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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? - Chemistry

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

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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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