Equilibria –Reversible Reactions Learning intention Learn how reversible reactions may reach equilibrium instead of completely converting reactants to products. Find out how dynamic equilibrium is defined in terms of reaction rates and concentrations of reactant and product.

Learning Outcomes/Success Criteria Unit 3 CfE Higher Chemistry 3.14 - Equilibrium Learn that many chemical reactions are reversible Learning Outcomes/Success Criteria Unit 3 CfE Higher Chemistry R A G I can explain what is meant by the term dynamic equilibrium   I can explain “Le Chatelier’s Principle” in terms of pressure, temperature and concentration and their effects on equilibrium I can understand that a catalyst has no effect on the position of the equilibrium I can use the Haber process as an example of an equilibrium reaction to examine changes in pressure, temperature and the use of a catalyst

Dynamic equilibrium Chemical reactions which take place in both directions are called reversible reactions. The following is an example of a reversible reaction - hydrogen and iodine reacting to form hydrogen iodide. H2(g) + I2(g)⇄ 2HI(g) The equilibrium can be arrived at from different starting points. The position of an equilibrium does not depend on the starting position.   Reactants Products

Dynamic equilibrium A reversible reaction attains a state of equilibrium when the rate of the forward reaction is equal to the rate of the reverse reaction. In the equilibrium mixture both the forward and reverse reactions are taking place. equilibrium Reaction rate Time Forward reaction Backward reaction Graph 1

Dynamic equilibrium Since the rates of the forward and reverse reactions are equal, the concentration of reactants and products remain constant, though not necessarily equal. The system is said to be at dynamic equilibrium. Graph 2 Concentration Forward (reactants) Backward (products) Time

Characteristics of a Dynamic Equilibrium Forward and backward rates are the same The equilibrium can be approached from either direction Can only happen in a closed system The position of the equilibrium can vary

Position of equilibrium At equilibrium, the concentration of the products and the reactants will remain constant The concentration of reactants will probably not equal the concentration of the products.

Position of equilibrium concentration products reactants time Equilibrium At equilibrium the concentration of products and reactants remains the same.

Equilibrium A reversible reaction can reach equilibrium in a closed system. N2 + 3H2 ⇄ 2NH3 A reaction reaches equilibrium when the rate of the forward reaction equals the rate of the reverse reaction.

Direction The equilibrium position will be the same whether we start with only the products or only the reactants Iodine dissolves in both cyclohexane and water/KI. The experiment shows one boiling tube set up with 100% iodine in cyclohexane and one with 100% iodine in water/KI.

2cm depth Iodine in Cyclohexane Iodine equilibrium Explanation The cyclohexane and water are immiscible so they form separate layers. When the iodine crystals are added they dissolve in the cyclohexane (like dissolves like). Iodine is only slightly soluble in water Over time the equilibrium is established and the iodine moves between the potassium iodine solution and the iodine aq in the cyclohexane. 2cm depth Iodine in Cyclohexane Equilibrium cyclohexane Iodine/KI solution

Iodine equilibrium A B cyclohexane Iodine/cyclohexane KI solution Iodine/KI solution https://www.youtube.com/watch?v=RjPhvifqMdg

Final equilibrium is the same A B

f) Altering Equilibrium Position Learning intention Learn how chemists alter the position of equilibrium to increase product yield, by changing factors such as concentration, pressure, temperature and also how the use of a catalyst can ensure equilibrium is reached more quickly.

Le Chatelier's Principle “For a system in equilibrium, alteration of one of the factors (pressure, temperature or concentration) will cause the position of equilibrium to shift to reduce the effects of the imposed conditions”.

Le Chateliers Principle An equilibrium will move to undo any change imposed upon it. If the forward reaction is favoured we say the equilibrium has moved to the right. If the reverse reaction is favoured we say the equilibrium has moved to the left

Shifting the equilibrium position The proportion of products to reactants in an equilibrium mixture is described as the equilibrium position. A + B ⇄ C + D If the conversion of A and B into C and D is small the position of equilibrium lies to the left, or to the side of the reactants. If the equilibrium mixture is largely composed of C and D, the position of equilibrium lies to the right, or to the side of the products.

Temperature may alter the position of equilibrium

Temperature may alter the position of equilibrium Heating a reversible reaction at equilibrium shifts the reaction in the direction of the ENDOTHERMIC REACTION Cooling a reversible reaction at equilibrium shifts the reaction in the direction of the EXOTHERMIC REACTION The equilibrium will move to undo any change imposed upon it.

Forward reaction is exothermic Exothermic reaction A + B ⇄ C + D + Energy -ΔH Forward reaction is exothermic Increasing temperature shifts equilibrium to the left. Conc product Low temp High temp - favours reactants so less product formed. faster

Forward reaction is endothermic Endothermic reaction Energy + A + B ⇄ C + D +ΔH Forward reaction is endothermic Increasing temperature shifts equilibrium to the right. Conc product High temp- favours products so more product formed. Low temp faster

N2O4 ⇄ 2NO2 Δ +ve Temperature may alter the position of equilibrium dinitrogen tetraoxide nitrogen dioxide (colourless) (dark brown) Higher Chemistry Eric Alan and John Harris

N2O4 ⇄ 2NO2 Δ +ve Temperature may alter the position of equilibrium dinitrogen tetraoxide nitrogen dioxide (colourless) (dark brown) Increasing the temperature will cause the equilibrium to move to lower the temperature. The forward reaction takes in energy so the equilibrium moves to the right producing more NO2 and less N2O4. So the colour becomes darker. Decreasing the temperature will cause the equilibrium to move to raise the temperature. The reverse reaction gives out energy so the equilibrium moves to the left producing more N2O4 and less NO2. So the colour becomes lighter.

Co(H2O)62+ + 4Cl- ⇄ CoCl42- + 6H2O Δ +ve Temperature may alter the position of equilibrium A mixture of cobalt chloride and conc HCl sets up the following equilibrium: Co(H2O)62+ + 4Cl- ⇄ CoCl42- + 6H2O Δ +ve Heat the equilibrium mixture turns blue Cool the equilibrium mixture turns pink SSERC

Co(H2O)62+ + 4Cl- ⇄ CoCl42- + 6H2O Δ +ve Temperature may alter the position of equilibrium A mixture of cobalt chloride and conc HCl sets up the following equilibrium: Co(H2O)62+ + 4Cl- ⇄ CoCl42- + 6H2O Δ +ve If the temperature is increased the equilibrium will favour the forward reaction because that will lower the temperature. The equilibrium move to the right, therefore the solution becomes blue in colour. If the temperature is reduced the equilibrium will favour the reverse, exothermic reaction because that will increase the temperature. The equilibrium move to the left, therefore the solution becomes pinker in colour.

Concentration may alter the position of equilibrium

Concentration may alter the position of equilibrium Consider the following reaction at equilibrium A + B ⇄ C + D An increase in concentration of A or B will speed up the forward reaction, thus increasing the concentration of C and D. A similar effect can be achieved by reducing the concentration of C or D.

Co(H2O)62+ + 4Cl- ⇄ CoCl42- + 6H2O Δ +ve Concentration may alter the position of equilibrium A mixture of cobalt chloride and conc HCl sets up the following equilibrium: Co(H2O)62+ + 4Cl- ⇄ CoCl42- + 6H2O Δ +ve Adding extra Cl- ions forces the equilibrium to try to remove these. The forward reaction is favoured because this uses up chloride ions. The equilibrium has moved to the right so the solution becomes blue in colour.

Fe 3+ + CNS - ⇄ [FeCNS]2+ yellow red Concentration may alter the position of equilibrium Add 10 cm3 iron (III) chloride to a test tube. Iron (III) ions are yellow. Add potassium thiocyanate solution until the solution goes orange. Red coloured iron thiocyanate ions form. The equilibrium position now lies in the middle, roughly equal amounts of both coloured ions are present. Fe 3+ + CNS - ⇄ [FeCNS]2+ yellow red

The equilibrium moves right and the solution becomes more red Concentration may alter the position of equilibrium Higher Chemistry Eric Alan and John Harris Fe 3+ + CNS - ⇄ [FeCNS]2+ yellow red In B, the Fe 3+ ions are added. The equilibrium moves to use them up, favouring the forward reaction. The equilibrium moves right and the solution becomes more red

The equilibrium moves right and the solution becomes more red Concentration may alter the position of equilibrium Higher Chemistry Eric Alan and John Harris Fe 3+ + CNS - ⇄ [FeCNS]2+ yellow red In C, CNS- ions are added. The equilibrium moves to remove these, favouring the forward reaction. The equilibrium moves right and the solution becomes more red

The equilibrium moves left and the solution becomes more yellow Concentration may alter the position of equilibrium Higher Chemistry Eric Alan and John Harris Fe 3+ + CNS - ⇄ [FeCNS]2+ yellow red In D, the Fe 3+ ions are removed. The equilibrium moves to replace this, favouring the reverse reaction. The equilibrium moves left and the solution becomes more yellow

ICl + Cl2 ⇄ ICl3 brown liquid yellow solid Concentration may alter the position of equilibrium ICl + Cl2 ⇄ ICl3 brown liquid yellow solid

ICl + Cl2 ⇄ ICl3 brown liquid yellow solid Concentration may alter the position of equilibrium ICl + Cl2 ⇄ ICl3 brown liquid yellow solid Increasing the concentration of a chemical will cause the equilibrium to move to use up the chemical. Increasing the concentration of chlorine will cause the equilibrium to move to use up the chlorine. The forward reaction uses up the chlorine so the equilibrium moves to the right producing more yellow solid and less brown liquid. http://chemistry.stackexchange.com/questions/16672/hcl-and-kmno%E2%82%84-reaction/16677 Decreasing the concentration of a chemical will cause the equilibrium to move to form the chemical. Decreasing the concentration of chlorine will cause the equilibrium to move to form the chlorine. The reverse reaction produces chlorine so the equilibrium moves to the left producing more brown liquid and less yellow solid.

Pressure may alter the position of equilibrium

Pressure may alter the position of an equilibrium The pressure exerted by a gas is caused by the freely moving molecules bombarding the walls of the container. An increase in the number of molecules will result in an increase in pressure, assuming the size of the container is kept constant. An increase in pressure will cause the equilibrium to counteract this effect i.e. it will reduce the pressure. By favouring the side with less gas molecules.

N2O2 (g) ⇄ 2NO2 (g) colourless brown 1mole 2moles Higher Chemistry Eric Alan and John Harris

Pressure may alter the position of an equilibrium N2O4 ⇄ 2NO2 colourless brown 1 mole, so fewer particles 2 moles Lowers the pressure Increasing the pressure will cause the equilibrium to move to decrease the pressure. The equilibrium will move to reduce the number of gas particles. The equilibrium moves to the left producing more N2O4 and less NO2 so the colour lightens.

Pressure may alter the position of an equilibrium N2O4 ⇄ 2NO2 colourless brown 1 mole, so fewer particles 2 moles increase the pressure Decreasing the pressure will cause the equilibrium to move to increase the pressure. The equilibrium will move to increase the number of gas particles. The equilibrium moves to the right producing more NO2 and less N2O4 so the colour darkens.

Effect of neutralisation Cl2 + H2O ⇄ 2H + (aq) + ClO- + Cl- e.g. NaOH Adding an alkali will remove hydrogen ions from the equilibrium which will move to the right to replace them The bleaching effect will be increased. + OH- (aq) ⇄ H2O (l)

Effect of precipitation Cl2 + H2O ⇄ 2H + + ClO- + Cl- AgNO3 Adding silver nitrate will remove chloride ions from the equilibrium as the precipitate silver chloride is formed. The equilibrium will move to the right to replace them so the bleaching effect will be increased. + Ag+  AgCl (s)

Effect of acid Cl2 + H2O ⇄ 2H + + ClO- + Cl- e.g. H2SO4 Adding an acid causes the equilibrium to move to use up H+ ions. The equilibrium moves to the left producing more toxic Cl2 This can be fatal and accidents caused by mixing bleach and acid are not unusual. + H+ ions

Catalysts Catalysts increase the rate at which an equilibrium is formed but do not effect the equilibrium position. The rate of the forward AND reverse reactions are speeded up equally.

7 principles of design process Availability of feedstocks Cost of the feedstock Sustainability of the feedstock Opportunities for recycling Energy requirements Marketability of by-products Product yield (or atom economy)

The Haber Process How does the Haber process illustrate the 7 principles of design for an industrial process? How is the equilibrium manipulated to move as far to the right as possible? Give 3 ways. How is the effect of temperature on rate and on yield balanced?

Haber process http://www.rmtech.net/uses_of_ammonia.htm The Haber Process combines nitrogen from the air with hydrogen derived mainly from natural gas (methane) into ammonia. The reaction is reversible and the production of ammonia is exothermic.                                                                                              http://www.rmtech.net/uses_of_ammonia.htm

The Haber Process and equilibrium N2 + 3H2 ⇄ 2 NH3 - ΔH In the reactor, ammonia is removed after each cycle of the gases through the reactor. What effect would that have on the equilibrium position?

The Haber Process and equilibrium N2 + 3H2 ⇄ 2 NH3 - ΔH Increasing the temperature would speed up the rate of reaction, but what would happen to the amount of ammonia produced in the equilibrium? The reacting gases are pumped around the reactor at a suitable ‘flow rate’

The Haber Process and equilibrium You need to shift the position of the equilibrium as far as possible to the right in order to produce the maximum possible yield of ammonia in the equilibrium mixture. The forward reaction (the production of ammonia) is exothermic.                                              According to Le Chatelier's Principle, this will be favoured if you lower the temperature. The system will respond by moving the position of equilibrium to counteract this - in other words by producing more heat. In order to get as much ammonia as possible in the equilibrium mixture, you need as low a temperature as possible. However, 400 - 450°C isn't a low temperature!

The Haber Process and equilibrium If the flow rate is adjusted so that the gases spend x time in the reactor, more ammonia is produced per day at the higher temperature, despite the lower yield at equilibrium Tonnes of ammonia 200 oC 400 oC time

The Haber Process and equilibrium The lower the temperature you use, the slower the reaction becomes. A manufacturer is trying to produce as much ammonia as possible per day. It makes no sense to try to achieve an equilibrium mixture which contains a very high proportion of ammonia if it takes several years for the reaction to reach that equilibrium. You need the gases to reach equilibrium within the very short time that they will be in contact with the catalyst in the reactor. The compromise 400 - 450°C is a compromise temperature producing a reasonably high proportion of ammonia in the equilibrium mixture (even if it is only 15%), but in a very short time.

The Haber Process and equilibrium N2 + 3H2 ⇄ 2 NH3 - ΔH What would be the effect of increasing the pressure? What might be the drawbacks, both technical and economic of a high pressure?

The Haber Process and equilibrium N2 + 3H2 ⇄ 2 NH3 - ΔH Notice that there are 4 molecules on the left-hand side of the equation, but only 2 on the right. According to Le Chatelier's Principle, if you increase the pressure the system will respond by favouring the reaction which produces fewer molecules. That will cause the pressure to fall again. In order to get as much ammonia as possible in the equilibrium mixture, you need as high a pressure as possible. 200 atmospheres is a high pressure, but not amazingly high.

The Haber Process and equilibrium Increasing the pressure brings the molecules closer together. In this particular instance, it will increase their chances of hitting and sticking to the surface of the catalyst where they can react. The higher the pressure the better in terms of the rate of a gas reaction.

The Haber Process and equilibrium Very high pressures are very expensive to produce on two counts. You have to build extremely strong pipes and containment vessels to withstand the very high pressure. That increases your capital costs when the plant is built. High pressures cost a lot to produce and maintain. That means that the running costs of your plant are very high.

The Haber Process and equilibrium The compromise 200 atmospheres is a compromise pressure chosen on economic grounds. If the pressure used is too high, the cost of generating it exceeds the price you can get for the extra ammonia produced.

The Haber Process The catalyst is iron, which is cheap. What is the effect a catalyst on a reversible reaction?

The Haber Process and equilibrium In the absence of a catalyst the reaction is so slow that virtually no reaction happens in any sensible time. The catalyst ensures that the reaction is fast enough for a dynamic equilibrium to be set up within the very short time that the gases are actually in the reactor. .

The Haber Process and equilibrium

The Haber Process and equilibrium When the gases leave the reactor they are hot and at a very high pressure. Ammonia is easily liquefied under pressure as long as it isn't too hot, and so the temperature of the mixture is lowered enough for the ammonia to turn to a liquid. The nitrogen and hydrogen remain as gases even under these high pressures, and can be recycled

The Haber Process and equilibrium http://www.freezeray.com/flashFiles/theHaberProcess.htm

In which of the following reactions would an increase in pressure cause the equilibrium position to move to the left? A CO(g) + H2O(g) →CO2(g) + H2(g) B CH4(g) + H2O(g) → CO(g) + 3H2(g) C Fe2O3(s) + 3CO(g) → 2Fe(s) + 3CO2(g) D N2(g) + 3H2(g) → 2NH3(g) B

If ammonia is added to a solution containing copper(II) ions an equilibrium is set up. Cu2+(aq) + 2OH–(aq) + 4NH3(aq) → Cu(NH3)4(OH)2(aq) (deep blue) If acid is added to this equilibrium system A the intensity of the deep blue colour will increase B the equilibrium position will move to the right C the concentration of Cu2+(aq) ions will increase D the equilibrium position will not be affected. C

D