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Chapter 15 Chemical Equilibrium

Chapter 15 Chemical Equilibrium. 15.1 The Concept of Equilibrium. Forward reaction: N 2 O 4 ( g )  2 NO 2 ( g ) Rate law: Rate = k f [N 2 O 4 ] Reverse reaction: 2 NO 2 ( g )  N 2 O 4 ( g ) Rate law: Rate = k r [NO 2 ] 2. [NO 2 ] 2 [N 2 O 4 ]. k f k r. =.

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Chapter 15 Chemical Equilibrium

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  1. Chapter 15Chemical Equilibrium

  2. 15.1 The Concept of Equilibrium

  3. Forward reaction: N2O4 (g) 2 NO2 (g) Rate law: Rate = kf [N2O4] • Reverse reaction: 2 NO2 (g) N2O4 (g) Rate law: Rate = kr [NO2]2

  4. [NO2]2 [N2O4] kf kr = • Therefore, at equilibrium Ratef = Rater kf [N2O4] = kr [NO2]2 • Rewriting this, it becomes

  5. [NO2]2 [N2O4] Keq = kf kr = • The ratio of the rate constants is a constant at that temperature, and the expression becomes

  6. 15.2 The Equilibrium Constant • Opposing reactions naturally lead to equilibrium regardless of complexity or the nature of the kinetic processes for the forward and reverse reactions. • Ex: Haber Process N2(g) + 3 H2(g) 2 NH3(g) • An equilibrium mixture is obtained regardless of the starting quantities of any of the reactants or products:

  7. 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. • The equilibrium condition can be reached from either direction. • Law of Mass Action: Expresses, for any reaction, the relationship between the concentrations of the reactants and the products present at equilibrium

  8. cC + dD aA + bB • A, B, D and E are chemical species involved and a, b, d, and e are the coefficients in the balanced equation • The equilibrium-constant expression (or equilibrium expression) for this reaction would be • The equilibrium-constant is expressed in terms of the concentrations of reactant products according to the stoichiometry of the reaction, not its mechanism. • Suppose we have the following general equilibrium equation

  9. Illustration of law of mass action • 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.

  10. Results from Experiments 3 and 4 show that equilibrium can be achieved starting with either N2O4 or NO2

  11. Writing Equilibrium-Constant Expressions

  12. (PC)c (PD)d (PA)a (PB)b Kp = Equilibrium Constants in Terms of Pressure, Kp • Because pressure is proportional to concentration for gases in a closed system, the equilibrium expression can also be written

  13. Relationship between Kc and Kp n = (moles of gaseous product) − (moles of gaseous reactant)

  14. 15.3 Interpreting and Working with Equilibrium Constants 1. The Magnitude of Equilibrium Constants • If K >> 1, the reaction is product-favored; product predominates at equilibrium. • The numerator of the equilibrium-constant expression must be much larger than the denominator.

  15. If K >> 1, the reaction is product-favored; product predominates at equilibrium. • If K << 1, the reaction is reactant-favored; reactant predominates at equilibrium.

  16. Kc = = 4.72 at 100C Kc = = 0.212 at 100C 2 NO2(g) N2O4(g) 2 NO2(g) N2O4(g) [NO2]2 [N2O4] [N2O4] [NO2]2 1 0.212 = 2. The Direction of the Chemical Equation and K • The direction in which we write the chemical equation for an equilibrium is arbitrary. • The equilibrium constant of a reaction in the reverse reaction is the reciprocal of the equilibrium constant of the forward reaction.

  17. Kc = = 0.212 at 100C Kc = = (0.212)2 = 0.0449 at 100C 2 NO2(g) N2O4(g) 4 NO2(g) 2 N2O4(g) [NO2]2 [N2O4] [NO2]4 [N2O4]2 3. Relating Chemical Equations and Equilibrium Constants • The equilibrium constant of a reaction that has been multiplied by a number is the equilibrium constant raised to a power that is equal to that number.

  18. The equilibrium constant for a net reaction made up of two or more steps is the product of the equilibrium constants for the individual steps.

  19. Equilibria in the same phase are called homogeneous equilibria. • In other cases the substances in equilibrium are in different phases, giving rise to heterogeneous equilibria. • Whenever a pure solid or a pure liquid is involved in a heterogeneous equilibrium, its concentration is not included in the equilibrium-constant expression for the reaction 15.4 Heterogeneous Equilibrium

  20. Even though a solid does not appear in the equilibrium-constant expression, it must be present for equilibrium to occur. • Why are they excluded? • The concentration of a pure solid or liquid has a constant value, regardless of the amount present.

  21. CaCO3 (s) CO2 (g) + CaO(s) • As long as some CaCO3 or CaO remain in the system, the amount of CO2 above the solid will remain the same.

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