OFB Chapter 7 Chemical Equilibrium

OFB Chapter 7 Chemical Equilibrium 7-1 Chemical Reactions in Equilibrium 7-2 Calculating Equilibrium Constants 7-3 The Reaction Quotient 7-4 Calculation of Gas-Phase Equilibrium 7-5 The effect of External Stresses on Equilibria: Le Châtelier s Principle 7-6 Heterogeneous Equilibrium 7-7 Extraction and Separation Processes 9/22/2004 OFB Chapter 7 1

Last Day to Drop Oct. 8 All exercises Homework: Chapter 7: -8, 14, 24, 42, 54 Mid-Term Grades Due Sept. 24 Satisfactory Note: this is all that s on WebCT Be familiar with equilibrium problems these are very important Unsatisfactory 9/24/2004 OFB Chapter 7 1

forward aa + bb cc + reverse dd 9/22/2004 OFB Chapter 7 3 ~220 K (-53 o C)

7-1 Chemical Reactions and Equilibrium The equilibrium condition for every reaction can be summed up in a single equation in which a number, the equilibrium constant (K) of the reaction, equals an equilibrium expression, a function of properties of the reactants and products. H 2 O(l) H 2 O(g) @ 25 o C Temperature ( o C) Vapor Pressure (atm) 15.0 0.01683 K = 0.03126 H 2 O(l) H 2 O(g) @ 30 o C K = 0.04187 17.0 0.01912 19.0 0.02168 21.0 0.02454 23.0 0.02772 25.0 0.03126 30.0 0.04187 50.0 0.1217 9/22/2004 OFB Chapter 7 4

Chemical Reactions and Equilibrium H 2 O (l) H 2 O (g) P H 2O P ref = K P ref is numerically equal to 1 The convention in the book is to express all pressures in atmospheres and to omit factors of P ref because their value is unity. An equilibrium constant K is a pure number. 9/22/2004 OFB Chapter 7 5

Chemical Reactions and Equilibrium 2 NO 2 (g) N 2 O 4 (g) N 2 O 4 (g) 2 NO 2 (g) An equilibrium reaction 9/22/2004 OFB Chapter 7 6

Chemical Reactions and Equilibrium 2 NO 2 (g) N 2 O 4 (g) N 2 O 4 (g) 2 NO 2 (g) T T P P 2 NO 2 (g) N 2 O 4 (g); K = 8.8 @ 25 o C 9/22/2004 OFB Chapter 7 7

Chemical Reactions and Equilibrium As the equilibrium state is approached, the forward and backward rates of reaction approach equality. At equilibrium the rates are equal, and no further net change occurs in the partial pressures of reactants or products. Four fundamental characteristics of equilibrium states in isolated systems: 1. They display no macroscopic evidence of change. 2. They are reached through spontaneous processes. 3. They show a dynamic balance of forward and backward processes. 4. They are the same regardless of the direction from which they are approached. forward aa + bb cc + reverse dd 9/22/2004 OFB Chapter 7 8

The Form of Equilibrium Expressions forward aa + bb cc + reverse dd In a chemical reaction in which a moles of species A and b moles of species B react to form c moles of species C and d moles of species D, the partial pressures at equilibrium are related through provided that all species are present as low-pressure gases. 9/22/2004 OFB Chapter 7 9

Exercise 7-1 Write equilibrium expressions for the reactions defined by the following equations: 3 H 2 (g) + SO 2 (g) H 2 S(g) + 2 H 2 O(g) 2 C 2 F 5 Cl(g) + 4 O 2 (g) Cl 2 (g) + 4 CO 2 (g) + 5 F 2 (g) 9/22/2004 OFB Chapter 7 10

Example 7-2 7-2 Calculating Equilibrium Constants Consider the equilibrium 4 NO 2 (g) 2 N 2 O(g) + 3 O 2 (g) The three gases are introduced into a container at partial pressures of 3.6 atm (for NO 2 ), 5.1 atm (for N 2 O), and 8.0 atm (for O 2 ) and react to reach equilibrium at a fixed temperature. The equilibrium partial pressure of the NO 2 is measured to be 2.4 atm. Calculate the equilibrium constant of the reaction at this temperature, assuming that no competing reactions occur. initial partial pressure (atm) change in partial pressure (atm) equilibrium partial pressure (atm) 4 NO 2 (g) 2 N 2 O(g) + 3 O 2 (g) 9/22/2004 OFB Chapter 7 11

4 NO 2 (g) 2 N 2 O(g) + 3 O 2 (g) initial partial pressure (atm) change in partial pressure (atm) equilibrium partial pressure (atm) 3.6-4x = 2.4 atm NO 2 ; x = 0.3 atm 5.1 + 2(0.3 atm) = 5.7 atm N 2 O 8.0 + 3(0.3 atm) = 8.9 atm O 2 K = (P N 2 O )2 (P O2 ) 3 (P NO2 ) 4 = (5.7)2 (8.9) 3 (2.4) 4 = 9/22/2004 OFB Chapter 7 12

Exercise 7-2 The compound GeWO 4 (g) forms at high temperature in the reaction 2 GeO (g) + W 2 O 6 (g) 2 GeWO 4 (g) Some GeO (g) and W 2 O 6 (g) are mixed. Before they start to react, their partial pressures both equal 1.000 atm. After their reaction at constant temperature and volume, the equilibrium partial pressure of GeWO 4 (g) is 0.980 atm. Assuming that this is the only reaction that takes place, (a) determine the equilibrium partial pressures of GeO and W 2 O 6, and (b) determine the equilibrium constant for the reaction. initial partial pressure (atm) change in partial pressure (atm) equilibrium partial pressure (atm) 2 GeO (g) + W 2 O 6 (g) 2 GeWO 4 (g) 9/22/2004 OFB Chapter 7 13

2 GeO(g) + W 2 O 6 (g) 2 GeWO 4 (g) initial partial pressure (atm) change in partial pressure (atm) equilibrium partial pressure (atm) 0 + 2x = 0.980 atm GeWO 4 ; x = 0.490 atm 1.000-2(0.490) = 0.020 atm GeO 1.000-0.490 = 0.510 atm W 2 O 6 K = (P GeWO2 ) 2 (P GeO ) 2 (P W2 O 6 ) = (0.980) 2 (0.020) 2 (0.510) 9/22/2004 OFB Chapter 7 14

Relationships Among the K s of Related Reactions Rule 1: The equilibrium constant for a reverse reaction is always the reciprocal of the equilibrium constant for the corresponding forward reaction. #1 #2 2 H 2 (g) + O 2 (g) 2 H 2 O(g) 2 H 2 O(g) 2 H 2 (g) + O 2 (g) (P H 2O )2 (P H 2 )2 (P O 2 ) = K 1 (P H 2 )2 (P O 2 ) (P H 2O )2 = K 2 K 1 = 1/K 2 9/22/2004 OFB Chapter 7 15

Relationships Among the K s of Related Reactions Rule 2: When the coefficients in a balanced chemical equation are all multiplied by a constant factor, the corresponding equilibrium constant is raised to a power equal to that factor. #1 #3 2 H 2 (g) + O 2 (g) 2 H 2 O(g) Rxn 1 H 2 (g) + ½ O 2 (g) H 2 O(g) Rxn 3 (P H 2O ) (P H 2 )(P O2 )½ = K 3 K 3 = K 1½ = K 1 9/22/2004 OFB Chapter 7 16

Relationships Among the K s of Related Reactions Rule 3: when chemical equations are added to give a new equation, their equilibrium constants are multiplied to give the equilibrium constant associated with the new equation. 2 BrCl(g) Br 2 (g) + Cl 2 (g) Br 2 (g) + I 2 (g) 2 IBr(g) (P Br 2 )(P Cl2 ) (P BrCl ) 2 = K 1 = 0.45 @ 25 o C (P IBr ) 2 (P Br 2 ) (P I2 ) = K 2 = 0.051 @ 25o C = K 1 K 2 = (0.45)(0.051) =0.023 @ 25 o C 9/22/2004 OFB Chapter 7 17

( ) c P ( P ) D ( P ) a ( P ) 7-3 The Reaction Quotient forward aa + bb cc + reverse d C = b A B K ( ) c P ( P ) D ( P ) a ( P ) d C = b A B dd Note that K (the Equilibrium Constant) uses equilibrium partial pressures Note that Q (the reaction quotient) uses prevailing partial pressures, not necessarily at equilibrium Q 9/22/2004 OFB Chapter 7 18

The Reaction Quotient forward aa + bb cc + reverse c ( P ) ( PD ) a ( P ) ( P ) d C = b A B Q dd If Q < K, reaction proceeds in a forward direction (toward products); If Q > K, reaction proceeds in a backward direction (toward reactants); If Q = K, the reactions stops because the reaction is in equilibrium. 9/22/2004 OFB Chapter 7 19

Exercise 7-4 The equilibrium constant for the reaction P 4 (g) 2 P 2 (g) is 1.39 at 400 o C. Suppose that 2.75 mol of P 4 (g) and 1.08 mol of P 2 (g) are mixed in a closed 25.0 L container at 400 o C. Compute Q (init) (the Q at the moment of mixing) and state the direction in which the reaction proceeds. P 2 P 2 = K = 1 P4 ( ) P ( ) 1.39 K = 1.39 @ 400 o C; n P4 (init) = 2.75 mol; n P 2 (init) P P 4(init) = n P 4(init)RT/V Q =? K PV = nrt = 1.08 mol P P 2(init) = n P 2(init)RT/V Q = (2.39) 2 /(6.08) = 0.939 0.939 < 1.39; Q < K 9/22/2004 OFB Chapter 7 20

7-4 Calculations of Gas-Phase Equilibria Exercise 7-5 Carbon monoxide reacts with water to give hydrogen: CO (g) +H 2 O (g) CO 2 (g) + H 2 (g) At 900 K, the equilibrium constant for this reaction, the socalled shift reaction, equals 0.64. suppose the partial pressures of three gases at equilibrium at 900 K are P CO = 2.00 atm, P CO2 = 0.80 atm, and P H2 =0.48 atm Calculate the partial pressure of water under these conditions. 9/22/2004 OFB Chapter 7 21

7-4 Calculations of Gas-Phase Equilibria Exercise 7-5 Carbon monoxide reacts with water to give hydrogen: CO (g) +H 2 O (g) CO 2 (g) + H 2 (g) At 900 K, the equilibrium constant for this reaction, the so-called shift reaction, equals 0.64. suppose the partial pressures of three gases at equilibrium at 900 K are P CO = 2.00 atm, P CO2 = 0.80 atm, and P H2 =0.48 atm Calculate the partial pressure of water under these conditions. 9/22/2004 OFB Chapter 7 22

Solving quadratic equations 9/22/2004 OFB Chapter 7 23

7-5 Effects of External Stresses on Equilibria: Le Châtelier s Principle A system in equilibrium that is subjected to a stress reacts in a way that counteracts the stress. Le Châtelier s Principle provides a way to predict the response of an equilibrium system to an external perturbation, such as Effects of Adding or Removing Reactants or Products Effects of Changing the Volume of the System Effects of Changing the Temperature 9/22/2004 OFB Chapter 7 24

Effects of Adding or Removing Reactants or Products PCl 5 (g) PCl 3 (g) + Cl 2 (g) add extra PCl 5 (g) add extra PCl 3 (g) remove some PCl 5 (g) remove some PCl 3 (g) ( )( 1 ) P P PCl3 Cl2 ( ) 1 P PCl 5 1 = Q 9/22/2004 OFB Chapter 7 25

Effects of Changing the Volume of the System PCl 5 (g) PCl 3 (g) + Cl 2 (g) Let s decrease the volume of the reaction container Let s increase the volume of the reaction container 9/22/2004 OFB Chapter 7 26

V reactants > V products V reactants < V products V reactants = V products Volume Decreased (Pressure Increased) Equilibrium shift right (toward products) Equilibrium Shifts left (toward reactants) Equilibrium not affected Volume Increased (Pressure Decreased) Equilibrium Shifts left (toward reactants) Equilibrium shift right (toward products) Equilibrium not affected 2 P 2 (g) P 4 (g) PCl 5 (g) PCl 3 (g) + Cl 2 (g) CO (g) +H 2 O (g) CO 2 (g) + H 2 (g) 9/22/2004 OFB Chapter 7 27

Effects of Changing the Temperature PCl 5 (g) PCl 3 (g) + Cl 2 (g) Let s increase the temperature of the reaction Let s decrease the temperature of the reaction 9/22/2004 OFB Chapter 7 28

Endothermic Reaction (absorb heat) Temperature Raised Equilibrium shift right (toward products) Temperature Lowered Equilibrium Shifts left (toward reactants) Exothermic Reaction (liberate heat) Equilibrium Shifts left (toward reactants) Equilibrium shift right (toward products) If a forward reaction is exothermic, Then the reverse reaction must be endothermic 9/22/2004 OFB Chapter 7 29

Effects of Changing the Temperature PCl 5 (g) PCl 3 (g) + Cl 2 (g) K = 11.5 @ 300 o C = Q Let s increase the temperature of the reaction Let s decrease the temperature of the reaction 9/22/2004 OFB Chapter 7 30

Driving Reactions to Completion Industrial Synthesis of Ammonia N 2 (g) + 3H 2 (g) 2NH 3 (g) Forward reaction exothermic V reactants > V products Volume Decreased (Pressure Increased) Equilibrium shift right (toward products) Volume Increased (Pressure Decreased) Equilibrium Shifts left (toward reactants) Exothermic Reaction (liberate heat) Temperature Raised Equilibrium Shifts left (toward reactants) Temperature Lowered Equilibrium shift right (toward products) 9/22/2004 OFB Chapter 7 31

( ) c P ( P ) D ( P ) a ( P ) A Review: Chemical Equilibrium forward aa + bb cc + reverse B d b ( ) c P ( P ) D ( P ) a ( P ) d C = b A B Q dd C = K Q =? K 1. Effects of Adding or Removing Reactants or Products 2. Effects of Changing the Volume (or Pressure) of the System 3. Effects of Changing the Temperature 9/22/2004 OFB Chapter 7 32

Exercise 7-10 State the effect of an increase in temperature and also of a decrease in volume on the equilibrium yield of the products in each of the following reactions. a) CH 3 OCH 3 (g) + H 2 O (g) 2 CH 4 (g) + O 2 (g) Increase Temp? Decrease Volume? b) H 2 O (g) + CO (g) HCOOH (g) Increase Temp? Decrease Volume? 9/22/2004 OFB Chapter 7 33

Heterogeneous Equilibrium Solids CaCO 3 (s) CaO(s) + CO 2 (g) Liquids H 2 O(l) H 2 O(g) Dissolved species I 2 (s) I 2 (aq) 9/22/2004 OFB Chapter 7 34

Law of Mass Action 1. Gases enter equilibrium expressions as partial pressures, in atmospheres. E.g., P CO2 2. Dissolved species enter as concentrations, in moles per liter. E.g., [Na + ] 3. Pure solids and pure liquids are represented in equilibrium expressions by the number 1 (unity); a solvent taking part in a chemical reaction is represented by unity, provided that the solution is dilute. E.g., I 2 K ( s) = [ I [ I 2 ( aq)] ( s)] 2 I 2 ( aq) = [ I ( aq)] 1 9/22/2004 OFB Chapter 7 35 2 = [ I 2 ]

Law of Mass Action 4. Partial pressures and concentrations of products appear in the numerator and those of the reactants in the denominator. Each is raised to a power equal to its coefficient in the balanced chemical equation. aa + bb cc + dd 9/22/2004 OFB Chapter 7 36

Exercise 7-11 Write equilibrium-constant equations for the following Equilibria: (a) Si 3 N 4 (s) + 4 O 2 (g) 3 SiO 2 (s) + 2 N 2 O(g) (b) O 2 (g) + 2 H 2 O(l) 2 H 2 O 2 (aq) (c) CaH 2 (s) + 2 C 2 H 5 OH(l) Ca(OC 2 H 5 ) 2 (s) + 2 H 2 (g) 9/22/2004 OFB Chapter 7 37

Exercise 7-12 A vessel holds pure CO (g) at a pressure of 1.282 atm and a temperature of 354K. A quantity of nickel is added, and the partial pressure of CO (g) drops to an equilibrium value of 0.709 atm because of the reaction Ni (s) + 4CO (g) Ni(CO) 4 (g) Compute the equilibrium constant for this reaction at 354K. P Ni(CO) 4 Ni(CO) 4 K = = 4 4 (PCO) [Ni(s)] (PCO) P (1) initial partial pressure (atm) change in partial pressure (atm) equilibrium partial pressure (atm) P CO (atm) P Ni(CO)4 (atm) 9/22/2004 OFB Chapter 7 39

7-7 Extraction and Separation Processes Extraction Partitioning of a solute between two immiscible solvents Chromatography Solute is partitioned between a mobile phase and a stationary phase Column Chromatography Gas-liquid Chromatography 9/22/2004 OFB Chapter 7 40