Consider the reaction of glucose with oxygen:

[tex]\[ C_6H_{12}O_6(s) + 6O_2(g) \rightarrow 6CO_2(g) + 6H_2O(l) \][/tex]

The enthalpy of formation $\left(\Delta H_f\right)$ for $ C_6H_{12}O_6(s) $ is $-1,273.02 \, \text{kJ/mol}$, $\Delta H_f$ for $ CO_2(g) $ is $-393.5 \, \text{kJ/mol}$, and $\Delta H_f$ for $ H_2O(l) $ is $-285.83 \, \text{kJ/mol}$.

What is [tex]$\Delta H_f$[/tex] for [tex]$ O_2(g) $[/tex]?

To determine the enthalpy of formation for $ O_2(g) $ and the enthalpy change of the reaction, we can follow a detailed, step-by-step process:

1. Identify the Given Data:
- Enthalpy of formation ($\Delta H_f$) for glucose ($ C_6H_{12}O_6(s) $): $-1273.02 \ \text{kJ/mol}$
- Enthalpy of formation ($\Delta H_f$) for carbon dioxide ($ CO_2(g) $): $-393.5 \ \text{kJ/mol}$
- Enthalpy of formation ($\Delta H_f$) for water ($ H_2O(l) $): $-285.83 \ \text{kJ/mol}$

2. Write the Balanced Chemical Equation:
[tex]\[ C_6H_{12}O_6(s) + 6O_2(g) \rightarrow 6CO_2(g) + 6H_2O(l) \][/tex]

3. Calculate the Enthalpy of the Reaction ($\Delta H_{reaction}$):

The enthalpy change for the reaction ($\Delta H_{reaction}$) can be calculated using the formula:
[tex]\[ \Delta H_{reaction} = \left[ \sum \Delta H_f (\text{products}) \right] - \left[ \sum \Delta H_f (\text{reactants}) \right] \][/tex]

For this reaction, the products are $ 6CO_2(g) $ and $ 6H_2O(l) $, and the reactants are $ C_6H_{12}O_6(s) $ and $ 6O_2(g) $.

4. Determine the Enthalpy of Formation of Products:
[tex]\[ 6 \Delta H_f (CO_2) + 6 \Delta H_f (H_2O) = 6 \times (-393.5 \ \text{kJ/mol}) + 6 \times (-285.83 \ \text{kJ/mol}) = -2361.0 \ \text{kJ} + (-1714.98 \ \text{kJ}) = -4075.98 \ \text{kJ} \][/tex]

5. Determine the Enthalpy of Formation of Reactants:
[tex]\[ \Delta H_f (C_6H_{12}O_6) + 6 \Delta H_f (O_2) = -1273.02 \ \text{kJ} + 6 \times 0 \ \text{kJ} = -1273.02 \ \text{kJ} \][/tex]

Note: The enthalpy of formation ($\Delta H_f$) of $ O_2(g) $ in its standard state is zero ($0 \ \text{kJ/mol}$).

6. Calculate the Enthalpy Change for the Reaction:
[tex]\[ \Delta H_{reaction} = \big( -4075.98 \ \text{kJ} \big) - \big( -1273.02 \ \text{kJ} \big) = -4075.98 \ \text{kJ} + 1273.02 \ \text{kJ} = -2802.96 \ \text{kJ} \][/tex]

7. Conclusion:
- The enthalpy change for the reaction ($\Delta H_{reaction}$) is $-2802.96 \ \text{kJ}$.
- The enthalpy of formation ($\Delta H_f$) for $ O_2(g) $ is $ 0 \ \text{kJ/mol} $ (since it is an element in its standard state).

Therefore, the enthalpy of formation for [tex]$ O_2(g) $[/tex] is [tex]$ 0 \ \text{kJ/mol} $[/tex].