Equilibrium and Energy Relationships

One kcal/mol is actually a lot

This article discusses the relationship between the equilibrium constant K and the difference in energy ΔG, and explores some applications of the equation towards various examples in organic chemistry.

• The difference in Gibbs energy ΔG between two species in equilibrium is related to the equilibrium constant K according to the equation

Δ G = –RT ln K

• where R is the gas constant (0.001987 kcal K^-1 mol^-1) (we will use kcal/mol here – 1 kcal/mol = 4.184 kJ/mol)
• and T is the temperature (in Kelvin)

Two species A and B that are present as a 50:50 ratio at equilibrium corresponds to an equilibrium constant  K = 1 and an energy difference  ΔG of 0 kcal/mol.

Not very exciting!

However, graphing a range of  Δ G values versus the equilibrium constant K, allows us to see that a relatively small quantity of energy (1 kcal/mol)  at room temperature (298 K) is sufficient to provide a 84:16 ratio of products to reactants. So 1 kcal/mol is actually a lot! 

Conversely, if you know the energy difference ΔG, you can calculate the equilibrium constant K through rearranging the equation to give:

K = e^(–ΔG/RT)

The article below has both discussion of these consequences and also some quizzes where you can practice doing various calculations involving K and ΔG at various temperatures.

Table of Contents

1. Calculating Product Ratios For a Product At Equilibrium
2. Calculating Energy Differences (ΔG) from Equilibrium Constants
3. 1 kcal/mol Is Actually Quite A Lot
4. Calculating Equilibrium Constant K When You Have The Energy Difference (ΔG)
5. Better Selectivity “For Free” By Lowering The Temperature
6. Application – Cyclohexane “A-Values”
7. Application – Alkene Reactions
8. Application – Substitution Reaction
9. Application – Keto Enol Tautomerism
10. Summary
11. Notes
12. Quiz Yourself!
13. (Advanced) References and Further Reading

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1. Calculating Product Ratios For A Reaction at Equilibrium

Let’s say we have a reversible process [Note 1] between a reactant A and a product B.

• The equilibrium constant K is equal to [ B ] / [A] where the square brackets refer to the concentrations of A and B (in mol/L) at equilibrium.
• For example, an equilibrium constant K of 5.4 means that there is a 5.4 : 1 ratio of products to starting materials at equilibrium.
• Since K is ultimately just a ratio, we can calculate the amount of product by dividing 5.4 by (5.4 + 1) to give 0.84 as the percentage composition of B, and therefore the amount of A is just whatever is left over (0.16).
• We can calculate the contribution of product B to the overall mixture by dividing the equilibrium constant K by (K + 1)

When the concentrations of A and B are equal, the equilibrium constant K is equal to 1 and the amount of B in the mixture is equal to K / (K+1) which in this case is  1 / [1+ 1] or just 0.5

2. Calculating Energy Differences From Equilibrium Constants

The equilibrium constant K is related to the difference in energy Δ G between two species by the equation:

ln K = – Δ G / RT

• where R is the gas constant (0.001987 kcal K^-1 mol^-1) (Recall that 1 kcal – 4.184 kJ . We tend to use kcal/mol here because kcal units are just so darn convenient for organic chemistry [See article – Why do Organic Chemists Use Kilocalories [Note 2])
• and T is the temperature (in Kelvin)

We can now rearrange this equation to get

Δ G = –RT ln K

This means that if you can measure the equilibrium constant for a given transformation at a specific temperature, you can use it to calculate the corresponding energy difference Δ G . That’s very useful!

At room temperature (298 K), making a table of these values looks like this:

Practice doing some calculations in the quiz below.

3. 1 Kcal/mol Is Actually Quite A Lot

From the table above, it’s worth pointing out that a 1 kcal/mol difference in energy gives you quite a bit of selectivity.

Graphing the product ratio versus delta G at 298 K,  gives us the following plot:

A mere 1 kcal/mol difference in energy gets you an 84:16 mixture of products. That’s pretty good selectivity for such a small energy difference.

Trust me when I say that there’s a lot of times that chemists would be happy to take an 84:16 ratio of products!

When you consider that 1 kcal/mol is just the energy cost for an H-H eclipsing interaction in ethane [See post – Staggered vs. Eclipsed Conformations in Ethane] this seems like a pretty good deal! 

If we bump up our energy difference a little bit more to 3 kcal/mol, we get over 99:1 selectivity for the product. That’s excellent.   [Note 3]

4. Calculating Equilibrium Constants From Energy Differences

We can use this equation to work in the opposite direction too.

If we know the energy difference Δ G, we can calculate the equilibrium constant K through rearranging the equation to give:

K = e^(–ΔG/RT)

When the difference in energy between two species is zero, this corresponds to an equilibrium constant K = 1

When the difference in energy is a mere 1 kcal/mol, the equilibrium constant at room temperature (298 K) is 5.4 .

This skyrockets to K = 5000 for an energy difference of 5 kcal/mol at 298 K.

A useful rule of thumb is that each change in energy of about 1.36 kcal/mol results in an order of magnitude (10× ) increase in the equilibrium constant K. [Note 4]

5. Better Selectivity “For Free” By Lowering The Temperature

Up until now we’ve been examining the relationship between free energy and equilibrium constant at room temperature (298 K)

What happens to the equilibrium constant when the temperature is changed?

K = e^(–ΔG/RT)

As the temperature becomes smaller,  the equilibrium constant K goes up. (assuming a spontaneous process (negative ΔG) )

In other words, for a given energy difference (e.g. 1 kcal/mol) the selectivity for the product increases.

Compare the trend in equilibrium constants for a process with ΔG 1 kcal/mol at room temperature (298 K), the freezing point of water (273 K) and the temperature of the readily available dry ice – acetone cold bath (–78°C or 195 K).

Running the reaction at extremely low temperature (-78°C) gives us an equilibrium constant K = 13 providing a 93:7 ratio of products, versus 84:16 at room temperature.

Contrast this with performing the same process at the temperature of the boiling point of water (100°C or 373 K), which gives a product ratio of 79:21 and equilibrium constant K of 3.8.

So all else being equal, if you want to maximize the selectivity of an equilibrium for a desired product, run it at the lowest temperature possible that will still give you an acceptable rate.

So what’s the catch? The last part of the previous sentence provides a clue. The price we pay for running a reaction at lower temperature is that it will be slower.

Another factor, important to note, is that running a reaction at low enough temperature might shut down the equilibrium between the two species. This is not necessarily a problem for selectivity (See article – Thermodynamic vs. Kinetic Control) but just note that using the Gibbs equation assumes the two species are in equilibrium. 

6. Application – Cyclohexane “A-values”

So if “1 kcal/mol really is a lot after all”,  this still may leave many of you with a pressing and important question:

So what?

OK, OK.  Let’s make this a bit more concrete by following up with some specific examples.

First up: cyclohexane chair flips.

Cyclohexanes adopt the “chair” conformation and interconvert between two “flipped” chair forms, where all axial substituents become equatorial and all equatorial substituents become axial. These two conformations are in equilibrium.  [See article – The Cyclohexane Chair Flip]

When a substituent is present on cyclohexane, the conformation with the substituent in the equatorial position is lower in energy than the  conformation where the substituent is axial (largely due to “gauche interactions” – steric interactions between the substituent and C-H bonds on the cyclohexane ring)

Measuring the ratio of the two conformers (NMR – Nuclear Magnetic Resonance – is a handy tool for this) allows us to calculate the equilibrium constant K, and from there we can calculate the difference in energy ΔG.

For example, at room temperature 1-methylcyclohexane exists as a 95:5 ratio of equatorial and axial conformers, which corresponds to an equilibrium constant of 17.6 and an energy difference of 1.7 kcal/mol.

This value of 1.7 kcal/mol is often referred to as the “A-value” of the CH₃ group. (See article – Cyclohexane A-Values).

Many A-values (energy differences) for various substituents on cyclohexane have been measured and tabulated. For example, the A-value of the O–CH₃ group is 0.60 (less of a preference for the equatorial position than CH₃) and the A-value of the t-butyl group is 4.9 (extremely strong preference for the equatorial position).

In the quiz below, fill out the complete tables with the information provided:

7. Application – Alkene Reactions

Here’s anoth