In the last section, we referred to enthalpy as a state function of the system, meaning that its value was determined by several state variables (e.g. T, P) and not on how the system was prepared. Being a state function has practical as well as conceptual importance. One thing it allows us to do is to use a relatively small thermodynamics database or set of tables to represent thermodynamic data for a much larger number of reactions. It also allows us to calculate enthalpy changes for reactions that are impractical or even impossible to measure directly.

Here's how it works. Say you are interested in determining the enthalpy of the reaction:

H_{2 (g)} + 1/2 O_{2 (g)} --> H₂O_(l)

Since it's a pretty common reaction, you might guess that someone else has probably already measured ΔH for it and you could simply look it up. Or if you wanted, you could measure it by calorometric methods, which we will discuss in the next couple of sections below. Either way you should find that to within the error of the measurement &Delta_rH^o = -285.83 kJ at 298 K and 1 bar for the reaction as written, i.e. for the production of one mole of liquid water.

You might have noticed that we have added some symbols to the enthalpy term. The subscript "r" labels the term as an enthalpy of reaction as opposed, say, to an enthalpy of vaporization which would be labeled Δ_vapH^o. The superscript "o" indicates the reaction has been carried out at standard state, meaning at a pressure of 1 bar (101.32 kPa) with the compounds in their pure form. Since enthalpy depends upon temperature, we must also specify the temperature of the reaction, here 298 K, and should never use enthalpies measured or tabulated at one temperature for reactions at other temperatures without some form of correction. IUPAC has recently set the standard state pressure at 1 bar (101.3 kPa); however earlier references, and some current citations, use the pressure of one atmosphere (0.987 atmospheres = 1 bar). While giving "standard state" values at 1 atmosphere rather than 1 bar is not strictly correct, the pressure difference is small and the effect on the thermodynamic quantities is usually negligible.

The physical state of the chemicals in the equation is also important, and if we had instead investigated the reaction that produced water vapor at 298 K

H_{2 (g)} + 1/2 O_{2 (g)} --> H₂O_(g)

we would obtain a value of Δ_rH^o = -241.82 kJ. Why does burning hydrogen to produce water vapor give off less heat than burning hydrogen to produce liquid water? Because it takes energy to vaporize water

H₂O_(l) --> H₂O_(g) Δ_vapH^o = 44.01 kJ/mole at 298 K

To get this value of the enthalpy of vaporization at 298 K, I did not look up the value in a thermodynamic table for enthalpies of vaporization because these are generally given at the normal boiling point, which for water is 373 K at 1 bar. Rather, I made use of the fact that enthalpy is a state function. As long as I start with the same reactants at the same reactants' temperature and pressure and end with the same products at their same temperature and pressure, I can pick any route I want between these two points and still get the same ΔH. One possibility is the direct route where we simply vaporize all the water:

H₂O_(l) --> H₂O_(g) Δ_vapH^o = 44.01 kJ/mole at 298 K

Another occurs in two stages. First we take liquid water and decompose it into hydrogen and oxygen gas:

(1)H₂O_(l) --> H_{2 (g)} + 1/2 O_{2 (g)} &Delta_rH^o = +285.83 kJ at 298 K

Since this inverts the reaction to produce liquid water given above, the sign of the enthalpy also inverts. In the next step we take the hydrogen and oxygen we made in the first reaction and recombine it, but this time to produce gaseous water:

(2) H_{2 (g)} + 1/2 O_{2 (g)} --> H₂O_(g) Δ_rH^o = -241.82 kJ at 298 K

But the two stage reaction sums to

H₂O_(l)+ H_{2 (g)}+1/2 O_{2 (g)}--> H_{2 (g)}+ 1/2 O_{2 (g)}+ H₂O_(g)

with Δ_rH^o = (+285.83 -241.82) kJ

When we remove the chemicals that appear on both sides of the equation and sum the enthalpies from the two steps, we get

H₂O_(l) --> H₂O_(g) Δ_vapH^o = 44.01 kJ/mole

We, of course, did not actually do the vaporization in two steps. But because the enthalpy is a state function, we can equate the enthalpy change for this process to the one step vaporization of water at the same temperature and pressure. Any number of steps can be added in this way, regardless of whether the reactions are practical or even possible to perform.

Remember that enthalpy is an extrinsic property and must be multiplied by the correct stoichiometric value. If you double the reaction coefficients, remember to double the value of the enthalpy too.

Enthalpies of reaction may be used to calculate the energy produced from a specific mass or number of moles of material. Use this site to practice calculating the energy from the chemical equation and enthalpy.

Thermodynamic tables are listed in terms of a specific type of reaction, the enthalpy of formation, Δ_fH^o. The enthalpy of formation is a reaction enthalpy in which the compound of interest is formed from elements in their standard state at the temperature of interest. For example, the enthalpy of formation reaction for ammonia gas is

1/2 N_{2 (g)} + 3/2 H_{2 (g)} --> NH_{3 (g)}

and at 298 K Δ_fH^o for NH_{3 (g)} is +294.1 kJ/mole. By definition, Δ_fH^o of any element is zero if it is in its standard state. For example, for hydrogen, the enthalpy of formation reaction is

H_{2 (g)} --> H_{2 (g)}

which represents no change and therefore no change in enthalpy. Δ_fH^o for H_{2 (g)} is 0 at 1 bar and any temperature for which dihydrogen is the most stable form of the element.

Be careful that you have chosen the correct (most thermodynamically) stable form of the element in formulating reaction enthalpies. For example, at 1 bar and 298 K the most stable form of carbon is graphite, not diamond! Diamond is kinetically stable relative to graphite, meaning the reaction is thermodynamically allowed but occurs so slowly as to be negligible over geological times. At 1 bar and 298 K, Δ_fH^o for graphite is 0. At 1 bar and 298 K, Δ_fH^o for diamond is 1.895 kJ/mole and the enthalpy of formation reaction is

C_graphite --> C_diamond

Tables of heats of formation are listed by compound only, because the reaction should be evident from the choice of compound. To use heat of formation to determine a reaction enthalpy, you merely need to add all the enthalpies of formation for the products and subtract all the enthalpies of formation for the reactants corrected for their coefficients in the stoichiometrically balanced reaction equation.

The above use of heats of formation is just one possible example of a general principle called Hess's law. Because enthalpies are state functions, any sort of chemical reaction can be summed with any other to find the enthalpy that would result from carrying out the net summed reaction. Hess's law is particularly useful for obtaining information on reactions that are impractical or even impossible for carrying out directly in the laboratory. For example, take a look at the combustion of sucrose and the Born-Harber cycle in the web page accessed by clicking here.

For practice at calculations using Hess's Law, this site give problems and answers.