A reaction is at equilibrium at 100^(@)C...

A reaction is at equilibrium at `100^(@)C` and the enthalpy change for the reaction is `"42.6 kJ mol"^(-1)`. What will be the value of `DeltaS` in `"J K"^(-1)"mol"^(-1)`?

120

426.2

373.1

114.2

Text Solution

Verified by Experts

The correct Answer is:

`DeltaG=DeltaH-TDeltaS`
At equilibiurm, `DeltaS=(DeltaH)/(T)=("42600 J mol"^(-1))/("373 K")="114.2 J K"^(-1)"mol"^(-1)`

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