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Calculate the free energy change when one mole of sodium chloride is dissolved in water at 298 K. (Given : Lattice energy of NaCl `= - 777.8 kJ mol^(-1)`, Hydration energy of NaCl ` = 774.1 kJ mol^(-1) and DeltaS " at" 298 K = 0.043 kJ mol^(-1)`.

A

`-9.114 KJ mol^(-1)`

B

`-11.4 KJ mol^(-1)`

C

`-5.2 KJ mol^(-1)`

D

`-4.5 KJ mol^(-1)`

Text Solution

Verified by Experts

The correct Answer is:
A
`DeltaH_(Hyd).=DeltaH_(Crys).+DeltaH_(Sol)`
`-774.1=-778.8+DeltaH_(Sol)`.
`DeltaG=DeltaH-TDeltaS`
`=4.7-(298xx0.043)`
`-9.114KJmol^(-1)`
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