1.

Calculate the entropy change of a process possessing ∆Ht = 2090 J mole-1

Answer»

Given,

Sn(α 13°C) - Sn(β 13°C)

∆Ht = 2090 J mol-1

∆S= ∆H/ Tt

∆S= 2090 / 286

∆St = 7.307 J K-1 mol-1



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