Question

Given the following values, ∆Ho = 6.01kJmol-1 and ∆So = 22.1 J K-1mol-1. Evaluate if this reaction would be spontaneous at -40°C.

Answer 1

$∆G=∆H-T∆S$

First convert 6.01KJ to joules and -40°C to Kelvin.

$∆G=(6.01×1000)-(273-40)22.1=+860.7joules$

Since ∆G is positive, the forward reaction is non spontaneous