Problem 1

Production of propene using heterogeneous catalysts. Propene is one of the most valuable chemicals for the petrochemical industry. It can be synthesized by direct dehydrogenation of propane over a heterogeneous catalyst,

\ce{C3H8(g) <=> C3H6(g) + H2(g)}

, but the reaction is not economically feasible. Use the following average bond enthalpy relations:

H_{\text{bond}}(\ce{C=C}) = 1.77\,H_{\text{bond}}(\ce{C-C})

H_{\text{bond}}(\ce{H-H}) = 1.05\,H_{\text{bond}}(\ce{C-H})

, and

H_{\text{bond}}(\ce{C-H}) = 1.19\,H_{\text{bond}}(\ce{C-C})

. (a) What is the enthalpy change of the direct dehydrogenation of propane? Express the answer in terms of

H_{\text{bond}}(\ce{C-C})

. (b) It is difficult to increase the amount of propene by raising the pressure at constant temperature — which law or principle best explains this? (c) For this reaction at equilibrium, what are the correct signs of

\Delta H

\Delta S

, and the change of

\Delta G

when the temperature is raised to a higher value

T^*

relative to the initial temperature?

Step 3 of 3: Signs of

\Delta H

\Delta S

and

\Delta G

trend

← Previous Next →

\Delta H > 0,\quad \Delta S > 0,\quad \Delta G = \Delta H - T\Delta S \Rightarrow \Delta G \text{ decreases as } T \text{ rises}

Analysis

The dehydrogenation is endothermic ( $\Delta H > 0$ , from part a) and increases the number of gas molecules ( $\Delta S > 0$ ). Hence $\Delta G = \Delta H - T\Delta S$ becomes more negative as temperature rises: the reaction is driven forward at higher $T^*$ .