Ethyne ( [tex]$C _2 H _2$[/tex] (g), [tex]$\Delta H_f=226.77 kJ / mol$[/tex] ) undergoes complete combustion in the presence of oxygen to produce carbon dioxide ([tex]$CO _2(g), \Delta H_f=-393.5 kJ / mol$[/tex]) and water ([tex]$H _2 O ( g ), \Delta H_f=-241.82 kJ / mol$[/tex]) according to the equation below.
[tex]$2 C_2 H_2(g)+5 O_2(g) \rightarrow 4 CO_2(g)+2 H_2 O(g)$[/tex]
What is the enthalpy of combustion (per mole) of [tex]$C _2 H _2(g)$[/tex]?
Use [tex]$\Delta H_{r \times n}=\sum\left(\Delta H_{f, \text { products }}\right)-\sum\left(\Delta H_{f, \text { reactants }}\right)$[/tex].
A. -2511.2 kJ / mol
B. -1255.6 kJ / mol
C. -862.1 kJ / mol
D. -431.0 kJ / mol
Answer (1)
Calculate the total enthalpy of formation for the products: 4 × ( − 393.5 kJ/mol ) + 2 × ( − 241.82 kJ/mol ) = − 2057.64 kJ/mol .
Calculate the total enthalpy of formation for the reactants: 2 × ( 226.77 kJ/mol ) + 5 × ( 0 kJ/mol ) = 453.54 kJ/mol .
Calculate the enthalpy of reaction: − 2057.64 kJ/mol − 453.54 kJ/mol = − 2511.18 kJ/mol .
Divide the enthalpy of reaction by 2 to find the enthalpy of combustion per mole of ethyne: 2 − 2511.18 kJ/mol = − 1255.6 kJ/mol .
Explanation
Problem Setup and Formula We are given the balanced chemical equation for the combustion of ethyne ( C 2 H 2 ) and the enthalpies of formation ( Δ H f ) for each compound involved. Our goal is to calculate the enthalpy of combustion per mole of ethyne 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, we need to calculate the total enthalpy of formation for the products. The products are 4 moles of C O 2 ( g ) and 2 moles of H 2 O ( g ) .
∑ ( Δ H f , p ro d u c t s ) = 4 × Δ H f ( C O 2 ) + 2 × Δ H f ( H 2 O ) ∑ ( Δ H f , p ro d u c t s ) = 4 × ( − 393.5 kJ/mol ) + 2 × ( − 241.82 kJ/mol ) ∑ ( Δ H f , p ro d u c t s ) = − 1574 kJ/mol − 483.64 kJ/mol = − 2057.64 kJ/mol
Calculating Enthalpy of Reactants Next, we calculate the total enthalpy of formation for the reactants. The reactants are 2 moles of C 2 H 2 ( g ) and 5 moles of O 2 ( g ) .
∑ ( Δ H f , re a c t an t s ) = 2 × Δ H f ( C 2 H 2 ) + 5 × Δ H f ( O 2 ) ∑ ( Δ H f , re a c t an t s ) = 2 × ( 226.77 kJ/mol ) + 5 × ( 0 kJ/mol ) ∑ ( Δ H f , re a c t an t s ) = 453.54 kJ/mol + 0 kJ/mol = 453.54 kJ/mol
Calculating Enthalpy of Reaction Now, we can calculate the enthalpy of reaction 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 ) Δ H r x n = − 2057.64 kJ/mol − 453.54 kJ/mol = − 2511.18 kJ/mol
Enthalpy of Combustion per Mole The balanced equation represents the combustion of 2 moles of ethyne. To find the enthalpy of combustion per mole of ethyne, we divide the calculated Δ H r x n by 2: Δ H co mb u s t i o n = 2 − 2511.18 kJ/mol = − 1255.59 kJ/mol Therefore, the enthalpy of combustion per mole of C 2 H 2 ( g ) is approximately -1255.6 kJ/mol.
Final Answer The enthalpy of combustion (per mole) of C 2 H 2 ( g ) is − 1255.6 kJ/mol .
Examples
Understanding the enthalpy of combustion is crucial in various real-world applications, such as designing efficient engines and heating systems. For instance, when engineers develop a new type of combustion engine, they need to know the precise amount of energy released by the fuel to optimize the engine's performance and fuel efficiency. Similarly, in designing heating systems for homes or industrial facilities, knowing the enthalpy of combustion helps in selecting the right fuel and designing a system that provides the required heat output while minimizing fuel consumption and environmental impact. This ensures both cost-effectiveness and sustainability in energy usage.
