1 / 46

Equilibrium

Equilibrium. Chapter 12. Reactions are reversible. A + B C + D ( forward) C + D A + B (reverse) Initially there is only A and B so only the forward reaction is possible As C and D build up, the reverse reaction speeds up while the forward reaction slows down.

avani
Télécharger la présentation

Equilibrium

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Equilibrium Chapter 12

  2. Reactions are reversible • A + B C + D ( forward) • C + D A + B (reverse) • Initially there is only A and B so only the forward reaction is possible • As C and D build up, the reverse reaction speeds up while the forward reaction slows down. • Eventually the rates are equal

  3. Forward Reaction Reaction Rate Equilibrium Reverse reaction Time

  4. What is equal at Equilibrium? • Rates are equal • Concentrations are not. • Rates are determined by concentrations and activation energy. • The concentrations do not change at equilibrium. • or if the reaction is very slow, no significant change will occur.

  5. Law of Mass Action • For any reaction • aA + bB cC + dD • K = [C]c[D]d PRODUCTSpower [A]a[B]b REACTANTSpower • K is called the equilibrium constant. • is how we indicate a reversible reaction

  6. calculating K • If we write the reaction in reverse. • cC + dD aA + bB • Then the new equilibrium constant is • K’ = [A]a[B]b = 1/K[C]c[D]d

  7. Calculating with K • If we multiply the equation by a constant • naA + nbB ncC + ndD • Then the equilibrium constant is • K’ = [A]na[B]nb= ([A] a[B]b)n= Kn[C]nc[D]nd ([C]c[D]d)n

  8. Adding K • When we have two equations which we are adding to get a resulting equation • We will need to multiply the K from each equation. • K1 X K2 = K3 (sum of equations 1 +2)

  9. The units for K • Are determined by the various powers and units of concentrations. • They depend on the reaction. • Will usually cancel out to leave no units.

  10. K is a CONSTANT • At each temp, the reaction has a unique K that won’t change for that temp.. • Temperature affects rate and K. • The equilibrium concentrations won’t be the same, only K. • Equilibrium position is a set of concentrations at equilibrium. • There are an unlimited number equilibrium positions.

  11. Calculate K • N2 + 3H2 2NH3 • Initially At Equilibrium • [N2]0 =1.000 M [N2] = 0.921M • [H2]0 =1.000 M [H2] = 0.763M • [NH3]0 =0 M [NH3] = 0.157M

  12. Calculate K • N2 + 3H2 2NH3 • Initial At Equilibrium • [N2]0 = 0 M [N2] = 0.399 M • [H2]0 = 0 M [H2] = 1.197 M • [NH3]0 = 1.000 M [NH3] = 0.203M • K is the same no matter what the amount of starting materials

  13. Equilibrium and Pressure • Some reactions are gaseous • PV = nRT • P = (n/V)RT • P = CRT • C is a concentration in moles/Liter • C = P/RT

  14. Equilibrium and Pressure • 2SO2(g) + O2(g) 2SO3(g) • Kp = (PSO3)2 (PSO2)2 (PO2) • K = [SO3]2 [SO2]2 [O2]

  15. Equilibrium and Pressure • K =(PSO3/RT)2 (PSO2/RT)2(PO2/RT) • K =(PSO3)2 (1/RT)2 (PSO2)2(PO2) (1/RT)3 • K = Kp (1/RT)2= Kp RT (1/RT)3

  16. General Equation • aA + bB cC + dD • Kp= (PC)c(PD)d=(CCxRT)c(CDxRT)d(PA)a(PB)b(CAxRT)a(CBxRT)b • Kp=(CC)c(CD)dx(RT)c+d(CA)a (CB)bx(RT)a+b • Kp =K (RT)(c+d)-(a+b) = K (RT)Dn • Dn=(c+d)-(a+b)=Change in moles of gas

  17. Homogeneous Equilibria • So far every example dealt with reactants and products where all were in the same phase. • We can use K in terms of either concentration or pressure.

  18. Heterogeneous Equilibria • If the reaction involves pure solids or pure liquids the concentration of the solid or the liquid doesn’t change. • So, we can leave them out of the equilibrium expression.

  19. For Example • H2(g) + I2(s) 2HI(g) • K = [HI]2 [H2][I2] • But the concentration of I2 does not change. • K= [HI]2 [H2]

  20. The Reaction Quotient (Q) • Tells you the directing the reaction will go to reach equilibrium • Calculated the same as the equilibrium constant, but for a system not at equilibrium • Q = [Products]coefficient [Reactants] coefficient • Compare value to equilibrium constant

  21. What Q tells us • If K>Q Not enough products Shift to right • If K<Q Too many products Shift to left • If Q=K system is at equilibrium

  22. Example • for the reaction • 2NOCl(g) 2NO(g) + Cl2(g) • K = 1.55 x 10-5 M at 35ºC • In an experiment 0.10 mol NOCl, 0.0010 mol NO(g) and 0.00010 mol Cl2 are mixed in 2.0 L flask. • Which direction will the reaction proceed to reach equilibrium?

  23. Solving Equilibrium Problems • Given the starting concentrations and one equilibrium concentration. • Use stoichiometry to figure out other concentrations and K.

  24. Consider the following reaction at 600ºC • 2SO2(g) + O2(g) 2SO3(g) • In a certain experiment 2.00 mol of SO2, 1.50 mol of O2 and 3.00 mol of SO3 were placed in a 1.00 L flask. At equilibrium 3.50 mol of SO3 were found to be present. Calculate • The equilibrium concentrations of O2 and SO2, K

  25. Consider the same reaction at 600ºC • In a different experiment .500 mol SO2 and .350 mol SO3 were placed in a 1.000 L container. When the system reaches equilibrium 0.045 mol of O2 are present. • Calculate the final concentrations of SO2 and SO3 and K

  26. What if you’re not given an equilibrium concentration? • The size of K will determine what approach to take. • First let’s look at the case of a LARGE value of K ( >100). • Allows us to make simplifying assumptions.

  27. Example • H2(g) + I2(g) 2HI(g) • K = 7.1 x 102 at 25ºC • Calculate the equilibrium concentrations if a 5.00 L container initially contains 15.9 g of H2 294 g I2 . • [H2]0 = (15.7g/2.02)/5.00 L = 1.56 M • [I2]0 = (294g/253.8)/5.00L = 0.232 M • [HI]0 = 0

  28. Q= 0, which is <K so more product will be formed. • Assume that since K is large, then the reaction will go to completion. • Set up table of initial, final and equilibrium in concentrations.

  29. H2(g) I2(g) HI(g) initial 1.56 M 0.232 M 0 M change -X -X +2Xequil. • Using to stoichiometry we can find • Change in H2 = 1.56M -X • Change in HI = twice change in H2 • Change in HI = 0 +2X

  30. H2(g) I2(g) HI(g) initial 1.56 M 0.232 M 0 M change -X -X +2X equil 1.56-X 0.232-X +2X • Now plug these values into the equilibrium expression • K = (2X)2 = 7.1 x 102 (1.56-X)(0.232-X)

  31. When we solve for X we get 0.232 • So we can find the other concentrations • I2 = 0 M • H2 = 1.328 M • HI = 0.464 M

  32. Practice • For the reaction Cl2 + O2 2ClO(g) K = 156 • In an experiment 0.100 mol ClO, 1.00 mol O2 and 0.0100 mol Cl2 are mixed in a 4.00 L flask. • If the reaction is not at equilibrium, which way will it shift? • Calculate the equilibrium concentrations.

  33. For example • For the reaction 2NOCl 2NO +Cl2 • K= 1.6 x 10-5 • If 1.20 mol NOCl, 0.45 mol of NO and 0.87 mol Cl2 are mixed in a 1 L container • What are the equilibrium concentrations • Q = [NO]2[Cl2] = (0.45)2(0.87) = 0.15 M [NOCl]2 (1.20)2

  34. 2NOCl 2NO Cl2 Initial 1.20 0.45 0.87 Change Equil

  35. 2NOCl 2NO Cl2 Initial 1.20 0.45 0.87 Change -2X +2X +X Equil 1.20-2X 0.45+2X 0.87+X

  36. 2NOCl 2NO Cl2 Initial 1.20 0.45 0.87 Change -2X +2X +X Equil 1.20-2X 0.45+2X 0.87+X • Now we can write equilibrium constant • K = (.45 + 2X)2 (0.87 + X) (1.20 – 2X)2 = 1.6E-5

  37. Practice Problem • For the reaction 2ClO(g) Cl2 (g) + O2 (g) • K = 6.4 x 10-3 • In an experiment 0.100 mol ClO(g), 1.00 mol O2 and 1.00 x 10-2 mol Cl2 are mixed in a 4.00 L container. • What are the equilibrium concentrations.

  38. Example • H2(g) + I2 (g) 2 HI(g) K=38.6 • What is the equilibrium concentrations if 1.800 mol H2, 1.600 mol I2 and 2.600 mol HI are mixed in a 2.000 L container?

  39. Problems Involving Pressure • Solved exactly the same, with same rules for choosing X depending on KP • For the reaction N2O4(g) 2NO2(g) KP = .131 atm. What are the equilibrium pressures if a flask initially contains 1.000 atm N2O4?

  40. Le Chatelier’s Principle • If a stress is applied to a system at equilibrium, the position of the equilibrium will shift to reduce the stress. • 3 Types of stress

  41. Change amounts of reactants and/or products • Adding product makes K<Q • Removing reactant makes K<Q • Adding reactant makes K>Q • Removing product makes K>Q • Determine the effect on Q, will tell you the direction of shift

  42. Change Pressure • By changing volume • System will move in the direction that has the least moles of gas. • Because partial pressures (and concentrations) change a new equilibrium must be reached. • System tries to minimize the moles of gas.

  43. Change in Pressure • By adding an inert gas • Partial pressures of reactants and product are not changed • No effect on equilibrium position

  44. Change in Temperature • Affects the rates of both the forward and reverse reactions. • Doesn’t just change the equilibrium position, changes the equilibrium constant. • The direction of the shift depends on whether it is exo- or endothermic

  45. Exothermic • DH<0 • Releases heat • Think of heat as a product • Raising temperature push toward reactants. • Shifts to left.

  46. Endothermic • DH>0 • Produces heat • Think of heat as a reactant • Raising temperature push toward products. • Shifts to right.

More Related