Which combination corresponds to a thermally favorable (exothermic) but entropically unfavorable process?

Which combination corresponds to a thermally favorable (exothermic) but entropically unfavorable process?

• A. +ΔH and +ΔS
• B. -ΔH and -ΔS
• C. -ΔH and +ΔS
• D. +ΔH and -ΔS

Answer: (B)

When a process releases heat, ΔH is negative. When it becomes more ordered (fewer microstates), ΔS is negative. So the combination that is thermally favorable (exothermic) but entropically unfavorable is negative ΔH with negative ΔS. The overall spontaneity is governed by ΔG = ΔH − TΔS. With ΔH negative and ΔS negative, ΔG becomes negative minus a negative term (which is plus T|ΔS|). At low temperatures, the negative ΔH tends to dominate, making ΔG negative and the process spontaneous. As temperature increases, the positive contribution from −TΔS can offset the negative ΔH, potentially making ΔG less favorable. This aligns with the idea of an exothermic process that loses entropy. An example is condensation, where heat is released and the water molecules become more ordered.

When a process releases heat, ΔH is negative. When it becomes more ordered (fewer microstates), ΔS is negative. So the combination that is thermally favorable (exothermic) but entropically unfavorable is negative ΔH with negative ΔS. The overall spontaneity is governed by ΔG = ΔH − TΔS. With ΔH negative and ΔS negative, ΔG becomes negative minus a negative term (which is plus T|ΔS|). At low temperatures, the negative ΔH tends to dominate, making ΔG negative and the process spontaneous. As temperature increases, the positive contribution from −TΔS can offset the negative ΔH, potentially making ΔG less favorable. This aligns with the idea of an exothermic process that loses entropy. An example is condensation, where heat is released and the water molecules become more ordered.