2019 · Q80
For a certain reaction, ΔG0 = – 45 kJ/mol and ΔH0 = – 90 kJ/mol at 0°C. What is the minimum temperature at which the reaction will become spontaneous, assuming that ΔH0 and ΔS0 are independent of temperature?
General KnowledgeChemistry
- A273 K
- B298 K
- C546 K
- D596 K
Explanation
From delta-G = delta-H - T*delta-S, the entropy change is delta-S = (delta-H - delta-G)/T = (-90 + 45)/273 = -0.1648 kJ/mol-K. The reaction becomes non-spontaneous when delta-G = 0, i.e. at T = delta-H/delta-S = (-90)/(-0.1648), about 546 K. Because delta-S is negative, the reaction is spontaneous below this temperature, making 546 K the boundary temperature. Hence option C is correct.