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Chemistry II Chapter 14 Chemical Equilibrium Speed of a chemical reaction is determined by kinetics How fast a reaction goes Extent of a chemical reaction is determined by thermodynamics How far a reaction goes Reaction Dynamics • forward reaction: reactants products therefore the [reactant] decreases and the [product] increases as [reactant] decreases, the forward reaction rate decreases • reverse reaction: products reactants assuming the products are not allowed to escape as [product] increases, the reverse reaction rate increases • processes that proceed in both the forward and reverse direction are said to be reversible reactants ⇌ products Reaction Dynamics Rate Initially, only the Because As Once the equilibrium forward the reactant reaction isforward established, concentration proceeds Eventually, the reaction proceeds reaction takes place. itin decreases, the makes forward products the and forward reverse and uses reaction reactions reactants. the reverse direction as fast asslows. As the products at in thethe same accumulate, rate,direction. sothe the itproceed proceeds forward reverse concentrations reaction ofspeeds all materials At this time equilibrium isup. established. stay constant. Rate Forward Rate Reverse Time Concentration H2(g) + I2(g) ⇌ 2 HI(g) Since the Asreactant the reaction proceeds, the [H2] and concentrations [I2]are decrease and the decreasing, [HI] increases the forward reaction And sincerate the slows down product concentration is increasing, the reverse reaction rate speeds up Since the [HI] at equilibrium is larger than the [H2] or [I2], we say the position of equilibrium favors products At Once equilibrium, equilibrium the is forward established, reaction the rate is the concentrations same as theno reverse longer reaction change rate Time Equilibrium Established Dynamic Equilibrium • some reactions reach equilibrium only after almost all the reactant molecules are consumed – equilibrium favors the products • other reactions reach equilibrium when only a small percentage of the reactant molecules are consumed – equilibrium favors the reactants An Analogy: Pwad(left) ⇌ Pwad(right) Rules: 1. Each student wads up two paper wads. 2. You must start and stop as the timekeeper says. 3. Throw only one paper wad at a time. 4. If a paper wad lands next to you, you must throw it back. # of paper wads on right K eq # of paper wads on left Equal Number of Students on Each Side of the Classroom Time (s) 0 Left Pwad (left) Right K eq All 0 Pwad (right) # of paper wads on right # of paper wads on left 70 5 10 15 Number of Paper Wads 60 50 40 30 20 10 0 0 20 5 10 Time (s) 15 20 Most Students on the Left Side– 2 Students on the Right Side Time (s) 0 Left Pwad (left) ⇌ Pwad (right) Right K eq 0 All # of paper wads on right # of paper wads on left 70 5 10 15 Number of Paper Wads 60 50 40 30 20 10 0 0 20 5 10 Time (s) 15 20 Most Students on the Left Side– 2 Students on the Right Side Time (s) 0 Left Pwad (left) ⇌ Pwad (right) Right K eq All 0 # of paper wads on right # of paper wads on left 70 5 10 15 Number of Paper Wads 60 50 40 30 20 10 0 0 5 10 Time (s) 20 15 20 Common Misconceptions • Equilibrium means equal amounts of reactant and product. No - A reaction can be at equilibrium and have more paper wads on the product side of the room • A reaction at equilibrium has stopped. No - The paper wads keep flying in both directions even after equilibrium is achieved • Equilibrium can only be achieved by starting with reactants. No - Equilibrium can also be achieved when starting with all of the paper wads on the product side of the room. The Equilibrium Constant • [reactants] and [products] are not equal at equilibrium, there is a relationship between them Law of Mass Action: aA + bB ⇌ cC + dD, Keq (or Kc) is called the equilibrium constant Units vary from reaction to reaction, unitless in this case C D K eq a b A B c d The Equilibrium Constant so for the reaction: 2 N2O5 ⇌ 4 NO2 + O2 NO2 O 2 K eq 2 N 2O5 4 Units: mol3/L3 The Equilibrium Constant Significance of Keq • Keq >> 1, when the reaction reaches equilibrium there will be more product than reactant molecules the position of equilibrium favors products • Keq << 1, when the reaction reaches equilibrium there will be more reactant molecules than product molecules the position of equilibrium favors reactants A Large Equilibrium Constant Remember: does not tell us about how fast, only how far A Small Equilibrium Constant Relationships between K and Chemical Equations • reaction is written backwards, the equilibrium constant is inverted for the reaction aA + bB ⇌ cC + dD the equilibrium constant expression is: for the reaction cC + dD ⇌ aA + bB the equilibrium constant expression is: C D a b A B A B K backward c d C D c K forward d K backward a 1 Kforward b Relationships between K and Chemical Equations • coefficients of an equation are multiplied by a factor, K is raised to that factor for the reaction aA + bB ⇌ cC equilibrium constant expression is: C K original Aa Bb c Knew K for the reaction 2aA + 2bB ⇌ 2cC equilibrium constant expression is: 2c C K new A2 a B2 b 2 c C a b A B n original Relationships between K and Chemical Equations • when you add equations to get a new equation, Keq for the new equation is the product of the equilibrium constants of the old equations for the reactions (1) aA ⇌ bB and (2) bB ⇌ cC the K expressions are: B K1 Aa b for the overall reaction aA ⇌ cC the K expression is: C K2 b B c Knew K1 K2 c C K new a A b c B C a b A B Ex 14.2 – Compute the equilibrium constant at 25°C for the reaction NH3(g) ⇌ 0.5 N2(g) + 1.5 H2(g) Given: for N2(g) + 3 H2(g) ⇌ 2 NH3(g), Keq = 3.7 x 108 at 25°C Find: Keq for NH3(g) ⇌ 0.5N2(g) + 1.5H2(g), at 25°C Concept Plan: K K’ Relationships: Kbackward = 1/Kforward, Knew = Koldn Solution: N2(g) + 3 H2(g) 2 NH3(g) 2 NH3(g) ⇌ N2(g) + 3 H2(g) NH3(g) ⇌ 0.5 N2(g) + 1.5 H2(g) K1 = 3.7 x 108 1 1 K2 K1 3.7 108 K ' K 2 1 2 1 8 3.7 10 K ' 5.2 10 5 1 2 Equilibrium Constant in Terms of Pressure Reactions Involving Gases • the concentration of a gas in a mixture is proportional to its partial pressure aA(g) + bB(g) ⇌ cC(g) + dD(g) C D Kc a b A B c d PC PD Kp a b PA PB c or d Kc and Kp • when calculating Kp, partial pressures are always in atm • the values of Kp and Kc are not necessarily the same because of the difference in units • the relationship between them is: Kp Kc RT n Δn = c + d – (a + b) or no. moles of reactants minus no. Moles products When n = 0, Kp = Kc Ex 14.3 – Find Kc for the reaction 2 NO(g) + O2(g) ⇌ 2 NO2(g), given Kp = 2.2 x 1012 @ 25°C Given: Kp = 2.2 x 1012 atm-1 Find: Kc Concept Plan: Relationships: Kp KP Kc RT n K p Kc RT n Solution: 2 NO(g) + O2(g) ⇌ 2 NO2(g) n = 2 3 = -1 Check: Kc Kc Kp RT n 2.2 1012 0.08206 atm L mol K 298 K 1 5.4 1013 K has units L mol-1 since there are more moles of reactant than product, Kc should be larger than Kp, and it is Heterogeneous Equilibria: Reactions Involving Solids and Liquids aA(s) + bB(aq) ⇌ cC(l) + dD(aq) D Kc b B d • Concentrations of solids and liquids doesn’t change its amount can change, but the amount of it in solution doesn’t because it isn’t in solution • solids and liquids are not included in the Keq expression Heterogeneous Equilibria CO2 Kc 2 CO Kp PCO 2 2 PCO The amount of C is different, amounts of CO and CO2 remain the same. C has no effect on the position of equilibrium. Calculating Equilibrium Constants from Measured Equilibrium Concentrations • To find K measure the amounts of reactants and products in a mixture at equilibrium • may have different amounts of reactants and products, but the value of the equilibrium constant will always be the same as long as the temperature is kept constant Initial and Equilibrium Concentrations for H2(g) + I2(g) ⇌ 2HI(g) @ 445°C Equilibrium Initial [H2] 0.50 0.0 0.50 1.0 [I2] 0.50 [HI] [H2] 0.0 [I2] 0.11 0.11 [HI] 0.78 Equilibrium Constant Keq [HI] 2 [H 2 ][I 2 ] [0.78]2 50 [0.11][0.11] 0.50 0.055 0.055 0.39 [0.39]2 50 [0.055][0.055] 0.50 0.50 0.165 0.165 1.17 [1.17]2 50 [0.165][0.165] 0.0 0.5 0.0 0.53 0.033 0.934 [0.934]2 50 [0.53][0.033] 37 Calculating Equilibrium Constants from Measured Equilibrium Concentrations • Stoichiometry can be used to determine the equilibrium [reactants] and [products] if you know initial concentrations and one equilibrium concentration e.g. 2A(aq) + B(aq) ⇌ 4C(aq) with initial concentrations [A] = 1.00 M, [B] = 1.00 M, and [C] = 0. You then measure the equilibrium concentration of C as [C] = 0.50 M. initial molarity [A] [B] [C] 1.00 1.00 0 change in concentration -½(0.50) -¼(0.50) +0.50 equilibrium molarity 0.75 0.88 0.50 Calculating Equilibrium Concentrations e.g. 2A(aq) + B(aq) ⇌ 4C(aq) initial molarity [A] [B] [C] 1.00 1.00 0 change in concentration -½(0.50) -¼(0.50) +0.50 equilibrium molarity 0.75 0.88 0.50 Referred to as ICE table (initial, change, equilibrium) Keq = [C]4/[A]2[B] = 0.13 Ex 14.6 Find the value of Kc for the reaction 2 CH4(g) ⇌ C2H2(g) + 3 H2(g) at 1700°C if the initial [CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M Construct an ICE table [CH4] [C2H2] [H2] for the reaction initial 0.115 0.000 0.000 change +0.035 for the substance whose equilibrium equilibrium 0.035 concentration is known, calculate the change in concentration Ex 14.6 Find the value of Kc for the reaction 2 CH4(g) ⇌ C2H2(g) + 3 H2(g) at 1700°C if the initial [CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M use the known change [CH4] [C2H2] [H2] to determine the change in the other initial 0.115 0.000 0.000 materials -2(0.035) +0.035 +3(0.035) change add the change to the initial concentration to 0.105 equilibrium 0.035 0.045 get the equilibrium concentration in each column use the equilibrium concentrations to calculate Kc C 2 H 2 H 2 Kc 2 CH 4 3 0.0350.105 2 2 0 . 020 mol L 2 0.045 3 The Reaction Quotient • reaction mixture not at equilibrium; how can we determine which direction it will proceed? • the answer is to compare the current concentration ratios to the equilibrium constant reaction quotient, Q for the gas phase reaction aA + bB ⇌ cC + dD the reaction quotient is: c d C D Qc Aa Bb PC PD Qp a b PA PB c d The Reaction Quotient: Predicting the Direction of Change • if Q > K, the reaction will proceed fastest in the reverse direction Q must decrease, the [products] will decrease and [reactants] will increase • if Q < K, the reaction will proceed fastest in the forward direction Q must increase, the [products] will increase and [reactants] will decrease • if Q = K, the reaction is at equilibrium the [products] and [reactants] will not change • if a reaction mixture contains just reactants, Q = 0, and the reaction will proceed in the forward direction • if a reaction mixture contains just products, Q = ∞, and the reaction will proceed in the reverse direction Q, K, and the Direction of Reaction 47 Ex 14.7 – For the reaction below, which direction will it proceed if PI2 = 0.114 atm, PCl2 = 0.102 atm & PICl = 0.355 atm Given: for I2(g) + Cl2(g) ⇌ 2 ICl(g), Kp = 81.9 Find: direction reaction will proceed Concept Plan: PI2, PCl2, PICl Q Relationships: If Q = K, equilibrium; If Q < K, forward; If Q > K, reverse Solution: I2(g) + Cl2(g) ⇌ 2 ICl(g) Kp = 81.9 PICl 0.355 Qp PI 2 PCl 2 0.114 0.102 2 2 Qp 10.8 since Q (10.8) < K (81.9), the reaction will proceed to the right Ex 14.8 If [COF2]eq = 0.255 M and [CF4]eq = 0.118 M, and Kc = 2.00 @ 1000°C, find the [CO2]eq for the reaction given. Given: Sort: You’re given the 2 COF2 ⇌ CO2 + CF4 reaction and Kc. You’re also [COF2]eq = 0.255 M, [CF4]eq = 0.118 M given the [X]eq of all but one Find: [CO2]eq of the chemicals Concept Strategize: You can calculate [CO2] Plan: K, [COF2], [CF4] the missing concentration by using the equilibrium CO2 CF4 constant expression Relationships: Kc COF2 2 Solve: Solve the equilibrium constant expression for the unknown quantity by substituting in the given amounts Check: Round to 1 sig fig and substitute back in Solution: Check: CO2 CF4 COF2 2 2 COF2 CO2 K c CF4 2 0.255 2.00 1.10 M 0.118 Kc Units & Magnitude OK Finding Equilibrium Concentrations When Given the Equilibrium Constant and Initial Concentrations or Pressures • first decide which direction the reaction will proceed compare Q to K • define the changes of all materials in terms of x use the coefficient from the chemical equation for the coefficient of x the x change is + for materials on the side the reaction is proceeding toward the x change is for materials on the side the reaction is proceeding away from • solve for x for 2nd order equations, take square roots of both sides or use the quadratic formula may be able to simplify and approximate answer for very large or small equilibrium constants Ex 14.11 For the reaction I2(g) + Cl2(g) ⇌ 2 ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures are all 0.100 atm, find the equilibrium concentrations Construct an ICE table for the reaction determine the direction the reaction is proceeding [I2] 0.100 initial change equilibrium [Cl2] 0.100 [ICl] 0.100 PICl 0.100 Qp PI2 PCl 2 0.100 0.100 2 2 Qp 1 since Qp(1) < Kp(81.9), the reaction is proceeding forward Ex 14.11 For the reaction I2(g) + Cl2(g) ⇌ 2 ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures are all 0.100 atm, find the equilibrium concentrations represent the change in the partial pressures in terms of x sum the columns to find the equilibrium concentrations in terms of x substitute into the equilibrium constant expression and solve for x [I2] 0.100 x [Cl2] 0.100 x [ICl] 0.100 +2x initial change equilibrium 0.100x 0.100x 0.100+2x 2 PICl Kp PI2 PCl 2 2 2 0.100 2 x 0.100 2 x 81.9 0.100 x 0.100 x 0.100 x 2 Ex 14.11 For the reaction I2(g) + Cl2(g) ⇌ 2 ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures are all 0.100 atm, find the equilibrium concentrations substitute into the equilibrium constant expression and solve for x [I2] 0.100 x [Cl2] 0.100 x [ICl] 0.100 +2x initial change equilibrium 0.100x 0.100x 0.100+2x 0.100 2 x 2 0.100 2 x 81.9 2 0.100 x 0.100 x 81.9 0.100 x 0.100 2 x 81.9 0.100 81.9 x 0.100 2 x 81.9 0.100 0.100 2 x 81.9 x 81.9 0.100 0.100 2 x 81.9 x 0.805 11.05 x 0.0729 x Ex 14.11 For the reaction I2(g) + Cl2(g) ⇌ 2 ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures are all 0.100 atm, find the equilibrium concentrations substitute x into the equilibrium concentration definition and solve [I2] [Cl2] 0.100 0.100 x x -0.0729 -0.0729 [ICl] 0.100 +2x 2(-0.0729) initial change 0.027 0.100x 0.027 0.100+2x 0.246 equilibrium 0.100x 0.0729 x PI2 0.100 x 0.100 0.0729 0.027 atm PCl 2 0.100 x 0.100 0.0729 0.027 atm PICl 0.100 2 x 0.100 20.0729 0.246 atm Disturbing and Re-establishing Equilibrium • once a reaction is at equilibrium, the concentrations • • of all the reactants and products remain the same however if the conditions are changed, the concentrations of all the chemicals will change until equilibrium is re-established the new concentrations will be different, but the equilibrium constant will be the same unless you change the temperature Le Châtelier’s Principle • Le Châtelier's Principle: if a system at equilibrium is disturbed, the position of equilibrium will shift to minimize the disturbance disturbances all involve making the system open Concentration, temperature, volume, pressure An Analogy: Population Changes The result will be people moving from Country B into Country A faster than people moving from Country A into Country B. This will continue until a new equilibrium between the populations is When the populations of Country A and Country B established, however the new When an influx ofthe population enters Country B the are in equilibrium, emigration rates between populations have different from outside Country A, itwill disturbs the two somewhere states are equal so the populations stay constant. numbers of people the B. old equilibrium established between Country A and than Country ones. The Effect of Concentration Changes on Equilibrium • Adding reactant : aA + bB ⇌ cC + dD System shifts in a direction to minimize the disturbance • increases the amount of products until a new equilibrium is found that has the same K C D K eq a b A B c d The Effect of Concentration Changes on Equilibrium • Removing product: aA + bB ⇌ cC + dD System shifts in a direction to minimize the disturbance • will increase the amounts of products and decrease the amounts of reactants you can use this to drive a reaction to completion! C D K eq a b A B c d The Effect of Adding a Gas to a Gas Phase Reaction at Equilibrium • adding a gaseous reactant increases its partial pressure, • causing the equilibrium to the right increasing its partial pressure increases its concentration does not increase the partial pressure of the other gases in the mixture adding an inert gas to the mixture has no effect on the position of equilibrium does not effect the partial pressures of the gases in the reaction Tro, Chemistry: A Molecular Approach 86 The Effect of Concentration Changes on Equilibrium Describe the effect of adding more NO2(g) to the following reaction: N2O4(g) ⇌ NO2(g) The Effect of Concentration Changes on Equilibrium Q=K When NO2 is added, some of it combines to make more N2O4 Q>K Q=K The Effect of Concentration Changes on Equilibrium Describe the effect of adding more N2O4(g) to the following reaction: N2O4(g) ⇌ NO2(g) The Effect of Concentration Changes on Equilibrium N2O4(g) ⇌ NO2(g) When N2O4 is added, reaction shifts to make more NO2 Summarizing If a chemical system is in equilibrium: • Inc. [reactant] (Q < K) reaction shifts to right • Inc. [product] (Q > K) reaction shifts to left • Dec. [reactant] (Q > K) reaction shifts to left • Dec. [product] (Q < K) reaction shifts to right Effect of Volume Change on Equilibrium • decreasing the size of the container increases the concentration of • • • • all the gases in the container increases their partial pressures if their partial pressures increase, then the total pressure in the container will increase according to Le Châtelier’s Principle, the equilibrium should shift to remove that pressure the way the system reduces the pressure is to reduce the number of gas molecules in the container when the volume decreases, the equilibrium shifts to the side with fewer gas molecules The Effect of Volume Changes on Equilibrium Since When there theare pressure more gas is decreased molecules by increasing on the the volume, reactantsthe side position of the of equilibrium reaction,shifts when toward the the side pressure with the is increased greater number the ofposition molecules of equilibrium – the reactant shifts towardside. the products. 93 Disturbing Equilibrium: Reducing the Volume • for solids, liquids, or solutions, changing the size of the container has no effect • • • • on the concentration, therefore no effect on the position of equilibrium decreasing the container volume will increase the total pressure Boyle’s Law if the total pressure increases, the partial pressures of all the gases will increase Dalton’s Law of Partial Pressures decreasing the container volume increases the concentration of all gases same number of moles, but different number of liters, resulting in a different molarity since the total pressure increases, the position of equilibrium will shift to decrease the pressure by removing gas molecules shift toward the side with fewer gas molecules at the new equilibrium position, the partial pressures of gaseous reactants and products will be such that the value of the equilibrium constant is the same Summarizing If a chemical system is in equilibrium: • Dec. volume causes reaction to shift in direction that has fewer moles of gas particles • Inc. volume causes reaction to shift in direction that has the greater number of moles of gas particles • If equal moles of gas on both sides, no effect • Adding an inert gas has no effect The Effect of Temperature Changes on Equilibrium Position • exothermic reactions release energy and endothermic • reactions absorb energy if we write Heat as a product in an exothermic reaction or as a reactant in an endothermic reaction, it will help us use Le Châtelier’s Principle to predict the effect of temperature changes even though heat is not matter and not written in a proper equation The Effect of Temperature Changes on Equilibrium for Exothermic Reactions • for an exothermic reaction, heat is a product • increasing the temperature is like adding heat • according to Le Châtelier’s Principle, the equilibrium will • shift away from the added heat adding heat to an exothermic reaction will decrease the [products] and increase the [reactants] adding heat to an exothermic reaction will decrease the value of K how will decreasing the temperature affect the system? aA + bB ⇌ cC + dD + Heat c d C D Kc a b A B The Effect of Temperature Changes on Equilibrium for Endothermic Reactions • for an endothermic reaction, heat is a reactant • increasing the temperature is like adding heat • according to Le Châtelier’s Principle, the equilibrium will shift away from the added heat adding heat to an endothermic reaction will decrease the [reactants] and increase the [products] adding heat to an endothermic reaction will increase the value of K • how will decreasing the temperature affect the system? Heat + aA + bB ⇌ cC + dD C D Kc a b A B c d The Effect of Temperature Changes on Equilibrium Not Changing the Position of Equilibrium - Catalysts • catalysts provide an alternative, more efficient mechanism • works for both forward and reverse reactions • affects the rate of the forward and reverse reactions by the • same factor therefore catalysts do not affect the position of equilibrium Practice - Le Châtelier’s Principle 2 SO2(g) + O2(g) ⇌ 2 SO3(g) H° = -198 kJ How will each of the following changes affect the equilibrium concentrations of each gas once equilibrium is re-established? adding more O2 to the container condensing and removing SO3 compressing the gases cooling the container doubling the volume of the container warming the mixture adding the inert gas helium to the container adding a catalyst to the mixture Practice - Le Châtelier’s Principle 2 SO2(g) + O2(g) ⇌ 2 SO3(g) adding more O2 to the container condensing and removing SO3 compressing the gases cooling the container doubling the volume of the container warming the mixture adding helium to the container adding a catalyst to the mixture shift to SO3 shift to SO3 shift to SO3 shift to SO3 shift to SO2 shift to SO2 no effect no effect