Chemical Equilibrium

Chemical Equilibrium 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 • But as with all reactions that go forward, the reaction can also go backward • We write these two equations as one like so • 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

Enthalpy DH and Equilibrium Constant Kc Ea DH = -|DH| Here DH < 0 and let’s assume for now that Af Ab “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

Initial and Equilibrium Concentrations forH2(g) + I2(g) 2HI(g) @ 445°C

Equilibrium: Multistep Reaction At equilibrium ratef = ratebfor every step Overall

Equilibrium: Multistep Reaction At equilibrium, K1, K2 and K3 are all constant so therefore K1 x K2 x K3 = Keq = Kcis also a constant 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 KandChemical Equations • when the reaction is written backwards, the equilibrium constant is inverted for aA + bBcC + dD for cC + dDaA + bB

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

Relationships between KandChemical 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 aAcC for (1) aAbB (2) bBcC

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 Relationshipbetween Kp and Kc

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

Kcand 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

HeterogeneousEquilibria • 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!

HeterogeneousEquilibria • 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

HeterogeneousEquilibria • 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:

HeterogeneousEquilibria 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.

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? -½(0.50) -¼(0.50) +0.50 0.88 0.75

Find the value of Kc for the reaction2 CH4(g) C2H2(g) + 3 H2(g) at 1700°C if the initial [CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M + + -2(0.035) +0.035 +3(0.035) 0.045 0.105

The Reaction Quotient For the gas phase reaction aA + bBcC + dD the reaction quotient is: • 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

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

PI2, PCl2, PICl 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 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

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 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 -x -x +2x 0.100+2x 0.100-x 0.100-x

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 -0.0729 -0.0729 2(0.0729) 0.027 0.027 0.246 Kp(calculated) = Kp(given) within significant figures

For the reactionI2(g) 2 I(g) Kc = 3.76 x 10-5 at 1000 KIf 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. 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

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]?

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]? 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. 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

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]? • Is the approximation valid? • Yes

Disturbing and Re-establishingEquilibrium • 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

Chemical Equilibrium