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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.
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