Thermochemical equations obey several simple rules that make computation of the enthalpy change in a reaction easy.

1. The magnitude of DH is directly proportional to the amount of reactants used in the reaction
2. DH for a reaction is equal in magnitude but opposite in sign to the reverse reaction
3. Hess' Law: The value of DH for a reaction is the same no matter what path is used to get from reactants to products

The first rule is intuitively obvious: if you react 2 moles of A and 2 moles of B in a reaction A + B -> C, you generate twice as much heat as if you only reacted one mole of A and B.

Example: If it takes 9.84 kJ to melt 1 mole of benzene, how much heat does it take to melt 3.50 moles of benzene?

Solution: 9.84 kJ/mole * 3.50 moles = 34.4 kJ.

The second rule is also fairly obvious. If it takes 6kJ of heat to melt a sample of ice, freezing the resulting water back into ice gives off 6kJ of heat. When melting the ice, heat flows from the surroundings to the system, when freezing the heat flows from the system to the surroundings. Thus, the two processes have the same magnitude for DH, but different signs.

The third rule simply states that it doesn't matter how you get there, the value for DH is the same. This means that if you have a way of getting the products in one step and a different way of getting the products in two steps, the final value for DH is the same. If we can break up a reaction into multiple steps, the final enthalpy change is the sum of the enthalpy changes for each reaction:

DH = DH₁+ DH₂+ ...
The third rule is very important for computing the value of DH for reactions we do not know through tables of the standard molar enthalpy of formation

Example:. It takes 6.00 kJ/mole to melt water and 40.7 kJ/mole to boil a sample of liquid water. What is the total enthalpy change if one mole of ice at 0^oC is turned into 1 mole of water vapor at 100^oC?

Solution: DH for the reaction is the sum of the various pieces. First, we have to convert the water ice to liquid, then heat the liquid to 100 degrees, then convert the liquid into vapor. The total enthalpy change is thus
DH = DH_melting+ DH_heating+ DH_vaporization

DH_melting = 6.00 kJ/mole * 1 mole = 6.00kJ.
DH_vaporization = 40.7 kJ/mole * 1 mole = 40.7kJ.
DH_heating = m*c*DT = 18.0 g*4.18 J/g^oC*(100-0)^oC = 7.52x10³ J = 7.52 kJ

DH = 6.00+7.52+40.7 = 54.2 kJ

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