Copyright 2011 Pearson Education, Inc. Chapter 14 Chemical Equilibrium Roy Kennedy Massachusetts Bay Community College Wellesley Hills, MA

Copyright 2011 Pearson Education, Inc.

Chapter 14Chemical

Equilibrium

Roy KennedyMassachusetts Bay Community College

Wellesley Hills, MA


Arrow Conventions• Chemists commonly use two

kinds of arrows in reactions to indicate the degree of completion of the reactions

• A single arrow indicates all the reactant molecules are converted to product molecules at the end

• A double arrow indicates the reaction stops when only some of the reactant molecules have been converted into products in these notes

2Tro: Chemistry: A Molecular Approach, 2/e


Reaction Dynamics• When a reaction starts, the reactants are consumed

and products are made forward reaction = reactants products therefore the reactant concentrations decrease and the

product concentrations increase as reactant concentration decreases, the forward reaction rate

decreases• Eventually, the products can react to re-form some of

the reactants reverse reaction = products reactants assuming the products are not allowed to escape as product concentration increases, the reverse reaction rate

increases• Processes that proceed in both the forward and reverse

direction are said to be reversible reactants Û products



Hypothetical Reaction2 Red Û Blue

The reaction slows over time,

but the Red molecules never run out!

At some time between 100 and 110 sec,the concentrations of both the Red andthe Blue molecules no longer change –equilibrium has been established.

Notice that equilibrium does not meanthat the concentrations are equal!

Once equilibrium is established, the rateof Red molecules turning into Blue is thesame as the rate of Blue molecules turning into Red



Hypothetical Reaction2 Red Û Blue



Eventually, the reaction proceedsin the reverse direction as fast asit proceeds in the forward direction.At this time equilibrium is established.

Reaction Dynamics

Time

Rate

Rate Forward

Rate Reverse

Initially, only the forwardreaction takes place

As the forward reaction proceedsit makes products and uses reactants

Because the reactant concentration decreases, the forward reaction slows.As the products accumulate, thereverse reaction speeds up.

Once equilibrium is established,the forward and reverse reactions proceed at the same rate, so theconcentrations of all materialsstay constant



Dynamic Equilibrium• As the forward reaction slows and the reverse

reaction accelerates, eventually they reach the same rate

• Dynamic equilibrium is the condition wherein the rates of the forward and reverse reactions are equal

• Once the reaction reaches equilibrium, the concentrations of all the chemicals remain constantbecause the chemicals are being consumed and

made at the same rate7Tro: Chemistry: A Molecular Approach, 2/e


Equilibrium Equal• The rates of the forward and reverse

reactions are equal at equilibrium• But that does not mean the concentrations of

reactants and products are equal• Some reactions reach equilibrium only after

almost all the reactant molecules are consumed – we say the position of equilibrium favors the products

• Other reactions reach equilibrium when only a small percentage of the reactant molecules are consumed – we say the position of equilibrium favors the reactants



When Country A citizens feel overcrowded, some will emigrate to Country B

An Analogy: Population Changes

9

However, after a time, emigration will occur inboth directions at the same rate, leading topopulations in Country A and Country B that areconstant, though not necessarily equal

Tro: Chemistry: A Molecular Approach, 2/e


Equilibrium Constant• Even though the concentrations of reactants and

products are not equal at equilibrium, there is a relationship between them

• The relationship between the chemical equation and the concentrations of reactants and products is called the Law of Mass Action

• For the general equation aA + bB Û cC + dD, the Law of Mass Action gives the relationship below the lowercase letters represent the coefficients of the

balanced chemical equation always products over reactants

• K is called the equilibrium constant unitless



Writing Equilibrium Constant Expressions

• For the reaction aA(aq) + bB(aq) Û cC(aq) + dD(aq) the equilibrium constant expression is

•So for the reaction 2 N2O5(g) Û 4 NO2(g) + O2(g)

the equilibrium constant expression is



What Does the Value of Keq Imply?

• When the value of Keq >> 1, we know that when the reaction reaches equilibrium there will be many more product molecules present than reactant molecules the position of equilibrium favors products

• When the value of Keq << 1, we know that when the reaction reaches equilibrium there will be many more reactant molecules present than product molecules the position of equilibrium favors reactants



A Large Equilibrium Constant



A Small Equilibrium Constant



Practice – Write the equilibrium constant expressions, K, and predict the position of

equilibrium for the following

2 SO2(g) + O2(g) Û 2 SO3(g) K = 8 x 1025

N2(g) + 2 O2(g) Û 2 NO2(g) K = 3 x 10−17

favors products

favors reactants



Relationships between K and Chemical Equations

• When the reaction is written backward, the equilibrium constant is inverted

For the reaction aA + bB Û cC + dDthe equilibrium constant expression is

For the reaction cC + dD Û aA + bBthe equilibrium constant expression is




• When the coefficients of an equation are multiplied by a factor, the equilibrium constant is raised to that factor

For the reaction aA + bB Û cC the equilibrium constant expression is

For the reaction 2aA + 2bB Û 2cC the equilibrium constant expression is




• When you add equations to get a new equation, the equilibrium constant of the new equation is the product of the equilibrium constants of the old equations

For the reactions (1) aA Û bB and (2) bB Û cC the equilibrium constant expressions are

For the reaction aA Û cC the equilibrium constant expression is



Kbackward = 1/Kforward, Knew = Koldn

Example 14.2: Compute the equilibrium constant at 25 °C for the reaction NH3(g) 0.5 N2(g) + 1.5 H2(g)

19

for N2(g) + 3 H2(g) Û 2 NH3(g), K = 3.7 x 108 at 25 °C

K for NH3(g) 0.5N2(g) + 1.5H2(g), at 25 °C

Solution:

Conceptual Plan:

Relationships:

Given:

Find:

N2(g) + 3 H2(g) Û 2 NH3(g) K1 = 3.7 x 108

K’K

2 NH3(g) Û N2(g) + 3 H2(g)

NH3(g) Û 0.5 N2(g) + 1.5 H2(g)



2 B Û A K3 = 1/Krxn 1 = 6.25 x 10−5

Z Û 2 B K4 = 1/Krxn 2 = 2Z Û A K3K4 = 1.25 x 10−4

3 Z Û 3 A (K3K4)3 = 2.0 x 10−12

Practice – When the reaction A(aq) Û 2 B(aq) reaches equilibrium [A] = 1.0 x 10−5 M and [B] = 4.0 x 10−1 M. When the reaction 2 B(aq) Û Z(aq) reaches equilibrium

[B] = 4.0 x 10−3 M and [Z] = 2.0 x 10−6 M. Calculate the equilibrium constant for each of these

reactions and the equilibrium constant for the reaction 3 Z(aq) Û 3 A(aq)

Krxn 1 = 1.6 x 104

Krxn 2 = 5.0 x 10−1



Equilibrium Constants for Reactions Involving Gases

• The concentration of a gas in a mixture is proportional to its partial pressure

• Therefore, the equilibrium constant can be expressed as the ratio of the partial pressures of the gases

• For aA(g) + bB(g) Û cC(g) + dD(g) the equilibrium constant expressions are

or



Kc and Kp

• In calculating Kp, the partial pressures are always in atm

• The values of Kp and Kc are not necessarily the samebecause of the difference in unitsKp = Kc when Dn = 0

• The relationship between them is

Dn is the difference between the number of moles of reactants and moles of products



Deriving the Relationshipbetween Kp and Kc



Deriving the RelationshipBetween Kp and Kc

for aA(g) + bB(g) cC(g) + dD(g)

substituting



Example 14.3: Find Kc for the reaction 2 NO(g) + O2(g) Û 2 NO2(g), given Kp = 2.2 x 1012 @ 25 °C

25

K is a unitless numberbecause there are more moles of reactant than product, Kc

should be larger than Kp, and it is

Kp = 2.2 x 1012

Kc

Check:

Solution:

Conceptual Plan:

Relationships:

Given:

Find:

Kp Kc

2 NO(g) + O2(g) 2 NO2(g)Dn = 2 3 = −1



Practice – Calculate the value of Kp or Kc for each of the following at 27 °C

2 SO2(g) + O2(g) Û 2 SO3(g) Kc = 8 x 1025

N2(g) + 2 O2(g) Û 2 NO2(g) Kp = 3 x 10−17



Heterogeneous Equilibria• Pure solids and pure liquids are materials whose

concentration doesn’t change during the course of a reaction its amount can change, but the amount of it in solution

doesn’t because it isn’t in solution

• Because their concentration doesn’t change, solids and liquids are not included in the equilibrium constant expression

• For the reaction aA(s) + bB(aq) Û cC(l) + dD(aq) the equilibrium constant expression is



Heterogeneous EquilibriaThe amount of C is different, but the amounts of CO and CO2 remain the same. Therefore the amount of C has no effect on the position of equilibrium.



HNO2(aq) + H2O(l) Û H3O+(aq) + NO2

−(aq) K = 4.6 x 10−4

Ca(NO3)2(aq) + H2SO4(aq) Û CaSO4(s) + 2 HNO3(aq) K = 1 x 104

Practice – Write the equilibrium constant expressions, K, and predict the position of

equilibrium for the following



Calculating Equilibrium Constants from Measured Equilibrium Concentrations

• The most direct way of finding the equilibrium constant is to measure the amounts of reactants and products in a mixture at equilibrium

• The equilibrium mixture may have different amounts of reactants and products, but the value of the equilibrium constant will always be the sameas long as the temperature is kept constant the value of the equilibrium constant is independent of

the initial amounts of reactants and products



Initial and Equilibrium Concentrations forH2(g) + I2(g) Û 2HI(g) @ 445 °C

Con

Initialcentra tion

EqCon

uilibricentr

umation

EquilibriumConstant

[H2] [I2] [HI] [H2] [I2] [HI]

0.50 0.50 0.0 0.11 0.11 0.78

0.0 0.0 0.50 0.055 0.055 0.39

0.50 0.50 0.50 0.165 0.165 1.17

1.0 0.5 0.0 0.53 0.033 0.934



Calculating Equilibrium Concentrations

• Stoichiometry can be used to determine the equilibrium concentrations of all reactants and products if you know initial concentrations and one equilibrium concentration

• Use the change in the concentration of the material that you know to determine the change in the other chemicals in the reaction



Because Product C increased by 0.50 mol/L, you must have used ½ that amount of Reactant A, according to the stoichiometry.

Calculating Equilibrium Concentrations:An Example

• Suppose you have a reaction 2 A(aq) + B(aq) Û 4 C(aq) with initial concentrations [A] = 1.00 M, [B] = 1.00 M, and [C] = 0. You then measure the equilibrium concentration of C as [C] = 0.50 M.

+0.50-¼(0.50) -½(0.50)

0.75 0.8833

Because Product C increased by 0.50 mol/L, you must have used ¼ that amount of Reactant B, according to the stoichiometry.

2 A(aq) + B(aq) Û 4 C(aq) Product C increased by 0.50 mol/L



Example 14.6: Find the value of Kc for the reaction2 CH4(g) Û C2H2(g) + 3 H2(g) at 1700 °C if the initial

[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M

34

+0.035



Example 14.6: Find the value of Kc for the reaction2 CH4(g) Û C2H2(g) + 3 H2(g) at 1700 °C if the initial

[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M

35

+0.035−2(0.035) +3(0.035)

0.045 0.105



Practice –The following data were collected for the reaction 2 NO2(g) Û N2O4(g) at 100 °C. Complete the

table and determine values of Kp and Kc.



Practice – A second experiment was done and the following data were collected for the reaction 2 NO2(g) Û N2O4(g) at 100 °C. Complete the table and determine values of Kp and

Kc. Compare them to the first experiment.



The Reaction Quotient• If a reaction mixture, containing

both reactants and products, is not at equilibrium; how can we determine in which direction it will proceed?

• The answer is to compare the current concentration ratios to the equilibrium constant

• The concentration ratio of the products (raised to the power of their coefficients) to the reactants (raised to the power of their coefficients) is called the reaction quotient, Q

for the gas phase reaction aA + bB Û cC + dDthe reaction quotient is:



The Reaction Quotient:Predicting the Direction of Change

• If Q > K, the reaction will proceed fastest in the reverse direction the [products] will decrease and [reactants] will increase

• If Q < K, the reaction will proceed fastest in the forward direction the [products] will increase and [reactants] will decrease

• If Q = K, the reaction is at equilibrium the [products] and [reactants] will not change

• If a reaction mixture contains just reactants, Q = 0, and the reaction will proceed in the forward direction

• If a reaction mixture contains just products, Q = ∞, and the reaction will proceed in the reverse direction



Q, K, and the Direction of Reaction



If Q = K, equilibrium; If Q < K, forward; If Q > K, reverse

Example 14.7: For the reaction below, which direction will it proceed if PI2 = 0.114 atm, PCl2 = 0.102 atm. & PICl = 0.355 atm

41

for I2(g) + Cl2(g) Û 2 ICl(g), Kp = 81.9

direction reaction will proceed

Solution:

Conceptual Plan:

Relationships:

Given:

Find:

I2(g) + Cl2(g) Û 2 ICl(g) Kp = 81.9

QPI2, PCl2, PICl

because Q (10.8) < K (81.9), the reaction will proceed to the rightTro: Chemistry: A Molecular Approach, 2/e


Practice – For the reaction CH4(g) + 2 H2S(g) Û CS2(g) + 4 H2(g) Kc = 3.59 at 900 °C. For each of the following measured

concentrations determine whether the reaction is at equilibrium. If not at equilibrium, in which direction will the reaction proceed to

get to equilibrium?



Example 14.8: If [COF2]eq = 0.255 M and [CF4]eq = 0.118 M, and Kc = 2.00 @ 1000 °C, find the [CO2]eq for the reaction given.

43

Check:Check: substitute approximations back in to see if the value agrees

Solution:Solve: Solve the equilibrium constant expression for the unknown quantity by substituting in the given amounts

Conceptual Plan:

Relationships:

Strategize: You can calculate the missing concentration by using the equilibrium constant expression

2 COF2 Û CO2 + CF4

[COF2]eq = 0.255 M, [CF4]eq = 0.118 M

[CO2]eq

Given:

Find:

Sort: You’re given the reaction and Kc. You’re also given the [X]eq of all but one of the chemicals

K, [COF2], [CF4] [CO2]



Practice – A sample of PCl5(g) is placed in a 0.500 L container and heated to 160 °C. The PCl5 is decomposed into PCl3(g) and Cl2(g). At equilibrium, 0.203 moles of PCl3

and Cl2 are formed. Determine the equilibrium concentration of PCl5 if Kc = 0.0635

PCl5 Û PCl3 + Cl2

44

0.203 mol

0.500 L

0.203 mol

0.500 L



Finding Equilibrium Concentrations When Given the Equilibrium Constant

and Initial Concentrations or PressuresStep 1: decide which direction the reaction will proceed

compare Q to K

Step 2: define the changes of all materials in terms of x use the coefficient from the chemical equation as the

coefficient of x the x change is + for materials on the side the reaction is

proceeding toward the x change is for materials on the side the reaction is

proceeding away from

Step 3: solve for x for 2nd order equations, take square roots of both sides or use the

quadratic formula may be able to simplify and approximate answer for very large or

small equilibrium constants



Example 14.11: For the reaction I2(g) + Cl2(g) Û 2 ICl(g) @ 25 °C, Kp = 81.9. If the initial partial

pressures are all 0.100 atm, find the equilibrium concentrations

46

Qp(1) < Kp(81.9), so the reaction is proceeding forward



Example 14.11: For the reaction I2(g) + Cl2(g) Û 2 ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures are

all 0.100 atm, find the equilibrium concentrations

47

+2xxx0.100x 0.100x 0.100+2x



Example 14.11: For the reaction I2(g) + Cl2(g) Û 2 ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures are

all 0.100 atm, find the equilibrium concentrations

48

+2xxx0.100x 0.100x 0.100+2x



Example 14.11: For the reaction I2(g) + Cl2(g) Û 2 ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures are all

0.100 atm, find the equilibrium concentrations

49

+2xxx0.100x 0.100x 0.100+2x0.027 0.027 0.246

−0.0729 +2(0.0729)−0.0729

0.0729 x

P

I20.100 x 0.100 0.0729 0.027 atm

P

Cl20.100 x 0.100 0.0729 0.027 atm

P

ICl0.100 2x 0.100 2 0.0729 0.246 atm



−0.0729

Example 14.11: For the reaction I2(g) + Cl2(g) Û 2 ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures are all

0.100 atm, find the equilibrium concentrations

50

0.027 0.027 0.246

−0.0729 2(0.0729)

Kp(calculated) = Kp(given) within significant figures Tro: Chemistry: A Molecular Approach, 2/e


Practice – For the reaction I2(g) Û 2 I(g) the value of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is

placed into a 2.00 L flask and heated, what will be the equilibrium concentrations of [I2] and [I]?

(Hint: you will need to use the quadratic formula to solve for x)





because [I]initial = 0, Q = 0 and the reaction must proceed forward



Practice – For the reaction I2(g) Û 2 I(g) the value of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is




Practice – For the reaction I2(g) Û 2 I(g) the value of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is


x = 0.00216

0.500 0.00216 = 0.498[I2] = 0.498 M

2(0.00216) = 0.00432[I] = 0.00432 M



Approximations to Simplify the Math

• When the equilibrium constant is very small, the position of equilibrium favors the reactants

• For relatively large initial concentrations of reactants, the reactant concentration will not change significantly when it reaches equilibriumthe [X]equilibrium = ([X]initial ax) [X]initial

we are approximating the equilibrium concentration of reactant to be the same as the initial concentration

assuming the reaction is proceeding forward



Checking the Approximation and Refining as Necessary

• We can check our approximation by comparing the approximate value of x to the initial concentration

• If the approximate value of x is less than 5% of the initial concentration, the approximation is valid



Example14.13: For the reaction 2 H2S(g) Û 2 H2(g) + S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask initially

containing 1.25 x 10−4 mol H2S is heated to 800 °C, find the equilibrium concentrations.

57

because no products initially, Qc = 0, and the reaction is proceeding forward



Example 14.13: For the reaction 2 H2S(g) Û 2 H2(g) + S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask

initially containing 1.25 x 10−4 mol H2S is heated to 800 °C, find the equilibrium concentrations.

58

+x+2x2x2.50E−4

2x 2x x





59

+x+2x2x2.50E−4

2x2x x2.50E−4





60

+x+2x2x

2x x2.50E–4

the approximation is not valid!!





61

+x+2x2x2.50E−42x

2x xxcurrent = 1.38 x 10−5





62

+x+2x2x2.50E−42x

2x x

xcurrent = 1.27 x 10−5

since xcurrent = xnew, approx. OK Tro: Chemistry: A Molecular Approach, 2/e




63

+x+2x2x2.50E−42x

2x x

xcurrent = 1.28 x 10−5

2.24E−4 2.56E−5 1.28E−5

H

2S 2.5010 4 2x 2.5010 4 2 1.2810 5 2.2410 4 M

H

2 2x 2 1.2810 5 2.5610 5 M

S

2 x 1.28 10 5 M





64

+x+2x2x

2.24E−4 2.56E−5 1.28E−5

Kc(calculated) = Kc(given) within significant figures Tro: Chemistry: A Molecular Approach, 2/e


Practice – For the reaction I2(g) Û 2 I(g) the value of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is placed into a 2.00 L flask and

heated, what will be the equilibrium concentrations of [I2] and [I]?

(use the simplifying assumption to solve for x)





because [I]initial = 0, Q = 0 and the reaction must proceed forward





the approximation is valid!!





x = 0.00217

0.500 0.00217 = 0.498[I2] = 0.498 M

2(0.00217) = 0.00434[I] = 0.00434 M



Practice – For the reaction N2O4(g) Û 2 NO2(g) the value of Kc = 1.07 x 10−5 . If the initial concentration

of N2O4 is 0.0125 M, what will be the equilibrium concentrations of [N2O4] and [NO2]?





because [NO2]initial = 0, Q = 0 and the reaction must proceed forward





without approximation with approximation

the approximation is valid!!

71

x = 1.82 x 10−4





x = 1.83 x 10−4

[N2O4]= 0.0125 1.83 x 10−4

[N2O4] = 0.0123 M

[NO2]=2(1.83 x 10−4)[NO2] = 3.66 x 10−4 M



Disturbing and RestoringEquilibrium

• Once a reaction is at equilibrium, the concentrations of all the reactants and products remain the same

• However if the conditions are changed, the concentrations of all the chemicals will change until equilibrium is restored

• The new concentrations will be different, but the equilibrium constant will be the sameunless you change the temperature



Le Châtelier’s Principle

• Le Châtelier's Principle guides us in predicting the effect various changes in conditions have on the position of equilibrium

• It says that if a system at equilibrium is disturbed, the position of equilibrium will shift to minimize the disturbance disturbances all involve making the system open



The result will be people moving from Country B into Country A faster than people moving from Country A into Country B. This will continue until a new equilibrium between the populations is established; the new populations will have different numbers of people than the old ones.

An Analogy: Population Changes

75

When an influx of population enters Country Bfrom somewhere outside Country A, it disturbs the

equilibrium established between Country A and Country B.

When the populations of Country A and Country Bare in equilibrium, the emigration rates between thetwo countries are equal so the populations stay constant.



Disturbing Equilibrium:Adding or Removing Reactants

• After equilibrium is established, a reactant is added as long as the added reactant is included in the equilibrium

constant expression i.e., not a solid or liquid

• How will this affect the rate of the forward reaction? • How will it affect the rate of the reverse reaction? • How will it affect the value of K?



Disturbing Equilibrium: Adding Reactants• Adding a reactant initially increases the rate of the

forward reaction, but has no initial effect on the rate of the reverse reaction

• The reaction proceeds to the right until equilibrium is restored

• At the new equilibrium position, you will have more of the products than before, less of the non-added reactants than before, and less of the added reactant but not as little of the added reactant as you had before

the addition• At the new equilibrium position, the

concentrations of reactants and products will be such that the value of the equilibrium constant is the same 77Tro: Chemistry: A Molecular Approach, 2/e


Disturbing Equilibrium:Adding or Removing Reactants

• After equilibrium is established, a reactant is removed as long as the added reactant is included in the equilibrium

constant expression i.e., not a solid or liquid

• How will this affect the rate of the forward reaction? • How will it affect the rate of the reverse reaction? • How will it affect the value of K?



Disturbing Equilibrium: Removing Reactants• Removing a reactant initially decreases the rate of

the forward reaction, but has no initial effect on the rate of the reverse reaction. so the reaction is going faster in reverse

• The reaction proceeds to the left until equilibrium is restored

• At the new equilibrium position, you will have less of the products than before, more of the non-removed reactants than before, and more of the removed reactant but not as much of the removed reactant as you had

before the removal• At the new equilibrium position, the concentrations

of reactants and products will be such that the value of the equilibrium constant is the same



The Effect of Concentration Changes on Equilibrium

• Adding a reactant will decrease the amounts of the other reactants and increase the amount of the products until a new position of equilibrium is found that has the same K

• Removing a product will increase the amounts of the other products and decrease the amounts of the reactants.you can use this to drive a reaction to completion!

• Equilibrium shifts away from side with added chemicals or toward side with removed chemicalsRemember, adding more of a solid or liquid does not

change its concentration – and therefore has no effect on the equilibrium




When NO2 is added, some of it combines to make more N2O4




When N2O4 is added, some of itdecomposes to make more NO2



The Effect of Adding a Gas to a Gas Phase Reaction at Equilibrium

• Adding a gaseous reactant increases its partial pressure, causing the equilibrium to shift to the rightincreasing its partial pressure increases its

concentrationdoes not increase the partial pressure of the other

gases in the mixture• Adding an inert gas to the mixture has no effect

on the position of equilibriumdoes not affect the partial pressures of the gases in

the reaction



Effect of Volume Change on Equilibrium• Decreasing the volume of the container increases

the concentration of all the gases in the container increases their partial pressures It does not change the concentrations of solutions!!

• If their partial pressures increase, then the total pressure in the container will increase

• According to Le Châtelier’s Principle, the equilibrium should shift to remove that pressure

• The way the system reduces the pressure is to reduce the number of gas molecules in the container

• When the volume decreases, the equilibrium shifts to the side with fewer gas molecules



Disturbing Equilibrium:Changing the Volume

• After equilibrium is established, the container volume is decreased

• How will it affect the concentration of solids, liquid, solutions, and gases?

• How will this affect the total pressure of solids, liquid, and gases?

• How will it affect the value of K?



Disturbing Equilibrium: Reducing the Volume• Decreasing the container volume increases the

concentration of all gases, but not solids, liquid, or solutions same number of moles, but different number of liters, resulting

in a different molarity• Decreasing the container volume will increase the total

pressure Boyle’s Law if the total pressure increases, the partial pressures of all the

gases will increase – Dalton’s Law of Partial Pressures• Because the total pressure increases, the position of

equilibrium will shift to decrease the pressure by removing gas molecules shift toward the side with fewer gas molecules

• At the new equilibrium position, the partial pressures of gaseous reactants and products will be such that the value of the equilibrium constant is the same



The Effect of Volume Changes on Equilibrium

87

Because there are more gas molecules on the reactants side of the

reaction, when the pressure is increased the position of

equilibrium shifts toward the side with fewer molecules to

decrease the pressure

When the pressure is decreased by increasing the

volume, the position of equilibrium shifts toward the side with the greater number of molecules – the reactant

side



The Effect of Temperature Changes on Equilibrium Position

• Exothermic reactions release energy and endothermic reactions absorb energy

• If we write Heat as a product in an exothermic reaction or as a reactant in an endothermic reaction, it will help us use Le Châtelier’s Principle to predict the effect of temperature changeseven though heat is not matter and not written in a

proper equation



The Effect of Temperature Changes on Equilibrium for Exothermic Reactions

• For an exothermic reaction, heat is a product• Increasing the temperature is like adding heat• According to Le Châtelier’s Principle, the

equilibrium will shift away from the added heat• Adding heat to an exothermic reaction will

decrease the concentrations of products and increase the concentrations of reactants

• Adding heat to an exothermic reaction will decrease the value of K

• How will decreasing the temperature affect the system?

aA + bB cC + dD + Heat

89

Kc

C c D

d

A a B

b



The Effect of Temperature Changes on Equilibrium for Endothermic Reactions• For an endothermic reaction, heat is a reactant• Increasing the temperature is like adding heat• According to Le Châtelier’s Principle, the

equilibrium will shift away from the added heat• Adding heat to an endothermic reaction will

decrease the concentrations of reactants and increase the concentrations of products

• Adding heat to an endothermic reaction will increase the value of K

• How will decreasing the temperature affect the system?

Heat + aA + bB cC + dD

90

Kc

C c D

d

A a B

b



The Effect of Temperature Changes on Equilibrium



Not Changing the Position of Equilibrium: The Effect of Catalysts

• Catalysts provide an alternative, more efficient mechanism

• Catalysts work for both forward and reverse reactions

• Catalysts affect the rate of the forward and reverse reactions by the same factor

• Therefore, catalysts do not affect the position of equilibrium



Practice – Le Châtelier’s Principle• The reaction 2 SO2(g) + O2(g) Û 2 SO3(g) with

DH° = −198 kJ is at equilibrium. How will each of the following changes affect the equilibrium concentrations of each gas once equilibrium is restored?adding more O2 to the containercondensing and removing SO3

compressing the gasescooling the containerdoubling the volume of the containerwarming the mixtureadding the inert gas helium to the containeradding a catalyst to the mixture



Practice – Le Châtelier’s Principle• The reaction 2 SO2(g) + O2(g) Û 2 SO3(g) with

DH° = −198 kJ is at equilibrium. How will each of the following changes affect the equilibrium concentrations of each gas once equilibrium is restored?adding more O2 to the container shift to

SO3

condensing and removing SO3 shift to SO3

compressing the gases shift to SO3

cooling the container shift to SO3

doubling the volume of the container shift to SO2

warming the mixture shift to SO2

adding helium to the container no effectadding a catalyst to the mixture no effect94Tro: Chemistry: A Molecular Approach, 2/e

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Copyright 2011 Pearson Education, Inc. Chapter 14 Chemical Equilibrium Roy Kennedy Massachusetts Bay Community College Wellesley Hills, MA