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THE EQUILIBRIUM CONSTANT

THE EQUILIBRIUM CONSTANT. SECTION 7.3. LAW OF CHEMICAL EQUILIBRIUM. At equilibrium there is a constant ratio between the concentrations of products and reactants in any change As the rate of the forward reaction increases, the rate of the reverse reaction decreases

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THE EQUILIBRIUM CONSTANT

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  1. THE EQUILIBRIUM CONSTANT SECTION 7.3

  2. LAW OF CHEMICAL EQUILIBRIUM • At equilibrium there is a constant ratio between the concentrations of products and reactants in any change • As the rate of the forward reaction increases, the rate of the reverse reaction decreases • At equilibrium kf= kr

  3. The Concept of Equilibrium • As a system approaches equilibrium, both the forward and reverse reactions are occurring. • At equilibrium, the forward and reverse reactions are still occurring but are proceeding at the sameRATE.

  4. A System at Equilibrium • Before equilibrium the concentrations of the reactants and products are changing. • At equilibrium, the concentration of each reactant and product remains constant.

  5. AT EQUILIBRIUM…

  6. For ex: N2O4(l) ↔ 2NO2(g) Chemical equilibrium occurs when the forward and reverse reaction proceed at the same rate.

  7. THE EQUILIBRIUM CONSTANT • A state of dynamic equilibrium • Occurs when the rate of the forward reaction exactly equals the rate of the reverse reaction. • Consider the reaction where reactants A and B form products C and D (lower case letters are reaction coefficients): • aA + bB cC + dD

  8. aA + bB cC + dD • Forward reaction rate law: kf [A]a[B]b • Reverse reaction rate law: kr[C]c[D]d • At equilibrium: • forward rate = reverse rate kf [A]a[B]b = kr[C]c[D]d Rearrange: kf = [C]c[D]d kr [A]a[B]b

  9. THE EQUILIBRIUM CONSTANT • The ratio of rate constants (forward divided by reverse) is constant at that temperature and known as the equilibrium constant, Keq: Keq = [C]c[D]d [A]a[B]b

  10. THE EQUILIBRIUM CONSTANT • The square brackets indicate molar concentrations (mol/L) so the subscript “c” is used instead of “eq” in the constant Kc = [C]c[D]d [A]a[B]b NOTE: T he units for the equilibrium constant, K, are typically ignored

  11. EXAMPLE • Write the equilibrium law expression for the conversion of dinitrogen tetroxide gas to nitrogen dioxide • N2O4(g) 2NO2(g) ANSWER: Kf = [NO2]2 Kr [N2O4]

  12. EQUILIBRIUM CONSTANT AND TEMPERATURE • Recall: changing the temperature of a reacting mixture changes the rate of the forward and the reverse reactions by different amounts because the f and r reactions have different activationenergies • Thus, the K (eq constant) changes at different temperatures • Kc is calculated using the concentration values AT EQUILIBRIUM

  13. TRY THESE… • SnO2(s) + 2CO(g) ↔ Sn(s) + 2CO2(g) • CaCO3(s) ↔ CaO(s) + CO2(g) • Zn(s) + Cu+2(aq) ↔ Cu(s) + Zn2+(aq)

  14. Equilibrium Can Be Reached from Either Direction As you can see, the ratio of [NO2]2 to [N2O4] remains constant at this temperature no matter what the initial concentrations of NO2 and N2O4 are.

  15. Equilibrium Can Be Reached from Either Direction This is the data from the last two trials from the table on the previous slide.

  16. Equilibrium Can Be Reached from Either Direction It does not matter whether we start with N2 and H2 or whether we start with NH3. We will have the same proportions of all three substances at equilibrium.

  17. What Does the Value of KMean? • If K>>1, the reaction is product-favored; or the product predominates at equilibrium. • If K<<1, the reaction is reactant-favored; or the reactant predominates at equilibrium.

  18. EXAMPLE • Mixture of nitrogen and chlorine gas were kept in a 5.0L flask. At equilibrium, there was 0.0070 mol N2 and 0.022 mol Cl2 and 0.95 mol NCl3. Calculate the equilibrium constant for the reaction. • N2 + 3Cl2 2NCl3

  19. NCl3 • n= 0.95mol • V=5L • C=0.95mol • 5L • C=0.19 • N2 • n= 0.0070mol • V=5L • C=0.0070mol • 5L • C=1.4 X 10-3 • Cl2 n= 0.0022mol V=5L C=0.0022mol 5L C=4.4 X 10-4 [NCl3]2 [N2][Cl]3 Kc= = = 3.03 X 1011 [0.19]2 [1.4 X 10-3][4.4 X 10-4]3

  20. EQUILIBRIUM CALCULATIONS

  21. H2 (g) + I2 (g)  2 HI (g) Equilibrium Calculations A closed system initially containing 1.000 x 10−3 M H2 and 2.000 x 10−3 M I2 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 Kc at 448C for the reaction taking place, which is

  22. What Do We Know?

  23. [HI] Increases by 1.87 x 10-3M

  24. Stoichiometry tells us [H2] and [I2]decrease by half as much

  25. We can now calculate the equilibrium concentrations of all three compounds…

  26. Kc= [HI]2 [H2] [I2] (1.87 x 10-3)2 (6.5 x 10-5)(1.065 x 10-3) = = 51 …and, therefore, the equilibrium constant

  27. MORE PRACTICE…

  28. CO(g) + H2O(g) ↔ H2(g) + CO2(g) • This reaction has been studied at different temperatures to find the optimum conditions. At 700K, the equilibrium constant is 8.3. Suppose that you start with 1.0 mol of CO and 1.0mol of H2O in a 5.0L container. What amount of each substance will be present in the container when the gases are at equilibrium, at 700K?

  29. H2(g) + I2(g) ↔ 2HI(g) • The above reaction has an equilibrium constant of 25.0 at 1100K. • 2.00 mol of H2 and 3.00mol of I2 are placed in a 1.00L reaction vessal at 1100K. What is the equilibrium concentration of each gas?

  30. N2(g) + O2(g) ↔ 2NO(g) • 0.085ml of N2 and 0.038 mol of O2 are in a 1.0L rigid cylinder. The value of Kc is 4.2 X10-8. What is the concentration of NO?

  31. TRY… • Page 336 # 1-5 • Page 338 # 6-10 • Page 347-348 #11-15 • Page 349 #16-20 • Page 352 #21-25 • Page 353 ##1-4 • Quiz tom!

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