Equilibrium AS

All reversible reactions reach an dynamic equilibrium state. Many reactions are reversible N2 + 3H2 2NH3 Two features of Dynamic Equilibrium 1. Forward and backward reactions are occurring at equal rates. 2. The concentrations of reactants and products stays constant 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 and decrease the concentration of OH- ions. 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 Both methanol and ethanol can be used as fuels If the carbon monoxide used to make methanol in the above reaction was extracted from the atmosphere then it could be classed as carbon neutral The term carbon neutral refers to “an activity that has no net annual carbon (greenhouse gas) emissions to the atmosphere” It would only be carbon neutral, however, if the energy required to carry out the reaction was not made by combustion of fossil fuels Learn the equations for the production of methanol and ethanol below 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

Equilibrium Equations 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= The unit of Kc changes and depends on the equation. Working out the unit of Kc Put the unit of concentration (mol dm-3 ) into the Kc equation [NH3 (g)]2 [N2 (g) ] [H2 (g)]3 Kc= [mol dm-3 ]2 [mol dm-3 ] [mol dm-3 ]3 Unit = Cancel out units 1 [mol dm-3 ] Unit = 2 Unit = [mol dm-3 ] -2 Unit = mol-2 dm+6 Example 2: writing Kc expression H2 (g) +Cl2 (g) 2HCl (g) [mol dm-3 ]2 [mol dm-3 ] [mol dm-3 ] Kc= [HCl (g)]2 [H2 (g) ] [Cl2 (g)] Unit Kc= Working out the unit = no unit Calculating Kc Most questions first involve having to work out the equilibrium moles and then concentrations of the reactants and products. Usually the question will give the initial amounts (moles) of the reactants, and some data that will help you work out the equilibrium amounts. moles of reactant at equilibrium = initial moles – moles reacted moles of product at equilibrium = initial moles + moles formed Calculating the moles at equilibrium In a container of volume 600cm3 there were initially 0.5mol of H2 and 0.6 mol of Cl2 . At equilibrium there were 0.2 moles of HCl. Calculate Kc H2 (g) +Cl2 (g) 2HCl (g) Example 3 For the following equilibrium H2 Cl2 HCl Initial moles 0.5 0.6 0 Equilibrium moles 0.2 It is often useful to put the mole data in a table. Work out the moles at equilibrium for the reactants Using the balanced equation if 0.2 moles of HCl has been formed it must have used up 0.1 of Cl2 and 0.1 moles of H2 (as 1:2 ratio) moles of reactant at equilibrium = initial moles – moles reacted moles of hydrogen at equilibrium = 0.5 – 0.1 = 0.4 moles of chlorine at equilibrium = 0.6 – 0.1 = 0.5 Work out the equilibrium concentrations H2 Cl2 HCl Initial moles 0.5 0.6 0 Equilibrium moles 0.4 0.5 0.2 Equilibrium concentration (M) 0.4/0.6 =0.67 0.5/0.6 =0.83 0.2/0.6 =0.33 conc = moles/ vol (in dm3 ) = 0.332 0.67×0.83 Kc = 0.196 no unit Finally put concentrations into Kc expression If the Kc has no unit then there are equal numbers of reactants and products. In this case you do not have to divide by volume to work out concentration and equilibrium moles could be put straight into the kc expression Kc= [HCl (g)]2 [H2 (g) ] [Cl2 (g)] N Goalby chemrevise.org 3 4 Initially there were 1.5 moles of N2 and 4 mole of H2, in a 1.5 dm3 container. At equilibrium 30% of the Nitrogen had reacted. Calculate Kc N2 (g) + 3H2 (g ) 2 NH3 (g) Example 4 For the following equilibrium N2 H2 NH3 Initial moles 1.5 4.0 0 Equilibrium moles Work out the moles at equilibrium for the reactants and products 30% of the nitrogen had reacted = 0.3 x1.5 = 0.45 moles reacted. Using the balanced equation 3 x 0.45 moles of H2 must have reacted and 2x 0.45 moles of NH3 must have formed moles of reactant at equilibrium = initial moles – moles reacted moles of nitrogen at equilibrium = 1.5 – 0.45 = 1.05 moles of hydrogen at equilibrium =4.0 – 0.45 x3 = 2.65 Work out the equilibrium concentrations N2 H2 NH3 Initial moles 1.5 4.0 0 Equilibrium moles 1.05 2.65 0.9 Equilibrium concentration (M) 1.05/1.5 =0.7 2.65/1.5 =1.77 0.9/1.5 =0.6 conc = moles/ vol (in dm3 ) = 0.62 0.7×1.773 Kc Finally put concentrations into Kc expression moles of product at equilibrium = initial moles + moles formed moles of ammonia at equilibrium = 0 + (0.45 x 2) = 0.9 [NH3 (g)]2 [N2 (g) ] [H2 (g)]3 Kc= = 0.0927 mol-2 dm+6 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 Effect of changing conditions on value of Kc Effect of Temperature on position of equilibrium and Kc Both the position of equilibrium and the value of Kc will change it temperature is altered N2 (g) + 3H2 (g ) 2 NH3 (g) In this equilibrium which is exothermic in the forward direction If temperature is increased the reaction will shift to oppose the change and move in the backwards endothermic direction. The position of equilibrium shifts left. The value of Kc gets smaller as there are fewer products. Effect of Pressure on position of equilibrium and Kc The position of equilibrium will change it pressure is altered but the value of Kc stays constant as Kc only varies with temperature N2 (g) + 3H2 (g ) 2 NH3 (g) In this equilibrium which has fewer moles of gas on the product side If pressure is increased the reaction will shift to oppose the change and move in the forward direction to the side with fewer moles of gas. The position of equilibrium shifts right. The value of Kc stays the same though as only temperature changes the value of Kc. Catalysts have no effect on the value of Kc or the position of equilibrium as they speed up both forward and backward rates by the same amount. N Goalby chemrevise.org N Goalby chemrevise.org 5 Calculating the amounts of the equilibrium mixture from Kc Using algebra it is possible to work out the amounts of each component in an equilibrium mixture using the value of Kc H2 (g) + Br2 (l) 2HBr (g) Example 5 For the following equilibrium If 0.200 mol of H2 and 0.200 mol of Br2 are mixed and allowed to reach equilibrium. If Kc = 0.210 what are the equilibrium amounts of each substance. Kc= [HBr (g)]2 [H2 (g) ] [Br2 (g)] Make x = moles of H2 that have reacted at equilibrium V = volume of container This reaction is equimolar (same number of moles of reactant as products) so it is possible to cancel out the volume V (2x/V)2 (0.2-x)/V . (0.2-x)/V (2x)2 (0.2-x) . (0.2-x) 0.21 = 0.21 = (2x)2 (0.2-x)2 0.21 = Square root both sides 2x 0.2-x √0.21 = Rearrange to give x 0.458(0.2-x) = 2x 0.0917 = 2x + 0.458x x = 0.0917/2.458 x = 0.0373 So at equilibrium Moles of H2 = 0.2000- 0.0373 = 0.163 mol Moles of Br2 = 0.2000- 0.0373 = 0.163mol Moles of HBr = 2x 0.0373 = 0.0746mol