# Maximum force that hydrogen-hydrogen bond can withstand

For $\ce{H-H}$:

• bond length is $74~\mathrm{pm}$
• bond energy is $436~\mathrm{kJ~mol^{-1}}$

(Source: article chemical bond on Wikipedia.)

The University of Waterloo page that the article cites says:

The amount of energy required to break a bond is called bond dissociation energy or simply bond energy.

and

The bond energy is essentially the average enthalpy change for a gas reaction to break all the similar bonds.

So I'm assuming that as the average enthalpy change increases, so does the force that stresses the bond (and thus the forces on each hydrogen atom).

For a single molecule of $\ce{H-H}$, what is the maximum force that the bond can withstand before breaking?

• You do understand that the bond dissociation energy is an energy. The magnitude of the force does not matter. Only the work done matters. You can (theoretically) use the tidal force from the black-hole itself on a hydrogen molecule for a tiny fraction of a second and it would not break.
– DHMO
Oct 24 '16 at 10:19
• @DHMO - I would disagree. Unless you're somehow applying a black hole symmetrically to the $\ce{H2}$ molecule, the high force is almost certainly applied differently to each atom, and will greatly exceed the force constant of the potential energy curve. There's a whole (new) area of mechanochemistry and we can clearly break bonds with applied forces. Oct 24 '16 at 15:47
• DHMO seems to be saying that if you apply a huge force for a short enough time, the molecular bond can survive because not enough energy will have been placed into the atoms to break the bond. That is correct. Aug 16 '17 at 20:42

The force and the bond enthalpy are not necessarily directly related.

The bond dissociation energy is the amount of energy released when the bond is formed, or the energy required to break the bond.

The force constant of the bond derives from the curvature of the potential energy curve. For example, if we assume a harmonic oscillator (spring) for a bond, then:

$$V(x) \approx \frac{1}{2}k_f \; x^2$$

Real bonds aren't purely parabolic, but clearly the second derivative of the potential energy curve near the equilibrium bond length will give the needed force constant.

Here I've plotted two fictional bonds with the same length. The blue curve gives a bond with a higher bond enthalpy and force constant (i.e., the curvature is higher than the red curve). It should be clear that this isn't necessarily true. We can easily imagine some bond that has a shallow curvature but a deep enthalpy.

I should indicate that in many cases, these are related - stronger bonds typically have higher force constants and are harder to distort, but the point here is that the bond dissociation energy does not directly tell us anything about the mechanical strength of a bond.

• One comment on the question - as the enthalpy increases, the molecule simply goes to a vibrationally excited state - increasing the bond length. I'm not sure I'd want to imply an intuition of forces occurring on the atoms. The enthalpy simply ensures the vibrational amplitude increases - it stretches more. Oct 24 '16 at 15:49
• Hmm. So, to directly answer the question, would that mean that the force required is simply $k_\mathrm{f}\cdot x$ where $x$ is the displacement (away from the equilibrium bond length) that one wants to stretch the H-H bond length to? (Assuming harmonicity, of course.) Oct 24 '16 at 18:17
• @orthocresol - Well yes, by unit analysis, force times displacement = energy. So yes, if you have a way to pull on a molecule with a set force over a displacement, you're putting energy into the bond and can break it. (For example, there are AFM "pulling" experiments.) Oct 24 '16 at 19:41