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Chapter 14 Chemical Equilibrium

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). Dynamic Equilibrium. The net result of a dynamic equilibrium is that no

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Chapter 14 Chemical Equilibrium

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  1. Chapter 14Chemical 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. 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. Dynamic Equilibrium Open System (No Equilibrium) Evaporation Evaporation Liquid Gas Liquid Gas Liquid Gas (No Equilibrium) (Equilibrium) (No Equilibrium)

  4. Dynamic Equilibrium Ag + + Cl - AgCl (s) Ag + Cl - Chemical Equilibrium Cl - Ag + AgCl (s) Rate of Precipitation = Rate of Dissolving HC2H3O2 (aq) H + + C2H3O2- Rate of dissociation (ionization) = Rate of Association HC2H3O2 H + H + C2H3O2- C2H3O2- HC2H3O2

  5. CHEM 1108 Lab Experiment Orange Red HC2H3O2 H+ + C2H3O2- Pink Blue [CoCl4]2- + 6 H2O (l) Co(H2O)62+ + 4 Cl- Colorless Solution White NH4Cl (s) NH4+ + Cl- You can actually “see” the equilibrium shift!

  6. Reversible Reactions R1 N2O4 (g) 2 NO2 (g) R2 2 NO2 (g) N2O4 (g) R1 N2O4 (g) 2 NO2 (g) [R1 = R2] R2 Homogeneous Equilibrium

  7. Reversible Reactions [N2O4]i [NO2]i [N2O4]eq [NO2]eq 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 N2O4 (g) 2 NO2 (g) Reaction Quotient Equilibrium Constant QC = [NO2]2 [N2O4] KC = [NO2]2eq [N2O4]eq

  8. 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 KC = [0.009 66]2/[0.0202] = 0.004 62 M Experiment 2 KC = [0.008 23]2/[0.0146] = 0.004 64 M Experiment 3 KC = [0.006 54]2/[0.009 23] = 0.004 63 M aA + bB cC + dD KC = [C]c[D]d [A]a[B]b

  9. Equilibrium Constants H2 (g) + I2 (g) 2 HI (g) KC = [HI]2eq [H2]eq[I2]eq QC = [HI]2 [H2][I2] • Exp. [H2]eq [I2]eq [HI]eq KC • 1 0.00291 0.00171 0.01648 54.58 • 2 0.00356 0.00125 0.01559 54.62 • 3 0.00225 0.00234 0.01685 53.93 • 4 0.00183 0.00313 0.01767 54.51 • 5 0.00114 0.00114 0.00841 54.42 • 0.00050 0.00050 0.00366 53.58 • Median KC = 54.47

  10. Equilibrium Constants 4 NH3 (g) + 3 O2 (g) 2 N2 (g) + 6 H2O (g) QC = [N2]2[H2O]6 [NH3]4[O2]3 [N2]eq2[H2O]6 KC = [NH3]eq4[O2]eq3

  11. Reaction Quotient vs. Equilibrium Constant Class Problem 14.1 -The concentration of N2O4 = concentration of NO2 = 0.0125 M in a reaction vessel. The equilibrium constant for N2O4 (g) = 2 NO2 (g) is 0.004 63. Calculate QC and state which direction the reaction will go. Class Problem 14.2 -If [O2] = 0.21 M and [O3] = 6.0 x 10-8 M, what is the value of the KC for the equilibrium, 2 O3 (g) = 3 O2 (g)?

  12. 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 KC expression. C (s, graphite) + CO2 (g) = 2 CO (g) KC = [CO]2 [C][CO2] Keq = [CO]2 [CO2]

  13. Chemical Equilibrium Class Problem 14.3 -A mixture that was initially 0.005 00 M in H2 (g) and 0.012 50 M in I2 (g), and contained no HI (g), was heated at 425.4oC until equilibrium was reached. The resulting equilibrium concentration of I2 (g) was found to be 0.007 72 M. What is the value of the KC for this equilibrium at 425.4oC? Construct an “ICE” Table: 425.4oC Equation: H2 (g) + I2 (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 + 0.009 56 - 0.000 22

  14. Chemical Equilibrium Calculate KC: [HI]2 (0.009 56)2 [H2][I2] (0.000 22)(0.007 72) = 54 KC = = Class Problem 14.4a. -When 1.000 mol each of H2O (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 H2O (g). Find KC for H2O (g) + CO (g)  H2 (g) + CO2 (g) Construct an “ICE” Table:

  15. Chemical Equilibrium 959oC Equation: H2O (g) + CO (g)  H2 (g) + CO2 (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 [H2][CO2] [H2O][CO] (0.578)2 (0.422)2 Kc = = = 1.88 Class Problem 14.4b. -Suppose that [H2O]I = 2.00 M and [CO]I = 4.00 M? What are the equilibrium concentrations of the four species?

  16. Chemical Equilibrium 959oC Equation: H2O (g) + CO (g)  H2 (g) + CO2 (g) I 2.00 4.00 0.00 0.00 C - x - x + x + x E 2.00 - x 4.00 - x xx [H2][CO2] [H2O][CO] x2 Kc = = = 1.88 (2.00 – x)(4.00 – x) x2 = 1.88(2.00 – x)(4.00 – x) = 1.88(8.00 - 6x – x2) x2 = 15.0 – 11.3x +1.88x2 0 = 0.88x2 – 11.3x +15.0

  17. 0 = 0.88 x2 – 11.3 x + 15.0 (ax2 + 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. Chemical Equilibrium What is ‘x’? It is the concentration of H2 and CO2 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. Chemical Equilibrium x = 1.5 M Therefore: [H2O]eq = 0.5 M [CO]eq = 2.5 M [H2]eq = [CO2]eq = 1.5 M Check: Kc = (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. Use Successive Approximation!!

  21. Class Exercise 14.5: Consider the following reaction for the decomposition of hydrogen sulfide: 2 H2S  2 H2 (g) + S2 (g) KC = 1.67 x 10-7 800o C A 0.500-L vessel initially contains 1.25 x 10-1 mol of H2S. Find the equilibrium concentrations of H2 and S2. Equation: 2 H2S (g)  2 H2 (g) + S2 (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. [H2]2[S2] (2x)2x (2.50 x 10-1 – 2x)2 Kc = = H2S 4x3 (2.50 x 10-1 – 2x)2 = = 1.67 x 10-7 Assume x is NEGLIGIBLE compared to 2.50 x 10-1 M. Then: 4x3 (2.50 x 10-1)2 4x3 6.25 x 10-2 = ~ 1.67 x 10-7 (6.25 x 10-2) (1.67 x 10-7) = 4x3 = 1.04 x 10-8 x3 = 2.61 x 10-9 x = 1.38 x 10-3 M

  23. Is x NEGLIGIBLE compared to 1.38 x 10-3 M? Plug it back in to check: 4x3 4x3 (2.50 x 10-1 – 2x)2 = [(2.50 x 10-1) - 2(1.38 x 10-3)]2 4x3 = 1.67 x 10-7 4.95 x 10-4 x3 = 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. Complex example of Successive Approximation

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

  26. Class Exercise 14.5: In an experiment starting with [N2O4]I = 0.020 00 M and [NO2]I = 0.000 00 M, [N2O4]eq = 0.004 52 M. (a) What is [NO2]eq? (b) What is the value for Kc? Equation: N2O4 (g)  2 NO2 (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 [NO2]2 [N2O4] KC = = (0.030 96)2/(0.004 52) = 0.212

  27. What does the value of Kc MEAN? The larger KC is, the closer to completion the rxn is! N2 (g) + O2 (g)  2 NO (g) KC = 1 x 10-30 2 NH3 (g)  N2 (g) + 3 H2 (g) KC = 9.5 H2 (g) + Cl2 (g)  2 HCl (g) KC = 1.33 x 1034

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