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.


2 Answers 2


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 $\endgroup$ Commented Dec 13, 2020 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. $\endgroup$ Commented Dec 13, 2020 at 13:39
  • $\begingroup$ I have taken word as used in a physics text. $\endgroup$ Commented Dec 13, 2020 at 13:39

I think each vibrational degree of freedom contributes kT to the energy because vibrational energy has two quadratic terms, each contributing kT/2 (one for kinetic and one for potential). So the correct reasoning is c. 4 vibrational degrees of freedom multiplied by kT per each one of them which is equal to 4kT contribution to energy from vibration.


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