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Calculate the Gibb's energy change when ...

Calculate the Gibb's energy change when 1 mole of NaCI is dissolved in water at `25^(@)C`. Littice energy of `NaCI=777.8 KJ mol^(-1),DeltaS` for dissolution `= 0.043 KJ mol^(-1)` and hydration energy of `NaCI=-774.1 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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