Equilibrium

1. Equilibrium

2. Brief Outline • What is reversible reaction? • Examples of reversible reaction • Dynamic Equilibrium • Le Chatelier’s Principle • The Haber Process

3. What is reversible reaction? • A reversible reaction is a chemical reaction in which the products can be converted back to reactants under suitable conditions. • A reversible reaction is shown by the sign ( ) a half-arrow to the right (forward reaction) and a half-arrow to the left (reverse reaction).

4. Reversibe Reactions HgO HgO HgO Hg HgO HgO HgO O2 HgO Hg HgO HgO Hg Hg HgO O2 2 HgO (s) 2 Hg (l) + O2 (g) 2 Hg (l) + O2 (g) 2HgO (g) Mercury and oxygen combine to form mercury oxide just as fast asmercury oxide decomposes into mercury and oxygen

5. 2 HgO (s) 2 Hg (l) + O2 (g) • Both reactions continue to occur, but there is no net change in the composition of the system. • The amounts of mercury(II)oxide, mercury(Hg), and oxygen (O2) remain constant. There is a state of equilibrium between two chemical reactions.

6. Dynamic Equilibrium Open System (No Equilibrium) Evaporation Closed System Evaporation Liquid Gas Liquid Gas Liquid Gas (No Equilibrium) (Equilibrium) (No Equilibrium)

7. Example For the decomposition, CaCO3 CaO + CO2 Would the reaction be the same when temperature is kept constantly high in an open and closed system?

8. Dynamic Equilibrium Dynamic equilibrium is the state in which the forward and reverse reactions are occurring at the same rate. The forward and reverse reactions both continue. The concentration of all reactants and products remains constant) Ag + Cl - Ag + + Cl - AgCl (s) Cl - Ag + Chemical Equilibrium AgCl (s) Rate of Precipitation = Rate of Dissolving HC2H3O2 (aq) H+ + C2H3O2- Rate of Dissociation (ionization) = Rate of Association HC2H3O2 H + H + C2H3O2- HC2H3O2 C2H3O2-

9. Static Equilibrium Dynamic Equilibrium Dynamic equilibrium vsStatic Equilibrium

10. Characteristics of Dynamic Equilibrium • The concentrations of the reactants and products (macroscopic properties) remain the same but the reactions don't stop! • The reactants are still reacting to form the product and the product is still being converted back to the reactants (microscopic processes).

11. H2 + I22 HI Synthesis of Hydrogen Iodide 2 HI H2+ I2Dissociation of Hydrogen Iodide These two sets represent the same chemical reaction system, but with the reactions occurring in opposite directions. Horizontal part : the concentrations of reactants and products remain constantwhen the system reaches equilibrium.

12. H2 + I2 2HI The equilibrium state is independent of the direction from which it is approached. Whether we start with an equimolar mixture of H2 and I2 (left) or a pure sample of hydrogen iodide (shown on the right, using twice the initial concentration of HI to keep the number of atoms the same), the composition after equilibrium is attained (shaded regions on the right) will be the same. For more information, Click Here

13. The adjacent graph shows the changes in the reaction rates of the forward and backward reactions: A + B C + D • Initially (t = 0), [A] and [B] were maximum, while [C] and [D]were zero. • The rate of the forwardreaction decreasesas A and B are used up. • The rate of the reverse reactionincreases as C and D are formed. • Equilibrium is attained when the two rates become equal • [A], [B], [C], and [D] remain constant at equilibrium.

14. If NO2 is reddish brown and N2O4 is colorless: What is happening here? What properties are changing? What is happening over time? After a long time? Chemical Equilibrium Consider this reaction:

15. aA + bB cC + dD [C]c[D]d [A]a[B]b Kc = The Equilibrium Constant (homogeneous reactions) For a reaction of the type the following is a CONSTANT (at a given T) The equilibrium expression for this reaction would be Kc : equilibrium constant [C] and [D] : concentration of products [A] and [B] : concentration of reactants • Kc provides information about how far a reaction proceeds at a particular temperature. (can predict the conc. of reactants & products) • Only write the Kc for homogeneous reversible reactions(all in the same physical state)

16. Your Turn Equilibrium mix What is the equilibrium expression for the reaction? .

17. Your Turn Kc ~ 92/(1 x 1) = 81

18. Your Turn Equilibrium mix What is the equilibrium expression for the reaction? . .

19. Your Turn Equilibrium mix . Kc ~ 22/(6 x 4) = 0.17

20. What Does the Value of K Mean? • If K >> 1, the reaction proceeds almost totally towards the products. • If K << 1, the reaction hardly proceeds at all towards the products.

21. What Does the Value of K Mean? • If K >> 1, the conc of the product is greater than the conc of the reactants at equilibrium. • If K << 1, the conc of the product is less than the conc of the reactants at equilibrium.

22. Reactants Products If K=1, the conc of the product is = the conc of the reactants at equilibrium. The reaction goes about half way to completion.

23. The equilibrium constant • can also be expressed as partial pressure of gases. (Keq) • when aqueous solutions are used, concentrations are used (Kc)

24. PbCl2 (s) Pb2+ (aq) + 2 Cl−(aq) The Equilibrium Constant (heterogeneous reactions) The concentrations of solids and liquids are constant. They do not appear in the equilibrium expression. H2O Keq= [Pb2+] [Cl−]2 Water is not included • Liquid water is in lower entropy phase than ions (ions are more spread out) • Concentration of water is a constant– factored into the Keq

25. CaCO3(s)  CaO(s) + CO2(g) Keq = [CO2] CO2 (g) is in the highest entropy NH3(g) + HCl(g)  NH4Cl(s) 1 [NH3] [HCl] Keq=

26. CaCO3 (s) CO2 (g) + CaO(s) As long as some CaCO3 or CaO remain in the system, the amount of CO2 above the solid will remain the same.

27. What Are the Equilibrium Expressions for these equilibria? • 2SO2(g) + O2(g) 2SO3(g) • Fe3+(aq) + SCN-(aq) [Fe(SCN)]2+(aq) • 4NH3(g) + 5O2(g) 4NO(g) + 6H2O(g) Page 283 Qn 4

28. Example Give the expression of the equilibrium constant N2 (g) + 3 H2 (g) 2 NH3 (g) Given [N2]=0.1M,[H2]=0.125M, [NH3]=0.11M, find the equilibrium constant for the reaction.

29. Example Find the relationship between the equilibrium constants of the forward and backward reaction for: H2 (g) + I2 (g) 2HI (g)

30. Example The following reaction is an esterification reaction producing ethyl ethanoate. CH3CO2H(l) + C2H5OH(l) CH3CO2C2H5(l) + H2O(l) The value of Kc at 250C is 4.0. This equilibrium can be approached experimentally from the opposite direction. What is the value for Kc for the reaction, the hydrolysis of ethyl ethanoate. CH3CO2C2H5(l) + H2O(l) CH3CO2H(l) + C2H5OH(l)

31. How far will a reaction go? • Kc tells us how far a reaction has gone towards completion under given conditions. E.g. H2 (g) + I2 (g) 2HI (g) Kc= 2 at 2770C H2 (g) + Cl2 (g) 2HCl (g) Kc= 1018 at 2770C The large difference in magnitude for Kc values illustrates how stable HCl is at 2770C compared to HI.The bondingin HI is weaker and therefore the reverse reaction is more evident with an equilibrium being established.

32. Example The following reaction takes place at 460ºC, where the equilibrium constant K has a value of 85. SO2(g) + NO2(g) NO (g) + SO3 (g) At a certain moment, the concentrations of the reactant and products were measured to be: [SO2] = 0.04, [NO2] = 0.5M, [NO] = 0.3M, [SO3] = 0.02M (a) Is this system at equilibrium? (b) If not, in which direction must the reaction go to reach equilibrium? Note: Use reaction quotient, Qc since we are unsure if it is in equilibrium. Since Qc < Kc , the reaction will move to the right to generate more product to reach equilibrium.

33. Relationship between Qc & Kc • If Qc < Kc , the reaction will move to the right to generate more product to reach equilibrium. • If Qc = Kc , the reaction is at equilibrium. • If Qc > Kc , the reaction will move to the left to generate more reactants to reach equilibrium. The value of Qc in relation to Kc indicates the direction in which the net reaction must proceed as the system moves towards its equilibrium.

34. Le Chatelier’s Principle • if a system at equiibrium is subjected to some changes, the position of equilibrium will shift to minimise the change. • The possible changes in conditions include • Changes in concentration o f either the reactants or products. • Changes in pressure for gas phase reactions • Changes in temperature • Presence of a catayst • Has industrial significance as it allows chemists to alter the reaction conditions to produce an increased amount of product, hence increase the profitability of the chemical process.

35. Concentration PCl5 (g) PCl3(g) + Cl 2(g) Kc = Qc If we double the conc. of PCl5 , what happens? [PCl3][Cl2] [PCl5] Kc =

36. Concentration • If one of the products is removed from the mixture, the equilibrium will shift to the right to replace the substance removed. In the reaction below, what happens when • water is removed / added ? • ethanol , C2H5OH is added / removed ? CH3CO2H(l) + C2H5OH(l) CH3CO2C2H5(l) + H2O(l)

37. Concentration Consider the reaction below: 2[CrO4]2-(aq) + 2H+(aq) [Cr2O7]2-(aq) + H2O(l) chromium(VI)ion from acid dichromate(VI) ion When acid (H+ ) is added, acording to Le Chetelier’s principle, equilibrium shifts to the right to use up the acid. The solution changes to orange. When alkali (OH- ) is added, the equilibrium shifts to the left because the OH- reacts with H+ to form water. This reduces the concentration of H+ ions. The equilibrium shifts to the left to reduce the effect by replacing the H+ ions. The solution changes to yellow.

38. Pressure Example • An increase in pressure will shift the position of the equilibrium towards the side that involves fewer molecules. In the reaction below, what happens when pressure is (i) increased ? (ii) decreased ? N2O4(g) 2NO2(g) colourless brown As the pressure is increased, the colour is initially darker as the same number of molecules are squeezed into a smaller space – the conc is increased . The mixture then becomes more colourless . According to Le Chatelier’s principle, the equilibrium shifts to the left , the side with fewer molecules. The colour lightens to almost colourless.

39. Temperature • An increase in temperature will always favour the endothermic process and will shift the equilibrium towards that direction. Lowering the temperature will favour the exothermic process. • A change in temperature will alter the Kc. Example In the reaction below, what happens when the temperature is (i) increased ? (ii) decreased ? N2O4(g) 2NO2(g) ΔHƟ = +57kJmol-1 colourless brown Increase in temp (putting in a hot water bath) cause the colour to be darker as the equilibrium shifts to the right that is endothermic in order to reduce the effect of the higher temperature. Decrease in temp (putting in ice bath) will cause the colour to be lighter.

40. N2O42NO2 Low T High T Get time progression Check silberberg • The final state depends on: • The chemical nature of the reactants and products • The conditions of the system (temperature, pressure, volume).

41. Kc = Kc = =1.46 moldm-3 Kc = Kc = =2.92 moldm-3 [1.71]2 [2.00] [3.42]2 [4.00] Effect of pressure on Kc At 400K, N2O4 2NO2 2.00M 1.71M At equilibrium , Kc = 1.46 moldm-3 at 400K. With pressure doubled,conc is doubled, [N2O4]= 4.00M , [NO2 ]=3.42M The system is not in equilibrium. NO2 must react to produce more N2O4 until the Kcbecomes equal. At higher pressure, the percentage of N2O4 is greater => the position of the equilibrium will shift to the left ( the side with fewer molecules of gas) K is not affected by a change in pressure.

42. Kc = Kc = =4 Kc = Kc = =8 [5.76][5.76] [2.88][2.88] [5.22][10.4] [2.61][2.61] Effect of concentration on Kc CH3COOH + C2H5OH CH3COOC2H5 + H2O At equilibrium, 2.88M 2.88M 5.76M 5.76M With water added, conc. of water increases but conc. of other species will drop as the total volume increased. 2.61M 2.61M 5.22M 10.4M The system is not in equilibrium. Ethyl ethanoate must react with water to use up the excess water. At higher concentration, the position of the equilibrium will shift to the left in order to maintain the same Kc K is not affected by a change in concentration. The equilibrium will shift to the side so as to restore the value of K.

43. [CH3OH(g) ] [CO(g)][H2(g)]2 Kc = Effect of temperature on Kc Consider the reaction CO(g) + 2H2(g) CH3OH(g) ΔHƟ = -90kJmol-1 In an exothermic reaction, according to Le Chatelier’s principle, the equilibirum will shift to the side that is endothermic, that is to the left. • the concentration of the top of Kc will derease ad the concentration of the bottom will increase. • Kc decreases.

44. [CH3OH(g) ] [CO(g)][H2(g)]2 Kc = Exothermic Reaction CO(g) + 2H2(g) CH3OH(g) ΔHƟ = -90kJmol-1 For exothermic reaction, the increase in temp decreases the Kc value.

45. [NO(g) ] 2 [N2(g)][O2(g)] Kc = Endothermic reaction N2 (g) + O2(g) 2NO(g) ΔHƟ = +181kJmol-1 For endothermic reaction, the increase in temp increases the Kc value.

46. Catalyst • A catalyst has no effect on the position of an equilibrium at a particular temperature as it increases the rate of the forward and the reverse reaction equaly. • It reduces the time it takes the system to reach equilibrium.

47. The equilibrium constant and rate • Kc provides information about the extent of a reaction • It does not provide information on the rate of reaction. Page 287 Qn 5

48. Example Draw a table showing how the position of equilibrium in the reaction A,B and C would be affected by the followng changes: • Increased temperature • Increased pressure Reaction A : interconversion of oxygen and ozone 3O2(g) 2O3(g) ΔHƟ = +284 kJmol-1 Reaction B : the reaction between sulfur dioxide and oxygen in the presence of a platinium/rhodium catalyst. SO2(g) + 2O3(g) SO3(g) ΔHƟ = -197 kJmol-1 Reaction C : the reaction between hydrogen and carbon dioxide. CO2(g) + 2H2(g) CO(g) + H2O(g) ΔHƟ = +41 kJmol-1

49. Conditions affecting the position of Equilibrium An equilibrium system opposes changes

50. Haber Process • N₂ (g) + 3H₂ (g) ↔ 2NH₃ (g) from air from natural gas, CH4 • Conditions • Pressure: 20000kPa (200 atm) • Temperature: about 500°C • Catalyst: Iron According to Le Chatelier’s principle, at higher pressure, there should be greater yield of ammonia, however, there will be an increased cost in building a plant that can operate at a much higher pressure, and other safety and maintenance issues. Hence, a pressure of 200 atm with about 55.8% yield of ammonia is ideal.