I am trying to think about radical reactions using zero-point energies (from the harmonic oscillator model). I have read that substitution of a bond with isotopes can alter the rate of a reaction.

For example, consider the reaction:

$\ce{CH4 + Br2 -> CH3Br + HBr }$

Suppose the 3 non reacted hydrogens are instead deuterated. The new reaction will be:

$\ce{CD3H + Br2 -> CD3Br + HBr}$

Now, based on the harmonic oscillator model, the zero point energy of the C-H bond in $\ce{CH4}$ can be calculated from $E_0=\frac{1}{2}h\frac{1}{2\pi}\sqrt{\frac{k}{\mu}}$ where $k$=the force constant for the bond and $\mu$ is the reduced mass.

I read that "isotopic substitution does not change the potential energy surface of molecules" so the value of $k$ is unaffected by isotopic substitution. Thus in a kinetic isotope effect analysis based on zero-point energies, we need only consider the change in reduced mass.

Now my question is about the specifics of handling the listed reaction above. For calculating the reduced mass of the C-H bond, would you use the mass of the whole molecule (1) or just the mass of the carbon atom that the hydrogen is bonded to (2)?

In case the wording is confusing, in the example reaction the difference between the reduced masses would be:

$\mu=\frac{m_1 m_2}{m_1+m_2}$

(1) $\mu=\frac{15\cdot2}{15+2}=1.76$

This version of the calculation takes $m_1$ as the mass of the rest of the molecule ($\ce{CD3}$) beyond the C-H bond.

(2) $\mu=\frac{12\cdot2}{12+2}=1.714$

This version takes $m_1$ as the mass of only the carbon atom in the bond.

Which would be the correct way to carry out the calculation?

I kept the rest of the molecule simple in the calculation, but if the rest of the molecule had a much larger molar mass (say 200), then the reduced mass would become closer to 2, so I feel like the difference between the two methods could significantly affect the results of calculations on more complex molecules.

  • $\begingroup$ I'm not an expert but I dimly recall that $m_1$ and $m_2$ refer to the two "point" masses at either end of the "vibrating" bond. Since the (complex) vibrational mode of the bond in question includes all the atoms, not just C, my guess is the first way you did it is right. $\endgroup$
    – Curt F.
    Apr 11, 2019 at 16:28


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