Using the information in the table, calculate the enthalpy of combustion of 1 mole of acetylene for the reaction:
[tex]2 C_2 H_2 + 5 O_2 \rightarrow 4 CO_2 + 2 H_2 O[/tex]
| Compound | [tex]\Delta H _{ f } (kJ/mol)[/tex] |
| :------------------------- | :------------------------ |
| methane ([tex]CH_4[/tex]) | -74.8 |
| acetylene ([tex]C_2 H_2[/tex]) | 226.8 |
| ethanol ([tex]C_2 H_6 O[/tex]) | -235.1 |
| carbon dioxide ([tex]CO_2[/tex]) | -393.5 |
| water ([tex]H_2 O[/tex]) | -241.8 |
Answer (2)
Calculate the enthalpy change for the reaction using the formula: Δ H re a c t i o n = ∑ Δ H f ( p ro d u c t s ) − ∑ Δ H f ( re a c t an t s ) .
Substitute the given values into the equation: Δ H re a c t i o n = [ 4 × ( − 393.5 ) + 2 × ( − 241.8 )] − [ 2 × ( 226.8 ) + 5 × ( 0 )] .
Calculate the value of Δ H re a c t i o n : Δ H re a c t i o n = − 2511.2 k J .
Divide the result by 2 to find the enthalpy of combustion for 1 mole of acetylene: Δ H co mb u s t i o n = 2 − 2511.2 = − 1255.6 k J / m o l . The final answer is − 1255.6 .
Explanation
Problem Setup and Given Data We are given the balanced chemical equation for the combustion of acetylene: 2 C 2 H 2 + 5 O 2 → 4 C O 2 + 2 H 2 O We are also given the standard enthalpies of formation ( Δ H f ) for acetylene ( C 2 H 2 ), carbon dioxide ( C O 2 ), and water ( H 2 O ). We need to calculate the enthalpy of combustion for 1 mole of acetylene.
Enthalpy of Reaction Formula The standard enthalpy of reaction can be calculated using the formula: Δ H re a c t i o n = ∑ Δ H f ( p ro d u c t s ) − ∑ Δ H f ( re a c t an t s ) In this case, the products are carbon dioxide and water, and the reactants are acetylene and oxygen.
Expanding the Formula We can expand the formula as follows: Δ H re a c t i o n = [ 4 × Δ H f ( C O 2 ) + 2 × Δ H f ( H 2 O )] − [ 2 × Δ H f ( C 2 H 2 ) + 5 × Δ H f ( O 2 )] We know that the standard enthalpy of formation for any element in its standard state (like O 2 ) is 0 kJ/mol.
Substituting Values and Calculating Now, we substitute the given values into the equation: Δ H re a c t i o n = [ 4 × ( − 393.5 ) + 2 × ( − 241.8 )] − [ 2 × ( 226.8 ) + 5 × ( 0 )] Δ H re a c t i o n = [ − 1574 − 483.6 ] − [ 453.6 + 0 ] Δ H re a c t i o n = − 2057.6 − 453.6 Δ H re a c t i o n = − 2511.2 k J This is the enthalpy change for the reaction as written, which involves 2 moles of acetylene.
Calculating Enthalpy of Combustion Since we want the enthalpy of combustion for 1 mole of acetylene, we divide the result by 2: Δ H co mb u s t i o n = 2 Δ H re a c t i o n = 2 − 2511.2 = − 1255.6 k J / m o l Therefore, the enthalpy of combustion of 1 mole of acetylene is -1255.6 kJ/mol.
Final Answer The enthalpy of combustion of 1 mole of acetylene for the reaction is − 1255.6 kJ/mol.
Examples
Understanding the enthalpy of combustion is crucial in various real-world applications. For instance, when designing engines or power plants that use fuels like acetylene, knowing the energy released during combustion helps engineers optimize efficiency and reduce emissions. In welding, acetylene's high heat output makes it ideal for cutting and joining metals. By calculating the enthalpy change, we can predict the amount of heat produced, ensuring safe and effective use in industrial processes and daily applications.
The enthalpy of combustion of 1 mole of acetylene, calculated using the standard enthalpies of formation of reactants and products, is -1255.6 kJ/mol. This means that the combustion process releases this amount of energy per mole of acetylene consumed. The final calculation involved determining the total enthalpy of products and subtracting the total enthalpy of reactants, followed by adjusting for the stoichiometry of the reaction.
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