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# Chemical Equilibrium

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1. Chemical Equilibrium AP Chemistry J.M.Soltmann

2. A Brief Review Question • Write out the theoretical rate law for the reaction: • 3 H2 (g) + N2 (g) <--> 2 NH3 (g)

3. A Brief Review Question • Write out the theoretical rate law for the reaction: • 3 H2 (g) + N2 (g) <--> 2 NH3 (g) • Rate = k[H2]3[N2]

4. What about this? • Write out the theoretical rate law for the reaction: • 2 NH3 (g) <--> 3 H2 (g) + N2 (g)

5. What about this? • Write out the theoretical rate law for the reaction: • 2 NH3 (g) <--> 3 H2 (g) + N2 (g) • Rate = k[NH3]2

6. 3 H2 (g) + N2 (g) <--> 2 NH3 (g) Rate = k[H2]3[N2] If we mix hydrogen and nitrogen, we can produce ammonia. 2 NH3 (g) <--> 3 H2 (g) + N2 (g) Rate = k[NH3]2 If we heat ammonia, we can decompose it to hydrogen and nitrogen. Let’s consider that…

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.

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.

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: ratef=rater. • Ratef = kf[H2]3[N2] • Rater = kr[NH3]2 • kf[H2]3[N2] = kr[NH3]2 • kf/kr = [NH3]2/[H2]3[N2] • Keq = [NH3]2/[H2]3[N2]

11. More on Keq • Keq = [products]/[reactants] • Keq is not given units! • Keq 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 Keq value.

12. Write out the expression for Keq in the following: • H2 (g) + I2 (s) <--> 2 HI (s) • 2 H2 (g) + O2 (g) <--> 2 H2O (l) • 2C2H6 (g) + 7O2 (g) <--> 4CO2 (g) + 6H2O (g)

13. Write out the expression for Keq in the following: • H2 (g) + I2 (s) <--> 2 HI (s) • Keq = 1/[H2] • 2 H2 (g) + O2 (g) <--> 2 H2O (l) • Keq = 1/[H2]2[O2] • 2C2H6 (g) + 7O2 (g) <--> 4CO2 (g) + 6H2O (g) • Keq = [CO2]4[H2O]6/[C2H6]2[O2]7

14. Calculating Keq • 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 Keq, 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: • [C2H6] = .300 atm [O2] = .450 atm • [CO2] = 4.4 atm [H2O] = 6.6 atm • What is the value of Keq?

16. We already looked at the RXN • 2C2H6 (g) + 7O2 (g)< --> 4CO2 (g) + 6H2O (g) • Keq = [CO2]4[H2O]6/[C2H6]2[O2]7 • Keq = [4.4]4[6.6]6/[.30]2[.45]7 • Keq = 9.21 x 1010 • The fact that Keq 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: • [HCO3-] = .200 M [H+] = .250 M • [CO2] = .6 atm [H2O] = 55.6 M • What is the value of Keq?

18. Write out the net ionic equation • 2NaHCO3 (aq) + H2SO4 (aq) <--> 2 H2O (l) + 2 CO2 (g) + Na2SO4 (aq) • HCO3- (aq) + H+ (aq) <--> H2O (l) + CO2 (g) • Keq = [CO2]/[HCO3-][H+] • Keq = [.6]/[.2][.25] • Keq = 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 Keq without some extra experimental information. • Let’s look at an example of this type of scenario.

20. Finding Keq 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 Keq?

21. ICE time • Here’s what was given: • C3H8 (g) + 5 O2 (g) <--> 3 CO2 (g) + 4 H2O (g) • I: 2.00 8.00 0 0 • C: • E: .500

22. ICE time • Now we work on the change of O2: • C3H8 (g) + 5 O2 (g) <--> 3 CO2 (g) + 4 H2O (g) • I: 2.00 8.00 0 0 • C: -7.50 • E: .500

23. ICE time • Now do the Stoichiometry: • C3H8 (g) + 5 O2 (g) <--> 3 CO2 (g) + 4 H2O (g) • I: 2.00 8.00 0 0 • C: -1.50 -7.50 +4.50 +6.00 • E: .500

24. ICE time • Next, bring it on down: • C3H8 (g) + 5 O2 (g) <--> 3 CO2 (g) + 4 H2O (g) • I: 2.00 8.00 0 0 • C: -1.50 -7.50 +4.50 + 6.00 • E: .50 .500 4.50 6.00

25. Now we can calculate Keq • Keq = [CO2]3[H2O]4/[C3H8][O2]5 • Keq = [4.5]3[6.0]4/[.5][.5]5 • Keq = [CO2]3[H2O]4/[C3H8][O2]5 • Keq = 7558272 or 7.56 x 106

26. Going the other way • Generally more useful to Chemists, we also can use a known value of Keq 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 HNO3 (aq) <--> H2O (l) + N2O5 (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 HNO3 (aq) --> H2O (l) + N2O5 (g) • I: 5.00 M 0 0

29. The Process • We don’t know changes, so we will • Assume the change in N2O5 is x: • 2 HNO3 (aq) --> H2O (l) + N2O5 (g) • I: 5.00 M 0 0 • C: -2x +x +x

30. The Process • Now we bring it down: • 2 HNO3 (aq) --> H2O (l) + N2O5 (g) • I: 5.00 M 0 0 • C: -2x +x +x • E: 5 - 2x x x

31. The Process • But wait, water is a liquid, so we don’t • need to include it. • 2 HNO3 (aq) --> H2O (l) + N2O5 (g) • I: 5.00 M - 0 • C: -2x - +x • E: 5 - 2x - x

32. The Process • Write out the equation for Keq • and substitute in, then solve. • 2 HNO3 (aq) <--> H2O (l) + N2O5 (g) • Keq = [N2O5]/[HNO3]2 • .1 = (x)/(5-x)2 = x/(25-10x+x2) • 2.5-x+.1x2 = x • .1x2 - 2x + 2.5 = 0

33. The Process • .1x2 - 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 Keq of Reversible Reactions • Think about this. Let’s assume that the rxn A + 2B--> C has Keq =5. • What is the Keq of the rxn C-->A + 2B?

35. Manipulating Keq of Reversible Reactions • A + 2B--> C has Keq =5. • So Keq1 = 5 = [C]/[A][B]2 • Now look at C-->A + 2B? • Keq2 = [A][B]2/[C] • The two equations are reciprocals: • [C]/[A][B]2 = ([A][B]2/[C])-1 • So, Keq2 = (Keq1)-1 = 5-1 = .2

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

37. Manipulating Keq of Mulitplied Reactions • A + 2B--> C has Keq =5. • So Keq1 = 5 = [C]/[A][B]2 • For the rxn 2A + 4B --> 2C? • Keq2 = [A]2[B]4/[C]2 • [A]2[B]4/[C]2 = ([C]/[A][B]2)2 • Thus Keq2 = (Keq1)2 = 5 2 = 25

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

39. Manipulating Keq of Additive Reactions • X + Q --> J • Keq1 = 12 = [J]/[X][Q] • J + Q --> D • Keq2 = 2 = [D]/[J][Q] • X + 2Q--> D • Keq3 = [D]/[X][Q]2 • [D]/[X][Q]2 =([J]/[X][Q])*([D]/[J][Q]) • So, Keq3 = Keq1 * Keq2 = 12*2 = 24

40. Can you put it all together? • Given: • N2 (g) + O2 (g)  2 NO (g) Keq = 6.00 • 2 NO (g) + 2 O2 (g)  2 NO2 (g) Keq = .500 • 2 N2O (g)  O2 (g) + 2 N2 (g) Keq = 15.0 • Calculate Keq for the reaction: • N2O (g) + NO2 (g)  3 NO (g)

41. It’s like Hess’s Law all over again!! • N2 (g) + O2 (g)  2 NO (g) Keq = 6.00 • This rxn and Keq stays the same. • 2 NO (g) + 2 O2 (g)  2 NO2 (g) Keq = .500 • This reaction is reversed and halved, so Keq = (.5)-.5 • 2 N2O (g)  O2 (g) + 2 N2 (g) Keq = 15.0 • This reaction is halved, so Keq = (15).5 • Now multiply all 3 Keq values. • N2O (g) + NO2 (g)  3 NO (g) Keq = 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:

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 H2 (g) + N2 (g) <--> 2NH3 (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 H2 (g) + N2 (g) <--> 2NH3 (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.

46. If we increase the concentration of a substance • 3 H2 (g) + N2 (g) <--> 2NH3 (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 [N2] is noticeably less and the [NH3] is noticeably greater.

47. 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. If we add a catalyst

48. Applying LeChatelier • 2C2H6 (g) + 7O2 (g) <--> 4CO2 (g) + 6H2O (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