Chemistry

Chemistry

Problem

What signs of ΔH⊖\Delta H^\ominus and ΔS⊖\Delta S^\ominus result in a complete reaction at all temperatures? a. ΔH⊖\Delta H^\ominus: –, ΔS⊖\Delta S^\ominus: – b. ΔH⊖\Delta H^\ominus: +, ΔS⊖\Delta S^\ominus: – c. ΔH⊖\Delta H^\ominus: –, ΔS⊖\Delta S^\ominus: + d. ΔH⊖\Delta H^\ominus: +, ΔS⊖\Delta S^\ominus: +
Intuitive explanation

The key is understanding how ΔG⊖=ΔH⊖−TΔS⊖\Delta G^\ominus = \Delta H^\ominus - T\Delta S^\ominus behaves across the entire temperature range. For a reaction to be complete (spontaneous) everywhere, ΔG⊖\Delta G^\ominus must always be negative, no matter how large or small TT becomes. Think about which combination of signs ensures the −TΔS⊖-T\Delta S^\ominus term always reinforces a negative ΔH⊖\Delta H^\ominus, rather than fighting against it at extreme temperatures.

Intuitive explanation

The key is understanding how ΔG⊖=ΔH⊖−TΔS⊖\Delta G^\ominus = \Delta H^\ominus - T\Delta S^\ominus behaves across the entire temperature range. For a reaction to be complete (spontaneous) everywhere, ΔG⊖\Delta G^\ominus must always be negative, no matter how large or small TT becomes. Think about which combination of signs ensures the −TΔS⊖-T\Delta S^\ominus term always reinforces a negative ΔH⊖\Delta H^\ominus, rather than fighting against it at extreme temperatures.

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