# Understanding vibrational mode of a molecule and its contribution to average energy

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.

• Thanks. Does each vibrational mode contributes 0.5kT or kT? I think it is kT Commented Dec 13, 2020 at 13:29
• 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. Commented Dec 13, 2020 at 13:39
• I have taken word as used in a physics text. 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.