Question at position 4 For a reaction to be spontaneous at all temperatures, the signs of ΔHº and ΔSº must be__________ and __________, respectively. negative, negativepositive, negativenegative, positivepositive, positivepositive, zero单项选择题
A
negative, negative
B
positive, negative
C
negative, positive
D
positive, positive
E
positive, zero

Question

Question at position 4 For a reaction to be spontaneous at all temperatures, the signs of ΔHº and ΔSº must be and , respectively. negative, negativepositive, negativenegative, positivepositive, positivepositive, zero单项选择题

A

negative, negative

B

positive, negative

C

negative, positive

D

positive, positive

E

positive, zero

BlackTom AI · Accepted Answer

Gibbs free energy ΔGº = ΔHº − TΔSº must be negative for a process to be spontaneous. For spontaneity at all temperatures (all T ≥ 0), the temperature-dependent term cannot change the sign of ΔGº. 1) Relevant concepts - ΔGº < 0 indicates a spontaneous process. - ΔGº depends on ΔHº and ΔSº via ΔGº = ΔHº − TΔSº. - Temperature T (in K) multiplies ΔSº, so the sign of ΔSº controls how ΔGº changes with T. 2) Step-by-step reasoning We require ΔGº = ΔHº − TΔSº < 0 for every T. Consider the sign of ΔSº: - If ΔSº > 0, then −TΔSº is always negative for all T > 0, so ΔGº = ΔHº + (negative term). To guarantee ΔGº < 0 for every T, ΔHº must also be ≤ 0; specifically ΔHº < 0 ensures ΔGº is strictly negative at T = 0 and remains negative as T increases. - If ΔSº ≤ 0, then −TΔSº is zero or positive, so even a negative ΔHº can be overwhelmed at sufficiently large T (making ΔGº positive). Thus ΔSº cannot be nonpositive. 3) Intermediate calculation/example Take ΔHº = −10 kJ/mol and ΔSº = +0.05 kJ/(mol·K) (+50 J/(mol·K)): ΔGº = −10 − T(0.05). At T = 0 K: ΔGº = −10 kJ/mol < 0. At T = 1000 K: ΔGº = −10 − 50 = −60 kJ/mol < 0. ΔGº remains negative for all T > 0. 4) Why the correct answer is correct ΔHº negative provides a favorable enthalpy at low T, and ΔSº positive makes the TΔSº term favorable at high T. Together they force ΔGº < 0 for every temperature. 5) Why other options are incorrect (brief) - negative, negative: ΔGº = negative + T|ΔSº| can become positive at high T. - positive, negative: both terms make ΔGº positive; never spontaneous. - positive, positive: spontaneous only at sufficiently high T, not at all T. - positive, zero: ΔGº = positive; never spontaneous. 6) Conclusion Therefore ΔHº must be negative and ΔSº must be positive. This matches the provided answer: negative, positive.

Question at position 4 For a reaction to be spontaneous at all temperatures, the signs of ΔHº and ΔSº must be and , respectively. negative, negativepositive, negativenegative, positivepositive, positivepositive, zero单项选择题

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