Chemical equilibrium

3.2.3 Chemical Equilibrium All reversible reactions reach an dynamic equilibrium state. Many reactions are reversible N2 + 3H2 2NH3 Dynamic equilibrium occurs when forward and backward reactions are occurring at equal rates. The concentrations of reactants and products stays constant and the reaction is continuous. We use the expression ‘position of equilibrium’ to describe the composition of the equilibrium mixture. If the position of equilibrium favours the reactants (also described as “towards the left”) then the equilibrium mixture will contain mostly reactants. Le Chatelier’s Principle We use Le Chatelier’s principle to work out how changing external conditions such as temperature and pressure affect the position of equilibrium Le Chatelier’s principle states that if an external condition is changed the equilibrium will shift to oppose the change (and try to reverse it). Effect of Temperature on equilibrium If temperature is increased the equilibrium will shift to oppose this and move in the endothermic direction to try and reduce the temperature by absorbing heat. N2 + 3H2 2NH3 ∆H = -ve exo If temperature is increased the equilibrium will shift to oppose this and move in the endothermic, backwards direction to try to decrease temperature. The position of equilibrium will shift towards the left, giving a lower yield of ammonia. Exam level answer : must include bold points Typical Exam question: What effect would increasing temperature have on the yield of ammonia? If temperature is decreased the equilibrium will shift to oppose this and move in the exothermic direction to try and increase the temperature by giving out heat. And its reverse Low temperatures may give a higher yield of product but will also result in slow rates of reaction. Often a compromise temperature is used that gives a reasonable yield and rate Effect of Pressure on equilibrium Increasing pressure will cause the equilibrium to shift towards the side with fewer moles of gas to oppose the change and thereby reduce the pressure. CO (g) + 2H2(g) CH3OH (g) If pressure is increased the equilibrium will shift to oppose this and move towards the side with fewer moles of gas to try to reduce the pressure . The position of equilibrium will shift towards the right because there are 3 moles of gas on the left but only 1 mole of gas on the right, giving a higher yield of methanol. Exam level answer : must include bold points Typical Exam question: What effect would increasing pressure have on the yield of methanol? Decreasing pressure will cause the equilibrium to shift towards the side with more moles of gas to oppose the change and thereby increase the pressure. And its reverse If the number of moles of gas is the same on both sides of the equation then changing pressure will have no effect on the position of equilibrium H2 + Cl2 2HCl Increasing pressure may give a higher yield of product and will produce a faster rate. Industrially high pressures are expensive to produce ( high electrical energy costs for pumping the gases to make a high pressure) and the equipment is expensive (to contain the high pressures) Effect of Concentration on equilibrium Increasing the concentration OHions causes the equilibrium to shift to oppose this and move in the forward direction to remove OHions. The position of equilibrium will shift towards the right, giving a higher yield of I- and IO- . ( The colour would change from brown to colourless) Adding H+ ions reacts with the OHions and reduces their concentration so the equilibrium shifts back to the left giving brown colour. Effect of Catalysts on equilibrium A catalyst has no effect on the position of equilibrium, but it will speed up the rate at which the equilibrium is achieved. It does not effect the position of equilibrium because it speeds up the rates of the forward and backward reactions by the same amount.

Importance of equilibrium to industrial processes Common examples Haber process T= 450oC, P= 200 – 1000 atm, catalyst = iron Low temp gives good yield but slow rate: compromise temp used High pressure gives good yield and high rate: too high a pressure would lead to too high energy costs for pumps to produce the pressure N2 + 3H2 2NH3 ∆H = -ve exo Contact process Stage 1 S (s) + O2 (g)
SO2 (g) Stage 2 SO2 (g) + ½ O2 (g) SO3 (g) ∆H = -98 kJ mol-1 T= 450oC, P= 10 atm, catalyst = V2O5 Low temp gives good yield but slow rate: compromise moderate temp used High pressure gives slightly better yield and high rate: too high a pressure would lead to too high energy costs for pumps to produce the pressure Hydration of ethene to produce ethanol CH2 =CH2 (g) + H2O (g) CH3CH2OH(l) ∆H = -ve T= 300oC, P= 70 atm, catalyst = conc H3PO4 Low temp gives good yield but slow rate: compromise temp used High pressure gives good yield and high rate: too high a pressure would lead to too high energy costs for pumps to produce the pressure High pressure also leads to unwanted polymerisation of ethene to poly(ethene) Production of methanol from CO CO (g) + 2H2(g) CH3OH (g) ∆H = -ve exo T= 400oC, P= 50 atm, catalyst = chromium and zinc oxides Low temp gives good yield but slow rate: compromise temp used High pressure gives good yield and high rate: too high a pressure would lead to too high energy costs for pumps to produce the pressure Recycling unreacted reactants back into the reactor can improve the overall yields of all these processes In all cases high pressure leads to too high energy costs for pumps to produce the pressure and too high equipment costs to have equipment that can withstand high pressures. In all cases catalysts speeds up the rate allowing lower temp to be used (and hence lower energy costs) but have no effect on equilibrium

For a generalised reaction mA + nB pC + qD m,n,p,q are the stoichiometric balancing numbers A,B,C,D stand for the chemical formula Equilibrium constant Kc [ C]p [D]q [ A]m [B]n Kc= [ ] means the equilibrium concentration Kc = equilibrium constant Example 1 N2 (g) + 3H2 (g) 2 NH3 (g) [NH3 (g)] 2 [N2 (g)] [H2 (g)] 3 Kc= Calculating Kc In a container of volume 600cm3 at equilibrium the concentrations of the substances were 0.67 mol dm-3 of H2 and 0.83 mol dm-3 of Cl2 and 0.33 mol dm-3 HCl. Calculate Kc H2 (g) +Cl2 (g) 2HCl (g) Example For the following equilibrium = 0.332 0.67×0.83 Kc = 0.196 no unit Kc= [HCl (g)] 2 [H2 (g) ] [Cl2 (g)] N Goalby chemrevise.org 3 The larger the Kc the greater the amount of products. If Kc is small we say the equilibrium favours the reactants Kc only changes with temperature. It does not change if pressure or concentration is altered. A catalyst also has no effect on Kc