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 energy factor `T DeltaS` changes appreciably. Thus, spontaneity of a process depends very much on temperature.
The dissolution of `CaCl_(2).6H_(2)O` in a large volume of water is endothermic to the extent of `3.5 kcal mol^(-1)`. For the reaction. `CaCl_(2)(s) +6H_(2)O(l) rarrCaCl_(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

Options

A. (a)`"log"_(e)(T_(2))/(T_(1))`
B. (b)`((1)/(T_(2)^(2))-(1)/(T_(1)^(2)))`
C. (c)`((1)/(T_(2)^(3))-(1)/(T_(1)^(3)))`
D. (d)`((1)/(T_(2)^(4))-(1)/(T_(1)^(4)))`

Answer

C

Explanation

As the body is isolated so, the rate of cooling
`-ms(d theta)/(dt)=Asigmatheta^(4)or,(d theta)/(theta^(4))=-(Asigma)/(ms)dt,tprop(1)/(T_(2)^(3))-(1)/(T_(1)^(3))`