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  1. Equilibrium Chapter 15

  2. Have you ever tried to maintain your balance as you walked across a narrow ledge? In a chemical reaction balance or equilibrium is also maintained. You can think of the yields  sign as the ledge Equilibrium

  3. Equilibrium systems exist in ocean water, blood, urine, and many other biological systems Chemical reactions for the most part are reversible You can think of the yield sign  as the ledge in a chem system

  4. I. Chemical Equilibrium Concept • Chemical Equilibrium occurs when opposing reactions are proceeding at equal rates • At equil rate forward = rate reverse • AB [] indicates a molar conc. • Fr AB rate kf [A] kf [A] = kr [B] • Rr BA rate kr [B] [B] = kf cons= Kc • Rearranging formula [A] kr

  5. Individual molecules are undergoing change but there is no net exchange in the concentration of reactants and products Does not mean that the concentrations are not changing just that the ratio equals a definite value Look at Habber reaction figure 15.6 text. Dihydrogen monoxide: is also known as hydric acid, and is the major component of acid rain. contributes to the "greenhouse effect." may cause severe burns. contributes to the erosion of our natural landscape. accelerates corrosion and rusting of many metals. may cause electrical failures and decreased effectiveness of automobile brakes. has been found in excised tumors of terminal cancer patients. A dynamic equilibrium

  6. Law of Mass Action • Equilibrium can be reached from either direction • Concentrations of reactants and products are expressed as aA + bB  pP + qQ equil constant Kc = [P]p [Q]q products [A]a [Q]b reactants

  7. Equilibrium Constant mass action • Kc depends on the stoichiometry not on mechanics • Doesn’t depend on the initial conc of reactants and products • Doesn’t depend on other added sub as long as they do not react • Varies with tempt • Catalyst do not effect just speeds – reaching eq.

  8. Putting it Together

  9. Writing equilibrium expressions • 2O3(g) 3O2(g) • 2NO (g) + Cl2(g)  2NOCl (aq) • AgCl (s)  Ag+(aq) + Cl-(aq) Kc = [O2]3 Kc = [NOCl]2 [O3]2 [NO]2 [Cl2] Kc = [Ag+] [Cl-] Pure solids and liquids do not effect the equilibrium because their conc remain unchanged

  10. Eq expressed as pressure • C3H8(g) + O2(g) CO2(g) + H2O(l) • Kp = CO2p3 C3H8p O2p5 • P = partial pressure of the gas • Kp = kc(RT)delta n

  11. What does kc tell you? • Ex CO (g) + Cl2(g) COCl2 (g) • Kc = [COCl2] = 4.57 X 10^9 [CO] [Cl2] Kc>>>1 larger numerator reaction goes almost totally to products – eq lies right –favors products Kc<<<1 larger denominator

  12. If a system at eq is disturbed by a change in temperature, pressure, or the concentration of one of the components, the system will shift it’s eq pos so as to counter the effects of the distrubance Henri-Louis Le Chatelier (1858-1936) Le Chatelier’s Principle

  13. Change in Concentration • Le Châtelier's principle states that if the concentration of one of the components of the reaction (either product or reactant) is changed, the system will respond in such a way as to counteract the effect • If a substance (either reactant or product) is removed from a system, the equilibrium will shift so as to produce more of that component (and once again achieve equilibrium)

  14. Change in concentration • If a substance (either reactant or product) is added to a system, the equilibrium will shift so as to consume more of that component (and once again achieve equilibrium)

  15. Putting it Together N2(g) + 3H2(g) <=> 2NH3(g)

  16. The reaction is driven "to the right" by the effects of added H2 The eq conc’s will not be identical to the original state. However, Kc will be the same. The new equilibrium state contains a slightly higher concentration of NH3(g), and slightly lower concentration of N2(g) (as well as a slightly higher concentration of H2(g).

  17. Change in Volume and Pressure • A chemical system in equilibrium can respond to the effects of pressure also. According to Le Châtelier's Rule, if the pressure is increased on a system, it will respond by trying to reduce the pressure. How does it do this? • We are primarily concerned with homogeneous gaseous reactions • The stoichiometry of the reaction may lead to a greater number of molecules on one side of the equation. • For example, in the Haber reaction, N2(g) + 3H2(g) <=> 2NH3(g) there are twice as many moles of reactants as products

  18. If the Haber reaction were in equilibrium, and the pressure was increased, the reaction would respond to oppose the increase in pressure. It could accomplish this by shifting the equilibrium to the right (producing NH3(g)) • This would reduce the overall number of moles in the reaction, and therefore, lower the pressure • Systems shift to the side with the fewest number of moles if both are the same then no change in eq con’s occurs

  19. Changes in Temperature • The intrinsic value of K does not change when we increase concentrations or pressures of components in a reaction. However, almost every equilibrium constant (K) changes in response to changes in temperature. • We will consider reaction conditions under which no work is done, and therefore all energy changes associated with reactions will be manifested by temperature changes)

  20. Temperature Changes • Exothermic reactions are associated with heat release when the reaction proceeds in theforward direction • Endothermic reactions are associated with heat release when the reaction proceeds in the reverse direction (i.e. heat is absorbed in the forward direction)

  21. These two types of reactions and their associated heat changes can be written as: • Exothermic: Reactants yield Products + Heat • Endothermic: Reactants + Heat yield Products • If temperature is increased, the equilibrium will shift so as to minimize the effect of the added heat • The reaction will shift in the appropriate direction such that the added heat is absorbed

  22. When heat is added to exothermic reactions at equilibrium, products will be consumed to produce reactants (shift to the LEFT)May also be written delta t is negative. • When heat is added to endothermic reactions at equilibrium, reactants will be consumed to produce products (shift to the RIGHT) May also be written delta t is positive.

  23. Based on this behavior, what is the effect of T upon K? • Assume K = 1.0 for an exothermic reaction at equilibrium. • Added heat causes the reaction to shift to the left. Reactants <= Products + Heat • Thus, 1.0 must represent a reaction quotient, Q, that is too large in comparison to the new value of K. • Thus, the effect of increasing temperature on an exothermic reaction is to lower the value of K. • Conversely, the effect of increasing temperature on an endothermic reaction is to increase the value of K

  24. Putting it Together Calc Delta H of formation for C3H8(g) + O2(g) CO2(g) + H2O(l)

  25. Calculations with eq K • Example calculating unknown concentrations using the eq constant • CO(g) + 3H2(g) <-> CH4(g) + H20(g) • At eq 0.3 mol of CO, 0.1 mol H2 and 0.02 mol of H20 are in 1.0 liter of a vessel at 1200 k kC is 3.92 what is the conc of CH4?

  26. Kc = [CH4] [H2O] [CO] [H2]3 3.93 = [CH4] (.020) (0.30) (0.10)3 [CH4] = (0.30)(0.10)3 3.93 (0.020) 0.059 mol/l

  27. Learning CheckPCl5(g)<-> PCl3(g) + Cl2(g) • A l.0 liter vessel has a unknown amount of PCl5 at eq Kc at 250 0C is 0.0415. Calc the unknow conc. if 0.02moles of PCl3 and Cl2 are in the container. (0.0096)

  28. Solving linear eq equasions • CO(g) + H20(g) <-> C02(g) + H2(g) Given 1.0 mol of CO2 and H20 in a 50.0 l vessel. How many moles are in an eq mix at 1000 oC Ec = 0.58 at 1000oC CO(g) + H20(g) <-> C02(g) + H2(g I 0.02 0.02 0 0 C -x -x +x +x E 0.02-x 0.02-x x x

  29. 0.58 = [CO2][H2] = X2 [CO] [H2O] (0.02-X)2 +,- 0.76 = X2 (0.02-X) (the neg one gives a neg answer x can’t be neg) _ +0.76(0.02-X)=X 0.0152-0.76X=X 0.0152= 1.76X X= 0.0086

  30. H2(g) + I2(g) <-> 2HI(g) • What is the eq comp of a reaction mixture starting with 0.500 mol each of H2 and I2 in a 1 l vessel? Kc = 49.7 at 458 oC (H2 &I2 = 0.11 mol/l HI = 0.78 mol/l

  31. Equil with quadratic expressions • Calc the conc. of the previous problem with 1.00 molar H2 and 2.00 molar I2 as the starting concentrations. H2(g) + I2(g) <-> 2HI(g) I 1.00 2.00 0 C -x -x 2x 49.9 = (2x)2 E 1.00-x 2.00-x 2x (1.00-x)(2.00-x)

  32. (1.00-x)(2.00-x)= (2x)2 49