Consider the following thermochemical equation at 1 atm pressure and 25oC: 2C6H6 (l) + 15O2 (g) ⟶ 12CO2 (g) + 6H2O(l) ∆H= –6534 kJ Standard heats of formation of CO2 (g), H2O(g), H2O(l), O3(g) are -393.5 kJ/mol, -242 kJ/mol, -285.8 kJ/mol, 143 kJ/mol, respectively. Calculate the standard enthalpy of formation of benzene (C6H6) in kJ/mol.
Consider the following thermochemical equation at 1 atm pressure and 25oC: 2C6H6 (l) + 15O2 (g) ⟶ 12CO2 (g) + 6H2O(l) ∆H= –6534 kJ Standard heats of formation of CO2 (g), H2O(g), H2O(l), O3(g) are -393.5 kJ/mol, -242 kJ/mol, -285.8 kJ/mol, 143 kJ/mol, respectively. Calculate the standard enthalpy of formation of benzene (C6H6) in kJ/mol.
Chemistry
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Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
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Chapter1: Chemical Foundations
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Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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Consider the following thermochemical equation at 1 atm pressure and 25oC:
2C6H6 (l) + 15O2 (g) ⟶ 12CO2 (g) + 6H2O(l) ∆H= –6534 kJ
Standard heats of formation of CO2 (g), H2O(g), H2O(l), O3(g) are -393.5 kJ/mol, -242 kJ/mol, -285.8
kJ/mol, 143 kJ/mol, respectively. Calculate the standard enthalpy of formation of benzene (C6H6) in
kJ/mol.
Expert Solution
Step 1
Given- Enthalpy of reaction (∆Hreaction) =
–6534 kJ
Standard heats of formation of
CO2(g)= -393.5 kJ/mol H20(g)= -242 kJ/mol. H20(l)= -285.8 kJ/mol. O3(g) = 143 kJ/mol
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