Question

The combustion of one mole of C6H6 takes place at 298 K and 1 atm. After combustion, CO2(g) and H20(l) are produced and 3267 KJ of heat is liberated. Calculate the ∆fH° of benzene. Standard enthalpy of formation of CO2 and H20 are -393.5 KJ/mol and -285.83 KJ/mol respectively



Please calculate this question by the formula
∆H= £fH° products - £fH° reactants

Answer 1

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Answer 2

Answer:

The standard enthalpy of formation of benzene is 6485.49 kJ/mol.

Explanation:

The balanced equation for the combustion of C6H6 is:

C6H6 + 15/2 O2 -> 6 CO2 + 3 H2O

To use the formula:

∆H = Σn∆fH°(products) - Σm∆fH°(reactants)

we need to determine the standard enthalpies of formation for the reactant, benzene (C6H6), and for the products, CO2 and H2O.

The enthalpy change for the combustion reaction is given as -3267 kJ/mol, which is the negative of the enthalpy change for the reaction:

C6H6 + 15/2 O2 -> 6 CO2 + 3 H2O (ΔH = -3267 kJ/mol)

Using the standard enthalpies of formation for CO2 and H2O from the problem, we can calculate the standard enthalpy of formation for benzene as follows:

∆H = Σn∆fH°(products) - Σm∆fH°(reactants)

∆H = (6 mol x (-393.5 kJ/mol)) + (3 mol x (-285.83 kJ/mol)) - ∆fH°(C6H6)

-3267 kJ/mol = (-2361 kJ/mol) + (-857.49 kJ/mol) - ∆fH°(C6H6)

∆fH°(C6H6) = (2361 kJ/mol) + (857.49 kJ/mol) + 3267 kJ/mol

∆fH°(C6H6) = 6485.49 kJ/mol

Therefore, the standard enthalpy of formation of benzene is 6485.49 kJ/mol.

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