$2 C_6 H_6(l)+15 O_2(g)-12 CO_2(g)+6 H_2 O(l) \quad \Delta H^{\circ}=?$
-5195.6 kJ
-6338.8 kJ
-1260.6 kJ
-6534.8 kJ
+1260.6 kJ
Answer (1)
Use the formula Δ H ∘ = ∑ n Δ H f ∘ ( p ro d u c t s ) − ∑ n Δ H f ∘ ( re a c t an t s ) to calculate the standard enthalpy change.
Find the standard enthalpies of formation for each compound: C O 2 ( g ) , H 2 O ( l ) , C 6 H 6 ( l ) , and O 2 ( g ) .
Substitute the values into the formula: Δ H ∘ = [ 12 ⋅ ( − 393.5 ) + 6 ⋅ ( − 285.8 )] − [ 2 ⋅ ( 49.1 ) + 15 ⋅ ( 0 )] .
Calculate the result: Δ H ∘ = − 6535 kJ , which is closest to the option − 6534.8 kJ .
Explanation
Understanding the Problem We are given the balanced chemical reaction for the combustion of benzene: 2 C 6 H 6 ( l ) + 15 O 2 ( g ) → 12 C O 2 ( g ) + 6 H 2 O ( l ) We need to calculate the standard enthalpy change, Δ H ∘ , for this reaction and choose the correct value from the given options.
Formula for Enthalpy Change The standard enthalpy change of a reaction can be calculated using the following formula: Δ H ∘ = ∑ n Δ H f ∘ ( p ro d u c t s ) − ∑ n Δ H f ∘ ( re a c t an t s ) where n represents the stoichiometric coefficients of the products and reactants, and Δ H f ∘ represents the standard enthalpies of formation of the products and reactants.
Applying the Formula to the Reaction We can expand the formula for our specific reaction: Δ H ∘ = [ 12 ⋅ Δ H f ∘ ( C O 2 ( g )) + 6 ⋅ Δ H f ∘ ( H 2 O ( l ))] − [ 2 ⋅ Δ H f ∘ ( C 6 H 6 ( l )) + 15 ⋅ Δ H f ∘ ( O 2 ( g ))] We need to find the standard enthalpies of formation for each compound.
Finding Standard Enthalpies of Formation The standard enthalpies of formation are:
Δ H f ∘ ( C O 2 ( g )) = − 393.5 kJ/mol
Δ H f ∘ ( H 2 O ( l )) = − 285.8 kJ/mol
Δ H f ∘ ( C 6 H 6 ( l )) = 49.1 kJ/mol
Δ H f ∘ ( O 2 ( g )) = 0 kJ/mol (since it is an element in its standard state)
Calculating the Enthalpy Change Now, substitute these values into the equation: Δ H ∘ = [ 12 ⋅ ( − 393.5 ) + 6 ⋅ ( − 285.8 )] − [ 2 ⋅ ( 49.1 ) + 15 ⋅ ( 0 )] Δ H ∘ = [ − 4722 − 1714.8 ] − [ 98.2 + 0 ] Δ H ∘ = − 6436.8 − 98.2 Δ H ∘ = − 6535 kJ The calculated value is approximately -6535 kJ.
Final Answer Comparing the calculated value with the given options, the closest value is -6534.8 kJ. Therefore, the standard enthalpy change for the reaction is − 6534.8 kJ .
Examples
Understanding enthalpy changes is crucial in many real-world applications. For instance, engineers use these calculations to design efficient engines and power plants. By knowing the energy released or absorbed during a chemical reaction, they can optimize processes to maximize energy output and minimize waste. In the context of combustion, like the burning of fuels, enthalpy calculations help determine the amount of heat produced, which is essential for designing heating systems and assessing the environmental impact of different fuels. This knowledge also aids in developing new, cleaner energy sources by comparing the energy efficiency and emissions of various reactions.
