1. Chemical Equilibrium AP Chemistry J.M.Soltmann AP Chemistry J.M.Soltmann

2. A Brief Review Question Write out the theoretical rate law for the reaction: 3 H 2 (g) + N 2 (g) 2 NH 3 (g) Write out the theoretical rate law for the reaction: 3 H 2 (g) + N 2 (g) 2 NH 3 (g)

3. A Brief Review Question Write out the theoretical rate law for the reaction: 3 H 2 (g) + N 2 (g) 2 NH 3 (g) Rate = k[H 2 ] 3 [N 2 ] Write out the theoretical rate law for the reaction: 3 H 2 (g) + N 2 (g) 2 NH 3 (g) Rate = k[H 2 ] 3 [N 2 ]

4. What about this? Write out the theoretical rate law for the reaction: 2 NH 3 (g) 3 H 2 (g) + N 2 (g) Write out the theoretical rate law for the reaction: 2 NH 3 (g) 3 H 2 (g) + N 2 (g)

5. What about this? Write out the theoretical rate law for the reaction: 2 NH 3 (g) 3 H 2 (g) + N 2 (g) Rate = k[NH 3 ] 2 Write out the theoretical rate law for the reaction: 2 NH 3 (g) 3 H 2 (g) + N 2 (g) Rate = k[NH 3 ] 2

6. Let’s consider that… 3 H 2 (g) + N 2 (g) 2 NH 3 (g) 3 H 2 (g) + N 2 (g) 2 NH 3 (g) Rate = k[H 2 ] 3 [N 2 ] Rate = k[H 2 ] 3 [N 2 ] If we mix hydrogen and nitrogen, we can produce ammonia. If we mix hydrogen and nitrogen, we can produce ammonia. 3 H 2 (g) + N 2 (g) 2 NH 3 (g) 3 H 2 (g) + N 2 (g) 2 NH 3 (g) Rate = k[H 2 ] 3 [N 2 ] Rate = k[H 2 ] 3 [N 2 ] If we mix hydrogen and nitrogen, we can produce ammonia. If we mix hydrogen and nitrogen, we can produce ammonia. 2 NH 3 (g) 3 H 2 (g) + N 2 (g) 2 NH 3 (g) 3 H 2 (g) + N 2 (g) Rate = k[NH 3 ] 2 Rate = k[NH 3 ] 2 If we heat ammonia, we can decompose it to hydrogen and nitrogen. If we heat ammonia, we can decompose it to hydrogen and nitrogen. 2 NH 3 (g) 3 H 2 (g) + N 2 (g) 2 NH 3 (g) 3 H 2 (g) + N 2 (g) Rate = k[NH 3 ] 2 Rate = k[NH 3 ] 2 If we heat ammonia, we can decompose it to hydrogen and nitrogen. If we heat ammonia, we can decompose it to hydrogen and nitrogen.

7. Reversible reactions The two reactions we looked at our opposites, and we call them reversible reactions. Not all reactions are reversible. If we put nitrogen and hydrogen into a closed container, they will make ammonia. However, over time the ammonia will decompose into hydrogen and nitrogen. At some point the rate of production of ammonia will be equal to the rate of consumption of ammonia and the reaction appears to stop. The two reactions we looked at our opposites, and we call them reversible reactions. Not all reactions are reversible. If we put nitrogen and hydrogen into a closed container, they will make ammonia. However, over time the ammonia will decompose into hydrogen and nitrogen. At some point the rate of production of ammonia will be equal to the rate of consumption of ammonia and the reaction appears to stop.

8. Why did you say “appears?” The reaction never stops on a molecular level, even at equilibrium. However if one molecule is made at the same time one molecule is broken the overall effect means that there is no change in concentration. The reaction never stops on a molecular level, even at equilibrium. However if one molecule is made at the same time one molecule is broken the overall effect means that there is no change in concentration.

9. Chemical Equilibrium In a closed system, when the rate of the forward reaction is equal to the rate of the reverse reaction the reaction appears to stop. It has reached equilibrium.

10. Equilibrium Constant Going back to the previous example, at equilibrium: rate f =rate r. Rate f = k f [H 2 ] 3 [N 2 ] Rate r = k r [NH 3 ] 2 k f [H 2 ] 3 [N 2 ] = k r [NH 3 ] 2 k f /k r = [NH 3 ] 2 /[H 2 ] 3 [N 2 ] K eq = [NH 3 ] 2 /[H 2 ] 3 [N 2 ] Going back to the previous example, at equilibrium: rate f =rate r. Rate f = k f [H 2 ] 3 [N 2 ] Rate r = k r [NH 3 ] 2 k f [H 2 ] 3 [N 2 ] = k r [NH 3 ] 2 k f /k r = [NH 3 ] 2 /[H 2 ] 3 [N 2 ] K eq = [NH 3 ] 2 /[H 2 ] 3 [N 2 ]

11. More on Keq K eq = [products]/[reactants] K eq is not given units! K eq is a constant for a given reaction, at a given temperature. Since solids and liquids have constant concentrations (called densities), they are not included in the products or reactants; solids and liquids are part of the K eq value. K eq = [products]/[reactants] K eq is not given units! K eq is a constant for a given reaction, at a given temperature. Since solids and liquids have constant concentrations (called densities), they are not included in the products or reactants; solids and liquids are part of the K eq value.

12. Write out the expression for K eq in the following: H 2 (g) + I 2 (s) 2 HI (s) 2 H 2 (g) + O 2 (g) 2 H 2 O (l) 2C 2 H 6 (g) + 7O 2 (g) 4CO 2 (g) + 6H 2 O (g) H 2 (g) + I 2 (s) 2 HI (s) 2 H 2 (g) + O 2 (g) 2 H 2 O (l) 2C 2 H 6 (g) + 7O 2 (g) 4CO 2 (g) + 6H 2 O (g)

13. Write out the expression for K eq in the following: H 2 (g) + I 2 (s) 2 HI (s) K eq = 1/[H 2 ] 2 H 2 (g) + O 2 (g) 2 H 2 O (l) K eq = 1/[H 2 ] 2 [O 2 ] 2C 2 H 6 (g) + 7O 2 (g) 4CO 2 (g) + 6H 2 O (g) K eq = [CO 2 ] 4 [H 2 O] 6 /[C 2 H 6 ] 2 [O 2 ] 7 H 2 (g) + I 2 (s) 2 HI (s) K eq = 1/[H 2 ] 2 H 2 (g) + O 2 (g) 2 H 2 O (l) K eq = 1/[H 2 ] 2 [O 2 ] 2C 2 H 6 (g) + 7O 2 (g) 4CO 2 (g) + 6H 2 O (g) K eq = [CO 2 ] 4 [H 2 O] 6 /[C 2 H 6 ] 2 [O 2 ] 7

14. Calculating K eq When we calculated the rate of reaction, we used the initial concentrations because we were really only interested in the initial rate. Now we are looking at the end, at equilibrium. When we calculate K eq, we need to know the equilibrium concentrations. When we calculated the rate of reaction, we used the initial concentrations because we were really only interested in the initial rate. Now we are looking at the end, at equilibrium. When we calculate K eq, we need to know the equilibrium concentrations.

15. For example, Ethane was combusted in a closed container with oxygen. At equilibrium the following information was found: [C 2 H 6 ] =.300 atm[O 2 ] =.450 atm [CO 2 ] = 4.4 atm[H 2 O] = 6.6 atm What is the value of K eq ? Ethane was combusted in a closed container with oxygen. At equilibrium the following information was found: [C 2 H 6 ] =.300 atm[O 2 ] =.450 atm [CO 2 ] = 4.4 atm[H 2 O] = 6.6 atm What is the value of K eq ?

16. We already looked at the RXN 2C 2 H 6 (g) + 7O 2 (g) 4CO 2 (g) + 6H 2 O (g) K eq = [CO 2 ] 4 [H 2 O] 6 /[C 2 H 6 ] 2 [O 2 ] 7 K eq = [4.4] 4 [6.6] 6 /[.30] 2 [.45] 7 K eq = 9.21 x 10 10 The fact that K eq is much greater than 1 tells us that products are heavily favored in this reaction. 2C 2 H 6 (g) + 7O 2 (g) 4CO 2 (g) + 6H 2 O (g) K eq = [CO 2 ] 4 [H 2 O] 6 /[C 2 H 6 ] 2 [O 2 ] 7 K eq = [4.4] 4 [6.6] 6 /[.30] 2 [.45] 7 K eq = 9.21 x 10 10 The fact that K eq is much greater than 1 tells us that products are heavily favored in this reaction.

17. Try this one Sodium bicarbonate is mixed with sulfuric acid. At equilibrium, the following concentrations were measured: [HCO 3 - ] =.200 M[H + ] =.250 M [CO 2 ] =.6 atm[H 2 O] = 55.6 M What is the value of K eq ? Sodium bicarbonate is mixed with sulfuric acid. At equilibrium, the following concentrations were measured: [HCO 3 - ] =.200 M[H + ] =.250 M [CO 2 ] =.6 atm[H 2 O] = 55.6 M What is the value of K eq ?

18. Write out the net ionic equation 2NaHCO 3 (aq) + H 2 SO 4 (aq) 2 H 2 O (l) + 2 CO 2 (g) + Na 2 SO 4 (aq) HCO 3 - (aq) + H + (aq) H 2 O (l) + CO 2 (g) K eq = [CO 2 ]/[HCO 3 - ][H + ] K eq = [.6]/[.2][.25] K eq = 12 Are products or reactants favored? 2NaHCO 3 (aq) + H 2 SO 4 (aq) 2 H 2 O (l) + 2 CO 2 (g) + Na 2 SO 4 (aq) HCO 3 - (aq) + H + (aq) H 2 O (l) + CO 2 (g) K eq = [CO 2 ]/[HCO 3 - ][H + ] K eq = [.6]/[.2][.25] K eq = 12 Are products or reactants favored?

19. What if we don’t know the equilibrium concentrations? If we know the initial values, we can’t determine K eq without some extra experimental information. Let’s look at an example of this type of scenario. If we know the initial values, we can’t determine K eq without some extra experimental information. Let’s look at an example of this type of scenario.

20. Finding K eq from initial concentrations 2.00 atm of propane are mixed in a closed vessel with 8.00 atm of oxygen. At equilibrium, the concentration of oxygen was.500 atm. What is the value of K eq ? 2.00 atm of propane are mixed in a closed vessel with 8.00 atm of oxygen. At equilibrium, the concentration of oxygen was.500 atm. What is the value of K eq ?

21. ICE time Here’s what was given: C 3 H 8 (g) + 5 O 2 (g) 3 CO 2 (g) + 4 H 2 O (g) I: 2.00 8.0000 C: E:.500 Here’s what was given: C 3 H 8 (g) + 5 O 2 (g) 3 CO 2 (g) + 4 H 2 O (g) I: 2.00 8.0000 C: E:.500

22. ICE time Now we work on the change of O 2 : C 3 H 8 (g) + 5 O 2 (g) 3 CO 2 (g) + 4 H 2 O (g) I: 2.00 8.0000 C: -7.50 E:.500 Now we work on the change of O 2 : C 3 H 8 (g) + 5 O 2 (g) 3 CO 2 (g) + 4 H 2 O (g) I: 2.00 8.0000 C: -7.50 E:.500

23. ICE time Now do the Stoichiometry: C 3 H 8 (g) + 5 O 2 (g) 3 CO 2 (g) + 4 H 2 O (g) I: 2.00 8.0000 C: -1.50 -7.50 +4.50 +6.00 E:.500 Now do the Stoichiometry: C 3 H 8 (g) + 5 O 2 (g) 3 CO 2 (g) + 4 H 2 O (g) I: 2.00 8.0000 C: -1.50 -7.50 +4.50 +6.00 E:.500

24. ICE time Next, bring it on down: C 3 H 8 (g) + 5 O 2 (g) 3 CO 2 (g) + 4 H 2 O (g) I: 2.00 8.0000 C: -1.50 -7.50 +4.50 + 6.00 E:.50.500 4.506.00 Next, bring it on down: C 3 H 8 (g) + 5 O 2 (g) 3 CO 2 (g) + 4 H 2 O (g) I: 2.00 8.0000 C: -1.50 -7.50 +4.50 + 6.00 E:.50.500 4.506.00

25. Now we can calculate K eq K eq = [CO 2 ] 3 [H 2 O] 4 /[C 3 H 8 ][O 2 ] 5 K eq = [4.5] 3 [6.0] 4 /[.5][.5] 5 K eq = [CO 2 ] 3 [H 2 O] 4 /[C 3 H 8 ][O 2 ] 5 K eq = 7558272 or 7.56 x 10 6 K eq = [CO 2 ] 3 [H 2 O] 4 /[C 3 H 8 ][O 2 ] 5 K eq = [4.5] 3 [6.0] 4 /[.5][.5] 5 K eq = [CO 2 ] 3 [H 2 O] 4 /[C 3 H 8 ][O 2 ] 5 K eq = 7558272 or 7.56 x 10 6

26. Going the other way Generally more useful to Chemists, we also can use a known value of K eq to help us determine the actual yield or the actual concentrations of a substance at equilibrium.

27. For Example: 5.00 M nitric acid is heated until it decomposes: 2 HNO 3 (aq) H 2 O (l) + N 2 O 5 (g) If Keq for this reaction is.100, then how what concentration of dinitrogen pentoxide will be formed? 5.00 M nitric acid is heated until it decomposes: 2 HNO 3 (aq) H 2 O (l) + N 2 O 5 (g) If Keq for this reaction is.100, then how what concentration of dinitrogen pentoxide will be formed?

28. The Process Let’s start our ICE table: 2 HNO 3 (aq) --> H 2 O (l) + N 2 O 5 (g) I: 5.00 M00 Let’s start our ICE table: 2 HNO 3 (aq) --> H 2 O (l) + N 2 O 5 (g) I: 5.00 M00

29. The Process We don’t know changes, so we will Assume the change in N 2 O 5 is x: 2 HNO 3 (aq) --> H 2 O (l) + N 2 O 5 (g) I: 5.00 M00 C: -2x+x+x We don’t know changes, so we will Assume the change in N 2 O 5 is x: 2 HNO 3 (aq) --> H 2 O (l) + N 2 O 5 (g) I: 5.00 M00 C: -2x+x+x

30. The Process Now we bring it down: 2 HNO 3 (aq) --> H 2 O (l) + N 2 O 5 (g) I: 5.00 M00 C: -2x+x+x E: 5 - 2xxx Now we bring it down: 2 HNO 3 (aq) --> H 2 O (l) + N 2 O 5 (g) I: 5.00 M00 C: -2x+x+x E: 5 - 2xxx

31. The Process But wait, water is a liquid, so we don’t need to include it. 2 HNO 3 (aq) --> H 2 O (l) + N 2 O 5 (g) I: 5.00 M- 0 C: -2x- +x E: 5 - 2x- x But wait, water is a liquid, so we don’t need to include it. 2 HNO 3 (aq) --> H 2 O (l) + N 2 O 5 (g) I: 5.00 M- 0 C: -2x- +x E: 5 - 2x- x

32. The Process Write out the equation for K eq and substitute in, then solve. 2 HNO 3 (aq) H 2 O (l) + N 2 O 5 (g) K eq = [N 2 O 5 ]/[HNO 3 ] 2.1 = (x)/(5-x) 2 = x/(25-10x+x 2 ) 2.5-x+.1x 2 = x.1x 2 - 2x + 2.5 = 0 Write out the equation for K eq and substitute in, then solve. 2 HNO 3 (aq) H 2 O (l) + N 2 O 5 (g) K eq = [N 2 O 5 ]/[HNO 3 ] 2.1 = (x)/(5-x) 2 = x/(25-10x+x 2 ) 2.5-x+.1x 2 = x.1x 2 - 2x + 2.5 = 0

33. The Process.1x 2 - 2x + 2.5 = 0 is a quadratic. To solve this, we will have to either graph it or use the quadratic equation. The choice is yours. You will get two answers: X = 1.34 M or X = 18.7 M Since 18.7 M is not chemically possible, we know that the equilibrium concentration of dinitrogen pentoxide is 1.34 M..1x 2 - 2x + 2.5 = 0 is a quadratic. To solve this, we will have to either graph it or use the quadratic equation. The choice is yours. You will get two answers: X = 1.34 M or X = 18.7 M Since 18.7 M is not chemically possible, we know that the equilibrium concentration of dinitrogen pentoxide is 1.34 M.

34. Manipulating K eq of Reversible Reactions Think about this. Let’s assume that the rxn A + 2B--> C has K eq =5. What is the K eq of the rxn C-->A + 2B? Think about this. Let’s assume that the rxn A + 2B--> C has K eq =5. What is the K eq of the rxn C-->A + 2B?

35. Manipulating K eq of Reversible Reactions A + 2B--> C has K eq =5. So K eq1 = 5 = [C]/[A][B] 2 Now look at C-->A + 2B? K eq2 = [A][B] 2 /[C] The two equations are reciprocals: [C]/[A][B] 2 = ([A][B] 2 /[C])-1 So, K eq2 = (K eq1 ) -1 = 5 -1 =.2 A + 2B--> C has K eq =5. So K eq1 = 5 = [C]/[A][B] 2 Now look at C-->A + 2B? K eq2 = [A][B] 2 /[C] The two equations are reciprocals: [C]/[A][B] 2 = ([A][B] 2 /[C])-1 So, K eq2 = (K eq1 ) -1 = 5 -1 =.2

36. Manipulating K eq of Mulitplied Reactions Now think about this. We will still say that the rxn A + 2B--> C has K eq =5. What is K eq for the rxn 2A + 4B--> 2 C ? Now think about this. We will still say that the rxn A + 2B--> C has K eq =5. What is K eq for the rxn 2A + 4B--> 2 C ?

37. Manipulating K eq of Mulitplied Reactions A + 2B--> C has K eq =5. So K eq1 = 5 = [C]/[A][B] 2 For the rxn 2A + 4B --> 2C? K eq2 = [A] 2 [B] 4 /[C] 2 [A] 2 [B] 4 /[C] 2 = ([C]/[A][B] 2 ) 2 Thus K eq2 = (K eq1 ) 2 = 5 2 = 25 A + 2B--> C has K eq =5. So K eq1 = 5 = [C]/[A][B] 2 For the rxn 2A + 4B --> 2C? K eq2 = [A] 2 [B] 4 /[C] 2 [A] 2 [B] 4 /[C] 2 = ([C]/[A][B] 2 ) 2 Thus K eq2 = (K eq1 ) 2 = 5 2 = 25

38. Manipulating K eq of Additive Reactions This time let’s say we have two rxns. X + Q --> J has K eq = 12. J + Q --> D has K eq = 2 What is K eq for the rxn X + 2Q--> D? This time let’s say we have two rxns. X + Q --> J has K eq = 12. J + Q --> D has K eq = 2 What is K eq for the rxn X + 2Q--> D?

39. Manipulating K eq of Additive Reactions X + Q --> J K eq1 = 12 = [J]/[X][Q] J + Q --> D K eq2 = 2 = [D]/[J][Q] X + 2Q--> D K eq3 = [D]/[X][Q] 2 [D]/[X][Q] 2 =([J]/[X][Q])*([D]/[J][Q]) So, K eq3 = K eq1 * K eq2 = 12*2 = 24 X + Q --> J K eq1 = 12 = [J]/[X][Q] J + Q --> D K eq2 = 2 = [D]/[J][Q] X + 2Q--> D K eq3 = [D]/[X][Q] 2 [D]/[X][Q] 2 =([J]/[X][Q])*([D]/[J][Q]) So, K eq3 = K eq1 * K eq2 = 12*2 = 24

40. Can you put it all together? Given: N 2 (g) + O 2 (g)  2 NO (g) Keq = 6.00 2 NO (g) + 2 O 2 (g)  2 NO 2 (g) Keq =.500 2 N 2 O (g)  O 2 (g) + 2 N 2 (g) Keq = 15.0 Calculate Keq for the reaction: N 2 O (g) + NO 2 (g)  3 NO (g) Given: N 2 (g) + O 2 (g)  2 NO (g) Keq = 6.00 2 NO (g) + 2 O 2 (g)  2 NO 2 (g) Keq =.500 2 N 2 O (g)  O 2 (g) + 2 N 2 (g) Keq = 15.0 Calculate Keq for the reaction: N 2 O (g) + NO 2 (g)  3 NO (g)

41. It’s like Hess’s Law all over again!! N 2 (g) + O 2 (g)  2 NO (g) Keq = 6.00 This rxn and K eq stays the same. 2 NO (g) + 2 O 2 (g)  2 NO 2 (g) Keq =.500 This reaction is reversed and halved, so K eq = (.5) -.5 2 N 2 O (g)  O 2 (g) + 2 N 2 (g) Keq = 15.0 This reaction is halved, so K eq = (15).5 Now multiply all 3 Keq values. N 2 O (g) + NO 2 (g)  3 NO (g) K eq = 32.9 N 2 (g) + O 2 (g)  2 NO (g) Keq = 6.00 This rxn and K eq stays the same. 2 NO (g) + 2 O 2 (g)  2 NO 2 (g) Keq =.500 This reaction is reversed and halved, so K eq = (.5) -.5 2 N 2 O (g)  O 2 (g) + 2 N 2 (g) Keq = 15.0 This reaction is halved, so K eq = (15).5 Now multiply all 3 Keq values. N 2 O (g) + NO 2 (g)  3 NO (g) K eq = 32.9

42. LeChatelier’s Principle When a change is made to a system at equilibrium, the system shifts the equilibrium position to counter the change (partially). Basically this means that the system wants to undo the stress applied to it. For example: When a change is made to a system at equilibrium, the system shifts the equilibrium position to counter the change (partially). Basically this means that the system wants to undo the stress applied to it. For example:

43. If we increase the pressure: (or decrease volume) The system responds by shifting towards the side with less pressure (this is generally the side with less gas moles). In the RXn 3 H 2 (g) + N 2 (g) 2NH 3 (g), which side is favored? The system responds by shifting towards the side with less pressure (this is generally the side with less gas moles). In the RXn 3 H 2 (g) + N 2 (g) 2NH 3 (g), which side is favored?

44. If we add energy: If energy is added, the system wants to use the energy, so it shifts equilibrium to the side of highest enthalpy. In other words, the endothermic reaction is favored. 3 H 2 (g) + N 2 (g) 2NH 3 (g) + heat If energy is added to this reaction, would equilibrium favor reactants or products? If energy is added, the system wants to use the energy, so it shifts equilibrium to the side of highest enthalpy. In other words, the endothermic reaction is favored. 3 H 2 (g) + N 2 (g) 2NH 3 (g) + heat If energy is added to this reaction, would equilibrium favor reactants or products?

45. If we increase the concentration of a substance: When more of a chemical is added to a reaction at equilibrium, the system tries to use up the extra substance, shifting equilibrium to the other side. Let’s look at that a bit more closely. When more of a chemical is added to a reaction at equilibrium, the system tries to use up the extra substance, shifting equilibrium to the other side. Let’s look at that a bit more closely.

46. If we increase the concentration of a substance 3 H 2 (g) + N 2 (g) 2NH 3 (g) + heat If I were to add extra hydrogen gas, there would be an increase in collisions between hydrogen and nitrogen. I would use up more hydrogen and more nitrogen, thus producing more ammonia. Overall the hydrogen changes little, but the [N 2 ] is noticeably less and the [NH 3 ] is noticeably greater. 3 H 2 (g) + N 2 (g) 2NH 3 (g) + heat If I were to add extra hydrogen gas, there would be an increase in collisions between hydrogen and nitrogen. I would use up more hydrogen and more nitrogen, thus producing more ammonia. Overall the hydrogen changes little, but the [N 2 ] is noticeably less and the [NH 3 ] is noticeably greater.

47. If we add a catalyst nothing happens to equilibrium. Why? Well, the point of a catalyst is to speed up a reaction, by lowering the activation energy. Since the forward and reverse reaction both happen faster, there is no net change. nothing happens to equilibrium. Why? Well, the point of a catalyst is to speed up a reaction, by lowering the activation energy. Since the forward and reverse reaction both happen faster, there is no net change.

48. Applying LeChatelier 2C 2 H 6 (g) + 7O 2 (g) 4CO 2 (g) + 6H 2 O (g) Explain what happens to the equilibrium concentrations of each chemical in the rxn if: The pressure is increased The volume is increased The temperature is increased More oxygen is added A palladium catalyst is used 2C 2 H 6 (g) + 7O 2 (g) 4CO 2 (g) + 6H 2 O (g) Explain what happens to the equilibrium concentrations of each chemical in the rxn if: The pressure is increased The volume is increased The temperature is increased More oxygen is added A palladium catalyst is used