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To find the enthalpy of formation ([tex]\(\Delta H_f\)[/tex]) for [tex]\(O_2(g)\)[/tex], we'll follow these steps:
1. Write the balanced chemical equation:
[tex]\[ C_6H_{12}O_6(s) + 6O_2(g) \rightarrow 6CO_2(g) + 6H_2O(l) \][/tex]
2. Understand the given enthalpies of formation:
[tex]\[ \Delta H_f \text{ for } C_6H_{12}O_6(s) = -1273.02 \, \text{kJ/mol} \][/tex]
[tex]\[ \Delta H_f \text{ for } CO_2(g) = -393.5 \, \text{kJ/mol} \][/tex]
[tex]\[ \Delta H_f \text{ for } H_2O(l) = -285.83 \, \text{kJ/mol} \][/tex]
3. Recall the enthalpy change for the reaction ([tex]\(\Delta H_{\text{reaction}}\)[/tex]) is calculated with the formula:
[tex]\[ \Delta H_{\text{reaction}} = \left( \sum \Delta H_f \text{ of products} \right) - \left( \sum \Delta H_f \text{ of reactants} \right) \][/tex]
4. Calculate the total enthalpy of the products:
[tex]\[ \Delta H_{\text{products}} = 6 \times \Delta H_f \text{ (CO}_2\text{)} + 6 \times \Delta H_f \text{ (H}_2\text{O)} \][/tex]
[tex]\[ \Delta H_{\text{products}} = 6 \times (-393.5 \, \text{kJ/mol}) + 6 \times (-285.83 \, \text{kJ/mol}) \][/tex]
[tex]\[ \Delta H_{\text{products}} = 6 \times (-393.5) + 6 \times (-285.83) \][/tex]
[tex]\[ \Delta H_{\text{products}} = -2361 \, \text{kJ/mol} + (-1714.98 \, \text{kJ/mol}) \][/tex]
[tex]\[ \Delta H_{\text{products}} = -4075.98 \, \text{kJ/mol} \][/tex]
5. Calculate the total enthalpy of the reactants:
[tex]\[ \Delta H_{\text{reactants}} = \Delta H_f \text{ (C}_6\text{H}_{12}\text{O}_6\text{)} + 6 \times \Delta H_f \text{ (O}_2\text{)} \][/tex]
Given that:
[tex]\[ \Delta H_f \text{ of O}_2\text{(g)} = 0 \text{ kJ/mol (by convention)} \][/tex]
[tex]\[ \Delta H_{\text{reactants}} = -1273.02 \, \text{kJ/mol} + 6 \times 0 \, \text{kJ/mol} \][/tex]
[tex]\[ \Delta H_{\text{reactants}} = -1273.02 \, \text{kJ/mol} \][/tex]
6. Now, calculate [tex]\(\Delta H_{\text{reaction}}\)[/tex]:
[tex]\[ \Delta H_{\text{reaction}} = \Delta H_{\text{products}} - \Delta H_{\text{reactants}} \][/tex]
[tex]\[ \Delta H_{\text{reaction}} = -4075.98 \, \text{kJ/mol} - (-1273.02 \, \text{kJ/mol}) \][/tex]
[tex]\[ \Delta H_{\text{reaction}} = -4075.98 \, \text{kJ/mol} + 1273.02 \, \text{kJ/mol} \][/tex]
[tex]\[ \Delta H_{\text{reaction}} = -2802.96 \, \text{kJ/mol} \][/tex]
Finally, since we are trying to find the [tex]\(\Delta H_f\)[/tex] for [tex]\(O_2(g)\)[/tex] and it's a standard convention that [tex]\(\Delta H_f\)[/tex] for [tex]\(O_2(g)\)[/tex] is exactly:
[tex]\[ \Delta H_f \text{ of } O_2(g) = 0 \text{ kJ/mol} \][/tex]
The answer to the question is [tex]\(0 \text{ kJ/mol}\)[/tex].
So, the correct answer is:
- exactly [tex]\(0 \text{ kJ/mol}\)[/tex].
1. Write the balanced chemical equation:
[tex]\[ C_6H_{12}O_6(s) + 6O_2(g) \rightarrow 6CO_2(g) + 6H_2O(l) \][/tex]
2. Understand the given enthalpies of formation:
[tex]\[ \Delta H_f \text{ for } C_6H_{12}O_6(s) = -1273.02 \, \text{kJ/mol} \][/tex]
[tex]\[ \Delta H_f \text{ for } CO_2(g) = -393.5 \, \text{kJ/mol} \][/tex]
[tex]\[ \Delta H_f \text{ for } H_2O(l) = -285.83 \, \text{kJ/mol} \][/tex]
3. Recall the enthalpy change for the reaction ([tex]\(\Delta H_{\text{reaction}}\)[/tex]) is calculated with the formula:
[tex]\[ \Delta H_{\text{reaction}} = \left( \sum \Delta H_f \text{ of products} \right) - \left( \sum \Delta H_f \text{ of reactants} \right) \][/tex]
4. Calculate the total enthalpy of the products:
[tex]\[ \Delta H_{\text{products}} = 6 \times \Delta H_f \text{ (CO}_2\text{)} + 6 \times \Delta H_f \text{ (H}_2\text{O)} \][/tex]
[tex]\[ \Delta H_{\text{products}} = 6 \times (-393.5 \, \text{kJ/mol}) + 6 \times (-285.83 \, \text{kJ/mol}) \][/tex]
[tex]\[ \Delta H_{\text{products}} = 6 \times (-393.5) + 6 \times (-285.83) \][/tex]
[tex]\[ \Delta H_{\text{products}} = -2361 \, \text{kJ/mol} + (-1714.98 \, \text{kJ/mol}) \][/tex]
[tex]\[ \Delta H_{\text{products}} = -4075.98 \, \text{kJ/mol} \][/tex]
5. Calculate the total enthalpy of the reactants:
[tex]\[ \Delta H_{\text{reactants}} = \Delta H_f \text{ (C}_6\text{H}_{12}\text{O}_6\text{)} + 6 \times \Delta H_f \text{ (O}_2\text{)} \][/tex]
Given that:
[tex]\[ \Delta H_f \text{ of O}_2\text{(g)} = 0 \text{ kJ/mol (by convention)} \][/tex]
[tex]\[ \Delta H_{\text{reactants}} = -1273.02 \, \text{kJ/mol} + 6 \times 0 \, \text{kJ/mol} \][/tex]
[tex]\[ \Delta H_{\text{reactants}} = -1273.02 \, \text{kJ/mol} \][/tex]
6. Now, calculate [tex]\(\Delta H_{\text{reaction}}\)[/tex]:
[tex]\[ \Delta H_{\text{reaction}} = \Delta H_{\text{products}} - \Delta H_{\text{reactants}} \][/tex]
[tex]\[ \Delta H_{\text{reaction}} = -4075.98 \, \text{kJ/mol} - (-1273.02 \, \text{kJ/mol}) \][/tex]
[tex]\[ \Delta H_{\text{reaction}} = -4075.98 \, \text{kJ/mol} + 1273.02 \, \text{kJ/mol} \][/tex]
[tex]\[ \Delta H_{\text{reaction}} = -2802.96 \, \text{kJ/mol} \][/tex]
Finally, since we are trying to find the [tex]\(\Delta H_f\)[/tex] for [tex]\(O_2(g)\)[/tex] and it's a standard convention that [tex]\(\Delta H_f\)[/tex] for [tex]\(O_2(g)\)[/tex] is exactly:
[tex]\[ \Delta H_f \text{ of } O_2(g) = 0 \text{ kJ/mol} \][/tex]
The answer to the question is [tex]\(0 \text{ kJ/mol}\)[/tex].
So, the correct answer is:
- exactly [tex]\(0 \text{ kJ/mol}\)[/tex].
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