Get reliable answers to your questions at Westonci.ca, where our knowledgeable community is always ready to help. Explore our Q&A platform to find in-depth answers from a wide range of experts in different fields. Join our Q&A platform to connect with experts dedicated to providing accurate answers to your questions in various fields.
Sagot :
To calculate the enthalpy change for the combustion of propane, we need to use the enthalpies of formation for the reactants and products involved in the balanced chemical equation. The balanced equation is:
[tex]\[ \text{C}_3\text{H}_8(g) + 5\text{O}_2(g) \rightarrow 3\text{CO}_2(g) + 4\text{H}_2\text{O}(g) \][/tex]
Given the standard enthalpies of formation ([tex]\( \Delta H_f \)[/tex]) in kilojoules per mole (kJ/mol):
- [tex]\( \Delta H_f (\text{C}_3\text{H}_8(g)) = -104.7 \)[/tex] kJ/mol
- [tex]\( \Delta H_f (\text{O}_2(g)) = 0 \)[/tex] kJ/mol (as it is in its standard state)
- [tex]\( \Delta H_f (\text{CO}_2(g)) = -393.5 \)[/tex] kJ/mol
- [tex]\( \Delta H_f (\text{H}_2\text{O}(g)) = -241.8 \)[/tex] kJ/mol
The enthalpy change for the reaction ([tex]\( \Delta H_{\text{reaction}} \)[/tex]) can be calculated using the following formula:
[tex]\[ \Delta H_{\text{reaction}} = \sum (\Delta H_f \text{ of products}) - \sum (\Delta H_f \text{ of reactants}) \][/tex]
1. Calculate the total enthalpy of the reactants:
[tex]\[ \Delta H_{\text{reactants}} = \Delta H_f (\text{C}_3\text{H}_8(g)) + 5 \cdot \Delta H_f (\text{O}_2(g)) \][/tex]
[tex]\[ \Delta H_{\text{reactants}} = -104.7 + 5 \cdot 0 \][/tex]
[tex]\[ \Delta H_{\text{reactants}} = -104.7 \text{ kJ/mol} \][/tex]
2. Calculate the total enthalpy of the products:
[tex]\[ \Delta H_{\text{products}} = 3 \cdot \Delta H_f (\text{CO}_2(g)) + 4 \cdot \Delta H_f (\text{H}_2\text{O}(g)) \][/tex]
[tex]\[ \Delta H_{\text{products}} = 3 \cdot (-393.5) + 4 \cdot (-241.8) \][/tex]
[tex]\[ \Delta H_{\text{products}} = -1180.5 + (-967.2) \][/tex]
[tex]\[ \Delta H_{\text{products}} = -2147.7 \text{ kJ/mol} \][/tex]
3. Calculate the enthalpy change for the reaction:
[tex]\[ \Delta H_{\text{reaction}} = \Delta H_{\text{products}} - \Delta H_{\text{reactants}} \][/tex]
[tex]\[ \Delta H_{\text{reaction}} = -2147.7 - (-104.7) \][/tex]
[tex]\[ \Delta H_{\text{reaction}} = -2147.7 + 104.7 \][/tex]
[tex]\[ \Delta H_{\text{reaction}} = -2043 \text{ kJ/mol} \][/tex] (approximately)
Thus, the enthalpy change for the combustion of propane is approximately [tex]\( -2043 \text{ kJ/mol} \)[/tex]. This negative value indicates that the reaction is exothermic, releasing that amount of energy.
[tex]\[ \text{C}_3\text{H}_8(g) + 5\text{O}_2(g) \rightarrow 3\text{CO}_2(g) + 4\text{H}_2\text{O}(g) \][/tex]
Given the standard enthalpies of formation ([tex]\( \Delta H_f \)[/tex]) in kilojoules per mole (kJ/mol):
- [tex]\( \Delta H_f (\text{C}_3\text{H}_8(g)) = -104.7 \)[/tex] kJ/mol
- [tex]\( \Delta H_f (\text{O}_2(g)) = 0 \)[/tex] kJ/mol (as it is in its standard state)
- [tex]\( \Delta H_f (\text{CO}_2(g)) = -393.5 \)[/tex] kJ/mol
- [tex]\( \Delta H_f (\text{H}_2\text{O}(g)) = -241.8 \)[/tex] kJ/mol
The enthalpy change for the reaction ([tex]\( \Delta H_{\text{reaction}} \)[/tex]) can be calculated using the following formula:
[tex]\[ \Delta H_{\text{reaction}} = \sum (\Delta H_f \text{ of products}) - \sum (\Delta H_f \text{ of reactants}) \][/tex]
1. Calculate the total enthalpy of the reactants:
[tex]\[ \Delta H_{\text{reactants}} = \Delta H_f (\text{C}_3\text{H}_8(g)) + 5 \cdot \Delta H_f (\text{O}_2(g)) \][/tex]
[tex]\[ \Delta H_{\text{reactants}} = -104.7 + 5 \cdot 0 \][/tex]
[tex]\[ \Delta H_{\text{reactants}} = -104.7 \text{ kJ/mol} \][/tex]
2. Calculate the total enthalpy of the products:
[tex]\[ \Delta H_{\text{products}} = 3 \cdot \Delta H_f (\text{CO}_2(g)) + 4 \cdot \Delta H_f (\text{H}_2\text{O}(g)) \][/tex]
[tex]\[ \Delta H_{\text{products}} = 3 \cdot (-393.5) + 4 \cdot (-241.8) \][/tex]
[tex]\[ \Delta H_{\text{products}} = -1180.5 + (-967.2) \][/tex]
[tex]\[ \Delta H_{\text{products}} = -2147.7 \text{ kJ/mol} \][/tex]
3. Calculate the enthalpy change for the reaction:
[tex]\[ \Delta H_{\text{reaction}} = \Delta H_{\text{products}} - \Delta H_{\text{reactants}} \][/tex]
[tex]\[ \Delta H_{\text{reaction}} = -2147.7 - (-104.7) \][/tex]
[tex]\[ \Delta H_{\text{reaction}} = -2147.7 + 104.7 \][/tex]
[tex]\[ \Delta H_{\text{reaction}} = -2043 \text{ kJ/mol} \][/tex] (approximately)
Thus, the enthalpy change for the combustion of propane is approximately [tex]\( -2043 \text{ kJ/mol} \)[/tex]. This negative value indicates that the reaction is exothermic, releasing that amount of energy.
Thanks for using our platform. We're always here to provide accurate and up-to-date answers to all your queries. Thank you for visiting. Our goal is to provide the most accurate answers for all your informational needs. Come back soon. We're glad you chose Westonci.ca. Revisit us for updated answers from our knowledgeable team.
