# Statistical Thermodynamics and Rate Theories/Sample problems

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## Problem 1

Calculate the probability of a molecule of N2 being in the ground vibrational state at 298 K.

The probability that a system occupies a given state at an instant of time and a specific temperature is given by the Boltzmann distribution.

$P_{i}={\frac {\exp \left({\frac {-E_{i}}{k_{B}T}}\right)}{\sum _{j}\exp \left({\frac {-E_{j}}{k_{B}T}}\right)}}$ $P_{i}={\frac {\exp \left({\frac {-E_{i}}{k_{B}T}}\right)}{Q}}$ where:

• i is the energy of the specific state, i, of interest
• kB is Boltzmann's constant, which equals $1.3806\times 10^{-34}$ JK-1
• T is the temperature in Kelvin

The denominator of this function is known as the partition function, Q, which corresponds to the total number of accessible states of the molecule.

The closed form of the molecular vibrational partition function is given by:

$q_{vib}={\frac {1}{1-e^{-h\nu /k_{B}T}}}$ where:

• $\nu$ is the fundamental vibrational frequency of N2 in s-1
• h is Planck's constant, which is $6.62607\times 10^{-34}$ Js

This is equivalent to Q since only the vibrational energy states are of interest and there is only one molecule of N2. The equation for determining the partition function Q, from molecular partition functions, q, is given by:

$Q={\frac {q^{N}}{N!}}$ where:

• N is the number of molecules

The fundamental vibrational frequency of N2 in wavenumbers, ${\tilde {\nu }}$ , is 2358.6cm-1 

The fundamental vibrational frequency in s-1 is given by:

$\nu ={\tilde {\nu }}\times c$ where

• c is the speed of light, which is $2.9979\times 10^{10}$ cm/s

For N2,

$\nu$ = (2358.6cm-1) \times (2.9979 \times 10^{10| cm/s) = 7.0708 \times 10^{13}[/itex]

For N2 at 298 K,

$q_{v}=\left({\frac {1}{1-e^{-(6.62607\times 10^{-34}Js\times 7.0708\times 10^{13}s^{-1})/(1.3806\times 10^{-23}JK^{-1}\times 298K)}}}\right)=1.000011333$ The vibrational energy levels follow that of a quantum mechanical harmonic oscillator. The energy levels are represented by:

$E_{n}=h\nu (n+{\frac {1}{2}})$ where:

• n is the quantum vibrational number, which equals 0, 1, 2,...

For the ground state (n=0), the energy becomes:

$E_{0}={\frac {1}{2}}h\nu$ Since the vibrational zero point energy is not zero, the energy levels are defined relative to the n=0 level. This is used in the molecular partition function above and therefore, the ground state is regarded as having zero energy.

For N2 the probability of being in the ground state at 298K is:

$P_{0}={\frac {e^{-E_{0}/k_{B}T}}{q_{v}}}$ $P_{0}={\frac {e^{(0J)/(1.3806\times 10^{-23}\times 298K)}}{1.000011333}}$ $P_{0}=0.999988667$ This means that at room temperature, the probability of a molecule of N2 being in the ground vibrational state is 99.9988667%.