2. Objectives Outline the characteristics of chemical and physical systems in a state of equilibrium Deduce the equilibrium constant expression from the equation for a homogeneous reaction.

3. Dynamic Equilibrium When wet clothes are hung on a washing line to dry, the water in them does not boil, because the temperature is still far below the normal boiling point of water. Some of the water molecules will be traveling much faster than the average, and they will have enough kinetic energy to escape the surface of the water and enter the atmosphere. The pressure they produce is known as the vapour pressure of water. The vapour pressure is very dependent upon the temperature. As the vapour leaves the wet clothes it will be blown away by the wind, and more water will evaporate to replace it, until eventually the clothes become dry. However what would happen if the water was held in a closed system?

4. Initially water will still vaporize, but once the air above the water becomes saturated with water vapour it can hold no more water. A state of equilibrium is established between the liquid water and the water vapour. Some water molecules will still be traveling fast enough to have enough kinetic energy to escape from the surface and enter the vapour phase. At the same time some of the water vapour molecules will slow down and condense back into the liquid state. Equilibrium is established when the rate of vaporization is equal to the rate of condensation: H 2 O (l) H 2 O (g)

5. Vapour Pressure Vapour pressure is the pressure of a vapour in equilibrium with its non-vapour phases. All liquids and solids have a tendency to evaporate to a gaseous form, and all gases have a tendency to condense back into their original form (either liquid or solid). At any given temperature, for a particular substance, there is a pressure at which the gas of that substance is in dynamic equilibrium with its liquid or solid forms. This is the vapour pressure of that substance at that temperature. The equilibrium vapour pressure is an indication of a liquid's evaporation rate. It relates to the tendency of molecules and atoms to escape from a liquid or a solid. A substance with a high vapour pressure at normal temperatures is often referred to as volatile.

6. This is an example of dynamic equilibrium. Can you think of another example of a dynamic equilibrium? At equilibrium the process taking place has not stopped, but the amounts on both sides of the process remain constant. The rate of the forward process is equal to the rate of the backward process

7. Explain how these diagrams demonstrate a dynamic equilibrium…

8. Closed Systems A dynamic equilibrium will only set up in a closed system. In a completely closed system neither matter nor energy can be lost or gained. When we use the term equilibrium we are usually referring to a closed system. Vapour Liquid vaporisation Water in a closed system

9. Reversible Reactions Some reactions go virtually to completion, however, many don’t. If a reaction does not go to completion it is said to be reversible. Once some products have been formed they start to react to re-form the reactants. The position of equilibrium can be reached either by starting with the reactants or by starting with the products.

10. Dissociation of dinitrogen tetroxide, N 2 O 4(g) Whether we start with dinitrogen tetroxide or with nitrogen dioxide, once equilibrium has been attained the composition of the mixture is constant. We can tell when equilibrium is established as the graph becomes horizontal 0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8 2.0 0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8 2.0 N2O4N2O4 N2O4N2O4 NO 2 At start 1.0 mol N 2 O 4 At start 2.0 mol NO 2 Concentration/mol dm -3 time

12. Homogeneous Chemical Equilibria A homogeneous reaction is one in which all the reactants and products are in the same phase. A phase is similar to a state such as solid, liquid or gas, but there is an important difference. There is a physically distinct boundary between two phases. For example: two immiscible liquids such as oil and water, both are in the liquid state, but there are two different phases, as there is a clear boundary where the two liquids meet.

13. If all the reactants and products are in the same phase, such as all in the gaseous state, all in the liquid phase or all in aqueous solution, we can write an equilibrium expression for the reaction: Consider the general reaction: aAaA+bBbBcCcCdDdD+ At equilibrium the concentrations of A, B, C and D are written as [A] eqm’ [B] eqm’ [C] eqm’ and [D] eqm’ respectively. KcKc = [C] c eqm x[D] d eqm [A] a eqm [B] b eqm x K c is known as the equilibrium constant, and it is a constant at a stated temperature

14. Lets practice Put the following equations into equilibrium expressions: C 2 H 5 OH (l) + CH 3 COOH (l) CH 3 COOC 2 H 5(l) + H 2 O (l) 3H 2(g) + N 2(g) 2NH 3(g) 2HI (g) H 2(g) + I 2(g) 2SO 2(g) + O 2(g) 2SO 3(g)

15. Activity: Read through pages 152- 155 of Calculations in AS/A Level Chemistry by Jim Clark 2000 Complete problem 1 on page 155, check your answers on page297.

16. Objectives Deduce the extent of a reaction from the magnitude of the equilibrium constant. Apply Le Châtelier’s principle to predict the qualitative effects of changes of, pressure concentration and temperature on the position of equilibrium and on the value of the equilibrium constant. State and explain the effect of a catalyst on an equilibrium reaction.

17. The magnitude of the equilibrium constant K c is essentially a measure of the amount of products in an equilibrium mixture compared with the amount of reactants. If K c > 1 then the position of equilibrium lies very much to the right, the mixture contains virtually all products, the reaction has almost gone to completion. If K c < 1, then very little product will be formed, the mixture contains virtually all reactants, the equilibrium lies to the left.

18. Le Châteliers Principle Henri Le Châtelier (1850-1936) was a French chemist who described what happens to the position of equilibrium if a change to the conditions is made. Le Châtelier’s Principle states that ”if a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium moves to counteract the change”

19. An example: Consider the reaction between ethanoic acid and ethanol to form ethyl ethanoate and water: If we remove some of the water from the equilibrium mixture, Le Châtelier’s Principle enables us to predict that more of the acid and alcohol will react to replace it. This concept of removing one of the products is a useful way to increase the amount of a desired product in an equilibrium reaction. C 2 H 5 OH (l) + CH 3 COOH (l) CH 3 COOC 2 H 5(l) + H 2 O (l)

20. The Conditions: Pressure in gaseous reactions Concentration for reactions in the liquid phase Temperature Using catalysts.

21. The effect of changing the pressure on gaseous reactions This only applies to reactions involving gases. An example of a gas reaction that reaches equilibrium is: N 2 O 4(g) 2NO 2(g) Dinitrogen tetraoxideNitrogen dioxide 1 mole2 moles colourlessbrown Increasing the pressure of a gas means that there are more molecules of it in a given volume – it is the equivalent to increasing the concentration of a solution.

22. If we increase the pressure on this system, Le Châtelier’s principle tells us that the position of equilibrium will move left because fewer molecules exert less pressure. In the same way if we decrease the pressure the equilibrium will move to the right. So how can this equilibrium mixture demonstrate this?

23. Dinitrogen tetraoxide is a colourless gas and nitrogen dioxide is brown. If we decrease the pressure which direction will the equilibrium lie, what would we see? If we left the mixture for a while what would we see? N 2 O 4(g) 2NO 2(g) Dinitrogen tetraoxide 1 mole colourless Nitrogen dioxide 2 moles brown

25. Equilibrium mixture For a moment the mixture becomes paler because we have reduced the concentration of brown NO 2 We can decrease the pressure by pulling out the syringe barrel. After a few moments the mixture becomes darker as the equilibrium moves to the right and more brown NO 2 is formed

26. However, if there are the same number of moles of gas on both sides of the equation, then pressure has no effect on the equilibrium mixture. For example: H 2(g) +I 2(g) 2HI (g) 2 moles The equilibrium position will not change in this reaction when the pressure is changed, so the proportions of the three gases will stay the same.

27. 1. In which of the following reactions will the position of equilibrium be affected by changing the pressure? Explain your answers. a. 2SO 2(g) + O 2(g) 2SO 3(g) b. CH 3 CO 2 H (aq) CH 3 CO 2 - (aq) + H + (aq) c. H 2(g) + CO 2(g) H 2 O (g) + CO (g)

28. Concentration If we increase the concentration of one of the reactants, Le Chateliers principle says that the equilibrium will shift in the direction that tends to reduce the concentration of this reactant. A (aq) + B (aq) C (aq) + D (aq) Suppose we add some extra A. This would increase the concentration of A. The only way that this system can reduce the concetration of A, is by A reacting with B and moves the equilibrium to the right

29. A(aq) + B(aq)C(aq) + D(aq) We end up with a greater proportion of products in the reaction than before we added A. The same thing would happen if we added more B. We could also remove C as it was formed. The equilibrium would move to the right to produce more C (and D). The same thing would happen if we removed D s soon as it was formed.

30. Temperature Reversible reactions that are exothermic (give out heat) in one direction are endothermic (take in heat) in the other direction. Suppose we increase the temperature of an equilibrium mixture that is exothermic in the forward direction. An example is: 2SO 2(g) + O 2(g) 2SO 3(g) ∆H ѳ = -197kJ mol -1

31. The negative sign of ∆H ѳ means that heat is given out when sulphur dioxide and oxygen react to form sulphur trioxide. Le Châteliers principle tells us that if we increase the temperature, the equilibrium moves in the direction that cools the system down. To do this it will move in the direction which absorbs heat (is endothermic) i.e. to the left. The equilibrium mixture will contain a greater proportion of sulphur dioxide and oxygen than before. In the same way, if we cool the mixture the equilibrium will move to the right and increase the proportion of sulphur trioxide.

32. Catalysts Catalysts have no effect on the position of equilibrium. They work in such a way that they affect both the forward and reverse reactions equally. Therefore they speed up the rate at which equilibrium is set up but do not alter the composition of the equilibrium mixture.

34. Objectives Apply the concepts of kinetics and equilibrium to industrial processes Haber Process Contact process

35. Ammonia, NH 3 Ammonia is an important chemical in industry. World production is over 140 million tonnes each year. Around 80% is used to make fertilisers like ammonium nitrate, ammonium sulphate and urea. The rest is used to make synthetic fibres (including nylon), dyes explosives and plastics like polyurethane

36. Making ammonia Nitrogen and hydrogen react together by a reversible reaction, which at equilibrium, forms a mixture of nitrogen, hydrogen and ammonia: N 2(g) +3H 2(g) 2NH 3(g) ∆H ѳ = -92 kJ mol -1 The percentage of ammonia obtained at equilibrium depends on temperature and pressure.

37. The graph shows that high pressure and low temperature would give close to 100% conversion, while low pressure and high temperature would give almost no ammonia. How could we use Le Châteliers Principle to predict the conditions for the best yield?

38. Le Châtelier’s Principle The reaction is exothermic from left to right. So, cooling the mixture will drive the equilibrium to the right (in the direction where heat is given out). There will be a greater percentage of ammonia in the equilibrium mixture. There are four molecules of gas on the left and two on the right. So, increasing the pressure will drive the equilibrium to the right (in the direction where the pressure is lower). There will be a greater percentage of ammonia in the reaction mixture.

39. The Haber Process In industry Ammonia is made by the Haber Process, in which the process of making ammonia is the key step. The Raw Materials are: AIRNATURAL GASWATER Provides the nitrogen (N 2 ) Provide the hydrogen by the following reaction CH 4(g) + H 2 O (g) → CO (g) + 3H 2(g)

40. The nitrogen and hydrogen are fed into a converter in the ratio of 1 : 3 and passed over an iron catalyst. Most plants run at a pressure of around 20000kPa (about 200 times atmospheric pressure) and a temperature of about 450K. This is a lower pressure and a higher temperature than would give the maximum conversion. These compromise conditions are used because: A high-pressure plant is expensive, both to build and maintain A lower temperature would slow down the rate of reaction At lower temperatures the catalyst lasts longer

41. Nitrogen and hydrogen flow continuously over the catalyst so the gases do not spend long enough in contact with the catalyst to reach equilibrium – there is about 15% conversion to ammonia. The ammonia is cooled so that it becomes liquid and is piped off. Any nitrogen and hydrogen that is not converted into ammonia is fed back into the reactor. The catalyst is iron in pea sized lumps (to increase the surface area). It lasts about 5 years before it becomes ‘poisoned’ by impurities in the gas stream and has to be replaced.

42. Natural gas CH 4 Steam H 2 O H 2, CO 2 Air N 2 + O 2 H 2, N 2 CO 2 Alkali Carbon dioxide by-product Reactor, Fe catalyst H2,N2H2,N2 H 2, N 2 NH 3, Cooler NH 3, Ammonia NH 3 H 2, N 2 recycled Raw Material Product Flow Chart of the Haber Process

43. The Contact Process The manufacture of sulphuric acid in the developed world has fallen in recent years. This is partly because the manufacture of phosphate fertilizers, which uses sulphuric acid, has declined. It is also because much of the industry has relocated to the developing world, where the plants are cheaper to build and operate, and where environmental restrictions can be less stringent.

44. The essential reaction is the conversion of sulphur dioxide to sulphur trioxide: 2SO 2(g) + O 2(g) 2SO 3(g) ∆H ѳ = -192kJ mol -1 The sources of sulphur dioxide include the combustion of the element sulphur itself and the roasting (combustion) of sulphide ores in air. As the reaction is exothermic and involves a reduction in volume, the same arguments as used for the Haber process can be applied. A high yield will be favoured by a high pressure and a low temperature.

45. The rate at which equilibrium is reached will be favoured by a high pressure, a high temperature, and the use of a suitable catalyst. As before, an optimum temperature must be used. In practice, plants operate at a temperature of about 450 o C, and the catalyst is vanadium (V) oxide. Although in theory a high pressure is required, in practice a pressure of only 2 atm is used as this will give a yield of over 98%. Operating the plant at higher pressure than this would not be cost-effective. 2SO 2(g) + O 2(g) 2SO 3(g) ∆H ѳ = -192kJ mol -1 V 2 O 5(s), 450 o C, 2atm 3 vol2 vol

46. The reaction between sulphur trioxide and water then produces sulphuric acid: SO 3(g) + H 2 O (l) H 2 SO 4(l) The sulphur trioxide produced cannot be added directly to water, because the heat evolved is too great. Instead it is absorbed by a concentrated solution of sulphuric acid to which small quantities of water are then added to keep the acid concentrated. The sulphuric acid produced has many uses, ranging from a chemical feedstock to paints and pigments, fertilizers, detergents and soaps, and as the electrolyte in car batteries.

47. Objectives Describe the equilibrium established between a liquid and its own vapour and how it is affected by temperature changes. Sketch graphs showing the relationship between vapour pressure and temperature and explain them in terms of the kinetic energy. State and explain the relationship between enthalpy of vaporization, boiling point and intermolecular forces

48. Vapour Pressure Vapour pressure is the pressure of a vapour in equilibrium with its non-vapour phases. All liquids and solids have a tendency to evaporate to a gaseous form, and all gases have a tendency to condense back into their original form (either liquid or solid). At any given temperature, for a particular substance, there is a pressure at which the gas of that substance is in dynamic equilibrium with its liquid or solid forms. This is the vapour pressure of that substance at that temperature. The equilibrium vapour pressure is an indication of a liquid's evaporation rate. It relates to the tendency of molecules and atoms to escape from a liquid or a solid. A substance with a high vapour pressure at normal temperatures is often referred to as volatile.

49. The vapour pressure of all liquids increases as the temperature increases. All liquids show this kind of exponential – shaped graph, and the boiling point of the liquid will be the temperature at which the external pressure is equal to the vapour pressure above the liquid. It can be seen from the graph that the vapour pressure rises rapidly at higher temperatures. Using the Maxwell-Boltzmann distribution, try and explain this phenomena.

50. Consider the velocities of the particles in the liquid. Not all particles are moving at the same velocity. If the temperature is increased Temperature T 1 Temperature T 2 T 2 >T 1 Average kinetic energy at T 1 Average kinetic energy at T 2 Minimum energy for particles to form vapour Particles traveling relatively slowly Particles traveling close to the average velocity Particles traveling much faster than the average velocity

51. At low temperatures only a few particles will be able to escape, and hence the vapour pressure is low. At higher temperatures many more of the particles will have the necessary kinetic energy to escape, resulting in a much higher vapour pressure If the system is open, then as the faster particles leave the system the average energy of the remaining liquid particles decreases, and the temperature of the liquid decreases. This is why a small amount of a highly volatile liquid, such as ether, (ethoxyethane, C 2 H 5 OC 2 H 5 ), on the back of your hand will feel cold to the touch.

52. Relationship between enthalpy of vaporization, boiling point and intermolecular forces. Boiling points are normally measured under a pressure of 1atm. The boiling point of a substance is the temperature at which the vapour pressure is the same as atmospheric pressure. Different liquids have different boiling points. When a liquid boils, the intermolecular forces of attraction between particles in the liquid state are completely broken, and in an ideal gas there no forces of attraction between the gaseous particles. Liquid (l) Vapour (g)

53. The stronger the forces of attraction, the higher the boiling point. The forces of attraction for liquids consisting of simple molecules depend on the polarity of the molecules: the more polar the molecule, the stronger the forces of attraction, and the higher the boiling point will be. The strength of the forces of attraction also affects the enthalpy of vaporization.

54. Water has a high enthalpy of vaporization due to the relatively strong hydrogen bonding that exists in liquid water. H 2 O (l) H 2 O (g) ∆H ѳ = +44 kJ mol -1 Compare this with the enthalpy of vaporization of silicon tetrachloride, which is non-polar. The molecules of silicon tetrachloride in the liquid state will be attracted to each other by only weak van der Waals’ forces. Even though molecules of silicon tetrachloride have a much greater mass, the enthalpy of vaporization is considerably lower. SiCl 4(l) SiCl 4(g) ∆H ѳ = +30 kJ mol -1

55. From the enthalpy of vaporization, we can deduce that the boiling point of silicon tetrachloride will be lower than the boiling point of water. The value is in fact 57.6 o C. Evidence of a close correlation between boiling point and enthalpy of vaporization is provided by comparing SiCl 4 with Br 2. Both have a similar relative molecular mass, and both are non-polar and both the boiling points and their enthalpies of vaporization are almost identical CompoundMrBp ( o C)Enthalpy of vaporization kJmol -1 SiCL 4 170.0957.630.0 Br 2 159.8058.830.7

56. Some compounds (e.g. 2-hydroxy propanoic acid), also known as lactic acid, CH 3 CH(OH)COOH) decompose below their normal boiling points. This could cause a problem if fractional distillation is being used to purify the liquid or separate the liquid from a mixture of liquids. However, the problem can usually be easily overcome by carrying out distillation under reduced pressure.

57. Objectives The equilibrium Law Solve homogeneous equilibrium problems using the expression for K c

58. Calculating Kc Consider the reaction between a moles of ethanoic acid and b moles of ethanol to produce x moles of ethyl ethanoate and x moles of water at equilibrium. Let the total volume for the liquid mixture be V dm 3 eqm conc /mol dm 3 CH 3 COOH (l) +C 2 H 5 OH (l) CH 3 COOC 2 H 5(l) H 2 O (l) + Initial amount /mol eqm amount /mol ab00 a - xb - x x x a - x V b - x V x V x V

59. Now comes the complicated part….. [CH 3 COOC 2 H 5(l) ] eqm x[H 2 O (l) ] eqm [C 2 H 5 OH (l) ] eqm x[CH 3 COOH (l) ] eqm KcKc = = (x/V)x [(a – x)/V][(b – x)/V] x = x2x2 (a – x)(b – x)

60. For example When 46g (1.0mol) of ethanol was reacted with 30.0g (0.50 mol) of ethanoic acid at 373K the equilibrium contained 37.0g (0.42 mol) of ethyl ethanoate. Calculate the value of K c at 373K to the nearest integer. So we saw before that…. = x2x2 (a – x)(b – x) = (0.42) 2 (1.0 – 0.42)(0.50 – 0.42) =4 KcKc now substitute the letters for the numbers

61. Working out the units The units of Kc vary from case to case, and you will have to work them out every time you calculate a value for Kc in a question. Write down the expression for Kc and then substitute units into it rather than numbers. Example: N 2 (g) + 3H 2(g) 2NH 3(g) Kc= [NH 3(g) ] 2 [N 2 (g) ][H 2(g) ] 3 = (mol dm -3 ) 2 (mol dm -3 ) (mol dm -3 ) 3 = (mol dm -3 ) 2 1 = mol 2 dm -6 1 =mol -2 dm 6

62. Activity Read through further examples of calculating units on page 155 in Calculations in AS/A Level Chemistry by Jim Clark 2000 Complete problems 2 – 7 on pages 160 -161. Answers can be found on page 297 - 298

63. Calculating the composition of a reaction mixture We can use the equilibrium law to calculate the composition of a reaction mixture that has reached equilibrium Lets use this example again: CH 3 COOH (l) +C 2 H 5 OH (l) CH 3 COOC 2 H 5(l) H 2 O (l) + Suppose Kc = 4 at the temperature of the experiment and we want to know how much ethyl ethanoate we could produce by mixing 1 mol of ethanol and 1 mol of ethanoic acid.

64. CH 3 COOH (l) +C 2 H 5 OH (l) CH 3 COOC 2 H 5(l) H 2 O (l) + Initial amount /mol eqm amount /mol 1 mol 00 (1 – x) mol x mol We do not know how many moles of ethyl ethanoate will be produced, so we call this x. The equation tells us that x mol of water will also be produced and in doing so x mol of both ethanol and ethanoic acid will be used up. So the amount of each of these remaining at equilibrium is (1 – x) mol

65. These figures are in moles, but we need concentration in mol dm -3 to substitute in the equilibrium law expression. CH 3 COOH (l) +C 2 H 5 OH (l) CH 3 COOC 2 H 5(l) H 2 O (l) + Initial amount /mol eqm amount /mol 1 mol 00 (1 – x) mol x mol Suppose the volume of the system at equilibrium Vdm -3. Then……..

66. These figures may now be put into the expression Kc: [CH 3 COOC 2 H 5(l) ] eqm = (1 – x)/V mol dm -3 [C 2 H 5 OH (l) ] eqm [H 2 O (l) ] eqm [CH 3 COOH (l) ] eqm = = = (1 – x)/V mol dm -3 x /V KcKc = (x/V)x (1 – x)/V x

67. KcKc = (x/V)x (1 – x)/V x The V’s cancel, so in this case we do not need to know the actual volume of the system. So…. 4= x2x2 (1 – x) 2 Taking the square root for both sides 2= x (1 – x) So…. x=2 (1 – x)x=2 - 2x 2 = 3x x = 2 3

68. So 2 / 3 mol ethyl ethanoate and 2 / 3 mol of water are produced when the reaction reaches equilibrium, so the composition of the equilibrium mixture would be: ethanol 1 / 3 mol, ethanoic acid 1 / 3 mol, ethyl ethanoate 2 / 3 mol, water 2 / 3 mol CH 3 COOH (l) +C 2 H 5 OH (l) CH 3 COOC 2 H 5(l) H 2 O (l) + Initial amount /mol eqm amount /mol 1 mol 00 1 / 3 mol 2 / 3 mol

69. Calculating the amount of a reactant needed We can also use Kc to find the amount of a reactant needed to give a required amount of product. For the following reaction in ethanol solution, Kc = 30.0 dm 3 mol -1 CH 3 COCH 3 +HCNCH 3 C(CN)(OH)CH 3 propanonehydrogen cyanide 2-hydroxy-2- methylpropanenitrile

70. Suppose we are carrying out this reaction in 2.00dm 3 of ethanol. How much of the hydrogen cyanide is required to produce 1.00mol of product if we start with 4.00mol of propanone? Let x be the number of moles of HCN required. At equilibrium, we want 1 mol of product. [CH 3 COCH 3 ] eqm [HCN] eqm [CH 3 C(CN)(OH)CH 3 ] eqm Kc==30.0dm 3 mol -1 EquationCH 3 COCH 3 +HCNCH 3 C(CN)(OH)CH 3 At start 4.00 mol x mol0 mol At eqm(4.00 – 1.00) mol (x – 1.00) mol1.00 mol 3.00 mol(x – 1.00) mol1.00 mol x

71. These are the number of moles but we need the concentrations to put in the equilibrium law expression. The volume of the solution is 2.00dm 3 and the units for concentration are mol dm -3 so we next divide each quantity by 2.00dm 3 So at equilibrium…. 3.00/2.00 mol dm -3 [CH 3 COCH 3 ] eqm [HCN] eqm [CH 3 C(CN)(OH)CH 3 ] eqm = = = (x - 1.00)/2.00 mol dm -3 1.00/2.00 mol dm -3

72. Putting the figures into the equilibrium law expression and cancel any volumes: 3.00/2.00 mol dm -3 [CH 3 COCH 3 ] eqm [HCN] eqm [CH 3 C(CN)(OH)CH 3 ] eqm = = = (x - 1.00)/2.00 mol dm -3 1.00/2.00 mol dm -3 30.0 dm 3 mol -1 = 1.00/2.00 mol dm -3 3.00/2.00 mol dm -3 x(x - 1.00)/2.00 mol dm -3 So...

73. Cancelling through and rearranging we have: 30 ( 3 / 2 ( x – 1 ) / 2 ) = 1 / 2 45 ( x - 1 ) = 1 45x = 46 x = 46 / 45 = 1.02 So, to obtain 1 mol of product we must start with 1.02 mol hydrogen cyanide, if the volume of the system is 2.0dm 3. In this example the volume of the system does make a difference, because this reaction does not have the same number of moles of products and reactants 30.0 dm 3 mol -1 = 1.00/2.00 mol dm -3 3.00/2.00 mol dm -3 x(x - 1.00)/2.00 mol dm -3

74. Activity Read through the examples on pages 161 – 164 from Calculations in AS/A level Chemistry by Jim Clark 2000 Complete problems 8 – 10 on page 165. Answers can be found on page 299.