Chemical Equilibrium PDF

3.2.3 Chemical Equilibrium Many reactions are reversible All reversible reactions reach an N2 + 3H2 2NH3 dynamic equilibrium state. We use the expression ‘position of Dynamic equilibr...

3.2.3 Chemical Equilibrium Many reactions are reversible All reversible reactions reach an N2 + 3H2 2NH3 dynamic equilibrium state. We use the expression ‘position of Dynamic equilibrium occurs when forward and equilibrium’ to describe the composition of the backward reactions are occurring at equal rates. The equilibrium mixture. concentrations of reactants and products stays constant and the reaction is continuous. 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 Le Chatelier’s principle states that if an external changing external conditions such as temperature condition is changed the equilibrium will shift to and pressure affect the position of equilibrium oppose the change (and try to reverse it). Effect of temperature on equilibrium Typical exam question: What effect would increasing If temperature is increased the equilibrium will temperature have on the yield of ammonia? shift to oppose this and move in the endothermic direction to try to reduce the N2 + 3H2 2NH3 H = -ve exo temperature by absorbing heat. Exam level answer: must include bold points And its reverse If temperature is increased the equilibrium will shift to oppose this and move in the endothermic, If temperature is decreased the equilibrium backwards direction to try to decrease will shift to oppose this and move in the temperature. The position of equilibrium will shift exothermic direction to try to increase the towards the left, giving a lower yield of ammonia. temperature by giving out heat. 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 Typical exam question: What effect would increasing Increasing pressure will cause the equilibrium to shift pressure have on the yield of methanol? towards the side with fewer moles of gas to oppose the change and thereby reduce the pressure. CO (g) + 2H2(g) CH3OH (g) And its reverse Exam level answer: must include bold points If pressure is increased the equilibrium will shift Decreasing pressure will cause the equilibrium to shift to oppose this and move towards the side with towards the side with more moles of gas to oppose fewer moles of gas to try to reduce the the change and thereby increase the pressure. pressure. The position of equilibrium will shift towards the right because there are 3 moles of If the number of moles of gas is the same on both gas on the left but only 1 mole of gas on the right, sides of the equation then changing pressure will have giving a higher yield of methanol. 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) N Goalby chemrevise.org 1 Effect of concentration on equilibrium Increasing the concentration OH- ions causes the equilibrium to shift to I2 + 2OH- I- + IO- + H2O oppose this and move in the forward direction to remove OH- ions. The position of equilibrium will shift towards the right, giving a higher yield brown colourless of I- and IO-. ( The colour would change from brown to colourless) Adding H+ ions reacts with the OH- ions 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 You should be able to apply the above ideas to given reactions Common examples Contact process Haber process Stage 1 S (s) + O2 (g)  SO2 (g) N2 + 3H2 2NH3 H = -ve exo Stage 2 SO2 (g) + ½ O2 (g) SO3 (g) H = -98 kJ mol-1 T= 450oC, P= 200 – 1000 atm, catalyst = iron T= 450oC, P= 1 to 2 atm, catalyst = V2O5 Low temp gives good yield but slow rate: Low temp gives good yield but slow rate: compromise compromise temp used moderate temp used High pressure gives good yield and high rate: High pressure gives slightly better yield and high rate: too too high a pressure would lead to too high high a pressure would lead to too high energy costs for energy costs for pumps to produce the pressure pumps to produce the pressure Hydration of ethene to produce ethanol Production of methanol from CO CH2=CH2 (g) + H2O (g) CH3CH2OH(l) H = -ve CO (g) + 2H2(g) CH3OH (g) H = -ve exo T= 300oC, P= 70 atm, catalyst = conc H3PO4 T= 400oC, P= 50 atm, catalyst = chromium and zinc oxides Low temp gives good yield but slow rate: compromise temp used. 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 High pressure gives good yield and high rate: too to produce the pressure. 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). In all cases catalysts speeds up the rate, allowing lower temperatures to be used (and hence lower energy costs) but have no effect on equilibrium. 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. Recycling unreacted reactants back into the reactor can improve the overall yields of all these processes. 2 N Goalby chemrevise.org Equilibrium constant Kc Kc = equilibrium constant Example 1 For a generalised reaction mA + nB pC + qD [ C]p [D]q N2 (g) + 3H2 (g) 2 NH3 (g) Kc = m,n,p,q are the stoichiometric balancing [ A]m [B]n numbers [NH3 (g)]2 Kc = A,B,C,D stand for the chemical formula [N2 (g)] [H2 (g)]3 [ ] means the equilibrium concentration Liquid and solid concentrations are constant, and are not included in heterogeneous Kc expressions. Calculating Kc Example For the following equilibrium H2 (g) +Cl2 (g) 2HCl (g) In a container of volume 600 cm3 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 Kc = [HCl (g)] 2 [H2 (g) ] [Cl2 (g)] Kc = 0.332 = 0.196 no unit 0.67x0.83 Effect of changing conditions on value of Kc Kc only changes with temperature. The larger the Kc the greater the amount of products. It does not change if pressure or If Kc is small we say the equilibrium favours the reactants concentration is altered. A catalyst also has no effect on Kc N Goalby chemrevise.org 3