Chemical Equilibrium Chapter 14 Copyright © The McGraw-Hill Companies, Inc.  Permission required for reproduction or display.

Contents and Concepts Describing Chemical Equilibrium Chemical Equilibrium—A Dynamic Equilibrium The Equilibrium Constant Heterogeneous Equilibria; Solvents in Homogeneous Equilibria Using the Equilibrium Constant Qualitatively Interpreting the Equilibrium Constant Predicting the Direction of Reaction Calculating Equilibrium Concentrations Copyright © Cengage Learning. All rights reserved.

Changing Reaction Conditions: Le Châtelier’s Principle Removing Products or Adding Reactants Changing the Pressure and Temperature Effect of a Catalyst Copyright © Cengage Learning. All rights reserved.

Chemical reactions often seem to stop before they are complete. Actually, such reactions are reversible. That is, the original reactants form products, but then the products react with themselves to give back the original reactants. When these two reactions—forward and reverse—occur at the same rate, a chemical equilibrium exists. Copyright © Cengage Learning. All rights reserved.

CO(g) + 3H2(g)  CH4(g) + H2O(g) The graph shows how the amounts of reactants and products change as the reaction approaches equilibrium. Copyright © Cengage Learning. All rights reserved.

Equilibrium is a state in which there are no observable changes as time goes by. Chemical equilibrium is achieved when: the rates of the forward and reverse reactions are equal and the concentrations of the reactants and products remain constant Physical equilibrium H2O (l) H2O (g) NO2 Chemical equilibrium N2O4 (g) 2NO2 (g)

CO(g) + 3H2(g)  CH4(g) + H2O(g) This graph shows how the rates of the forward reaction and the reverse reaction change as the reaction approaches equilibrium. Copyright © Cengage Learning. All rights reserved.

N2O4 (g) 2NO2 (g) equilibrium equilibrium equilibrium Start with NO2 Start with NO2 & N2O4

constant

N2O4 (g) 2NO2 (g) K = [NO2]2 [N2O4] = 4.63 x 10-3 aA + bB cC + dD K = [C]c[D]d [A]a[B]b Law of Mass Action

Write the equilibrium expressions, Kc, for each of the following reactions: 2H2S(g)  2H2 (g) + S2 (g) 2HCl (g) + 1/2O2 (g)  Cl2(g) + H2O(g)

Equilibrium Will K = [C]c[D]d [A]a[B]b aA + bB cC + dD K >> 1 Lie to the right Favor products K << 1 Lie to the left Favor reactants

We can apply stoichiometry to compute the content of the reaction mixture at equilibrium. Copyright © Cengage Learning. All rights reserved.

When heated PCl5, phosphorus pentachloride, forms PCl3 and Cl2 as follows: PCl5(g)  PCl3(g) + Cl2(g) When 1.00 mol PCl5 in a 1.00-L container is allowed to come to equilibrium at a given temperature, the mixture is found to contain 0.135 mol PCl3. What is the molar composition of the mixture? Copyright © Cengage Learning. All rights reserved.

Initially we had 1.00 mol PCl5 and no PCl3 or Cl2. We will organize this problem by using the chemical reaction to set up a table of initial, change, and equilibrium amounts. Initially we had 1.00 mol PCl5 and no PCl3 or Cl2. The change in each is stoichiometric: If x moles of PCl5 react, then x moles of PCl3 and x moles of Cl2 are produced. For reactants, this amount is subtracted from the original amount; for products, it is added to the original amount. Copyright © Cengage Learning. All rights reserved.

We can now find the amounts of the other substances. PCl5(g)  PCl3(g) + Cl2(g) Initial 1.00 mol Change –x +x Equilibrium 1.00 – x x We were told that the equilibrium amount of PCl3 is 0.135 mol. That means x = 0.135 mol. We can now find the amounts of the other substances. Copyright © Cengage Learning. All rights reserved.

(given with 3 significant figures) Moles PCl5 = 1.00 – 0.135 = 0.87 mol (2 decimal places) Moles PCl3 = 0.135 mol (given with 3 significant figures) Moles Cl2 = 0.135 mol Copyright © Cengage Learning. All rights reserved.

The Equilibrium Constant, Kc The equilibrium constant expression for a reaction is obtained by multiplying the concentrations of products, dividing by the concentrations of reactants, and raising each concentration term to a power equal to its coefficient in the balanced chemical equation. The equilibrium constant, Kc, is the value obtained for the Kc expression when equilibrium concentrations are substituted. Copyright © Cengage Learning. All rights reserved.

Homogenous equilibrium applies to reactions in which all reacting species are in the same phase. N2O4 (g) 2NO2 (g) Kp = NO2 P 2 N2O4 P Kc = [NO2]2 [N2O4] In most cases Kc  Kp aA (g) + bB (g) cC (g) + dD (g) Kp = Kc(RT)Dn Dn = moles of gaseous products – moles of gaseous reactants = (c + d) – (a + b)

Homogeneous Equilibrium CH3COOH (aq) + H2O (l) CH3COO- (aq) + H3O+ (aq) [CH3COO-][H3O+] [CH3COOH][H2O] Kc = ′ [H2O] = constant [CH3COO-][H3O+] [CH3COOH] = Kc [H2O] ′ Kc = General practice not to include units for the equilibrium constant.

What is the Kc expression for this reaction? Methanol (also called wood alcohol) is made commercially by hydrogenation of carbon monoxide at elevated temperature and pressure in the presence of a catalyst: 2H2(g) + CO(g)  CH3OH(g) What is the Kc expression for this reaction? Copyright © Cengage Learning. All rights reserved.

When we are given some information about equilibrium amounts, we are able to calculate the value of Kc. We need to take care to remember that the Kc expression uses molar concentrations. Copyright © Cengage Learning. All rights reserved.

The equilibrium concentrations for the reaction between carbon monoxide and molecular chlorine to form COCl2 (g) at 740C are [CO] = 0.012 M, [Cl2] = 0.054 M, and [COCl2] = 0.14 M. Calculate the equilibrium constants Kc and Kp. CO (g) + Cl2 (g) COCl2 (g) [COCl2] [CO][Cl2] = 0.14 0.012 x 0.054 Kc = = 220 Kp = Kc(RT)Dn Dn = 1 – 2 = -1 R = 0.0821 T = 273 + 74 = 347 K Kp = 220 x (0.0821 x 347)-1 = 7.7

The equilibrium constant Kp for the reaction is 158 at 1000K. What is the equilibrium pressure of O2 if the PNO = 0.400 atm and PNO = 0.270 atm? 2 2NO2 (g) 2NO (g) + O2 (g) Kp = 2 PNO PO PNO PO 2 = Kp PNO PO 2 = 158 x (0.400)2/(0.270)2 = 347 atm

What is the value of Kc at this temperature? Carbon dioxide decomposes at elevated temperatures to carbon monoxide and oxygen: 2CO2(g)  2CO(g) + O2(g) At 3000 K, 2.00 mol CO2 is placed into a 1.00-L container and allowed to come to equilibrium. At equilibrium, 0.90 mol CO2 remains. What is the value of Kc at this temperature? Copyright © Cengage Learning. All rights reserved.

We can find the value of x. 2.00 – 2x = 0.90 1.10 = 2x x = 0.55 mol 2CO2(g)  2CO(g) + O2(g) Initial 2.00 mol Change –2x +2x +x Equilibrium 2.00 – 2x 2x x 0.90 mol 1.10 mol 0.55 mol We can find the value of x. 2.00 – 2x = 0.90 1.10 = 2x x = 0.55 mol 2x = 2(0.55) = 1.10 mol Copyright © Cengage Learning. All rights reserved.

2CO2(g)  2CO(g) + O2(g) Copyright © Cengage Learning. All rights reserved.

Heterogenous equilibrium applies to reactions in which reactants and products are in different phases. CaCO3 (s) CaO (s) + CO2 (g) [CaO][CO2] [CaCO3] Kc = ′ [CaCO3] = constant [CaO] = constant [CaCO3] [CaO] Kc x ′ Kc = [CO2] = Kp = PCO 2 The concentration of solids and pure liquids are not included in the expression for the equilibrium constant.

In a heterogeneous equilibrium, in the Kc expression, the concentrations of solids and pure liquids are constant (due to these substances’ constant density). As a result, we incorporate those constants into the value of Kc, thereby making a new constant, Kc. In other words, equilibrium is not affected by solids and pure liquids as long as some of each is present. More simply, we write the Kc expression by replacing the concentration of a solid or pure liquid with 1. Copyright © Cengage Learning. All rights reserved.

H2O(g) + C(s)  CO(g) + H2(g) Write the Kc expression for the following reaction: H2O(g) + C(s)  CO(g) + H2(g) Copyright © Cengage Learning. All rights reserved.

does not depend on the amount of CaCO3 or CaO CaCO3 (s) CaO (s) + CO2 (g) PCO 2 = Kp PCO 2 does not depend on the amount of CaCO3 or CaO

Consider the following equilibrium at 295 K: The partial pressure of each gas is 0.265 atm. Calculate Kp and Kc for the reaction? NH4HS (s) NH3 (g) + H2S (g) Kp = P NH3 H2S P = 0.265 x 0.265 = 0.0702 Kp = Kc(RT)Dn Kc = Kp(RT)-Dn Dn = 2 – 0 = 2 T = 295 K Kc = 0.0702 x (0.0821 x 295)-2 = 1.20 x 10-4

′ ′ ′ ′ ′′ Kc = [C][D] [A][B] Kc = [E][F] [C][D] A + B C + D Kc Kc ′′ C + D E + F [E][F] [A][B] Kc = A + B E + F Kc Kc = Kc ′′ ′ x If a reaction can be expressed as the sum of two or more reactions, the equilibrium constant for the overall reaction is given by the product of the equilibrium constants of the individual reactions.

′ N2O4 (g) 2NO2 (g) 2NO2 (g) N2O4 (g) = 4.63 x 10-3 K = [NO2]2 [N2O4] = 216 When the equation for a reversible reaction is written in the opposite direction, the equilibrium constant becomes the reciprocal of the original equilibrium constant.

When the reaction is doubled: 2aA + 2bB  2cC + 2dD; K2 Given: aA + bB  cC + dD; K1 When the reaction is doubled: 2aA + 2bB  2cC + 2dD; K2 The equilibrium constant expression, K2 , is the square of the equilibrium constant expression, K1: K2 = Copyright © Cengage Learning. All rights reserved.

aA(g) + bB(g)  cC(g) + dD(g) For the reaction aA(g) + bB(g)  cC(g) + dD(g) The equilibrium constant expressions are Kc = and Kp = Copyright © Cengage Learning. All rights reserved.

From the ideal gas law, we know that How are these related? We know From the ideal gas law, we know that So, Copyright © Cengage Learning. All rights reserved.

In general, the value of Kp is different from that of Kc. When you express an equilibrium constant for a gaseous reaction in terms of partial pressures, you call it the equilibrium constant, Kp. In general, the value of Kp is different from that of Kc. Recall the ideal gas law and the relationship between pressure and molarity of a gas: Copyright © Cengage Learning. All rights reserved.

CO(g) + 3H2(g) CH4(g) + H2O(g) For catalytic methanation, CO(g) + 3H2(g) CH4(g) + H2O(g) the equilibrium expression in terms of partial pressures becomes and Copyright © Cengage Learning. All rights reserved.

Writing Equilibrium Constant Expressions The concentrations of the reacting species in the condensed phase are expressed in M. In the gaseous phase, the concentrations can be expressed in M or in atm. The concentrations of pure solids, pure liquids and solvents do not appear in the equilibrium constant expressions. The equilibrium constant is a dimensionless quantity. In quoting a value for the equilibrium constant, you must specify the balanced equation and the temperature. If a reaction can be expressed as a sum of two or more reactions, the equilibrium constant for the overall reaction is given by the product of the equilibrium constants of the individual reactions.

Chemical Kinetics and Chemical Equilibrium ratef = kf [A][B]2 A + 2B AB2 kf kr rater = kr [AB2] Equilibrium ratef = rater kf [A][B]2 = kr [AB2] kf kr [AB2] [A][B]2 = Kc =

The reaction quotient (Qc) is calculated by substituting the initial concentrations of the reactants and products into the equilibrium constant (Kc) expression. IF Qc > Kc system proceeds from right to left to reach equilibrium Qc = Kc the system is at equilibrium Qc < Kc system proceeds from left to right to reach equilibrium

Calculating Equilibrium Concentrations Express the equilibrium concentrations of all species in terms of the initial concentrations and a single unknown x, which represents the change in concentration. Write the equilibrium constant expression in terms of the equilibrium concentrations. Knowing the value of the equilibrium constant, solve for x. Having solved for x, calculate the equilibrium concentrations of all species.

ICE At 1280oC the equilibrium constant (Kc) for the reaction Is 1.1 x 10-3. If the initial concentrations are [Br2] = 0.063 M and [Br] = 0.012 M, calculate the concentrations of these species at equilibrium. Br2 (g) 2Br (g) Let x be the change in concentration of Br2 Br2 (g) 2Br (g) Initial (M) 0.063 0.012 ICE Change (M) -x +2x Equilibrium (M) 0.063 - x 0.012 + 2x [Br]2 [Br2] Kc = Kc = (0.012 + 2x)2 0.063 - x = 1.1 x 10-3 Solve for x

Kc = (0.012 + 2x)2 0.063 - x = 1.1 x 10-3 4x2 + 0.048x + 0.000144 = 0.0000693 – 0.0011x 4x2 + 0.0491x + 0.0000747 = 0 -b ± b2 – 4ac  2a x = ax2 + bx + c =0 x = -0.0105 x = -0.00178 Br2 (g) 2Br (g) Initial (M) Change (M) Equilibrium (M) 0.063 0.012 -x +2x 0.063 - x 0.012 + 2x At equilibrium, [Br] = 0.012 + 2x = -0.009 M or 0.00844 M At equilibrium, [Br2] = 0.062 – x = 0.0648 M

Nitrogen and oxygen form nitric oxide. N2(g) + O2(g)  2NO(g) If an equilibrium mixture at 25°C contains 0.040 M N2 and 0.010 M O2, what is the concentration of NO in this mixture? The equilibrium constant at 25°C is 1.0 × 10−30. Copyright © Cengage Learning. All rights reserved.

N2(g) + O2(g)  2NO(g) Copyright © Cengage Learning. All rights reserved.

When the initial concentration and the value of Kc are known, we return to the stoichiometric chart of initial, change, and equilibrium (ICE) amounts or concentrations to find the equilibrium concentrations. Copyright © Cengage Learning. All rights reserved.

Hydrogen iodide decomposes to hydrogen gas and iodine gas. 2HI(g)  H2(g) + I2(g) At 800 K, the equilibrium constant, Kc, for this reaction is 0.016. If 0.50 mol HI is placed in a 5.0-L flask, what will be the composition of the equilibrium mixture in molarities? Copyright © Cengage Learning. All rights reserved.

2HI(g)  H2(g) + I2(g) Initial 0.10 M Change +x Equilibrium 0.10 – 2x Change –2x +x Equilibrium 0.10 – 2x x Copyright © Cengage Learning. All rights reserved.

The Kc expression is, Kc = Substituting: Because the right side of the equation is a perfect square, we can take the square root of both sides. Copyright © Cengage Learning. All rights reserved.

Solving: 0.126(0.10 – 2x) = x 0.0126 – 0.252x = x 0.0126 = 1.252x x = 0.010 M Substituting: Copyright © Cengage Learning. All rights reserved.

When the Kc expression is not a perfect square, the equation must be rearranged to fit the quadratic format: ax2 + bx + c = 0 The solution is Copyright © Cengage Learning. All rights reserved.

Le Châtelier’s Principle When a system in chemical equilibrium is disturbed by a change in temperature, pressure, or concentration, the system shifts in equilibrium composition in a way that tends to counteract this change of variable. Copyright © Cengage Learning. All rights reserved.

Le Châtelier’s Principle If an external stress is applied to a system at equilibrium, the system adjusts in such a way that the stress is partially offset as the system reaches a new equilibrium position. Changes in Concentration N2 (g) + 3H2 (g) 2NH3 (g) Add NH3 Equilibrium shifts left to offset stress

Le Châtelier’s Principle Changes in Concentration continued Remove Add Add Remove aA + bB cC + dD Change Shifts the Equilibrium Increase concentration of product(s) left Decrease concentration of product(s) right Increase concentration of reactant(s) right Decrease concentration of reactant(s) left

The following reaction is at equilibrium: COCl2(g)  CO(g) + Cl2(g) a. Predict the direction of reaction when chlorine gas is added to the reaction mixture. b. Predict the direction of reaction when carbon monoxide gas is removed from the mixture. Copyright © Cengage Learning. All rights reserved.

Note: forward = right = . COCl2(g)  CO(g) + Cl2(g) When we add Cl2, the reaction will shift in the reverse direction to use it. Note: reverse = left = . When we remove CO, the reaction will shift in the forward direction to produce it. Note: forward = right = . Copyright © Cengage Learning. All rights reserved.

Le Châtelier’s Principle Changes in Volume and Pressure A (g) + B (g) C (g) Change Shifts the Equilibrium Increase pressure Side with fewest moles of gas Decrease pressure Side with most moles of gas Increase volume Side with most moles of gas Decrease volume Side with fewest moles of gas

c. COCl2(g)  CO(g) + Cl2(g) In which direction will each reaction shift when the volume of the reaction container is increased? a. CO(g) + 2H2(g)  CH3OH(g) b. 2SO2(g) + O2(g)  2SO3(g) c. COCl2(g)  CO(g) + Cl2(g) Copyright © Cengage Learning. All rights reserved.

This reaction shifts reverse = left =  2SO2(g) + O2(g)2SO3(g) When the container volume is increased, the total pressure is decreased. Each system will shift to produce more gas by shifting toward the side with more moles of gas. CO(g) + 2H2(g)  CH3OH(g) This reaction shifts reverse = left =  2SO2(g) + O2(g)2SO3(g) COCl2(g)  CO(g) + Cl2(g) This reaction shifts forward = right =  Copyright © Cengage Learning. All rights reserved.

Le Châtelier’s Principle Changes in Temperature Change Exothermic Rx Endothermic Rx Increase temperature K decreases K increases Decrease temperature K increases K decreases N2O4 (g) 2NO2 (g) colder hotter

Decreasing the temperature has the opposite effect. For an endothermic reaction, increasing the temperature increases the value of Kc. For an exothermic reaction, increasing the temperature decreases the value of Kc. Decreasing the temperature has the opposite effect. Copyright © Cengage Learning. All rights reserved.

In addition to the value of Kc, we can consider the direction in which the equilibrium will shift. When heat is added (temperature increased), the reaction will shift to use heat. When heat is removed (temperature decreased), the reaction will shift to produce heat. Copyright © Cengage Learning. All rights reserved.

Heat + 2H2O(g)  2H2(g) + O2(g) Given: 2H2O(g)  2H2(g) + O2(g); DH° = 484 kJ Would you expect this reaction to be favorable at high or low temperatures? We rewrite the reaction to include heat: Heat + 2H2O(g)  2H2(g) + O2(g) When heat is added, the reaction shifts forward = right = . The reaction is favorable at high temperatures. Copyright © Cengage Learning. All rights reserved.

8CO(g) + 17H2(g)  C8H18(g) + 8H2O(g) The Fischer–Tropsch process for the synthesis of gasoline consists of passing a mixture of carbon monoxide and hydrogen over an iron–cobalt catalyst. A typical reaction that occurs in the process is 8CO(g) + 17H2(g)  C8H18(g) + 8H2O(g) Suppose the reaction is at equilibrium at 200°C, then is suddenly cooled to condense the octane, and then the remaining gases are reheated to 200°C. In which direction will the equilibrium shift? Copyright © Cengage Learning. All rights reserved.

8CO(g) + 17H2(g) C8H18(g) + 8H2O(g) The Fischer–Tropsch process for the synthesis of gasoline consists of passing a mixture of carbon monoxide and hydrogen over an iron–cobalt catalyst. A typical reaction that occurs in the process is 8CO(g) + 17H2(g) C8H18(g) + 8H2O(g) Suppose the reaction is at equilibrium at 200°C, then is suddenly cooled to condense the octane, and then the remaining gases are reheated to 200°C. In which direction will the equilibrium shift? Copyright © Cengage Learning. All rights reserved.

This is essentially removing octane, a product This is essentially removing octane, a product. This change causes the reaction to produce octane by shifting forward = right = . Copyright © Cengage Learning. All rights reserved.

Le Châtelier’s Principle Adding a Catalyst does not change K does not shift the position of an equilibrium system system will reach equilibrium sooner Catalyst lowers Ea for both forward and reverse reactions. Catalyst does not change equilibrium constant or shift equilibrium.

Life at High Altitudes and Hemoglobin Production Chemistry In Action Life at High Altitudes and Hemoglobin Production Hb (aq) + O2 (aq) HbO2 (aq) Kc = [HbO2] [Hb][O2]

Chemistry In Action: The Haber Process N2 (g) + 3H2 (g) 2NH3 (g) DH0 = -92.6 kJ/mol

Le Châtelier’s Principle - Summary Change Shift Equilibrium Change Equilibrium Constant Concentration yes no Pressure yes* no Volume yes* no Temperature yes yes Catalyst no no *Dependent on relative moles of gaseous reactants and products

N2O4 decomposes to NO2. The equilibrium reaction in the gas phase is N2O4(g) 2NO2(g) At 100°C, Kc = 0.36. If a 1.00-L flask initially contains 0.100 M N2O4, what will be the equilibrium concentration of NO2? Copyright © Cengage Learning. All rights reserved.

Again, we begin with the table: N2O4(g)  2NO2(g) Initial 0.100 M Change –x +2x Equilibrium 0.100 – x 2x Copyright © Cengage Learning. All rights reserved.

The Kc expression is, Kc = Substitute: Rearrange: Copyright © Cengage Learning. All rights reserved.

Substitute to find the equilibrium concentration of NO2: Solve: We eliminate the negative value because it is impossible to have a negative concentration. Substitute to find the equilibrium concentration of NO2: Copyright © Cengage Learning. All rights reserved.

H2(g) + F2(g)  2HF(g); Kc = 1.15 × 102 Given: H2(g) + F2(g)  2HF(g); Kc = 1.15 × 102 3.000 mol of each species is put in a 1.500-L vessel. What is the equilibrium concentration of each species? First calculate the initial concentrations: Copyright © Cengage Learning. All rights reserved.

H2(g) + F2(g)  2HF(g) Initial 2.000 M Change +2x Equilibrium Copyright © Cengage Learning. All rights reserved.

Copyright © Cengage Learning. All rights reserved.

Now compute the equilibrium concentrations: Double-check by substituting these equilibrium concentrations into the Kc expression and solving. The answer should be the value of Kc. This is within round-off error. Copyright © Cengage Learning. All rights reserved.

The value of Kc at 227°C is 0.0952 for the following reaction: CH3OH(g)  CO(g) + 2H2(g) What is Kp at this temperature? Kp = 0.0952(RT)Dn T = 227 + 273 = 500. K R = 0.08206 L  atm/(mol  K) Dn = 2 Kp = 1.60 × 102 Copyright © Cengage Learning. All rights reserved.