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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.

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Equilibrium

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  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.

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