Calculate the free enegry change when `1mol` of `NaCI` is dissolved in water at `298K`. Given:
a. Lattice enegry of `NaCI =- 778 kJ mol^(-1)`
b. Hydration energy of `NaCI - 774.3 kJ mol^(-1)`
c. Entropy change at `298 K = 43 J mol^(-1)`

Answer

`-9.1555 kj`

Explanation

`= 3.7 kJ mol^(-1) = 3700 J mol^(-1)`
`Delta H_("dissolution") = Delta h_(("ionisation")) + Delta H_(("hydration")) = 778 - 774.3`
`Delta S_("disolution") = 43 J mol^(-1)`
`Delta G_("disolution") = Delta H - T Delta S = 3700 - 298 X 43 = -9114 J, Delta G = - 9.114 kJ`