Propane $(C_3H_8(g), \Delta H_f = -103.8 \text{ kJ/mol})$ reacts with oxygen to produce carbon dioxide $(CO_2, \Delta H_f = -393.5 \text{ kJ/mol})$ and water $(H_2O, \Delta H_f = -241.82 \text{ kJ/mol})$ according to the equation below:
$C_3H_8(g) + 5O_2(g) \rightarrow 3CO_2(g) + 4H_2O(g)$
What is the enthalpy of combustion (per mole) of $C_3H_8(g)$?
Use $\Delta H_{\text{rxn}} = \sum(\Delta H_{\text{f, products}}) - \sum(\Delta H_{\text{f, reactants}})$.
A. -2,044.0 kJ/mol
B. -531.5 kJ/mol
C. 531.5 kJ/mol
D. 2,044.0 kJ/mol
Answer (2)
Calculate the total enthalpy of formation for the products: ∑ Δ H f , p ro d u c t s = 3 × ( − 393.5 ) + 4 × ( − 241.82 ) = − 2147.78 kJ/mol .
Calculate the total enthalpy of formation for the reactants: ∑ Δ H f , re a c t an t s = − 103.8 + 5 × ( 0 ) = − 103.8 kJ/mol .
Calculate the enthalpy of combustion: Δ H co mb u s t i o n = − 2147.78 − ( − 103.8 ) = − 2043.98 kJ/mol .
The enthalpy of combustion of C 3 H 8 ( g ) is − 2044.0 kJ/mol .
Explanation
Understanding the Problem We are given the balanced chemical equation for the combustion of propane: C 3 H 8 ( g ) + 5 O 2 ( g ) → 3 C O 2 ( g ) + 4 H 2 O ( g ) We are also given the enthalpies of formation for each compound: Δ H f ( C 3 H 8 ( g )) = − 103.8 kJ/mol Δ H f ( C O 2 ( g )) = − 393.5 kJ/mol Δ H f ( H 2 O ( g )) = − 241.82 kJ/mol Δ H f ( O 2 ( g )) = 0 kJ/mol (since it is an element in its standard state). We need to calculate the enthalpy of combustion using the formula: Δ H r x n = ∑ Δ H f , p ro d u c t s − ∑ Δ H f , re a c t an t s
Calculating Enthalpy of Products First, calculate the total enthalpy of formation of the products: ∑ Δ H f , p ro d u c t s = 3 × Δ H f ( C O 2 ( g )) + 4 × Δ H f ( H 2 O ( g )) ∑ Δ H f , p ro d u c t s = 3 × ( − 393.5 kJ/mol ) + 4 × ( − 241.82 kJ/mol ) ∑ Δ H f , p ro d u c t s = − 1180.5 kJ/mol − 967.28 kJ/mol ∑ Δ H f , p ro d u c t s = − 2147.78 kJ/mol
Calculating Enthalpy of Reactants Next, calculate the total enthalpy of formation of the reactants: ∑ Δ H f , re a c t an t s = Δ H f ( C 3 H 8 ( g )) + 5 × Δ H f ( O 2 ( g )) ∑ Δ H f , re a c t an t s = − 103.8 kJ/mol + 5 × ( 0 kJ/mol ) ∑ Δ H f , re a c t an t s = − 103.8 kJ/mol
Calculating Enthalpy of Combustion Now, calculate the enthalpy of combustion: Δ H co mb u s t i o n = ∑ Δ H f , p ro d u c t s − ∑ Δ H f , re a c t an t s Δ H co mb u s t i o n = − 2147.78 kJ/mol − ( − 103.8 kJ/mol ) Δ H co mb u s t i o n = − 2147.78 kJ/mol + 103.8 kJ/mol Δ H co mb u s t i o n = − 2043.98 kJ/mol
Final Answer The enthalpy of combustion of propane is approximately − 2043.98 kJ/mol . The closest answer choice is − 2044.0 kJ/mol .
Examples
The enthalpy of combustion is crucial in various real-world applications, such as designing efficient engines and power plants. For instance, when engineers design a propane-fueled engine, knowing the precise amount of energy released during combustion helps them optimize the engine's performance and fuel consumption. This ensures the engine operates efficiently, minimizing waste and maximizing power output. Understanding the enthalpy of combustion also aids in assessing the environmental impact of using propane as a fuel source, allowing for informed decisions about its sustainability.
The enthalpy of combustion of propane is calculated by subtracting the total enthalpy of formation of the reactants from that of the products, yielding approximately -2044.0 kJ/mol. Therefore, the answer choice is A. -2044.0 kJ/mol.
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