1.

Find the value of `DeltaG^(@)` if the `DeltaH^(@)=-29.8` kcal and `DeltaS^(@)=-0.100` kcal `K^(-1)` is given at 298K.

Answer» Correct Answer - A
`DeltaG^(@)=DeltaH^(@)-TDeltaS^(@)=-29.8-298xx(-0.1)`
`=-29.8+29.8=0`


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