Calculate the free energy change on dissolving one mole of sodium chloride at `25^(@)C.` Lattice energy `=- 777.8 kJ mol^(-1)` hydration energy `=-774 kJ mol^(-1)` of `NaCl, Delta S` at `25 ^(@)C = 0.043 kJ mol ^(-1) K ^(-1).`

Explanation

`Delta H _("solution") = Delta H _("latice") + Delta H _("hydration") = 777.8 kJ mol ^(-1) + (-774.1 kJ mol ^(-1))`
`=777.8 kJ mol ^(-1) - 774.1kJ mol ^(-1) = 3.7 kJ mol ^(-1)`
According to Givvs -Helmholiz equation, `Delta G = Delta H - T Delta S`
`Delta G = 3.7 kJ mol ^(-1) - ( 273 + 25) K xx 0. 0 43 kJ mol ^(-1) K ^(-1)`
`= 3.7 kJ mol ^(-1) -298 xx 0. 0 43 kJ mol ^(-1) = 3.7 kJ mol ^(-1) - 12.814 kJ mol ^(-1) =- 9.114 kJ mol ^(-1)`