Calculate the standard Gibbs free energy change from the free energies of formation data for the following reaction:
`C_(6)H_(6)(l) +(15)/(2)O_(2)(g) rarr 6CO_(2)(g) +3H_(2)O(g)`
Given that `Delta_(f)G^(Theta) =[C_(6)H_(6)(l)] = 172.8 kJ mol^(-1)`
`Delta_(f)G^(Theta)[CO_(2)(g)] =- 394.4 kJ mol^(-1)`
`Delta_(f)G^(Theta) [H_(2)O(g)] =- 228.6 kJ mol^(-1)`

Explanation

`DeltaG^(Theta) = sum Delta_(f)G^(Theta) ("products") -sum Delta_(f)G^(Theta) ("reactants")`
`={6Delta_(f)G^(Theta) [CO_(2)(g)] +3Delta_(f)G^(Theta) [H_(2)O(g)]}`
`-{Delta_(f)G^(Theta) [C_(6)H_(6)(l)] +(15)/(2)Delta_(f)G^(Theta) [O_(2)(g)]}`
`= {(16xx -394.4)+3 xx(-228.6)} - {172.8 +0}`
`=- 3225 kJ mol^(-1)`.