When we say a conformer or compound is of "higher energy" than another, are we quantifying energy in terms of the bond strength? Or are we going off of how much energy it takes to break the bond? I don't think we can just assign energies to each compound... so what is this based off of?

One might answer that its based off of stability. Well, what is "Stability?" Is it how rigid/strong the bonds are? Is it how reactive the compound is?

Often times i find myself reading explanations with circular language on this topic. A higher energy conformer means lower stability or A stable compound is one that has low energy. I'm finding these definitions to be crazy ambiguous, and don't thoroughly define what these are.

If someone can clearly explain this in terms that a first year undergraduate student can understand, you have earned my gratitude.

  • $\begingroup$ There's more than one kind of stability... $\endgroup$
    – Zhe
    Commented Mar 15, 2019 at 2:41
  • $\begingroup$ Energy is generally a free energy. $\endgroup$
    – Zhe
    Commented Mar 15, 2019 at 2:42
  • 6
    $\begingroup$ These terms are well defined in (chemical) thermodynamics, however, chemists like to use them in a more casual way in high-school and early undergrad textbooks, too. It is not necessarily bad, as the intuitive meaning more or less correct, and helps the discussion. As @Zhe said, the energy is generally Gibbs free energy if we are talking about an ensemble, but in certain discussions like comparing conformers and potential energy surfaces it is actual energy (potential energy) of the system. $\endgroup$
    – Greg
    Commented Mar 15, 2019 at 5:00

5 Answers 5


As a chemist, I would agree that textbooks are not clear about the term stability and energy. It is not your fault and undergraduate organic chemistry books make the situation worse. I cannot recall the text, it was an old book, however it clearly said that stability of a compound does not mean anything. We should always ask, stability with respect to what? Are we talking about thermodynamic stability or kinetic stability? For example H2+O2 mixture is not stable. This statement is meaningless because we are not saying stability with respect to what? This mixture should form water right away but people actually waited for 30 years and did not find much water in a mixture of H2 and O2. Thermodynamically, this mixture is not stable and it will react violently once you provide the right activation energy. When you are reading about stability, ask this question yourself.

In the same vein, the term energy should also be clarified by authors. As a crude analogy is that you can think of energy as the potential energy of the whole molecule when the authors talk about conformers. Have a overall picture of the molecule. Imagine two springs, A) one which is less stretched and the other one (B) is more stretched. You can say spring B has more (potential) energy than (A). An eclipsed conformer has higher potential energy (overall, no need to think locally) than a staggered one.


Finding minimums: a general view

When minimizing energy you are searching (using numerical methods) for the minimum of Potential Energy). You are at the minimum when the derivative(Jacobian) is equal to zero. However, there are a couple unstable places that also have a derivative of zero, namely saddle points and maximums (top of a circle). We are after the minimum (bottom of a circle).

Think of a pendulum. It has two spots where the derivative is zero... when it is all the way up, and all the way down, but only when it is in the all the way down position is it stable. When the Hessian (second derivative) is positive definite, you are at the "bottom of the pendulum swing" and at a minimum. Pendulum up the derivative is zero, but this is a maximum.

There is also the problem that often in energy landscapes it looks like a rollercoaster... At a commenters request, a better image (taken from wikipedia, source in image description)

AimNature [CC BY-SA 3.0 (https://creativecommons.org/licenses/by-sa/3.0)]

We need to find not just the a minimum that has a derivative of zero,but the lowest of all minimums present. This is a real tough problem since numerical solvers stop once the derivative is zero. They don't know there are other minimums. There are methods to try to get around this, but it is a very active area of research and a huge prize awaits whoever can guarantee a global minimum is found.

A thought on energy

What is this energy we are minimizing? We are minimizing the energy compared to a reference point. Everything is always with respect to a reference point.

I am not a chemist, and I don't know much Quantum mechanics but I will try to give a chemistry answer...This is my impression of what is done in the scenario you are asking about. I perform minimizations on Free Energy, but that is related but not quite what you are after. In Quantum mechanics the reference point is when all nuclei and valence electrons are infinitely apart. Another common reference point in chemistry, physics and engineering is the ideal gas, which is when all molecules are infinitely apart from each other but atoms (and electrons) are still bonded to each other.

Given a reference point which we define to have "zero" energy, for each conformer we would calculate the energy to form all of the covalent bonds in the molecule, which is to say, calculate the energy to pull the valence electrons and nuclei in from infinitely far apart and put them in the exact geometry you want. Do this for many geometries. The one that took the lowest energy to form will be the stablest of the ones you tried (There may be others you did not try though! you may be in the wrong loop of the rollercoaster still). It takes different amounts of work to put two nuclei different distances apart etc. In practice, QM programs do all of this for us, and they are getting quite fast but it can still take a very long time. It is a hard thing to calculate.

Bond Energies

Generally in molecular mechanics bond energies are calculated using the harmonic spring approximation (hookes law), however the below image shows the QM calculated energies.

enter image description here

Angle Energies

Here is an image from the simualtion software Espresso's website which shows that the energy required to bend an angle is closely approximated by a quadratic function

enter image description here

Torsional Energies

Torsional energies are more complicated than the bond and angle potential energies because a torsional angle needs to be able to be turned 360 degrees and be back where it started. Here is the energy for different conformers as the geometry of the torsion angle is turned. Because of the different geometries, the nuclei and electrons are different distances apart from each other, and it takes a different amount of work (energy) to move them from infinitely far apart to each of those geometries. The image is from here.

enter image description here


In molecular mechanics we use functions (called force-fields) to approximate the above potential energy contributions. Here is a general set of the equations used. You can find the minimum of energy by changing bond lengths/angles/torsion parameters for a single molecule, and after each change, calculating the energy. The geometry with the lowest energy is the most stable because molecules, like people, always take the path of least resistance. I have copied the equations as an image from here

enter image description here

  • 1
    $\begingroup$ Could you label the axes? What are you plotting? $\endgroup$
    – ACR
    Commented Mar 15, 2019 at 4:09
  • $\begingroup$ It is from wikipedia. What I am plotting doesn't matter, it is a general take on numerical methods and finding a minimum. Since the OP is after Energy, pretend y-axis is Energy. Then pick a thing to minimize it with respect to... Bond length, angle, anything you want to minimize it with respect to can be the x axis. Bond lengths and angles look like quadratics. If OP wants I can put images for the Bond/Angle/Torsion potentials, but that is forcing the question in a specific direction. I tried to keep my answer ... general $\endgroup$
    – B. Kelly
    Commented Mar 15, 2019 at 4:19
  • $\begingroup$ Actually I liked your pendulum analogy and I thought the graph represented that! $\endgroup$
    – ACR
    Commented Mar 15, 2019 at 4:29
  • $\begingroup$ Since you have a theoretical background, I am an analytical chemist, what do theoreticians mean by energy when they show it on the vertical axis? Is it the sum of kinetic and potential energy of the electrons in a bound state at a given geometry? Your reference point is infinitely separated H and O atoms? $\endgroup$
    – ACR
    Commented Mar 15, 2019 at 4:42
  • 1
    $\begingroup$ For me, I am only interested in potential energies. My reference state is generally an ideal gas, rather than the separated nuclei/electrons. Kinetic energy is import too in simulations, it keeps the temperature correct, and also, you need to know the total energy to make sure you are conserving energy. I don't want to delve to much into the QM side of things since I am a novice in that area :) $\endgroup$
    – B. Kelly
    Commented Mar 15, 2019 at 4:59

The answer is complicated because there are different kinds of stability and when we talk about energy, we also can mean different kinds of energy.

When we speak of energy, we generally refer to a free energy. Your normal laboratory setup is an open-air, constant-pressure system, so we use Gibbs free energy. If you were working with bomb calorimeters, you'd want to use Helmholtz free energy for constant-volume processes.

Note importantly, that energies are relative. "What is the energy of this compound?" is a meaningless question. And when we speak of "high energy," we mean high relative to some reference point. That's precisely why most examinations of energy in chemistry focus on a change, most commonly $\Delta G$, the change in Gibbs free energy. Gibbs free energy is a tool for measuring changes in internal energy (most commonly bond-breaking and bond-making) while compensating for the entropic costs of those changes and pressure-volume work done by the system.

At standard state, the Gibbs free energy change for a reaction or transformation can describe the relative preference of the "reactants" versus the "products" via an equilibrium constant $K$.

$$\Delta G^{\circ} = -RT \ln K$$

This is a way to measure one kind of stability, thermodynamic stability.

Considering $\ce{A <=> B}$, $\ce{A}$ is thermodynamically stable relative to $\ce{B}$ if the free energy change of this process is positive (i.e., $\Delta G > 0$), and vice versa.

Unfortunately, thermodynamics only tells half the story for a reaction. There's also the kinetic piece, which determines how fast a reaction proceeds, that is, the rate of the reaction. Rates are determined by the relative energy of a fleeting transition state, captured by $\Delta G^{\ddagger}$. For a reaction that is highly spontaneous, if the reactant is kinetically stable (that is $\Delta G^{\ddagger}$ has a large value), then reaction will still proceed very slowly. This is kinetic stability.

  • $\begingroup$ Couldn't the stability of a single compound refer to its Gibbs free energy at standard conditions? Why we need to introduce the relative term? $\endgroup$
    – ado sar
    Commented Dec 20, 2020 at 13:41
  • 1
    $\begingroup$ Gibbs free energy at standard conditions is loose. The term is the standard Gibbs free energy of formation, which is relative to the component elements in their standard states. In that case, you are still referenced. $\endgroup$
    – Zhe
    Commented Dec 20, 2020 at 13:49
  • $\begingroup$ Ok but now we have a measure of stability of the compound which doesn't depend on the type of the reaction unlike reactivity which does. $\endgroup$
    – ado sar
    Commented Dec 20, 2020 at 13:52
  • $\begingroup$ I don't understand your comment... @adosar $\endgroup$
    – Zhe
    Commented Dec 20, 2020 at 14:57
  • $\begingroup$ A compound A can react fast with a compound B to give C (slow) and with compound D to give E (fast). The Gibbs free energy of formation of A is the same on both reactions but the reactivity (which is reflected on the activation energy) will depend on which coumpound A is reacting with (B or D) i.e. in the specific reaction. $\endgroup$
    – ado sar
    Commented Dec 20, 2020 at 23:06

As folks are saying in the comments, energy is used throughout chemistry and comes in many different flavors, so instead of attempting to answer the broad question "How is energy and stability defined throughout chemistry", I will focus on your two examples.

When we say a conformer is of "higher energy" than another, are we quantifying energy in terms of the bond strength?

Conformers are different shapes (3D structures) a given molecule assumes over time (usually by rotating about single bonds, i.e changing torsion angles). High energy conformers are rare, and low energy conformers are common. Conformational energy is a potential energy. It can be used to figure out the forces on the molecule as you change the torsion angles.

The reason that molecules don't just assume the lowest energy conformation is that there is kinetic (thermal) energy available. Other molecules keep bumping into our molecule of interest, giving it the energy to adopt higher energy conformations. On the other hand, when our molecule of interest goes from a higher energy conformation to a lower one, the potential energy is transformed into kinetic energy.

The reason that different conformers have different energies is because in some conformations, atoms that are not directly bonded come too close to each other ("steric clashes", associated with high conformational energy) or have favorable interactions (hydrogen bonds, dipole-dipole interactions etc). Also, some conformations don't allow optimal bonding. An extreme example would be a double bond, rotation about which would break the pi-bond and increase the potential energy a lot. Even with a single bond, some conformations have lower energy than others because the allow for a slightly stronger bond (this is called a stereo-electronic effect).

So the conformational energy depends on subtle changes in bond strength (stereo-electronic effects) and on non-bonded interactions (steric clashes, non-covalent interactions).

When we say a compound is of "higher energy" than another, are we quantifying energy in terms of the bond strength?

For a clean comparison of the "energy" of two compounds, they have to contain the same set of atoms, i.e. they should be isomers. So you could ask whether silver fulminate (AgCNO) or silver cyanate (AgOCN) has the lower energy. Or you could ask whether acetaldehyde occurs as aldehyde (CH3-CHO) or enol (CH2=CH-OH). Conceptually, we could add up all the bond energies in the two compounds to see which one has the lower potential energy.

In cases where the potential energy difference is very high, we can safely say that one compound will be common and the other rare or non-existent. You would say one is more stable than the other. If the potential energies are similar, we also have to consider how the compounds would interact with surrounding molecules (e.g. solvent), and we have to consider entropy (or, equivalently, we have to compare Gibbs free energy instead of potential energy only), which is beyond the scope of this answer (and some might argue beyond first year undergraduate study).


"When we say a conformer or compound is of "higher energy" than another, are we quantifying energy in terms of the bond strength? Or are we going off of how much energy it takes to break the bond?"

The two options are expressing the same and yes, we are doing that.

But for an analysis is generally convenient to think of the same bonds and consider other potential energies, which can be due to torsion, electrostatic repulsion / steryc hindrance, tension... Entering in the discussion of what is more conceptually appropriate might astray from the question, but often the above partition is surely motivated by the different energies involved.

Suppose to separately burn two conformers of an alkane. The products are totally insensitive to this, still the heat of combustion will be higher when burning what we then call the unstable conformer, right because its energy content was higher. Note that so was, independent of the way we thought to distribute the energy surplus in & around the molecule.

" don't think we can just assign energies to each compound... so what is this based off of?".

We can as we have just did above, assuming a reference of course. In the above example we could have taken the energy of the products to be zero, tough you may look for how standard states are defined.

Once you have standard states, you can order different molecules by their energy of formation, and this gives you an order of stability.

When we consider the free energy instead of entalpy, we can even compare different and differently sizes molecules. The order is done per mole. The fact that burning more matter gives more heat is of course trivial and does not tell itself about the stability of that matter. Enthalpy alone worked above because we had a) conformers and b) the same reaction.

It is not necessary to mention quantum chemistry, if not calculations or deeper insights are required. Thermochemistry is older and deal with this concept as well, as in the given example. While the core quantity in thermochemistry is heat of formation, combustion makes (made) easy to run the experiments and get things by comparison. This is to say that my example must not be taken as a definition of stability, but it was to show that there is no circular language. Thermodynamics quantities are of course real and quantifiable.

"Is it how reactive the compound is?"

As addressed by others reactivity should not be confused with thermodynamical instability. It is just matter of wording dictated by the fact that a system wich does not change is called "stable", both in daily chemistry and daily life. I have difficulties to think of a language which instead of still stones speak about stones in metastable equilibrium. However in chemical physics we have this term, as well the term activation energy. Once the activation energy is available to a system in a metastable state, then the fate of the latter is dictated by the thermodynamics, as above.


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