1. Equilibrium Constants

2. Recall: At equilibrium, the rate of the forward and reverse reactions are equal Equilibrium

3. The Equilibrium Constant, K eq/ K c For the reaction: aA + bB cC + dD At equilibrium: r fwd = r rev k fwd [A] a [B] b = k rev [C] c [D] d k fwd = [C] c [D] d k rev [A] a [B] b K eq = [C] c [D] d [A] a [B] b Sub in rate law equation Rearrange Sub in K eq for k fwd /k rev Equilibrium constant

4. Examples Write the equilibrium constant expression for the following two reactions

5. 5 The Equilibrium Constant The Magnitude of Equilibrium Constants An equilibrium can be approached from any direction. Example:

6. 6 The Equilibrium Constant The Magnitude of Equilibrium Constants However, The equilibrium constant for a reaction in one direction is the reciprocal of the equilibrium constant of the reaction in the opposite direction.

7. Example: Simple K eq Calculation Calculate the value of K eq for the following system At Equilibrium: [CO 2 ] = 0.0954 mol/L [H 2 ] = 0.0454 mol/L [CO] = [H 2 O] = 0.00460 mol/L

8. 8 Calculating Equilibrium Constants Steps to Solving Problems: 1.Write an equilibrium expression for the balanced reaction. 2.Write an ICE table. Fill in the given amounts. 3.Use stoichiometry (mole ratios) on the change in concentration line. 4.Deduce the equilibrium concentrations of all species. Usually, the initial concentration of products is zero. (This is not always the case.)

9. Equilibrium Calculations A closed system initially containing 1.000 x 10 −3 M H 2 and 2.000 x 10 −3 M I 2 At 448  C is allowed to reach equilibrium. Analysis of the equilibrium mixture shows that the concentration of HI is 1.87 x 10 −3 M. Calculate K c at 448  C for the reaction taking place, which is H 2 (g) + I 2 (g) 2 HI (g)

10. What Do We Know? [H 2 ], M[I 2 ], M[HI], M Initially1.000 x 10 -3 2.000 x 10 -3 0 Change At equilibrium 1.87 x 10 -3

11. [HI] Increases by 1.87 x 10 -3 M [H 2 ], M[I 2 ], M[HI], M Initially1.000 x 10 -3 2.000 x 10 -3 0 Change+1.87 x 10 -3 At equilibrium 1.87 x 10 -3

12. Stoichiometry tells us [H 2 ] and [I 2 ] decrease by half as much [H 2 ], M[I 2 ], M[HI], M Initially1.000 x 10 -3 2.000 x 10 -3 0 Change-9.35 x 10 -4 +1.87 x 10 -3 At equilibrium 1.87 x 10 -3

13. We can now calculate the equilibrium concentrations of all three compounds… [H 2 ], M[I 2 ], M[HI], M Initially1.000 x 10 -3 2.000 x 10 -3 0 Change-9.35 x 10 -4 +1.87 x 10 -3 At equilibrium 6.5 x 10 -5 1.065 x 10 -3 1.87 x 10 -3

14. …and, therefore, the equilibrium constant Kc =Kc = [HI] 2 [H 2 ] [I 2 ] = 51 = (1.87 x 10 -3 ) 2 (6.5 x 10 -5 )(1.065 x 10 -3 )

15. At 1280 0 C the equilibrium constant (K c ) for the reaction Is 1.1 x 10 -3. If the initial concentrations are [Br 2 ] = 0.063 M and [Br] = 0.012 M, calculate the concentrations of these species at equilibrium. Br 2 (g) 2Br (g) Let x be the change in concentration of Br 2 Initial (M) Change (M) Equilibrium (M) 0.0630.012 +x+x-2x 0.063 + x0.012 - 2x [Br] 2 [Br 2 ] K c = (0.012 - 2x) 2 0.063 + x = 1.1 x 10 -3 Solve for x 14.4

16. K c = (0.012 + 2x) 2 0.063 - x = 1.1 x 10 -3 4x 2 + 0.048x + 0.000144 = 0.0000693 – 0.0011x 4x 2 + 0.0491x + 0.0000747 = 0 ax 2 + bx + c =0 -b ± b 2 – 4ac  2a2a x = Br 2 (g) 2Br (g) Initial (M) Change (M) Equilibrium (M) 0.0630.012 -x-x+2x 0.063 - x0.012 + 2x x = -0.00178x = -0.0105 At equilibrium, [Br] = 0.012 + 2x = -0.009 Mor 0.00844 M At equilibrium, [Br 2 ] = 0.062 – x = 0.0648 M 14.4

17. Approximating If Keq is really small the reaction will not proceed to the right very far, meaning the equilibrium concentrations will be nearly the same as the initial concentrations of your reactants. 0.20 – x is just about 0.20 as x is really, really tiny. If the difference between Keq and initial concentrations is around 3 orders of magnitude or more, go for it. Otherwise, you have to use the quadratic.

18. Example Initial Concentration of I 2 : 0.50 mol/2.5L = 0.20 M I 2 2 I Initial change equil: 0.20 0 -x +2x 0.20-x 2x With an equilibrium constant that small, whatever x is, it’s near dink, and 0.20 minus dink is 0.20 (like a million dollars minus a nickel is still a million dollars). 0.20 – x is the same as 0.20 x = 3.83 x 10 -6 M More than 3 orders of mag. between these numbers. The simplification will work here.

19. Example Initial Concentration of I 2 : 0.50 mol/2.5L = 0.20 M I 2 2 I Initial change equil: 0.20 0 -x +2x 0.20-x 2x These are too close to each other... 0.20-x will not be trivially close to 0.20 here. Looks like this one has to proceed through the quadratic...

20. The Equilibrium Constant… food for thought Because pressure is proportional to concentration for gases in a closed system, the equilibrium expression can also be written K p = (P C ) c (P D ) d (P A ) a (P B ) b Relationship between K c and K p  From the ideal gas law we know that Rearranging it, we get PV = nRT P = RT nVnV

21. Relationship between K c and K p Plugging this into the expression for K p for each substance, the relationship between K c and K p becomes Where K p = K c (RT)  n  n = (moles of gaseous product) − (moles of gaseous reactant)

22. The equilibrium concentrations for the reaction between carbon monoxide and molecular chlorine to form COCl 2 (g) at 74 0 C are [CO] = 0.012 M, [Cl 2 ] = 0.054 M, and [COCl 2 ] = 0.14 M. Calculate the equilibrium constants K c and K p. CO (g) + Cl 2 (g) COCl 2 (g) Kc =Kc = [COCl 2 ] [CO][Cl 2 ] = 0.14 0.012 x 0.054 = 220 K p = K c (RT)  n  n = 1 – 2 = -1 R = 0.0821T = 273 + 74 = 347 K K p = 220 x (0.0821 x 347) -1 = 7.7 14.2

23. 23 The Equilibrium Constant Heterogeneous Equilibria When all reactants and products are in one phase, the equilibrium is homogeneous. If one or more reactants or products are in a different phase, the equilibrium is heterogeneous. Consider: –experimentally, the amount of CO 2 does not seem to depend on the amounts of CaO and CaCO 3. Why?

24. 24 Heterogeneous Equilibria Neither density nor molar mass is a variable, the concentrations of solids and pure liquids are constant. (You can’t find the concentration of something that isn’t a solution!) We ignore the concentrations of pure liquids and pure solids in equilibrium constant expressions. The amount of CO 2 formed will not depend greatly on the amounts of CaO and CaCO 3 present.

25. Magnitude of K c K c >> 1 At equilibrium there is more products than reactants. The reaction is product favoured K c = 1At equilibrium there is an equal amount of products and reactants K c << 1At equilibrium there is more reactants than products. The reaction is reactant favoured K c = [C] c [D] d [A] a [B] b

26. The Reaction Quotient (Q)  To calculate Q, one substitutes the initial concentrations on reactants and products into the equilibrium expression.  Q gives the same ratio the equilibrium expression gives, but for a system that is not at equilibrium.  We define Q, the reaction quotient, for a reaction at conditions NOT at equilibrium  where [A], [B], [P], and [Q] are molarities at any time.  Q = K only at equilibrium.

27. If Q = K, the system is at equilibrium.

28. If Q > K, there is too much product and the equilibrium shifts to the left.

29. If Q < K, there is too much reactant, and the equilibrium shifts to the right.

30. Example (p. 464) In a container at 450°C, N 2 and H 2 react to produce NH 3. K = 0.064. When the system is analysed, [N 2 ] = 4.0 mol/L, [H 2 ] = 2.0 X 10 -2 mol/L, and [NH 3 ] = 2.2 X 10 -4 mol/L. Is the system at equilibrium, if not, predict the direction in which the reaction will proceed.