Answer :
To determine the enthalpy of formation ([tex]\(\Delta H_f\)[/tex]) for [tex]\(O_2(g)\)[/tex] in the given reaction, let's go through the steps systematically.
### Given
- Enthalpy of formation of glucose ([tex]\(C_6H_{12}O_6(s)\)[/tex]) = [tex]\(-1273.02 \, \text{kJ/mol}\)[/tex]
- Enthalpy of formation of carbon dioxide ([tex]\(CO_2(g)\)[/tex]) = [tex]\(-393.5 \, \text{kJ/mol}\)[/tex]
- Enthalpy of formation of water ([tex]\(H_2O(l)\)[/tex]) = [tex]\(-285.83 \, \text{kJ/mol}\)[/tex]
The reaction is:
[tex]\[ C_6H_{12}O_6(s) + 6 O_2(g) \rightarrow 6 CO_2(g) + 6 H_2O(l) \][/tex]
### Formula
The enthalpy change of the reaction ([tex]\(\Delta H_{reaction}\)[/tex]) can be calculated using the formula:
[tex]\[ \Delta H_{reaction} = \sum \Delta H_f(\text{products}) - \sum \Delta H_f(\text{reactants}) \][/tex]
### Breakdown
For the given reaction:
Products:
- 6 moles of [tex]\(CO_2(g)\)[/tex]
- 6 moles of [tex]\(H_2O(l)\)[/tex]
Reactants:
- 1 mole of [tex]\(C_6H_{12}O_6(s)\)[/tex]
- 6 moles of [tex]\(O_2(g)\)[/tex]
### Calculation
Let's use the formula to find [tex]\(\Delta H_{reaction}\)[/tex]:
Step 1: Calculate [tex]\(\sum \Delta H_f(\text{products})\)[/tex]:
[tex]\[ \sum \Delta H_f(\text{products}) = 6 \times \Delta H_f(CO_2(g)) + 6 \times \Delta H_f(H_2O(l)) \][/tex]
[tex]\[ \sum \Delta H_f(\text{products}) = 6 \times (-393.5 \, \text{kJ/mol}) + 6 \times (-285.83 \, \text{kJ/mol}) \][/tex]
[tex]\[ \sum \Delta H_f(\text{products}) = -2361 \, \text{kJ} + (-1714.98 \, \text{kJ}) \][/tex]
[tex]\[ \sum \Delta H_f(\text{products}) = -4075.98 \, \text{kJ/mol} \][/tex]
Step 2: Calculate [tex]\(\sum \Delta H_f(\text{reactants})\)[/tex]:
[tex]\[ \sum \Delta H_f(\text{reactants}) = \Delta H_f(C_6H_{12}O_6(s)) + 6 \times \Delta H_f(O_2(g)) \][/tex]
[tex]\[ \sum \Delta H_f(\text{reactants}) = -1273.02 \, \text{kJ/mol} + 6 \times \Delta H_f(O_2(g)) \][/tex]
Using the fact that elemental oxygen ([tex]\(O_2(g)\)[/tex]) in its standard state has an enthalpy of formation of [tex]\(0 \, \text{kJ/mol}\)[/tex]:
### Conclusion
Therefore, the enthalpy of formation [tex]\(\Delta H_f\)[/tex] for [tex]\(O_2(g)\)[/tex] is:
[tex]\[ \Delta H_f(O_2(g)) = 0 \, \text{kJ/mol} \][/tex]
Hence, the correct answer is:
- [tex]\(\boxed{0 \, \text{kJ/mol}}\)[/tex]
### Given
- Enthalpy of formation of glucose ([tex]\(C_6H_{12}O_6(s)\)[/tex]) = [tex]\(-1273.02 \, \text{kJ/mol}\)[/tex]
- Enthalpy of formation of carbon dioxide ([tex]\(CO_2(g)\)[/tex]) = [tex]\(-393.5 \, \text{kJ/mol}\)[/tex]
- Enthalpy of formation of water ([tex]\(H_2O(l)\)[/tex]) = [tex]\(-285.83 \, \text{kJ/mol}\)[/tex]
The reaction is:
[tex]\[ C_6H_{12}O_6(s) + 6 O_2(g) \rightarrow 6 CO_2(g) + 6 H_2O(l) \][/tex]
### Formula
The enthalpy change of the reaction ([tex]\(\Delta H_{reaction}\)[/tex]) can be calculated using the formula:
[tex]\[ \Delta H_{reaction} = \sum \Delta H_f(\text{products}) - \sum \Delta H_f(\text{reactants}) \][/tex]
### Breakdown
For the given reaction:
Products:
- 6 moles of [tex]\(CO_2(g)\)[/tex]
- 6 moles of [tex]\(H_2O(l)\)[/tex]
Reactants:
- 1 mole of [tex]\(C_6H_{12}O_6(s)\)[/tex]
- 6 moles of [tex]\(O_2(g)\)[/tex]
### Calculation
Let's use the formula to find [tex]\(\Delta H_{reaction}\)[/tex]:
Step 1: Calculate [tex]\(\sum \Delta H_f(\text{products})\)[/tex]:
[tex]\[ \sum \Delta H_f(\text{products}) = 6 \times \Delta H_f(CO_2(g)) + 6 \times \Delta H_f(H_2O(l)) \][/tex]
[tex]\[ \sum \Delta H_f(\text{products}) = 6 \times (-393.5 \, \text{kJ/mol}) + 6 \times (-285.83 \, \text{kJ/mol}) \][/tex]
[tex]\[ \sum \Delta H_f(\text{products}) = -2361 \, \text{kJ} + (-1714.98 \, \text{kJ}) \][/tex]
[tex]\[ \sum \Delta H_f(\text{products}) = -4075.98 \, \text{kJ/mol} \][/tex]
Step 2: Calculate [tex]\(\sum \Delta H_f(\text{reactants})\)[/tex]:
[tex]\[ \sum \Delta H_f(\text{reactants}) = \Delta H_f(C_6H_{12}O_6(s)) + 6 \times \Delta H_f(O_2(g)) \][/tex]
[tex]\[ \sum \Delta H_f(\text{reactants}) = -1273.02 \, \text{kJ/mol} + 6 \times \Delta H_f(O_2(g)) \][/tex]
Using the fact that elemental oxygen ([tex]\(O_2(g)\)[/tex]) in its standard state has an enthalpy of formation of [tex]\(0 \, \text{kJ/mol}\)[/tex]:
### Conclusion
Therefore, the enthalpy of formation [tex]\(\Delta H_f\)[/tex] for [tex]\(O_2(g)\)[/tex] is:
[tex]\[ \Delta H_f(O_2(g)) = 0 \, \text{kJ/mol} \][/tex]
Hence, the correct answer is:
- [tex]\(\boxed{0 \, \text{kJ/mol}}\)[/tex]
