# Statistical Thermodynamics and Rate Theories/Sample problems

## 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%.

## Problem 2

Derive an equation for the population of rotational state i of a linear diatomic. Make a bar graph of the distribution of rotational states of N2 at 298 K.

1. Equation for the population of rotational state i

For a diatomic molecule, it can be approximated as a rigid rotor. Solving Schrödinger equation of rigid rotors gives the energy levels of the molecule at state J:

$E(J)=BJ(J+1)$

where $J$  is the quantum number for total rotational angular momentum; $B$  is the rotational constant in cm-1.

• $B={\frac {h}{8\pi ^{2}cI}}$ ,

where $h$  is Planck constant; $c$  is the vacuum light speed in cm/s; and $I$  is the moment of inertia.

• $I=\mu r^{2}$ ,

where $\mu$  is the reduced mass and $r$  is the bond length.

By the Maxwell–Boltzmann_distribution, the population of rotational state i comparing to ground state is:

${\frac {N_{i}}{N_{0}}}={\frac {g_{i}}{g_{0}}}\left({\frac {e^{-E_{i}/k_{B}T}}{e^{-E_{0}/k_{B}T}}}\right)$

where $g$  is the degeneracy of the state; $E$  is the energy of the state; $k_{B}$  is Boltzmann constant and $T$  is the temperature.

Substitute $E_{i}=hcBi(i+1)$  ,$E_{0}=hcB0(0+1)=0$ , and the degeneracy of the state i $g_{i}=2i+1$  to the equation, get:

${\frac {N_{i}}{N_{0}}}=(2i+1)\left({\frac {e^{-hcBi(i+1)/k_{B}T}}{e^{0/k_{B}T}}}\right)=(2i+1)e^{-hcBi(i+1)/k_{B}T}$

### 2. Distribution of rotational states of N2

For N2, the population of state i is:

${\frac {N_{i}}{N_{0}}}=(2i+1)e^{-hcBi(i+1)/k_{B}T}$

Combine the constant part, define constant $a=hcB/k_{B}T={\frac {h^{2}}{8\pi ^{2}k_{B}T\mu r^{2}}}$ .

The reduced mass μ of Nitrogen is 7.00D a=1.16×10-26 kg and the bond length r of N2 is 110 pm = 1.10×10-10m.

Substitute T = 298 K, $a={\frac {h^{2}}{8\pi ^{2}k_{B}T\mu r^{2}}}=9.60\times 10^{-3}$

By sum over states, qrot=1/a = 104, so the probability of N2 occupying the ground vibrational state at 298 K is 1/104 = 9.60×10-3

The bar graph of the distribution of rotational states of N2 at 298 K is:

## Problem 3

Estimate the number of translational states that are available to a molecule of N2 in 1 m3 container at 298 K.

The equation that is used to determine translational states of the molecule of N2 at 298 K is shown below.

$q_{trans}(V,T)=\left({\frac {2\pi mk_{B}T}{h^{2}}}\right)^{\frac {3}{2}}$ $V$

Where $q_{trans}$ (V,T) represents the partition function for translation, $m$  represents the mass of the particle in kilograms (kg), $k_{B}$  represents Boltzmann's constant $(1.38064852\times 10^{-23}J\times K^{-1})$ , $T$  represents the temperature in kelvins $(K)$ , $h^{2}$  represents Planck's constant $(6.626\times 10^{-34}J\times s)$  and $V$  represents the volume in 3 dimensions in $m^{3}$ .

The steps for solving this problem is shown below:

m = $28.0102$  amu

m = $(28.0102amu)(1.6605\times 10^{-27}kg)$

m = $4.6512\times 10^{-26}kg$

$q_{trans}(V,T)=\left({\frac {2\pi mk_{B}T}{h^{2}}}\right)^{3/2}V$

$q_{trans}(V,T)=\left({\frac {2\times 3.14159\times (4.6512\times 10^{-26}kg)\times (1.38064852\times 10^{-23}J/K)\times 298K}{{(6.626\times 10^{-34}J\times s})^{2}}}\right)^{3/2}(1.000m^{3})$

$q_{trans}(V,T)=1.4322\times 10^{32}$

Therefore, there should be $1.4322\times 10^{32}$  translational states for N2 at 298 K.

## Problem 4

Calculate the DeBroglie wavelength, rotational temperature, and vibrational temperature of N2 and Cl2 at 298 K.

### The DeBroglie Wavelength

$\Lambda =\left({\frac {2{\pi }mk_{B}T}{h^{2}}}\right)^{-1/2}$

where:

• m is the mass of the molecule.
• Boltzmann's constant kB = 1.3806488×10-23 J K-1
• Planck's constant h = 6.62606957×10-34 J s

For N2 at 298K,

$m=2\times 14.0067u$
$m=28.0134u\times {\frac {1.66054\times 10^{-27}kg}{u}}$
$m=4.65173712\times 10^{-26}kg$
$\Lambda =\left({\frac {2{\pi }(4.65173712\times 10^{-26}kg)(1.3806488\times 10^{-23}JK^{-1})(298K)}{(6.62606957\times 10^{-34}Js)^{2}}}\right)^{-1/2}$
$\Lambda =\left({\frac {1.20252611\times 10^{-45}kgJ}{4.39047979\times 10^{-67}J^{2}s^{2}}}\right)^{-1/2}$
$\Lambda =(2.738940097\times 10^{-21}kgJ^{-1}s^{-2})^{-1/2}$
$\Lambda =1.9107714\times 10^{-11}m$

For Cl2 at 298K,

$m=2\times 35.453u$
$m=70.906u\times {\frac {1.66054\times 10^{-27}kg}{u}}$
$m=1.17742249\times 10^{-25}kg$
$\Lambda =\left({\frac {2{\pi }(1.17742249\times 10^{-25}kg)(1.3806488\times 10^{-23}JK^{-1})(298K)}{(6.62606957\times 10^{-34}Js)^{2}}}\right)^{-1/2}$
$\Lambda =\left({\frac {3.04376893\times 10^{-45}kgJ}{4.39047979\times 10^{-67}J^{2}s^{2}}}\right)^{-1/2}$
$\Lambda =(6.932656726\times 10^{21}kgJ^{-1}s^{-2})^{-1/2}$
$\Lambda =1.20101976\times 10^{-11}m$

Unit Conversion for the DeBroglie Wavelength:

$(kgJ^{-1}s^{-2})^{-1/2}=\left({\frac {kg}{kgm^{2}s^{-2}s^{2}}}\right)^{-1/2}$
$(kgJ^{-1}s^{-2})^{-1/2}=(m^{-2})^{-1/2}$
$(kgJ^{-1}s^{-2})^{-1/2}=m$

### The Rotational Temperature

$\Theta _{r}={\frac {\hbar ^{2}}{2k_{B}\mu r_{e}^{2}}}$

where:

• Planck's constant $\hbar ={\frac {h}{2\pi }}={\frac {6.62606957\times 10^{-34}Js}{2\pi }}=1.054571726\times 10^{-34}Js$
• μ is the reduced mass.
• re is the bond length between two atoms in a molecule

For N2,

re = 1.09769Å = 1.09769×10-10m 
$\mu ={\frac {m_{N}m_{N}}{m_{N}+m_{N}}}={\frac {m_{N}^{2}}{2m_{N}}}={\frac {m_{N}}{2}}={\frac {14.00307400529u}{2}}=7.001537003u$
$\mu =7.001537003u\times {\frac {1.66054\times 10^{-27}kg}{u}}$
$\mu =1.16263323\times 10^{-26}kg$
$\Theta _{r}={\frac {(1.054571726\times 10^{-34}Js)^{2}}{2(1.3806488\times 10^{-23}JK^{-1})(1.16263323\times 10^{-26}kg)(1.09769\times 10^{-10}m)^{2}}}$
$\Theta _{r}={\frac {1.11212153\times 10^{-68}J^{2}s^{2}}{3.86825738\times 10^{-69}JK^{-1}kgm^{2}}}$
$\Theta _{r}=2.874993624Js^{2}Kkg^{-1}m^{-2}$
$\Theta _{r}=2.874993624K$

For Cl2,

re = 1.988Å = 1.988×10-10m 
$\mu ={\frac {m_{Cl}m_{Cl}}{m_{Cl}+m_{Cl}}}={\frac {m_{Cl}^{2}}{2m_{Cl}}}={\frac {m_{Cl}}{2}}={\frac {34.968852714u}{2}}=17.48442636u$
$\mu =17.4842636u\times {\frac {1.66054\times 10^{-27}kg}{u}}$
$\mu =2.90335893\times 10^{-26}kg$
$\Theta _{r}={\frac {(1.054571726\times 10^{-34}Js)^{2}}{2(1.3806488\times 10^{-23}JK^{-1})(2.90335893\times 10^{-26}kg)(1.988\times 10^{-10}m)^{2}}}$
$\Theta _{r}={\frac {1.11212153\times 10^{-68}J^{2}s^{2}}{3.16844888\times 10^{-68}JK^{-1}kgm^{2}}}$
$\Theta _{r}=0.350998729Js^{2}Kkg^{-1}m^{-2}$
$\Theta _{r}=0.350998729K$

Unit Conversion for the Rotational Temperature:

$Js^{2}Kkg^{-1}m^{-2}=J(kg^{-1}m^{-2}s^{2})K$
$Js^{2}Kkg^{-1}m^{-2}=JJ^{-1}K$
$Js^{2}Kkg^{-1}m^{-2}=K$

### The Vibrational Temperature

$\Theta _{v}={\frac {h\nu }{k_{B}}}$
$\Theta _{v}={\frac {h{\tilde {\nu }}c}{k_{B}}}$

where:

• ${\tilde {\nu }}$  the vibrational frequency of the molecule in wavenumbers.
• The speed of light c = 2.99792458×1010cm s-1

For N2,

${\tilde {\nu }}=2358.57cm^{-1}$  
$\Theta _{v}={\frac {(6.62606957\times 10^{-34}Js)(2358.57cm^{-1})(2.99792458\times 10^{10}cms^{-1})}{1.3806488\times 10^{-23}JK^{-1}}}$
$\Theta _{v}={\frac {4.6851712\times 10^{-20}J}{1.3806488\times 10^{-23}JK^{-1}}}$
$\Theta _{v}=3393.456175K$

For Cl2,

${\tilde {\nu }}=559.7cm^{-1}$  
$\Theta _{v}={\frac {(6.62606957\times 10^{-34}Js)(559.7cm^{-1})(2.99792458\times 10^{10}cms^{-1})}{1.3806488\times 10^{-23}JK^{-1}}}$
$\Theta _{v}={\frac {1.11181365\times 10^{-20}J}{1.3806488\times 10^{-23}JK^{-1}}}$
$\Theta _{v}=805.2834645K$

### References

1. a b Lide, D. R., (84th ed.). (2003-2004). Handbook of Chemistry and Physics. CRC Press. pg.9-85.
2. a b Lide, D. R., (84th ed.). (2003-2004). Handbook of Chemistry and Physics. CRC Press. pg.9-83.
Molecule N2 Cl2
$\Lambda$  1.91×10-11 m 1.20×10-11m
$\Theta _{r}$  2.87 K 0.351 K
$\Theta _{v}$  3.39×103 K 805 K

## Problem 5

At what temperature would the probability for N2 being the ground vibrational state be reduced to 50%?

50% of the N2 molecules are in the ground state, and 50% are in the excited state. So, the population can be expressed in the following equation:

$P_{0}=0.5$

The value for the ${\tilde {\nu }}$  wavenumber value for the N2 molecule, taken from the NIST Webbook, is 2358.57 cm-1

This can be converted into the fundamental vibrational frequency by the relation,

$\nu ={\tilde {\nu }}\times c$
where:
• ${\tilde {\nu }}$  is fundamental wavenumber for the molecule in cm-1.
• $c$  is the speed of light, 2.998x1010 cm/s By knowing this relation, the fundamental vibrational frequency can be calculated.
$\nu ={\tilde {\nu }}\times c=2358.57cm^{-1}\times 2.998\times 10^{10}cm/s=7.07099\times 10^{13}s^{-1}$
The population of a system can be expressed by the following equation:
$P_{0}={\frac {\exp \left({\frac {-E_{0}}{k_{B}T}}\right)}{q_{vib}}}$
where:
• $P_{0}$  is the population of the ground state molecules
• $E_{0}$  is the energy at the ground state
• $k_{B}$ is the Boltzmann constant, 1.38064853 \times 10-23 J/K
• $T$  is the temperature of the system in K
• $q_{vib}$  is the molecular vibrational partition function

$q_{vib}$  is related by the following equation:

$q_{vib}={\frac {1}{1-\exp \left({\frac {-h\nu }{k_{B}T}}\right)}}$
where:
• h is the Planck constant, 6.626*10-34 J*s
Knowing these relations, we can solve for the temperature of the system when the molecule is reduced to 50% ground state, which is when 50% of the molecules are in the ground state and 50% of the molecules are excited.
$P_{0}={\frac {\exp {\frac {-E_{0}}{k_{B}T}}}{q_{vib}}}=50\%=0.5$
$q_{vib}={\frac {\exp \left({\frac {-E_{0}}{k_{B}T}}\right)}{P_{0}}}$
$q_{vib}={\frac {\exp(0)}{0.5}}=2$
Knowing the value of $q_{vib}$ , we can solve the molecular vibrational partition function to obtain the exact value of T for when 50% of the molecules are in the ground state.
$q_{vib}={\frac {1}{1-e^{\frac {-h\nu }{k_{B}T}}}}$ $=2$
$2={\frac {1}{1-\exp ^{\frac {-h\nu }{k_{B}T}}}}$
$2(1-\exp ^{\frac {-h\nu }{k_{B}T}})=1$
$2-2\exp ^{\frac {-h\nu }{k_{B}T}}=1$
$-2\exp \left({\frac {-h\nu }{k_{B}T}}\right)=-1$
$\exp \left({\frac {-h\nu }{k_{B}T}}\right)=0.5$
$\ln(\exp \left({\frac {-h\nu }{k_{B}T}}\right)=\ln(0.5)$
${\frac {-h\nu }{k_{B}T}}=-0.6931472$
$T={\frac {h\nu }{0.6931472k_{B}}}$
Therefore, by substituting the Planck constant, fundamental vibrational frequency and the Boltzmann constant into the equation, we can obtain the temperature for when half of the molecules are in the ground state.
$T=6.626\times 10^{-34}J\cdot s\times 7.07099\times 10^{13}s^{-}1/(0.6931472\times 1.38064853\times 10^{-23}J/K)=4895.792994K$
The temperature at which 50% of the N2 molecules are in the ground vibrational state is 4895 K.