Ace Your Next Organic Chemistry Exam.

With these Downloadable PDF Study Guides

Our Study Guides

General Chemistry Review

By James Ashenhurst

From Gen Chem to Organic Chem, Pt. 10 – Hess’ Law

Last updated: March 21st, 2019

Probably nowhere else does the contrast between the largely quantitative nature of Gen Chem with (relatively) qualitative nature of Org Chem become more apparent than with the topic of thermodynamics. Whereas you probably spent a lot of time in Gen chem figuring out heats of formation, free energy differences, and electrochemical potentials, these types of challenges are largely absent in introductory org chem. However, just because certain types of questions don’t come up doesn’t mean that the key concepts have gone away. It’s still vitally important to understand the meaning of some key thermodynamic equations and to be able to apply them in an intuitive sense to reactions in organic chemistry.

Today we’ll talk about Hess’ Law and we’ll deal with some other key equations over the next few days as we wrap up this series with discussions of chemical kinetics and chemical equilibria.

Hess’ Law

Energy in the universe is a constant. It is netiher created nor destroyed, it merely changes form. The enthalpy (heat) given off (or absorbed)  in a reaction is equal to the enthalpy of formation of the products minus the enthalpy of formation of the reactions. This relationship is referred to as Hess’ Law. It’s nature’s version of the accounting equation: profit(loss) = income – expenses. However, when dealing with enthalpy, the release of heat energy (an exothermic process) is denoted with a negative sign, whereas processes which absorb energy (endothermic processes) maintain the positive sign. By the way, this also applies to the Gibbs free energy, delta G.

Δ H (reaction) = ΔH(products) – ΔH(reactants)

Enthalpy is what is known as a state function, which means the change in enthalpy is independent of the pathway taken. It’s intuitive to think of enthalpy as altitude, with each atom/molecule at a fixed “elevation”, which represents its standard heat of formation. The distance between my apartment in Jerusalem (2600 feet above sea level) which is where I live, and the Dead Sea about 90 minutes away (1200 feet below sea level) is a constant – 3600 feet. Going from Jerusalem to the Dead Sea, I descend 3600 feet, whether I do it by foot, by helicopter, or via a side trip to Mount Everest. It’s the net result that matters. You can draw an analogy with currency: bankruptcy courts don’t care if you used to be a millionaire, they care about how much money you have now.

Just as altitude is measured relative to sea level, enthalpy is not measured directly, but through the measurement of changes. For instance, you might not think that sodium chloride, NaCl, as very remarkable compound. It might be more exciting if you saw this (the reaction of molten sodium with chlorine gas). There’s a lot of energy bound up in your table salt! Alternatively, you could form NaCl any number of other, perhaps less excting ways – like through the reaction of solid sodium hydroxide with hydrochloric acid. However, no matter which path you take to making NaCl, the compound is identical in every respect. The enthalpy of formation of NaCl is independent of how it is made.

How do we apply these concepts to organic chemistry?

Here’s a fairly simple reaction that you’ll learn in Org 1, called hydrogenation. In this reaction we treat a very simple olefin (alkene, in this case ethene) with hydrogen gas over a catalyst (typically palladium on charcoal, which we depict as Pd/C)). The result is a saturated hydrocarbon (ethane). The heats of formation of both these compounds have been measured by intrepid scientists before us, so we can just look the values up in a table and use Hess’ Law to figure out the ΔH for the reaction as follows:

Δ H(reaction) = Hf (product) – Hf (reactants).

In the case of the hydrogenation of ethylene this would be –`140 kJ/mol.


Here’s where another facet comes in. A lot of times in organic chemistry we’ll be dealing with compounds that we don’t happen to know the heats of formation of. But there’s a useful way around this dilemma. The enthalpy of the whole molecule can be estimated by examining the enthalpies of the individual bonds. We can break down the thermodynamics of this process even further by applying a table of standard bond dissociation energies. For a pretty thorough table, click here. “Standard” bond dissociation energies refer to an average value of the measured bond energies of the same type of bond in different molecules. Since we’re measuring relative change in energy, we only care about the energies of the bonds that are breaking or forming (and make the relatively safe assumption that the other bonds don’t affect the process very much).  For instance, in our case we are breaking a C=C π bond (note – only breaking the double bond, not both bonds!) and an H-H single bond. The table tells us the values of these are roughly 264 kJ/mol and 436 kJ/mol respectively. This is what the reaction “costs”. What is “gained” is the energy of two C-H bonds (approximately 410 kJ/mol each).

We use the equation Δ H (reaction) = Δ H (bonds broken) – Δ H (bonds formed). We put the bonds broken term first because breaking bonds is an endothermic process – it requires energy. Forming bonds is an exothermic process (it releases energy) and thus has the negative sign in front of it.

When we do the math we have a value of –120 kJ/mol, or a release of  120 kJ/mol of heat energy in this process. Within ~15% of the standard value, so not a terrible estimate. The answer isn’t exact due to some small cooperative effects, but this is an excellent first-order way of figuring out if a reaction will be favorable. [a note of caution – bond dissociation energies generally represent homolytic bond cleavage (i.e. each atom obtains one electron), while many processes in organic chemistry go through heterolytic processes (one atom receives 2 and the other zero) – but the point stands]. If you find yourself breaking a really strong bond – C–F, for instance – something’s probably gone wrong.

Next Post – From Gen Chem to Organic Chem, Part 11 – The Second Law

Related Posts:


Comment section

5 thoughts on “From Gen Chem to Organic Chem, Pt. 10 – Hess’ Law

  1. Just having trouble understanding why an H-H bond was broken in the example above. I can see how, in ethene, a C=C(pi) bond was broken to form a C-H bond in ethane, but I fail to see how an H-H bond in ethene was broken.

    As well, what is the reasoning in assigning a negative to Hf(reactants)? I think I have an understanding of how energy must be applied to form the products and when I think about how to form the reactants, in order to create them, there would also need to be an energy input, too.

    1. The H-H bond was in H2. The H2 (hydrogen gas) is used to add 2 hydrogen at the ethene.
      The reactants will be destroyed, so the enegy associated with them will be used.

  2. Hi
    Bond energy is defined in the gas phase. It means that when you use bond energies to calculate enthalpy of reaction, all the reactants and the products should be in the gas phase.
    However textbooks often ignore that. They will give a rxn where the phase of a reactant or a product is in the liquid or solid phase , and do not use enthalpy of evaporation or sublimation in the calculations, but only bond energies. This can lead sometimes to a big difference in the calc. and it looks to me like they are not following Hess’s law.

    Do I miss anything or these textbooks are wrong? Tnx

Leave a Reply

Your email address will not be published.

This site uses Akismet to reduce spam. Learn how your comment data is processed.