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 P B /P 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  Obtaining the equilibrium constant from the measured equilibrium concentrations  Calculating the composition of the equilibrium system 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

23. Applications (II)  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

24. 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 

25. Applications (III)  Equilibrium is approachable from either side of the reaction.

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 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!!