Ch. 13. Fundamental Equilibrium Concepts

Creative Commons License Ch. 13: Fundamental Equilibrium Concepts  Images and tables in this file have been used from the following sources:  OpenStax: Creative Commons Attribution License 4.0. Movement of carbon dioxide through tissues and blood cells involves several equilibrium reactions. OpenStax: Chemistry 2 1 The Equilibrium State N2O4 (g) ⇌ 2NO2 (g) • Not all chemical reactions go to completion ─ there are many that are reversible and they attain a state of equilibrium in which a mixture of reactants and products is present. ➢ Note: a double arrow is used for an equilibrium reaction to show it is a reversible reaction. ➢ In the forward reaction each molecule of N2O4 breaks down to form two molecules of NO2. • Equilibrium defined: • Rate of the forward reaction equals the rate of the reverse reaction. • Reactant and product concentrations remain constant and no longer change. • It DOES NOT mean reactant and product concentrations are equal! ➢ In the reverse reaction two molecules of NO2 combine to form N2O4. ➢ Equilibrium occurs when the rate at which an N2O4 molecule breaks apart in the forward reaction is equal to the rate at which it is formed by the reverse reaction. 3 Equilibrium is Dynamic! 4 Chemical Equilibrium N2O4 (g) ⇌ 2NO2 (g) 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. ratef = kf[N2O4] and rater = kr[NO2]2 Starting with N2O4 5 Starting with NO2 6 1 Relationship Between Rate Constants and Equilibrium Constants Concentration and rate vs. time N2O4 (g) ⇌ 2NO2 (g) At equilibrium: rate forward = rate reverse Figure 13.2 kf[N2O4] = kr[NO2]2 N2O4 ⇌ 2NO2 equilibrium expression [NO 2 ]2 kf = Kc = kr [N 2 O 4 ] N2O4 = colorless NO2 = orange equilibrium constant Kc is the equilibrium constant, a numerical value. The ratio is the equilibrium constant expression. 7 The Equilibrium Constant, Kc The Equilibrium Constant Write the equilibrium constant expressions (Kc) for the following equations. a) CO(g) + 3H2(g) ⇌ CH4(g) + H2O(g) b) 2NH3(g) ⇌ N2(g) + 3H2(g) For a reaction: aA + bB ⇌ cC + dD Equilibrium Constant: Kc = [C]c [D]d [A]a [B]b 8 products reactants Kc is the ratio of the equilibrium concentrations of products over the equilibrium concentrations of reactants each raised to the power of their stoichiometric coefficients. a) b) Kc is a constant at a given temperature for a reaction at equilibrium (law of mass action). 9 Equilibrium Constant Kc Characteristics: The equilibrium concentrations of reactants and products may be different, but the value for Kc remains the same. Trial [N2O4]i, M [NO2]i, M [N2O4]eq, M [NO2]eq, M 0.0400 0.0000 0.0337 0.0125 4.64x10 -3 2 0.0000 0.0800 0.0337 0.0125 4.64x10 -3 3 0.0600 0.0000 0.0522 0.0156 4.66x10 -3 4 0.0000 0.0600 0.0246 0.0107 4.65x10 -3 5 0.0200 0.0600 0.0429 0.0141 4.63x10 -3 N2O4(g) ⇌ 2NO2(g) 2 [NO2 ] Kc = [N2O4 ] 1) Equilibrium can be attained starting with only reactants or products (or can start with a mixture of reactants and products). 2) Kc does not depend on the initial concentrations of reactants and products. 3) Kc does depend on temperature. 4) Kc values are listed without units. Kc 1 10  don't include units when calculating Kc. For Trial 1 Kc = 11 12 2 Kc tells us extent of reaction Calculating Kc • Large Kc: If Kc > 102, the reaction is nearly complete; products are favored. 2NH3(g) ⇌ N2(g) + 3H2(g) At 500 K, the following concentrations were measured: [N2] = 3.0 x 10-2 M, [H2] = 3.7 x 10-2 M, [NH3] = 1.6 x 10-2 M. What is Kc? Eq lies to right (more products) • Small Kc: If Kc < 10-2, the reaction hardly proceeds at all; reactants are favored. Eq lies to left (more reactants) • If Kc is in the range 10-2 - 102, appreciable amounts of reactants and products are present. Eq lies close to middle 13 14 Heterogenous Equilibria Interpreting Equilibrium Constants • Which system below (all at equilibrium) has the largest equilibrium constant? Smallest? Why? Homogeneous equilibria: all substances are in one phase (all gas, all solid, etc.) Heterogeneous equilibria: substances in 2 or more different phases. CaCO3(s) ⇌ CaO(s) + CO2(g) • Pure solids and liquids have constant concentrations so they are not included in the K expressions. • Omit solids and liquids from K expressions A2 + B2 ⇌ 2 AB A2 + C2 ⇌ 2 AC --only include gases and aqueous substances which have variable concentrations! A2 + D2 ⇌ 2 AD 15 Heterogeneous Equilibria 16 Answers Kc & Kp Expressions Kc – use [ ] terms to represent concentrations of (aq) and (g) substances. Kp – use P terms to represent the partial pressure of (g) substances. a) 2Cu2S(s) + 3O2(g) ⇌ 2Cu2O(s) + 2SO2(g) Write Kc and Kp expressions for the following equations : a) 2Cu2S(s) + 3O2(g) ⇌ 2Cu2O(s) + 2SO2(g) b) Hg(l) + Hg2+(aq) ⇌ Hg22+(aq) b) Hg(l) + Hg2+(aq) ⇌ Hg22+(aq) DON’T include solids and liquids in K expressions! 17 Always include charges for ions! 18 3 Example Kp Relationship between Kc and Kp Pressure is directly proportional to concentration. • PV = nRT → P = (n/V) RT → P = MRT where n/V = M CH4(g) + 2H2S(g) ⇌ CS2(g) + 4H2(g) • At 1000 K, the equilibrium pressure are: CH4 = 0.20 atm, H2S = 0.25 atm, CS2 = 0.52 atm, and H2 = 0.10 atm. • What is Kp? Kp = Kc(RT)n n = # product gas molecules - # reactant gas molecules R = 0.08206 L·atm / mol·K T = temperature in Kelvin 19 20 Example: Converting from Kc to Kp When does Kc = Kp? N2(g) + 3H2(g) ⇌ 2 NH3(g) At 375 C, Kc = 2.79x10-5 What is Kp? They are equal when there are the same # of gas molecules on both sides of equation. KP = Kc(RT)Δn Example. Does Kc = Kp? a) H2(g) + F2(g) ⇌ 2HF(g) b) 2SO2(g) + O2(g) ⇌ 2SO3(g) Note: while Kc and Kp don’t have units we must plug in values that have correct units or K values wouldn’t be a constant! 21 Manipulating Reactions & K values B. Adding Reactions: A. Reversing a Reaction 1) 2NO(g) + O2(g) ⇌ 2NO2(g) K1 = 22 2 1. 2NO2(g) ⇌ 2NO(g) + O2(g) K1 = [NO] [O22 ] [NO 2 ]2 [NO 2 ] [NO ]2 [O 2 ] 2. 2H2 (g) + O2(g) ⇌ 2H2O (g) K2 = [H2O] [H2 ]2 [O 2 ] ______________________ 2 2 2) 2NO2(g) ⇌ 2NO(g) + O2(g) K2 = [NO ] [O 2 ] [NO 2 ]2 2 2 3. 2NO2 + 2H2 ⇌ 2NO + 2H2O K3 = [NO] [2H2O]2 K2 is reciprocal of K1: Thus K3 = K1 * K2 [NO 2 ] [H2 ] If K1 = 1.5x10-6 and K2 = 2.4x1047 Then K3 = If K1 = 6.8x105, then K2 = 23 24 4 Reaction Quotient, Q Q (blue) versus K (red) Q is set up like K, but we plug in initial concentrations and compare Q to K to tell which direction (if any) a reaction must go to reach equilibrium. • If Qc < Kc, reaction shifts right (small Q means not enough products so shifts right to make more products). Q K • If Qc > Kc, reaction shifts left (big Q means too many products so shifts left to convert products into reactants). Q K K Q QK 25 Finding Q & Direction of Shift 26 Calculating Q for Example. For the reaction, B ⇌ 2A, Kc = 3. 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? B ⇌ 2A [A] = [B] = • Q= • Compare Q to Kc =3 • What direction does equilibria shift? • Does [B] increase or decrease? 27 Le Chatelier’s Principle 28 Concentration Changes Le Châtelier’s principle: When a stress is applied to a system at equilibrium, it will respond by shifting in the direction that counteracts the effect of the stress. • 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. Types of Stresses  A change in concentration (addition or removal of a reactant or product).  A change in pressure or volume (only important for gas phase reactions).  A change in temperature. 29 30 5 Changes in P and V Effect of Concentration Changes Addition of a reactant or removal of a product will cause equilibrium to shift to the right. • 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 Kc. Removal of a reactant or addition of a product or will cause equilibrium to shift to the left. Increase Increase Decrease N2 (g) + 3 H2 (g) ⇌ 2 NH3 (g) Decrease Decrease Increase 31 32 Changes in P and V Shifting Equilibria: Vol./Press. N2(g) + 3H2(g) ⇌ 2NH3(g) When volume is decreased, pressure is increased. Equilibrium is driven toward the side with fewer moles of gas. What happens when volume is increased? 1 + 3 = 4 moles (left) 2 moles (right) What will happen if we increase pressure? What will happen if we increase volume? N2O4 (g) ⇌ 2 NO2 (g) gray red 33 Changes in Temperature How T alters Kc N2O4 (g) ⇌ 2NO2 (g) Ho = +57.2 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! Low T 34 Changes in Conc., P, and V just shift equilibrium position to maintain a constant K. Both the equilibrium position and Kc will vary with temperature:  Kc is larger when the reaction shifts right (more products).  Kc is smaller when the reaction shifts left (more reactants). High T 35 Example. The temperature is decreased for the reaction: 2CO2 ⇌ 2CO + O2, ΔH = 566 kJ. a) Will the equilibrium shift left or right? b) Does Kc become larger or smaller? 36 6 Effect of Temperature Changes Endotheric or Exothermic Reaction? CoCl42- + 6H2O ⇌ Co(H2O)62+ + 4Clpink blue For an endothermic reaction, heat is a reactant: heat + reactants ⇌ products ΔH° > 0 kJ/mol • Adding heat shifts the reaction →, Kc ↑ • Removing heat shifts the reaction ←, Kc ↓ Endo likes it hot! For an exothermic reaction, heat is a product: reactants ⇌ products + heat ΔH° < 0 kJ/mol • Adding heat shifts the reaction ←, Kc ↓ • Removing heat shifts the reaction →, Kc ↑ Opposite of endo! Room Temperature Bunsen burner Ice water bath 37 38 Effect of Catalysts & Inert Gases Applying Le Chatelier’s Principle Catalysts – adding a catalyst reduces Ea which occurs to the same extent for both the forward and reverse reactions. • It reduces the time required to reach equilibrium but has no effect on Kc or the position of equilibrium. • Adding an inert gas also has no effect on position of Equilibrium. • Determine how the equilibrium will shift if the following changes are made: • 2Cl2(g) + 2H2O(g) ⇌ 4HCl(g) + O2(g) Ho = +113 kJ a) Temperature is increased b) Volume is increased c) Pressure is decreased d) HCl is added e) Ne (g) is added f) Cl2 is added g) H2O is removed 39 ICE method – use to find eq [ ]’s, given Kc and initial [ ] ‘s 40 Solving for Equil. Conc. • 3 methods for finding equilibrium concentrations:  Use perfect squares (e.g. H2 + I2 ⇌ 2HI) 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 ▪ 50.5 = (2x) 2 (1.00 − x)2  Assume x is much smaller than initial concentration; General rule: if K < 10-3, assume x is small ▪ Omit –x terms Note, values in ICE tables can be in terms of moles or Molarity (or atm for Kp), but values used in the Kc expression must be in terms of Molarity (or atm for Kp). 41  Use quadratic if Kc is too big to make assumption. ▪ get two values of x; one will give negative eq concentration(s) x= -b  b 2 − 4ac 2a 42 7 Equil. Conc. - Perfect Squares Perfect Square Cont Calculate the equilibrium concentrations of all species. H2 (g) + I2 (g) ⇌ 2HI (g) Kc = 50.5 Kc = [HI] 2 [H2 ][I2 ] Plug in eq terms from ICE: Given the initial concentrations of H2 and I2 are both 1.00 M. Set up ICE H2 (g) + I Kc = Set up Kc 1.00 I2 (g) ⇌ 2HI (g) 1.00 0 C -x -x +2x E 1.00 - x 1.00 - x +2x [HI] 2 [H2 ][I2 ] Solve for x: Plug x into eq [ ]’s: •plug eq concs. back into Kc to doublecheck! 43 Using Q to find Eq Concentrations Approximation Method- Assume small x A 5.00 L reaction vessel is filled with 2.50 moles of H2, 2.50 moles of I2, and 25.0 moles of HI. Kc = 81.0 at 670 K. H2(g) + I2(g) ⇌ 2HI (g), I (M) Is it at equilibrium? Use Q to tell direction it proceeds! Q= compare Q to K What are the equilibrium concentrations? H2 (g) + 44 Calculate the equilibrium concentrations for N2(g) + O2(g) ⇌ 2 NO(g) Kc = 1.0 x 10-5 Initial: 0.80 M 0.20 M • Set up Kc: • Set up ICE and plug eq concs into Kc: • Assume x is small (its much smaller than 0.80 or 0.20) because Kc is small (less than 10-3) I2 (g) ⇌ 2HI (g) I x= C [N2] = E 45 Quadratic Method , [O2] = , [NO] = • Check assumption: (x /initial concentration) * 100% < 5% 46 Quadratic Example H2(g) + F2(g) ⇌ 2HF(g), Kc = 1.15x102 3.000 moles of H2 and 6.000 moles of F2 are placed in a 3.000 L container Calculate the equilibrium concentrations of all species. x= [H2] = [F2] = [HF] = Assumption that x is small isn’t always valid! What to do? Quadratic formula!!! -b  b 2 − 4ac x= 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! 47 48 8 Solving for x and Kc In an experiment, 3.00 moles of CO and 5.00 moles of H2O were placed in a 2.00 L flask at 350 K. CO (g) + H2O (g) ⇌ CO2 (g) + H2 (g) At equilibrium, there were 2.00 mol of CO remaining. What is the value of the equilibrium constant for this reaction? Initial Change Equilibrium x= CO (g) + H2O (g) ⇌ CO2 (g) + H2 (g) 1.50 M 2.50 M 0 0 -x -x +x +x 1.00 M 2.00 M 0.500 M 0.500 M K= 49 9