Question 22.3: For O2(g), the following spectroscopic data is available: Ma...

The fraction of O _{2}( g ) in a specific state j is equivalent to the probability of finding O _{2}( g ) in that state and is given by:

f_{j}=p_{j}=\frac{\exp \left(-\beta \epsilon_{j}\right)}{q}                        (14)

Equation (14) shows that in order to calculate p_{j}, one has to determine \epsilon_{j} and q.
Contrary to Eq. (6), g_{j} does not appear in the numerator of Eq. (14). This is because f_{j} in Eq. (14) represents the fraction in a specific energy state, while f_{j} in Eq. (6) represents the fraction in a specific energy level.

Part (a)

As discussed in Part III, the ground-state energy for translational motion can be obtained from the expression:

\epsilon_{n_{x}, n_{y}, n_{z}}=\frac{h^{2}\left(n_{x}^{2}+n_{y}^{2}+n_{z}^{2}\right)}{8 m \underline{V}^{2 / 3}}, \quad \text { with } \quad n_{x}=n_{y}=n_{z}=1                      (15)

That is:

\epsilon_{111}=\frac{3 h^{2}}{8 m \underline{V}^{2 / 3}}                       (16)

The translational partition function is given by:

q_{t r a n s}=\left(\frac{2 \pi m k_{B} T}{h^{2}}\right)^{3 / 2} \underline{V}                       (17)

Using the given values of m and V in Eq. (16) yields:

\begin{aligned}\epsilon_{111}=& \frac{3\left(6.626 \times 10^{-34} Js \right)^{2}}{8\left(53.15 \times 10^{-27} kg \right)\left(1000 cm ^{3}\right)^{\frac{2}{3}}} \\& \epsilon_{111}=3.097 \times 10^{-40} J\end{aligned}

Next, we calculate q_{\text {trans }} using the values of h=6.626 \times 10^{-34} Js, m=53.15 \times 10^{-27} kg , k_{B}=1.381 \times 10^{-23} J K ^{-1}, T=298  K, and \underline{ V }=1000  cm ^{3} in Eq. (17). This yields:

q_{\text {trans }}=1.75 \times 10^{29}

From Eq. (14), the fraction of molecular oxygen in the ground translational state is given by:

f_{111}^{\text {trans }}=\frac{\exp \left(-\beta \epsilon_{j}\right)}{q_{\text {trans }}}                         (18)

Substituting the values of \epsilon_{111}=3.097 \times 10^{-40} J \text { and } q_{\text {trans }}=1.75 \times 10^{29} in Eq. (18) yields:

f_{111}^{\text {trans }}=5.7143 \times 10^{-30}

Because k_{B} T=4.115 \times 10^{-25} \gg \epsilon_{111}, most translational states are accessible. As a result, the probability of occupancy of any one of the available translational states is the same and is given by \frac{1}{q_{\text {trans }}} .

Part (b)

Similar to our approach in Part (a), we will first determine the ground-state rotational energy and the rotational partition function and then use Eq. (14) to calculate the fraction of O _{2}( g ) in the ground-rotational state. The energy levels of a rotational state are given by:

\epsilon_{J}=\frac{h^{2} J(J+1)}{2 I}                          (19)

The ground-state rotational energy corresponds to J = 0 in Eq. (19), such that:

\epsilon_{0}=0                            (20)

The rotational partition function of O _{2}( g ) is given by:

q_{r o t}=\frac{T}{\sigma \Theta_{r o t}}                           (21)

where in the case of O _{2}( g ), the symmetry number, σ = 2, and where:

\Theta_{r o t}=\frac{h}{8 \pi^{2} I k_{B}}                        (22)

In Eq. (22), the moment of inertia, I, is given by:

I=\mu d^{2}                             (23)

where μ is the reduced mass of the molecule and is given by:

\mu=\frac{m(O) m(O)}{m(O)+m(O)}=\frac{m(O)}{2}=\frac{m\left(O_{2}\right)}{4}                           (24)

Using the value of m \left( O _{2}\right)=53.15 \times 10^{-27}  kg in Eq. (24) yields:

\mu=\frac{m\left(O_{2}\right)}{4}=13.28 \times 10^{-27} kg                           (25)

Using the given value of the bond length, d, and \mu=13.28 \times 10^{-27} kg in Eq. (23) yields:

I=1.945 \times 10^{-46} kgm ^{2}

Using this value of I in Eq. (22), \Theta_{r o t} is given by:

\Theta_{r o t}=2.07  K                           (26)

Recall that Eq. (22) was derived under the assumption that \Theta_{r o t} \ll T \text {. } Indeed, at T = 298 K, we see that \Theta_{r o t} \ll T. We can now use \Theta_{r o t} in Eq. (26), and σ = 2, to calculate q_{\text {rot }} in Eq. (21). Specifically,

q_{\text {rot }}=\frac{298}{2(2.07)}=71.98

We can next calculate the fraction of O _{2}( g ) in the ground rotational state using Eq. (14), which yields:

f_{0}^{r o t}=\frac{\exp \left(-\beta \epsilon_{0}\right)}{q_{r o t}}

Substituting the values of \epsilon_{0}=0 from Eq. (20), and q_{ rot }=71.98 from Eq. (26), in the last result yields:

\begin{gathered}f_{0}^{r o t}=\frac{\exp (-\beta(0))}{71.98}=\frac{1}{71.98} \\f_{0}^{\text {rot }}=0.0139\end{gathered}

Part (c)

We will follow the same approach as in Parts (a) and (b) to determine the fraction of O _{2}( g ) in the ground vibrational state. That is, we will first calculate the ground-state energy, the vibrational partition function, and then use Eq. (14) to obtain the desired quantity.

The vibrational energies are given by:

\epsilon_{n}=\left(n+\frac{1}{2}\right) h \nu=\left(n+\frac{1}{2}\right) h c \widetilde{\nu}                          (27)

The ground-state vibrational energy corresponds to n = 0, and hence:

\epsilon_{0}=\frac{1}{2} h c \widetilde{\nu}                         (28)

The vibrational partition function is given by:

q_{v i b}=\frac{\exp \left(-\Theta_{v i b} / 2 T\right)}{1-\exp \left(-\Theta_{v i b} / T\right)}                         (29)

where

\Theta_{v i b}=\frac{h c \widetilde{\nu}}{k_{B}}                           (30)

Now that we derived expressions for \epsilon_{0} \text { and } q_{v i b}, let us compute the values of these quantities. Using the values of h =6.626 \times 10^{-34}  Js , c =2.998 \times 10^{10}  cm s ^{-1}, k _{ B }=1.381 \times 10^{-23}  J , \text { and } \widetilde{\nu}=1567  cm ^{-1} in Eqs. (28) and (30), \epsilon_{0} \text { and } \theta_{v i b} can be calculated as follows:

\epsilon_{0}=1.5564 \times 10^{-20}  J                          (31)

\Theta_{v i b}=2254  K                             (32)

Because \Theta_{v i b} \gg T, we anticipate that few vibrational states will be accessible besides the ground vibrational state. That is, we can expect to observe a high fraction of O _{2}( g ) in the ground vibrational state. Let us see if our calculation reflects that. Using \Theta_{v i b} from Eq. (32) in Eq. (29), we find:

q_{v i b}=0.2279                           (33)

We can now calculate the fraction of O _{2}( g ) in the ground vibrational state by substituting the values of \epsilon_{0}=1.5564 \times 10^{-20} J from Eq. (31) and q_{v i b}=0.2279 from Eq. (33) in Eq. (14). This yields:

f_{0}^{v i b}=\frac{\exp \left(-\beta \epsilon_{0}\right)}{q_{v i b}}=0.99956                           (34)

As expected, Eq. (34) clearly shows that O _{2}( g ) is essentially in the ground vibrational state.

Part (d)

Calculating the molecular enthalpy for O _{2}( g ) requires that we calculate the molecular internal energy of O _{2}( g ) using the expression derived in Part III for diatomic molecules and then use the thermodynamic relation:

H=U+P V                          (35)

where

U=k_{B} T\left\{\frac{3}{2}+\frac{2}{2}+\left[\frac{\Theta_{v i b}}{2 T}+\frac{\frac{\theta_{v i b}}{T}}{\exp \left(\frac{\Theta_{v i b}}{T}\right)-1}\right]\right\}-D_{e}                          (36)

In Eq. (35), because T = 298 K and P = 1 atm, oxygen can be modeled as an ideal gas, for which:

P V=k_{B} T                         (37)

Using Eqs. (36) and (37) in Eq. (35), we obtain:

H=k_{B} T\left\{\frac{3}{2}+\frac{2}{2}+1+\left[\frac{\theta_{v i b}}{2 T}+\frac{\frac{\theta_{v i b}}{T}}{\exp \left(\frac{\theta_{v i b}}{T}\right)-1}\right]\right\}-D_{e}                          (38)

Next, we will calculate the values of each of the terms in Eq. (38) separately and then will substitute them in Eq. (38) to determine H. From the calculation carried out in Part (c), we obtain:

\Theta_{v i b}=2254  K                       (39)

D_{e}=D_{0}+\frac{1}{2} h \nu=D_{0}+\frac{1}{2} h c \widetilde{\nu}                           (40)

\begin{aligned}D_{e}=& \frac{118 \times 4.184  Jmol ^{-1}}{6.023 \times 10^{23}  mol ^{-1}}+\frac{1}{2}\left(6.626 \times 10^{-34}  Js \right)\left(2.998 \times 10^{10}  cm  s ^{-1}\right) \\& \times\left(1567  cm ^{-1}\right)\end{aligned}                           (41)

In Eq. (41), we note that we converted the basis of the given value of D_{0} from moles to molecules by dividing by Avogadro’s number. Use of Eq. (41) yields:

D_{e}=8.5097 \times 10^{-19} J                          (42)

We are now ready to compute H in Eq. (39) using \Theta_{v i b} from Eq. (39) and D_{e} from Eq. (42). Substituting the values of \Theta_{v i b}=2254 K, D_{e}=8.5097 \times 10^{-19} J, and T = 298 K in Eq. (38), we obtain:

H=-8.2099 \times 10^{-19} J

Note: There is no need to be concerned about the negative value of H. This is because of the negative contribution by D_{e} to H and the fact that D_{e} has a negative contribution because of the choice of reference state for the electronic energy levels.