Describe the changes in equilibrium of ethanol production from ethene and water (enthalpy of reaction is ∆H = - 46 kJ/mol) when: (a) a high pressure is applied; (b) ethanol concentration is increased; (c) temperature is increased; (d) a catalyst is used.
The equation for the reaction (would be given for this type of question but I ran out of space): C2H4 (g) + H2O (g) ⇌ C2H5OH (g) ∆H = - 46 kJ/molQuestion is based on Le Chatelier's principle: if a system in dynamic equilibrium is exposed to a change, processes will occur to oppose that change.Answers: (a) High pressure applied: the equilibrium will shift to the direction where less gas molecules are produced - to the right. That is because we have 2 mol of gases (ethene and water) on the left and only 1 mol on the right. (b) Ethanol concentration increased: the equilibrium will shift to the left, so that production of ethanol is reduced. (c) Temperature increased: the reaction that occurs left to right is exothermic (negative ∆H) - heat is released. Hence the equilibrium will shift in the opposite direction, as the reaction that occurs right to left is then endothermic (absorbs heat) and therefore will oppose the increase in temperature, so the equilibrium will shift to the left. (d) Catalyst used: no change in the position of equilibrium! (common trick question)
