1. Chapter 14 Chemical Equilibrium “ Old Chemists Never Die; they just reach EQUILIBRIUM! ” All physical and chemical changes TEND toward a state of equilibrium. A (l) A (g) A (s) A (l)

2. 09/05/09 Dynamic Equilibrium The net result of a dynamic equilibrium is that no change in the system is evident. Le Chatelier’s Principle - If a change is made in a system at equilibrium, the equilibrium will shift in such a way so as to reduce the effect of the change. Apply Pressure Pressure applied to the system at equilibrium caused it to shift until a new equilibrium was established.

3. 09/05/09 Dynamic Equilibrium Evaporation Open System (No Equilibrium) Evaporation Liquid Gas (No Equilibrium) (No Equilibrium) Liquid Gas Liquid Gas (Equilibrium)

4. 09/05/09 Dynamic Equilibrium Ag + + Cl - AgCl (s) Chemical Equilibrium Ag + Cl - Cl - Ag + AgCl (s) Rate of Precipitation = Rate of Dissolving HC 2 H 3 O 2 (aq) H + + C 2 H 3 O 2 - Rate of dissociation (ionization) = Rate of Association HC 2 H 3 O 2 H + C 2 H 3 O 2 - HC 2 H 3 O 2 C 2 H 3 O 2 - H +

5. 09/05/09 CHEM 1108 Lab Experiment HC 2 H 3 O 2 H + + C 2 H 3 O 2 - Red Orange [CoCl 4 ] 2- + 6 H 2 O (l) Co(H 2 O) 6 2+ + 4 Cl - Pink Blue NH 4 Cl (s) NH 4 + + Cl - White Colorless Solution You can actually “see” the equilibrium shift!

6. 09/05/09 Reversible Reactions N 2 O 4 (g) 2 NO 2 (g) R 1 2 NO 2 (g) N 2 O 4 (g) R 2 N 2 O 4 (g) 2 NO 2 (g) R 1 R 2 [R 1 = R 2 ] Homogeneous Equilibrium

7. 09/05/09 Reversible Reactions Exp. 1 0.0250 M 0.0 M 0.0202 M 0.009 66 M Exp. 2 0.0150 M 0.0125 M 0.0146 M 0.008 23 M Exp. 3 0.0 M 0.0250 M 0.0923 M 0.006 54 M [N 2 O 4 ] i [NO 2 ] i [N 2 O 4 ] eq [NO 2 ] eq N 2 O 4 (g) 2 NO 2 (g) Q C = [NO 2 ] 2 [N 2 O 4 ] Reaction Quotient K C = [NO 2 ] 2 eq [N 2 O 4 ] eq Equilibrium Constant

8. 09/05/09 Equilibrium Constants Equilibrium Constant - When the rates of the forward and reverse reactions are equal, the system is “at equil- ibrium” and the reaction quotient = equilibrium constant . Experiment 1 K C = [0.009 66] 2 /[0.0202] = 0.004 62 M Experiment 2 K C = [0.008 23] 2 /[0.0146] = 0.004 64 M Experiment 3 K C = [0.006 54] 2 /[0.009 23] = 0.004 63 M aA + bB cC + dD K C = [C] c [D] d [A] a [B] b

10. 09/05/09 Equilibrium Constants 4 NH 3 (g) + 3 O 2 (g) 2 N 2 (g) + 6 H 2 O (g) Q C = K C = [NH 3 ] 4 [O 2 ] 3 [N 2 ] 2 [H 2 O] 6 [N 2 ] eq 2 [H 2 O] 6 [NH 3 ] eq 4 [O 2 ] eq 3

11. 09/05/09 Reaction Quotient vs. Equilibrium Constant Class Problem 14.1 - The concentration of N 2 O 4 = concentration of NO 2 = 0.0125 M in a reaction vessel. The equilibrium constant for N 2 O 4 (g) = 2 NO 2 (g) is 0.004 63. Calculate Q C and state which direction the reaction will go. Class Problem 14.2 - If [O 2 ] = 0.21 M and [O 3 ] = 6.0 x 10 -8 M, what is the value of the K C for the equilibrium, 2 O 3 (g) = 3 O 2 (g) ?

12. 09/05/09 Chemical Equilibrium Heterogeneous Reaction - A reaction that takes place in more than one phase or state. These reactions occur at the interface between phases - on the surface of liquids and solids. At a constant temperature, the concentration of a solid or liquid component remains constant in a heterogeneous equilibrium. WHY? Since the concentration is constant, it can be considered a part of the equilibrium constant and, thus, does NOT appear in the K C expression. C (s, graphite) + CO 2 (g) = 2 CO (g) K C = [CO] 2 [C][CO 2 ] K eq = [CO] 2 [CO 2 ]

13. 09/05/09 Chemical Equilibrium Class Problem 14.3 - A mixture that was initially 0.005 00 M in H 2 (g) and 0.012 50 M in I 2 (g), and contained no HI (g), was heated at 425.4 o C until equilibrium was reached. The resulting equilibrium concentration of I 2 (g) was found to be 0.007 72 M. What is the value of the K C for this equilibrium at 425.4 o C? Construct an “ICE” Table: 425.4 o C Equation: H 2 (g) + I 2 (g)  2 HI (g) Initial (I) conc., M 0.005 00 0.012 50 0.000 00 Change (C) in conc., M Equil. (E) conc., M 0.007 72 - 0.004 78 - 0.004 78 + 0.009 56 - 0.000 22 + 0.009 56

14. 09/05/09 Chemical Equilibrium Calculate K C : [HI] 2 (0.009 56) 2 [H 2 ][I 2 ] (0.000 22)(0.007 72) = K C = = 54 Class Problem 14.4a. - When 1.000 mol each of H 2 O (g) and CO (g) are introduced into an empty 1.000 L vessel at 959 K and allowed to come to equilibrium, the equilibrium mixture contains 0.422 mol H 2 O (g). Find K C for H 2 O (g) + CO (g)  H 2 (g) + CO 2 (g) Construct an “ICE” Table:

15. 09/05/09 959 o C Equation: H 2 O (g) + CO (g)  H 2 (g) + CO 2 (g) I 1.000 1.000 0.000 0.000 C - 0.578 - 0.578 + 0.578 + 0.578 E 0.422 0.422 0.578 0.578 Chemical Equilibrium K c = [H 2 ][CO 2 ] [H 2 O][CO] = 1.88 Class Problem 14.4b. - Suppose that [H 2 O] I = 2.00 M and [CO] I = 4.00 M? What are the equilibrium concentrations of the four species? (0.578) 2 (0.422) 2 =

16. 09/05/09 959 o C Equation: H 2 O (g) + CO (g)  H 2 (g) + CO 2 (g) I 2.00 4.00 0.00 0.00 C - x - x + x + x E 2.00 - x 4.00 - x x x Chemical Equilibrium [H 2 ][CO 2 ] [H 2 O][CO] K c = = 1.88 (2.00 – x)(4.00 – x) x 2 = x 2 = 1.88(2.00 – x)(4.00 – x) = 1.88(8.00 - 6x – x 2 ) x 2 = 15.0 – 11.3x +1.88x 2  0 = 0.88x 2 – 11.3x +15.0

17. 09/05/09 0 = 0.88 x 2 – 11.3 x + 15.0 (ax 2 + bx + c) Dust off the old Quadratic Formula: http://www.freemathhelp.com/algebra-help.html -(-11.3) ± [(-11.3) 2 – 4(0.88)(15.0)] 1/2 2(0.88) = 11 and 1.5 ! Which is RIGHT?

18. 09/05/09 Chemical Equilibrium What is ‘x’? It is the concentration of H 2 and CO 2 at equilibrium! But…you can’t have more hydrogen gas than you have of reactants to begin with! Thus, 11 M can’t be right! 1.5 M is the only sensible answer!

19. 09/05/09 Chemical Equilibrium x = 1.5 M Therefore: [H 2 O] eq = 0.5 M [CO] eq = 2.5 M [H 2 ] eq = [CO 2 ] eq = 1.5 M Check: K c = (1.5M) 2 /(0.5M)(2.5M) = 1.8 There are no units in this case! What if you don’t remember the quadratic formula??

20. 09/05/09 Use Successive Approximation!!

21. 09/05/09 Class Exercise 14.5: Consider the following reaction for the decomposition of hydrogen sulfide: 2 H 2 S  2 H 2 (g) + S 2 (g) K C = 1.67 x 10 -7 800 o C A 0.500-L vessel initially contains 1.25 x 10 -1 mol of H 2 S. Find the equilibrium concentrations of H 2 and S 2 . Equation: 2 H 2 S (g)  2 H 2 (g) + S 2 (g) Initial (M) 2.50 x 10 -1 0.00 0.00 Change (M) - 2x + 2x + x Equilibrium (M) (2.50 x 10 -1 – 2x ) 2x x

22. 09/05/09 K c = [H 2 ] 2 [S 2 ] H 2 S = (2x) 2 x (2.50 x 10 -1 – 2x) 2 = 4x 3 (2.50 x 10 -1 – 2x) 2 Assume x is NEGLIGIBLE compared to 2.50 x 10 -1 M. Then: 4x 3 (2.50 x 10 -1 ) 2 = 1.67 x 10 -7 ~ 1.67 x 10 -7 4x 3 6.25 x 10 -2 = (1.67 x 10 -7 ) (6.25 x 10 -2 ) = 4x 3 = 1.04 x 10 -8 x 3 = 2.61 x 10 -9 x = 1.38 x 10 -3 M

23. 09/05/09 Is x NEGLIGIBLE compared to 1.38 x 10 -3 M? Plug it back in to check: 4x 3 (2.50 x 10 -1 – 2x) 2 [(2.50 x 10 -1 ) - 2(1.38 x 10 -3 )] 2 4x 3 = 4x 3 4.95 x 10 -4 = 1.67 x 10 -7 x 3 = 2.07 x 10 -11 >>>>> x = 2.74 x 10 -4 There appears to be a mistake in these calculations! Please check carefully and see if you can see where it is!

24. 09/05/09 Complex example of Successive Approximation

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30. 09/05/09 Do NOT Panic! This is NOT a typical Problem! It is a Worst Case Scenario !!! Any Exam Problem will be MUCH Shorter!!

31. 09/05/09 Class Exercise 14.5: In an experiment starting with [N 2 O 4 ] I = 0.020 00 M and [NO 2 ] I = 0.000 00 M, [N 2 O 4 ] eq = 0.004 52 M. (a) What is [NO 2 ] eq ? (b) What is the value for K c ? Equation: N 2 O 4 (g)  2 NO 2 (g) I (M) 2.000 x 10 -2 0.000 00 C (M) E (M) 4.52 x 10 -3 -0.015 48 +0.030 96 0.030 96 [NO 2 ] 2 [N 2 O 4 ] K C = = (0.030 96) 2 /(0.004 52) = 0.212

32. 09/05/09 What does the value of K c MEAN? The larger K C is, the closer to completion the rxn is! N 2 (g) + O 2 (g)  2 NO (g) K C = 1 x 10 -30 2 NH 3 (g)  N 2 (g) + 3 H 2 (g) K C = 9.5 H 2 (g) + Cl 2 (g)  2 HCl (g) K C = 1.33 x 10 34