Question 10.4: Using the simplex model, show that the equilibrium constant ...
Using the simplex model, show that the equilibrium constant for the chemical reaction O_{2}+\frac{1}{2}N_{2}\rightleftarrows NO_{2} is given by
K_{p} = \kappa_{1}\Lambda^{{3}/{2}}_{N_{2}}\left(\frac{P_{\circ}}{kT}\right)^{{1}/{2}}\frac{Z_{NO_{2},int}}{Z_{O_{2},int }Z^{{1}/{2}}_{N_{2},int}}\exp\left(-\frac{\kappa_{2}}{T}\right),
where the thermal de Broglie wavelength is
\Lambda_{i} = \frac{h}{\sqrt{2\pi m_{i}kT}},
and the constants \kappa_{1} = 1.724 and \kappa_{2} = 4320 K.
From Eqs. (10.58) and (10.73), the equilibrium constant based on pressure is
\Delta h^{\circ}_{0} = -N_{A}\sum\limits_{i}{\nu_{i}D_{\circ i}} = N_{A}\left(D_{\circ,A_{2}}+D_{\circ,B_{2}}-2D_{\circ,AB} \right), (10.73)
K_{p} = \prod\limits_{i}{\left(\frac{Z_{i}}{N}\right)^{\nu_{i}}_{\circ}}\exp\left(\frac{\sum_{i}\nu_{i}D_{\circ i}}{kT} \right) = \prod\limits_{i}{\left(\frac{Z_{i}}{N}\right)^{\nu_{i}}_{\circ}}\exp\left(-\frac{\Delta h^{\circ}_{0}}{RT}\right),
where, from Eq. (10.53),
\left(\frac{Z_{i}}{N}\right)_{\circ} = \frac{\left(2\pi m_{i}\right)^{{3}/{2}}}{h^{3}P_{\circ}}(kT)^{{5}/{2}}Z_{i,int}, (10.53)
\left(\frac{Z_{i}}{N}\right)_{\circ} = \left(\frac{2\pi m_{i}kT}{h^{2}}\right)^{{3}/{2}}{\left(\frac{kT}{P_{\circ}}\right)}Z_{i,int} = \left(\frac{kT}{\Lambda^{3}_{i}P_{\circ}}\right)Z_{i,int}.
Applying these expressions to O_{2}+\frac{1}{2}N_{2}\rightleftarrows NO_{2}, we obtain
K_{p} = \frac{\left(Z_{NO_{2}}/N\right)_{\circ}}{\left(Z_{O_{2}}/N\right)_{\circ}\left(Z_{N_{2}}/N\right)^{{1}/{2}}_{\circ}}\exp\left(-\frac{\Delta h^{\circ}_{0}}{RT} \right),
so that, via substitution, we may verify
K_{p} = \kappa_{1}\Lambda^{{3}/{2}}_{N_{2}}\left(\frac{P_{\circ}}{kT}\right)^{{1}/{2}}\frac{Z_{NO_{2},int}}{Z_{O_{2},int} Z^{{1}/{2}}_{N_{2},int}}\exp\left(-\frac{\kappa_{2}}{T}\right),
where
\kappa_{1} = \frac{\Lambda^{3}_{O_{2}}}{\Lambda^{3}_{NO_{2}}}
\kappa_{2} = \frac{\Delta h^{\circ}_{0}}{R}.
On this basis, \kappa_{1} and \kappa_{2} may be evaluated as follows:
\kappa_{1} = \left(\frac{m_{NO_{2}}}{m_{O_{2}}}\right)^{{3}/{2}} = \left(\frac{46.008}{32.000}\right)^{{3}/{2}} = 1.724
\kappa_{2} = \frac{\Delta h^{\circ}_{0}}{R} = \frac{35927_\ J/mol}{8.3145 _\ J/K · mol} = 4320 K,
where here \Delta h^{\circ}_{0} is the standard enthalpy of formation for NO_{2} at absolute zero, as obtained from the appropriate JANAF table in Appendix E (E.1, E.2, E.3, E.4, E.5, E.6, E.7, E.8, E.9, E.10, E.11, E.12). Hence, we have verified the given numerical values for \kappa_{1} and \kappa_{2}.