Equilibrium

1. Equilibrium

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 verrrryslooooow.

5. Law of Mass Action • For any reaction • jA + kBlC + mD • K = [C]l[D]mPRODUCTSpower [A]j[B]kREACTANTSpower • K is called the equilibrium constant. • is how we indicate a reversible reaction • Only gases and aqueous solutions are included in the law of mass action

6. Playing with K • If we write the reaction in reverse. • lC + mDjA + kB • Then the new equilibrium constant is • K’ = [A]j[B]k = 1/K[C]l[D]m

7. Playing with K • If we multiply the equation by a constant • njA + nkBnlC + nmD • Then the equilibrium constant is • K’ = [A]nj[B]nk= ([A] j[B]k)n=Kn[C]nl[D]nm ([C]l[D]m)n

8. From a practical point of view, here is how K gets manipulated: If a reaction gets doubled, the value of K is squared. If it is tripled, K is raised to the third power. If the reaction gets cut in ½ then K is found by taking the square root of K. Manipulating K

9. The units for K • This one is simple. • K has no units.

10. K is CONSTANT • Temperature affects the equilibrium constant. • The equilibrium concentrations don’t have to be the same only K. • Equilibrium position is a set of concentrations at equilibrium. • There are an unlimited number of K values so the temperature needs to be stated.

11. Equilibrium Constant One for each Temperature

12. Calculate K • N2 + 3H2 3NH3 Initial 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

13. Calculate K • N2 + 3 H2 2 NH3 (all gases) 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] = 0.157M • K is the same no matter what the amount of starting materials

14. Equilibrium and Pressure • Don’t forget your gas laws, they are often needed to get some information for the problem. Especially PV = nRT • Some reactions are gaseous • P = (n/V)RT • P = CRT • C is a concentration in moles/Liter • C = P/RT

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

16. Equilibrium and Pressure: For those that are a glutton for punishment and want to see the full treatment. • 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

17. General Equation • jA + kBlC + mD • Kp= (PC)l (PD)m= (CC x RT)l (CD x RT)m (PA)j (PB)k (CA x RT)j(CB x RT)k • Kp=(CC)l (CD)m x (RT)l+m(CA)j(CB)k x (RT)j+k • Kp=K (RT)(l+m)-(j+k) = K (RT)Dn • Dn=(l+m)-(j+k) = Change in moles of gas

18. 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. • Units depend on reaction.

19. Heterogeneous Equilibria • If the reaction involves pure solids or pure liquids the concentration of the solid or the liquid doesn’t change. • As long as they are not used up we can leave them out of the equilibrium expression. • For an example, check out the next slide.

20. For Example • H2(g) + I2(s) 2 HI(g) • K = [HI]2 [H2][I2] • But the concentration of I2 does not change, so it factors out of the equilibrium expression and looks like this. • K = [HI]2 [H2]

21. Solving Equilibrium Problems Type 1

22. The Reaction Quotient • Tells you the direction 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

23. What Q tells us • If Q < K • Not enough products • Shift to right • If Q > K • Too many products • Shift to left • If Q = K system is at equilibrium. We like these ones – there is nothing to do !!

24. Example • For the reaction: • 2 NOCl(g) 2 NO(g) + Cl2(g) • K = 1.55 x 10-5 M at 35ºC • In an experiment 0.10 molNOCl, 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?

25. Solving Equilibrium Problems • Given the starting concentrations and one equilibrium concentration. • Use stoichiometry to figure out other concentrations and K. • Learn to create a table of initial and final conditions. • Check out the next slide for an actual problem.

26. Consider the following reaction at 600ºC • 2 SO2(g) + O2(g) 2 SO3(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 and KP

27. The ICE BOX The table we create is called the ICE Box. 2 SO2(g) + O2(g) 2 SO3(g) Here is the information given in the problem. The blank spaces are what we have to determine. Some are determined and others are calculated. Initial 2 1.5 3 Change Equilibrium 3.5

28. The ICE BOX The table we create is called the ICE Box. 2 SO2(g) + O2(g) 2 SO3(g) Here is the information given in the problem. The blank spaces are what we have to determine. Some are determined and others are calculated. Initial 2 1.5 3 Change - 2x - x + 2x Equilibrium 3.5

29. Where are we at ? The values in the CHANGE row are a combination of stoichiometry and equilibrium. Notice that the numbers are the same as the coefficients. We can’t forget stoichiometry. The [SO3] increases so it has to be a + 2x. This tells us it shifts to the right and that the value of the reactants has to decrease. We combine the + or – sign along with the stoichiometry to set up the equation.

30. The ICE BOX The table we create is called the ICE Box. 2 SO2(g) + O2(g) 2 SO3(g) Here is the information given in the problem. The blank spaces are what we have to determine. Some are determined and others are calculated. Initial 2 1.5 3 Change - 2x - x + 2x Equilibrium 2 – 2x 1.5 – x 3.5

31. Where are we at ? Part 2 The values of the equilibrium are known algebraically. Here is our Law of Mass Action. We can solve that x = .25 K = [ SO3] 2 = (3.5) 2 [ O2] [ SO2]2 (1.25) (1.5)2 K = 4.36

32. A New Problem • Consider the same reaction at 600ºC 2 SO2(g) + O2(g) 2 SO3(g) • 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.

33. Let’s set up the ICE Box 2 SO2(g) + O2(g) 2 SO3(g) Initial .5 0 .35 Change + 2x + x - 2x Notice the signs. Since there is no oxygen present at the start, the reaction must shift to the left and that means products consumed and reactants created. Equilibrium .5 +2x.045 .35 – 2x This tells us that x = .045 and therefore the concentrations must be as follows. [SO2] = .5 + 2x or .5 + .09 = .59 [SO3] = .35 – 2x or .35 - .09 = .26 K = (.26)2 / (.59)2 (.045) = 4.32

34. What if you’re not given 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). • We can make simplifying assumptions.

35. 5 % rule Now is as good a time as any to introduce the 5 % rule. Equilibrium concentrations are often a bit of an approximation. Therefore it’s allowable to simplify the math to make it more workable. The 5 % rule states that if x is small in relation to the stated amount, and the K, then it can be ignored in the calculation. Generally works if the Keq is less than 10-3 or greater than 103

36. Example • H2 (g) + I2 (g) 2 HI (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 and 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

37. Let’s set up the Ice Box !! • Q > K so more product will be formed. • Since K is large, rxn. will go to completion. • Another great clue is that when there is a component with a concentration of zero, the reaction must shift in that direction. • Stoichiometry tells us I2 is LR, it will be smallest at equilibrium let it be x • Set up table of initial, final and change in concentrations.

38. This equation needs manipulating H2(g) + I2(g)  2 HI(g) initial 1.56 M 0.232 M 0 M change - x - x + 2xequilibrium 1.56 - x .232 - x 2x • Choose x so it is small. • Since we know the I2 is going to lose most of its concentration, x is actually going to be a very high percentage of the amount of I2 • There is a way around this dilemma.

39. Here we go. In the work below, we are going to assume that the reaction goes to completion. Let’s combine the following facts and ideas and solve the problem. H2(g) + I2(g)  2 HI(g) initial 1.56 M .232 M 0 M change - .232 - .232 + .464 equilibrium 1.328 0 .464

40. Piecing it together 1. We didn’t change the conditions of the rxn. (ie. Temp. ) so the Keq remains unchanged. 2. The Keq is large so we know there will be more product than reactant at equilibrium. 3. Since we assumed the reaction went to completion we now have no Iodine and the reaction must move to the left. 4. X must be small because there needs to be more product to satisfy Keq. X will be negligible compared to the amounts given. 5. However, x can’t be ignored for the I2 because it is at zero and x is 100% of the total amount of I2

41. H2(g) + I2(g)  2 HI(g) initial 1.328 0 .464 change + x + x -2x equilibrium 1.328 + x x.464 – 2x Here is our new Law of Mass Action after we remove the ‘x’ according to the 5% rule. Remember we can’t remove the x in the I2 because that is all that there is of it. 710 = (.464)2 / (1.328) ( x ) = 2.28 x 10-4

42. Checking the assumption • The rule of thumb is that if the value of X is less than 5% of all the other concentrations, our assumption was valid. • If not we would have had to use the quadratic equation • More on this later. • Our assumption was valid.

43. 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? Q = ? • Calculate the equilibrium concentrations of all components.

44. Problems with small K K< .01

45. Process is the same • Set up table of initial, change, and final concentrations. Your ICE Box • Choose X to be small. • Sounds familiar so far, I hope !! • For this case it will be a product. • For a small K the product concentration is small.

46. Let’s start the set up. • For the reaction 2 NOCl 2 NO + Cl2 • K= 1.6 x 10-5 Make note the small K favors more reactant • 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 [NOCl]2 (1.20)2 • Q > K so it shifts to the left.

47. Here is the ICE Box set up Once again we can manipulate the components to make the math easier and we do this by assuming the reaction goes to completion according to stoichiometry. • We can now use the Equilibrium concentrations to rework the problem. Remember that the K does not change. K favors the reactants. Having [ NO ] = 0 means that the reaction will shift to the right. Combining the two means that the change will be small so x will be small and we can use the 5 % simplification rule. This is a Happy Day ! 2 NOCl  2 NO + Cl2 Initial 1.20 0.45 0.87 Change +.45 -.45 -.225 Equilibrium 1.65 0 .645

48. Where are we at ? We can now use the Equilibrium concentrations to rework the problem. Remember that the K does not change. K favors the reactants. Having [ NO ] = 0 means that the reaction will shift to the right. Combining the two means that the change will be small so x will be small and we can use the 5 % simplification rule. This is truly a Happy Day !

49. Here is the new ICE Box 2 NOCl  2 NO + Cl2 Initial 1.65 0 .645 Change - 2x +2x +x Equilibrium 1.65 – 2x 2x .645 + x Equilibrium (5%) 1.65 2x .645 Solving: 1.6 x 10-5 = (.645) (2x)2 (1.65)2 The algebra is trickier but doable x = .0042

50. Always verify the 5 % rule • If x = .0042, we compare that to the concentration. .0042 / .87 = .48 % • We are good !!