For the reaction,
`2NO_(g) + O_(2)(g) rarr 2NO_(2)(g)`
Calculate `DeltaG` at `700K` when enthalpy and entropy changes are `-113.0 kJ mol^(-1)` and `-145 J K^(-1) mol^(-1)`, respectively.

Explanation

We know, `DeltaG = DeltaH - T DeltaS`
Given `DeltaH =- 113 kJ mol^(-1) = - 113000 J mol^(-1)`
`DeltaS =- 145 J K^(-1) mol^(-1)`
`T = 700 K`
Substituting these values in the above equation, we get
`DeltaG =- 113000 - 700 xx (-45)`
`=- 11500 J mol^(-1) =- 11.5 k J mol^(-1)`