State the law of equipartition of energy.

Hint: Equipartition means total energy which is contributed equally over all directions. Total energy possesses translational, rotational and vibrational energy. The molecule which doesn't possess vibrational or rotational energy will only possess translational energy.

Complete step-by-step answer:
Consider a molecule in three dimensions. That is X, Y, Z.
Then kinetic energy of a single molecule in X, Y, Z dimension is given by,
$E=\dfrac{1}{2}mv_{x}^{2}+\dfrac{1}{2}mv_{y}^{2}+\dfrac{1}{2}mv_{z}^{2}$ ---------(1)
This is translational energy.
Consider two molecules having angular speed of about its own axis and are the corresponding moment of inertia.
Since it do moment in all direction therefore it has both energy translational and rotational energy which is given by,
\[\begin{align}
  & TotalEnergy(E)={{E}_{_{tr}}}+{{E}_{rot}} \\
 & Total Energy={{E}_{_{tr}}}+\dfrac{1}{2}{{I}_{1}}\omega _{1}^{2}+\dfrac{1}{2}{{I}_{2}}\omega _{2}^{2} \\
 & {{E}_{_{tr}}}+{{E}_{rot}}=\dfrac{1}{2}mv_{x}^{2}+\dfrac{1}{2}mv_{y}^{2}+\dfrac{1}{2}mv_{z}^{2}+\dfrac{1}{2}{{I}_{1}}\omega _{1}^{2}+\dfrac{1}{2}{{I}_{2}}\omega _{2}^{2} \\
\end{align}\]
But some molecules possess vibrational energy also. Even at moderate temperature, molecules possess vibrational motion like CO.
Therefore it contributes a vibrational energy to the total energy.
Therefore total energy is given by,
\[TotalEnergy(E)={{E}_{_{tr}}}+{{E}_{rot}}\]
\[{{E}_{vr}}\]is equal to $\dfrac{1}{2}m{{\left( \dfrac{dy}{dt} \right)}^{2}}+\dfrac{1}{2}k{{y}^{2}}$
Where k is force constant if the oscillator and y the vibrational coordinate.
In equilibrium the total energy is equally distributed in all possible energy modes, with each mode having an average energy equal to $\dfrac{1}{2}{{K}_{B}}T.$ This is known as equipartition of energy.

Note: Each translational and rotational degree of freedom has contributed only one squared term, but one vibrational mode contributes kinetic and potential energies. Therefore each translational and rotational degree of freedom of a molecule contributes to $\dfrac{1}{2}{{K}_{B}}T$ energy while each vibrational frequency contributes $2\times \dfrac{1}{2}{{K}_{B}}T={{K}_{B}}T$.