1. Equilibrium 940701....53slideshttp:\\academicstaff.kmu.ac.ir\aliasadipour1

2. Equilibrium 940701....53slideshttp:\\academicstaff.kmu.ac.ir\aliasadipour2 Chemical equilibrium (Mark=1.5)

3. Equilibrium The Concept of Equilibrium As a system approaches equilibrium, both the forward and reverse reactions are occurring. At equilibrium, the forward and reverse reactions are proceeding at the same rate. 3940701....53slideshttp:\\academicstaff.kmu.ac.ir\aliasadipour N 2 O 4 (g) 2 NO 2 (g)

4. Equilibrium A System at Equilibrium Once equilibrium is achieved, the amount of each reactant and product remains constant. 4940701....53slideshttp:\\academicstaff.kmu.ac.ir\aliasadipour N 2 O 4 (g) 2 NO 2 (g)

5. Equilibrium 5940701....53slideshttp:\\academicstaff.kmu.ac.ir\aliasadipour

6. Equilibrium The Equilibrium Constant N 2 O 4 (g)  2 NO 2 (g) Rate = k f [N 2 O 4 ] 2 NO 2 (g)  N 2 O 4 (g) Rate = k r [NO 2 ] 2 In Equilibrium Rate f = Rate r k f [N 2 O 4 ] = k r [NO 2 ] 2 6940701....53slideshttp:\\academicstaff.kmu.ac.ir\aliasadipour N 2 O 4 (g) 2 NO 2 (g) kfkrkfkr [NO 2 ] 2 [N 2 O 4 ] =Keq= Single step Stocumetric coefficient

7. Equilibrium Equilibrium Constants of multi steps reactions 2NO 2 Cl(g)  2NO 2 (g)+Cl 2 (g) R=K[NO2Cl] 1=NO2Cl  NO2+Cl 2=NO2Cl+Cl  NO2+Cl2 K eq =[NO 2 ] 2 [Cl 2 ]/[NO 2 Cl] 2 1=NO2Cl  NO2+Cl In equilibrium R1=R2============K1[NO2Cl]=K2 [NO2][Cl] K1/K2= [NO2][Cl]/[NO2Cl]=Keq1 7940701....53slideshttp:\\academicstaff.kmu.ac.ir\aliasadipour multisteps

8. Equilibrium Equilibrium Constants of multi steps reactions 2NO 2 Cl(g)  2NO 2 (g)+Cl 2 (g) 2=NO2Cl+Cl  NO2+Cl2 R3=R4=========K3[NO2Cl][Cl]=K4 [NO2][Cl2] K3/K4=[NO2][Cl2]/[NO2Cl][Cl]=Keq2 K1/K2 *K3/K4= [NO2][Cl]/[NO2Cl]*[NO2][Cl2]/[NO2Cl][Cl] K1/K2 *K3/K4 =Keq1*Keq2= K eq = [NO 2 ] 2 [Cl 2 ]/[NO 2 Cl] 2 8940701....53slideshttp:\\academicstaff.kmu.ac.ir\aliasadipour Stocumetric coefficient

9. Equilibrium The Equilibrium Constant To generalize this expression, consider the reaction The equilibrium expression for this reaction would be K eq = [C] c [D] d [A] a [B] b aA + bBcC + dD 9 940701....53slideshttp:\\academicstaff.kmu.ac.ir\aliasadipour

10. Equilibrium The Concept of Equilibrium Chemical equilibrium occurs when a reaction and its reverse reaction proceed at the same rate. 10940701....53slideshttp:\\academicstaff.kmu.ac.ir\aliasadipour

11. Equilibrium K eq in presence of Solids and Liquids The Concentrations of Solids and Liquids Are Essentially Constant [PbCl 2 ] × K eq = K c = [Pb 2+ ] [Cl − ] 2 PbCl 2 (s) Pb 2+ (aq) + 2 Cl − (aq) 11940701....53slideshttp:\\academicstaff.kmu.ac.ir\aliasadipour K eq = [Pb 2+ ] [Cl - ] 2 [PbCl 2 ] Therefore, the concentrations of solids and liquids do not appear in the equilibrium expression

12. Equilibrium What Are the Equilibrium Expressions K eq & K c for These Equilibria? 12940701....53slideshttp:\\academicstaff.kmu.ac.ir\aliasadipour K====K c

13. Equilibrium The Equilibrium Constant for gaseoues Because pressure is proportional to concentration for gases in a closed system, the equilibrium expression can also be written K p = (P C ) c (P D ) d (P A ) a (P B ) b 13940701....53slideshttp:\\academicstaff.kmu.ac.ir\aliasadipour aA (g) + bB (g) cC (g) + dD (g) K C = [C] c [D] d [A] a [B] b

14. Equilibrium Relationship between K c and K p From the ideal gas law we know that Rearranging it, we get PV = nRT P = RT nVnV 14940701....53slideshttp:\\academicstaff.kmu.ac.ir\aliasadipour [C]

15. Equilibrium Relationship between K c and K p Plugging this into the expression for K p for each substance, the relationship between K c and K p becomes K p = K c (RT) c+d-a-b  n = (moles of gaseous product) − (moles of gaseous reactant) 15940701....53slideshttp:\\academicstaff.kmu.ac.ir\aliasadipour P = CRT K p = (P C ) c (P D ) d (P A ) a (P B ) b K p = (C C RT) c (C D RT) d (C A RT) a (C B RT) b ( C C ) c (C D ) d ×(RT) c (RT) d (C A ) a (C B ) b ×(RT) a (RT) b = K c (RT)  n

16. Equilibrium Equilibrium Can Be Reached from Either Direction As you can see, the ratio of [NO 2 ] 2 to [N 2 O 4 ] remains constant at this temperature. No matter what the initial concentrations of NO 2 and N 2 O 4 are. 16940701....53slideshttp:\\academicstaff.kmu.ac.ir\aliasadipour N 2 O 4 (g) 2 NO 2 (g)

17. Equilibrium Equilibrium Can Be Reached from Either Direction It does not matter whether we start with N 2 and H 2 or whether we start with NH 3. We will have the same proportions of all three substances at equilibrium. 17940701....53slideshttp:\\academicstaff.kmu.ac.ir\aliasadipour N 2(g) + 3H 2(g)  2NH 3(g)

18. Equilibrium What Does the Value of K Mean? If K >> 1, the reaction is product-favored; product predominates at equilibrium. 18940701....53slideshttp:\\academicstaff.kmu.ac.ir\aliasadipour

19. Equilibrium What Does the Value of K Mean? If K >> 1, the reaction is product-favored; product predominates at equilibrium. If K << 1, the reaction is reactant-favored; reactant predominates at equilibrium. 19940701....53slideshttp:\\academicstaff.kmu.ac.ir\aliasadipour

20. Equilibrium K & Time of Equilibrium The size of K and the time required to reach equilibrium are not directly related! 940701....53slideshttp:\\academicstaff.kmu.ac.ir\aliasadipour20

21. Equilibrium 940701....53slideshttp:\\academicstaff.kmu.ac.ir\aliasadipour21

22. Equilibrium Manipulating Equilibrium Constants The equilibrium constant of a reaction in the reverse reaction is the reciprocal of the equilibrium constant of the forward reaction at 100  C. K c = = 0.212 [NO 2 ] 2 [N 2 O 4 ] N2O4 (g)N2O4 (g) 2 NO 2 (g) N2O4 (g)N2O4 (g) 22940701....53slideshttp:\\academicstaff.kmu.ac.ir\aliasadipour [N 2 O 4 ] [NO 2 ] 2 1 0.212 = K c =

23. Equilibrium Manipulating Equilibrium Constants The equilibrium constant of a reaction that has been multiplied by a number is the equilibrium constant, raised to a power that is equal to that number. K c = = 0.212 at 100  C [NO 2 ] 2 [N 2 O 4 ] N2O4 (g)N2O4 (g) 2 NO 2 (g) 2 N 2 O 4 (g) 4 NO 2 (g) 23940701....53slideshttp:\\academicstaff.kmu.ac.ir\aliasadipour K c = = (0.212) 2 at 100  C [NO 2 ] 4 [N 2 O 4 ] 2

24. Equilibrium As long as some CaCO 3 or CaO remain in the system, the amount of CO 2 above the solid will remain the same. CaCO 3 (s) CO 2 (g) + CaO (s) 24940701....53slideshttp:\\academicstaff.kmu.ac.ir\aliasadipour K = [CO 2 ]

25. Equilibrium Homogeneous Equilibria Homogeneous equilibria are equilibria in which all substances are in the same state. N 2(g) + 3H 2(g)  2NH 3(g) H 2(g) + F 2(g)  2HF (g) 25940701....53slideshttp:\\academicstaff.kmu.ac.ir\aliasadipour 2HOH (g)  2H 2(g) + O 2(g)

26. Equilibrium Heterogeneous Equilibria Write equilibrium expressions for the following: 2HOH (l)  2H 2(g) + O 2(g) K = [H 2 ] 2 [O 2 ] 26940701....53slideshttp:\\academicstaff.kmu.ac.ir\aliasadipour CuSO 4. 5 HOH (s)  CuSO 4(s) + 5HOH (g) K = [HOH] 5

27. Equilibrium 27940701....53slideshttp:\\academicstaff.kmu.ac.ir\aliasadipour

28. Equilibrium Equilibrium Calculations A closed system initially containing 1.000 x 10 −3 M H 2 and 2.000 x 10 −3 M I 2 At 448  C is allowed to reach equilibrium. Analysis of the equilibrium mixture shows that the concentration of HI is 1.87 x 10 −3 M. Calculate K c at 448  C for the reaction taking place, which is H 2 (g) + I 2 (g) 2 HI (g) 28940701....53slideshttp:\\academicstaff.kmu.ac.ir\aliasadipour

29. Equilibrium What Do We Know? [H 2 ], M[I 2 ], M[HI], M Initially1.000 x 10 -3 2.000 x 10 -3 0 Change At equilibrium 1.87 x 10 -3 29940701....53slideshttp:\\academicstaff.kmu.ac.ir\aliasadipour

30. Equilibrium [HI] Increases by 1.87 x 10 -3 M [H 2 ], M[I 2 ], M[HI], M Initially1.000 x 10 -3 2.000 x 10 -3 0 Change+1.87 x 10 -3 At equilibrium 1.87 x 10 -3 30940701....53slideshttp:\\academicstaff.kmu.ac.ir\aliasadipour

31. Equilibrium Stoichiometry tells us [H 2 ] and [I 2 ] decrease by half as much [H 2 ], M[I 2 ], M[HI], M Initially1.000 x 10 -3 2.000 x 10 -3 0 Change-9.35 x 10 -4 +1.87 x 10 -3 At equilibrium 1.87 x 10 -3 31940701....53slideshttp:\\academicstaff.kmu.ac.ir\aliasadipour

32. Equilibrium We can now calculate the equilibrium concentrations of all three compounds… [H 2 ], M[I 2 ], M[HI], M Initially1.000 x 10 -3 2.000 x 10 -3 0 Change-9.35 x 10 -4 +1.87 x 10 -3 At equilibrium 6.5 x 10 -5 1.065 x 10 -3 1.87 x 10 -3 32940701....53slideshttp:\\academicstaff.kmu.ac.ir\aliasadipour

33. Equilibrium …and, therefore, the equilibrium constant Kc =Kc = [HI] 2 [H 2 ] [I 2 ] = 51 = (1.87 x 10 -3 ) 2 (6.5 x 10 -5 )(1.065 x 10 -3 ) 33940701....53slideshttp:\\academicstaff.kmu.ac.ir\aliasadipour

34. Equilibrium 940701....53slideshttp:\\academicstaff.kmu.ac.ir\aliasadipour34

35. Equilibrium The Reaction Quotient (Q) To calculate Q, one substitutes the initial concentrations on reactants and products into the equilibrium expression. Q gives the same ratio the equilibrium expression gives, but for a system that is not at equilibrium. 35940701....53slideshttp:\\academicstaff.kmu.ac.ir\aliasadipour

36. Equilibrium If Q = K, Q>K & Q<K 36940701....53slideshttp:\\academicstaff.kmu.ac.ir\aliasadipour

37. Equilibrium 37940701....53slideshttp:\\academicstaff.kmu.ac.ir\aliasadipour

38. Equilibrium Le Châtelier’s Principle “If a system at equilibrium is disturbed by a change in temperature, pressure, or the concentration of one of the components, the system will shift its equilibrium position so as to counteract the effect of the disturbance.” 38940701....53slideshttp:\\academicstaff.kmu.ac.ir\aliasadipour

39. Equilibrium Effects of Changes on the System 1.Concentration : The system will shift away from the added component. 3.Temperature : K will change depending on the endothermic or exothermic reactions. 2.Pressure : a. Addition of inert gas b. Decreasing the volume 39940701....53slideshttp:\\academicstaff.kmu.ac.ir\aliasadipour a: does not affect the equilibrium position. b: shifts the equilibrium toward the side with fewer gaseous molecules.

40. Equilibrium Effect of Changes in Concentration on Equilibrium The addition of 1.000 M H 2 in position I → the position II 40940701....53slideshttp:\\academicstaff.kmu.ac.ir\aliasadipour N 2(g) + 3H 2(g)  2NH 3(g) position I 0.399M 1.197 M 0.203 M 1.348 M 1.044 M 0.304 M position II = 0.0602

41. Equilibrium Production of NH 3 If H 2 is added to the system, N 2 will be consumed and the two reagents will form more NH 3. 41940701....53slideshttp:\\academicstaff.kmu.ac.ir\aliasadipour N 2(g) + 3H 2(g)  2NH 3(g)

42. Equilibrium Changes in Concentration Predict the effect of the changes listed to this equilbrium: As 4 O 6(s) + 6C (s)  As 4(g) + 6CO (g) A) addition of CO (g) B) removal of some of CO (g) C) removal of all of CO (g) D) addition of C (s) or As 4 O 6(s) E) removal of some of C (s) or As 4 O 6(s) F ) removal of all of C (s) or As 4 O 6(s) A) Left B) Right C) Right D)None E)None F)Left 42940701....53slideshttp:\\academicstaff.kmu.ac.ir\aliasadipour

43. Equilibrium 940701....53slideshttp:\\academicstaff.kmu.ac.ir\aliasadipour43

44. Equilibrium The Effect of Container Volume on Equilibrium If the size of a container is changed, the concentration of the gases change. N 2(g) + 3H 2(g)  2NH 3(g). A smaller container shifts to the right Four gaseous molecules produce two gaseous molecules. A larger container shifts to the left two gaseous molecules produce four gaseous molecules. 44940701....53slideshttp:\\academicstaff.kmu.ac.ir\aliasadipour

45. Equilibrium The Effect of Container Volume on Equilibrium Predict the direction of shift for the following equilibrium systems when the volume is reduced: a) P 4(s) + 6Cl 2(g)  4PCl 3(l) b) PCl 3(g) + Cl 2(g)  PCl 5(g) c) PCl 3(g) + 3NH 3(g)  P(NH 2 ) 3(g) + 3HCl (g) a) right b) right c) none 45940701....53slideshttp:\\academicstaff.kmu.ac.ir\aliasadipour

46. Equilibrium The Effect of Changes in Pressure CaCO 3 (s)  CaO(s) + CO 2 (g) 46940701....53slideshttp:\\academicstaff.kmu.ac.ir\aliasadipour K = [CO 2 ]

47. Equilibrium The system of N 2, H 2, and NH 3 are initially at equilibrium. When the volume is decreased, the system shifts to the right -- toward fewer molecules. 47940701....53slideshttp:\\academicstaff.kmu.ac.ir\aliasadipour N 2(g) + 3H 2(g)  2NH 3(g) K=0.0602

48. Equilibrium Effect of Temperature on Equilbirum For an exothermic reaction, if the temperature increases, K decreases. N 2(g) + 3H 2(g)  2NH 3(g) + 92 kJ For an endothermic reaction, if the temperature increases, K increases. CaCO 3(s) + 556 kJ  CaO (s) + CO 2(g) 48940701....53slideshttp:\\academicstaff.kmu.ac.ir\aliasadipour

49. Equilibrium The Effect of Changes in Temperature Q+Co(H 2 O) 6 2+ (aq) + 4 Cl (aq) CoCl 4 (aq) + 6 H 2 O (l) 49940701....53slideshttp:\\academicstaff.kmu.ac.ir\aliasadipour +Q-Q

50. Equilibrium 50940701....53slideshttp:\\academicstaff.kmu.ac.ir\aliasadipour

51. Equilibrium Catalysts increase the rate of both the forward and reverse reactions. 51940701....53slideshttp:\\academicstaff.kmu.ac.ir\aliasadipour Equilibrium is achieved faster, but the equilibrium composition remains unaltered.

52. Equilibrium The Haber Process This apparatus helps push the equilibrium to the right by 52940701....53slideshttp:\\academicstaff.kmu.ac.ir\aliasadipour N 2(g) + 3H 2(g)  2NH 3(g) + 92 kJ 0 3 4 3 2 1 5 7 6 1) Heating 2) Catalyzing 3) Cooling 4) Pressure 5) Refrigerating 6) Removing

53. Equilibrium Offer to U 940701....53slideshttp:\\academicstaff.kmu.ac.ir\aliasadipour53