Hydrogen bonding as a function of $kT$

Question

The strength of a hydrogen bond in water at ambient conditions is about $8kT$, where $k$ is the Boltzmann constant and $T$ is the temperature. So why do hydrogen bonds form and break so readily in water when their energy is so much higher than $kT$? And what determines the timescale $(\sim\, 1 - 10\ \rm ps)$ of the bond making and breaking?

Answer 1

Using Transition state theory approximation for the rate of water molecule to cross over the bonding barrier, $E_b = 8KT \approx 21 KJ/mole$:

$$ \tag{1} \frac{1}{\tau} = \omega \exp\left\{-\frac{E_b}{KT}\right\}. $$

Now, we have to estimate the frequency prefactor $\omega$ in Eq.(1). Since there is no data available for the vibration between water-water molecule interaction (don't mess up with the vibration within the water molecule), we simply simulate a quadratic potential from the bond length and bond energy: \begin{align} \frac{1}{2} k x^2 =& E_b;\\ k =& \frac{2 E_b}{x^2};\,\,\,\text{using } E_b= 21 \text{ KJ/mole, and bond length } 3 \dot A;\\ \omega = & \sqrt{\frac{k}{m}}=\sqrt{\frac{2 E_b}{m x^2}} = \sqrt{\frac{2 (21\times 10^3)/(6\times10^{23})}{0.018/(6\times 10^{23}) \times (3\times 10^{-10})^2}} = 5\times 10^{12}\frac{1}{s} \end{align}

Using this estimation into Eq.(1), the rate of crossing the energy barrier: $$ \frac{1}{\tau} = 5\times 10^{12} \exp(-8) = 2\times 10^9. $$ Therefore the bond breaking time scale is about $$ \tau =\frac{1}{2} 10^{-9} sec \approx 0.5 \, ns. $$

In the article ACS, where they estimate the activation energy to be $11$ KJ/mole (approximate $4\,KT$) half of the binding energy I adopted, which brings a factor of $e^4=55$ to the estimation here. This will render a bond breaking time to $9$ ps. Their low activation energy is due to large fluctuation of the hydrogen bonding energy. It arises from the frequent drawn near of two oxygen molecules, which generate a pulse of repulsion force to break the bond, results in a rather low activation energy.