Calculation of the Standard Heat of Reaction as a Function of Temperature Compute the standard-state heat of reaction for the gas-phase reaction N2O4 = 2NO2 over the temperature range of 200 to 600 K.^13 Data: See Appendices A.II and A.IV.

Question

Calculation of the Standard Heat of Reaction as a Function of Temperature

Compute the standard-state heat of reaction for the gas-phase reaction [latex]N _{2} O _{4}=2 NO _{2}[/latex] over the temperature range of 200 to 600 [latex]K. ^{13}[/latex]

Data: See Appendices A.II and A.IV.

[latex]^{13}[/latex]Since, as we will see, [latex]\bar{H}_{ i }=\underline{H}_{ i }[/latex] for an ideal gas mixture, the standard state heat of reaction and the actual heat of reaction are identical in this case.

Accepted Answer

The heat of reaction at a temperature T can be computed from

[latex]\Delta_{ rxn } H^{\circ}(T)=\sum_{ i } u_{ i } \Delta_{ f } \underline{H}_{ i }^{\circ}(T)[/latex]

At [latex]T=25^{\circ} C[/latex] we find, from the data in Appendix A.IV, that for each mole of [latex]N _{2} O _{4}[/latex] reacted,

[latex]\Delta_{ rxn } H^{\circ}\left(T=25^{\circ} C ight)=[2 imes 33.18-9.16] kJ / mol =57.20 kJ / mol[/latex]

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, [latex]C_{ P , i }^{\circ}=C_{ P , i }^{*}[/latex]. Therefore,

[latex]\begin{aligned}\Delta_{ rxn } H^{\circ}(T, 1 bar ) &=\sum_{ i } u_{ i } \Delta_{ f } H_{ i }^{\circ}\left(25^{\circ} C , 1 bar ight)+\sum_{ i } u_{ i } \int_{T=25^{\circ} C }^{T} C_{ P , i }^{\circ} d T \&=\Delta_{ rxn } H^{\circ}\left(25^{\circ} C , 1 bar ight)+\sum_{ i } u_{ i } \int_{T=25^{\circ} C }^{T} C_{ P , i }^{\circ} d T\end{aligned}[/latex] (8.5-5)

[latex]\Delta_{ rxn } H^{\circ}(T)=\Delta_{ rxn } H^{\circ}\left(T=25^{\circ} C ight)+\int_{T=298.2 K }^{T} \sum_{ i } u_{ i } C_{ P , i }^{*} d T[/latex]

For the case here we have, from Appendix A.II,

[latex]\begin{aligned}\sum u_{ i } C_{ P , i }^{*} &=2 C_{ P , NO _{2}}^{*}-C_{ P , N _{2} O _{4}}^{*} \&=12.804-7.239 imes 10^{-2} T+4.301 imes 10^{-5} T^{2}+1.5732 imes 10^{-8} T^{3} \frac{ J }{ mol K }\end{aligned}[/latex]

Thus

[latex]\begin{aligned}\Delta_{ rxn } \underline{H}^{\circ}(T)=& 57200+\int_{298.2}^{T}\left(12.804-7.239 imes 10^{-2} T ight.\&\left.+4.301 imes 10^{-5} T^{2}+1.5732 imes 10^{-8} T^{3} ight) d T \=& 57200+12.804(T-298.15)-\frac{7.239}{2} imes 10^{-2}\left(T^{2}-298.15^{2} ight) \&+\frac{4.301}{3} imes 10^{-5}\left(T^{3}-298.15^{3} ight)+\frac{1.5732}{4} imes 10^{-8}\left(T^{4}-298.15^{4} ight) \=& 56189+12.804 T-3.619 imes 10^{-2} T^{2}+1.4337 imes 10^{-5} T^{3} \&+3.933 imes 10^{-9} T^{4} J / mol N _{2} O _{4}\end{aligned}[/latex]

Values of [latex]\Delta_{\operatorname{rxn}} \underline{H}^{\circ}[/latex] for various values of T are given in the following table.

T (K)	200	300	400	500	600
[latex]\Delta_{\operatorname{rxn}} \underline{H}^{\circ}[/latex] [latex]\left( kJ / mol N _{2} O _{4} ight)[/latex]	57.423	57.192	56.538	55.580	54.448

[Aspen Plus[latex]^R[/latex] 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	[latex]\left( kJ / mol N _{2} O _{4} ight)[/latex]
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[latex]^R[/latex]].

Comment

The heat of reaction can be, and in fact usually is, a much stronger function of temperature than is the case here.

Question 8.5.2: Calculation of the Standard Heat of Reaction as a Function o...

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