Question 9.3: Evaluate the rotational contribution to the entropy (J/K · m...
Evaluate the rotational contribution to the entropy (J/K · mol) for HF at 300 K.
From Appendix K.1, we find that for HF the rotational constant B_{e} = 20.956_\ cm^{-1}, so that the characteristic rotational temperature \theta_{r} = (1.4387 cm · K) (20.956 cm^{-1}) = 30.1494 K. Therefore, at the given temperature of 300 K, T/\theta_{r}= 9.9505. From Table 9.3,
Table 9.3 Protocol for evaluation
of the rotational partition function
| Condition | Equation |
| T/\theta_{r} ≤ 3 | 9.24 |
| 3 < T /\theta_{r} ≤ 30 | 9.25 |
| T/\theta_{r} > 30 | 9.26 |
we thus choose Eq. (9.25) and its related expressions to determine properly the rotational contribution to thermodynamic properties. In particular, from Eq. (8.15), the rotational contribution to the entropy is given by
Z_{rot} = \frac{T}{\sigma\theta_{r}}\left[1+\frac{1}{3}\left(\frac{\theta_{r}}{T}\right)+\frac{1}{15}\left(\frac{\theta_{r}}{T}\right)^{2}+\frac{4}{315}\left(\frac{\theta_{r}}{T}\right)^{3}+\cdot\cdot\cdot \right]. (9.25)
\left(\frac{s}{R}\right)_{int} = T\left(\frac{\partial\ln Z_{int}}{\partial T}\right)_{V}+\ln Z_{int} (8.15)
\left(\frac{s}{R}\right)_{rot} = T\left(\frac{\partial\ln Z_{rot}}{\partial T}\right)_{V}+\ln Z_{rot},
where Z_{rot} must be obtained from Eq. (9.25) and the first-derivative term must be obtained from Eq. (9.31). On this basis, we first evaluate the rotational correction term via Eq. (9.30) and the first-derivative rotational correction term via Eq. (9.33):
T\left(\frac{\partial\ln Z_{rot}}{\partial T}\right)_{V} = 1-\frac{Z^{\prime}_{rc}}{Z_{rc}} (9.31)
Z_{rc} = 1+\frac{1}{3}\left(\frac{\theta_{r}}{T}\right)+\frac{1}{15}\left(\frac{\theta_{r}}{T}\right)^{2}+\frac{4}{315}\left(\frac{\theta _{r}}{T}\right)^{3}
= 1 + 0.0335 + 0.0007 = 1.0342
Z^{\prime}_{rc} = \frac{1}{3}\left(\frac{\theta_{r}}{T}\right)+\frac{2}{15}\left(\frac{\theta _{r}}{T}\right)^{2}+\frac{12}{315} \left(\frac{\theta_{r}}{T}\right)^{3}
= 0.03350 + 0.00135 + 0.00004 = 0.03489.
We now determine the rotational partition function from Eq. (9.25) and the required partial derivative from Eq. (9.31), thus obtaining
Z_{rot} = \frac{T}{\sigma\theta_{r}}Z_{rc} = 9.9505 (1.0342) = 10.2908
T\left(\frac{\partial\ln Z_{rot}}{\partial T}\right)_{V} = 1-\frac{Z^{\prime}_{rc}}{Z_{rc}}= 1-\frac{0.03489}{1.0342} = 0.96626,
where σ = 1 for a heteronuclear diatomic molecule. The dimensionless entropy now becomes
\left(\frac{s}{R}\right)_{rot} = T\left(\frac{\partial\ln Z_{rot}}{\partial T}\right)_{V}+\ln Z_{rot} = 0.96626 + ln (10.2908) = 3.2976,
so that a final calculation gives for the rotational energy mode,
s_{rot} = 3.2976 (8.3145 J/K · mol) = 27.418 J/K · mol.