Question

It was stated that at room temperature (25°C) the majority of molecules are in the ground vibrational energy level (v = 0).
(a) Use the Boltzmann equation (Equation 8-1) to calculate the excited-state and ground-state population ratios for HCl: N(v =1)/N(v = 0). The fundamental vibrational frequency of HCl occurs at 2885 cm^-1.

(b) Use the results of part (a) to find N(v = 2)/N(v = 0)

Short answer

a) The excited state and ground state population ratio for HCL is :$N(v=1)/N(v=0)\text{is}8.829\times {10}^{-7}$

b) The excited state and f=ground state population ratio is : $N(v=2)/N(v=0)\text{is}7.795\times {10}^{-13}$

Step-by-step solution

Part(a) Step1-Given information

The fundamental vibrational frequency of HCl occurs at 2885 cm^-1.

Find the excited-state and ground-state population ratios for HCl: N(v=1)/N(v =0).

Part(a)  Step 2 Explanation

Use the concept:

When an electrons have sufficient energy to jump in to higher orbit we call an atom, molecule or electron in excited-state . The ground-state is the state of zero energy level.

In this state the electron does not have sufficient energy to jump from the orbital.

From the Boltzmann equation we can calculate the ratio of the exited-state and ground state.

The relation between the number of atoms at exited state and the ground state is given by the Boltzmann equation as below

$\frac{N_j}{N_0}=\frac{g_j}{g_0}{e}^{\left(\frac{-E_j}{kT}\right)}\dots \dots ..\left(\mathrm{l}\right)$

Where,
T : temperature
N₀ : number of atoms at ground state
N_j: number of atoms at exited state
E_j : energy difference between the exited and ground state

g_O: statistical factor for the ground state

g_i : the statistical factor for the exited state
k : Boltzmann constant

The energy difference between the exited state and the ground state is given as

$E_j=hc\overline{v}\dots \dots \text{(।)}$

Where,

$\overline{v}$: wave number is ,

h : plank's constant is

c : velocity of the light is
We know the plank's constant =6.626 x 10^-34 J.sec
The velocity of the light = 3.0 x 10⁸ m/sec

Now substitute the values we get

$$\begin{gathered} E_j=\left(6.626\times {10}^{-34}\mathrm{J}·\mathrm{s}\right)\left(3.0\times {10}^8\mathrm{m}/\mathrm{s}\right)\left(2885{\mathrm{cm}}^{-1}\right) \\ =\left(6.626\times {10}^{-34}\mathrm{J}·\mathrm{s}\right)\left(3.0\times {10}^8\mathrm{m}/\mathrm{s}\right)\left(2885{\mathrm{cm}}^{-1}\left(\frac{100\mathrm{cm}}{1\mathrm{m}}\right)\right) \\ =5.734\times {10}^{-20}\mathrm{J} \end{gathered}$$

The Boltzmann constant is 1.38 x 10^-23 J/K
The statistical factor for the ground state and the exited are unchanged
Convert the temperature from degree Celsius to Kelvin:
25^oC = 298K
Substitute values in equation (1)

$$\begin{gathered} \frac{N_j}{N_0}=\frac{g_j}{g_j}{e}^{\left(\frac{-5.734\times {10}^{-20}\mathrm{J}}{1.38\times {10}^{-23}\mathrm{Jk}(298\mathrm{K})}\right)} \\ =e^{\left(\frac{-5.734\times {10}^{-20}}{1.38\times {10}^{-23\left(298\right)}}\right)} \\ =e^{(-13.94)} \\ =8.829\times {10}^{-7} \end{gathered}$$

Part(b)-Step1-Explanation

The energy difference between the ground state and exited state for HCL is given by

$\frac{N_j}{N_0}=\frac{g_j}{g_0}{e}^{\left(\frac{-E_{g_2}}{kT}\right)}\dots \dots ..\text{(IV)}$

Substitute $5.734\times {10}^{-20}\mathrm{J}$ for E_j in Equation (III).

$$\begin{gathered} E_{j2}=2\left(5.734\times {10}^{-20}\mathrm{J}\right) \\ =11.468\times {10}^{-20}\mathrm{J} \end{gathered}$$

Now substitute values in equation (IV)

$$\begin{gathered} \frac{N_j}{N_0}=\frac{g_j}{g_j}{e}^{\left(\frac{111.468\times 10-{20}_j}{1.38\times {10}^{-23}\mathrm{J}/(298\mathrm{K})}\right)} \\ =e^{\left(\frac{-11.468\times {10}^{-20}}{1.38\times {10}^{-23}\left(298\right)}\right)} \\ =e^{(-27.88)} \\ =7.795\times {10}^{-13} \end{gathered}$$