Calculation of the Standard Heat of Reaction as a Function of Temperature
Compute the standard-state heat of reaction for the gas-phase reaction N _{2} O _{4}=2 NO _{2} over the temperature range of 200 to 600 K. ^{13}
Data: See Appendices A.II and A.IV.
^{13}Since, as we will see, \bar{H}_{ i }=\underline{H}_{ i } for an ideal gas mixture, the standard state heat of reaction and the actual heat of reaction are identical in this case.
The heat of reaction at a temperature T can be computed from
\Delta_{ rxn } H^{\circ}(T)=\sum_{ i } \nu_{ i } \Delta_{ f } \underline{H}_{ i }^{\circ}(T)
At T=25^{\circ} C we find, from the data in Appendix A.IV, that for each mole of N _{2} O _{4} reacted,
\Delta_{ rxn } H^{\circ}\left(T=25^{\circ} C \right)=[2 \times 33.18-9.16] kJ / mol =57.20 kJ / mol
To compute the heat of reaction at any temperature T, we start from Eq. 8.5-5 and note that since the standard state for each species is a low-pressure gas, C_{ P , i }^{\circ}=C_{ P , i }^{*}. Therefore,
\begin{aligned}\Delta_{ rxn } H^{\circ}(T, 1 bar ) &=\sum_{ i } \nu_{ i } \Delta_{ f } H_{ i }^{\circ}\left(25^{\circ} C , 1 bar \right)+\sum_{ i } \nu_{ i } \int_{T=25^{\circ} C }^{T} C_{ P , i }^{\circ} d T \\&=\Delta_{ rxn } H^{\circ}\left(25^{\circ} C , 1 bar \right)+\sum_{ i } \nu_{ i } \int_{T=25^{\circ} C }^{T} C_{ P , i }^{\circ} d T\end{aligned} (8.5-5)
\Delta_{ rxn } H^{\circ}(T)=\Delta_{ rxn } H^{\circ}\left(T=25^{\circ} C \right)+\int_{T=298.2 K }^{T} \sum_{ i } \nu_{ i } C_{ P , i }^{*} d T
For the case here we have, from Appendix A.II,
\begin{aligned}\sum \nu_{ i } C_{ P , i }^{*} &=2 C_{ P , NO _{2}}^{*}-C_{ P , N _{2} O _{4}}^{*} \\&=12.804-7.239 \times 10^{-2} T+4.301 \times 10^{-5} T^{2}+1.5732 \times 10^{-8} T^{3} \frac{ J }{ mol K }\end{aligned}
Thus
\begin{aligned}\Delta_{ rxn } \underline{H}^{\circ}(T)=& 57200+\int_{298.2}^{T}\left(12.804-7.239 \times 10^{-2} T\right.\\&\left.+4.301 \times 10^{-5} T^{2}+1.5732 \times 10^{-8} T^{3}\right) d T \\=& 57200+12.804(T-298.15)-\frac{7.239}{2} \times 10^{-2}\left(T^{2}-298.15^{2}\right) \\&+\frac{4.301}{3} \times 10^{-5}\left(T^{3}-298.15^{3}\right)+\frac{1.5732}{4} \times 10^{-8}\left(T^{4}-298.15^{4}\right) \\=& 56189+12.804 T-3.619 \times 10^{-2} T^{2}+1.4337 \times 10^{-5} T^{3} \\&+3.933 \times 10^{-9} T^{4} J / mol N _{2} O _{4}\end{aligned}
Values of \Delta_{\operatorname{rxn}} \underline{H}^{\circ} for various values of T are given in the following table.
| T (K) | 200 | 300 | 400 | 500 | 600 |
| \Delta_{\operatorname{rxn}} \underline{H}^{\circ} \left( kJ / mol N _{2} O _{4}\right) | 57.423 | 57.192 | 56.538 | 55.580 | 54.448 |
[Aspen Plus^R can be used to compute the standard heat of reaction using the folder Aspen Illustrations>Chapter 8>8.5-2 on Wiley website for this book following the procedure used in Illustration 8.5-1. One difference is that calculation is repeated for each of the temperatures with an isothermal reactor (reactor feed and exit at the same temperatures). The results are
| T (K) | Watts | \left( kJ / mol N _{2} O _{4}\right) |
| 200 | 15134.7 | 54.48 |
| 300 | 15920.5 | 57.31 |
| 400 | 16169.9 | 58.21 |
| 500 | 16087.3 | 57.91 |
| 600 | 15828.9 | 56.98 |
These results are in only approximate agreement with those in the illustration above. The differences are the result of differences in the databases in the appendices and in Aspen Plus^R].
Comment
The heat of reaction can be, and in fact usually is, a much stronger function of temperature than is the case here.