Statistical Thermodynamics and Rate Theories/Vibrational partition function of a diatomic molecule

The general form of the molecular partition function is an infinite sum which is open form, making it difficult to calculate, this is why the sum is approximated as a closed form which leads to an algebraic equation. The derivation of the closed form on the equation is as follows:

The open form of the vibrational partition function:$$q_{vib}=\sum_{j} g_j \exp \left( \frac{-\epsilon_j}{k_B T} \right)$$

solving for $$\epsilon_j$$

$$\epsilon_j$$$$=E_j -E_0$$

$$\epsilon_j$$$$=$$$$h\nu$$ $$(j+\frac{1}{2})$$$$-h\nu$$$$(0-\frac{1}{2})$$

$$\epsilon_j$$$$=h\nu$$$$j$$$$-$$ $$\frac{1}{2}$$ $$h\nu$$$$-$$$$\frac{1}{2}$$$$h\nu$$$$=h$$$$\nu$$$$j$$

(where j=0,1,2...)

and by substituting $$g_j$$$$=1$$  (i.e., singly degenerate) into the summation for $$q_{vib}$$ the resulting equation is:

$$q_{vib}=\sum_{j=0}^{\infin}$$$$\exp \left( \frac{-h\nu j}{k_BT} \right)=\sum_{j=0}^{\infin} [ \exp \left(\frac{-h\nu}{k_BT} \right)]^j$$ The j is taken outside the brackets by the common exponent rule: $$\exp (ab) =\exp(a) \exp(b) $$

Note that $$E_n$$$$=$$$$h\nu (n+\frac{1}{2})$$ is the equation that represents the energy levels of a harmonic oscillator which is used to approximate the vibrational molecular degree of freedom. The vibrational zero point energy is not negligible and must be defined at n=0.

Next, in order for the open system to be converted into the closed system, the equation must take the form of a geometric series identity, like in calculus. if x<1, $$\sum_{i=0}^{\infin}$$$$x^i$$$$=$$ $$\frac{1}{1-x}$$

This is done by first letting $$e^\left (\frac{-h\nu}{k_B T} \right)$$ $$=x$$

Then, $$q_{vib}=\sum_{j=0}^{\infin}$$ $$x^j$$

this converges when x<1, giving $$\frac{1}{1-x}$$ and by replacing x with the original expression, you have:

$$q_{vib}=$$$$\frac{1}{1-\exp \left( \frac{-h\nu j}{k_B T} \right)}$$

where $$\nu$$ is the vibrational frequency of the molecule, which can by found by the following equation: $$\nu$$=$$\frac{1}{2\pi}$$$$\left (\frac{k}{\mu} \right)^\frac{1}{2}$$

where k is the spring constant of the molecule and $$\mu$$ is the reduced mass

Example
Calculate the population of the ground vibrational state of $$N_2$$ at 298.15 K. ($$\nu=7.072\times 10^{13} s^{-1}$$)


 * $$q_{vib}=\frac{1}{1-\exp \left( \frac{-h\nu j}{k_B T} \right)}$$

$$=\frac{1}{1-\exp \left( \frac{-(6.626068\times 10^{-34} Js \times 7.072 \times 10^{13} s^{-1}}{1.3807\times 10^{-23} JK^{-1} \times 298.15 K} \right)}=1.000011$$

Next the Probability of the ground state can be calculated: $$P_0=\frac{\exp \left( \frac{-E_0}{k_B T} \right)}{q_{vib}}=\frac{\exp (0)}{1.000011} =0.999998$$

This means that 99.9998% of all $$N_2$$ molecules are in the ground vibrational state at 298.15 K.