Equilibrium and Reversible Rxns. CHAPTER 14 Chemical Equilibrium. What happens? Stoichiometry

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1 CHAPTER 14 Chemical Equilibrium 17-1 What happens? Stoichiometry How Fast? Kinetics applies to the speed of a reaction, the concentration of product that appears (or of reactant that disappears) per unit time. To What Extent? Equilibrium applies to the extent of a reaction, the concentrations of reactant product present after an unlimited time, or once no further change occurs Equilibrium and Reversible Rxns. double arrows = "reversible reaction"

2 Equilibrium A system at equilibrium is dynamic on the molecular level; no further net change is observed because changes in one direction are balanced by changes in the other. At equilibrium: rate forward = rate reverse aa + bb k f k r cc + dd rate of forward reaction = k f [A] a [B] b rate of reverse reaction = k r [C] c [D] d ([..] = concentration (M) at equilibrium ) k f [A] a [B] b = k r [C] c [D] d k forward k reverse [products] n = = K the equilibrium constant [reactants] m This is also known as the LAW OF MASS ACTION

3 Approach to equilibrium.. Equilibrium achieved 17-7 Equilibrium Constants for Heterogeneous Systems -Pure solids and pure liquids are materials whose concentration doesn t change during the course of a reaction its amount can change, but the amount of it in solution doesn t because it isn t in solution -Because their concentration doesn t change, solids and liquids are not included in the equilibrium constant expression -For the reaction aa(s) + bb(aq) cc(l) + dd(aq) the equilibrium constant expression is 17-8 Figure 17.4 The Equilibrium Constant for a heterogeneous system. solids do not change their concentrations

4 Write the equilibrium equation for each of the following reactions: (a) CO 2 (g) + C(s) 2 CO(g) (b) Hg(l) + Hg 2+ (aq)!!hg 2 2+ (aq) (c) 2 Fe(s) + 3 H 2 O(g)! Fe 2 O 3 (s) + 3 H 2 (g) (d) 2 H 2 O(l) 2 H 2 (g) + O 2 (g) Figure 17.2 The range of equilibrium constants. small K < 10-3 large K > intermediate 10-3 < K < Product- or Reactant-Favored. Product-favored K > 1 Reactant-favored K <

5 Calculating Variations on K aa + bb cc + dd K c = [C] c [D] d [A] a [B] b cc + dd aa + bb K' = 1/K c n aa + bb cc + dd K c ' = (K c ) n For a sequence of equilibria, K overall = K 1 x K 2 x K 3 x K 1 = [SO 2 ] [O 2 ] Net K 2 = [SO 3 ] [SO 2 ][O 2 ] 1/ Multiply their respective K's K net = [SO 3 ] [O 2 ] 3/2 K net = K 1 K 2 Writing and Manipulating K Expressions K new = [SO 3] 2 [O 2 ] 3 = (K old ) 2 K = [SO 3 ] [O 2 ] 3/2 K new = [SO 3] 2 [O 2 ]

6 Writing and Manipulating K Expressions K = [SO 2] [O 2 ] K new = [O 2] [SO 2 ] K new = [O 2] [SO 2 ] = 1 K old Q - The Reaction Quotient At any time, t, the system can be sampled to determine the amounts of reactants and products present. A ratio of products to reactants, calculated in the same manner as K tells us whether the system has come to equilibrium (Q = K) or whether the reaction has to proceed further from reactants to products (Q < K) or in the reverse direction from products to reactants (Q > K). For a general reaction aa + bb cc + dd where a, b, c, and d are the numerical coefficients in the balanced equation, Q (and K) can be calculated as (C) Q = c (D) d e.g., initial concentrations (A) a (B) b

7 Figure 17.5 Reaction direction and the relative sizes of Q and K. Reaction Progress Reaction Progress Reactants Products Equilibrium: no net change reactants products Figure 17.6 Steps in solving equilibrium problems. PRELIMINARY SETTING UP 1. Write the balanced equation. 2. Write the Eq. Constant, K. 3. Convert all of the amounts into the correct units (M or atm). SOLVING FOR X AND EQUILIBRIUM QUANTITIES 6. Substitute the quantities into K. 7. To simplify the math, assume that x is negligible. 8. [A] ini - x = [A] eq = [A] init 9. Solve for x. 10. Find the equilibrium quantities. Check that assumption is justified (<5% error). If not, solve quadratic equation for x. Check to see that calculated values give the known K Determining K

8 Another Example of Calculating K PROBLEM: Place 1.00 mol each of H 2 and I 2 in a 1.00 L flask. Calc. equilibrium concentrations. K c = [HI] 2 [H 2 ][I 2 ] = Nitrogen Dioxide Equilibrium N 2 O 4 (g) 2 NO 2 (g) K c = [NO 2 ]2 [N 2 O 4 ] = at 298 K When a chemical system at equilibrium is subjected to a stress, the system will return to equilibrium by shifting to reduce the stress. If the concentration increases, the system reacts to consume some of it. If the concentration decreases, the system reacts to produce some of it

9 The effect of a change in concentration no change in K only the equilibrium composition changes. The Effect of Added Cl 2 on the PCl 3 -Cl 2 -PCl 5 System Concentration (I) PCl 3 (g) + Cl 2 (g) PCl 5 (g) Original equilibrium Disturbance New initial Change x -x +x New equilibrium x x x (0.637)* *Experimentally determined value Figure 17.8 The effect of added Cl 2 on the PCl 3 -Cl 2 -PCl 5 system. PCl 3 (g) + Cl 2 (g) PCl 5 (g)

10 Figure 17.9 The effect of pressure (volume) on a system at equilibrium. + lower P (higher V) more moles of gas higher P (lower V) fewer moles of gas K c = [NO 2 ]2 [N 2 O 4 ] Increase P in the system by reducing the volume (at constant T) = at 298 K The Effect of a Change in Temperature on an Equilibrium Only temperature changes can alter K. Consider heat as a product or a reactant. In an exothermic reaction, heat is a product. In an endothermic reaction, heat is a reactant. A temperature rise will increase K for a system with a positive ΔH o rxn. A temperature rise will decrease K for a system with a negative ΔH o rxn What will decreases in temperature do to K? 10

11 K c = [NO 2 ]2 [N 2 O 4 ] K = [isobutane] = 2.5 [butane] At equilibrium with [iso] = 1.25 M and [butane] = 0.50 M. K = 2.5. Add 1.50 M butane. When the system comes to equilibrium again, what are [iso] and [butane]?

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