Chapter 14 Chemical Equilibrium Roy Kennedy Massachusetts Bay Community College Wellesley Hills, MA

Arrow Conventions Chemists commonly use two kinds of arrows in reactions to indicate the degree of completion of the reactions A single arrow indicates all the reactant molecules are converted to product molecules at the end A double arrow indicates the reaction stops when only some of the reactant molecules have been converted into products  in these notes Tro: Chemistry: A Molecular Approach, 2/e

Reaction Dynamics When a reaction starts, the reactants are consumed and products are made forward reaction = reactants  products therefore the reactant concentrations decrease and the product concentrations increase as reactant concentration decreases, the forward reaction rate decreases Eventually, the products can react to re-form some of the reactants reverse reaction = products  reactants assuming the products are not allowed to escape as product concentration increases, the reverse reaction rate increases Processes that proceed in both the forward and reverse direction are said to be reversible reactants Û products Tro: Chemistry: A Molecular Approach, 2/e

Hypothetical Reaction 2 Red Û Blue The reaction slows over time, but the Red molecules never run out! At some time between 100 and 110 sec, the concentrations of both the Red and the Blue molecules no longer change – equilibrium has been established. Notice that equilibrium does not mean that the concentrations are equal! Once equilibrium is established, the rate of Red molecules turning into Blue is the same as the rate of Blue molecules turning into Red Tro: Chemistry: A Molecular Approach, 2/e

Hypothetical Reaction 2 Red Û Blue Tro: Chemistry: A Molecular Approach, 2/e

Reaction Dynamics Once equilibrium is established, the forward and reverse reactions proceed at the same rate, so the concentrations of all materials stay constant Eventually, the reaction proceeds in the reverse direction as fast as it proceeds in the forward direction. At this time equilibrium is established. Because the reactant concentration decreases, the forward reaction slows. As the products accumulate, the reverse reaction speeds up. As the forward reaction proceeds it makes products and uses reactants Initially, only the forward reaction takes place Rate Rate Forward Rate Reverse Time Tro: Chemistry: A Molecular Approach, 2/e

Dynamic Equilibrium As the forward reaction slows and the reverse reaction accelerates, eventually they reach the same rate Dynamic equilibrium is the condition wherein the rates of the forward and reverse reactions are equal Once the reaction reaches equilibrium, the concentrations of all the chemicals remain constant because the chemicals are being consumed and made at the same rate Tro: Chemistry: A Molecular Approach, 2/e

Equilibrium  Equal The rates of the forward and reverse reactions are equal at equilibrium But that does not mean the concentrations of reactants and products are equal Some reactions reach equilibrium only after almost all the reactant molecules are consumed – we say the position of equilibrium favors the products Other reactions reach equilibrium when only a small percentage of the reactant molecules are consumed – we say the position of equilibrium favors the reactants Tro: Chemistry: A Molecular Approach, 2/e

An Analogy: Population Changes However, after a time, emigration will occur in both directions at the same rate, leading to populations in Country A and Country B that are constant, though not necessarily equal When Country A citizens feel overcrowded, some will emigrate to Country B Tro: Chemistry: A Molecular Approach, 2/e

Equilibrium Constant Even though the concentrations of reactants and products are not equal at equilibrium, there is a relationship between them The relationship between the chemical equation and the concentrations of reactants and products is called the Law of Mass Action For the general equation aA + bB Û cC + dD, the Law of Mass Action gives the relationship below the lowercase letters represent the coefficients of the balanced chemical equation always products over reactants K is called the equilibrium constant unitless Tro: Chemistry: A Molecular Approach, 2/e

Writing Equilibrium Constant Expressions For the reaction aA(aq) + bB(aq) Û cC(aq) + dD(aq) the equilibrium constant expression is So for the reaction 2 N2O5(g) Û 4 NO2(g) + O2(g) the equilibrium constant expression is Tro: Chemistry: A Molecular Approach, 2/e

What Does the Value of Keq Imply? When the value of Keq >> 1, we know that when the reaction reaches equilibrium there will be many more product molecules present than reactant molecules the position of equilibrium favors products When the value of Keq << 1, we know that when the reaction reaches equilibrium there will be many more reactant molecules present than product molecules the position of equilibrium favors reactants Tro: Chemistry: A Molecular Approach, 2/e

A Large Equilibrium Constant Tro: Chemistry: A Molecular Approach, 2/e

A Small Equilibrium Constant Tro: Chemistry: A Molecular Approach, 2/e

Practice – Write the equilibrium constant expressions, K, and predict the position of equilibrium for the following 2 SO2(g) + O2(g) Û 2 SO3(g) K = 8 x 1025 N2(g) + 2 O2(g) Û 2 NO2(g) K = 3 x 10−17 favors products favors reactants Tro: Chemistry: A Molecular Approach, 2/e

Relationships between K and Chemical Equations When the reaction is written backward, the equilibrium constant is inverted For the reaction aA + bB Û cC + dD the equilibrium constant expression is For the reaction cC + dD Û aA + bB the equilibrium constant expression is Tro: Chemistry: A Molecular Approach, 2/e

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 the reaction aA + bB Û cC the equilibrium constant expression is For the reaction 2aA + 2bB Û 2cC the equilibrium constant expression is Tro: Chemistry: A Molecular Approach, 2/e

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 the reactions (1) aA Û bB and (2) bB Û cC the equilibrium constant expressions are For the reaction aA Û cC the equilibrium constant expression is Tro: Chemistry: A Molecular Approach, 2/e

Example 14.2: Compute the equilibrium constant at 25 °C for the reaction NH3(g)  0.5 N2(g) + 1.5 H2(g) Given: Find: for N2(g) + 3 H2(g) Û 2 NH3(g), K = 3.7 x 108 at 25 °C K for NH3(g)  0.5N2(g) + 1.5H2(g), at 25 °C Conceptual Plan: Relationships: Kbackward = 1/Kforward, Knew = Koldn K’ K Solution: N2(g) + 3 H2(g) Û 2 NH3(g) K1 = 3.7 x 108 2 NH3(g) Û N2(g) + 3 H2(g) NH3(g) Û 0.5 N2(g) + 1.5 H2(g) Tro: Chemistry: A Molecular Approach, 2/e

Practice – When the reaction A(aq) Û 2 B(aq) reaches equilibrium [A] = 1.0 x 10−5 M and [B] = 4.0 x 10−1 M. When the reaction 2 B(aq) Û Z(aq) reaches equilibrium [B] = 4.0 x 10−3 M and [Z] = 2.0 x 10−6 M. Calculate the equilibrium constant for each of these reactions and the equilibrium constant for the reaction 3 Z(aq) Û 3 A(aq) Krxn 1 = 1.6 x 104 Krxn 2 = 5.0 x 10−1 2 B Û A K3 = 1/Krxn 1 = 6.25 x 10−5 Z Û 2 B K4 = 1/Krxn 2 = 2 Z Û A K3K4 = 1.25 x 10−4 3 Z Û 3 A (K3K4)3 = 2.0 x 10−12 Tro: Chemistry: A Molecular Approach, 2/e

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) Û cC(g) + dD(g) the equilibrium constant expressions are or Tro: Chemistry: A Molecular Approach, 2/e

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 Dn = 0 The relationship between them is Dn is the difference between the number of moles of reactants and moles of products Tro: Chemistry: A Molecular Approach, 2/e

Deriving the Relationship between Kp and Kc Tro: Chemistry: A Molecular Approach, 2/e

Deriving the Relationship Between Kp and Kc for aA(g) + bB(g)  cC(g) + dD(g) substituting Tro: Chemistry: A Molecular Approach, 2/e

Example 14.3: Find Kc for the reaction 2 NO(g) + O2(g) Û 2 NO2(g), given Kp = 2.2 x 1012 @ 25 °C Conceptual Plan: Relationships: Kp Kc Solution: 2 NO(g) + O2(g)  2 NO2(g) Dn = 2  3 = −1 Check: K is a unitless number because there are more moles of reactant than product, Kc should be larger than Kp, and it is Tro: Chemistry: A Molecular Approach, 2/e

2 SO2(g) + O2(g) Û 2 SO3(g) Kc = 8 x 1025 Practice – Calculate the value of Kp or Kc for each of the following at 27 °C 2 SO2(g) + O2(g) Û 2 SO3(g) Kc = 8 x 1025 N2(g) + 2 O2(g) Û 2 NO2(g) Kp = 3 x 10−17 Tro: Chemistry: A Molecular Approach, 2/e

Heterogeneous Equilibria Pure solids and pure liquids are materials whose concentration doesn’t change during the course of a reaction its amount can change, but the amount of it in solution doesn’t because it isn’t in solution Because their concentration doesn’t change, solids and liquids are not included in the equilibrium constant expression For the reaction aA(s) + bB(aq) Û cC(l) + dD(aq) the equilibrium constant expression is Tro: Chemistry: A Molecular Approach, 2/e

Heterogeneous Equilibria The amount of C is different, but the amounts of CO and CO2 remain the same. Therefore the amount of C has no effect on the position of equilibrium. Tro: Chemistry: A Molecular Approach, 2/e

Practice – Write the equilibrium constant expressions, K, and predict the position of equilibrium for the following HNO2(aq) + H2O(l) Û H3O+(aq) + NO2−(aq) K = 4.6 x 10−4 Ca(NO3)2(aq) + H2SO4(aq) Û CaSO4(s) + 2 HNO3(aq) K = 1 x 104 Tro: Chemistry: A Molecular Approach, 2/e

Calculating Equilibrium Constants from Measured Equilibrium Concentrations The most direct way of finding the equilibrium constant is to measure the amounts of reactants and products in a mixture at equilibrium The equilibrium mixture 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 the value of the equilibrium constant is independent of the initial amounts of reactants and products Tro: Chemistry: A Molecular Approach, 2/e

Initial and Equilibrium Concentrations for H2(g) + I2(g) Û 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 Tro: Chemistry: A Molecular Approach, 2/e

Calculating Equilibrium Concentrations Stoichiometry can be used to determine the equilibrium concentrations of all reactants and products if you know initial concentrations and one equilibrium concentration Use the change in the concentration of the material that you know to determine the change in the other chemicals in the reaction Tro: Chemistry: A Molecular Approach, 2/e

Calculating Equilibrium Concentrations: An Example 2 A(aq) + B(aq) Û 4 C(aq) Suppose you have a reaction 2 A(aq) + B(aq) Û 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. Product C increased by 0.50 mol/L Because Product C increased by 0.50 mol/L, you must have used ½ that amount of Reactant A, according to the stoichiometry. Because Product C increased by 0.50 mol/L, you must have used ¼ that amount of Reactant B, according to the stoichiometry. -½(0.50) -¼(0.50) +0.50 0.75 0.88 Tro: Chemistry: A Molecular Approach, 2/e

Example 14.6: Find the value of Kc for the reaction 2 CH4(g) Û C2H2(g) + 3 H2(g) at 1700 °C if the initial [CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M +0.035 Tro: Chemistry: A Molecular Approach, 2/e

Example 14.6: Find the value of Kc for the reaction 2 CH4(g) Û 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 Tro: Chemistry: A Molecular Approach, 2/e

Practice –The following data were collected for the reaction 2 NO2(g) Û N2O4(g) at 100 °C. Complete the table and determine values of Kp and Kc. Tro: Chemistry: A Molecular Approach, 2/e

Practice – A second experiment was done and the following data were collected for the reaction 2 NO2(g) Û N2O4(g) at 100 °C. Complete the table and determine values of Kp and Kc. Compare them to the first experiment. Tro: Chemistry: A Molecular Approach, 2/e

The Reaction Quotient If a reaction mixture, containing both reactants and products, is not at equilibrium; how can we determine in 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 Û cC + dD the reaction quotient is: Tro: Chemistry: A Molecular Approach, 2/e

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 Tro: Chemistry: A Molecular Approach, 2/e

Q, K, and the Direction of Reaction Tro: Chemistry: A Molecular Approach, 2/e

If Q = K, equilibrium; If Q < K, forward; If Q > K, reverse Q Example 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: Find: for I2(g) + Cl2(g) Û 2 ICl(g), Kp = 81.9 direction reaction will proceed Conceptual Plan: Relationships: If Q = K, equilibrium; If Q < K, forward; If Q > K, reverse Q PI2, PCl2, PICl Solution: I2(g) + Cl2(g) Û 2 ICl(g) Kp = 81.9 because Q (10.8) < K (81.9), the reaction will proceed to the right Tro: Chemistry: A Molecular Approach, 2/e

Practice – For the reaction CH4(g) + 2 H2S(g) Û CS2(g) + 4 H2(g) Kc = 3.59 at 900 °C. For each of the following measured concentrations determine whether the reaction is at equilibrium. If not at equilibrium, in which direction will the reaction proceed to get to equilibrium? Tro: Chemistry: A Molecular Approach, 2/e

Example 14. 8: If [COF2]eq = 0. 255 M and [CF4]eq = 0 Example 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: Find: Sort: You’re given the reaction and Kc. You’re also given the [X]eq of all but one of the chemicals 2 COF2 Û CO2 + CF4 [COF2]eq = 0.255 M, [CF4]eq = 0.118 M [CO2]eq Conceptual Plan: Relationships: Strategize: You can calculate the missing concentration by using the equilibrium constant expression K, [COF2], [CF4] [CO2] Solution: Solve: Solve the equilibrium constant expression for the unknown quantity by substituting in the given amounts Check: substitute approximations back in to see if the value agrees Check:  Tro: Chemistry: A Molecular Approach, 2/e

Practice – A sample of PCl5(g) is placed in a 0 Practice – A sample of PCl5(g) is placed in a 0.500 L container and heated to 160 °C. The PCl5 is decomposed into PCl3(g) and Cl2(g). At equilibrium, 0.203 moles of PCl3 and Cl2 are formed. Determine the equilibrium concentration of PCl5 if Kc = 0.0635 PCl5 Û PCl3 + Cl2 if in addition you calculate Kp from Kc you find that it is 2.26, slightly greater than 1 – showing that the equilibrium constant may be unreliable for predicting the position of equilibrium when it is close to 1 Tro: Chemistry: A Molecular Approach, 2/e

Finding Equilibrium Concentrations When Given the Equilibrium Constant and Initial Concentrations or Pressures Step 1: decide which direction the reaction will proceed compare Q to K Step 2: define the changes of all materials in terms of x use the coefficient from the chemical equation as 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 Step 3: 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 Tro: Chemistry: A Molecular Approach, 2/e

Example 14.11: For the reaction I2(g) + Cl2(g) Û 2 ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures are all 0.100 atm, find the equilibrium concentrations Qp(1) < Kp(81.9), so the reaction is proceeding forward Tro: Chemistry: A Molecular Approach, 2/e

Example 14.11: For the reaction I2(g) + Cl2(g) Û 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x 0.100x 0.100+2x Tro: Chemistry: A Molecular Approach, 2/e

Example 14.11: For the reaction I2(g) + Cl2(g) Û 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x 0.100x 0.100+2x Tro: Chemistry: A Molecular Approach, 2/e

Example 14.11: For the reaction I2(g) + Cl2(g) Û 2 ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures are all 0.100 atm, find the equilibrium concentrations −0.0729 x −0.0729 x +2(0.0729) +2x 0.100x 0.027 0.100x 0.027 0.100+2x 0.246 Tro: Chemistry: A Molecular Approach, 2/e

Example 14.11: For the reaction I2(g) + Cl2(g) Û 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 Tro: Chemistry: A Molecular Approach, 2/e

Practice – For the reaction I2(g) Û 2 I(g) the value of Kc = 3 Practice – For the reaction I2(g) Û 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]? (Hint: you will need to use the quadratic formula to solve for x) Tro: Chemistry: A Molecular Approach, 2/e

Practice – For the reaction I2(g) Û 2 I(g) the value of Kc = 3 Practice – For the reaction I2(g) Û 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]? because [I]initial = 0, Q = 0 and the reaction must proceed forward Tro: Chemistry: A Molecular Approach, 2/e

Practice – For the reaction I2(g) Û 2 I(g) the value of Kc = 3 Practice – For the reaction I2(g) Û 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]? Tro: Chemistry: A Molecular Approach, 2/e

Practice – For the reaction I2(g) Û 2 I(g) the value of Kc = 3 Practice – For the reaction I2(g) Û 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   Tro: Chemistry: A Molecular Approach, 2/e

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 assuming the reaction is proceeding forward Tro: Chemistry: A Molecular Approach, 2/e

Checking the Approximation and Refining as Necessary We can check our approximation 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 Tro: Chemistry: A Molecular Approach, 2/e

Example14.13: For the reaction 2 H2S(g) Û 2 H2(g) + S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask initially containing 1.25 x 10−4 mol H2S is heated to 800 °C, find the equilibrium concentrations. because no products initially, Qc = 0, and the reaction is proceeding forward Tro: Chemistry: A Molecular Approach, 2/e

Example 14.13: For the reaction 2 H2S(g) Û 2 H2(g) + S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask initially containing 1.25 x 10−4 mol H2S is heated to 800 °C, find the equilibrium concentrations. 2x +2x +x 2.50E−4 2x 2x x Tro: Chemistry: A Molecular Approach, 2/e

Example 14.13: For the reaction 2 H2S(g) Û 2 H2(g) + S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask initially containing 1.25 x 10−4 mol H2S is heated to 800 °C, find the equilibrium concentrations. 2x +2x +x 2.50E−4 2x 2x 2.50E−4 x Tro: Chemistry: A Molecular Approach, 2/e

the approximation is not valid!! Example 14.13: For the reaction 2 H2S(g) Û 2 H2(g) + S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask initially containing 1.25 x 10−4 mol H2S is heated to 800 °C, find the equilibrium concentrations. 2x +2x +x 2x x 2.50E–4 the approximation is not valid!! Tro: Chemistry: A Molecular Approach, 2/e

Example 14.13: For the reaction 2 H2S(g) Û 2 H2(g) + S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask initially containing 1.25 x 10−4 mol H2S is heated to 800 °C, find the equilibrium concentrations. 2x +2x +x 2.50E−4 2x 2x x xcurrent = 1.38 x 10−5 Tro: Chemistry: A Molecular Approach, 2/e

Example 14.13: For the reaction 2 H2S(g) Û 2 H2(g) + S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask initially containing 1.25 x 10−4 mol H2S is heated to 800 °C, find the equilibrium concentrations. 2x +2x +x 2.50E−4 2x 2x x xcurrent = 1.27 x 10−5 since xcurrent = xnew, approx. OK Tro: Chemistry: A Molecular Approach, 2/e

Example 14.13: For the reaction 2 H2S(g) Û 2 H2(g) + S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask initially containing 1.25 x 10−4 mol H2S is heated to 800 °C, find the equilibrium concentrations. 2x +2x +x 2.50E−4 2x 2x x 2.24E−4 2.56E−5 1.28E−5 xcurrent = 1.28 x 10−5 Tro: Chemistry: A Molecular Approach, 2/e

Kc(calculated) = Kc(given) within significant figures Example 14.13: For the reaction 2 H2S(g) Û 2 H2(g) + S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask initially containing 1.25 x 10−4 mol H2S is heated to 800 °C, find the equilibrium concentrations. 2x +2x +x 2.24E−4 2.56E−5 1.28E−5 Kc(calculated) = Kc(given) within significant figures Tro: Chemistry: A Molecular Approach, 2/e

Practice – For the reaction I2(g) Û 2 I(g) the value of Kc = 3 Practice – For the reaction I2(g) Û 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]? (use the simplifying assumption to solve for x) Tro: Chemistry: A Molecular Approach, 2/e

Practice – For the reaction I2(g) Û 2 I(g) the value of Kc = 3 Practice – For the reaction I2(g) Û 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]? because [I]initial = 0, Q = 0 and the reaction must proceed forward Tro: Chemistry: A Molecular Approach, 2/e

the approximation is valid!! Practice – For the reaction I2(g) Û 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]? the approximation is valid!! Tro: Chemistry: A Molecular Approach, 2/e

Practice – For the reaction I2(g) Û 2 I(g) the value of Kc = 3 Practice – For the reaction I2(g) Û 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]? 0.500  0.00217 = 0.498 [I2] = 0.498 M 2(0.00217) = 0.00434 [I] = 0.00434 M x = 0.00217   Tro: Chemistry: A Molecular Approach, 2/e

Practice – For the reaction N2O4(g) Û 2 NO2(g) the value of Kc = 1 Practice – For the reaction N2O4(g) Û 2 NO2(g) the value of Kc = 1.07 x 10−5 . If the initial concentration of N2O4 is 0.0125 M, what will be the equilibrium concentrations of [N2O4] and [NO2]? Tro: Chemistry: A Molecular Approach, 2/e

Practice – For the reaction N2O4(g) Û 2 NO2(g) the value of Kc = 1 Practice – For the reaction N2O4(g) Û 2 NO2(g) the value of Kc = 1.07 x 10−5 . If the initial concentration of N2O4 is 0.0125 M, what will be the equilibrium concentrations of [N2O4] and [NO2]? because [NO2]initial = 0, Q = 0 and the reaction must proceed forward Tro: Chemistry: A Molecular Approach, 2/e

the approximation is valid!! Practice – For the reaction N2O4(g) Û 2 NO2(g) the value of Kc = 1.07 x 10−5 . If the initial concentration of N2O4 is 0.0125 M, what will be the equilibrium concentrations of [N2O4] and [NO2]? without approximation with approximation x = 1.82 x 10−4 the approximation is valid!! Tro: Chemistry: A Molecular Approach, 2/e

Practice – For the reaction N2O4(g) Û 2 NO2(g) the value of Kc = 1 Practice – For the reaction N2O4(g) Û 2 NO2(g) the value of Kc = 1.07 x 10−5 . If the initial concentration of N2O4 is 0.0125 M, what will be the equilibrium concentrations of [N2O4] and [NO2]? [N2O4]= 0.0125  1.83 x 10−4 [N2O4] = 0.0123 M [NO2]=2(1.83 x 10−4) [NO2] = 3.66 x 10−4 M x = 1.83 x 10−4   Tro: Chemistry: A Molecular Approach, 2/e

Disturbing and Restoring 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 restored The new concentrations will be different, but the equilibrium constant will be the same unless you change the temperature Tro: Chemistry: A Molecular Approach, 2/e

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 disturbances all involve making the system open Tro: Chemistry: A Molecular Approach, 2/e

An Analogy: Population Changes When an influx of population enters Country B from somewhere outside Country A, it disturbs the equilibrium established between Country A and Country B. 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 established; the new populations will have different numbers of people than the old ones. When the populations of Country A and Country B are in equilibrium, the emigration rates between the two countries are equal so the populations stay constant. Tro: Chemistry: A Molecular Approach, 2/e

Disturbing Equilibrium: Adding or Removing Reactants After equilibrium is established, a reactant is added as long as the added reactant is included in the equilibrium constant expression i.e., not a solid or liquid How will this affect the rate of the forward reaction? How will it affect the rate of the reverse reaction? How will it affect the value of K? Tro: Chemistry: A Molecular Approach, 2/e

Disturbing Equilibrium: Adding Reactants Adding a reactant initially increases the rate of the forward reaction, but has no initial effect on the rate of the reverse reaction The reaction proceeds to the right until equilibrium is restored At the new equilibrium position, you will have more of the products than before, less of the non-added reactants than before, and less of the added reactant but not as little of the added reactant as you had before the addition At the new equilibrium position, the concentrations of reactants and products will be such that the value of the equilibrium constant is the same Tro: Chemistry: A Molecular Approach, 2/e

Disturbing Equilibrium: Adding or Removing Reactants After equilibrium is established, a reactant is removed as long as the added reactant is included in the equilibrium constant expression i.e., not a solid or liquid How will this affect the rate of the forward reaction? How will it affect the rate of the reverse reaction? How will it affect the value of K? Tro: Chemistry: A Molecular Approach, 2/e

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 restored 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 Tro: Chemistry: A Molecular Approach, 2/e

The Effect of Concentration Changes on Equilibrium Adding a reactant will decrease the amounts of the other reactants and increase the amount of the products until a new position of equilibrium is found that has the same K Removing a product will increase the amounts of the other products and decrease the amounts of the reactants. you can use this to drive a reaction to completion! Equilibrium shifts away from side with added chemicals or toward side with removed chemicals Remember, adding more of a solid or liquid does not change its concentration – and therefore has no effect on the equilibrium Tro: Chemistry: A Molecular Approach, 2/e

The Effect of Concentration Changes on Equilibrium When NO2 is added, some of it combines to make more N2O4 Tro: Chemistry: A Molecular Approach, 2/e

The Effect of Concentration Changes on Equilibrium When N2O4 is added, some of it decomposes to make more NO2 Tro: Chemistry: A Molecular Approach, 2/e

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 shift 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 affect the partial pressures of the gases in the reaction Tro: Chemistry: A Molecular Approach, 2/e

Effect of Volume Change on Equilibrium Decreasing the volume of the container increases the concentration of all the gases in the container increases their partial pressures It does not change the concentrations of solutions!! 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 Tro: Chemistry: A Molecular Approach, 2/e

Disturbing Equilibrium: Changing the Volume After equilibrium is established, the container volume is decreased How will it affect the concentration of solids, liquid, solutions, and gases? How will this affect the total pressure of solids, liquid, and gases? How will it affect the value of K? Tro: Chemistry: A Molecular Approach, 2/e

Disturbing Equilibrium: Reducing the Volume Decreasing the container volume increases the concentration of all gases, but not solids, liquid, or solutions same number of moles, but different number of liters, resulting in a different molarity 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 Because 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 Tro: Chemistry: A Molecular Approach, 2/e

The Effect of Volume Changes on Equilibrium Because there are more gas molecules on the reactants side of the reaction, when the pressure is increased the position of equilibrium shifts toward the side with fewer molecules to decrease the pressure 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 Tro: Chemistry: A Molecular Approach, 2/e

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 Tro: Chemistry: A Molecular Approach, 2/e

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 concentrations of products and increase the concentrations of 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 Tro: Chemistry: A Molecular Approach, 2/e

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 concentrations of reactants and increase the concentrations of 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 Tro: Chemistry: A Molecular Approach, 2/e

The Effect of Temperature Changes on Equilibrium Tro: Chemistry: A Molecular Approach, 2/e

Not Changing the Position of Equilibrium: The Effect of Catalysts Catalysts provide an alternative, more efficient mechanism Catalysts work for both forward and reverse reactions Catalysts affect the rate of the forward and reverse reactions by the same factor Therefore, catalysts do not affect the position of equilibrium Tro: Chemistry: A Molecular Approach, 2/e

Practice – Le Châtelier’s Principle The reaction 2 SO2(g) + O2(g) Û 2 SO3(g) with DH° = −198 kJ is at equilibrium. How will each of the following changes affect the equilibrium concentrations of each gas once equilibrium is restored? 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 Tro: Chemistry: A Molecular Approach, 2/e

Practice – Le Châtelier’s Principle The reaction 2 SO2(g) + O2(g) Û 2 SO3(g) with DH° = −198 kJ is at equilibrium. How will each of the following changes affect the equilibrium concentrations of each gas once equilibrium is restored? adding more O2 to the container shift to SO3 condensing and removing SO3 shift to SO3 compressing the gases shift to SO3 cooling the container shift to SO3 doubling the volume of the container shift to SO2 warming the mixture shift to SO2 adding helium to the container no effect adding a catalyst to the mixture no effect Tro: Chemistry: A Molecular Approach, 2/e