Chemical Equilibrium 𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑡𝑠⇋𝑝𝑟𝑜𝑑𝑢𝑐𝑡𝑠 If the absorption of O2 by hemoglobin was not reversible we would suffocate, O2 absorption illustrates the principle of chemical equilibrium Reactions go forwards and backwards

Dynamic Equilibrium as the forward reaction slows and the reverse reaction accelerates, eventually they reach the same rate dynamic equilibrium is the condition where the rates of the forward and reverse reactions are equal once the reaction reaches equilibrium, the concentrations of all the chemicals remain constant

Dynamic Equilibrium: H2 + I2  2HI This reaction occurs in one elementary step The rate law is connected to the molecularity 𝐻 2 + 𝐼 2 →2HI 𝑟𝑎𝑡𝑒 𝑓 = 𝑘 𝑓 𝐻 2 𝐼 2 But as with all reactions that go forward, the reaction can also go backward 2HI→𝐻 2 + 𝐼 2 𝑟𝑎𝑡𝑒 𝑏 = 𝑘 𝑏 𝐻𝐼 2 We write these two equations as one like so 𝐻 2 + 𝐼 2 ⇌2HI At equilibrium the rate to go forward = rate to go backward kf kb

Reaction Dynamics

Reaction Dynamics } } Rate we use up H2 Rate we make H2

Reaction Dynamics

Reaction Dynamics 𝑘 𝑓 𝑘 𝑏 = 𝐻𝐼 𝑒𝑞 2 𝐻 2 𝑒𝑞 𝐼 2 𝑒𝑞 = 𝐾 𝑐

Enthalpy DH and Equilibrium Constant Kc 𝐾 𝑐 = 𝑘 𝑓 𝑘 𝑏 = 𝐴 𝑓 𝐴 𝑏 ∙ 𝑒 − 𝐸 𝑎 𝑅𝑇 𝑒 − (𝐸 𝑎 −Δ𝐻) 𝑅𝑇 𝑘 𝑓 𝑘 𝑏 = 𝐻𝐼 𝑒𝑞 2 𝐻 2 𝑒𝑞 𝐼 2 𝑒𝑞 = 𝐾 𝑐 𝑖𝑓 𝐴 𝑓 = 𝐴 𝑏 𝐾 𝑐 = 𝑘 𝑓 𝑘 𝑏 = 𝑒 − Δ𝐻 𝑅𝑇 Ea 𝑖𝑓 𝐴 𝑓 ≠ 𝐴 𝑏 𝐾 𝑐 = 𝑘 𝑓 𝑘 𝑏 = 𝑒 − Δ𝐻 𝑅𝑇 + Δ𝑆 𝑅 DH = -|DH| Here DH < 0 and let’s assume for now that Af ≈ Ab 𝐾 𝑐 ≈ 𝑒 |Δ𝐻| 𝑅𝑇 “The speed of a chemical reaction is determined by kinetics, but the extent of the reaction is determined by the relative stability of reactants and product – this is the dominion of thermodynamics” As T gets big Kc  1 As T gets small Kc gets big

Equilibrium: Multistep Reaction Overall

Equilibrium: Multistep Reaction Overall

Law of Mass Action The overall equilbrium constant is the same as you would get if you treated the overall reaction as occuring in one elementary reaction

Equilibrium Constant for a Reaction aA(aq) + bB(aq) cC(aq) + dD(aq) 2 N2O5 4 NO2 + O2

Interpreting Keq Keq >> 1 more product molecules present than reactant molecules the position of equilibrium favors products Keq << 1 more reactant molecules present than product molecules the position of equilibrium favors reactants

Keq >> 1 Lots of HBr but very little H2 or Br2

Keq << 1 Lots of N2 and O2 but very little NO

Relationships between K and Chemical Equations when the reaction is written backwards, the equilibrium constant is inverted for aA + bB cC + dD for cC + dD aA + bB

Relationships between K and Chemical Equations when the reaction is written backwards, the equilibrium constant is inverted for aA + bB cC + dD for cC + dD aA + bB

Relationships between K and Chemical Equations when the coefficients of an equation are multiplied by a factor, the equilibrium constant is raised to that factor for aA + bB cC the equilibrium constant expression is: for 2aA + 2bB 2cC the equilibrium constant expression is:

Relationships between K and Chemical Equations when you add equations to get a new equation, the equilibrium constant of the new equation is the product of the equilibrium constants of the old equations for (1) aA bB (2) bB cC for aA cC

Equilibrium Constants for Reactions Involving Gases the concentration of a gas in a mixture is proportional to its partial pressure therefore, the equilibrium constant can be expressed as the ratio of the partial pressures of the gases for aA(g) + bB(g) cC(g) + dD(g) or

Deriving the Relationship between Kp and Kc

Deriving the Relationship Between Kp and Kc for aA(g) + bB(g) cC(g) + dD(g) substituting

Kc and Kp in calculating Kp, the partial pressures are always in atm the values of Kp and Kc are not necessarily the same because of the difference in units Kp = Kc when Δn = 0 the relationship between them is: Δn is the difference between the number of moles of reactants and moles of products

Heterogeneous Equilibria Consider The reaction occurs at the surface where the solid and water meet Instead of the concentration of NaCl(s) the rate of dissolving is related to the surface area ANaCl K is independent of the amount of solid!

Heterogeneous Equilibria Consider The Equilibrium constant would be But what is [H2O]? What is [H+]? The [H2O] is effectively a constant for changes in Equilibrium and an be folded into the equilibrium constant so the rate of the backward reaction doesn’t change

Heterogeneous Equilibria pure liquids are materials whose concentration doesn’t change (much) during the course of a reaction and is essentially constant compared to other reagents For solids, the forward and backward reactions depend on the surface area making the equilibrium constant independent of the amount of solid for the reaction aA(s) + bB(aq) cC(l) + dD(aq) the equilibrium constant expression is:

Heterogeneous Equilibria The amount of C is different, but the amounts of CO and CO2 remains the same. Therefore the amount of C has no effect on the position of equilibrium.

Initial and Equilibrium Concentrations for H2(g) + I2(g) 2HI(g) @ 445°C Constant [H2] [I2] [HI] 0.50 0.0 0.11 0.78 0.055 0.39 0.165 1.17 1.0 0.5 0.53 0.033 0.934

Calculating Equilibrium Concentrations Stoichiometry can be used to determine the equilibrium concentrations of all reactants and products if you know .. Initial concentrations One equilibrium concentration Example: suppose you have a reaction 2 A(aq) + B(aq) 4 C(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. What are [A] and [B] at equilibrium? [A] [B] [C] Initial molarity 1.00 Change in concentration Equilibrium molarity 0.50 -½(0.50) -¼(0.50) +0.50 0.75 0.88

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 [CH4] [C2H2] [H2] initial 0.115 0.000 change equilibrium 0.035 use the known change to determine the change in the other materials add the change to the initial concentration to get the equilibrium concentration in each column use the equilibrium concentrations to calculate Kc + + -2(0.035) +0.035 +3(0.035) 0.045 0.105

The Reaction Quotient if a reaction mixture, containing both reactants and products, is 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 the concentration ratio of the products (raised to the power of their coefficients) to the reactants (raised to the power of their coefficients) is called the reaction quotient, Q For the gas phase reaction aA + bB cC + dD the reaction quotient is:

The Reaction Quotient: Predicting the Direction of Change if Q > K, the reaction will proceed fastest in the reverse direction the [products] will decrease and [reactants] will increase if Q < K, the reaction will proceed fastest in the forward direction 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

If Q = K, equilibrium; If Q < K, forward; If Q > K, reverse Q For the reaction below, which direction will it proceed if PI2 = 0.114 atm, PCl2 = 0.102 atm & PICl = 0.355 atm? Given: Find: for I2(g) + Cl2(g) 2 ICl(g), Kp = 81.9 direction reaction will proceed Concept Plan: Relationships: If Q = K, equilibrium; If Q < K, forward; If Q > K, reverse Q PI2, PCl2, PICl Solution: I2(g) + Cl2(g) 2 ICl(g) Kp = 81.9 since Q (10.8) < K (81.9), the reaction will proceed to the right

Finding Equilibrium Concentrations: 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

since Qp(1) < Kp(81.9), the reaction is proceeding forward 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] [Cl2] [ICl] Initial 0.100 Change Equilibrium since Qp(1) < Kp(81.9), the reaction is proceeding forward

For the reaction I2(g) + Cl2(g) 2 ICl(g) @ 25°C, Kp = 81. 9 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] [Cl2] [ICl] initial 0.100 change equilibrium -x -x +2x 0.100-x 0.100-x 0.100+2x

For the reaction I2(g) + Cl2(g) 2 ICl(g) @ 25°C, Kp = 81. 9 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 [I2] [Cl2] [ICl] initial 0.100 change equilibrium substitute into the equilibrium constant expression and solve for x -x -x +2x 0.100-x 0.100-x 0.100+2x

For the reaction I2(g) + Cl2(g) 2 ICl(g) @ 25°C, Kp = 81. 9 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 [I2] [Cl2] [ICl] initial 0.100 change equilibrium check by substituting the equilibrium concentrations back into the equilibrium constant expression and comparing the calculated K to the given K -0.0729 -0.0729 2(0.0729) 0.027 0.027 0.246 Kp(calculated) = Kp(given) within significant figures

For the reaction I2(g) 2 I(g) Kc = 3. 76 x 10-5 at 1000 K If 1 For the reaction I2(g) 2 I(g) Kc = 3.76 x 10-5 at 1000 K If 1.00 moles of I2 is placed into a 2.00 L flask and heated, what will be the equilibrium concentrations of [I2] and [I]? (Hint: you will need to use the quadratic formula to solve for x)

For the reaction I2(g) 2 I(g) the value of Kc = 3. 76 x 10-5 at 1000 K For the reaction I2(g) 2 I(g) the value of Kc = 3.76 x 10-5 at 1000 K. If 1.00 moles of I2 is placed into a 2.00 L flask and heated, what will be the equilibrium concentrations of [I2] and [I]? since [I]initial = 0, Q = 0 and the reaction must proceed forward [I2] [I] initial 0.500 change -x +2x equilibrium 0.500- x 2x

For the reaction I2(g) 2 I(g) the value of Kc = 3. 76 x 10-5 at 1000 K For the reaction I2(g) 2 I(g) the value of Kc = 3.76 x 10-5 at 1000 K. If 1.00 moles of I2 is placed into a 2.00 L flask and heated, what will be the equilibrium concentrations of [I2] and [I]? [I2] [I] initial 0.500 change -x +2x equilibrium 0.500- x 2x

For the reaction I2(g) 2 I(g) the value of Kc = 3. 76 x 10-5 at 1000 K For the reaction I2(g) 2 I(g) the value of Kc = 3.76 x 10-5 at 1000 K. If 1.00 moles of I2 is placed into a 2.00 L flask and heated, what will be the equilibrium concentrations of [I2] and [I]? [I2] [I] initial 0.500 change -x +2x equilibrium 0.498 0.00432 0.500 - 0.00216 = 0.498 [I2] = 0.498 M 2(0.00216) = 0.00432 [I] = 0.00432 M x = 0.00216 

Approximations to Simplify the Math when the equilibrium constant is very small, the position of equilibrium favors the reactants for relatively large initial concentrations of reactants, the reactant concentration will not change significantly when it reaches equilibrium the [X]equilibrium = ([X]initial ± ax) ≈ [X]initial we are approximating the equilibrium concentration of reactant to be the same as the initial concentration

Checking the Approximation Refining as Necessary we can check our approximation afterwards by comparing the approximate value of x to the initial concentration if the approximate value of x is less than 5% of the initial concentration, the approximation is valid

For the reaction I2(g) 2 I(g) the value of Kc = 3. 76 x 10-5 at 1000 K For the reaction I2(g) 2 I(g) the value of Kc = 3.76 x 10-5 at 1000 K. If 1.00 mole of I2 is placed into a 2.00 L flask and heated, what will be the equilibrium concentrations of [I2] and [I]? Since [I]initial = 0, Q = 0 and the reaction must proceed forward Kc is very small so equilibrium stays to the left and x is small [I2] [I] initial 0.500 change -x +2x equilibrium 0.500- x 2x

Is the approximation valid? For the reaction I2(g) 2 I(g) the value of Kc = 3.76 x 10-5 at 1000 K. If 1.00 moles of I2 is placed into a 2.00 L flask and heated, what will be the equilibrium concentrations of [I2] and [I]? [I2] [I] initial 0.500 change -x +2x equilibrium 0.500- x 2x Is the approximation valid? Yes

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 guides us in predicting the effect various changes in conditions have on the position of equilibrium it says that if a system at equilibrium is disturbed, the position of equilibrium will shift to minimize the disturbance

The Effect of Concentration Changes on Equilibrium

Disturbing Equilibrium: Adding Reactants Adding a reactant increases the rate of the forward reaction [Reactant] decreases and [Product] increases Equilibrium shifts toward the product side (Equilibrium shifts right) you can use this to drive a reaction to completion! Remember, adding more of a solid or liquid does not change its concentration – and therefore has no effect on the equilibrium

Disturbing Equilibrium: Adding Products Adding a product increases the rate of the backward reaction [Reactant] increases and [Product] decreases Equilibrium shifts toward the reactant side (Equilibrium shifts to the left)

Disturbing Equilibrium: Removing Reactants Removing a reactant initially decreases the rate of the forward reaction, but has no initial effect on the rate of the reverse reaction so the reaction is going faster in reverse The reaction proceeds to the left until equilibrium is re-established at the new equilibrium position, you will have less of the products than before, more of the non-removed reactants than before, and more of the removed reactant but not as much of the removed reactant as you had before the removal at the new equilibrium position, the concentrations of reactants and products will be such that the value of the equilibrium constant is the same

Disturbing Equilibrium: Removing Products Removing a product initially decreases the rate of the backward reaction, but has no initial effect on the rate of the forward reaction so the reaction is going faster in the forward direction The reaction proceeds to the right until equilibrium is re-established At the new equilibrium position, you will have less of the reactants than before, more of the non-removed products than before, and more of the removed product but not as much of the removed product as you had before the removal at the new equilibrium position, the concentrations of reactants and products will be such that the value of the equilibrium constant is the same

Effect of Volume Change on Gas- Phase Equilibria Decreasing the size of the container increases the concentration of all the gases in the container and their partial pressures 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

Effect of Volume Change on Gas- Phase Equilibria Consider Decrease V P increase equilibrium shift to reduce moles of gas Increase V P decreases equilibrium shifts to increase the moles of gas

Disturbing Equilibrium: Changing the Volume How will decreasing the container volume affect the total pressure of solids, liquid, and gases? How will it affect the concentration of solids, liquid, solutions, and gases? What will this cause? How will it affect the value of K?

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 increases the concentration of all gases since the total pressure increases, the position of equilibrium will shift to decrease the pressure by removing 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

Volume Changes: Equilibrium Since there are more gas molecules on the reactants side of the reaction, when the pressure is increased the position of equilibrium shifts toward the products. When the pressure is decreased by increasing the volume, the position of equilibrium shifts toward the side with the greater number of molecules – the reactant side.

Temperature and Equilibrium 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 Temperature changes K

Temperature and Equilibrium

Temperature Changes: Exothermic Reactions for an exothermic reaction, heat, dq, 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 concentrations of products and increase the concentrations of reactants (shift left) adding heat to an exothermic reaction will decrease the value of K

Temperature Changes: 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 concentrations of reactants and increase the concentrations of products (shifts right) adding heat to an endothermic reaction will increase the value of K

Temperature and 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 The reaction 2 SO2(g) + O2(g) 2 SO3(g) with ΔH° = -198 kJ at equilibrium. 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 right right right right left left No change no change