Question

Consider the dehydrogenation of ethane to form ethene (ethylene) and hydrogen, LaTeX: \text{C}_2\text{H}_6\rightleftharpoons\text{C}_2\text{H}_4+\text{H}_2C 2 H 6 ⇌ C 2 H 4 + H 2 The standard enthalpies of formation and the standard entropies at 25 °C and 1 atm are given below, as pulled from Oxtoby, Gillis, and Nachtrieb, Principles of Modern Chemistry, Fourth Edition; Fort Worth: Saunders, 1999. Substance LaTeX: \Delta H^\circ_fΔ H f ∘ (kJ/mol) LaTeX: S^\circS ∘ (J/mol/K) Carbon (s, graphite) 5.74 Hydrogen (g) 130.57 Ethylene (g) 52.26 219.45 Ethane (g) −84.68 229.49 Assuming your goal is the production of hydrogen and/or ethylene, would it be beneficial to carry out this reaction at high pressure or low pressure, assuming kinetics are not an issue (i.e., that equilibrium is achieved reasonably quickly at any pressure)?

Answer 1

It would be better to carry out the reaction at low pressure

Step-by-step explanation:

[\tex]C_2H_6 (g)\rightleftharpoons C_2H_4 (g)+H_2 (g)[/tex]

Assuming that the reaction above is in equilibrium and getting based on the principles of Le Chatellier we can say that:

If we increase the pressure in the system, the reaction will go Leftward because it will try to reduce the pressure by consuming moles of gas.

If we reduce the pressure in the system, the reaction will go rightward because it will try to increase the pressure by producing moles of gas.

Given that our goal is to produce ehtylene, we want the reaction to go rightward and to do so is better to work at low pressure.