CHAPTER 14 CHEMICAL EQUILIBRIUM Vanessa N PrasadPermaul Valencia

CHAPTER 14: CHEMICAL EQUILIBRIUM Vanessa N. Prasad-Permaul Valencia Community College CHM 1046 1

Introduction 1. How far does a reaction proceed toward completion before it reaches a state of chemical equilibrium? 2. Chemical equilibrium a) The state reached when the concentrations of reactants and products remain constant over time b) A state in which the concentration of reactants and products no longer change (net) 3. Equilibrium mixture A mixture of reactants and products in the equilibrium state 2

Introduction 4. What are we interested in? a) What is the relationship between the concentration of reactants and products in an equilibrium mixture? b) How can we determine equilibrium concentrations from initial concentrations? c) What factors can be exploited to alter the composition of an equilibrium mixture? 3

The Equilibrium State In previous chapters we have generally assumed that chemical reactions result in complete conversion of reactants to products Many reactions do not go to completion!! Example 1: 4

The Equilibrium State 5

The Equilibrium State The two experiments demonstrate that the interconversion of N 2 O 4 and NO 2 is reversible and that the same equilibrium state is reached starting from either substance. 1. This is why we use a instead of 2. Since both NO 2 and N 2 O 5 are products and reactants we will call the chemical on the left reactants and on the right products. 3. All chemical reactions are reversible 6

The Equilibrium State 4. We call a reaction irreversible when it proceed nearly to completion a. Equilibrium mixture contains almost all products and almost no reactants b. Reverse reaction is too slow to be detected 5. In an equilibrium state the reaction does not stop at particular concentrations of reactants and products, the rates of the forward and reverse reactions become equal. Important: reaction does not stop 7

The Equilibrium State 6. Chemical equilibrium is a dynamic state in which forward and reverse reactions continue at equal rates so that there is no net conversion of reactants to products 8

APPLYING STOICHIOMETRY TO AN EQUILIBRIUM MIXTURE EXAMPLE 14. 1: CO(g) + 3 H 2(g) CH 4(g) + H 2 O(g) When 1. 000 mol CO & 3. 000 mol H 2 are placed in a 10. 00 L vessel @ 927 o. C and allowed to come to equilibrium; the mixture is found to contain 0. 387 mol H 2 O. What is the molar composition of the equilibrium mixture? (How many moles of each substance are present? ) Starting Change Equilibrium CO(g) + 3 H 2(g) 1. 000 -x 1. 000 - x 3. 000 -3 x 3. 000 - 3 x CH 4(g) + H 2 O(g) 0 +x x = 0. 387 CO = 1. 000 – x = 1. 000 – 0. 387 = 0. 613 mol H 2 = 3. 000 – x = 3. 000 – 3(0. 387) = 1. 839 mol CH 4 = x = 0. 387 mol

APPLYING STOICHIOMETRY TO AN EQUILIBRIUM MIXTURE EXERCISE 14. 1: CO(g) + H 2 O(g) CO 2(g) + H 2(g) Suppose there is a mixture containing 1. 00 mol CO and 1. 00 mol H 2 O. When equilibrium is reached at 1000 o. C, the mixture contains 0. 43 mol H 2. What is the molar composition of the equilibrium mixture?

The Equilibrium Constant, Kc a A + b. B c. C + d. D Kc = [C]c [D]d [A]a [B]b If we write the equation in the reverse direction c. C + d. D a. A + b. B K’c = [A]a [B]b = 1 [C]c [D]d kc 11

The Equilibrium Constant, Kc General equation: a. A + b. B c. C + d. D Equilibrium equation: Kc = [C]c [D]d products [A]a [B]b reactants The substances in the equilibrium equation must be gases or molecules and ions in solution: NO SOLIDS! NO PURE LIQUIDS! Kc units are omitted but you must say at what temperature! 12

WRITING EQUILIBRIUM-CONSTANT EXPRESSIONS EXAMPLE 14. 2: Write the equilibrium-constant expression Kc for catalytic methylation CO(g) + 3 H 2(g) CH 4(g) + H 2 O(g) Kc = [CH 4][H 2 O] [CO][H 2]3 Write the equilibrium-constant expression Kc for the reverse reaction CH 4(g) + H 2 O(g) Kc = CO(g) + 3 H 2(g) [CO][H 2]3 [CH 4][H 2 O]

WRITING EQUILIBRIUM-CONSTANT EXPRESSIONS EXAMPLE 14. 2: Write the equilibrium-constant expression Kc the synthesis of NH 3 N 2(g) + 3 H 2(g) Kc = 2 NH 3(g) [NH 3]2 [N 2][H 2]3 Write the equilibrium-constant expression Kc for the following rxn. 1/2 N 2(g) + 3/2 H 2(g) Kc = NH 3(g) [NH 3] [N 2]1/2 [H 2]3/2

WRITING EQUILIBRIUM-CONSTANT EXPRESSIONS EXERCISE 14. 2: Write the equilibrium-constant expression Kc for the following reaction: 2 NO 2(g) + 7 H 2(g) 2 NH 3(g) + 4 H 2 O(g) Write the equilibrium-constant expression Kc for the following reaction: NO 2(g) + 7/2 H 2(g) NH 3(g) + 2 H 2 O(g)

The Equilibrium Constant Kp Kp = equilibrium constant with respect to partial pressures of reactants and products a A + b. B c. C + d. D Kp = (PC)c (PD)d (PA)a (PB)b Relationship between Kc and Kp Kp = Kc(RT) n 16

Heterogeneous Equilibria Introduction 1. So far we have been talking about homogeneous equilibria, in which all reactants and products are in a single phase (gas or solution) 2. Heterogeneous equilibria are those in which reactants and products are present in more than one phase 17

Using the Equilibrium Constant Introduction Knowing the value of the equilibrium constant for a chemical reaction lets us: 1. Judge the extent of the reaction 2. Predict the direction of the reaction 3. Calculate the equilibrium concentrations from any initial concentrations 18

Using the Equilibrium Constant The numerical value of the equilibrium constant for a reaction indicates the extent to which reactants are converted to products 1. Large value for Kc > 103 reaction proceeds essentially to 100% (mostly products) 2. Small value for Kc < 10 -3 reaction proceeds hardly at all before equilibrium is reached (mostly reactants) 3. If a reaction has an intermediate value of Kc = 103 to 10 -3 a. Appreciable concentrations of both reactants and products are present in the equilibrium mixture 19

OBTAINING AN EQUILIBRIUM CONSTANT FROM REACTION COMPOSITION EXAMPLE 14. 3: What is the value of Kc for the decomposition of HI at room temp. ? 2 HI(g) Starting Change Equilibrium 0. 800 M -2 x 4. 00 mol – 2 x H 2(g) + I 2(g) 0 +x x Kc = [H 2] [I 2] [HI]2 CONC 4. 00 mol/5. 00 L = 0. 800 M 0. 442 mol/5. 00 L = 0. 0884 M (0. 800 – 2(0. 0884))M = 0. 623 M = [0. 0884 M] [0. 623 M]2 = 0. 0201

OBTAINING AN EQUILIBRIUM CONSTANT FROM REACTION COMPOSITION EXERCISE 14. 3: Hydrogen sulfide, a colorless gas with a foul odor, dissociates on heating: 2 H 2 S(g) 2 H 2(g) + S 2(g) When 0. 100 mol H 2 S was put into a 10. 0 L vessel and heated to 1132 o. C, the equilibrium mixture contained 0. 0285 mol H 2. What is the value of Kc at this temperature?

WRITING Kc FOR A REACTION WITH PURE SOLIDS OR LIQUIDS EXAMPLE 14. 4: Quicklime (calcium oxide, Ca. O), is prepared by heating a source of calcium carbonate Ca. CO 3 such as limestone or seashells. Ca. CO 3(s) Ca. O(s) + CO 2(g) Write the expression for Kc. Kc = [CO 2] An equilibrium constant can also be written for a physical equilibrium. H 2 O(l) H 2 O(g) Write the expression for Kc for the vaporization of water. Kc = [H 2 O(g)]

WRITING Kc FOR A REACTION WITH PURE SOLIDS OR LIQUIDS EXERCISE 14. 4: The Mond process for purifying nickel involves the formation of nickel tetracarbonyl Ni(CO)4, a volatile liquid, from nickel metal and carbon monoxide. Carbon monoxide is passed over impure nickel to form nickel carbonyl vapor, which, when heated, decomposes and deposits pure nickel. Ni(s) + 4 CO(g) Ni(CO)4(g) Write the expression for Kc for this reaction

Predicting the direction of Reaction Quotient = Qc 1. Not necessarily equilibrium concentrations, at some time, t, snapshot of reaction 2. As time passes, Qc changes toward the value of Kc 3. When the equilibrium state is reached Qc = Kc 4. Qc allows us to predict the direction of reaction by comparing the values of Kc and Qc a) If Qc< Kc, net reaction goes from left to right, (reactant to products) b) If Qc > Kc, net reaction goes from right to left, (products to reactants) c) If Qc = Kc, no net reaction occurs 24

USING THE REACTANT QUOTIENT EXAMPLE 14. 5: A 50. 0 L reaction vessel contains 1. 00 mol N 2, 3. 00 mol H 2, and 0. 500 mol NH 3. Will more ammonia be formed or will it dissociate when a mixture goes to equilibrium @ 400 o. C? Kc is 0. 500 @ 400 o. C. N 2(g) + 3 H 2(g) Qc = = 2 NH 3(g) [NH 3]2 [N 2][H 2]3 [0. 500 mol/50. 0 L]2 [1. 00 mol/50. 0 L][3. 00 mol/50. 0 L]3 = [0. 0100]2 [0. 0200][0. 0600]3 Qc = 23. 1 which is greater than Kc = 0. 500 the reaction will go to the left or ammonia will dissociate.

USING THE REACTANT QUOTIENT EXAMPLE 14. 5: A 10. 0 L reaction vessel contains 0. 0015 mol CO 2 and 0. 10 mol CO. If a small amount of carbon is added to this vessel and the temperature is raised to 1000 o. C, will more CO form? CO 2(g) + C(s) The value of Kc is 1. 17 at 1000 o. C 2 CO(g)

OBTAINING EQUILIBRIUM CONCENTRATION EXAMPLE 14. 6: A gaseous mixture contains 0. 30 mol CO, 0. 10 mol H 2 and 0. 020 mol H 2 O plus an unknown amount of CH 4, in each liter. The mixture is at equilibrium @ 1200 K. What is the concentration of CH 4 in this mixture? Kc = 3. 92 CO(g) + 3 H 2(g) Kc = 3. 92 = CH 4(g) + H 2 O(g) [CH 4][H 2 O] [CO][H 2]3 [CH 4][0. 020 mol/1. 0 L] [0. 30 mol/1. 0 L][0. 10 mol/1. 0 L]3 3. 92[3. 00 x 10 -4 M] = [CH 4] = [0. 020 M] = [CH 4][0. 020 M] [0. 30 M][0. 10 M]3 0. 059 M

OBTAINING EQUILIBRIUM CONCENTRATION EXAMPLE 14. 6: Phosphorus pentachloride gives an equilibrium mixture of PCl 5, PCl 3, and Cl 2 when heated. PCl 5(g) PCl 3(g) + Cl 2(s) A 1. 00 L vessel contains an unknown amount of PCl 5 and 0. 020 mol of each PCl 3 and Cl 2 at equilibrium at 250 o. C. How many moles of PCl 5 are in the vessel if Kc for this reaction is 0. 0415 @ 250 o. C?

Altering an Equilibrium Mixture: Changes in Pressure and Volume In general Le Chatelier’s Principle predicts that: 1. An increase in pressure by reducing the volume will bring about net reaction in the direction that decreases the number of moles of gas 2. A decrease in pressure by enlarging the volume will bring about net reaction in the direction that increases the number of moles of gas. 29

Factors that Alter the Composition of an Equilibrium Mixture Introduction One of the principal goals of chemical synthesis is to maximize the conversion of reactants to products while minimizing the expenditure of energy. 1. Can be achieved if the reaction goes nearly to completion at mild temperatures and pressures. 2. If the equilibrium mixture is high in reactants and poor in products, the experimental conditions must be changed. 3. Several factors can be exploited to alter the composition of an equilibrium mixture. A. The concentration of reactants or products B. The pressure and volume C. The temperature 30

Le Chatelier’s Principle If a stress is applied to a reaction mixture at equilibrium, net reaction occurs in the direction that relieves the stress 1. Stress means a change in the concentration, pressure, volume, or temperature that disturbs the original equilibrium 2. Reaction then occurs to change the composition of the mixture until a new state of equilibrium is reached 3. The direction that the reaction takes (reactants to products or products to reactants) is the one that reduces the stress 31

Altering an Equilibrium Mixture: Changes in Concentration In general, when an equilibrium is disturbed by the addition or removal of any reactant or product, Le Chatelier’s principle predicts that: 1. The concentration stress of an added reactant or product is relieved by net reaction in the direction that consumes the added substance 2. The concentration stress of a removed reactant or product is relieved by net reaction in the direction that replenishes the removed substance 32

LE CHATLIER’S PRINCIPLE EXAMPLE 14. 6: Predict the direction of reaction when H 2 is removed from a mixture (lowering the concentration) in which the following equilibrium has been established: H 2(g) + I 2(g) 2 HI(g) When H 2 is removed from the reaction mixture, HI will dissociate to partially restore the H 2 that was removed The reaction will go to the left

LE CHATLIER’S PRINCIPLE EXAMPLE 14. 6: Consider each of the following equilibria, which are disturbed as indicated. Predict the direction of the reaction: The following equilibria is disturbed by increasing pressure of carbon dioxide: Ca. CO 3(s) Ca. O(s) + CO 2(g) The following equilibrium is disturbed by increasing the concentration of hydrogen: 2 Fe(s) + 3 H 2 O(g) Fe 2 O 3(s) + 3 H 2(g)

LE CHATLIER’S PRINCIPLE WITH ALTERED PRESSURE EXAMPLE 14. 7: Will the products increase, decrease or have no effect if the pressure is increased: CO(g) + Cl 2(g) COCl 2(g) Reaction decreases the number of molecules of gas an increase in pressure increases the amount of product and pushes the reaction to the right. 2 H 2 S(g) 2 H 2(g) + S 2(g) Reaction increases the number of molecules of gas an increase of pressure decreases the amount of products and the reaction shifts to the left. C(s) + S 2(g) CS 2(g) Reaction does not change the number of molecules. Only look at gas volumes when decide the effect of pressure change on equilibrium composition. Pressure change has no effect.

LE CHATLIER’S PRINCIPLE WITH ALTERED PRESSURE EXERCISE 14. 7: Can the amount of product be increased in each of the following reactions by increasing the pressure? explain: CO 2(g) + H 2(g) 4 Cu. O(s) 2 SO 2(g) CO(g) + H 2 O(g) 2 Cu 2 O(s) + O 2(g) 2 SO 3(g)

Altering the Equilibrium Mixture: Changes in Temperature In general, the temperature dependence of the equilibrium constant depends on the sign of H for the reaction 1. The equilibrium constant for an exothermic reaction (negative H ) decreases as the temperature increases 2. The equilibrium constant for an endothermic reaction (positive H ) increases as the temperature increases. 3. H = standard enthalpy of reaction, enthalpy change measured under standard conditions 4. Standard conditions = most stable form of a substance at 1 atm pressure and at a specified temperature, usually 25 C; 1 M concentration for all substances 37

Altering the Equilibrium Mixture: Changes in Temperature Le Chatelier’s Principle says that if heat is added to an equilibrium mixture (increasing the temperature) net reaction occurs in the direction that relieves the stress of the added heat. 1. For an endothermic reaction heat is absorbed by reaction in the forward direction. The equilibrium shifts to the right at the higher temperatures, Kc increases with increasing temperature 2. For an exothermic heat is absorbed by net reaction in the reverse direction, so Kc decreases with temperature, and the reaction would flow to the left (reactants) 38

LE CHATLIER’S PRINCIPLE WITH ALTERED TEMPERATURE EXAMPLE 14. 8: Carbon monoxide is formed when carbon dioxide reacts with solid carbon: CO 2(g) + C(s) 2 CO(g) ; Ho = 172. 5 k. J Is a high or low temperature more favorable to the formation of carbon monoxide? Increase products need to shift equilibrium to the right Endothermic The temperature needs to be raised High temperature is more favorable to the formation of carbon monoxide.

LE CHATLIER’S PRINCIPLE WITH ALTERED TEMPERATURE EXAMPLE 14. 8: Carbon monoxide is formed when carbon dioxide reacts with solid carbon: CO 2(g) + H 2(g) CO(g) + H 2 O(g) Is a high or a low temperature more favorable to the production of carbon monoxide? Explain.

The Effect of a Catalyst on Equilibrium A catalyst increases the rate of a chemical reaction by making available a new, lower-energy pathway for conversion of reactants to products. 1. Since the forward and reverse reaction pass through the same transition state, a catalyst lowers the activation energy for both 2. The rates of the forward and reverse reactions increase by the same factor 3. Catalyst accelerates the rate at which equilibrium is reached 4. Catalyst does not affect the composition of the equilibrium mixture 41

The Effect of a Catalyst on Equilibrium 42

The Link Between Chemical Equilibrium and Chemical Kinetics A + B C + D Assuming that the forward and reverse reactions occur in a single bimolecular step, elementary reactions, we can write the following rate laws Rate of forward reaction = kf [A] [B] Rate of reverse reaction = kr [C] [D] When t=0 [C] = [D] = 0 As A and B are converted to C and D the rate of the forward reaction decreases and the rate of the reverse reaction is increasing, until they are equal, chemical equilibrium kf [A] [B] = kr [C] [D] 43

The Link Between Chemical Equilibrium and Chemical Kinetics kf = [C] [D] kr [A] [B] The right side of this equation is the equilibrium constant expression for the forward reaction, which equals the equilibrium constant Kc Kc = [C] [D] [A] [B] Therefore the equilibrium constant is simply the ratio of the rate constants for the forward and reverse reactions: Kc = kf kr 44

Example 1: Which of the following is correct? 1. Some reactions are truly not reversible 2. All reactants go to all products in all reactions 3. All reactions are reversible to some extent 4. The rates of the forward and reverse reactions will never be equal 45

Example 2: Write the equilibrium equation for each of the following reaction: a) N 2(g) + 3 H 2(g) 2 NH 3(g) b) 2 NH 3(g) N 2(g) + 3 H 2(g) 46

Example 3: The oxidation of sulfur dioxide to give sulfur trioxide is an important step in the industrial process for synthesis of sulfuric acid. Write the equilibrium equation for each of the following reactions: a) 2 SO 2(g) + O 2(g) 2 SO 3(g) b) 2 SO 3(g) 2 SO 2(g) + O 2(g) The following equilibrium concentrations were measured at 800 K: [SO 2] = 3. 0 x 10 -3 M [O 2] = 3. 5 x 10 -3 M [SO 3] = 5. 0 x 10 -2 M Calculate the equilibrium constant at 800 K a and b 47

Example 4: In the industrial synthesis of hydrogen, mixtures of CO and H 2 O are enriched in H 2 by allowing the CO to react with steam. The chemical equation for this so-called water-gas shift reaction is: CO(g) + H 2 O(g) CO 2(g) + H 2(g) What is the value of Kp at 700 K if the partial pressures in an equilibrium mixture at 700 K are 1. 31 atm of CO, 10. 0 atm of H 2 O, 6. 12 atm of CO 2, and 20. 3 atm H 2? 48

Example 5: When will kc = kp ? 1. 2 SO 2(g) + O 2(g) 2 SO 3(g) 2. CO(g) + H 2 O(g) CO 2(g) + H 2(g) 3. N 2(g) + 3 H 2(g) 2 NH 3(g) 49

Example 6: Nitric oxide reacts with oxygen to give nitrogen dioxide, an important reaction in the Ostwald process for the industrial synthesis of nitric acid: 2 NO(g) + O 2(g) 2 NO 2(g) a) If Kc = 6. 9 x 105 @ 227 C, what is the value of Kp @ 227 C? b) If Kp = 1. 3 x 10 -2 @ 1000 K, what is the value of Kc @ 1000 K? 50

Example 7: For each of the following reactions, write the equilibrium constant expression for Kc a) 2 Fe(s) + 3 H 2 O(g) Fe 2 O 3(s) + 3 H 2(g) b) 2 H 2 O(l) 2 H 2(g) + O 2(g) c) Si. Cl 4(g) + 2 H 2(g) Si(s) + 4 HCl(g) d) Hg 22+(aq) + 2 Cl-(aq) Hg 2 Cl 2(s) 51

Example 8: Which of the following has a Heterogeneous equilibria? 1. 2 SO 2(g) + O 2(g) 2 SO 3(g) 2. CO(g) + H 2 O(g) CO 2(g) + H 2(g) 3. Si. Cl 4(g) + 2 H 2(g) Si(s) + 4 HCl(g) 52

Example 9: The equilibrium constant for the reaction 2 NO(g) + O 2(g) 2 NO 2(g) is 6. 9 x 105 @ 500 K. A 5. 0 L reaction vessel at this temperature was filled with 0. 060 mol of NO, 1. 0 mol O 2, and 0. 80 mol NO 2. a) Is the reaction mixture at equilibrium? If not, in which direction does the net reaction proceed? b) What is the direction of the net reaction if the initial amounts are 5. 0 x 10 -3 mol of NO, 0. 20 mol of O 2 and 4. 0 mol of NO 2? 53

Example 10: Consider the equilibrium for the water-gas shift reaction: CO(g) + H 2 O(g) CO 2(g) + H 2(g) Use Le Chatelier’s principle to predict how the concentration of H 2 will change and what direction the reaction will flow when the equilibrium is disturbed by: 1. Adding CO 2. Adding CO 2 3. Removing H 2 O 4. Removing CO 2 54

Example 11: In the following reaction, if I take away CO, which direction will the reaction proceed to equilibrium? CO 2(g) + H 2(g) CO(g) + H 2 O(g) 1. Products 2. Reactants 55

Example 12: Which direction will the reaction flow if the following equilibria is subjected to an increase in pressure by decreasing the volume? 1. CO(g) + H 2 O(g) CO 2(g) + H 2(g) 2. 2 CO(g) C(s) + CO 2(g) 3. N 2 O 4(g) 2 NO 2(g) 56

Example 13: If I increase the pressure by decreasing the volume, which direction will the reaction flow to reach equilibrium? C(s) + CO 2(g) 2 CO(g) 1. Products 2. Reactants 57

Example 14: When air is heated at very high temperatures in an automobile engine, the air pollutant nitric oxide is produced by the reaction N 2(g) + O 2(g) 2 NO(g) H = 180. 5 k. J 1. How does the equilibrium amount of NO vary with an increase in temperature? 2. What direction is the net reaction flowing? 58

Example 15: A platinum catalyst is used in automobile catalytic converters to hasten the oxidation of carbon monoxide: 2 CO(g) + O 2(g) 2 CO 2(g) H = -566 k. J Suppose that you have a reaction vessel containing an equilibrium mixture. Will the amount of CO increase, decrease, or remain the same when: 1. 2. 3. 4. 5. A platinum catalyst is added The temperature is increased The pressure is increased by decreasing the volume The pressure is increased by adding argon gas The pressure is increased by adding O 2 gas 59

Example 16: Nitric oxide emitted from the engines of supersonic transport planes can contribute to the destruction of stratospheric ozone: NO(g) + O 3(g) NO 2(g) + O 2(g) This reaction is highly exothermic ( E = -200 k. J), and its equilibrium constant Kc is 3. 4 x 1034 at 300 K 1. Which rate constant is larger, kf or kr? 2. The value of kf at 300 K is 8. 5 x 106 M-1 s-1. What is the value of kr at the same temperature? 3. A typical temperature in the stratosphere is 230 K. Do the values of k f, kr, and Kc increase or decrease when the temperature is lowered from 300 K to 230 K? 60

Example 17: If I increase the temperature of reaction which way will the reaction flow to equilibrium? NO 2(g) + O 2(g) NO(g) + O 3(g) H = 200 k. J 1. Products 2. Reactants 61