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At what temperature does the reaction become feasible? When ΔH = 492.7 kJmol^-1 and ΔS = 542.6 JK^-1mol^-1.

When a reaction is feasible ΔG = 0. Rearrange the equation ΔG = ΔH - TΔS. So that when ΔG = 0,       T = ΔH / ΔS.

Must convert ΔS from JK^-1mol^-1 to kJK^-1mol^-1 by dividing it by 1000.

Then plug in numbers given so T = (492.7 kJmol^-1) / (0.5426 kJK^-1mol^-1) 

T = 908 K.

Lily W. GCSE Chemistry tutor, A Level Chemistry tutor

11 months ago

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