Question 10.2: Calculate the equilibrium constant of the reaction I2(g)↽ ⇀2...

We can compute the equilibrium constant K by means of the formula

To compute the standard molecular partition function of the iodine atom, we only have to calculate its translational and electronic contributions, as there are no rotational or vibrational states. The electronic contribution is the number of degeneracy; $q_{\mathrm{I,el}}^{\ominus}$ = 4. The translational contribution can be calculated using (10.67), in which m is the mass of an iodine atom:

$q^{trans} = (\frac{2πmkT}{h^{2} } )^{3/2}\cdot V,$                                    (10.67)

$m=\frac{M_{\mathrm{I}}}{N_{\mathrm{A}}}=\frac{126.90447 \mathrm{ ~gmol}^{-1}}{6.0220 \times 10^{23} \mathrm{~mol}^{-1}}=2.10734 \times 10^{-25} \mathrm{~kg}$,

and the ideal gas molar volume V can be obtained by using the equation of state of the ideal gas:
$V=\frac{R T}{P}=\frac{8.314 \mathrm{ ~Jmol}^{-1} \mathrm{~K}^{-1} \cdot 800 \mathrm{~K}}{101325 \mathrm{~Pa}}=0.06564 \mathrm{~m}^{3} / \mathrm{mol}.$

Thus,

$q_{\mathrm{I}, \text { trans }}^{\ominus} =\left(\frac{2 \pi m k T}{h^{2}}\right)^{3 / 2} \cdot V \\$
=$\left(\frac{2 \pi \cdot 2.10734 \times 10^{-25} \mathrm{~kg} \cdot 1.3807 \times 10^{-23} \mathrm{JK}^{-1} \cdot 800 \mathrm{~K}}{\left(6.626 \times 10^{-34} \mathrm{Js}\right)^{2}}\right)^{3 / 2}$ ×0.06564m³.

The result is $q_{\mathrm{I}, \text { trans }}^{\ominus}=3.992 \times 10^{32}$, which multiplied by the electronic contribution gives $q_{\mathrm{I}}^{\ominus}=1.597 \times 10^{33}.$

To compute the standard molecular partition function of the $I_{2}$ molecule, we have to calculate all four contributions. The electronic contribution is simple, as there is only a single electronic state available; thus, $q_{\mathrm{I_{2},el }}^{\ominus}$ = 1. The translational contribution can be calculated similarly to the case of the iodine atom; the only difference is the molar mass, which is twice that of the atom. The result of the calculation is accordingly $2^{3/2}$ times that of $q_{\mathrm{I,trans }}^{\ominus}$ ; that is, $q_{\mathrm{I_{2},trans}}^{\ominus}= 1.129 × 10^{34}.$

The rotational contribution can be calculated using (10.85), but with a division by 2, as the $I_{2}$ molecule is a homonuclear rotor, with a rotational symmetry factor of σ = 2:

$q_{\mathrm{l in}}^{R} = \frac{kT}{hcB}$               (10.85)

When substituting the actual values, we should be careful with units. The rotational constant B is measured in $cm^{-1}$ units, which should be multiplied by hc to yield energy. Thus, either hcB should be given in SI energy units, or kT should be given in $cm^{-1}$ units. Choosing the latter, we can express the Boltzmann constant as 0.69503 $cm^{-1}/K$. (As a matter of fact, a division by hc is also included.) The resulting rotational partition function is $q_{\mathrm{I_{2},rot}}^{\ominus}$ = 7454.1.

The vibrational contribution can be calculated using (10.90).

$q^{V} = \frac{1}{1-e^{-\frac{hv}{KT} } }$                (10.90)

Again, we should be careful using units. The energy of the vibration is given as the wavenumber in $cm^{–1}$ units, thus we can use again the Boltzmann constant given above as k = 0.69503 $cm^{–1}$/K and consider the vibrational wavenumber as energy, instead of hv. The resulting vibrational partition function is $q_{\mathrm{I_{2},vib}}^{\ominus}$ = 3.126.

Finally, we can multiply the four contributions to get the molecular partition function $q_{\mathrm{I_{2}}}^{\ominus} = 2.632 ×10^{37}.$

When calculating the equilibrium constant, we should again take into account the units of energy. As the zero-point reaction energy is given in eV units, it is useful to use the according units of the gas constant, which is then given as R = 5.189×$10^{19} eV K^{-1} mol^{-1}.$ The resulting equilibrium constant is 3.104×$10^{-5}.$