1. Chemical Equilibrium Chemistry 100

2. The concept A condition of balance between opposing physical forces A state in which the influences or processes to which a thing is subject cancel one another and produce no overall change Oxford English Dictionary

3. Static and Dynamic A book sitting on a desk is in static equilibrium;  The book remains at rest; its position is constant. The moon circles the earth.  There is movement but the (average) distance between the two is unaltered. This is dynamic equilibrium.

4. Equilibrium The molecules of A are able to turn into molecules of B The rate at which this happens is proportional to [A]. Rate for = k for [A] Likewise, if B can turn into A, then Rate rev = k rev [B]

5. The Equilibrium Condition Start with pure A.  [A] decreases and [B] increases as A turns into B What happens to the rate at which A turns into B, and the rate at which B turns into A?  The rate of A  B decreases, while B  A increases What eventually happens?  Rate of A  B = Rate B  A Rate for = Rate rev  k for [A] = k rev [B]

6. The Equilibrium Condition #2

7. And then what? We have as an equilibrium condition k for [A] = k rev [B] K eq  the thermodynamic equilibrium constant

8. The Meaning of [B]/[A] = K K is a constant number such as 2.3, 0.65, etc What the equilibrium expression means is: No matter how much A or B we start with, when the system reaches equilibrium

9. Reversible reactions If these two reactions are possible A  B and B  A,  we have a reversible reaction A ⇌ B Here is a real reversible reaction N 2 (g) +3H 2 (g) ⇌ 2NH 3 (g)

10. Equilibrium can be reached from either side At start P H 2 = 3; P N 2 = 1; P NH 3 = 0 At start P H 2 = 0; P N 2 = 0; P NH 3 =2

11. Law of Mass Action Expression for K For the reaction aA (g) + bB (g) ⇌ pP (g) + qQ (g)

12. Examples of K eq N 2 (g) +3H 2 (g) ⇌ 2NH 3 (g) Br 2 (g) +Cl 2 (g) ⇌ 2BrCl(g) SO 2 (g) +½O 2 (g) ⇌ SO 3 (g)

13. Magnitude of K eq 2 HI(g) ⇌ H 2 (g) + I 2 (g)K eq = 0.016 The magnitude (size) of K eq provides information  K >> 1 the products are favoured  K << 1 the reactants are favoured CO(g) + Cl 2 (g) ⇌ COCl 2 (g) K eq = 4.57  10 9  Equilibrium lies far to the right - there is very little CO and Cl 2 in the equilibrium mixture.

14. Heterogeneous Equilibrium When the substances in the reaction are in the same phase (e.g., all gases), reactions are termed homogeneous equilibria. When different phases are present, we speak of heterogeneous equilibrium. We will look at reactions involving gases and solids, and gases and liquids

16. Solids do not appear in K eq Examine the reaction CaCO 3 (s)  CaO(s) + CO 2 (g) For a pure solid X (or liquid X) [X] = density/molar mass. Note  molar mass and density are intensive properties!!  [X] = constant

17. Heterogeneous Equilibrium At a given temperature, the equilibrium between CaCO 3 (s), CaO(s), and CO 2 (g) yields the same concentration (same partial pressure) of CO 2 (g). True as long as all three components are present. Note that it does not matter how much of the two solids are present; we just need some.

18. More heterogeneous equilibria CO 2 (g) + H 2 (g) ⇄ CO(g) + H 2 O(l) SnO 2 (s) + 2CO(g) ⇄ Sn(s) + 2CO 2 (g)

19. K eq values for forward and reverse reactions For the reaction 2 HI(g) ⇌ H 2 (g) + I 2 (g), K eq = 0.016 What is K eq for H 2 (g) + I 2 (g) ⇌ 2HI(g) ? Call the first reaction (F) and the second (R)

20. Forward and Reverse (II)

21. An aside For the equilibrium H 2 O(l) ⇌ H 2 O(g), write down the expression for K eq. When liquid water and water vapour are in equilibrium the vapour has a fixed pressure at a given temperature!

22. Applications For a given reaction, K eq has a set value for a given temperature Q depends on the experimental conditions Q = K eq at equilibrium

23. Summary of the Q and K eq story When Q > K eq reaction shifts left  When Q = K eq equilibrium When Q < K eq reaction shifts right 

24. Applications (II) Equilibrium is approachable from either side of the reaction.

25. Other Applications Obtaining the equilibrium constant from the measured equilibrium concentrations Calculating the composition of the equilibrium system (i.e., concentration of products and reactants at equilibrium)  have the concentration of all but one component at equilibrium and the value of K eq  given initial amounts of reactants and the equilibrium constant

26. Le Châtelier’s Principle Perturb a system at equilibrium  Change in temperature, pressure, or the concentration of a component The system will shift its equilibrium position so as to counteract the disturbance. The effect of the last two disturbances (change in pressure and change in concentration) can also be be predicted by the Law of Mass Action

27. Changing concentration (I) Examine the system 2 NO 2 Cl(g) ⇌ 2 NO 2 (g) + Cl 2 (g) Introduce a small amount of substance X that reacts with Cl 2 to make XCl.  The value of P Cl 2 has been decreased Le Châtelier’s Principle predicts that more NO 2 Cl will react to increase P Cl 2

28. Changing concentration (II) At equilibrium Remove Cl 2 - the new value of P Cl 2 is 0.05 atm Q is now smaller than K eq. The reaction moves to the right to increase Q eq. Same prediction!

29. Changing concentration (III) CaCO 3 (s) ⇌ CaO(s) + CO 2 (g) If we have this system in equilibrium and add either CaCO 3 (s) or CaO(s), there will be no effect on the equilibrium.

30. Changing the concentration

31. Changing pressure (I) 2 NO 2 Cl(g) ⇌ 2 NO 2 (g) + Cl 2 (g) Increase the pressure in the system by making the vessel smaller. Note that there are a total of 3 moles on the right of the reaction and 2 on the left. The left “takes up less space” Le Châtelier’s Principle predicts that the species on the left will react to form more NO 2 Cl.

32. Changing pressure (II) 2 NO 2 Cl (g)  2 NO 2 (g) + Cl 2 (g) System is initially at equilibrium. Increase the pressure by making the vessel smaller. We could use the K eq expression to predict what happens but Le Châtelier’s Principle is much easier to use!

33. Cautionary Note Le Châtelier’s Principle predicts what occurs when we change the partial pressure of one or more of the species in the reaction Change the total pressure by adding/removing an inert gas (not involved in the reaction) NO EFFECT ON THE EQUILIBRIUM

34. More pressure changes Predict what happens N 2 (g) + 3H 2 (g) ⇌ 2NH 3 (g); total pressure is decreased Reaction shifts to left; more moles on left H 2 (g) + I 2 (g) ⇌ 2HI(g); total pressure is increased No effect; 2 moles on each side

35. What is Heat - not a substance! Some textbooks  Heat is treated as a chemical reagent when applying Le Châtelier’s Principle to change of temperature problems. Treating heat as a substance can lead to confusion. There is a better way!

36. Changing the temperature An exothermic reaction causes an increase in temperature. The reverse causes cooling. Warm up an exothermic reaction Le Châtelier’s Principle predicts the system will move in the direction that will bring the temperature back down  The direction that cools. So the reverse reaction (  ) occurs

37. Changing Temperature N 2 O 4 (g)  2 NO 2 (g)  H = 58.0 kJ The forward reaction is endothermic  Absorbs heat. Decrease the temperature - reaction shifts to the left  Brings T back up. If we increase the temperature, opposite effect  Reaction takes in heat and lower the temperature

38. Temperature and K eq Endothermic reactions – increasing temperature increases the value of the equilibrium constant! Exothermic reactions – increasing temperature decreases the value of the equilibrium constant! Temperature changes are the only stresses on the systems that change the numerical values of K eq

39. Temperature and K eq (II) Co(H 2 O) 6 2+ (aq)+ 4 Cl - (aq) ⇌ CoCl 4 2- (aq) + 6 H 2 O (l) ∆H  > 0

40. Catalyst does NOT change K A catalyst speeds up a reaction by providing and alternate reaction pathway with a lower E a. Reversible reaction  the forward and backward reactions have their E a ’s changed by the same amount.  K eq is not altered. A catalyst cannot alter K!! Otherwise we would be able to build a perpetual motion machine!!