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Gibbs-Helmoholtz equation relates the free energy change to the enthalpy and entropy changes of the process as
`(DeltaG)_(PT) = DeltaH - T DeltaS`
The magnitude of `DeltaH` does not change much with the change in temperature but the enrgy factor `T DeltaS` changes appreciably. Thus, spontaneity of a process depends very much on temperature.
The dissolution of `CaCI_(2).6H_(2)O` in a large volume of water is endothermic to the extent of `3.5 kcal mol^(-1)`. For the reaction. `CaCI_(2)(s) +6H_(2)O(l) rarrCaCI_(2).6H_(2)O(s) DeltaH is -23.2 kcal`. The heat of solution of anhydrous `CaCI_(2)` in large quantity of water will be
A. `-16.7 kcal mol^(-1)`
B. `-19.7 kcal mol^(-1)`
C. `19.7 kcal mol^(-1)`
D. `16.7 kcal mol^(-1)`

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Best answer
`DeltaG = G_(P) - G_(R )`
`=- 19.7 kcal mol^(-1)`

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