Calculation of the Standard Heat of Reaction at 25°C

Compute the standard heat of reaction for the hydrogenation of benzene to cyclohexane,
$C _{6} H _{6}+3 H _{2} \longrightarrow C _{6} H _{12}$
from the standard-heat-of-combustion data.

Question

Calculation of the Standard Heat of Reaction at 25°C Compute the standard heat of reaction for the hydrogenation of benzene to cyclohexane,

[latex]C _{6} H _{6}+3 H _{2} \longrightarrow C _{6} H _{12}[/latex]

from the standard-heat-of-combustion data.

Accepted Answer

The standard heat of reaction can, in principle, be computed from Eq. 8.5-3; however, for illustration, we will use the heat of combustion for cyclohexane. From the standard heat-of-combustion data in Appendix A.V, we have [latex]\Delta_{ c } \underline{H}^{\circ}=-3919906[/latex] J/mol of cyclohexane for the following reaction:

[latex]\Delta_{ rxn } H^{\circ}(T, P=1 bar )=\sum_{ i } u_{ i } \Delta_{ f } \underline{H}_{ i }^{\circ}(T, P=1 bar )[/latex] (8.5-3)

[latex]C _{6} H _{12}( l )+9 O _{2} ightarrow 6 CO _{2}+6 H _{2} O ( l )[/latex]

Thus

[latex]\Delta_{ c } \underline{H}_{ C _{6} H _{12}}^{\circ}=6 \underline{H}_{ CO _{2}}+6 \underline{H}_{ H _{2} O }-\underline{H}_{ C _{6} H _{12}}-9 \underline{H}_{ O _{2}}=-3919906 J / mol[/latex]

or

[latex]\underline{H}_{ C _{6} H _{12}}=-\Delta_{ c } \underline{H}_{ C _{6} H _{12}}^{\circ}-9 \underline{H}_{ O _{2}}+6 \underline{H}_{ CO _{2}}+6 \underline{H}_{ H _{2} O }[/latex]

Similarly,

[latex]\underline{H}_{ C _{6} H _{6}}=-\Delta_{ c } \underline{H}_{ C _{6} H _{6}}^{\circ}-7 \frac{1}{2} \underline{H}_{ O _{2}}+6 \underline{H}_{ CO _{2}}+3 \underline{H}_{ H _{2} O }[/latex]

and

[latex]3 \underline{H}_{ H _{2}}=-3 \Delta_{ c } \underline{H}_{ H _{2}}^{\circ}-1 \frac{1}{2} \underline{H}_{ O _{2}}+3 \underline{H}_{ H _{2} O }[/latex]

Therefore,

[latex]\begin{aligned}\Delta_{ rxn } \underline{H}^{\circ}=&-\Delta_{ c } \underline{H}_{ C _{6} H _{12}}^{\circ}-9 \underline{H}_{ O _{2}}+6 \underline{H}_{ CO _{2}}+6 \underline{H}_{ H _{2} O }+\Delta_{ c } \underline{H}_{ C _{6} H _{6}}^{\circ} \&+7 \frac{1}{2} \underline{H}_{ O _{2}}-6 \underline{H}_{ CO _{2}}-3 \underline{H}_{ H _{2} O }+3 \Delta_{ c } \underline{H}_{ H _{2}}^{\circ}+1 \frac{1}{2} \underline{H}_{ O _{2}}-3 \underline{H}_{ H _{2} O } \=&-\Delta_{ c } \underline{H}_{ C _{6} H _{12}}^{\circ}+\Delta_{ c } \underline{H}_{ C _{6} H _{6}}^{\circ}+3 \Delta_{ c } \underline{H}_{ H _{2}}=-\sum_{ i } u_{ i } \Delta_{ c } \underline{H}_{ i }^{\circ} \=& 3919906-3267620-3 imes 285840=-205234 \frac{ J }{ ext { mol benzene }} \=&-205.23 \frac{ kJ }{ ext { mol benzene }}\end{aligned}[/latex]

Comment

Note that in the final equation, [latex]\Delta_{ rxn } H^{\circ}=-\sum u_{ i } \Delta_{ c } \underline{H}_{ i }^{\circ}[/latex] the enthalpies of the reference-state atomic species cancel, as they must due to conservation of atomic species on chemical reaction. Aspen Plus[latex]^R[/latex] can be used to solve this illustration as shown inWiley website for this book in the folder Aspen Illustrations>Chapter 8>8.5-1. The calculation is done there using the RStoic reactor block in the Simulation mode with the ideal gas model. Using 1 kmol/hr flow rate of benzene with an isothermal 25◦C reactor. Looking the Results Summary>Model gives in a heat duty of −57272.2

[latex]\begin{aligned} ext { Watts } &=-57272.2 \frac{ J }{ S } imes 3600 \frac{ S }{ hr } imes \frac{1 hr }{1 kmol ext { benzene }} imes 1 \frac{ kmol }{10^{3} mol } imes \frac{1 kJ }{10^{3} mol } \&=-206.18 \frac{ kJ }{ mol ext { benzene }}\end{aligned}[/latex]

This is in agreement with the result above.

Question 8.5.1: Calculation of the Standard Heat of Reaction at 25°C Compute...

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