Question 9.4: Employing the simplex model, calculate the internal energy (...
Employing the simplex model, calculate the internal energy (kJ/mole) for AlO at 2000 K.
From Section 8.4, we recognize that the internal energy can be determined by summing the contributions from all four energy modes:
\frac{u}{RT} = \left(\frac{u}{RT}\right)_{tr}+\left(\frac{u}{RT}\right)_{el}+\left(\frac{u}{RT}\right)_{rot}+\left(\frac{u}{RT}\right)_{vib} .
The translational contribution, from Eq. (9.6), is
\left(\frac{u}{RT}\right)_{tr} = T\left(\frac{\partial \ln Z_{tr} }{\partial T}\right)_{V} = \frac{3}{2}. (9.6)
\left(\frac{u}{RT}\right)_{tr} = \frac{3}{2}.
The electronic contribution can be obtained from the data of Appendix K.2, as evaluated based on the first two energy levels in the following table. The next level is at 20,689 cm^{−1}, which is much too high to have any further influence on the internal energy at the given temperature. Because of our choice for a zero of energy, we must convert from the T_{e}-formulation provided by typical spectroscopic tables to the required T_{\circ}-formulation. From Fig. 9.1, we recognize that, under the harmonic-oscillator approximation, the appropriate conversion can be made via
T_{\circ} = T_{e}+\frac{1}{2}\left(\omega^{\prime}_{e}-\omega^{\prime\prime}_{e}\right).
| T_{e}(cm^{-1}) | \omega_{e}(cm^{-1}) | T_{\circ}(cm^{-1}) | g_{j} | \varepsilon_{j}/kT | g_{j}e^{-\varepsilon_{j}/kT} | g_{j}(\varepsilon_{j}/kT)e^{-\varepsilon_{j}/kT} |
| 0 | 979.2 | 0 | 2 | 0 | 2 | 0 |
| 5406 | 728.5 | 5281 | 4 | 3.799 | 0.0896 | 0.3404 |
| 2.0896 | 0.3404 |
Consequently, employing the above table, we can obtain the electronic contribution from Eq. (9.18), thus giving
\left(\frac{u}{RT}\right)_{el} = \left(\frac{h}{RT}\right)_{el} = \frac{Z^{\prime}_{el}}{Z_{el}} (9.18)
\left(\frac{u}{RT}\right)_{el} = \frac{Z^{\prime}_{el}}{Z_{el}} = \frac{0.3404}{2.0896} = 0.1629.
The rotational contribution, from the rigid-rotor approximation, is based solely on the rotational constant, B_{e} = 0.6414 cm^{−1}, as obtained from Appendix K.2. As a result, T/θ_{r} = (2000)/(1.4387)(0.6414) = 2167, so that, from Table 9.3, we find that Eq. (9.27) applies; hence,
Table 9.3 Protocol for evaluation
of the rotational partition function
| Condition | Equation |
| T/θ_{r} ≤ 3 | 9.24 |
| 3 < T/θ_{r} ≤ 30 | 9.25 |
| T/θ_{r} > 30 | 9.26 |
\left(\frac{u}{RT}\right)_{rot} = \left(\frac{h}{RT}\right)_{rot} = 1 (9.27)
\left(\frac{u}{RT}\right)_{rot} = 1.
Finally, the vibrational contribution, employing the harmonic-oscillator approximation, depends primarily on the characteristic vibrational temperature, which is θ_{ν} = (1.4387 cm · K)(979.2 cm^{−1}) = 1408.8 K. Therefore, from Eq. (9.48), we obtain
\left(\frac{u}{RT}\right)_{vib} = \left(\frac{h}{RT}\right)_{vib} = \frac{\theta_{\nu}/T}{e^{\theta_{\nu}/T}-1} (9.48)
\left(\frac{u}{RT}\right)_{vib} = \frac{\theta_{\nu}/T}{e^{\theta_{\nu}/T}-1} = 0.6888.
Having now evaluated the contributions from all four energy modes, we find that
\frac{u}{RT} = \left(\frac{u}{RT}\right)_{tr}+\left(\frac{u}{RT}\right)_{el}+\left(\frac{u}{RT}\right)_{rot}+\left(\frac{u}{RT}\right)_{vib}
= 1.5000 + 1.0000 + 0.6888 + 0.1629 = 3.3517.
Therefore, a final calculation gives, for the total internal energy,
u = 3.3517(8.3145 J/K · mol) (2000 K) = 55.735 kJ/mol.
