I'm very confused about this please help! (Hess Law)

Question

Using the equations 2 C₆H₆ (l) + 15 O₂ (g) → 12 CO₂ (g) + 6 H₂O (g)∆H° = -6271 kJ/mol 2 H₂ (g) + O₂ (g) → 2 H₂O (g) ∆H° = -483.6 kJ/mol C (s) + O₂ (g) → CO₂ (g) ∆H° = -393.5 kJ/mol Determine the enthalpy (in kJ/mol) for the reaction 6 C (s) + 3 H₂ (g) → C₆H₆ (l).

Answer 1

Hess' Law

You use the information provided and rearrange as needed in order to arrive at the target equation being asked about.

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

Given:

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

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

eq.3: C (s) + O₂ (g) → CO₂ (g) ∆H° = -393.5 kJ/mol

Procedure:

copy eq.3 & x 6: 6C (s) +6 O₂ (g) ==> 6CO₂ (g) ∆H° = -393.5 kJ/mol x 6 = -2361

reverse eq.1 & ÷ 2: 6CO2(g) + 3H2O(g) ==> C6H6(l) + 7.5 O2(g) ∆Hº = +3135.5

copy eq.2 x 1.5: 3H2(g) + 1.5 O2(g) ==> 3H2O(g) ∆Hº = -725.4

Add them up:

6C (s) +6 O₂ (g) + 6CO2(g) + 3H2O(g) + 3H2(g) + 1.5 O2(g) => 6CO₂ (g) + C6H6(l) + 7.5 O2(g) + 3H2O(g)

Combine and cancel like terms to get...

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

Now just add up the associated values for ∆Hº

∆Hº = -2361 + 3135.5 - 724.4

∆Hº = 47.1 kJ