The purpose of this article is to show you how to present Hess’s Law in a logical manner, with essentially no reliance on memorizing.1 Students will understand that the different ways of calculating ΔHr using Hess’s Law are ALL THE SAME.
Hess’s Law of Heat Summation is an important topic in AP Chemistry. It is based on the fact that ΔHr is a state function2. Simply put, the enthalpy change, ΔHr, for a chemical reaction is independent of the pathway taken between reactants and products3.
An excellent analogy to begin teaching Hess’s Law involves cycling where there are changes in elevation. Figure 1 illustrates a trip from point A, (sea level) to point B (200 m elevation). Whether one cycles the direct route from A to B, or the indirect route through point C (250 m elevation), the net difference in elevation (gravitational potential energy) between points A and B is the same.
Figure 1: A bicycle journey can analogize Hess’s Law. One can equate ΔPE, the change of gravitational potential energy, with ΔHr.
And so for Hess’s Law of Heat Summation: ΔHr is a state function, depending only on the T, P of the reactants and products. The pathway taken for reactants to form products is irrelevant.4
I recommend beginning the study of Hess’s Law with a lab. An empirical foundation for Hess’s Law makes the concept easier to “sell”—what you teach is backed-up with what students have seen5.
Aside: Prior to studying Hess’s Law, my students have already done a simple calorimetry lab using a coffee-cup calorimeter: the determination of ΔHr for the Reaction of Mg with (excess) HCl(aq)6, represented by:
Mg(s) + 2 HCl(aq) → MgCl2(aq) + H2(g) Equation 1
This ensures that students are familiar with the use of an inexpensive, yet very good constant-pressure calorimeter, and the attendant calculations. This is, I think, important before students go “all-in” to accurately determine Hess’s Law.
Figure 2: An inexpensive—and excellent coffee-cup calorimeter. Note that two nested Styrofoam cups are used, and that these sit in a beaker; the cups rest on the rim of the beaker. This allows air, an excellent insulator, to surround the coffee cups, even on the bottom. A robust Styrofoam lid completes the apparatus.7
There are many decent laboratory activities on Hess’s Law8; all rely on the Algebraic Combination of Thermochemical Equations—see #2 below.
With a successful lab completed, you can explain that ΔHr can now be calculated with Hess’s Law, using any of the following methods, and that all three methods are variations on a theme.
1. Bond Energy Values
ΔHr = ∑bond energy of bonds broken – ∑bond energy of bonds formed9 Equation 2
2. Algebraic Addition of Thermochemical Equations
If you begin this topic with a lab, students will likely be familiar with this before you teach it.
3. Using Tabulated Standard Heat of Formation Data and the equation
ΔHr = ∑ n* ΔHof (products) – ∑ m* ΔHof (reactants)10, Equation 3
where n and m represent stoichiometric coefficients of each reactant and product, obtained from the balanced chemical equation.
These methods of calculating ΔHr are the same. This can be illustrated using the complete combustion of methane, represented by the balanced chemical equation:
CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(g) Equation 4
Bond Energy Calculations
In order to use a table of average bond energies to calculate ΔHr, we must draw the Lewis structure of each reactant and product. This allows one to see the quantity and type of each bond that is broken or formed.
Figure 3: Pertinent Lewis structures for equation 4.
The use of bond energy calculations assumes that the reactants can be broken down to their constituent atoms, and that the “loose” atoms bond in new ways to form products.
Bond breaking is endothermic11 -- think of sawing a board in half; bond making is exothermic, and so
ΔHr = ∑ (bond energy for bonds broken) – ∑ (bond energy for bonds formed) Equation 2
Table 1: Energetics of bond-making and bond-breaking for the complete combustion of CH4
Bonds Broken | Average bond energy (kJ·mol-1)12 | Bonds formed | Average bond energy (kJ·mol-1) |
4(C-H) | 4(415) | 4(C=O) | 2(805) |
2(O=O) | 2(498) | 4(O-H) | 4(464) |
Totals | 2656 kJ | 3466 kJ |
ΔHr = 2656 kJ – 3466 kJ = –810 kJ·(mol CH4)-1
This shows that the complete combustion of methane is exothermic, consistent with students’ experience with a laboratory burner, or with a natural gas (CH4)-fuelled stove.
To help with the logic behind (equation 2), students can think of the energy required for bond breaking as the “investment”—think $$; the energy released in bond making is the “return”. If the “return” is greater than the “investment”, one has an “exo$$” business, or an exothermic chemical reaction.
As you know, this method of calculating ΔHr has limitations:
a) Lewis structures are required, which restricts this to covalent molecules for which correct Lewis structures can be drawn;
b) Tabulated bond energy values are averages; not all C–H bonds, for example, are of equal strength;
c) This method does not take into account the physical states of the reactants and products, which is important.
That said, using average bond energy values is incredibly useful—one can calculate ΔHr for literally thousands of chemical reactions to a decent level of accuracy, with only a modest table of bond energy values.
We can proceed to look at this example in terms of thermochemical equations:
Considering the bond-breaking and bond-making in (equation 4), with data taken from Table 1, we have:
Figure 4: Thermochemical Equations representing Bond-breaking and Bond-making in the complete combustion of methane
Algebraic combination of these equations gives the overall equation for the complete combustion of methane. This simple example illustrates calculating ΔHr using average bond energies is the same as the combination of Thermochemical Equations.
As for the use of Standard Heat of Formation data13, we have:
Figure 5: Thermochemical equations pertaining to the formation of reactants and products in (equation 3)
We manipulate the thermochemical equations in Figure 3 to give (equation 3) as follows:
Figure 6: Combining formation equations for the complete combustion of methane (equation 3)
The ΔHr value obtained here is highly accurate; it compares favorably to that obtained using Average Bond Energies. These values won’t be exactly the same; it’s the nature of the beast.
We have simply subtracted the sum of the mole-adjusted ΔHf values of the reactants from that of the products, which is the same as
ΔHr = ∑ n* ΔHof (products) – ∑ m* ΔHof (reactants), Equation 3
These examples illustrate that the three ways of using Hess’s Law of Heat Summation to calculate ΔHr are essentially the same. This should make your teaching—and your students’ understanding (and learning) much easier.
Notes and References
- That said, if students memorize the formulas—I hate that word—who are we to stop them?
- State vs. Path Functions - Chemistry LibreTexts
- 3.9: Hess' Law - Chemistry LibreTexts
- See footnotes 1 and 2
- I am, in most cases, opposed to “verification” labs; concepts should be taught based on empirical evidence, not with a laboratory activity tacked on.
- See supplementary material
- My students verified experimentally that two cups are better than one, and that three cups are no better than two. A ca. 1 cm-thick Styrofoam lid, with a hole for a thermometer or temperature probe, was shown to make this calorimeter a better insulator.
- See supplementary material
- Reference E&G
- Reference E&G
- I recently had a student insist that bond breaking is exothermic. She learned in Biology class that when “high energy” bonds are broken, energy was released. This misconception was fertile ground for a conversation . . .
- Average bond energy values taken from: 9.4: Bond Strength and Energy - Chemistry LibreTexts & The Value for the C=O bond energy in CO2 taken from: Bond Enthalpies - Chemistry LibreTexts
- T1: Standard Thermodynamic Quantities - Chemistry LibreTexts