1. UNIT 3 CHEMICAL EQUILIBRIUM

2. Introduction to Chemical Equilibrium  Focus has always been placed upon chemical reactions which are proceeding in one direction (forward)  We assume that chem. rxn. proceed until one of the reactants is entirely consumed  However many rxns are reversible, i.e. they can occur in either direction  Products regenerate the original reactants

3. For a given forward reaction aA + bB ---> cC + dD One can conceive of the reverse process cC + dD ---> aA + bB  product molecules (C & D) are not present at the beginning of a reaction, thus reverse reaction is non-existent  as [products]  the reverse rxn. is initiated

4. For a given forward reaction aA + bB ---> cC + dD  the rate of the forward rxn.  b/c [reactant] is reduced by conversion to products & the rate of the reverse reaction  (as product concentrations build up)  the rate of  of the reverse rxn. will be proportional to some product of the concentrations of C and D

5. For a given forward reaction aA + bB ---> cC + dD  eventually a point in time is reached when the Rate Forward Rxn. = Rate Reverse Rxn.  This is the point of CHEMICAL EQUILIBRIUM  unless the system is disturbed by a temp. change or by adding excess reactant or product, the equality of RATES is maintained  CONCENTRATIONS of all chemical species reach "equilibrium" (i.e. []’s remain constant, but NOT the same)

6. The variation of reaction rates with time is illustrated graphically as follows:

7. Chemical Equilibrium = Dynamic Equil.  once equilibrium is reached, the rxn. seems to be stopped  we cannot visibly see any changes  in reality the reactions are still taking place; chem. equil. is a dynamic or changing process  the observable properties and [] of reactants and products are constant

8. A chemical system is said to be in a state of equilibrium if it meets the following criteria: 1. The system is closed. 2. The observable macroscopic properties are constant. 3. The reaction is sufficiently reversible so that observable properties change and then return to the original rate when a factor that affects the rate of the reaction is varied and then restored to its original value.

9. The Equilibrium Constant Consider the forward and reverse processes for some general "atom exchange" reaction aA + bB ---> cC + dD a, b, c & d = coefficients for substances A, B, C & D respectively

10. Assuming the forward rxn. occurs by a one-step mechanism, then… Rate Forward Rxn. = R f = k f [A][B] Similarly, assuming the reverse rxn. is also a one-step mechanism, then Rate Reverse Rxn. = R r = k r [C][D] "Principle of Microscopic reversibility”  activated complex in the reverse rxn. is identical to the activated complex for the forward rxn. Thus, the energy profile for the forward and reverse rxns can be illustrated by the same plot:

11. E f = activation energy of forward rxn. E r = activation energy of reverse rxn. The activated complex is identical for both directions, thus ∆H = E r - E f NOTE: If ∆H = 0, then E r = E f

12. At equilibrium R f = R r Thus, k f [A][B] = k r [C][D] By rearrangement we can obtain the expression [C] c [D] d = k f = K eq (equil. constant) [A] a [B] b k r  K eq describes the relative []’s of reactants & products @ equilibrium  K eq ratio is always a constant value for a given rxn. regardless of []’s of reactants and products @ equil.  Every reversible rxn. obeys this relationship & has a specific K eq

13. “Law of chemical equilibrium”  every reversible rxn. proceeds to a state of equil. & has a specific ratio of the [reactant] : [products] expressed by K eq Examples: Write the equil. expressions for the following: 1. 2CO (g) + O 2(g) 2CO 2(g) K eq = [ CO 2 ] 2. [CO] 2 [O 2 ] 2. CO (g) + 3H 2(g) CH 4(g) + H 2 O (g) K eq = [ CH 4 ][ H 2 O ]. [CO][H 2 ] 3

14. Examples: 3. At 100°C 1.0L of gas at equilibrium contains 0.0045 mol N 2 O 4 and 0.030 mol NO 2. Calculate the K eq. 2NO 2 N 2 O 4 K eq = [ N 2 O 4 ] = (0.0045) = 5.0 [ NO 2 ] 2 (0.030) 2 NOTE: K eq has no units

15. What is indicated by the magnitude of the K eq value?  K eq does NOT tell you anything about the time it takes to reach equil.  K eq tells you the extent to which a rxn. proceeds to completion

16. K eq >> 1 (“>>” means significantly)  numerator (products) must be much larger than denominator (reactants)  @ equilibrium this rxn. system consists mainly of products and is considered to proceed to completion “Equilibrium lies to the right” (toward product side)

17. K eq << 1  a very small value, as the denominator (reactants) is much larger than the numerator (products)  system mostly reactants  reaction barely gets going “Equilibrium lies to the left” (toward reactant side)

18. In general… K eq > 1products favored @ equilibrium (spontaneous rxn.) K eq < 1reactants favored @ equilibrium Assignment  Pg. 419 – 420, # 11 – 15, Pg. 425 – 426, # 43 – 46, # 48 – 50

19. Calculation of Concentrations at Equilibrium Use “ICE” to help you find the concentrations of each substance at equilibrium. I nitial C hange E quilibrium

20. Examples: 1. When 0.40 moles of PCl 5 is heated in a 10.0 L container, an equilibrium is established in which 0.25 moles of Cl 2 is present. a) What is the number of moles of PCl 5 and PCl 3 at equilibrium? b) What are the equilibrium concentration of all three components?

21. a) PCl 5 PCl 3 + Cl 2 Initial 0.40 mol ------ ------- Change  0.25 mol +0.25 mol +0.25 mol Equil. 0.15 mol 0.25 mol 0.25 mol b) [PCl 5 ] = 0.15 moles = 0.015 moles/L 10.0 L [PCl 3 ] = [Cl 2 ] = 0.25 moles = 0.025 moles/L 10.0 L

22. Example: 2. The equilibrium constant for the reaction represented below is 50 at 448°C H 2(g) + I 2(g) 2 HI (g) a) How many moles of HI are present at equilibrium when 1.0 moles of H 2 is mixed with 1.0 moles of I 2 in a 0.50 L container and allowed to react at 448°C? b) How many moles of H 2 and I 2 are left unreacted? c) If the conversion of H 2 and I 2 to HI is essentially complete, how many moles of HI would be present? d) What is the percent yield of the equilibrium mixture?

23. a) The concentrations at the start are: [H 2 ] = 1.0 mol/0.50 L = 2.0 mol/L [I 2 ] = 1.0 mol/0.50 L = 2.0 mol/L [HI] = 0 moles therefore 0 mol/L

24. Fill in the numbers under the reaction for the starting concentrations. Let 'x' = number of moles of H 2 consumed per litre H 2 + I 2 2 HI I 2.0 M 2.0 M 0 M C  x  x 2(+x) E 2.0  x 2.0  x 2x

25. Fill in the finish concentrations into the Keq equation. and solve for 'x‘. 7.1 = 2x. 2.0 - x 50 = (2x) 2. 14.2 - 7.1x = 2x (2.0 - x)(2.0 - x)x = 1.56 mol/L 50 = (2x) 2. (2.0 - x) 2 You are solving for 'x' which is in moles/L (M) therefore you can automatically fill in the units for any value of 'x'.

26. Now to answer the question. What are the final reactant and product concentrations? [H 2 ] = 2.0 M  x = 2.0 M  1.56 M = 0.44 M [I 2 ] = 2.0 M  x = 2.0 M  1.56 M = 0.44 M [HI] = 2x = 2 1.56 M = 3.12 M

27. b) From part a) you get the M of H 2 and I 2 left unreacted. This is the concentration not the answer to the question. You are asked for the number of moles left unreacted therefore moles = concentration volume [H 2 ] = [I 2 ] = 0.44 mol/L 0.5 L= 0.22 moles Therefore 0.22 moles of H 2 and 0.22 moles of I 2 are left unreacted.

28. c) If you look at the equation again you can see that there is exactly the right amount of H 2 to react with I 2. Theoretically if both the H 2 and I 2 where all used up then we should make 2 moles of HI. This is the theoretical yield because it is what could be made in theory if everything went to completion.

29. d) The percentage yield is calculated by using the actual yield from part a) and the theoretical yield from part b). You may use either moles or moles/L but make sure that they are both the same. Percentage Yield = Actual Yield 100 Theoretical Yield = 1.56 M 100 2.00 M = 78% yield (note that the moles/L units cancel)

30. Example: (Challenging) 3. How many moles of HI are present at equilibrium when 2.0 moles of H 2 is mixed with 1.0 moles of I 2 in a 0.50 L container and allowed to react at 448°C. At this temperature K eq = 50. H 2(g) + I 2(g) 2 HI (g) The concentrations at the start are: [H 2 ] = 2.0 mol/0.50 L = 4.0 mol/L [I 2 ] = 1.0 mol/0.50 L = 2.0 mol/L [HI] = 0 moles therefore 0 mol/L

31. Let 'x' = number of moles of H 2 consumed per litre H 2 + I 2 2 HI I 4.0 M 2.0 M 0 M C  x  x 2(+x) E 4.0  x 2.0  x 2x

32. 50 = (2x) 2. (4.0 - x)(2.0 - x) 50 = 4x 2. 8.0 - 6.0x + x 2 400 - 300x + 50x 2 = 4x 2 46x 2 - 300x + 400 = 0 (standard form) Use the quadratic equation to solve for 'x'.

33. In this case a = +46, b = -300 and c = +400. We only started with 4 mols/L of H 2 to begin with, so 4.7 is to large. It is therefore the unreal root. So let x = 1.9 mol/L.

34. Substitute 'x =1.9 mol/L' back up into the original concentration calculations to see how much is left after reaction. [H 2 ] = 4.0 M  x = 4.0 M  1.9 M = 2.1 M [I 2 ] = 2.0 M  x = 2.0 M  1.9 M = 0.1 M [HI] = 2x = 2 1.9 M = 3.8 M The moles of HI present at equilibrium would be: moles = concentration volume = 3.8 moles/L 0.5 L = 1.9 moles

35. Homogenous & Heterogeneous Equilibria Homogenous Equilibria  equilibrium conditions for rxns in which all the reactants & products are in the same state eg. 2CO (g) + O 2(g) 2CO 2(g) Heterogenous Equilibria  equilibrium conditions for rxns that involve substances in more than 1 state eg. NH 4 Cl (s) NH 3(g) + HCl (g)

36.  Earlier, we expressed [] as M (for gases)  [] of solids & liquids in a chem. rxn. do not change substantially and so are not included in the equilibrium constant [solid & liquid]  change during a rxn, left out of K eq  eg. C (s) + H 2 O (g) CO (g) + H 2(g) K eq = [CO][H 2 ]. [H 2 O]

37. The Reaction Quotient  It is hard to tell if a rxn. has reached equilibrium  The “reaction quotient” ( Q ) is used to determine if a rxn. is @ equilibrium  Q is calculated like K eq except the []’s are taken @ the time of measurement, not necessarily @ equilibrium  Thus we can compare the value of Q to the value of K eq to see how the rxn. is progressing  Rxn. is @ equilibrium when K eq = Q

38. Example: At 473°C, nitrogen gas is reacted with hydrogen gas to produce ammonia. A measurement was taken as the reaction proceeded and it was found that there was 0.15 M of nitrogen gas, 0.0020 M of hydrogen gas, and 0.15 M of ammonia. Given that the K eq is 0.105, is the reaction at equilibrium?

39. N 2(g) + 3H 2(g) 2NH 3(g) temp. = 473°C[NH 3 ] = 0.15 M K eq = 0.105[N 2 ] = 0.15 M [H 2 ] = 0.0020 M K eq = [NH 3 ] 2 Q = (0.15) 2. [N 2 ][H 2 ] 3 (0.15)(0.0020) 3 = 1.9 x 10 7 K eq  Q this rxn. is NOT @ equilibrium

40. Which direction will rxns proceed to reach equilibrium? Q < K eq Q is less than K eq  denominator is too large and numerator is too small  too much of reactants, too little of products  rxn. will proceed to the right (in the direction of products)

41. Q > K eq Q is greater than K eq  denominator is too small and numerator is too large  too much products, too little reactants  rxn. will proceed to the left (in the direction of reactants)

42. Example: COCl 2(g) CO (g) + Cl 2(g) K eq = 170 At the time of measurement: [CO] = 0.15 M [Cl 2 ] = 0.15 M [COCl 2 ] = 1.1 x 10  3 M Is the rxn. @ equilibrium? If not, which direction will it proceed?

43. Q = [CO][Cl 2 ] = (0.15)(0.15). = 20. [COCl 2 ] (1.1 x 10  3 ) Q < K eq thus rxn. is NOT @ equilibrium  there is too much reactants  rxn. will shift right in the direction of products