Chemical Equilibrium

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]. Ratefor = kfor[A] • Likewise, if B can turn into A, then Raterev = krev[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 Ratefor = Raterev  kfor[A] = krev[B]

6. The Equilibrium Condition #2

7. We have as an equilibrium condition kfor[A] = krev[B] And then what? Keq the thermodynamic equilibrium constant

8. The Meaning of PB/PA= 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 N2(g) +3H2(g) ⇌ 2NH3(g)

10. Equilibrium can be reached from either side At start PH2 = 3; PN2 = 1; PNH3 = 0 At start PH2 = 0; PN2 = 0; PNH3 =2

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

12. Examples of Keq N2(g) +3H2(g) ⇌2NH3(g) Br2(g) +Cl2(g) ⇌ 2BrCl(g) SO2(g) +½O2(g) ⇌ SO3(g)

13. Magnitude of Keq • 2 HI(g) ⇌H2(g) + I2(g) Keq = 0.016 • The magnitude (size) of Keq provides information • K >> 1 the products are favoured • K << 1 the reactants are favoured • CO(g) + Cl2(g) ⇌COCl2(g)Keq= 4.57109 • Equilibrium lies far to the right - there is very little CO and Cl2 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 Keq • Examine the reaction CaCO3(s) ⇌CaO(s) + CO2(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 CaCO3(s), CaO(s), and CO2(g) yields the same concentration (same partial pressure) of CO2(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 CO2(g) + H2(g) ⇄CO(g) + H2O(l) SnO2(s) + 2CO(g) ⇄ Sn(s) + 2CO2(g)

19. Keq values for forward and reverse reactions • For the reaction 2 HI(g) ⇌H2(g) + I2(g), Keq = 0.016 • What is Keq for H2(g) + I2(g) ⇌2HI(g) ? • Call the first reaction (F) and the second (R)

20. Forward and Reverse (II)

21. An aside • For the equilibrium H2O(l) ⇌ H2O(g), write down the expression for Keq. 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 Keq • given initial amounts of reactants and the equilibrium constant

23. Applications (II) • For a given reaction, Keq has a set value for a given temperature • Q depends on the experimental conditions • Q = Keq at equilibrium

24. Summary of the Q and Keq story • When Q > Keq • reaction shifts left  • When Q = Keq • equilibrium • When Q < Keq • reaction shifts right 

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

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 NO2Cl(g) ⇌ 2 NO2(g) + Cl2(g) • Introduce a small amount of substance X that reacts with Cl2 to make XCl. • The value of PCl2 has been decreased • Le Châtelier’s Principle predicts that more NO2Cl will react to increase PCl2

28. Changing concentration (II) At equilibrium Remove Cl2 - the new value of PCl2 is 0.05 atm Q is now smaller than Keq. The reaction moves to the right to increase Qeq. Same prediction!

29. Changing concentration (III) CaCO3(s) ⇌ CaO(s) + CO2(g) If we have this system in equilibrium and add either CaCO3(s) or CaO(s), there will be no effect on the equilibrium.

30. Changing the concentration

31. Changing pressure (I) 2 NO2Cl(g) ⇌ 2 NO2(g) + Cl2(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 NO2Cl.

32. Changing pressure (II) 2 NO2Cl (g)  2 NO2 (g) + Cl2 (g) • System is initially at equilibrium. • Increase the pressure by making the vessel smaller. • We could use the Keq 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 • N2(g) + 3H2(g) ⇌ 2NH3(g); total pressure is decreased Reaction shifts to left; more moles on left • H2(g) + I2(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 N2O4(g)  2 NO2(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 Keq • 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 Keq

39. Temperature and Keq (II) Co(H2O)62+(aq)+ 4 Cl- (aq) ⇌ CoCl42- (aq) + 6 H2O (l) ∆H> 0

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