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

it can be observed without changing the identity of the object.

it depends on what substance you're talking about as well as many other factors. in quantum mechanics, the observation of particles affects the observers ability to measure certain properties. for example: you can know the velocity of a particle, but you can't know it's position; or you can know it's position, but not it's velocity. it's like trying to find out where a grain of sand is by throwing other grains of sand at it. once you find out where it is, it's been moved. it's also like trying to find an answer to the question of whether or not a tree falling in a forest makes a sound if nobody is there to hear it. paradoxically, the only way to find out is to be there and hear it.