Determine the enthalpy for the reaction

Question

Using the equations 2 C₆H₆ (l) + 15 O₂ (g) → 12 CO₂ (g) + 6 H₂O (g)∆H° = -6271 kJ/mol C (s) + O₂ (g) → CO₂ (g) ∆H° = -393.5 kJ/mol 2 H₂ (g) + O₂ (g) → 2 H₂O (g) ∆H° = -483.6 kJ/mol Determine the enthalpy for the reaction

6 C (s) + 3 H₂ (g) → C₆H₆ (l).

Answer 1

And yet another problem for Hess' Law.

Target equation:

6C(s) + 3H₂(g) ==> C₆H₆(l)

eq. 1.: 2C₆H₆(l) + 15O₂(g) ==> 12CO₂(g) + 6H₂O(g) ∆H = -6271 kJ/mol

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

eq. 3.: 2H₂(g) + O₂(g) ==> 2H₂O ∆H = -483.6 kJ/mol

reverse eq.1 and divide it by 2: 6CO₂ + 3H₂O ==> C₆H₆ + 7.5 O₂ ∆H = +6271/2 = +3136 kJ

copy eq. 2 and multiply it by 6: 6C + 6O₂ ==> 6CO₂ ∆H = -393.5 x 6 = -2361 kJ

copy eq. 3 and multiply it by 1.5: 3H₂ + 1.5O₂ ==> 3H₂O ∆H = -483.6 x 1.5 = -725.4 kJ

_____________________________________________ add the three equations together....

6CO₂ + 3H₂O + 6C + 6O₂ + 3H₂ + 1.5O₂ ==> C₆H₆ + 7.5O₂ + 6CO₂ + 3H₂O cancel like terms to get...

6C + 3H₂ ==> C₆H₆ and adding the ∆H values (3136 - 2361 -725) = 50 kJ = ∆H