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Question
For a certain reaction, ∆H = − 50 kJ and ∆S = − 80 J K-1, at what temperature does the
reaction turn from spontaneous to non-spontaneous?
(A) 6.25 K
(B) 62.5 K
(C) 625 K
(D) 6250 K
Solution
625 K
The reaction turns from spontaneous to non-spontaneous when ΔG = 0.
∴ 0 = ΔG = ΔH − T Δ S
T = `"ΔH"/"ΔS" = (-50xx10^3 J)/(-80JK^(-1))` = 625 K
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