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Determining the Enthalpy of Reaction from Standard Enthalpies of Formation

Propane [tex]\(\left( C_3H_8 \right)\)[/tex] burns according to the following balanced equation:
[tex]\[ C_3H_8(g) + 5O_2(g) \rightarrow 3CO_2(g) + 4H_2O(g) \][/tex]

Calculate [tex]\(\Delta H_{rxn}^{\circ}\)[/tex] for this reaction using standard enthalpies of formation. (The standard enthalpy of formation of gaseous propane is [tex]\(-103.9 \, \text{kJ/mol}\)[/tex].) Express the enthalpy in kilojoules to four significant figures.

Hint:

The enthalpy of the reaction is found by subtracting the enthalpies of formation of the reactants (because of decomposition) from the enthalpies of formation of the products. To account for the reaction stoichiometry, the coefficients from the balanced equation are multiplied by the corresponding enthalpies of formation.

[tex]\[
\Delta H_{rxn}^{\circ} = -2040 \, \text{kJ}
\][/tex]

Sagot :

GinnyAnsweridn
To determine the standard enthalpy of reaction (ΔH°) for the combustion of propane ([tex]\( C_3H_8 \)[/tex]), we'll use the standard enthalpies of formation for the reactants and products given in the balanced chemical equation:

[tex]\[ C_3H_8 (g) + 5 O_2 (g) \rightarrow 3 CO_2 (g) + 4 H_2O (g) \][/tex]

The standard enthalpy of formation ([tex]\(\Delta H_f^\circ\)[/tex]) values are provided as follows:
- For [tex]\( C_3H_8 \)[/tex]: [tex]\(\Delta H_f^\circ = -103.9 \, \text{kJ/mol} \)[/tex]
- For [tex]\( CO_2 \)[/tex]: [tex]\(\Delta H_f^\circ = -393.5 \, \text{kJ/mol} \)[/tex]
- For [tex]\( H_2O \)[/tex]: [tex]\(\Delta H_f^\circ = -241.8 \, \text{kJ/mol} \)[/tex]

1. Enthalpy Contribution of Reactants:
- For [tex]\( C_3H_8 \)[/tex]: [tex]\(1 \times (-103.9 \, \text{kJ/mol}) = -103.9 \, \text{kJ} \)[/tex]
- Note that the enthalpy of formation for [tex]\( O_2 \)[/tex] is zero because it is an elemental form ([tex]\(\Delta H_f^\circ = 0 \, \text{kJ/mol}\)[/tex]).

Therefore, the total enthalpy of reactants ([tex]\( H_{\text{reactants}} \)[/tex]):
[tex]\[ H_{\text{reactants}} = -103.9 \, \text{kJ} \][/tex]

2. Enthalpy Contribution of Products:
- For [tex]\( CO_2 \)[/tex]: [tex]\(3 \times (-393.5 \, \text{kJ/mol}) = -1180.5 \, \text{kJ} \)[/tex]
- For [tex]\( H_2O \)[/tex]: [tex]\(4 \times (-241.8 \, \text{kJ/mol}) = -967.2 \, \text{kJ} \)[/tex]

Therefore, the total enthalpy of products ([tex]\( H_{\text{products}} \)[/tex]):
[tex]\[ H_{\text{products}} = -1180.5 \, \text{kJ} + -967.2 \, \text{kJ} = -2147.7 \, \text{kJ} \][/tex]

3. Calculate the Standard Enthalpy of Reaction (ΔH°):
The enthalpy of reaction ([tex]\(\Delta H_{\text{reaction}}^\circ\)[/tex]) is calculated by subtracting the total enthalpy of reactants from the total enthalpy of products:
[tex]\[ \Delta H_{\text{reaction}}^\circ = H_{\text{products}} - H_{\text{reactants}} \][/tex]
[tex]\[ \Delta H_{\text{reaction}}^\circ = -2147.7 \, \text{kJ} - (-103.9 \, \text{kJ}) \][/tex]
[tex]\[ \Delta H_{\text{reaction}}^\circ = -2147.7 \, \text{kJ} + 103.9 \, \text{kJ} \][/tex]
[tex]\[ \Delta H_{\text{reaction}}^\circ = -2043.8 \, \text{kJ} \][/tex]

Therefore, the standard enthalpy of reaction (ΔH°) for the combustion of propane to four significant figures is:
[tex]\[ \Delta H_{\text{reaction}}^\circ = -2043.8 \, \text{kJ} \][/tex]
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