS°. So ΔH° = −20 + (300 × 0.05) = −20 + 15 = −5 kJ/mol. Wait, no choices match. Recalculating: 300×50 J = 15 kJ. −20 = ΔH° − 15 → ΔH° = −5 kJ/mol. No option given. Error in data? Let’s re-calculate.