Calculate the `Delta_(r)G^(@)` for the reaction:
`C_(6)H_(12)O_(6)(s) + 6O_(2)(g) to 6CO_(2)(g) + 6H_(2)O(l)`
`Delta_(f)G^(@)` values (kJ `mol^(-1)`) are :
`C_(12)H_(12)O_(6)(s) = -910.2, CO_(2)(g) = -394.4, H_(2)O(l) = -237.2`

Explanation

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

\[ \text{C}_6\text{H}_{12}\text{O}_6(s) + 6\text{O}_2(g) \rightarrow 6\text{CO}_2(g) + 6\text{H}_2\text{O}(l) \]

we will use the standard Gibbs free energy of formation (Δ_fG°) values provided:

- Δ_fG° for C₆H₁₂O₆(s) = -910.2 kJ/mol
- Δ_fG° for CO₂(g) = -394.4 kJ/mol
- Δ_fG° for H₂O(l) = -237.2 kJ/mol
- Δ_fG° for O₂(g) = 0 kJ/mol (since it is in its standard state)

### Step-by-Step Solution:

1. **Write the Gibbs Free Energy Change Equation:**
The standard Gibbs free energy change for the reaction can be calculated using the formula:
\[ \Delta_rG° = \sum (\Delta_fG° \text{ of products}) - \sum (\Delta_fG° \text{ of reactants}) \]

2. **Calculate the Gibbs Free Energy of Products:**
For the products, we have:
- 6 moles of CO₂:
\[ 6 \times (-394.4 \text{ kJ/mol}) = -2366.4 \text{ kJ} \]
- 6 moles of H₂O:
\[ 6 \times (-237.2 \text{ kJ/mol}) = -1423.2 \text{ kJ} \]
- Total Gibbs Free Energy for products:
\[ -2366.4 \text{ kJ} + (-1423.2 \text{ kJ}) = -3789.6 \text{ kJ} \]

3. **Calculate the Gibbs Free Energy of Reactants:**
For the reactants, we have:
- 1 mole of C₆H₁₂O₆:
\[ -910.2 \text{ kJ} \]
- 6 moles of O₂ (which is 0):
\[ 6 \times 0 \text{ kJ} = 0 \text{ kJ} \]
- Total Gibbs Free Energy for reactants:
\[ -910.2 \text{ kJ} + 0 \text{ kJ} = -910.2 \text{ kJ} \]

4. **Substitute into the Gibbs Free Energy Change Equation:**
Now substitute the values into the equation:
\[ \Delta_rG° = (-3789.6 \text{ kJ}) - (-910.2 \text{ kJ}) \]
\[ \Delta_rG° = -3789.6 + 910.2 = -2879.4 \text{ kJ} \]

5. **Final Result:**
The standard Gibbs free energy change for the reaction is:
\[ \Delta_rG° = -2879.4 \text{ kJ/mol} \]