AP* Chapter 13. Chemical Equilibrium

AP* Chapter 13 Chemical Equilibrium

Section 13.1 The Equilibrium Condition Chemical Equilibrium The state where the concentrations of all reactants and products remain constant with time. On the molecular level, there is frantic activity. Equilibrium is not static, but is a highly dynamic situation. Copyright Cengage Learning. All rights reserved 2

Section 13.1 The Equilibrium Condition Equilibrium Is: Macroscopically static Microscopically dynamic May favor either products or reactants If products are favored, the equilibrium position of the reaction lies far to the right If reactants are favored, the equilibrium position of the reaction lies far to the left Copyright Cengage Learning. All rights reserved 3

Section 13.1 The Equilibrium Condition Chemical Equilibrium Concentrations reach levels where the rate of the forward reaction equals the rate of the reverse reaction. Copyright Cengage Learning. All rights reserved 5

Section 13.1 The Equilibrium Condition CONCEPT CHECK! Consider an equilibrium mixture in a closed vessel reacting according to the equation: H 2 O(g) + CO(g) H 2 (g) + CO 2 (g) You add more H 2 O(g) to the flask. How does the concentration of each chemical compare to its original concentration after equilibrium is reestablished? Justify your answer. Copyright Cengage Learning. All rights reserved 7

Section 13.1 The Equilibrium Condition CONCEPT CHECK! Consider an equilibrium mixture in a closed vessel reacting according to the equation: H 2 O(g) + CO(g) H 2 (g) + CO 2 (g) You add more H 2 to the flask. How does the concentration of each chemical compare to its original concentration after equilibrium is reestablished? Justify your answer. Copyright Cengage Learning. All rights reserved 8

Section 13.1 The Equilibrium Condition Factors Determining Equilibrium Position of a Reaction Initial concentrations Relative energies of reactants and products Relative degree of organization of reactants and products

Section 13.2 The Equilibrium Constant Consider the following reaction at equilibrium: ja + kb lc + md K = [C] l [A] j [D] [B] m k A, B, C, and D = chemical species. Square brackets = concentrations of species at equilibrium. j, k, l, and m = coefficients in the balanced equation. K = equilibrium constant (given without units). Copyright Cengage Learning. All rights reserved 10

Section 13.2 The Equilibrium Constant Interactive Example 13.1 - Solution Applying the law of mass action gives Coefficient of NO 2 4 6 NO H O 2 2 K = NH O 4 7 3 2 Coefficient of NH 3 Coefficient of H 2 O Coefficient of O 2

Section 13.2 The Equilibrium Constant Interactive Example 13.2 - Calculating the Values of K The following equilibrium concentrations were observed for the Haber process for synthesis of ammonia at 127 C: 2 [NH 3] = 3.1 10 mol/l 1 [N 2 ] = 8.5 10 mol/l 3 [H 2 ] = 3.1 10 mol/l

Section 13.2 The Equilibrium Constant Interactive Example 13.2 - Calculating the Values of K (Continued) a) Calculate the value of K at 127 C for this reaction b) Calculate the value of the equilibrium constant at 127 C for the following reaction: 2HN 3( g) N 2( g) + 3H 2 ( g) c) Calculate the value of the equilibrium constant at 127 C given by the following equation: 1 3 N 2 ( g) + H 2 ( g) NH 3( g) 2 2

Section 13.2 The Equilibrium Constant Interactive Example 13.2 - Solution (a) The balanced equation for the Haber process is Thus, K 2 N H N 2( g) + 3H 2( g) 2HN 3( g) NH (3.1 10 ) = = = 3.8 10 (8.5 10 ) (3.1 10 ) 2 2 2 2 3 4 3 1 3 3 Note that K is written without units

Section 13.2 The Equilibrium Constant Interactive Example 13.2 - Solution (b) To determine the equilibrium expression for the dissociation of ammonia, the reaction is written in the reverse order This leads to the following expression: K' NH 3 N H 1 1 = = = = 2.6 10 K 3.8 10 2 2 5 2 4 3

Section 13.2 The Equilibrium Constant Interactive Example 13.2 - Solution (c) Determine the equilibrium constant using the law of mass action NH 3 K'' = 1 3 N 2 2 H 2 2 Compare the above expression to the one obtained in solution (a)

Section 13.2 The Equilibrium Constant Interactive Example 13.2 - Solution (c) (Continued) NH 3 3 1 3 3 2 2 N H 2 2 2 2 N H NH 1 2 2 Thus, K'' = K 1 2 1 1 2 4 2 2 K'' = K = (3.8 10 ) = 1.9 10

Section 13.2 The Equilibrium Constant Equilibrium Position versus Equilibrium Constant Equilibrium position Refers to each set of equilibrium concentrations There can be infinite number of positions for a reaction Depends on initial concentrations Equilibrium constant One constant for a particular system at a particular temperature Remains unchanged Depends on the ratio of concentrations

Section 13.2 The Equilibrium Constant Table 13.1 - Synthesis of Ammonia at Different Concentrations of Nitrogen and Hydrogen

Section 13.2 The Equilibrium Constant Equilibrium Expression - Conclusions Consider the following reaction: ja + kb lc + md The equilibrium expression is K = l C D j A B Reversing the original reaction results in a new expression j k l C D m A B 1 K' = = m K k

Section 13.2 The Equilibrium Constant Equilibrium Expression - Conclusions (Continued) Multiplying the original reaction by the factor n gives nja + n kb nlc + nmd The equilibrium expression becomes K'' C nl D = = nj nk A B nm K n

Section 13.2 The Equilibrium Constant Conclusions About the Equilibrium Expression Equilibrium expression for a reaction is the reciprocal of that for the reaction written in reverse. When the balanced equation for a reaction is multiplied by a factor of n, the equilibrium expression for the new reaction is the original expression raised to the nth power; thus K new = (K original ) n. K values are usually written without units. Copyright Cengage Learning. All rights reserved 22

Section 13.2 The Equilibrium Constant K always has the same value at a given temperature regardless of the amounts of reactants or products that are present initially. For a reaction, at a given temperature, there are many equilibrium positions but only one equilibrium constant, K. Equilibrium position is a set of equilibrium concentrations. Copyright Cengage Learning. All rights reserved 23

Section 13.2 The Equilibrium Constant

Section 13.3 Equilibrium Expressions Involving Pressures Example N 2 (g) + 3H 2 (g) 2NH 3 (g) Equilibrium pressures at a certain temperature: P P P NH N H 2 2 3 2 = 2.9 10 atm 1 = 8.9 10 atm 3 = 2.9 10 atm Copyright Cengage Learning. All rights reserved 27

Section 13.3 Equilibrium Expressions Involving Pressures Example N 2 (g) + 3H 2 (g) 2NH 3 (g) K K = 2 PNH P N PH p 3 3 2 2 2 = 2.9 10 8.9 10 2.9 10 p 3 1 3 2 K p = 3.9 10 Copyright Cengage Learning. All rights reserved 28 4

Section 13.3 Equilibrium Expressions Involving Pressures The Relationship Between K and K p K p = K(RT) Δn Δn = sum of the coefficients of the gaseous products minus the sum of the coefficients of the gaseous reactants. R = 0.08206 L atm/mol K T = temperature (in Kelvin) Copyright Cengage Learning. All rights reserved 29

Section 13.3 Equilibrium Expressions Involving Pressures Interactive Example 13.5 - Calculating K from K p Using the value of K p obtained in example 13.4, calculate the value of K at 25 C for the following reaction: 2NO( g) + Cl 2 ( g) 2NOCl( g) Kp 1.9 10 3

Section 13.3 Equilibrium Expressions Involving Pressures Interactive Example 13.5 - Solution The value of K p can be used to calculate K using the formula K p = K(RT) Δn T = 25 + 273 = 298 K Δn = 2 (2+1) = 1 Thus, Sum of product coefficients Sum of reactant coefficients Δn Kp ( ) 1 = K RT = K RT

Section 13.3 Equilibrium Expressions Involving Pressures Interactive Example 13.5 - Solution (Continued) Therefore, K = K ( RT ) p 3 = (1.9 10 )(0.08206)(298) = 4.6 10 4

Section 13.3 Equilibrium Expressions Involving Pressures Example N 2 (g) + 3H 2 (g) 2NH 3 (g) Using the value of K p (3.9 10 4 ) from the previous example, calculate the value of K at 35 C. K p = 24 4 3.9 10 = K 0.08206 L atm/mol K 308K K K RT = 2.5 10 n 7 Copyright Cengage Learning. All rights reserved 33

Section 13.4 Heterogeneous Equilibria Homogeneous Equilibria Homogeneous equilibria involve the same phase: N 2 (g) + 3H 2 (g) 2NH 3 (g) HCN(aq) H + (aq) + CN - (aq) Copyright Cengage Learning. All rights reserved 34

Section 13.4 Heterogeneous Equilibria Heterogeneous Equilibria Heterogeneous equilibria involve more than one phase: 2KClO 3 (s) 2KCl(s) + 3O 2 (g) 2H 2 O(l) 2H 2 (g) + O 2 (g) Copyright Cengage Learning. All rights reserved 35

Section 13.4 Heterogeneous Equilibria The position of a heterogeneous equilibrium does not depend on the amounts of pure solids or liquids present. The concentrations of pure liquids and solids are constant. 2KClO 3 (s) 2KCl(s) + 3O 2 (g) K = O 3 2 Copyright Cengage Learning. All rights reserved 36

Section 13.4 Heterogeneous Equilibria Interactive Example 13.6 - Equilibrium Expressions for Heterogeneous Equilibria Write the expressions for K and K p for the following processes: a. Solid phosphorus pentachloride decomposes to liquid phosphorus trichloride and chlorine gas b. Deep blue solid copper(ii) sulfate pentahydrate is heated to drive off water vapor to form white solid copper(ii) sulfate

Section 13.4 Heterogeneous Equilibria Interactive Example 13.6 - Solution (a) The balanced equation for the reaction is PCl 5 ( s) PCl 3( l) + Cl 2( g) The equilibrium expressions are K= [ Cl ] and K = P 2 p Cl 2 In this case neither the pure solid PCl 5 nor the pure liquid PCl 3 is included in the equilibrium expressions

Section 13.4 Heterogeneous Equilibria Interactive Example 13.6 - Solution (b) The balanced equation for the reaction is CuSO 5H O( s) CuSO ( s) + 5H O( g) 4 2 4 2 The equilibrium expressions are The solids are not included K= [ H O ] and K = ( P ) 5 5 2 p H O 2

Section 13.4 Heterogeneous Equilibria Uses of the Equilibrium Constant Helps in predicting features of reactions, such as determining: Tendency (not speed) of a reaction to occur Whether a given set of concentrations represent an equilibrium condition Equilibrium position that will be achieved from a given set of initial concentrations

Section 13.4 Heterogeneous Equilibria Uses of the Equilibrium Constant - Example (Continued 4) Let x be the number of molecules that need to disappear so that the system can reach equilibrium

Section 13.5 Applications of the Equilibrium Constant The Extent of a Reaction A value of K much larger than 1 means that at equilibrium the reaction system will consist of mostly products the equilibrium lies to the right. Reaction goes essentially to completion. Copyright Cengage Learning. All rights reserved 42

Section 13.5 Applications of the Equilibrium Constant The Extent of a Reaction A very small value of K means that the system at equilibrium will consist of mostly reactants the equilibrium position is far to the left. Reaction does not occur to any significant extent. Copyright Cengage Learning. All rights reserved 43

Section 13.5 Applications of the Equilibrium Constant Size of K and Time Required to Reach Equilibrium Not directly related Time required depends on the rate of the reaction Determined by the size of the activation energy Size of K is determined by thermodynamic factors Example - Energy difference between products and reactants Copyright Cengage Learning. All rights reserved 44

Section 13.5 Applications of the Equilibrium Constant CONCEPT CHECK! If the equilibrium lies to the right, the value for K is. large (or >1) If the equilibrium lies to the left, the value for K is. small (or <1) Copyright Cengage Learning. All rights reserved 45

Section 13.5 Applications of the Equilibrium Constant Reaction Quotient, Q Used when all of the initial concentrations are nonzero. Apply the law of mass action using initial concentrations instead of equilibrium concentrations. Copyright Cengage Learning. All rights reserved 46

Section 13.5 Applications of the Equilibrium Constant Reaction Quotient, Q Q = K; The system is at equilibrium. No shift will occur. Q > K; The system shifts to the left. Consuming products and forming reactants, until equilibrium is achieved. Q < K; The system shifts to the right. Consuming reactants and forming products, to attain equilibrium. Copyright Cengage Learning. All rights reserved 47

Section 13.5 Applications of the Equilibrium Constant Interactive Example 13.7 - Using the Reaction Quotient For the synthesis of ammonia at 500 C, the equilibrium constant is 6.0 10 2 Predict the direction in which the system will shift to reach equilibrium in the following case: [NH 3 ] 0 = 1.0 10 3 M [N 2 ] 0 = 1.0 10 5 M [H 2 ] 0 = 2.0 10 3 M

Section 13.5 Applications of the Equilibrium Constant Interactive Example 13.7 - Solution Calculate the value of Q 2 3 NH 1.0 10 3 0 Q = = N H 1.0 10 2.0 10 2 0 2 0 = 1.3 10 7 Since K = 6.0 10 2, Q is much greater than K 3 3 5 3 2

Section 13.5 Applications of the Equilibrium Constant Interactive Example 13.7 - Solution (Continued) To attain equilibrium: The concentrations of the products must be decreased The concentrations of the reactants must be increased Therefore, the system will shift to the left N ( g) + 3H ( g) 2NH ( g) 2 2 3

Section 13.5 Applications of the Equilibrium Constant EXERCISE! Consider the reaction represented by the equation: Fe 3+ (aq) + SCN - (aq) FeSCN 2+ (aq) Trial #1: 6.00 M Fe 3+ (aq) and 10.0 M SCN - (aq) are mixed at a certain temperature and at equilibrium the concentration of FeSCN 2+ (aq) is 4.00 M. What is the value for the equilibrium constant for this reaction? Copyright Cengage Learning. All rights reserved 51

Section 13.5 Applications of the Equilibrium Constant K = 0.333 Copyright Cengage Learning. All rights reserved 52 Set up ICE Table Fe 3+ (aq) + SCN (aq) FeSCN 2+ (aq) Initial 6.00 10.00 0.00 Change 4.00 4.00+4.00 Equilibrium 2.00 6.00 4.00 2 FeSCN 4.00 = M K = Fe 3 SCN 2.00 M 6.00 M

Section 13.5 Applications of the Equilibrium Constant EXERCISE! Consider the reaction represented by the equation: Fe 3+ (aq) + SCN - (aq) FeSCN 2+ (aq) Trial #2: Initial: 10.0 M Fe 3+ (aq) and 8.00 M SCN (aq) (same temperature as Trial #1) Equilibrium:? M FeSCN 2+ (aq) 5.00 M FeSCN 2+ Copyright Cengage Learning. All rights reserved 53

Section 13.5 Applications of the Equilibrium Constant EXERCISE! Consider the reaction represented by the equation: Fe 3+ (aq) + SCN - (aq) FeSCN 2+ (aq) Trial #3: Initial: 6.00 M Fe 3+ (aq) and 6.00 M SCN (aq) Equilibrium:? M FeSCN 2+ (aq) 3.00 M FeSCN 2+ Copyright Cengage Learning. All rights reserved 54

Section 13.5 Applications of the Equilibrium Constant Interactive Example 13.8 - Calculating Equilibrium Pressures I Dinitrogen tetroxide in its liquid state was used as one of the fuels on the lunar lander for the NASA Apollo missions In the gas phase, it decomposes to gaseous nitrogen dioxide: N O g 2NO 2 4 2 g

Section 13.5 Applications of the Equilibrium Constant Interactive Example 13.8 - Calculating Equilibrium Pressures I (Continued) Consider an experiment in which gaseous N 2 O 4 was placed in a flask and allowed to reach equilibrium at a temperature where K p = 0.133 At equilibrium, the pressure of N 2 O 4 was found to be 2.71 atm Calculate the equilibrium pressure of NO 2 (g)

Section 13.5 Applications of the Equilibrium Constant Interactive Example 13.8 - Solution The equilibrium pressures of the gases NO 2 and N 2 O 4 must satisfy the following relationship: K p 2 = = 0.133 2 4 Solve for the equilibrium pressure of NO 2 P P NO NO 2 P = K ( P ) = (0.133)(2.71) = 0.360 2 NO p N O 2 2 4

Section 13.5 Applications of the Equilibrium Constant Interactive Example 13.8 - Solution (Continued) Therefore, P NO 2 = 0.360 = 0.600

Section 13.5 Applications of the Equilibrium Constant Interactive Example 13.9 - Calculating Equilibrium Pressures II At a certain temperature, a 1.00 L flask initially contained 0.298 mole of PCl 3 (g) and 8.70 10 3 mole of PCl 5 (g) After the system had reached equilibrium 2.00 10 3 mole of Cl 2 (g) was found in the flask Gaseous PCl 5 decomposes according to the reaction PCl 5( g) PCl 3( g) + Cl 2 ( g) Calculate the equilibrium concentrations of all species and the value of K

Section 13.5 Applications of the Equilibrium Constant Interactive Example 13.9 - Solution The equilibrium expression for this reaction is K = Cl 2 PCl3 PCl 5 To find the value of K: Calculate the equilibrium concentrations of all species Substitute the derived quantities into the equilibrium expression

Section 13.5 Applications of the Equilibrium Constant Interactive Example 13.9 - Solution (Continued 1) Determine the initial concentrations [Cl ] = 0 2 0 0.298 mol [PCl 3] 0= = 0.298 M 1.00 L 3 8.70 10 mol [PCl 5] 0= = 8.70 10 1.00 L 3 M

Section 13.5 Applications of the Equilibrium Constant Interactive Example 13.9 - Solution (Continued 2) Determine the change required to reach equilibrium PCl ( g) PCl ( g) + Cl ( g) 5 3 2 3 3 3 2.00 10 mol 2.00 10 mol + 2.00 10 mol Net amount of PCl decomposed 5 Net amounts of products formed Apply these values to the initial concentrations

Section 13.5 Applications of the Equilibrium Constant Interactive Example 13.9 - Solution (Continued 3) Determine the equilibrium concentrations 3 2.00 10 mol [Cl 2] = 0 + = 2.00 10 1.00 L 3 M [Cl 2 ] 0 3 2.00 10 mol [PCl 3] = 0.298 M + = 0.300 1.00 L M [PCl 3 ] 0 3 2.00 10 mol 3 [PCl 5] = 8.70 10 M = 6.70 10 1.00 L [PCl 5 ] 0 3 M

Section 13.5 Applications of the Equilibrium Constant Interactive Example 13.9 - Solution (Continued 4) Determine the value of K Substitute the equilibrium concentrations into the equilibrium expression K 2 3 3 Cl PCl (2.00 10 )(0.300) = = PCl 6.70 10 5 = 8.96 10 2 3

Section 13.5 Applications of the Equilibrium Constant Interactive Example 13.11 - Calculating Equilibrium Concentrations II Assume that the reaction for the formation of gaseous hydrogen fluoride from hydrogen and fluorine has an equilibrium constant of 1.15 10 2 at a certain temperature In a particular experiment, 3.000 moles of each component were added to a 1.500 L flask Calculate the equilibrium concentrations of all species

Section 13.5 Applications of the Equilibrium Constant Interactive Example 13.11 - Solution The balanced equation for this reaction is H 2( g) + F 2( g) 2HF( g) The equilibrium expression is K 2 = 1.15 10 = The initial concentrations are 2 HF H F 2 2 3.000 mol HF = H 0 2 = F 0 2 = = 2.000 M 0 1.500 L

Section 13.5 Applications of the Equilibrium Constant Interactive Example 13.11 - Solution (Continued 1) The value of Q is Q 2 2 HF 2.000 0 = = = 1.000 H F 2.000 2.000 2 0 2 0 Since Q is much less than K, the system must shift to the right to reach equilibrium To determine what change in concentration is necessary, define the change needed in terms of x

Section 13.5 Applications of the Equilibrium Constant Interactive Example 13.11 - Solution (Continued 2) Let x be the number of moles per liter of H 2 consumed to reach equilibrium The stoichiometry of the reaction shows that x mol/l F 2 also will be consumed and 2x mol/l HF will be formed H ( g) + F ( g) 2HF( g) 2 2 x mol/l + x mol/l 2 x mol/l

Section 13.5 Applications of the Equilibrium Constant Interactive Example 13.11 - Solution (Continued 3) Determine the equilibrium concentrations in terms of x

Section 13.5 Applications of the Equilibrium Constant Interactive Example 13.11 - Solution (Continued 4) The concentrations can be expressed in a shorthand table as follows: To solve for the value of x, substitute the equilibrium concentrations into the equilibrium expression

Section 13.5 Applications of the Equilibrium Constant Interactive Example 13.11 - Solution (Continued 5) K 2 = 1.15 10 = 2 2 2 2 HF 2.000 + 2x H F 2.000 x 2 The right side of this equation is a perfect square, so taking the square root of both sides gives 2 2.000 + 2x 1.15 10 2.000 x Therefore, x = 1.528

Section 13.5 Applications of the Equilibrium Constant Interactive Example 13.11 - Solution (Continued 6) The equilibrium concentrations are Reality check H = F = 2.000 M x = 0.472 M 2 2 HF = 2.000 M + 2 x = 5.056 M Checking the values by substituting them into the equilibrium expression gives the same value of K

Section 13.6 Solving Equilibrium Problems Solving Equilibrium Problems 1) Write the balanced equation for the reaction. 2) Write the equilibrium expression using the law of mass action. 3) List the initial concentrations. 4) Calculate Q, and determine the direction of the shift to equilibrium. Copyright Cengage Learning. All rights reserved 73

Section 13.6 Solving Equilibrium Problems Solving Equilibrium Problems 5) Define the change needed to reach equilibrium, and define the equilibrium concentrations by applying the change to the initial concentrations. 6) Substitute the equilibrium concentrations into the equilibrium expression, and solve for the unknown. 7) Check your calculated equilibrium concentrations by making sure they give the correct value of K. Copyright Cengage Learning. All rights reserved 74

Section 13.6 Solving Equilibrium Problems Interactive Example 13.12 - Calculating Equilibrium Pressures Assume that gaseous hydrogen iodide is synthesized from hydrogen gas and iodine vapor at a temperature where the equilibrium constant is 1.00 10 2 Suppose HI at 5.000 10 1 atm, H 2 at 1.000 10 2 atm, and I 2 at 5.000 10 3 atm are mixed in a 5.000 L flask Calculate the equilibrium pressures of all species

Section 13.6 Solving Equilibrium Problems Interactive Example 13.12 - Solution The balanced equation for this process is H 2( g) + I 2( g) 2HI( g) The equilibrium expression in terms of pressure is K p P 2 HI = 1.00 10 H I P P 2 2 2

Section 13.6 Solving Equilibrium Problems Interactive Example 13.12 - Solution (Continued 1) The initial pressures provided are P HI0 = 5.000 10 1 atm P H20 = 1.000 10 2 atm P I20 = 5.000 10 3 atm The value of Q for this system is Q 0 2 1 2 HI 3 P 0 0 P H PI 2 2 5.000 10 atm 2 3 1.000 10 atm 5.000 10 atm = = = 5.000 10

Section 13.6 Solving Equilibrium Problems Interactive Example 13.12 - Solution (Continued 2) Since Q is greater than K, the system will shift to the left to reach equilibrium Use pressures for a gas-phase system at constant temperature and volume Pressure is directly proportional to the number of moles P RT = n V Constant if constant T and V

Section 13.6 Solving Equilibrium Problems Interactive Example 13.12 - Solution (Continued 3) Let x be the change in pressure (in atm) of H 2 as the system shifts left toward equilibrium This leads to the following equilibrium pressures:

Section 13.6 Solving Equilibrium Problems Interactive Example 13.12 - Solution (Continued 4) Determine the value of K p K p P HI P H PI 2 2 5.000 10 2x 2 3 1.000 10 x 5.000 10 x 2 1 2 Multiply and collect terms that yield the quadratic equation where a = 9.60 10 1, b = 3.5, and c = 2.45 10 1 1 2 x x 1 9.60 10 + 3.5 2.45 10 = 0

Section 13.6 Solving Equilibrium Problems Interactive Example 13.12 - Solution (Continued 5) From the quadratic formula, the correct value for x is 3.55 10 2 atm The equilibrium pressures can now be calculated from the expressions involving x P 1 2 1 HI = 5.000 10 atm 2 3.55 10 atm = 4.29 10 atm P 2 2 2 H = 1.000 10 atm + 3.55 10 atm = 4.55 10 atm 2 2 3 2 2 P I = 5.000 10 atm + 3.55 10 atm = 4.05 10 atm

Section 13.6 Solving Equilibrium Problems Interactive Example 13.12 - Solution (Continued 6) Reality check P H P 2 1 2 HI 2 2 PI 2 2 (4.29 10 ) = = 99.9 (4.55 10 ) (4.05 10 ) This agrees with the given value of K (1.00 10 2 ) Thus, the calculated equilibrium concentrations are correct

Section 13.6 Solving Equilibrium Problems Treating Systems That Have Small Equilibrium Constants Consider the decomposition of gaseous NOCl at 35 C with an equilibrium constant of 1.6 10 5 The following steps determine the equilibrium concentrations of NOCl, NO, and Cl 2 when one mole of NOCl is placed in a 2.0 L flask: The balanced equation is 2NOCl( g) 2NO( g) + Cl 2( g)

Section 13.6 Solving Equilibrium Problems Treating Systems that have Small Equilibrium Constants (Continued 1) The equilibrium expression is 2 [NO] [Cl 2] 2 K = 1.6 10 [NOCl] The initial concentrations are 1.0 mol [NOCl] 0 = = 0.50 M 2.0 L [NO] = 0 0 [Cl ] = 0 2 0 5

Section 13.6 Solving Equilibrium Problems Treating Systems that have Small Equilibrium Constants (Continued 2) Since there are no products initially, the system will move to the right to reach equilibrium Let x be the change in concentration of Cl 2 needed to reach equilibrium The changes in the concentrations can be obtained from the following balanced equitation: 2NOCl( g) 2NO( g) + Cl ( g) 2 x 2 x + x 2

Section 13.6 Solving Equilibrium Problems Treating Systems that have Small Equilibrium Constants (Continued 3) The concentrations can be summarized as follows: The equilibrium concentrations must satisfy the following equilibrium expression 2 2 5 [NO] [Cl 2 ( ) 2 ] x x 2 2 K 1.6 10 = [NOCl] 0.50 2x

Section 13.6 Solving Equilibrium Problems Treating Systems that have Small Equilibrium Constants (Continued 4) In this situation, K is so small that the system will not proceed far to the right to reach equilibrium x represents a relatively small number, so when x is small 0.50 2 x 0.50 Simplify the equilibrium expression using this approximation 2 2 3 5 (2 x) ( x) 2 x ( x) 4x 1.6 10 = = 2 2 2 (0.50 2 x) 0.50 (0.50)

Section 13.6 Solving Equilibrium Problems Treating Systems that have Small Equilibrium Constants (Continued 5) Solving for x 3 gives x 5 2 3 6 x Test the validity of the approximation If x = 1.0 10 2, then (1.6 10 )(0.50) = = 1.0 10 4 2 = 1.0 10 2 0.50 2 x = 0.50 2(1.0 10 ) = 0.48

Section 13.6 Solving Equilibrium Problems Treating Systems that have Small Equilibrium Constants (Continued 6) The difference between 0.50 and 0.48 is 0.02 This discrepancy will have little effect on the outcome Since 2x is very small compared with 0.50, the value of x obtained should be very close to the exact value Use the approximate value of x to calculate equilibrium concentrations [ NOCl ] 0.50 2 x 0.50 M [NO] = 2 = 2(1.0 10 ) = 2.0 10 2 2 2 x M M [Cl ] = x = 1.0 10 2 M

Section 13.6 Solving Equilibrium Problems Treating Systems that have Small Equilibrium Constants (Continued 7) Reality check 2 2 2 2 [NO] [Cl 2] (2.0 10 ) (1.0 10 ) = 1.6 10 2 2 [NOCl] (0.50) Since the given value of K is the same, these calculations are correct The small value of K and the resulting small shift to the right to reach equilibrium allowed simplification 5

Section 13.6 Solving Equilibrium Problems Critical Thinking You have learned how to treat systems that have small equilibrium constants by making approximations to simplify the math What if the system has a very large equilibrium constant? What can you do to simplify the math for this case? Use the example from the text, change the value of the equilibrium constant to 1.6 10 5, and rework the problem Why can you not use approximations for the case in which K = 1.6?

Section 13.6 Solving Equilibrium Problems EXERCISE! Consider the reaction represented by the equation: Fe 3+ (aq) + SCN - (aq) FeSCN 2+ (aq) Fe 3+ SCN - FeSCN 2+ Trial #1 9.00 M 5.00 M 1.00 M Trial #2 3.00 M 2.00 M 5.00 M Trial #3 2.00 M 9.00 M 6.00 M Find the equilibrium concentrations for all species. Copyright Cengage Learning. All rights reserved 92

Section 13.6 Solving Equilibrium Problems EXERCISE! Answer Trial #1: [Fe 3+ ] = 6.00 M; [SCN - ] = 2.00 M; [FeSCN 2+ ] = 4.00 M Trial #2: [Fe 3+ ] = 4.00 M; [SCN - ] = 3.00 M; [FeSCN 2+ ] = 4.00 M Trial #3: [Fe 3+ ] = 2.00 M; [SCN - ] = 9.00 M; [FeSCN 2+ ] = 6.00 M Copyright Cengage Learning. All rights reserved 93

Section 13.6 Solving Equilibrium Problems CONCEPT CHECK! A 2.0 mol sample of ammonia is introduced into a 1.00 L container. At a certain temperature, the ammonia partially dissociates according to the equation: NH 3 (g) N 2 (g) + H 2 (g) At equilibrium 1.00 mol of ammonia remains. Calculate the value for K. K = 1.69 Copyright Cengage Learning. All rights reserved 94

Section 13.6 Solving Equilibrium Problems CONCEPT CHECK! A 1.00 mol sample of N 2 O 4 (g) is placed in a 10.0 L vessel and allowed to reach equilibrium according to the equation: N 2 O 4 (g) 2NO 2 (g) K = 4.00 10-4 Calculate the equilibrium concentrations of: N 2 O 4 (g) and NO 2 (g). Concentration of N 2 O 4 = 0.097 M Concentration of NO 2 = 6.32 10-3 M Copyright Cengage Learning. All rights reserved 95

Section 13.7 Le Châtelier s Principle If a change is imposed on a system at equilibrium, the position of the equilibrium will shift in a direction that tends to reduce that change. Helps in the qualitative prediction of the effects of changes in concentration, pressure, and temperature on a system at equilibrium Copyright Cengage Learning. All rights reserved 96

Section 13.7 Le Châtelier s Principle Effects of Changes on the System 1. Concentration: The system will shift away from the added component. If a component is removed, the opposite effect occurs. 2. Temperature: K will change depending upon the temperature (endothermic energy is a reactant; exothermic energy is a product). Copyright Cengage Learning. All rights reserved 97

Section 13.7 Le Châtelier s Principle Effects of Changes on the System 3. Pressure: a) The system will shift away from the added gaseous component. If a component is removed, the opposite effect occurs. b) Addition of inert gas does not affect the equilibrium position. c) Decreasing the volume shifts the equilibrium toward the side with fewer moles of gas. Copyright Cengage Learning. All rights reserved 98

Section 13.7 Le Châtelier s Principle Interactive Example 13.13 - Using Le Châtelier s Principle I Arsenic can be extracted from its ores by first reacting the ore with oxygen (called roasting) to form solid As 4 O 6, which is then reduced using carbon As O s + 6C s As g + 6CO g 4 6 4

Section 13.7 Le Châtelier s Principle Interactive Example 13.13 - Using Le Châtelier s Principle I (Continued) Predict the direction of the shift of the equilibrium position in response to each of the following changes in conditions: a. Addition of carbon monoxide b. Addition or removal of carbon or tetraarsenic hexoxide (As 4 O 6 ) c. Removal of gaseous arsenic (As 4 )

Section 13.7 Le Châtelier s Principle Interactive Example 13.13 - Solution a. Le Châtelier s principle predicts that the shift will be away from the substance whose concentration is increased Equilibrium position will shift to the left when carbon monoxide is added b. The amount of a pure solid has no effect on the equilibrium position

Section 13.7 Le Châtelier s Principle Interactive Example 13.13 - Solution (Continued) Changing the amount of carbon or tetraarsenic hexoxide will have no effect c. If gaseous arsenic is removed, the equilibrium position will shift to the right to form more products In industrial processes, the desired product is often continuously removed from the reaction system to increase the yield

Section 13.7 Le Châtelier s Principle The Effect of a Change in Pressure - Key Points Addition of an inert gas increases the total pressure Does not affect the concentrations or partial pressures of the reactants or products When the volume of the container holding a gaseous system is reduced, the system responds by reducing its own volume Total number of gaseous molecules is reduced

Section 13.7 Le Châtelier s Principle The Effect of a Change in Pressure - Key Points (Continued) Rearranging the ideal gas law gives V RT P n At constant temperature (T) and pressure (P), V n Equilibrium position shifts toward the side of the reaction that involves smaller number of gaseous molecules in the balanced equation

Section 13.7 Le Châtelier s Principle Interactive Example 13.14 - Using Le Châtelier s Principle II Predict the shift in equilibrium position that will occur during the preparation of liquid phosphorus trichloride P 4 ( s) + 6Cl 2( g) 4PCl 3( l) Assume that the volume is reduced

Section 13.7 Le Châtelier s Principle Interactive Example 13.14 - Solution Since P 4 and PCl 3 are a pure solid and a pure liquid, respectively, we need to consider only the effect of the change in volume on Cl 2 Volume is decreased, so the position of the equilibrium will shift to the right Reactant side contains six gaseous molecules, and the product side has none

Section 13.7 Le Châtelier s Principle The Effect of a Change in Temperature Value of K changes with the temperature Consider the synthesis of ammonia, an exothermic reaction N ( g) + 3H ( g) 2NH ( g) + 92 kj 2 2 3 According to Le Châtelier s principle, the shift will be in the direction that consumes energy Concentration of NH 3 decreases and that of N 2 and H 2 increases, thus decreasing the value of K Copyright Cengage Learning. All rights reserved 107

Section 13.7 Le Châtelier s Principle The Effect of a Change in Temperature (Continued) Consider the decomposition of calcium carbonate, an endothermic reaction 556 kj + CaCO 3( s) CaO( s) + CO 2( g) Increase in temperature causes the equilibrium to shift to the right Value of K increases Copyright Cengage Learning. All rights reserved 108

Section 13.7 Le Châtelier s Principle To play movie you must be in Slide Show Mode PC Users: Please wait for content to load, then click to play Mac Users: CLICK HERE Copyright Cengage Learning. All rights reserved 109

Section 13.7 Le Châtelier s Principle Equilibrium Decomposition of N 2 O 4 To play movie you must be in Slide Show Mode PC Users: Please wait for content to load, then click to play Mac Users: CLICK HERE Copyright Cengage Learning. All rights reserved 110