Presentation on theme: "Zumdahl’s Chapter 13 Chemical Equilibrium Chapter Contents Equilibrium’s Hallmarks The Equilibrium Constant, K C Expressions for Pressure Equilibria,"— Presentation transcript:
Zumdahl’s Chapter 13 Chemical Equilibrium
Chapter Contents Equilibrium’s Hallmarks The Equilibrium Constant, K C Expressions for Pressure Equilibria, K P Heterogeneous Equilibria Applications –Reaction Quotient, Q –Extent of Reaction, –Finding Equilibrium Extreme Equilibria Le Châtlier’s Principle –Varying concentration –Varying pressure –Varying temperature
Equilibrium’s Hallmarks When bulk concentrations of all species no longer change with time. [species] eq = fixed –But reaction and unreaction still proceed (at equal rates), equilibrium is dynamic not static. –Same arrival point whether the initial conditions are reactants or products! Rate forward = Rate reverse for A+B C+D –I.e., k f [A] eq [B] eq = k r [C] eq [D] eq from kinetics
The Equilibrium Constant aA + bB cC + dD for this elementary reaction at equilibrium, Rate f = Rate r is k f [A] eq a [B] eq b = k r [C] eq c [D] eq d & becomes K = k f / k r = [C] eq c [D] eq d / [A] eq a [B] eq b –Scaling reaction by factor ±n scales all exponents and thus K new = K old ±n Negative exponents denote reversed reactions. This Law of Mass Action holds whether the reaction is elementary or not!
Pressure Equilibria Expressions If aA + bB cC + dD involves gases, instead of K C we have K P where K P = ( P C c P D d ) / ( P A a P B b ) with partial pressures in place of concentrations. Of course [A] = n A / V = P A / RT by ideal gas law so P A a = [A] a (RT) a or K P = K C (RT) n = c+d–(a+b) Although it appears as if both K C and K P have dimensions (if n 0), neither do!
Heterogeneous Equilibrium Greek: heteros- “other” and –genos “kind” Chemistry: more than one physical phase. While gases appear as partial pressures and solutes appear as concentrations, pure liquids and solids vanish from K. Because densities of solutes and gases vary but those of pure condensed phases do not! Same reasoning eliminates [ H 2 O ] (fixed at 55.5 M)
Reaction Quotient, Q While K C uses concentration at equilibrium exclusively, we can construct another mass action expression away from equilibrium. Q = [C] c [D] d / [A] a [B] b (arbitrary concentrations) –Is Q = K ? We’re at equilibrium! Rejoice? –Is Q < K ? Rxn. runs forward to equilibrium. –Is Q > K ? Rxn. runs backward to equilibrium.
Extent of Reaction, As reaction proceeds, varies from 0 to 1. But common practice dictates a simpler x indicating a change in [A] or P A. Each such change must obey reaction stoichiometry. –Example: 2 NOCl(g) 2 NO(g) + Cl 2 (g) from an intial P NOCl = 0.5 atm gives, at equilibrium: –K P = P(NO) 2 P(Cl 2 ) / P(NOCl) 2 –K P = (x) 2 (½ x) / ( 0.5 – x ) 2 = 1.6 10 –5 at 35ºC
Finding Equilibrium Solve the extent of reaction expression that renders Q = K. Try 2 NOCl 2 NO + Cl 2 For (x) 2 (½ x) / ( 0.5 – x ) 2 = 1.6 10 –5 find x ½ x 3 = 1.6 10 –5 (0.5 – x) 2 is a (painful) cubic While cubics are solvable, we hope the small K will yield a small enough x so 0.5 – x 0.5, whereupon ½ x 3 4 10 –6 or x 3 8 10 –6 or x 0.02 Is it OK? Test: (0.02) 2 (½ 0.02) / (0.50 – 0.02) 2 = 1.7 10 –5 K OK! Equil. Pressures are 0.48, 0.02, and 0.01 atm
Extreme Equilibria For very small K, equilibrium lies virtually with the reactants with negligible products –As it was with the NOCl decomposition. For very large K, equilibrium lies virtually with the products with negligible reactants. –Start with products, and move back by x. At either extreme, x can be presumed tiny, and equilibrium algebra simplifies.
Le Châtlier’s Principle “Equilibrium shifts to minimize its perturbation.” –In other words, if you impose a change, you render Q K, and the equilibrium shifts to restore Q=K. E.g., removing a product drives rxn forward And increasing pressure drives rxn to the side with fewer gaseous molecules, relieving P total. Heating an exothermic rxn drives it backwards!