The method of calorimetry we have developed works well provided that the reaction is easily carried out in a way that we can capture the energy transfer in a known quantity of water. But many reactions of great interest are very difficult to carry out under such controlled circumstances. Many biologically important chemical reactions may require the conditions and enzymes only available inside a cell. For example, conversion of glucose C_{6}H_{12}O_{6} to lactic acid CH_{3}CHOHCOOH is one of the primary means of providing energy to cells:

C
6
H
12
O
6
→ 2 CH
3
CHOHCOOH
C
6
H
12
O
6
→ 2 CH
3
CHOHCOOH

(3)Measuring the energy of this reaction is important to understanding the biological process of energy transfer in cells. However, we can’t simply put glucose in a beaker and wait for it to turn into lactic acid. Very specific conditions and enzymes are required. We need to develop a different method for measuring the energy of this reaction, and this requires more experimentation.

To begin our observations, we will work with a few reactions for which we can measure the energy change. Hydrogen gas, which is of potential interest nationally as a clean fuel, can be generated by the reaction of carbon (coal) and water:

C(s) + 2 H
2
O(g) → CO
2
(g) + 2 H
2
(g)
C(s) + 2 H
2
O(g) → CO
2
(g) + 2 H
2
(g)

(3)Calorimetry reveals that this reaction requires the __input__ of 90.1 kJ of heat for every mole of C(s) consumed.

It is interesting to ask where this input energy goes when the reaction occurs. One way to answer this question is to consider the fact that Reaction (3) converts one fuel, C(s), into another, H_{2}(g). To compare the energy available in each fuel, we can measure the heat evolved in the combustion of each fuel with one mole of oxygen gas. We observe that

C(s) + O
2
(g) → CO
2
(g)
C(s) + O
2
(g) → CO
2
(g)

(4)produces 393.5 kJ for one mole of carbon burned; hence q = -393.5 kJ. The reaction

2 H
2
(g) + O
2
(g) → 2 H
2
O(g)
2 H
2
(g) + O
2
(g) → 2 H
2
O(g)

(5)produces 483.6 kJ for two moles of hydrogen gas burned, so q = -483.6 kJ. Therefore, more energy is available from the combustion of hydrogen fuel than from the combustion of carbon fuel. Because of this, we should not be surprised that conversion of the carbon fuel to hydrogen fuel requires the input of energy.

We can stare at the numbers for the heats of these reactions, 90.1 kJ, -393.5 kJ, -483.6 kJ, and notice a surprising fact. The first number 90.1 kJ, is exactly equal to the __difference__ between the second number -393.5 kJ and the third number -483.6 kJ. In other words, the heat input in Equation 3 is exactly equal to the __difference__ between the heat evolved in the combustion of carbon (Equation 4) and the heat evolved in the combustion of hydrogen (Equation 5). This might seem like an odd coincidence, but the numbers from the data are too precisely equal for this to be a coincidence. Just as we do anytime we see a fascinating pattern in quantitative data, we should develop a model to understand this.

It is interesting that the energy of Equation 3 is the difference between the energies in Equation 4 and Equation 5, and not the sum. One way to view this is to remember that the energy of a reaction running in reverse must be the negative of the energy of the same reaction running forward. In other words, if we convert reactants to products with some change in energy, and then convert the products back to the reactants, the change in energy of the reverse process must be the negative of the change in energy in the forward process. If this were not true, we could carry out the reaction many times and generate energy, which would violate the Law of Conservation of Energy.

Let’s list all three reactions together, but in doing so, let’s __reverse__ Equation 5:

C(s) + 2 H
2
O(g) → CO
2
(g) + 2 H
2
(g)
q = 90.1kJ
C(s) + 2 H
2
O(g) → CO
2
(g) + 2 H
2
(g)
q = 90.1kJ

(6)
C(s) + O
2
(g) → CO
2
(g)
q = -393.5kJ
C(s) + O
2
(g) → CO
2
(g)
q = -393.5kJ

(7)
2 H
2
O(g) → 2 H
2
(g) + O
2
(g)
q = 483.6kJ
2 H
2
O(g) → 2 H
2
(g) + O
2
(g)
q = 483.6kJ

(8)What if we do __both__ Equation 4 and Equation 8 at the same time?

C(s) + O
2
(g) + 2 H
2
O(g) → CO
2
(g) + 2 H
2
(g) + O
2
(g)
C(s) + O
2
(g) + 2 H
2
O(g) → CO
2
(g) + 2 H
2
(g) + O
2
(g)

(9)Since O2 (g) is on both sides of this reaction, it is net neither a reactant nor product, so we could remove it without losing anything. Equation 9 is then equivalent to Equation 3, except that we carried it out by doing two separate reactions. What if we do Equation 9 in two steps, first doing Equation 4 and then doing Equation 8? That would be the same outcome as Equation 9 so it must not matter whether we do Equation 4 and Equation 8 at the same time or one after the other. The reactants and products are the same either way.

What would be the energy of doing Equation 4 and Equation 5 either at the same time or in sequence? It must be the energy of Equation 4 plus the energy of Equation 5, of course. So if we add these together using the above numbers, the energy of doing both reactions together is -393.5kJ + 483.6kJ = 90.1kJ. This is the energy of Equation 9 and it is experimentally exactly the same as the energy of Equation 3.

This is a new observation. We have found that the energy of a single reaction (Equation 3) is equal to the sum of the energies of two reactions (Equation 4 and Equation 8), which together give the same total reaction as the single reaction. When the separate reactions add up to an overall reaction, the energies of the separate reactions add up to the energy of the overall reaction.

We could study many reactions in a similar manner to see if this pattern holds up. We find experimentally that it does. The pattern is therefore a new natural law called Hess’ Law. Stated generally (but wordily), the energy of a reaction is equal to the sum of the energies of any set of reactions which, when carried out in total, lead from the same reactants to the same products. This is a powerful observation! (It is important to note here that we have omitted something from our observations. Hess’ Law requires that all reactions considered proceed under similar conditions like temperature and pressure: we will consider all reactions to occur at constant pressure.)

Why would Hess’ Law be true? Why doesn’t it matter to the energy whether we carry out a reaction in a single step or in a great many steps which produce the same products? We might have guessed that more steps somehow require more energy or somehow waste more energy. But if so, our guess would be wrong. As such, it would be helpful to develop a model to account for this law to improve our intuition about reaction energies.

Figure 1 gives us a way we can make progress based on the work we have already done. This is a just a picture of Hess' Law showing all of Equation 4, Equation 8, and Equation 9. (Remember that the total reaction in Equation 9 is exactly the same as the original reaction in Equation 3.) In Figure 1, the reactants C(s) + 2 H2O (g) + O_{2}(g) are placed together in a box, representing the reactant __state__ of the matter before Equation 9 occurs. The products CO2(g) + 2H2(g) + O_{2}(g) are placed together in a second box representing the product __state__ of the matter after Equation 9. The reaction arrow connecting these boxes is labeled with the heat of Equation 3 (which is also the heat of Equation 9), since that is the energy absorbed when the matter is transformed chemically from reactants to products in a single step.

Also in Figure 1, we have added a box in which we place the same matter as in the reactant box but showing instead the products of carrying out Equation 4. In other words, we will first do Equation 4, producing CO_{2}(g) but leaving the H_{2}O(g) unchanged. Notice that the reaction arrow is labeled with the energy of Equation 4. Now we can also add a reaction arrow to connect this box to the product box, because that reaction is just Equation 8, producing H_{2}(g) and O_{2}(g) from the H_{2}O(g). And we can label this reaction arrow with the energy of Equation 8.

This picture of Hess' Law makes it clear that the energy of the reaction along the "path" directly connecting the reactant state to the product state is exactly equal to the total energy of the same reaction along the alternative "path" consisting of two steps which connect reactants to products. (This statement is again subject to our restriction that all reactions in the alternative path must occur under the same conditions, e.g. constant pressure conditions.)

Now let’s take a slightly different view of Figure 1. Visually make a loop by beginning at the reactant box and following a complete circuit through the other boxes leading back to the reactant box, summing the total energy of reaction as you go. If you go “backwards” against a reaction arrow, then reverse the sign of the energy, since a reverse reaction has the negative energy of the forward reaction. When we complete a loop and do the sum, we discover that the net energy transferred around the loop starting with reactants and ending with reactants is exactly zero. This makes a lot of sense when we remember the Law of Conservation of Energy: surely we cannot extract any energy from the reactants by a process which simply recreates the reactants. If this were not the case, in other words if the sum didn’t equal zero, we could endlessly produce unlimited quantities of energy by following the circuit of reactions which continually reproduce the initial reactants. Experimentally, this never works since energy is conserved.

However, we do get a clearer understanding of why adding the reaction energies together gives the total energy of the overall reaction. Hess’ Law is a consequence of the Law of Conservation of Energy.