Westonci.ca offers fast, accurate answers to your questions. Join our community and get the insights you need now. Discover reliable solutions to your questions from a wide network of experts on our comprehensive Q&A platform. Explore comprehensive solutions to your questions from knowledgeable professionals across various fields on our platform.
Sagot :
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].
Visit us again for up-to-date and reliable answers. We're always ready to assist you with your informational needs. We hope this was helpful. Please come back whenever you need more information or answers to your queries. Thank you for visiting Westonci.ca. Stay informed by coming back for more detailed answers.
