Heat of formation

Question

If the heat of combustion of benzene is -3267.6 kJ/mol and the delta H for CO2 is -393.5 kJ/mol with the delta H for H20 being -285.8 kJ/mol, find the heat of formation of benzene, C6H6.

Answer 1

C₆H₆ = benzene

2C₆H₆ + 15O₂ ==> 12CO₂ + 6H₂O .. ∆H = -3267.6 kJ/mole

To find the ∆H_f for benzene, subtract the sum of ∆H_f of reactants, from that of products, i.e.

∆Hrxn = ∑∆H_fproducts - ∑∆H_freactants

-3267.6 = [(12 x - 393.5) + (6 x -285.8)] - [(2 x ∆H_C6H6) + (15 x 0)]

-3267.6 = (-4722 + -1715) - 2∆H_C6H6

-3267.6 = -6437 - 2∆H_C6H6

∆H_f C₆H₆ = - 1585 kJ/mol

Answer 2

Use Hess' Law

C₆H₆ + 9/2 O₂ -> 6CO₂ + 3H₂O : -3267.6 kj/mol

H₂ + 1/2 O₂ -> H₂O : -285.8 kj/mol

C + O₂ -> CO₂ : -393.5 kj/mol

Rearrange and balance equations to get the following:

(6CO₂ + 3H₂O ) + 3 ( H₂ + 1/2 O₂ ) + 6(C + O₂ ) -> 6(CO₂ ) + 3( H₂O ) + (C₆H₆ + 9/2 O₂ )

This turns out to be

3H₂ + 6C -> C₆H₆

or

-delH _benzene +3 del H _water + 6 del H_{carbon dioxide}

The rest is trivial