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.
"\u2206G=\u2206H-T\u2206S"
First convert 6.01KJ to joules and -40°C to Kelvin.
"\u2206G=(6.01\u00d71000)-(273-40)22.1=+860.7joules"
Since ∆G is positive, the forward reaction is non spontaneous
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