I'm facing difficulty understanding how vibrational energy modes contribute to a molecule's average energy (or heat capacity). What I know is : For a polyatomic non-linear molecule, there are $3N-6$ normal modes of vibration ($3N-5$ for linear molecule). Also, in a one-dimensional harmonic oscillator, $(1/2)kT$ contribution to energy comes from potential energy, and $(1/2)kT$ from kinetic energy. So that in a system of many oscillators, $kT$ is avg. energy per oscillator, in Equilibrium.

Now consider linear triatomic molecule $CO_2$. It has $3$ translational and $2$ rotational d.o.f. There are two ways I can think of its vibrational d.o.f. -

  • $(a)$ It has $2$ C-O double bonds which are like $2$ oscillators. So vibrations contribute $2 \times kT $ to molecule energy. Therefore, the total energy of the molecule $E=(3+2)\times(1/2)kT + 2kT=(9/2)kT$.

  • $(b)$ It has $4$ normal modes of vibration. Considering each normal mode as one d.o.f., $E=(3+2+4)\times (1/2)kT=(9/2)kT$

  • $(c)$ I think considering each normal mode as an oscillator makes more sense. In that case, vibrational contribution $= 4kT$ and $E=(13/2)kT$.

Which of these is correct way (if either) to approach for any general molecule? I can't find enough examples that talk about all degrees of freedom active at high temperatures that could help me clear my doubt.


Your second reasoning is Correct! As suggested in Here

A molecule with $N$ atoms has more complicated modes of molecular vibration, with $3N − 5$ vibrational modes for a linear molecule and $3N − 6$ modes for a nonlinear molecule.

Now to understand How you get $4$ normal modes for $CO_2$ molecule need a little bit knowledge of theory of small oscillations. Here I will try to give a short way to understand this:

Recall that Equipartition theorem says that

Each quadratic dependence of the system( called mode) of the system the system contributes an amount of energy equal to $1/2 (k_bT)$ to total mean energy of the system

The translation and rotational are not concern us here that I consider you already know.

Let's see How the vibrational mode look like :

We consider a system consisting of a particle of mass $\mu$ situated midway between two particles of mass unity. To specify the configuration of the system we introduce a set of Cartesian coordinate axes with the z axis along the line joining the particle. The coordinate $x_1,y_1$ measure the displacement of the first of the two particles of unit mass away from the $z-axis$, and $z_1$ measures the displacement along the $z-axis$ away from the position of equilibrium. $x_2,y_2,z_2$ do the same for the other particle of unit mass and $x_3,y_3,z_3$ describe the particle of mass $\mu$.

We place the origin of the coordinate system at the center of mass, in this case at center(carbon) atom. $$\dot{x_1}+\dot{x_2}+\mu\dot{x_3}=0$$ and similarly for $y$ and $z$ so that $$x_3=-\frac{x_1+x_2}{\mu} , \mathrm{etc.}$$

We shall consider the case in which the angular momentum about the center of mass is also zero. $$l_x=a(\dot{y}_1-\dot{y}_2)=0$$ $$l_y=a(\dot{x}_2-\dot{x}_1)=0$$ where $a$ is the equillibrium separation of the particle. Thus $$\dot{y}_1=\dot{y}_2, \ \ \ \dot{x}_1=\dot{x}_2$$ and $$y_1=y_2 \ \ \ \ \ x_1=x_2$$

$$x_3=-\frac{2x_1}{\mu}, \ \ \ \ y_3=-\frac{2y_1}{\mu}$$ The potential energy $$\mathcal{V}=\frac{1}{2}k\left[(z_1-z_3)^2+(z_2-z_3)^2\right]\frac{1}{2}\kappa \left[(x_1-x_3)^2+(x_2-x_3)^2+(y_1-y_3)^2+(y_2-y_3)^2\right]+\frac{1}{2}k'(z_2-z_1)^2$$ Little bit of substitution and algebra lead you to $$\mathcal{V}=\frac{1}{2}k\left[\left(\cdots\right)(z_1^2+z_2^2)+\left(\cdots\right)z_1z_2\right]+\frac{1}{2}\kappa \left[\left(\cdots\right)(x_1^2+y_1^2)\right]+\frac{1}{2}k'(z_1-z_2)^2$$

Only four coordinate remain in $\mathcal{V}$. The matrix $\mathcal{V}$ is $4\times 4$ matrix with $4$ eigen value and $4$ normal modes as promised.

  • $\begingroup$ Thanks. Does each vibrational mode contributes 0.5kT or kT? I think it is kT $\endgroup$ – aneet kumar Dec 13 '20 at 13:29
  • $\begingroup$ As I stated each mode or degree of freedom or quadratic term in energy contributes $ 1/2(k_BT)$ to the total energy of the system. Note that In chemistry, it is taken to be a different meaning. $\endgroup$ – Young Kindaichi Dec 13 '20 at 13:39
  • $\begingroup$ I have taken word as used in a physics text. $\endgroup$ – Young Kindaichi Dec 13 '20 at 13:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.