1. DP Chemistry EQUILIBRIUM

4.  Thornley Equilibrium page (7 SL Videos to help you): https://www.youtube.com/playlist?list=PL816Qsrt2Os394LaHIen _URp8ZFC90Kpg  Chem Guide (great UK resource) http://www.chemguide.co.uk/physical/equilibmenu.html  Another useful website with a great powerpoint http://mrjdfield.edublogs.org/2014/03/11/topic-7-equilibrium/ RESOURCES YOU SHOULD USE:

5.  Many important reaction are reversible and exist in a state of equilibrium – with both the forward and reverse reactions occurring simultaneously.  An understanding of chemical equilibrium allows chemists to control reaction conditions and maximise the yield of the desired product.  Example: hot water in a sealed storage container – some water molecules will have enough energy to move from the liquid phase to the gaseous phase and some gaseous water molecules will lose energy and return to the liquid phase (condensation).  Eventually the rate of evaporation and the rate of condensation will be equal – the system is in equilibrium. EQUILIBRIUM

6.  Look at the example below with Bromine EQUILIBRIUM EXAMPLE

7.  The rates of evaporation and condensation as equilibrium is established in a closed system.  The rate of evaporation is constant, while the rate of condensation increases with increasing concentration of vapour. EQUILIBRIUM EXAMPLE

8.  As seen in the last example, when the rate of condensation and evaporation are equal the system is at equilibrium – the forward and reverse reactions are occurring at equal rates.  The concentration of liquid and gas does not change, but molecules are constantly being interconverted between the two phases.  This is called a dynamic equilibrium (continued random movement of particles with no overall concentration change) DYNAMIC EQUILIBRIUM

9.  There will at first be an increase in the colour purple owing to the production of iodine gas  After a while this will stop and it may appear that the reaction has stopped – but eventually they reach equilibrium (forward and reverse reactions are equal)  HI, H 2 and I 2 would all be present and their concentrations would remain constant  Note: same equilibrium mixture reached when starting with H 2 and I 2 or HI DYNAMIC EQUILIBRIUM

10.  Reversible – both reactants to products / left to right (known as forward reaction) and products to reactants / backward (reverse reaction)  Technically all reactions can be considered as equilibrium reactions but many consist almost entirely of products so are considered to have gone to completion (use  symbol)  Other reaction there could be so little product formed that it is almost undetectable (considered not to have happened) THE EQUILIBRIUM STATE HAS CHARACTERISTICS

12.  Equilibrium does NOT mean that the concentrations of reactant and product are equal – they are constant.  The concentration of reactant or product may be much higher than the other (example below)  The proportion of reactant and product in the equilibrium mixture is referred to as its equilibrium position.  Reactions where the mixture contains mostly products are said to ‘lie to the right’ and visa versa.  This can also be done mathematically……. EQUILIBRIUM POSITION

13. LEARNING CHECK

14.  The equilibrium constant K c can be predicted from a reaction’s stoichiometry  Consider the reaction: H 2 (g) + I 2 (g) = 2HI(g)  If we carry out a series of experiments on this reaction with different starting concentrations of H 2, I 2 and HI (at 440 o C), we could wait until each reaction reached equilibrium and then measure the composition of each equilibrium mixture  What is the relationship? THE EQUILIBRIUM CONSTANT K C

15.  Clearly this way of processing the equilibrium data produces a constant value (within the limits of experimental accuracy)  This constant is known as the equilibrium constant, K c  It has a fixed value for this reaction at a specified temperature EQUILIBRIUM CONSTANT K C

16.  Every reaction has its own particular value of K c, which can be derived in a similar way – first we use the balanced reaction equation to write the equilibrium constant expression:  Has the concentrations of products in the numerator and the concentrations of reactants in the denominator  Each concentration is raised to the power of its coefficient in the balanced equation  Where there is more than one reactant or product the terms are multiplied together EQUILIBRIUM CONSTANT EXPRESSION

18. LEARNING CHECK

19.  Different reactions have different K c values  As the equilibrium constant puts products on the numerator and reactants on the denominator, a high value of K c will mean there are more products than reactants.  And a low K c must mean there are proportionally less products with respect to reactants so the equilibrium mixture has barely reacted.  Consider the following reactions at 550 K  Reaction between H 2 and Cl 2 has completed the most at 550 K THE MAGNITUDE OF K C GIVES INFORMATION ON THE EXTENT OF REACTION

20. General rule:  If K c >> 1, the reaction is considered to go almost to completion (very high conversion of reactants to products)  If K c << 1, the reaction hardly proceeds at this temperature  Note: the magnitude of K c does not give information on the rate of the reaction MAGNITUDE OF K C

21.  Remember the value of K c is calculated from substituting the equilibrium concentrations of reactants and products into the equilibrium constant expression.  But what if the reaction is not at equilibrium?  If we take the concentrations of the reactants and products at one moment when the reaction is not at equilibrium and put them in to the equilibrium constant expression, we obtain a value known as the reaction quotient, Q.  So the value of Q changes in the direction of K c THE REACTION QUOTIENT, Q, ENABLES THE PREDICTION OF DIRECTION OF REACTION

22.  Example: H 2 (g) + I 2 (g) = 2HI(g)(at 440 o C)  The table below shows concentrations of the reaction components from experiments I and II at a time when the reaction mixture was not at equilibrium  Experiment I: Q< K c so Q must increase as the reaction moves towards equilibrium. This means that the net reaction must be to the right, in favour of products. REACTION QUOTIENT

23.  Experiment II, Q > K c  So Q must decrease as the reaction moves towards equilibrium. This means that the net reaction must be to the left, in favour or reactants Summary:  If Q = K c, the reaction is at equilibrium  If Q < K c, the reaction proceeds to the right in favour of products  If Q > K c, the reaction proceeds to the left in favour of reactants REACTION QUOTIENT

25.  As K c is defined with products on the numerator and reactants on the denominator, each raised to the power of their stoichiometric coefficients in the balanced equation, we can manipulate its value according to changes made to these terms RELATIONSHIPS BETWEEN K C FOR DIFFERENT EQUATIONS OF A REACTION

27. SUMMARY

29. LEARNING CHECK

30.  A system remains at equilibrium so long as the rate of the forward reaction equals the rate of the backward reaction.  But as soon as this balance is disrupted by any change in conditions that unequally affects the rates of these reactions, the equilibrium condition will no longer be met.  And the equilibria respond in a predictable way – based on the Le Chatelier principle.  This states that a system at equilibrium when subjected to a change will respond in such a way as to minimise the effect of the change WHEN EQUILIBRIUM IS DISRUPTED

31.  This means that system whatever we do to a system at equilibrium, the system will respond in the opposite way.  Add something and the system will react to remove it  Remove something and the system will react to replace it  After a while, a new equilibrium will be established and this will have a different composition from the earlier equilibrium mixture. Typical changes:  Changes in concentration  Changes in pressure  Changes in temperature LE CHATELIER PRINCIPLE

32.  Suppose an equilibrium is disrupted by an increase in the concentration of the reactants.  This will cause the rate of the forward reaction to increase while the backward reaction will not be affected, so the reaction rates will no longer be equal  When equilibrium is re-established, the mixture will have new concentrations of all reactants and products, and the equilibrium will have increased in favour of the products  The value of K c will be unchanged – this follows Le Chatelier’s principle: addition of reactant causes the system to respond by removing reactant (favours the forward reaction and so shifts the equilibrium to the right)  Similarly, the equilibrium could be disrupted by a decrease in the concentration of product by removing it – this shifts the equilibrium to the right CHANGES IN CONCENTRATION

33.  After H 2 is added, the concentration of N 2 and H 2 decrease, while the concentration of NH 3 rises as the forward reaction increases  The new equilibrium has a higher proportion of products  Removal of NH 3 from the equilibrium mixture causes a decrease in the concentrations of N 2 and H 2 as they react to form more NH 3 CHANGES IN CONCENTRATION

34.  Equilibria involving gases will be affected by a change in pressure if the reaction involves a change in the number of molecules.  This is because there is a direct relationship between the number of gas molecules & the pressure exerted by a gas in a fixed volume  So if there is an increase in pressure, the system responds to decrease this pressure by favouring the side with the smaller number of molecules.  Conversely, a decrease in pressure will cause a shift to the side with the larger number of molecules.  A different equilibrium position will result but the value of K c will be unchanged  An increase in pressure favours the side of the equilibrium reaction that has the smaller number of molecules CHANGES IN PRESSURE

35. Example 1: CO(g) + 2H 2 (g) = CH 3 OH(g)  3 molecules on the left and one on the right – so high pressure will shift the equilibrium to the right, which increases the yield of CH 3 OH Example 2: H 2 (g) + I 2 (g) = 2HI(g)  Does not involve a change in the number of molecules so a change in pressure does not have an affect the position of the equilibrium or the K c (but will affect the reaction rate) CHANGES IN PRESSURE

36.  Changing the temperature will affect Kc – but you must check the enthalpy changes of the forward and backward reactions to see how it will change it Remember: exothermic releases energy (-ve ΔH) and visa versa  Enthalpy changes of the forward and reverse reactions are equal and opposite  The –ve sign tells us that the forward reaction is exothermic  If there is a temperature decrease, the system will respond by producing heat by favouring the forward reaction  So the reaction will shift to the right, in favour of the product  A new equilibrium mixture will be achieved and a higher K c value  Here we get a higher yield of products at a lower temperature CHANGES IN TEMPERATURE

37.  This table shows the effect of temperature on K c for this reaction  Even though a lower temperature will produce a higher proportion of products, it will also cause a lower rate of reaction CHANGES IN TEMPERATURE

38.  Consider: N 2 (g) + O 2 (g) = 2NO(g) ΔH = +181kJmol - 1  Forward reaction is endothermic so absorbs heat – the effect of a decreased temperature will be to favour the backward exothermic reaction  Therefore the equilibrium will shift to the left, in favour of reactants, and K c will decrease.  At higher temperatures, the forward endothermic reaction is favoured, so the equilibrium shifts to the right and Kc will decrease CHANGES IN TEMPERATURE

39.  Adding a catalyst speeds up the rate of the reaction by providing an alternative pathway by lowering the E a  This increases the number of particles with sufficient energy without raising the temperature  The E a is lowered by the same amount for the forward & reverse reaction as they both pass through the same transition state  The catalyst will therefore have no effect on the position of equilibrium or on the value of K c – it will not increase the yield but will increase the speed of reaching the equilibrium state ADDITION OF A CATALYST

40. SUMMARY

41. LEARNING CHECK

43. In industry, yield of a reaction and the rate are clearly significant Haber Process: the production of ammonia, NH 3 The equation gives you this information:  All reactants and products are gases  There is a change in the number of molecules (4 vs 2)  Forward reaction is exothermic so releases heat EQUILIBRIUM IN INDUSTRY: HABER PROCESS

44. Applying Le Chatelier’s principle means:  Concentration: ammonia is removed as it forms, pulling the equilibrium to the right and increasing the yield  Pressure: as the forward reaction involves a decrease in the number of gas molecules, it will be favoured by high pressure  Temperature: as the forward reaction is exothermic, it will be favoured by a lower temperature, however, too low a temperature would cause it to be very slow, so a moderate 450 o C is used  Catalyst: speeds up the rate of production (mostly iron is used, plus small amounts of Mg and Al)  Conversion of only 10-20% but reactants are recycled to obtain an overall yield of 95% HABER PROCESS

45.  Involves 3 simple reactions to produce H 2 SO 4  The overall rate of the process depends on on step ii EQUILIBRIUM IN INDUSTRY: THE CONTACT PROCESS

46. EQUILIBRIUM IN INDUSTRY PRODUCTION OF METHANOL

47. LEARNING CHECK