Answered

Westonci.ca is the premier destination for reliable answers to your questions, brought to you by a community of experts. Connect with a community of experts ready to help you find accurate solutions to your questions quickly and efficiently. Discover detailed answers to your questions from a wide network of experts on our comprehensive Q&A platform.

MISSED THIS? Walch Enthalpies of Formation

Enthalpies of Formation Section 79. You can click on the Review link to access the section in your text.

Propane [tex]\left( C_3 H_8 \right)[/tex] burns according to the following balanced equation:

[tex]
C_3 H_8(g) + 5 O_2(g) \rightarrow 3 CO_2(g) + 4 H_2O(g)
[/tex]

Sagot :

GinnyAnswer
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.
Thank you for your visit. We are dedicated to helping you find the information you need, whenever you need it. We hope you found what you were looking for. Feel free to revisit us for more answers and updated information. We're dedicated to helping you find the answers you need at Westonci.ca. Don't hesitate to return for more.