Hess's Law Paper

Question

Calculate the standard enthalpy of formation of benzene (C₆H₆) given the following standard enthalpies of combustion.

C₆H₆ (l) = -3273 kJ mol^-1

C (s) = -393.5 kJ mol^-1

H₂ (g) = -285.5 kJ mol^-1

Answer 1

The standard enthalpy of formation for C6H6 would be from the following reaction:

6C(s) + 3H₂(g) ==> C₆H₆(l) TARGET EQUATION

Given:

(1). C₆H₆ + 15/2 O₂ ==> 6CO₂ + 3H₂O ∆H = -3273 kJ/mol

(2). C + O₂ ==> CO₂ ∆H = -393.5 kJ/mol

(3). 2H₂ + O₂ ==> 2H₂O ∆H = -285.5 kJ/mol

Procedure:

reverse eq.1: 6CO₂ + 3H₂O ==> C₆H₆ + 15/2 O₂ ∆H = +3273 kJ

eq 2 x 6: 6C + 6O₂ ==> 6CO₂ ∆H = -2362 kJ

eq 3 x 3/2: 3H₂ + 3/2 O₂ ==> 3H₂O ∆H = -428.3 kJ

------------------------------------------------------------- add them and combine/cancel like terms to get....

6CO₂ + 3H₂O + 6C + 6O₂ + 3H₂ + 3/2 O₂ ==> C₆H₆ + 15/2 O₂ + 6CO₂ + 3H₂O

6C + 3H₂ ==> C₆H₆ TARGET EQUATION

∆H_f = 3273 -2362 - 428 = 483 kJ