Chemistry Labs

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, CX3HX8(g)⇌CX3HX6(g)+HX2(g)\ce{C3H8(g) <=> C3H6(g) + H2(g)}, but the reaction is not economically feasible. Use the following average bond enthalpy relations: Hbond(C=C)=1.77 Hbond(C−C)H_{\text{bond}}(\ce{C=C}) = 1.77\,H_{\text{bond}}(\ce{C-C}), Hbond(H−H)=1.05 Hbond(C−H)H_{\text{bond}}(\ce{H-H}) = 1.05\,H_{\text{bond}}(\ce{C-H}), and Hbond(C−H)=1.19 Hbond(C−C)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 Hbond(C−C)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 ΔH\Delta H, ΔS\Delta S, and the change of ΔG\Delta G when the temperature is raised to a higher value T∗T^* relative to the initial temperature?
Step 2 of 3: Pressure effect on the equilibrium
← PreviousNext →
Le Chatelier’s principle: raising p shifts the equilibrium toward fewer gas moles (propane)\text{Le Chatelier's principle: raising } p \text{ shifts the equilibrium toward fewer gas moles (propane)}
Analysis

The reaction produces 2 moles of gas from 1: raising pressure shifts equilibrium backward (fewer moles), so propene yield drops — a direct application of Le Chatelier's principle.