Discover a wealth of knowledge at Westonci.ca, where experts provide answers to your most pressing questions. Discover a wealth of knowledge from experts across different disciplines on our comprehensive Q&A platform. Join our Q&A platform to connect with experts dedicated to providing accurate answers to your questions in various fields.
Sagot :
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]
We hope this information was helpful. Feel free to return anytime for more answers to your questions and concerns. Thank you for your visit. We're dedicated to helping you find the information you need, whenever you need it. Westonci.ca is your trusted source for answers. Visit us again to find more information on diverse topics.