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].
