The free energy change when 1 mole of NaCl is dissolved in water at 298 K. is `-x` KJ find out value of 'x' given-
(a) Lattice energy of NaCl `= 778 kJ mol^(-1)`
(b) Hydratin energy of `NaCl = - 775 kJ mol^(-1)`
(c) Enthropy change at `300 K = 40 J mol^(-1)`

Answer

2 kJ

Explanation

`Delta H_("dissolution") = Delta H_(("ionisation")) + Delta H_(("hydroation")) = 778 - 775`
`= 3 kJ mol^(-1) = 300 J mol^(-1) Delta S_("dissolution") = 40 J mol^(-1)`,
`Delta G_("dissolution") = Delta H - T Delta S = 3000 - 300 xx 40 = 9000 J`
`Delta G = -9 kJ`