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 1 of 3: Bonds broken minus bonds formed
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ΔHrxn=[2H(C−C)+8H(C−H)]−[H(C=C)+H(C−C)+6H(C−H)+H(H−H)]=+0.360 Hbond(C−C)\Delta H_{\text{rxn}} = [2H(\ce{C-C}) + 8H(\ce{C-H})] - [H(\ce{C=C}) + H(\ce{C-C}) + 6H(\ce{C-H}) + H(\ce{H-H})] = +0.360\,H_{\text{bond}}(\ce{C-C})
Analysis

Propane has 2 C–C and 8 C–H bonds; propene has 1 C=C, 1 C–C and 6 C–H, plus 1 H–H. So ΔH=[2+8×1.19]−[1.77+1+6×1.19+1.05×1.19]=(2+9.52)−(1.77+1+7.14+1.25)=11.52−11.16=+0.36 H(C−C)\Delta H = [2 + 8\times1.19] - [1.77 + 1 + 6\times1.19 + 1.05\times1.19] = (2 + 9.52) - (1.77 + 1 + 7.14 + 1.25) = 11.52 - 11.16 = +0.36\,H(\ce{C-C}): the reaction is endothermic.

Common pitfall. Count the C–H bonds carefully: propane has 8 (3+4+1... i.e. CHX3−CHX2−CHX3\ce{CH3-CH2-CH3} gives 6+2 = 8), propene only 6 — the two removed hydrogens form HX2\ce{H2}.