For a reaction at `25^(@)C` enthalpy change `(DeltaH)` and entropy change `(DeltsS)` are `-11.7 K J mol^(-1)` and `-105J mol^(-1) K^(-1)`, respectively. Find out whther this reaction is spontaneous or not?

Explanation

Given, `DeltaH = - 11.7 xx 10^(3) J mol^(-1)`
`DeltaS =- 105 J mol^(-1) K^(-1), T = 298 K`
`DeltaG = DeltaH -T DeltaS =- 11700 - 298 xx (-105)`
`= 19590J =+ 19.59kJ`
Since `DeltaG = +ve`
Therefore, reaction is not spontaneous.