The Equilibrium State. Chapter 13 - Chemical Equilibrium. The Equilibrium State. Equilibrium is Dynamic! 5/29/2012

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1 Chapter 13 - Chemical Equilibrium The Equilibrium State Not all chemical reactions go to completion; instead they attain a state of equilibrium. When you hear equilibrium, what do you think of? Example: weather patterns: ocean water evaporates at the same rate that it rains. They are in equilibrium. The Equilibrium State Equilibrium is Dynamic! Equilibrium: a state in which the rates of the forward and reverse reactions are equal and the concentrations of reactants and products remain constant over time. Most reactions are reversible. A + B C + D A and B react to make C and D But C and D can also react to make A and B The conversions of reactants to products and products to reactants are still going on, although there is no net change in the number of reactant and product molecules. N 2 O 4 (g) 2NO 2 (g) Exp data for N 2 O 4 (g) 2NO 2 (g) In the forward reaction each molecule of N 2 O 4 breaks down to form two molecules of NO 2. Figure 13.1 What s the same/different? In the reverse reaction two molecules of NO 2 combine to form N 2 O 4. Equilibrium occurs when the rate at which an N 2 O 4 molecule breaks apart in the forward reaction is equal to the rate at which it is formed by the reverse reaction. 1

2 The Equilibrium Constant Example. Write the equilibrium constant, K c, for N 2 O 4 (g) 2NO 2 (g) The Equilibrium Constant, K c, tells us which side of the reaction is favored. For a reaction: aa + bb cc + dd K c is the equilibrium constant, a value The fraction is the equilibrium constant expression Law of Mass Action Reaction: N 2 O 4 (g) 2NO 2 (g) The value of the equilibrium constant, K c, is constant at a given temperature for a reaction at equilibrium. The equilibrium concentrations of reactants and products may be different, but the value for K c remains the same. K c Characteristics: The Equilibrium Constant, K c 1) Equilibrium can be approached from either direction. 2) K c does not depend on the initial concentrations of reactants and products. 3) K c does depend on temperature. 4) K c values are listed without units don't include units when calculating K c. Write the equilibrium constant expressions (K c ) for the following equations. (Note: Expressions don t include solids or liquids!) a) CO(g) + 3H 2 (g) CH 4 (g) + H 2 O(g) b) 2NH 3 (g) N 2 (g) + 3H 2 (g) a) b) 2

3 Worked Example 13.1 Calculating K c Group Quiz #4 2NH 3(g) N 2(g) + 3H 2(g) At 500 K, the following concentrations were measured: [N 2 ] = 3.0 x 10-2 M, [H 2 ] = 3.7 x 10-2 M, [NH 3 ] = 1.6 x 10-2 M. What is K c? N 2 O 4(g) 2NO 2(g) At 100 o C, the following concentrations are measured: [N 2 O 4 ] = M, [NO 2 ] = M. What is K c at this temperature? What is K c if we reverse this reaction? K c versus K p Can measure gas pressures instead of molarity. Pressure is directly proportional to concentration. PV = nrt P = (n/v) RT P = MRT where n/v = M 2NH 3(g) N 2(g) + 3H 2(g) K c N 2 H2 2 NH 3 3 K p P N P H 2 2 P 2 NH3 K p = K c (RT) n n = # product gas molecules - # reactant gas molecules R = L atm / mol K T = temperature in Kelvin 3 When does K c = K p? They are equal when there are the same # of gas molecules on both sides of equation. Example. Does K c = K p? a) H 2 (g) + F 2 (g) 2HF(g) b) 2SO 2 (g) + O 2 (g) 2SO 3 (g) Worked Example 13.5 CH 4(g) + 2H 2 S (g) CS 2(g) + 4H 2(g) At 1000 K, the equilibrium pressure are: CH 4 = 0.20 atm, H 2 S = 0.25 atm, CS 2 = 0.52 atm, and H 2 = 0.10 atm. What is K p? Heterogenous Equilibria Homogeneous equil.: all substances are in one phase (all gas, all solid, etc.) Heterogeneous equil.: substances in 2 or more different phases CaCO 3(s) CaO (s) + CO 2(g) Solids and liquids have constant concentrations. If you increase the amount of CaCO 3, you also increase its volume. 3

4 Heterogeneous Equilibria The Equilibrium Constant, K c Write K c and K p expressions for the following equations: a) CaCO 3 (s) CaO(s) + CO 2 (g) b) 2Cu 2 S(s) + 3O 2 (g) 2Cu 2 O(s) + 2SO 2 (g) c) Hg(l) + Hg 2+ (aq) Hg 2 2+ (aq) Worked Example 13.7, Problem 13.7 Usually one side of the equation is favored (reactants or products). For Example, 2 H 2(g) + O 2(g) 2 H 2 O (g) If K c = 2.4 x (e.g., a large number) what does this tell you about the equilibrium of the reaction? A. There are more reactants than products B. There are more products than reactants C. More information is needed. How can you figure this out? The Equilibrium Constant, K c Interpreting the Equil. Constant 2 HBr (g) H 2 (g) + Br 2 (g) K c = 2 x (e.g., a small number), what does this tell you about the equilibrium of the reaction? A. There are more reactants than products B. There are more products than reactants C. More information is needed. If K c > 10 3, products are favored over reactants; reaction goes nearly to completion. If K c < 10-3, reactants are favored over products; reaction hardly proceeds at all. If K c is in the range , appreciable concentration of reactants and products are present. Problem 13.8 Interpreting the Equil. Constant Which system below (all at equilibrium) has the largest equilibrium constant? Why? Using the Equil. Constant 2H 2(g) + O 2(g) 2H 2 O (g) K c = 2.4 x at 500K What is the concentration of H 2 at equilibrium if [H 2 O] eq = 1.0 M and [O 2 ] eq = 1.0 x M? 3H 2(g) + N 2(g) 2NH 3(g) Calculate K c at 510 K: [H 2 ] eq = M, [N 2 ] eq = M, [NH 3 ] eq = M Using the K c found, calculate P NH 3. Given: P H2 = 1.24 atm, P N2 = 2.17 atm 4

5 Group Quiz #5 Reaction Quotient, Q ClF 3(g) ClF (g) + F 2(g) K p = Calculate K c at 700 K Q is set up like K but the reaction mix may not be at equilibrium. Calculating Q tells us which direction a reaction must go to reach equilibrium. H 2(g) + I 2(g) 2HI (g) K c = 57.0 at 700 K If [H 2 ] = M, [I 2 ] = 0.20 M, and [HI] = 0.40 M, is this system at equilibrium? What needs to happen to reach equilibrium? Using Q to predict shift K c = 57.0 Q = Compare Q to K: What happens? Reaction Quotient, Q If Q c < K c, reaction shifts right (small Q means not enough products so shifts right to make more products) If Q c > K c, reaction shifts left (big Q means too many products so shifts left to convert products into reactants) If Q c = K c, reaction is at equilibrium (no shifting) Reaction Quotient, Q Finding Q & Direction of Shift Example. For the reaction, B 2A, K c = 2. Suppose 3.0 moles of A and 3.0 moles of B are introduced into a 2.00 L flask. a) Is this system at equilibrium? b) In which direction will the reaction proceed to reach equilibrium? c) Does the concentration of B increase, decrease or remain the same as the system moves towards equilibrium? 5

6 Group Quiz #6: CO (g) + 3H 2(g) CH 4(g) + H 2 O (g) K c = M 0.20 M 1.0 M 0.50 M Is this system at equilibrium? In which direction will the reaction proceed to reach equilibrium? ICE method use to find eq [ ] s, given K c and initial [ ] s 1. I = initial concentration: Initial concentration of reactants are usually given; initial [Product]'s are assumed to be 0 unless otherwise specified. 2. C = change in concentration: Assign change as the variable x; use the stoichiometry of the reaction to assign changes for all species. 3. E = equilibrium concentration: E = I + C Note, values in ICE tables can be in terms of moles or Molarity (or atm for K p ), but values used in the K c expression must be in terms of Molarity (or atm for K p ). Setting up ICE table N 2 (g) + 3H 2 (g) 2NH 3 (g) K c = 2.5 x 10-5 A reaction mixture initially contains 1.00 M N 2 and 1.00 M H 2. What are the equilibrium concentrations for N 2, H 2 and NH 3? Set up ICE table. N 2(g) 3H 2(g) 2NH 3(g) Initial 1.00 M 1.00 M 0 M Change -x -3x +2x Equilibrium x x 2x Solving for Equil. Conc. 3 methods for finding equilibrium concentrations: Use perfect squares (e.g. H 2 + I 2 2HI) = (2x) 2 (1.00 x) Assume x is much smaller than initial concentration; General rule: if K < 10-3, assume x is small Omit x terms Use quadratic if K c is too big to make assumption. get two values of x; one will give x = -b b2 4ac negative eq concentration(s) 2a Equil. Conc. - Perfect Squares Equil. Conc. - Assume small x H 2 (g) + I 2 (g) HI (g) K c = M 1.00 M Calculate the equilibrium concentrations of all species. Calculate the equilibrium concentrations for N 2 (g) + O 2 (g) NO(g) K c = 1.0 x M 0.20 M 6

7 Group Quiz #7 Solving for Equil. Conc. N 2 (g) + H 2 (g) NH 3 (g) K c = 2.5 x M 1.00 M What are the equilibrium concentrations? Hints: What do we do first? Can we assume x is small? Assumption that x is small isn t always valid! What to do? Quadratic formula!!! x = -b b2 4ac 2a Gives two values of x: If obtain +x and x, pick +x! If both x s are +, pick smaller x! Can eq concentrations (M) be values? Note: only one x should give you all + eq concs! Solving for Equil. Conc. Factors Affecting Equil. H 2 (g) + F 2 (g) 2HF(g), K c = 1.15x moles of H 2 and moles of F 2 are placed in a L container Calculate the equilibrium concentrations of all species. Now we look at what factors can cause shifts in equilibrium. These can alter expenses in industrial settings where a shift toward creation of more products is often more economical. Concentrations of reactants or products can be changed. Pressure and volume of the system can be changed (only important for gas phase reactions). Temperature can be changed. Addition of a catalyst. Addition of an inert gas (e.g. Noble gas) Le Chatelier s Principle Concentration Changes If a stress is applied to a reaction mixture at equilibrium, net reaction occurs in the direction that relieves the stress. Stress means a change in one of the factors mentioned on the previous slide. Adding a reactant or product, the equilibria shifts away from the increase in order to consume part of the added substance. Removing a reactant or product, the equilibria shifts toward the decrease to replace part of the removed species. 7

8 Changes in Concentration Changes in Concentration A B K = 4 At equilibrium [A] = 5 M and [B] = 20 M What happens if B is removed & [B] drops to 15 M? Is this still at equilibrium? If not, what is Q? Which way does this system need to shift? What happens if A is added and [A] increases to 10 M? What is Q? Which direction does it shift? N 2(g) + 3 H 2(g) 2 NH 3(g) K c = at 700 K At equilibrium, [N 2 ] = 0.50 M, [H 2 ] = 3.00 M, [NH 3 ] = 2.00 M Verify K c Adding Reactant, N 2 Changes in P and V What will happen if we add N 2 to the system? Figure 13.8 Increase in volume: more space = more gases allowed. Shift to side with more moles of gas. Increase in pressure (same as decrease in volume): less volume = less gas allowed. Shift to side with fewer moles of gas. If same number of moles on both sides, P and V don t affect equilibrium. Adding an inert gas (e.g. a noble gas) also doesn t cause an equilibrium shift or affect K c. Changes in P and V Changes in Temperature N 2 (g) + 3 H 2 (g) 2 NH 3 (g) There are 4 moles of reactant and 2 moles of product. What will happen if we increase pressure? What will happen if we increase volume? Example and Problem13.17 Changes in Conc., P, and V just shift equilibrium to maintain a constant K. Changes in Temperature will affect the value of K. We can look at heat exchange (enthalpy) of a reaction to predict shift. Endothermic: heat is a reactant Exothermic: heat is a product 8

9 Changes in Temperature How T affects K c N 2 O 4 (g) 2NO 2 (g) H o = kj Endothermic, heat is a reactant, so adding heat helps the reaction proceed forward. Heat + reactants products heat, eq shifts heat, eq shifts Endo likes it hot! Opposite shifts for exo! Low T High T Both K c and the position of the equilibrium system will vary with temperature: K c is larger when the reaction shifts right. K c is smaller when the reaction shifts left. Example. The temperature is decreased for the reaction: 2CO 2 2CO + O 2, ΔH = 566 kj. a) Will the equilibrium shift left or right? b) Does K c become larger or smaller? Adding a Catalyst Le Chatelier s Principle A catalyst speeds up a reaction so it attains equilibrium faster. If a catalyst is added to a system that is already at equilibrium, it won t affect the system at all! The catalyst does not affect the equilibrium concentrations of reactants and products in the equilibrium mixture; thus, the K c value does not change. Determine how the equilibrium will shift if the following changes are made: 2Cl 2(g) + 2H 2 O (g) 4HCl (g) + O 2(g) H o = +113 kj a) Temperature is increased b) Volume is increased c) Pressure is decreased d) HCl is added e) Ne (g) is added f) Cl 2 is added g) H 2 O is removed Group Quiz #8 Relationship between k & K c Determine how the equilibrium will shift if the following changes are made: 2H 2 (g) + O 2 (g) 2H 2 O (g) H o = kj a. Temperature is increased b. Volume is decreased c. Pressure is decreased d. H 2 is added e. O 2 is removed f. H 2 O is removed A + B C + D Forward, rate f = k f [A][B] Reverse, rate r = k r [C][D] Since forward and reverse rates are the same at equilibrium, rate f = rate r k f [A][B] = k r [C][D] k f /k r = [C][D]/[A][B] = K c (equil. cons.) Thus K c is just the ratio of the forward and reverse rate constants! 9