Chemical Equilibrium - Equilibrium Constant Notes, Lecture notes of Chemistry

Download Chemical Equilibrium - Equilibrium Constant Notes and more Lecture notes Chemistry in PDF only on Docsity! 1 Learning goals and key skills:  Explain what is meant by chemical equilibrium and how it relates to reaction rates  Write the equilibrium-constant expression for any reaction  Convert Kc to Kp and vice versa  Relate the magnitude of an equilibrium constant to the relative amounts of reactants and products present in an equilibrium mixture.  Manipulate the equilibrium constant to reflect changes in the chemical equation  Write the equilibrium-constant expression for a heterogeneous reaction  Calculate an equilibrium constant from concentration measurements  Predict the direction of a reaction given the equilibrium constant and the concentrations of reactants and products  Calculate equilibrium concentrations given the equilibrium constant and all but one equilibrium concentration  Calculate equilibrium concentrations given the equilibrium constant and the starting concentrations  Use Le Chatelier’s principle to predict how changing the concentrations, volume, or temperature of a system at equilibrium affects the equilibrium position. Chapter 15 Chemical Equilibrium The Concept of Equilibrium Chemical equilibrium occurs when a reaction and its reverse reaction proceed at the same rate. 2 Concept of Equilibrium • As a system approaches equilibrium, both the forward and reverse reactions are occurring. • At equilibrium, the forward and reverse reactions are proceeding at the same rate. • Once equilibrium is achieved, the amount of each reactant and product remains constant. The same equilibrium is reached whether we start with only reactants (N2 and H2) or with only product (NH3). Equilibrium is reached from either direction. 5 Relationship Between Kc and Kp Plugging this into the expression for Kp for each substance, the relationship between Kc and Kp becomes where Kp = Kc (RT)n n = (moles of gaseous product) - (moles of gaseous reactant) From the Ideal Gas Law we know that: PV = nRT and P = (n/V)RT = [A]RT What Does the Value of K Mean? • If K>>1, the reaction is product-favored; product predominates at equilibrium. • If K<<1, the reaction is reactant-favored; reactant predominates at equilibrium. *When 10-3 < K < 103, the reaction is considered to contain a significant amount of both reactants and products at equilibrium. 6 Direction of Chemical Equation and K The equilibrium constant of a reaction in the reverse direction is the reciprocal of the equilibrium constant of the forward reaction. Kc = = 0.212 at 100 C [NO2]2 [N2O4] N2O4 (g) 2 NO2 (g) Kc = = 4.72 at 100 C [N2O4] [NO2]2 N2O4 (g)2 NO2 (g) ⇌ ⇌ Kc = = 0.212 at 100 C [NO2]2 [N2O4] Kc = = (0.212)2 at 100 C [NO2]4 [N2O4]2 4 NO2(g)2 N2O4(g) N2O4(g) 2 NO2(g)⇌ ⇌ 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. Stoichiometry and K 7 2 NOBr ⇌ 2 NO + Br2 K1 = 0.014 Br2 + Cl2 ⇌ 2 BrCl K2 = 7.2 2 NOBr + Cl2 ⇌ 2 NO + 2 BrCl K3 = K1 × K2 = 0.014 × 7.2 = 0.10 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. Multiple equilibria and K Example Consider the following reactions at 1200 K. CO(g) + 3 H2(g) ⇌ CH4(g) + H2O(g) Kc,1 = 3.92 CH4(g) + 2 H2S(g) ⇌ CS2(g) + 4 H2(g) Kc,2 = 3.3x104 Use the above reactions to determine the equilibrium constant (Kc) for the following reaction at 1200 K. CO(g) + 2 H2S(g) ⇌ H2O(g) + CS2(g) + H2(g) 10 What Do We Know? [H2], M [I2], M [HI], M Initially 1.000 x 10-3 2.000 x 10-3 0 Change Equilibrium 1.87 x 10-3 [HI] Increases by 1.87 x 10-3 M [H2], M [I2], M [HI], M Initially 1.000 x 10-3 2.000 x 10-3 0 Change +1.87 x 10-3 Equilibrium 1.87 x 10-3 11 Stoichiometry tells us [H2] and [I2] decrease by half as much. [H2], M [I2], M [HI], M Initially 1.000 x 10-3 2.000 x 10-3 0 Change -9.35 x 10-4 -9.35 x 10-4 +1.87 x 10-3 Equilibrium 1.87 x 10-3 Calculate the equilibrium concentrations of all three compounds… [H2], M [I2], M [HI], M Initially 1.000 x 10-3 2.000 x 10-3 0 Change -9.35 x 10-4 -9.35 x 10-4 +1.87 x 10-3 Equilibrium 6.5 x 10-5 1.065 x 10-3 1.87 x 10-3 Kc = [HI]2 [H2] [I2] = 51= (1.87 x 10-3)2 (6.5 x 10-5)(1.065 x 10-3) 12 Example Phosphorus pentachloride gas partially decomposes to phosphorus trichloride gas and chlorine gas. 1.20 mol PCl5 is placed in a 1.00 L container at 200 °C. At equilibrium 1.00 mol PCl5 remains. Calculate Kc and Kp at 200 °C. The Reaction Quotient (Q) • Q gives the same ratio the equilibrium expression gives, but for a system that is not at equilibrium. • To calculate Q, substitute the (initial) concentrations of reactants and products into the equilibrium expression. aA + bB cC + dD C D A B c d a b [ ] [ ] [ ] [ ] Q ⇌ 15 Example Problem: Finding equilibrium concentrations from initial concentrations and the equilibrium constant. N2O4(g) ⇌ 2NO2(g) Kc = 0.36 at 100 °C A reaction mixture at 100 °C initially contains [NO2] = 0.100 M. Find the equilibrium concentrations of the reactants and products at this temperature. If a system at equilibrium is disturbed by a change in temperature, pressure or the concentration of one of the components, the system will shift its equilibrium position so as to counteract the effect of the disturbance. Changing concentration Temperature Changing volume/pressure Le Châtelier’s Principle 16 N2 (g) + 3 H2 (g)⇌ 2 NH3 (g) Kp = 0.0214 at 540 K Example: Le Châtelier’s Principle At equilibrium PH2 = 2.319 atm PNH3 = 0.454 atm PN2 = 0.773 atm What happens upon addition of 1 atm of H2? 17 The Haber Process This apparatus helps push the equilibrium to the right by removing the ammonia (NH3) from the system as a liquid. Change in Volume or Pressure If gases are involved in an equilibrium, a change in pressure or volume will affect K: • Higher volume or lower pressure favors the side of the equation with more moles of gas (and vice-versa).