Chapter 13 Chemical Equilibrium Chemical Equilibrium The state

Chapter 13 Chemical Equilibrium

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

Equilibrium Is: • Macroscopically static • Microscopically dynamic Copyright © Cengage Learning. All rights reserved 3

Chemical Equilibrium • When compounds react, they eventually form a mixture of products and unreacted reactants, in a dynamic equilibrium. A dynamic equilibrium consists of a forward reaction, in which substances react to give products, and a reverse reaction, in which products react to give the original reactants. Chemical equilibrium is the state reached by a reaction mixture when the rates of the forward and reverse reactions have become equal.

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 H 2 O (l) H 2 O (g) Chemical equilibrium N 2 O 4 (g) 2 NO 2 (g)

Chemical Equilibrium Much like water in a U-shaped tube, there is constant mixing back and forth through the lower portion of the tube. “reactants” “products” It’s as if the forward and reverse “reactions” were occurring at the same rate. The system appears to static (stationary) when, in reality, it is dynamic (in constant motion).

N 2 O 4 (g) 2 NO 2 (g) equilibrium Start with NO 2 Start with N 2 O 4 Start with NO 2 & N 2 O 4

Chemical Equilibrium The Haber process for producing ammonia from N 2 and H 2 does not go to completion: It establishes an equilibrium state where all three species are present.

A Problem to Consider • Applying Stoichiometry to an Equilibrium Mixture. Suppose we place 1. 000 mol N 2 and 3. 000 mol H 2 in a reaction vessel at 450 o. C and 10. 0 atmospheres of pressure. The reaction is What is the composition of the equilibrium mixture if it contains 0. 080 mol NH 3?

A Problem to Consider Using the information given, set up the following table. Initial 1. 000 3. 000 Change -x -3 x Equilibrium 1. 000 - x 3. 000 - 3 x 0 +2 x 2 x = 0. 080 mol The equilibrium amount of NH 3 was given as 0. 080 mol. Therefore, 2 x = 0. 080 mol NH 3 (x = 0. 040 mol).

A Problem to Consider Using the information given, set up the following table. Initial 1. 000 3. 000 Change -x -3 x Equilibrium 1. 000 - x 3. 000 - 3 x 0 +2 x 2 x = 0. 080 mol Equilibrium amount of N 2 = 1. 000 - 0. 040 = 0. 960 mol N 2 Equilibrium amount of H 2 = 3. 000 - (3 x 0. 040) = 2. 880 mol H 2 Equilibrium amount of NH 3 = 2 x = 0. 080 mol NH 3

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 12

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 13

The Equilibrium Constant • Every reversible system has its own “position of equilibrium” under any given set of conditions. The ratio of products produced to unreacted reactants for any given reversible reaction remains constant under constant conditions of pressure and temperature. The numerical value of this ratio is called the equilibrium constant for the given reaction.

The Equilibrium Constant • 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 to a power equal to its coefficient in the balanced chemical equation. For the general equation above, the equilibrium-constant expression would be:

The Equilibrium Constant • 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 to a power equal to its coefficient in the balanced chemical equation. The molar concentration of a substance is denoted by writing its formula in square brackets.

Write equilibrium expressions for the following equilibria: 1. 2 HI(aq) 2. Ba. Cl 2(s) 3. Ca. CO 3(s) H 2(g) + I 2(g) Ba 2+(aq) + 2 Cl-(aq) Ca. CO(s) + CO 2(g)

The Equilibrium Constant • The equilibrium constant, Kc, is the value obtained for the equilibrium-constant expression when equilibrium concentrations are substituted. A large Kc indicates large concentrations of products at equilibrium. A small Kc indicates large concentrations of unreacted reactants at equilibrium.

Interpreting Equilibrium Constants • The numerical value of the equilibrium constant tells us the extent to which reactants have converted to products. 1. Keq greater than 1 x 102. • Large value of Keq: numerator > denominator. • At equilibrium mostly product present.

Keq less than 1 x 10 -2. • Which is larger, numerator or denominator? • Ans: Denominator. • At equilibrium which is present in the larger quantity? Reactants or Products? • Ans: Reactants. Keq between 1 x 10 -2 and 1 x 102 • Equilibrium mixture contains significant concentration of both reactants and products.

constant

N 2 O 4 (g) K= [NO 2]2 [N 2 O 4] a. A + b. B K= [C]c[D]d [A]a[B]b 2 NO 2 (g) = 4. 63 x 10 -3 c. C + d. D Law of Mass Action

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

The Equilibrium Constant • The law of mass action states that the value of the equilibrium constant expression Kc is constant for a particular reaction at a given temperature, whatever equilibrium concentrations are substituted. Consider the equilibrium established in the Haber process.

The Equilibrium Constant – The equilibrium-constant expression would be Note that the stoichiometric coefficients in the balanced equation have become the powers to which the concentrations are raised.

Equilibrium: A Kinetics Argument • If the forward and reverse reaction rates in a system at equilibrium are equal, then it follows that their rate laws would be equal. Consider the decomposition of N 2 O 4, dinitrogen tetroxide. However, once some NO 2 is produced it can recombine to form N 2 O 4.

Equilibrium: A Kinetics Argument kf kr – Call the decomposition of N 2 O 4 the forward reaction and the formation of N 2 O 4 the reverse reaction. These are elementary reactions, and you can immediately write the rate law for each. Here kf and kr represent the forward and reverse rate constants.

Equilibrium: A Kinetics Argument kf kr – Ultimately, this reaction reaches an equilibrium state where the rates of the forward and reverse reactions are equal. Therefore,

Equilibrium: A Kinetics Argument kf kr – Combining the constants you can identify the equilibrium constant, Kc, as the ratio of the forward and reverse rate constants.

Obtaining Equilibrium Constants for Reactions • Equilibrium concentrations for a reaction must be obtained experimentally and then substituted into the equilibrium-constant expression in order to calculate Kc.

Obtaining Equilibrium Constants for Reactions Consider the reaction below Suppose we started with initial concentrations of CO and H 2 of 0. 100 M and 0. 300 M, respectively.

Obtaining Equilibrium Constants for Reactions Consider the reaction below When the system finally settled into equilibrium we determined the equilibrium concentrations to be as follows. Reactants [CO] = 0. 0613 M Products [CH 4] = 0. 0387 M [H 2] = 0. 1893 M [H 2 O] = 0. 0387 M

Obtaining Equilibrium Constants for Reactions Consider the reaction below The equilibrium-constant expression for this reaction is:

Obtaining Equilibrium Constants for Reactions Consider the reaction below If we substitute the equilibrium concentrations, we obtain:

Obtaining Equilibrium Constants for Reactions Consider the reaction below Regardless of the initial concentrations (whether they be reactants or products), the law of mass action dictates that the reaction will always settle into an equilibrium where the equilibrium-constant expression equals Kc.

Obtaining Equilibrium Constants for Reactions Consider the reaction below As an example, let’s repeat the previous experiment, only this time starting with initial concentrations of products: [CH 4]initial = 0. 1000 M and [H 2 O]initial = 0. 1000 M

Obtaining Equilibrium Constants for Reactions Consider the reaction below We find that these initial concentrations result in the following equilibrium concentrations. Reactants [CO] = 0. 0613 M Products [CH 4] = 0. 0387 M [H 2] = 0. 1893 M [H 2 O] = 0. 0387 M

Obtaining Equilibrium Constants for Reactions Consider the reaction below Substituting these values into the equilibrium-constant expression, we obtain the same result. Whether we start with reactants or products, the system establishes the same ratio.

The Equilibrium Constant, Kp • In discussing gas-phase equilibria, it is often more convenient to express concentrations in terms of partial pressures rather than molarities. It can be seen from the ideal gas equation that the partial pressure of a gas is proportional to its molarity.

The Equilibrium Constant, Kp • If we express a gas-phase equilibria in terms of partial pressures, we obtain Kp. Consider the reaction below. The equilibrium-constant expression in terms of partial pressures becomes:

The Equilibrium Constant, Kp • In general, the numerical value of Kp differs from that of Kc. From the relationship n/V=P/RT, we can show that where Dn is the sum of the moles of gaseous products in a reaction minus the sum of the moles of gaseous reactants.

A Problem to Consider • Consider the reaction Kc for the reaction is 2. 8 x 102 at 1000 o. C. Calculate Kp for the reaction at this temperature.

A Problem to Consider • Consider the reaction We know that From the equation we see that Dn = -1. We can simply substitute the given reaction temperature and the value of R (0. 08206 L. atm/mol. K) to obtain Kp.

A Problem to Consider • Consider the reaction Since

Homogenous equilibrium applies to reactions in which all reacting species are in the same phase. N 2 O 4 (g) Kc = [NO 2 2 NO 2 (g) ]2 Kp = [N 2 O 4] 2 PNO 2 PN 2 O 4 In most cases Kc K p a. A (g) + b. B (g) c. C (g) + d. D (g) Kp = Kc(RT)Dn Dn = moles of gaseous products – moles of gaseous reactants = (c + d) – (a + b)

Homogeneous Equilibrium CH 3 COOH (aq) + H 2 O (l) [CH 3 COO-][H 3 O+] Kc‘ = [CH 3 COOH][H 2 O] CH 3 COO- (aq) + H 3 O+ (aq) [H 2 O] = constant [CH 3 COO-][H 3 O+] = Kc‘ [H 2 O] Kc = [CH 3 COOH] General practice not to include units for the equilibrium constant.

The equilibrium concentrations for the reaction between carbon monoxide and molecular chlorine to form COCl 2 (g) at 740 C are [CO] = 0. 012 M, [Cl 2] = 0. 054 M, and [COCl 2] = 0. 14 M. Calculate the equilibrium constants Kc and Kp. CO (g) + Cl 2 (g) COCl 2 (g) [COCl 2] 0. 14 = = 220 Kc = [CO][Cl 2] 0. 012 x 0. 054 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 2 NO 2 (g) 2 NO (g) + O 2 (g) is 158 at 1000 K. What is the equilibrium pressure of O 2 if the PNO 2 = 0. 400 atm and PNO = 0. 270 atm? Kp = 2 PNO PO 2 2 PNO 2 PO 2 = Kp 2 PNO 2 2 PNO PO 2 = 158 x (0. 400)2/(0. 270)2 = 347 atm

Heterogenous equilibrium applies to reactions in which reactants and products are in different phases. Ca. CO 3 (s) [Ca. O][CO 2] Kc‘ = [Ca. CO 3] Kc = [CO 2] = Kc‘ x [Ca. O] Ca. O (s) + CO 2 (g) [Ca. CO 3] = constant [Ca. O] = constant Kp = PCO 2 The concentration of solids and pure liquids are not included in the expression for the equilibrium constant.

Heterogeneous Equilibria Consider the reaction below. The equilibrium-constant expression contains terms for only those species in the homogeneous gas phase…H 2 O, CO, and H 2.

Ca. CO 3 (s) Ca. O (s) + CO 2 (g) PCO 2 = Kp PCO 2 does not depend on the amount of Ca. CO 3 or Ca. O

Consider the following equilibrium at 295 K: NH 4 HS (s) NH 3 (g) + H 2 S (g) The partial pressure of each gas is 0. 265 atm. Calculate Kp and Kc for the reaction? Kp = PNH PH S = 0. 265 x 0. 265 = 0. 0702 3 2 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

A+B C+D Kc‘ C+D E+F K‘c‘ A+B E+F Kc [C][D] Kc‘ = [A][B] [E][F] Kc‘‘ = [C][D] [E][F] Kc = [A][B] Kc = Kc‘ x K‘c‘ 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.

N 2 O 4 (g) K= [NO 2]2 [N 2 O 4] 2 NO 2 (g) = 4. 63 x 10 -3 2 NO 2 (g) N 2 O 4 (g) [N 2 O 4] 1 = 216 = K‘ = 2 K [NO 2] When the equation for a reversible reaction is written in the opposite direction, the equilibrium constant becomes the reciprocal of the original equilibrium constant.

Equilibrium Constant for the Sum of Reactions • For example, nitrogen and oxygen can combine to form either NO(g) or N 2 O (g) according to the following equilibria. (1) Kc = 4. 1 x 10 -31 (2) Kc = 2. 4 x 10 -18 Using these two equations, we can obtain Kc for the formation of NO(g) from N 2 O(g): (3) Kc = ?

Equilibrium Constant for the Sum of Reactions To combine equations (1) and (2) to obtain equation (3), we must first reverse equation (2). When we do we must also take the reciprocal of its Kc value. (1) Kc = 4. 1 x 10 -31 (2) Kc = (3)

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 Knew = (Koriginal)n. • K values are usually written without units. Copyright © Cengage Learning. All rights reserved 57

• 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 58

• K involves concentrations. • Kp involves pressures. Copyright © Cengage Learning. All rights reserved 59

Predicting the Direction of Reaction • How could we predict the direction in which a reaction at non-equilibrium conditions will shift to reestablish equilibrium? To answer this question, substitute the current concentrations into the reaction quotient expression and compare it to Kc. The reaction quotient, Qc, is an expression that has the same form as the equilibrium-constant expression but whose concentrations are not necessarily at equilibrium.

Predicting the Direction of Reaction • For the general reaction the Qc expresssion would be:

Predicting the Direction of Reaction • For the general reaction If Qc > Kc, the reaction will shift left…toward reactants. • If Qc < Kc, the reaction will shift right… toward products. • If Qc = Kc, then the reaction is at equilibrium.

A Problem to Consider the following equilibrium. A 50. 0 L vessel contains 1. 00 mol N 2, 3. 00 mol H 2, and 0. 500 mol NH 3. In which direction (toward reactants or toward products) will the system shift to reestablish equilibrium at 400 o. C? Kc for the reaction at 400 o. C is 0. 500.

A Problem to Consider First, calculate concentrations from moles of substances. 1. 00 mol 50. 0 L 3. 00 mol 50. 0 L 0. 500 mol 50. 0 L

A Problem to Consider First, calculate concentrations from moles of substances. 0. 0200 M 0. 0600 M 0. 0100 M The Qc expression for the system would be:

A Problem to Consider First, calculate concentrations from moles of substances. 0. 0200 M 0. 0600 M 0. 0100 M Substituting these concentrations into the reaction quotient gives:

A Problem to Consider First, calculate concentrations from moles of substances. 0. 0200 M 0. 0600 M 0. 0100 M Because Qc = 23. 1 is greater than Kc = 0. 500, the reaction will go to the left (toward reactants) as it approaches equilibrium.

Calculating Equilibrium Concentrations • Once you have determined the equilibrium constant for a reaction, you can use it to calculate the concentrations of substances in the equilibrium mixture.

Calculating Equilibrium Concentrations For example, consider the following equilibrium. Suppose a gaseous mixture contained 0. 30 mol CO, 0. 10 mol H 2, 0. 020 mol H 2 O, and an unknown amount of CH 4 per liter. What is the concentration of CH 4 in this mixture? The equilibrium constant Kc equals 3. 92.

Calculating Equilibrium Concentrations First, calculate concentrations from moles of substances. 0. 30 mol 0. 10 mol 1. 0 L ? ? 0. 020 mol 1. 0 L

Calculating Equilibrium Concentrations First, calculate concentrations from moles of substances. 0. 30 M 0. 10 M ? ? 0. 020 M The equilibrium-constant expression is:

Calculating Equilibrium Concentrations First, calculate concentrations from moles of substances. 0. 30 M 0. 10 M ? ? 0. 020 M Substituting the known concentrations and the value of Kc gives:

Calculating Equilibrium Concentrations First, calculate concentrations from moles of substances. 0. 30 M 0. 10 M ? ? 0. 020 M You can now solve for [CH 4]. The concentration of CH 4 in the mixture is 0. 059 mol/L.

Calculating Equilibrium Concentrations Suppose we begin a reaction with known amounts of starting materials and want to calculate the quantities at equilibrium.

Calculating Equilibrium Concentrations Consider the following equilibrium. Suppose you start with 1. 000 mol each of carbon monoxide and water in a 50. 0 L container. Calculate the molarity of each substance in the equilibrium mixture at 1000 o. C. Kc for the reaction is 0. 58 at 1000 o. C.

Calculating Equilibrium Concentrations First, calculate the initial molarities of CO and H 2 O. 1. 000 mol 50. 0 L

Calculating Equilibrium Concentrations First, calculate the initial molarities of CO and H 2 O. 0. 0200 M 0 M 0 M The starting concentrations of the products are 0. We must now set up a table of concentrations (starting, change, and equilibrium expressions in x).

Calculating Equilibrium Concentrations Let x be the moles per liter of product formed. Initial 0. 0200 Change -x Equilibrium 0. 0200 -x 0 +x x The equilibrium-constant expression is: 0 +x x

Calculating Equilibrium Concentrations Solving for x. Initial 0. 0200 Change -x -x Equilibrium 0. 0200 -x 0 +x x Substituting the values for equilibrium concentrations, we get: 0 +x x

Calculating Equilibrium Concentrations Solving for x. Initial 0. 0200 Change -x -x Equilibrium 0. 0200 -x Or: 0 +x x

Calculating Equilibrium Concentrations Solving for x. Initial 0. 0200 Change -x -x Equilibrium 0. 0200 -x 0 +x x Taking the square root of both sides we get:

Calculating Equilibrium Concentrations Solving for x. Initial 0. 0200 Change -x -x Equilibrium 0. 0200 -x Rearranging to solve for x gives: 0 +x x

Calculating Equilibrium Concentrations Solving for equilibrium concentrations. Initial 0. 0200 Change -x -x Equilibrium 0. 0200 -x 0 +x x If you substitute for x in the last line of the table you obtain the following equilibrium concentrations. 0. 0114 M CO 0. 0114 M H 2 O 0. 0086 M CO 2 0. 0086 M H 2

Calculating Equilibrium Concentrations • The preceding example illustrates the three steps in solving for equilibrium concentrations. 1. Set up a table of concentrations (starting, change, and equilibrium expressions in x). 2. Substitute the expressions in x for the equilibrium concentrations into the equilibriumconstant equation. 3. Solve the equilibrium-constant equation for the values of the equilibrium concentrations.

Calculating Equilibrium Concentrations • In some cases it is necessary to solve a quadratic equation to obtain equilibrium concentrations. • The next example illustrates how to solve such an equation.

At 12800 C the equilibrium constant (Kc) for the reaction Br 2 (g) 2 Br (g) Is 1. 1 x 10 -3. If the initial concentrations are [Br 2] = 0. 063 M and [Br] = 0. 012 M, calculate the concentrations of these species at equilibrium. Let x be the change in concentration of Br 2 Initial (M) Change (M) Equilibrium (M) [Br]2 Kc = [Br 2] Br 2 (g) 2 Br (g) 0. 063 0. 012 -x +2 x 0. 063 - x 0. 012 + 2 x (0. 012 + 2 x)2 = 1. 1 x 10 -3 Kc = 0. 063 - x Solve for x

(0. 012 + 2 x)2 = 1. 1 x 10 -3 Kc = 0. 063 - x 4 x 2 + 0. 048 x + 0. 000144 = 0. 0000693 – 0. 0011 x 4 x 2 + 0. 0491 x + 0. 0000747 = 0 -b ± b 2 – 4 ac 2 x= ax + bx + c =0 2 a x = -0. 0105 x = -0. 00178 Initial (M) Change (M) Equilibrium (M) Br 2 (g) 2 Br (g) 0. 063 0. 012 -x +2 x 0. 063 - x 0. 012 + 2 x At equilibrium, [Br] = 0. 012 + 2 x = -0. 009 M or 0. 00844 M At equilibrium, [Br 2] = 0. 062 – x = 0. 0648 M

Review: Writing Equilibrium Constant Expressions 1. 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. 2. The concentrations of pure solids, pure liquids and solvents do not appear in the equilibrium constant expressions. 3. The equilibrium constant is a dimensionless quantity. 4. In quoting a value for the equilibrium constant, you must specify the balanced equation and the temperature. 5. 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.

Le Chatelier’s Principle • Obtaining the maximum amount of product from a reaction depends on the proper set of reaction conditions. Le Chatelier’s principle states that when a system in a chemical equilibrium is disturbed by a change of temperature, pressure, or concentration, the equilibrium will shift in a way that tends to counteract this change.

Removing Products or Adding Reactants • Let’s refer back to the illustration of the U-tube in the first section of this chapter. “reactants” “products” It’s a simple concept to see that if we were to remove products (analogous to dipping water out of the right side of the tube) the reaction would shift to the right until equilibrium was reestablished.

Removing Products or Adding Reactants • Let’s refer back to the illustration of the U-tube in the first section of this chapter. “reactants” “products” Likewise, if more reactant is added (analogous to pouring more water in the left side of the tube) the reaction would again shift to the right until equilibrium is reestablished.

Le Châtelier’s Principle • Changes in Concentration continued a. A + b. B c. C + d. D Change Shifts the Equilibrium Increase concentration of product(s) Decrease concentration of product(s) Increase concentration of reactant(s) Decrease concentration of reactant(s) left right left

Effects of Pressure Change • A pressure change caused by changing the volume of the reaction vessel can affect the yield of products in a gaseous reaction only if the reaction involves a change in the total moles of gas present

Effects of Pressure Change If the products in a gaseous reaction contain fewer moles of gas than the reactants, it is logical that they would require less space. So, reducing the volume of the reaction vessel would, therefore, favor the products. Conversely, if the reactants require less volume (that is, fewer moles of gaseous reactant), then decreasing the volume of the reaction vessel would shift the equilibrium to the left (toward reactants).

Effects of Pressure Change Literally “squeezing” the reaction will cause a shift in the equilibrium toward the fewer moles of gas. It’s a simple step to see that reducing the pressure in the reaction vessel by increasing its volume would have the opposite effect. In the event that the number of moles of gaseous product equals the number of moles of gaseous reactant, vessel volume will have no effect on the position of the equilibrium.

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

Effect of Temperature Change • Temperature has a significant effect on most reactions Reaction rates generally increase with an increase in temperature. Consequently, equilibrium is established sooner. In addition, the numerical value of the equilibrium constant Kc varies with temperature.

Effect of Temperature Change Let’s look at “heat” as if it were a product in exothermic reactions and a reactant in endothermic reactions. We see that increasing the temperature is analogous to adding more product (in the case of exothermic reactions) or adding more reactant (in the case of endothermic reactions). This ultimately has the same effect as if heat were a physical entity.

Effect of Temperature Change For example, consider the following generic exothermic reaction. Increasing temperature would be analogous to adding more product, causing the equilibrium to shift left. Since “heat” does not appear in the equilibriumconstant expression, this change would result in a smaller numerical value for Kc.

Effect of Temperature Change For an endothermic reaction, the opposite is true. Increasing temperature would be analogous to adding more reactant, causing the equilibrium to shift right. This change results in more product at equilibrium, amd a larger numerical value for K c.

Effect of Temperature Change In summary: For an endothermic reaction (DH positive) the amounts of products are increased at equilibrium by an increase in temperature (Kc is larger at higher temperatures). • For an exothermic reaction (DH is negative) the amounts of reactants are increased at equilibrium by an increase in temperature (Kc is smaller at higher temperatures).

Le Châtelier’s Principle • Changes in Temperature Change Increase temperature Decrease temperature Exothermic Rx Endothermic Rx K decreases K increases K decreases colder hotter

Effect of a Catalyst • A catalyst is a substance that increases the rate of a reaction but is not consumed by it. It is important to understand that a catalyst has no effect on the equilibrium composition of a reaction mixture A catalyst merely speeds up the attainment of equilibrium.

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 uncatalyzed Catalyst lowers Ea for both forward and reverse reactions. Catalyst does not change equilibrium constant or shift equilibrium.

Le Châtelier’s Principle Change Shift Equilibrium Change Equilibrium Constant Concentration yes no Pressure yes no Volume yes no Temperature yes Catalyst no no

Worked Example 14. 1

Worked Example 14. 2

Worked Example 14. 5

Worked Example 14. 8

Worked Example 14. 9

Worked Example 14. 10 b

Worked Example 14. 11

Worked Example 14. 12

Worked Example 14. 13 b