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).

The enthalpy : 49.1 kJ/mol

The change in enthalpy in the formation of 1 mole of the elements is called enthalpy of formation  

The enthalpy of formation measured in standard conditions (25 ° C, 1 atm) is called the standard enthalpy of formation (ΔHf °)  

Based on the principle of Hess's Law, the change in enthalpy of a reaction will be the same even though it is through several stages or ways  

Reaction

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

Reverse

12 CO₂ (g) + 6 H₂O (g) ⇒ 2C₆H₆ (l) + 15 O₂ (g) ∆H° = 6271 kJ/mol : 2

6CO₂ (g) + 3H₂O (g) ⇒ C₆H₆ (l) + 15/2 O₂ (g) ∆H° = 3135.5 kJ/mol

2. 2 H₂ (g) + O₂ (g) → 2 H₂O (g) ∆H° = -483.6 kJ/mol  x 3/2

3H₂ (g) + 3/2O₂ (g) → 3H₂O (g) ∆H° = -725.4 kJ/mol

3. C (s) + O₂ (g) → CO₂ (g) ∆H° = -393.5 kJ/mol x 6

6C (s) + 6O₂ (g) → 6CO₂ (g) ∆H° = -2361 k/j/mol

6 C (s) + 3 H₂ (g) → C₆H₆ (l) ∆H° = 49.1 kJ/mol

We add up and the same compound that is on different sides we eliminate

58 km per hour : d