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CHAPTER 10 : (GENERAL EQUILIBRIUM). MR BELLAND PRESENTS - HOW MY SON AND I PLAY.

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Presentation on theme: "CHAPTER 10 : (GENERAL EQUILIBRIUM). MR BELLAND PRESENTS - HOW MY SON AND I PLAY."— Presentation transcript:

1 CHAPTER 10 : (GENERAL EQUILIBRIUM)

2 MR BELLAND PRESENTS - HOW MY SON AND I PLAY

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8 Chemical equilibrium is when two opposing reactions occur simultaneously and at the same rate. For example: A + B → C + D andC + D → A + B This is usually written as a reversible equation using double arrows: A + B ⇌ C + D

9 What equilibrium is not: equal amounts of products and reactants permanent instantly establish a stop in the chemical reaction for reactions that go to completion (like dissociation of a strong acid) the same for every condition of T

10 Example case 1: in a 1 L vessel R2SO 2 + O 2 ⇌ 2SO 3 I0.400 mol 0.200 mol 0 C-0.056 mol -0.028 mol + 0.056 mol E0.344 mol 0.172 mol 0.056 mol in 1 L vessel the above mols are also M Describe what occurred here.

11 Example case 2: in a 1 L vessel R2SO 2 + O 2 ⇌ 2SO 3 I0 0 0.500 mol C +0.424 mol +0.212 mol -0.424 mol E0.424 mol 0.212 mol 0.076 mol in 1 L vessel the above mols are also M Describe what occurred here.

12 Suppose we want a rate law: for case 1 (forward reaction) rate f = k f [SO 2 ] 2 [O 2 ] for case 2 (reverse reaction) rate r = k r [SO 3 ] 2 (for only 1 step it is the slow step, so the coefficients = order) At equilibrium rate f = rate r (by definition)

13 If rate f = rate r then k f [SO 2 ] 2 [O 2 ] = k r [SO 3 ] 2 rearranging to group terms… What is on top, and what is on bottom, in relationship to the forward reaction? K c is called the equilibrium (concentration) constant (capitol K) and the c is sometimes left off kfkf = K c = [SO 3 ] 2 krkr [SO 2 ] 2 [O 2 ]

14 What is the value of the equilibrium constant, K c, for case 1? K c is unitless, so be sure the units on all your numbers match! K c = (0.056) 2 = 0.15 (0.344) 2 (0.172)

15 What is the value of the equilibrium constant, K c, for case 2? What do you notice? K c = (0.076) 2 = 0.15 (0.424) 2 (0.212)

16 Did you notice for both cases we used the forward reaction? If we used the reverse reaction, we would have gotten a different K, but it would still be related to the forward K, it is just the inverse. So K forward = 1/K reverse or K reverse = 1/K forward Let’s try it…

17 Example case 2: in a 1 L vessel R 2SO 3 ⇌ 2SO 2 + O 2 I0.500 mol 0 0 C -0.424 mol +0.424 mol +0.212 mol E0.076 mol 0.424 mol 0.212 mol What is K c and what is 1/K c for this reaction?

18 One other thing, if perhaps many equilibrium reactions are involved in a mechanism, and you are given the K for each step, the total can be found as… K total = K 1 K 2 K 3 …

19 In general for aA + bB ⇌ cC + dD How would this change if the number of products or reactants was different? Remember, the assumption is this is the only step in the mechanism or is the slow step, otherwise the coefficients don’t necessarily match the exponents. K c = [C] c [D] d [A] a [B] b

20 So K c is a measure of how much the reaction has occurred (the activity of the reaction). The bigger the K c the more products and less reactants. Explain. What would be K c for no reaction? What would be K c for a strong acid/base?

21 Practice: in a 5 L vessel RN 2 + 3H 2 ⇌ 2NH 3 I C E3.01 mol 2.10 mol 0.565 mol Find K c

22 Practice: in a 2 L vessel R2N 2 O ⇌ 2N 2 + O 2 I10.0 mol 0 0 C E2.20 mol Find K c

23 In the previous example the reactant and products were gases, but we used Molarity anyway. Sometimes when all the substances are gases it is easier to use their partial pressures instead of concentrations, so… Why call it K p ? What would the units be? K p = (P C ) c (P D ) d (P A ) a (P B ) b

24 K p is not the same as K c, because [] is only dependent on the moles and volume where P is also dependent on temperature and the gas constant, but if these are accounted for, then… K p = K c (RT) Δn where Δn = n gas products - n gas reactants When would K p = K c ?

25 What if we don’t know if a reaction is at equilibrium yet? We then determine the reaction quotient, Q if Q < K c, then it is not yet at equilibrium forward reaction will still be faster if Q > K c, then it is past equilibrium reverse reaction will now be faster if Q = K c, then it is at equilibrium Q = [C] c [D] d [A] a [B] b

26 Practice: For 2HI ⇌ H 2 + I 2 the K c = 65.0 if E 0.500M 0.280 M 3.40 M is this at equilibrium?

27 Typically K c is used to find the concentration of reactants and products at equilibrium. For example, what is the equilibrium concentrations if K c = 49.0 R A + B ⇌ C + D I0.200 M 0.200 M 0 0

28 What is the equilibrium concentrations if K c = 49.0 R A + B ⇌ C + D I0.300 M 0.100 M 0 0

29 Changes to equilibrium: if [] or pressure/volume changes or a catalyst is added, then Q changes, but K c will not change, so the reaction rates will change to get Q = K c again (LeChatelier’s can tell you how it will change, shifting right or left means speeding up forward or reverse reaction) if T changes then K c will change, so Q = K c will happen, but it will be a new K c

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32 Let’s practice using LeChatelier’s Principle If 2A(g) + 3B(g) ⇌ C(g) + 2D(g) Then for each change, state how it will shift (forward or reverse) and why it will shift (Q less than K or Q greater than K)

33 If 2A(g) + 3B(g) ⇌ C(g) + 2D(g) Pressure is increased: What affect does that have? How will the equilibrium shift? Why will the equilibrium shift?

34 If 2A(g) + 3B(g) ⇌ C(g) + 2D(g) Pressure is decreased: What affect does that have? How will the equilibrium shift? Why will the equilibrium shift?

35 If 2A(g) + 3B(g) ⇌ C(g) + 2D(g) Volume is increased: What affect does that have? How will the equilibrium shift? Why will the equilibrium shift?

36 If 2A(g) + 3B(g) ⇌ C(g) + 2D(g) Volume is decreased: What affect does that have? How will the equilibrium shift? Why will the equilibrium shift?

37 Explain why pressure and volume can only affect K p and not K c. What change would affect K c ?

38 If 2A(g) + 3B(g) ⇌ C(g) + 2D(g) [A] is increased: What affect does that have? How will the equilibrium shift? Why will the equilibrium shift?

39 If 2A(g) + 3B(g) ⇌ C(g) + 2D(g) [A] is decreased: What affect does that have? How will the equilibrium shift? Why will the equilibrium shift?

40 If 2A(g) + 3B(g) ⇌ C(g) + 2D(g) [D] is increased: What affect does that have? How will the equilibrium shift? Why will the equilibrium shift?

41 If 2A(g) + 3B(g) ⇌ C(g) + 2D(g) [D] is decreased: What affect does that have? How will the equilibrium shift? Why will the equilibrium shift?

42 Why would catalysts not change the equilibrium? Why would the addition of inert gases not change the equilibrium? What affect does changing temperature have on equilibrium? What can we do with LeChatelier's principle?

43 If 2A(g) + 3B(g) ⇌ C(g) + 2D(g) + Heat Temperature is increased: What affect does that have? How will the equilibrium shift? Why will the equilibrium shift?

44 If 2A(g) + 3B(g) ⇌ C(g) + 2D(g) + Heat Temperature is decreased: What affect does that have? How will the equilibrium shift? Why will the equilibrium shift?

45 If 2A(g) + 3B(g) + Heat ⇌ C(g) + 2D(g) Temperature is increased: What affect does that have? How will the equilibrium shift? Why will the equilibrium shift?

46 If 2A(g) + 3B(g) + Heat ⇌ C(g) + 2D(g) Temperature is decreased: What affect does that have? How will the equilibrium shift? Why will the equilibrium shift?

47 If 2A(g) + 3B(g) ⇌ C(g) + 2D(g) ∆H = + 76 kJ/mol Temperature is decreased: What affect does that have? How will the equilibrium shift? Why will the equilibrium shift?

48 If 2A(g) + 3B(g) ⇌ C(g) + 2D(g) ∆H = - 76 kJ/mol Temperature is increased: What affect does that have? How will the equilibrium shift? Why will the equilibrium shift?

49 Practice: in a 1 L vessel, H 2 + I 2 ⇌ 2HI at equilibrium [HI] = 0.490 M, [H 2 ] = 0.080 M and [I 2 ] = 0.060 M then 0.300 more moles of HI is added, what will be the new equilibrium concentrations?

50 First, find the K c

51 Next, find the new [] with the addition and find Q (we are sure we are not at equilibrium!)

52 Then, we set us a RICE Table with the new initial [] and the change that Q vs K c told us would happen, finally solve for x

53 There are other things we can do with K: 1) Remember ΔG 0 rxn ? ΔG 0 rxn is the energy change as ALL reactants react to produce ONLY products. In an equilibrium we still have some reactants left over, so this can be accounted for… ΔG 0 rxn = -RTlnK (can be K c or K p or Q) Why would ΔG 0 rxn be less if there are still some reactants left over?

54 Another form, on the equation pages: ∆G = ∆G° + RTlnQ Why Q, not K? Notice! For both of these equations R is used in Joules but G is typically measured in kiloJoules. Be careful that the units match!

55 There are other things we can do with K: 2)The Nernst equation should actually look like this: E cell = E° cell - lnQ Why Q, and not K? R would be in volt coulombs mol -1 k -1 ! RT nF

56 There are other things we can do with K: 3) As mentioned earlier, K varies with Temperature, but thanks to van’t Hoff’s equation we can calculate the K at a new temperature if ΔH 0 is known. ln ( K T2 ) = ΔH0ΔH0 ( 1 - 1 ) K T1 RT1T1 T2T2


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