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# Chemical equilibrium

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### Chemical equilibrium

1. 1. 19/01/2011 TOPIC 4: CHEMICAL EQUILIBRIUM Lecturer and contacts Mr. Vincent Madadi Department of Chemistry, University of Nairobi P. O. Box 30197-00100, Nairobi, Kenya Chemistry Dept. Rm 114 Tel: 4446138 ext 2185 Email: vmadadi@uonbi.ac.ke, madadivin2002@yahoo.com Website: http://www.uonbi.ac.ke/staff/vmadadi 19/01/2011 mov 1 Introduction• Chemical equilibrium is based on the fact that reversible reactions do not go to completion• Chemical equilibrium exists when two opposing reactions occur simultaneously at the same rate.• A chemical equilibrium is a reversible reaction where the forward reaction rate is equal to the reverse reaction rate.• Thus, Molecules are continually reacting, even though the overall composition of the reaction mixture does not change. 19/01/2011 mov 2 1
2. 2. 19/01/2011 Introduction cont.• The RATE of the forward reaction equals the RATE of the reverse reaction• Symbolically, this is represented as:• aA(g) + bB (g) ⇌ cC (g) + dD (g)• The equilibrium constant can be expressed as:19/01/2011 mov 3 Introduction cont.• Graphically, this is a representation of the rates for the forward and reverse reactions for this general reaction.• aA(g) + bB (g) ⇌ cC (g) + dD (g)19/01/2011 mov 4 2
3. 3. 19/01/2011 Law of mass action• At constant temperature, the rate of chemical reaction is proportional to active mass of the reacting substances• Active mass = is a thermodynamic quantity defined as a = fC,• Where a = active mass; f = activity coefficient, C = molar concentration• For gaseous systems at low pressure or dilute solutions, f = 1, thus a = C 19/01/2011 mov 5 Equilibrium law • For the reaction , aA(g) + bB (g) → cC (g) + dD (g) • Rate of forward reaction R1 = k1[A]a[B]b • Rate of backward reaction R2 = k2[C]c[D]d • At equilibrium, k1[A]a[B]b = k2[C]c[D]d Thus, K1/k2 = [C]c[D]d/[A]a[B]b = Kc 19/01/2011 mov 6 3
4. 4. 19/01/2011 Introduction cont. • Therefore , for the general reaction: a A(g) + bB (g) ⇌ c C (g) + d D (g) • Eqiliburium law can define a constantKc is defined for a reversible reaction at a given temperature as theproduct of the equilibrium concentrations (in M) of the products,each raised to a power equal to its stoichiometric coefficient in thebalanced equation, divided by the product of the equilibriumconcentrations (in M) of the reactants, each raised to a powerequal to its stoichiometric coefficient in the balanced equation. 7 19/01/2011 mov Introduction cont.• For gaseous reactions, concentration can be expressed in terms of partial pressure• Where, aA + Bb ⇌ lL + mM,• If PA, PB, PL and PM are the partial pressures of the gaseous species, then at equilibrium• Kp = (PL)i(PM)m/(PA)a(PB)b• If concentration is in mole fraction (X), the equilibrium constant can be expressed in form of Kx 19/01/2011 mov 8 4
5. 5. 19/01/2011 Introduction cont. • Where Kx = (XL)I(XM)m/(XA)a(XB)b 19/01/2011 mov 9 Relationship between Kc, Kp and Kx• For ideal gas, partial pressure of the component in the mixture is given by,• Pi = (m/v)RT• Where m = number of moles of the gas occupying volume V,• Thus m/v = Ci = Molar concentration• Therefore, Pi = CiRT,• Substituting the value of Pi into the expression for Kp 19/01/2011 mov 10 5
6. 6. 19/01/2011Relationship between Kc, Kp and Kx ...• Kp = (CLRT)l(CMRT)m/(CART)a(CBRT)b• Kp = (CL)l(CM)m (RT)l(RT)m (CA)a(CB)b(RT)a(RT)b• Kp = (CL)l(CM)m (RT)l+m (CA)a(CB)b(RT)a+b• ButThus Kp = Kc (RT)Δn19/01/2011 mov 11Relationship between Kc, Kp and Kx ...• From Kp = Kc (RT)Δn• When Δn = +ve, the number of moles of reactants are less than the number of moles of products , thus Kp >Kc• When Δn = -ve, the number of moles of reactants are more than the number of moles of products , thus Kp <Kc• When Δn = 0, the number of moles of reactants equals the number of moles of products , thus Kp =Kc19/01/2011 mov 12 6
7. 7. 19/01/2011Relationship between Kc, Kp and Kx ...• For Kp and Kx,• Note that the mole fraction of a component of a given gas mixture is given by Xi = Pi/P• Or Pi = XiP• Where P = total pressure of the gaseous mixture• Substituting the value of partial pressure into the expression for Kp• Then, Kp = (PL)l(PM)m/(PA)a(PB)b• Or Kp = (XLP)l(XMP)m/(XAP)a(XBP)b19/01/2011 mov 13Relationship between Kc, Kp and Kx ...• This can be expressed as, Kp = (XL)l (XM)m(P)l (P)m (XA)a (XB)b(P)a (P)b• Or Kp = (XL)l (XM)m(P)(l+m) (XA)a (XB)b(P)(a+b)This can be reduced to,• Or Kp = Kx (P)(l+m) = Kx PΔn19/01/2011 (P)(a+b) mov 14 7
8. 8. 19/01/2011Relationship between Kc, Kp and Kx ...• Therefore, Kp = Kx PΔn = Kc (RT)Δn• Hence, Kc = Kx (P/RT) Δn• Equivalent to, Kc = Kx (n/v) Δn19/01/2011 mov 15Factors That Affect Chemical Equilibrium1) concentration2) pressure3) volume4) temperature5) Catalysts- have no effect on position of equilibrium19/01/2011 mov 16 8
9. 9. 19/01/2011 Le Chatelier’s Principle• If an external stress is applied to a system at equilibrium, the system adjusts itself in such a way that the stress is partially offset.1) Changes in Concentrations• Increase the yield of product by – increasing concentration of reactant – removing the product from the equilibrium 19/01/2011 mov 17 Le Chatelier’s Principle cont.• Eaxample N2 (g) + 3H2 (g) 2NH3 (g) 0.683 M 8.80 M 1.05 M• Increase the concentration of NH3 to 3.65 M, the position of equilibrium shifts to the left Q= [NH3]2 = ( 3.65 )2 = 0.0286 [N2] [H2]3 (.683) (8.80)3 But Kc = [NH3]2 = ( 1.05 )2 = [N2] [H2]3 (.683) (8.80)3• 19/01/2011>Kc Q mov 18 9
10. 10. 19/01/2011 Le Chatelier’s Principle cont. BaSO4 (s) Ba2+(aq) + SO42- (aq) Kc = [Ba2+] [SO42-]• By adding Ba2+(aq), [SO4 2-] decreases, but BaSO4 (s) increases• By add SO42- (aq), [Ba2+] decreases but BaSO4 (s) increases• Add BaSO4 (s) no change 19/01/2011 mov 19 Changes in Volume and Pressure• Little effect on reactions in solution• Effect can be large on reactions in the gas phase • Increase in pressure shifts the equilibrium to the side with the fewer moles of gas• Example:• How does the position of equilibrium change as the pressure is increased? five moles of gaseous reactant two moles of gaseous product• Increase in pressure causes an increase in products at the expense of reactants 19/01/2011 mov 20 10
11. 11. 19/01/2011 Changes in Temperature • Changes the value of the equilibrium constant – A temperature increase favours an endothermic reaction, and a temperature decrease favours an exothermic reaction • Example: 1) Endothermic reactions • Shifts in the equilibrium position for the reaction: 58 kJ + N2O4 (g) 2NO2 (g) • Increase temperature: Favours forward reaction 19/01/2011 mov 21 The Effect of a Catalyst• Increases rate, but has no effect on position of equilibrium effect on forward and reverse processes is the same a catalyst increases both the rate of both the forward and reverse reactions 19/01/2011 mov 22 11
12. 12. 19/01/2011 The Haber Process1.• Maximize the yield of ammonia by: carrying out the reaction at high pressure2. ΔHo = -92.6 kJ/mol• Maximize the yield of ammonia by: carrying out the reaction at low temperatures 19/01/2011 mov 23 The Haber Process • In practice the reaction is carried out at 500 ºC because the rate is too slow at lower temperatures 19/01/2011 mov 24 12
13. 13. 19/01/2011 Disruption and restoration of equilibriumAt the left, the concentrations of the three components do not change with timebecause the system is at equilibrium.By adding more hydrogen to the system, disrupts the equilibrium. A net reaction thenensues that moves the system to a new equilibrium state (right) in which the quantityof hydrogen iodide has increased; in the process, some ofthe I2 and H2 are consumed.The new equilibrium state contains more hydrogen than did the initial state, but not asmuch as was added; as the LeChâtelier principle predicts, the change wemade (addition of H2) has been partially counteracted by the "shift to the right". 19/01/2011 mov 25 Equilibrium constant calculations • One litre of equilibrium mixture from the following system at a high temperature was found to contain 0.172 mole of phosphorus trichloride, 0.086 mole of chlorine, and 0.028 mole of phosphorus pentachloride. Calculate Kc for the reaction. • Ans: Reaction PCl5 PCl3 + Cl2 • Expression for Kc • Substitute the values 19/01/2011 mov 26 13
14. 14. 19/01/2011 Equilibrium constant calculations cont.• The decomposition of PCl5 was studied at another temperature. One mole of PCl5 was introduced into an evacuated 1.00 litre container. The system was allowed to reach equilibrium at the new temperature. At equilibrium 0.60 mole of PCl3 was present in the container. Calculate the equilibrium constant at this temperature. Ans: Reaction PCl5 ⇌ PCl3 + Cl2 19/01/2011 mov 27 Equilibrium constant calculations cont. • Kc = [PCl3][Cl2] [PCl5], • Kc = (0.60)(0.60) = 0.90 • (0.40)Significance of Kc• Large equilibrium constants indicate that most of the reactants are converted to products.• Small equilibrium constants indicate that only small amounts of products are formed. Kc > 1 mostly products Kc < 1 mostly reactants 19/01/2011 1 equal amounts of products and reactants Kc ~ mov 28 14
15. 15. 19/01/2011 Equilibrium constant calculations cont.• The mass action expression or reaction quotient has the symbol Q.• Q has the same form as Kc, but the difference between them is that the concentrations used in Q are not necessarily equilibrium values.• The Reaction Quotient aA + bB ⇌ cC + dD• Q will help us predict how the equilibrium will respond to an applied stress. 19/01/2011 mov 29 Equilibrium constant calculations cont.• To make this prediction we compare Q with Kc. Q = [C]c[D]d [A]a[B]b• Q < K Products are formed• Q = K System is at equilibrium• Q > K Reactants are formed• Example• The equilibrium constant for the following reaction is 49 at 450°C. If 0.22 mole of I2, 0.22 mole of H2, and 0.66 mole of HI were put into an evacuated 1.00-liter container, would the system be at equilibrium? 19/01/2011 mov 30 15
16. 16. 19/01/2011 Equilibrium constant calculations cont.• Ans. H2(g) + I2(g) → 2 HI(g) 0.22 M 0.22 M 0.66 M• The Reaction Quotient• Q= [HI]2 = (0.66)2 = 9.0 H2 [ ] [I2] (0.22)(0.22)• Q = 9.0 but Kc = 49• Thus Q < Kc 19/01/2011 mov 31 Equilibrium constant calculations cont.• The equilibrium constant, Kc, is 3.00 for the following reaction at a given temperature. If 1.00 mole of SO2 and 1.00 mole of NO2 are put into an evacuated 2.00 L container and allowed to reach equilibrium, what will be the concentration of each compound at equilibrium? SO2(g) + NO2 (g) ⇌ SO3(g) + NO (g) 19/01/2011 mov 32 16
17. 17. 19/01/2011 Equilibrium constant calculations cont. 19/01/2011 mov 33 Equilibrium constant calculations cont.Equilibrium 0.0800+ x 0.0600+ x 0.790 -2x 19/01/2011 mov 34 17
18. 18. 19/01/2011 Equilibrium constant calculations cont.19/01/2011 mov 35 Types of chemical equlibria• There are two types of chemical equilibria:1) Homogeneous equilibrium• All the reactants and products are in the same phase2) Heterogeneous equilibrium• The reactants and products are in two or more different phases19/01/2011 mov 36 18
19. 19. 19/01/2011 Types of chemical equiliburia cont.• Heterogeneous equilibria have more than one phase present.• For example, a gas and a solid or a liquid and a gas. CaCO3(s) ⇌ CaO(s) + CO2(g)• How does the equilibrium constant differ for heterogeneous equilibria?• Pure solids and liquids have activities of unity.• Solvents in very dilute solutions have activities that are essentially unity. 19/01/2011 mov 37 Heterogeneous equilibrium• The thermodynamic equilibrium, Ka is given by: Ka = (aCaO )(aCO2 ) Eq.1 (aCaCO3)• Where a = activity of various species• For pure solids and liquids, activity are taken as unity at all temperatures• Thus, = aCaCO3 = 1 = aCaO• The equation reduces to Ka = aCO2 Eq. 2 19/01/2011 mov 38 19
20. 20. 19/01/2011 Heterogeneous equilibrium cont.• If the gas behaves ideally, activity reduces to pressure. Thus the equation 2 becomes: Kp = P CO2• The equilibrium constant for heterogeneous reactions is independent of the amount of a pure solid (liquid) phase• Thus equilibrium constant involves gaseous constituents but does not include any term for concentration of either pure solid or liquid• Thus Kp of heterogeneous reaction is generally called condensed equilibrium mov 39 Example- calculations involving Heterogeneous equilibrium • At a certain temperature, Kp for the dissociation of the solid calcium carbonate is 4 x 10-2 atm and for the reaction Cs + CO2 ⇌ 2CO is 2 atm respectively. Calculate the pressure of CO at this temperature when solid carbon, CaO and CaCO3 are mixed and allowed to attain equilibrium • Ans. CaCO3(s) ⇌ CaO(s) + CO2 (g) eq. 1 • Cs + CO2(g) ⇌ 2CO(g) eq.2 • Kp for the reaction 1 is: 19/01/2011 mov 40 20
21. 21. 19/01/2011 Example- calculations involving Heterogeneous equilibrium cont. • Kp = aCaO.aCO2/aCaCO3 = P CO2 • K’p for reaction 2 K’p = (aCO)2/(aC)(aCO2) = P2 CO/P CO2 K’pKp = P CO2. P2CO/P CO2 = P2 CO Thus, P CO = (KpK’p)½ = ((4.0 x 10-2)(2.0))½ = 0.28 atm 19/01/2011 mov 41Example- calculations involving Heterogeneous equilibrium cont.• A solid NH4HS is placed in a flask containing 1.50 atmospheric ammonia. What is the partial pressures of ammonia and H2S when equilibrium is attained? (Kp = 0.11)• Ans: Reaction, NH4HS(S) ⇌ NH3 + H2S P NH3 = 1.5 + P H2S Kp = (P NH3) (P H2S) Thus 0.11 = (1.50 + P H2S) (P H2S) This implies, P H2S = 0.06 atm or -1.56 atm 19/01/2011 mov 42 21
22. 22. 19/01/2011 Example- calculations involving Heterogeneous equilibrium cont. • But 0.06 atm is more correct, • Hence P NH3 = 1.50+0.06 = 1.56 atm 19/01/2011 mov 43 Example involving Kc, Kp & Kx• At 21.5 oC and total pressure of 0.787 atm, N2O4 is 48.3% converted into NO2. Calculate the value of Kp and Kc for the reaction• Ans:• Reaction N2O4 (g) ⇌ 2NO2 (g)• Initial: a 0• At t= t, a-x 2x• Let a= 100, then x = 48.3 19/01/2011 mov 44 22
23. 23. 19/01/2011 Example involving Kc, Kp & Kx• Therefore, (a-x) = 100 - 48.3 = 51.7Thus 2x = 48.3 x 2 = 96.6Hence (nT)aq. = 51.7 + 96.6 = 148.3PNO2 = (96.6/148.3) X PT = 96.6X 0.0787/148.3PN2O4 = 51.7x 0.0787/148.3Kp = (PNO2)2 = 96.6x0.0787/148.3 PN2O4 51.7 x0.0787/148.319/01/2011 mov 45 Example involving Kc, Kp & Kx• Kp = 9.58 x 10-3 atm• From th first principles,• Kp = Kc(RT)Δn but Δn = 2-1 = 1• Thus Kc = Kp/RT (R = 8.31 J/mol K = 0.0821 Latmmol-1K-1) = 9.58x10-3/0.082 x (273 +21.5) = 0.00396 molL-119/01/2011 mov 46 23
24. 24. 19/01/2011 Shifts in chemical equilibrium 19/01/2011 mov 47 Example: Le Chatelier’s principle in biological systems• Carbon monoxide poisoning. Carbon monoxide, a product of incomplete combustion that is present in automotive exhaust and cigarette smoke, binds to hemoglobin 200 times more tightly than does O2.• This blocks the uptake and transport of oxygen by setting up a competing equilibrium O2-hemoglobin hemoglobin CO- hemoglobin• Air that contains as little as 0.1 percent carbon monoxide can tie up about half of the hemoglobin binding sites, reducing the amount of O2 reaching the tissues to fatal levels. 19/01/2011 mov 48 24
25. 25. 19/01/2011 Le Chatelier’s principle in biological systems cont.• Carbon monoxide poisoning is treated by administration of pure O2 which promotes the shift of the above equilibrium to the left.• This can be made even more effective by placing the victim in a hyperbaric chamber in which the pressure of O2 can be made greater than 1 atm. 19/01/2011 mov 49 Examples of chemical equilibrium calculations• The commercial production of hydrogen is carried out by treating natural gas with steam at high temperatures and in the presence of a catalyst (“steam reforming of methane”): CH4 + H2O ⇌ CH3OH + H2• Given the following boiling points: CH4 (methane) = –161°C, H2O = 100°C, CH3OH = 65°, H2 = –253°C, predict the effects of an increase in the total pressure on this equilibrium at 50°, 75° and 120°C.Ans: 19/01/2011 mov 50 25
26. 26. 19/01/2011 Examples of chemical equilibrium calculations cont.Comment: This net reaction describes the dissolution of limestone by acid; it isresponsible for the eroding effect of acid rain on buildings and statues. The firstreaction is “driven” by a second reaction having a large equilibrium constant.From the standpoint of the LeChâtelier principle, the first reaction is “pulled to theright” by the removal of carbonate by the hydrogen ion. “Coupled” reactions of thistype are widely encountered in all areas of chemistry, and especially in biochemistry,in which a dozen or so reactions may be linked in this way. 19/01/2011 mov 51 Examples of chemical equilibrium calculations cont. 19/01/2011 mov 52 26
27. 27. 19/01/2011 Examples of chemical equilibrium calculations cont.19/01/2011 mov 53 Examples of chemical equilibrium calculations cont.19/01/2011 mov 54 27
28. 28. 19/01/2011 Examples of chemical equilibrium calculations cont.19/01/2011 mov 55 Examples of chemical equilibrium calculations cont.19/01/2011 mov 56 28
29. 29. 19/01/2011 Examples of chemical equilibrium calculations cont.19/01/2011 mov 57 Examples of chemical equilibrium calculations cont.19/01/2011 mov 58 29
30. 30. 19/01/2011 Examples of chemical equilibrium calculations cont.19/01/2011 mov 59 Examples of chemical equilibrium calculations cont.19/01/2011 mov 60 30