The enthalpy change of a reaction is the amount of heat absorbed or released as the reaction takes place, if it happens at a constant pressure. You complete the calculation in different ways depending on the specific situation and what information you have available. For many calculations, Hess’s law is the key piece of information you need to use, but if you know the enthalpy of the products and the reactants, the calculation is much simpler.

The precise definition of enthalpy (H) is the sum of the internal energy (U) plus the product of pressure (P) and volume (V). In symbols, this is:

H = U + PV

A change in enthalpy (∆H) is therefore:

∆H = ∆U + ∆P∆V

Where the delta symbol (∆) means “change in.” In practice, the pressure is held constant and the above equation is better shown as:

∆H = ∆U + P∆V

However, for a constant pressure, the change in enthalpy is simply the heat (q) transferred:

∆H = q

If (q) is positive, the reaction is endothermic (i.e., absorbs heat from its surroundings), and if it is negative, the reaction is exothermic (i.e., releases heat into its surroundings). Enthalpy has units of kJ/mol or J/mol, or in general, energy/mass. The equations above are really related to the physics of heat flow and energy: thermodynamics.

The most basic way to calculate enthalpy change uses the enthalpy of the products and the reactants. If you know these quantities, use the following formula to work out the overall change:

∆H = H_products − H_reactants

The addition of a sodium ion to a chloride ion to form sodium chloride is an example of a reaction you can calculate this way. Ionic sodium has an enthalpy of −239.7 kJ/mol, and chloride ion has enthalpy −167.4 kJ/mol. Sodium chloride (table salt) has an enthalpy of −411 kJ/mol. Inserting these values gives:

∆H = −411 kJ/mol – (−239.7 kJ/mol −167.4 kJ/mol)

= −411 kJ/mol – (−407.1 kJ/mol)

= −411 kJ/mol + 407.1 kJ/mol = −3.9 kJ/mol

So the formation of salt releases almost 4 kJ of energy per mole.

When a substance changes from solid to liquid, liquid to gas or solid to gas, there are specific enthalpies involved in these changes. The enthalpy (or latent heat) of melting describes the transition from solid to liquid (the reverse is minus this value and called the enthalpy of fusion), the enthalpy of vaporization describes the transition from liquid to gas (and the opposite is condensation) and the enthalpy of sublimation describes the transition from solid to gas (the reverse is again called the enthalpy of condensation).

For water, the enthalpy of melting is ∆H_melting = 6.007 kJ/mol. Imagine that you heat ice from 250 Kelvin until it melts, and then heat the water to 300 K. The enthalpy change for the heating parts is just the heat required, so you can find it using:

∆H = nC∆T

Where (n) is the number of moles, (∆T) is the change in temperatue and (C) is the specific heat. The specific heat of ice is 38.1 J/K mol and the specific heat of water is 75.4 J/K mol. So the calculation takes place in a few parts. First, the ice has to be heated from 250 K to 273 K (i.e., −23 °C to 0°C). For 5 moles of ice, this is:

∆H = nC∆T

= 5 mol × 38.1 J/K mol × 23 K

= 4.382 kJ

Now multiply the enthalpy of melting by the number of moles:

∆H = n ∆H_melting

= 5 mol × 6.007 kJ/mol

= 30.035 kJ

Calculations for vaporization are the same, except with the vaporization enthalpy in place of the melting one. Finally, calculate the final heating phase (from 273 to 300 K) in the same way as the first:

∆H = nC∆T

= 5 mol × 75.4 J/K mol × 27 K

= 10.179 kJ

Sum these parts to find the total change in enthalpy for the reaction:

∆H_total = 10.179 kJ + 30.035 kJ + 4.382 kJ

= 44.596 kJ

Hess’s law is useful for when the reaction you’re considering has two or more parts and you want to find the overall change in enthalpy. It states that the enthalpy change for a reaction or process is independent of the route through which it occurs. This means that if reaction transforms on substance into another, it doesn’t matter if the reaction occurs in one step (reactants become products immediately) or whether it goes through many steps (reactants become intermediaries and then become products), the resulting enthalpy change is the same in both cases.

It usually helps to draw a diagram (see Resources) to help you use this law. One example is if you start with six moles of carbon combined with three of hydrogen, they combust to combine with oxygen as an intermediary step and then form benzene as an end-product.

Hess’ law states that the change in enthalpy of the reaction is the sum of the changes in enthalpy of both parts. In this case, the combustion of one mole of carbon has ∆H = −394 kJ/mol (this happens six times in the reaction), the change in enthalpy for the combustion of one mole of hydrogen gas is ∆H = −286 kJ/mol (this happens three times) and the carbon dioxide and water intermediaries become benzene with an enthalpy change of ∆H = +3,267 kJ/mol.

Take the sum of these changes to find the total enthalpy change, remembering to multiply each by the number of moles needed in the first stage of the reaction:

∆H_total = 6×(−394) + 3×(−286) +3,267

= 3,267 − 2,364 - 858

= 45 kJ/mol