Calculate the standard Gibbs enegry change for the combustion of `alpha-D` glucose at `300K`.
`C_(6)H_(12)O_(6)(s) +6O_(2)(g) rarr 6CO_(2)(g) +6H_(2)O(l)`
Given the standard enthalpies of formation `(kJ mol^(-1))`
`C_(6)H_(12)O_(6) =- 1274.5, CO_(2) =- 393.5, H_(2)O =- 285.8`.
Entropies `(J K mol^(-1))`
`C_(6)H_(12)O_(6) = 212.1, O_(1) = 205.0, CO_(2) =213, H_(2)O = 69.9`

Explanation

`DeltaG^(Theta) = DeltaH^(Theta) - T DeltaS^(Theta)`
`DeltaH^(Theta) =[6 Delta_(f)H^(Theta)(CO_(2)) +6Delta_(f)H^(Theta)(H_(2)O)]`
`-[6Delta_(f)H^(Theta)(C_(6)H_(12)O_(6)) +6Delta_(f)H^(Theta)(O_(2))`
`= [6 xx (-393.5) +6 xx (-285.8)]`
`- 6 xx (-1274.5)+6xx0] [ :' Delta_(f)H^(Theta)(O_(2)) =0]`
`=- 2361.0 - 1714.8 +1274.5 =- 2801.3 kJ`
`DeltaS^(Theta) =[6S^(Theta)(CO_(2)) +6S^(Theta) (H_(2)O)] -[S^(Theta)(C_(6)H_(12)O_(6)) +6S^(Theta)(O_(2))]`
`=[6xx(213.6) +6xx (69.9)] -[(212.1) +6xx(205.0)]`
`= 258.9 J K^(-1)`
`DeltaG^(Theta) = DeltaH^(Theta) - T DeltaS^(Theta)`
`DeltaH^(Theta) =- 2801.3 kJ = 2801300 J mol^(-1)`,
`DeltaS^(Theta) = 258.9 T = 298K`
`:. DeltaG^(Theta) =- 2801300 -300 xx (258.9) =- 2878970J`
or `-2878.97 kJ`