Propane $(C_3 H_8(g), \Delta H_{ f }=-103.8 kJ / mol)$ reacts with oxygen to produce carbon dioxide $(CO_2, \Delta H_{ f }=-393.5 kJ / mol)$ and water ($H_2 O , \Delta H_f=-241.82 kJ / mol$) according to the equation below.
$C_3 H_8(g)+5 O_2(g) \rightarrow 3 CO_2(g)+4 H_2 O(g)$
What is the enthalpy of combustion (per mole) of $C_3 H_8(g)$ ?
Use $\Delta H_{\text {pxn }}=\sum(\Delta H_{\text {t, products }})-\sum(\Delta H_{\text {t, reactants }})$.
A. $-2,044.0 kJ / mol$
B. $-531.5 kJ / mol$
C. $531.5 kJ / mol$
D. $2,044.0 kJ / mol$
Answer (1)
Calculate the total enthalpy of formation for the products: ∑ Δ H f ( products ) = 3 × ( − 393.5 kJ/mol ) + 4 × ( − 241.82 kJ/mol ) = − 2147.78 kJ/mol .
Calculate the total enthalpy of formation for the reactants: ∑ Δ H f ( reactants ) = − 103.8 kJ/mol + 5 × ( 0 kJ/mol ) = − 103.8 kJ/mol .
Apply the formula for enthalpy of reaction: Δ H r x n = ∑ Δ H f ( products ) − ∑ Δ H f ( reactants ) .
Substitute the values and calculate: Δ H r x n = − 2147.78 kJ/mol − ( − 103.8 kJ/mol ) = − 2043.98 kJ/mol . The closest option is − 2 , 044.0 kJ/mol .
Explanation
Problem Setup and Formula We are given the balanced chemical equation for the combustion of propane: C 3 H 8 ( g ) + 5 O 2 ( g ) i g h t ha r p oo n u p 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 ( Δ H r x n ) using the formula: Δ H r x n = ∑ Δ H f ( products ) − ∑ Δ H f ( reactants )
Calculating Enthalpy of Formation for Products First, let's calculate the total enthalpy of formation for the products: ∑ Δ H f ( products ) = 3 × Δ H f ( C O 2 ) + 4 × Δ H f ( H 2 O ) ∑ Δ H f ( products ) = 3 × ( − 393.5 kJ/mol ) + 4 × ( − 241.82 kJ/mol ) ∑ Δ H f ( products ) = − 1180.5 kJ/mol − 967.28 kJ/mol ∑ Δ H f ( products ) = − 2147.78 kJ/mol
Calculating Enthalpy of Formation for Reactants Next, let's calculate the total enthalpy of formation for the reactants: ∑ Δ H f ( reactants ) = Δ H f ( C 3 H 8 ) + 5 × Δ H f ( O 2 ) ∑ Δ H f ( reactants ) = − 103.8 kJ/mol + 5 × ( 0 kJ/mol ) ∑ Δ H f ( reactants ) = − 103.8 kJ/mol
Calculating Enthalpy of Combustion Now, we can calculate the enthalpy of reaction: Δ H r x n = ∑ Δ H f ( products ) − ∑ Δ H f ( reactants ) Δ H r x n = − 2147.78 kJ/mol − ( − 103.8 kJ/mol ) Δ H r x n = − 2147.78 kJ/mol + 103.8 kJ/mol Δ H r x n = − 2043.98 kJ/mol
Final Answer The enthalpy of combustion (per mole) of C 3 H 8 ( g ) is approximately − 2043.98 kJ/mol . Among the given options, the closest value is − 2044.0 kJ/mol .
Examples
Understanding enthalpy of combustion is crucial in various real-world applications, especially in energy production and chemical engineering. For instance, when designing a propane-fueled heating system for a home, knowing the enthalpy of combustion helps engineers calculate how much propane is needed to generate a specific amount of heat. This ensures the system is efficient and cost-effective. Similarly, in the automotive industry, understanding the combustion enthalpy of fuels like propane is essential for optimizing engine performance and reducing emissions. By accurately determining the energy released during combustion, engineers can design more efficient and environmentally friendly vehicles.
