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

To calculate the standard Gibbs free energy change (ΔG°) for the reaction:

\[ C_{6}H_{6}(l) + \frac{15}{2}O_{2}(g) \rightarrow 6CO_{2}(g) + 3H_{2}O(g) \]

we will use the standard Gibbs free energies of formation (ΔG°f) for each species involved in the reaction. The formula for calculating ΔG° for a reaction is:

\[
\Delta G^{\circ}_{reaction} = \sum (m \cdot \Delta G^{\circ}_{f, products}) - \sum (n \cdot \Delta G^{\circ}_{f, reactants})
...