Question 9.6: Evaluate the entropy for NO2 (J/K · mol) at 500 K and 1 bar.
Evaluate the entropy for NO_{2} (J/K · mol) at 500 K and 1 bar.
From Section 8.4, we recognize that the entropy can be determined by summing the various contributions from all four energy modes, i.e.,
\frac{s}{R}=\left(\frac{s}{R}\right)_{t r}+\left(\frac{s}{R}\right)_{e l}+\left(\frac{s}{R}\right)_{r o t}+\left(\frac{s}{R}\right)_{v i b}.
The translational contribution, from Eq.(9.11), is
\left(\frac{s}{R}\right)_{t r}=\ln \left[\frac{(2 \pi m)^{3 / 2}(k T)^{5 / 2}}{h^{3} P}\right]+\frac{5}{2}so that
\left(\frac{s}{R}\right)_{t r}=\ln \left\{\frac{\left[2 \pi(46.0055)\left(1.6605 \times 10^{-27}\right)\right]^{3 / 2}\left[\left(1.3807 \times 10^{-23}\right)(500)\right]^{5 / 2}}{\left(6.6261 \times 10^{-34}\right)^{3}\left(1.0 \times 10^{5}\right)}\right\}+\frac{5}{2}= 20.1280.
A more direct procedure is to use Eq. (9.12), i.e., the Sackur–Tetrode equation, thus obtaining
\left(\frac{s}{R}\right)_{t r}=\frac{5}{2} \ln T+\frac{3}{2} \ln M-\ln P-1.1516=\frac{5}{2} \ln (500)+\frac{3}{2} \ln (46.0055)-1.1516= 20.1280.
From Eq. (8.15), the contribution from all internal energy modes is
\left(\frac{s}{R}\right)_{i n t}=T\left(\frac{\partial \ln Z_{i n t}}{\partial T}\right)_{V}+\ln Z_{i n t} (8.15)
\frac{s_{i n t}}{R}=T\left(\frac{\partial \ln Z_{i n t}}{\partial T}\right)_{V}+\ln Z_{i n t} .Hence, the electronic contribution, as obtained by using Eqs. (9.77) and (9.78) with the NO_{2} term symbol ({ }^{2} A_{1}) from Appendix K.3, becomes
Z_{e l}=\sum\limits_{j} g_{j} e^{-\varepsilon_{j} / k T} \simeq g_{0}, (9.77)
g_{el} = 2S + 1, (9.78)
\left(\frac{s}{R}\right)_{e l}=\ln g_{0}=\ln 2=0.6931.
The rotational contribution can be determined from Eqs. (9.82) and (9.84); the result is
Z_{r o t}=\frac{1}{\sigma} \sqrt{\frac{\pi T^{3}}{\theta_{r x} \theta_{r y} \theta_{r z}}}, (9.82)
\left(\frac{u}{R T}\right)_{r o t}=T\left(\frac{\partial \ln Z_{r o t}}{\partial T}\right)_{V}=\frac{3}{2} (9.84)
\left(\frac{s}{R}\right)_{r o t}=\frac{3}{2}+\ln \left[\frac{1}{\sigma} \sqrt{\frac{\pi T^{3}}{\theta_{r x} \theta_{r y} \theta_{r z}}}\right].
From Appendix K.3, the three rotational constants for NO_{2} follow: B_{e 1}=8.0012_\ cm ^{-1},B_{e 2}=0.4336_\ cm ^{-1} \text {, and } B_{e 3}=0.4104_\ cm ^{-1}. The resulting characteristic rotational temperatures, as obtained from Eq. (9.83), are \theta_{r x} = 11.511 K, \theta_{r y} = 0.6238 K, and \theta_{r z} = 0.5904 K. Hence, the rotational contribution to the entropy becomes
\theta_{r i}=\frac{h c}{k} B_{e i}=\frac{h^{2}}{8 \pi^{2} k I_{e i}}. (9.83)
\left(\frac{s}{R}\right)_{r o t}=\frac{3}{2}+\ln \left[\frac{1}{2} \sqrt{\frac{\pi(500)^{3}}{(11.511)(0.6238)(0.5904)}}\right]=9.9789,
where σ = 2 from the V-shaped molecular structure of O–N–O. Finally, the vibrational contribution, from Eqs. (9.86) and (9.88), is
Z_{v i b}=\prod\limits_{i=1}^{m}\frac{1}{1-e^{-\theta_{v i} / T}}, (9.86)
\left(\frac{u}{R T}\right)_{v i b}=T\left(\frac{\partial \ln Z_{v i b}}{\partial T}\right)_{V}=\sum\limits_{i=1}^{m} \frac{\theta_{v i} / T}{e^{\theta_{v i} / T}-1} (9.88)
\left(\frac{s}{R}\right)_{v i b}=\sum\limits_{i=1}^{3}\left\{\frac{\theta_{v i} / T}{e^{\theta_{v i} / T}-1}-\ln \left(1-e^{-\theta_{v i} / T}\right)\right\}.
The three vibrational frequencies, extracted from Appendix K.3, are \omega_{e 1}=1616.8_\ cm ^{-1}, \omega_{e 2}=1319.7_\ cm ^{-1}, \text { and } \omega_{e 3}=749.65_\ cm ^{-1}. The resulting characteristic vibrational temperatures, as obtained from Eq. (9.87), are \theta_{v 1} = 2326.1 K, \theta_{v 2} = 1898.7 K, and \theta_{v 3} = 1078.5 K. Therefore, the vibrational contribution to the entropy from the three normal modes of NO_{2} becomes
\theta_{v i}=\frac{h c}{k} \omega_{e i}, (9.87)
\left(\frac{s}{R}\right)_{v i b} = (0.0448 + 0.0871 + 0.2821) + (0.0096 + 0.0227 + 0.1229) = 0.5692.
As we have now evaluated the contributions from all four energy modes, we find that
\frac{s}{R}=\left(\frac{s}{R}\right)_{t r}+\left(\frac{s}{R}\right)_{e l}+\left(\frac{s}{R}\right)_{r o t}+\left(\frac{s}{R}\right)_{v i b}
= 20.1280 + 0.6931 + 9.9789 + 0.5692 = 31.3692.
Therefore, a final calculation gives, for the total entropy,
s = 31.3692 (8.3145 J/K · mol) = 260.819 J/K · mol.