14 chemical equilibrium contents 14-1 introduction to chemical equilibria 14-2 equilibrium constants...

14 Chemical Equilibrium Contents

14-1 Introduction to Chemical Equilibria14-2 Equilibrium Constants and their Expression14-3 Determination of Values of Equilibrium Constants 14-4 Calculations Involving Equilibrium Constants 14-5 Using Le Chatelier’s Principle to Predict Shifts in Chemical Equilibria

14-6 Some Industrially Important Chemical Equilibria

Reversible ReactionsIn a reversible reaction, there is both a forward and a

reverse reaction. Suppose SO2 and O2 are present initially. As they

collide, the forward reaction begins.

2SO2(g) + O2(g) 2SO3 (g)

As SO3 molecules form, they also collide in the reverse reaction that forms reactants. The reversible reaction is written with a double arrow.

forward

2SO2(g) + O2 (g) 2SO3(g) reverse

14-1 Introduction to Chemical Equilibria

Chemical Equilibrium

At equilibrium

The rate of the forward reaction becomes equal to the rate of the reverse reaction.

The forward and reverse reactions continue at equal rates in both directions.

Copyright © 2007 by Pearson Education, Inc.Publishing as Benjamin Cummings

Chemical Equilibrium

When equilibrium is

reached There is no further

change in the amounts of reactant and product.

Copyright © 2007 by Pearson Education, Inc.Publishing as Benjamin Cummings

Equilibrium

At equilibrium The forward reaction of N2 and O2 forms NO.

The reverse reaction of 2NO forms N2 and O2.

The amounts of N2, O2, and NO remain constant.

N2(g) + O2(g) 2NO(g)

Copyright © 2007 by Pearson Education, Inc. Publishing as Benjamin Cummings

Learning Check

Write the forward and reverse reactions for the

following:

CH4(g) + 2H2S(g) CS2(g) + 4H2(g)

The two single-headed arrows, one pointing to the right

and the other to the left, indicate an equilibrium.

Solution

Write the forward and reverse reactions for the

following:

Forward:

CH4(g) + 2H2S(g) CS2(g) + 4H2(g)

Reverse:

CS2(g) + 4H2(g) CH4(g) + 2H2S(g)

Or

CH4(g) + 2H2S(g) CS2(g) + 4H2(g)

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

• Dynamic equilibria

15.1

• HC2H3O2(aq) H+(aq) + C2H3O2-(aq)

Acetic acid: weak acid; weak eletrolyte

When HC2H3O2 is dissolved in water, some ions are formed, a

lthough most of the acetic acid remains in the form of mole

cules.

(0.1M HAc solution at 25℃: 1.3%H+, 1.3%Ac-, 98.7%HAc)

• HCl (aq) H+ + Cl-

Hydrochloric acid: strong acid; strong eletrolyte

HCl is assumed to be ionized completely in aqueous solution.

Reaching equilibrium on the macroscopic and molecular levels.

N2O4(g) 2NO2(g)

N2O4 (g; colorless) 2NO2 (g; reddish brown)

Start with NO2 Start with N2O4 Start with NO2 & N2O4

equilibrium

equilibrium

equilibrium

15.1

15.1

constantThe NO2-N2O4 System at 25ºC

N2O4 (g) 2NO2 (g)

= 4.63 x 10-3K = [NO2]2

[N2O4]

aA + bB cC + dD

K = [C]c[D]d

[A]a[B]b

K >> 1

K << 1

Lie to the right Favor products

Lie to the left Favor reactants

Equilibrium Will

14.1

14-2 Equilibrium Constants and Equilibrium

Homogeneous Equilibria: Equilibria that involve only a single phase.

H2(g) +I2(g) 2HI(g)

Equilibrium Constants expression:

eqeq

eq

cK22

2

IH

HI

[ ] : Stand for concentrati[ ] : Stand for concentration in molarityon in molarity

eq : indicate an equilibrium eq : indicate an equilibrium concentration, usually omi concentration, usually omi

tted.tted.

Kc is equal to the product of the equilibrium molarities of the produKc is equal to the product of the equilibrium molarities of the products of reaction, each raised to a power given by the coefficient in tcts of reaction, each raised to a power given by the coefficient in the equation, divided by the product of the equilibrium molarities of he equation, divided by the product of the equilibrium molarities of the reactants of reaction, each raised to a power given by the coefthe reactants of reaction, each raised to a power given by the coef

ficient in the equation.ficient in the equation.

The Equilibrium ConstantThe Equilibrium Constant

• No matter the starting composition of reactants and products, the same ratio of concentrations is achieved at equilibrium.

• For a general reaction

the equilibrium constant expression is

where Kc is the equilibrium constant.

aA + bB(g) pP + qQ

ba

qp

cKBA

QP

14-2 Equilibrium Constants and Equilibrium

The Equilibrium ConstantThe Equilibrium Constant

• Kc is based on the molarities of reactants and products

at equilibrium.

• We generally omit the units of the equilibrium

constant.

• Note that the equilibrium constant expression has

products over reactants.

The Equilibrium ConstantThe Equilibrium ConstantThe Equilibrium Constant in Terms of PressureThe Equilibrium Constant in Terms of Pressure• If KP is the equilibrium constant for reactions

involving gases, we can write:

• KP is based on partial pressures measured in atmospheres.

• We can show that

PA = [A](RT)

ba

qp

PPP

PPK

BA

QP

The Equilibrium ConstantThe Equilibrium ConstantThe Equilibrium Constant in Terms of PressureThe Equilibrium Constant in Terms of Pressure

PA = [A](RT)

• This means that we can relate Kc and KP:

where n is the change in number of moles of gas.• It is important to use:

n = ngas(products) - ngas(reactants)

ncP RTKK

The Equilibrium ConstantThe Equilibrium ConstantThe Magnitude of Equilibrium ConstantsThe Magnitude of Equilibrium Constants• The equilibrium constant, K, is the ratio of products

to reactants.• Therefore, the larger K the more products are present

at equilibrium.• Conversely, the smaller K the more reactants are

present at equilibrium.• If K >> 1, then products dominate at equilibrium and

equilibrium lies to the right.• If K << 1, then reactants dominate at equilibrium and

the equilibrium lies to the left.

The Equilibrium ConstantThe Equilibrium ConstantThe Magnitude of Equilibrium ConstantsThe Magnitude of Equilibrium Constants

The range of equilibrium constants.

small K

large K

intermediate K

N2(g) + O2(g) 2NO(g)

K=1×10-30

Barely proceed;

no reaction

2CO (g) + O2(g) 2CO2(g)

K=2.2×1022

Go to completion

2BrCl(g) Br2(g) + Cl2(g) K=5

Significant amounts of both reactant and product are present at equilibrium.

The Equilibrium ConstantThe Equilibrium ConstantThe Magnitude of Equilibrium ConstantsThe Magnitude of Equilibrium Constants• An equilibrium can be approached from any

direction.• Example:

• has

N2O4(g) 2NO2(g)

212.0

ONNO

42

22 cK

The Equilibrium ConstantThe Equilibrium ConstantThe Magnitude of Equilibrium ConstantsThe Magnitude of Equilibrium Constants• However,

• has

• The equilibrium constant for a reaction in one direction is the reciprocal of the equilibrium constant of the reaction in the opposite direction.

2NO2(g) N2O4(g)

c

c kK

172.4

212.0

1

NO

ON' 2

2

42

The Equilibrium ConstantThe Equilibrium ConstantHeterogeneous EquilibriaHeterogeneous Equilibria• When all reactants and products are in one phase, the eq

uilibrium is homogeneous.• If one or more reactants or products are in a different p

hase, the equilibrium is heterogeneous. Equilibria that involve more than one phase.

• Consider:

– experimentally, the amount of CO2 does not seem to depend on the amounts of CaO and CaCO3. Why?

• The concentration of a solid or pure liquid is its density divided by molar mass.

CaCO3(s) CaO(s) + CO2(g)

The Equilibrium ConstantThe Equilibrium ConstantHeterogeneous EquilibriaHeterogeneous Equilibria

The Equilibrium ConstantThe Equilibrium ConstantHeterogeneous EquilibriaHeterogeneous Equilibria• Neither density nor molar mass is a variable, the conc

entrations of solids and pure liquids are constant.

• For the decomposition of CaCO3:

• We ignore the concentrations of pure liquids and pure solids in equilibrium constant expressions.

• The amount of CO2 formed will not depend greatly on the amounts of CaO and CaCO3 present.

2

223

COconstant

.COconstantCOCaCO

CaO

cc

c

KK

K

Heterogenous equilibrium applies to reactions in which reactants and products are in different phases.

CaCO3 (s) CaO (s) + CO2 (g)

Kc =‘[CaO][CO2]

[CaCO3][CaCO3] = constant[CaO] = constant

Kc = [CO2] = Kc x‘[CaCO3][CaO]

Kp = PCO2

The concentration of solids and pure liquids are not included in the expression for the equilibrium constant.

14.2

PCO 2= Kp

CaCO3 (s) CaO (s) + CO2 (g)

PCO 2 does not depend on the amount of CaCO3 or CaO

14.2

Homogenous equilibrium applies to reactions in which all reacting species are in the same phase.

N2O4 (g) 2NO2 (g)

Kc = [NO2]2

[N2O4]Kp =

NO2P2

N2O4P

In most cases

Kc Kp

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

14.2

Kp = Kc(RT)n

n = moles of gaseous products – moles of gaseous reactants

= (c + d) – (a + b)

Ways Different States of Matter Can Appear in the Equilibrium Constant, K

Molarity Partial Pressure

gas, (g) YES YES

aqueous, (aq) YES ---

liquid, (l) --- ---

solid, (s) --- ---

Writing Equilibrium Constant Expressions

• 1. Write an equation for the equilibrium.

• 2. Put the product of the equilibrium concentrations of the

products in the numerator. Omit pure solids, pure liquids and

solvents in dilute solutions (concentration around 0.1M or

less).

• 3. Write the product of the equilibrium concentrations of the

reactants in the denominator. Omit pure solids, pure liquids

and solvents in dilute solutions.

• 4 The exponent of each concentration should be the same as

the coefficient of the species in the equation. 14.2

• Notice:

• 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.

• The concentrations of pure solids, pure liquids and solvents in

dilute solutions do not appear in the equilibrium constant

expressions.

• The equilibrium constant is a dimensionless quantity.

• In quoting a value for the equilibrium constant, you must specify

the balanced equation and the temperature.

33

Copyright (c) 1999 by Harcourt Brace & CompanyAll rights reserved

The Meaning of KThe Meaning of KThe Meaning of KThe Meaning of K

1. Can tell if a reaction is product-favored or 1. Can tell if a reaction is product-favored or reactant-favored.reactant-favored.

For NFor N22(g) + 3 H(g) + 3 H22(g) (g) 2 NH2 NH33(g)(g)

34



1.1. Can tell if a reaction is product-favored or Can tell if a reaction is product-favored or reactant-favored.reactant-favored.

For NFor N22(g) + 3 H(g) + 3 H22(g) (g) 2 NH2 NH33(g)(g)

Kc = [NH3 ]2

[N2 ][H2]3 = 3.5 x 108Kc =

[NH3 ]2

[N2 ][H2]3 = 3.5 x 108

35




For NFor N22(g) + 3 H(g) + 3 H22(g) (g) 2 NH 2 NH33(g)(g)

Conc. of products is Conc. of products is much greatermuch greater than that of than that of reactants at equilibrium. reactants at equilibrium.

Kc = [NH3 ]2

[N2 ][H2]3 = 3.5 x 108Kc =

[NH3 ]2

[N2 ][H2]3 = 3.5 x 108

36




For NFor N22(g) + 3 H(g) + 3 H22(g) (g) 2 NH 2 NH33(g)(g)

Conc. of products is Conc. of products is much greatermuch greater than that of than that of reactants at equilibrium. reactants at equilibrium.

The reaction is strongly The reaction is strongly product-favoredproduct-favored..

Kc = [NH3 ]2

[N2 ][H2]3 = 3.5 x 108Kc =

[NH3 ]2

[N2 ][H2]3 = 3.5 x 108

37



For AgCl(s) For AgCl(s) AgAg++(aq) + Cl(aq) + Cl--(aq)(aq)

KKcc = [Ag = [Ag++] [Cl] [Cl--] = 1.8 x 10] = 1.8 x 10-5-5

Conc. of products is Conc. of products is much lessmuch less tha tha

n that of reactants at equilibrium. n that of reactants at equilibrium.

The reaction is strongly The reaction is strongly reactant-fareactant-fa

voredvored.. AgAg++(aq) + Cl(aq) + Cl--(aq) (aq) AgCl(s)AgCl(s)is product-favored.is product-favored.

AgAg++(aq) + Cl(aq) + Cl--(aq) (aq) AgCl(s)AgCl(s)is product-favored.is product-favored.

38



2. Can tell if a reaction is at equilibrium. If not, 2. Can tell if a reaction is at equilibrium. If not, which way it moves to approach equilibrium.which way it moves to approach equilibrium.

H

H

H

H

H

H

H

H

H H H

CH

HHH H

H—C—C—C—C—H H—C—C—C—H

K =

n-butane iso-butane

[iso][n]

= 2.5

39



If [iso] = 0.35 M and [n] = 0.15 M, are you at equilibriIf [iso] = 0.35 M and [n] = 0.15 M, are you at equilibrium? um?

Which way does the reaction “shift” to approach equWhich way does the reaction “shift” to approach equilibrium?ilibrium?

H

H

H

H

H

H

H

H

H H H

CH

HHH H

H—C—C—C—C—H H—C—C—C—H

K =

n-butane iso-butane

[iso][n]

= 2.5

40


Predicting the Direction of ReactionPredicting the Direction of Reaction• We define Q, the reaction quotient, for a general reaction

as

where [A], [B], [P], and [Q] are molarities at any time.

• Q = K only at equilibrium.

aA + bB(g) pP + qQ

ba

qpQ

BA

QP

14-4 Calculations Involving Equilibrium Constants

41


Applications of Equilibrium ConstantsApplications of Equilibrium Constants

Predicting the Direction of ReactionPredicting the Direction of Reaction

• If Q > K then the reverse reaction must occur to reach

equilibrium (i.e., products are consumed, reactants are

formed, the numerator in the equilibrium constant

expression decreases and Q decreases until it equals K).

• If Q < K then the forward reaction must occur to reach

equilibrium.

42


Reaction Quotient

1. Q=K, equilibrium

2. Q>K, the system shifts to the left

3. Q<K, the system shifts to the right

N2+3H2 2NH330202

20

3

][][

][

HN

NHQ

43


The reaction quotient (Qc) is calculated by substituting the initial concentrations of the reactants and products into the equilibrium constant (Kc) expression.

IF

• Qc > Kc system proceeds from right to left to reach equilibrium

• Qc = Kc the system is at equilibrium

• Qc < Kc system proceeds from left to right to reach equilibrium

14.4

44


Applications of Equilibrium ConstantsApplications of Equilibrium ConstantsCalculation of Equilibrium ConcentrationsCalculation of Equilibrium Concentrations• The same steps used to calculate equilibrium constants

are used.

• Generally, we do not have a number for the change in concentration line.

• Therefore, we need to assume that x mol/L of a species is produced (or used).

• The equilibrium concentrations are given as algebraic expressions.

45



In general, all reacting chemical systems are In general, all reacting chemical systems are

characterized by their characterized by their REACTION QUOTIENT, QREACTION QUOTIENT, Q..

If Q = K, then system is at equilibrium.If Q = K, then system is at equilibrium.

Q = product concentrationsreactant concentrations


46



In general, all reacting chemical systems are characterizeIn general, all reacting chemical systems are characterized by their d by their REACTION QUOTIENT, QREACTION QUOTIENT, Q..


If [iso] = 0.35 M and [n] = 0.15 M, are you at equilibrium?If [iso] = 0.35 M and [n] = 0.15 M, are you at equilibrium?

Q (2.3) < K (2.5). Q (2.3) < K (2.5).



Q = conc. of isoconc. of n

= 0.350.15

= 2.3Q = conc. of isoconc. of n

= 0.350.15

= 2.3

47



In general, all reacting chemical systems are characIn general, all reacting chemical systems are characterized by their terized by their REACTION QUOTIENT, QREACTION QUOTIENT, Q..


Q (2.33) < K (2.5). Q (2.33) < K (2.5).

Reaction is NOT at equilibrium, so [Iso] must becoReaction is NOT at equilibrium, so [Iso] must become ________ and [n] must ____________. me ________ and [n] must ____________.



Q = conc. of isoconc. of n

= 0.350.15

= 2.3Q = conc. of isoconc. of n

= 0.350.15

= 2.3

14-3 Determination of Values of Equilibrium Constants14-3 Determination of Values of Equilibrium Constants

The numerical values of equilibrium constants can be found by experiment.

See Sample problem 14.4.

Initial concentration: the concentrations at the beginning of the experiment, [ ]0

Equilibrium concentration, [ ]equ

Stoichiometry problem

Changes in concentration due to reaction

The equilibrium concentrations for the reaction between ca

rbon monoxide and molecular chlorine to form COCl2 (g) at

740C are [CO] = 0.012 M, [Cl2] = 0.054 M, and [COCl2] = 0.

14 M. Calculate the equilibrium constants Kc and Kp.

CO (g) + Cl2 (g) COCl2 (g)

Kc = [COCl2]

[CO][Cl2]=

0.140.012 x 0.054

= 220

Kp = Kc(RT)n

n = 1 – 2 = -1 R = 0.0821 T = 273 + 74 = 347 K

Kp = 220 x (0.0821 x 347)-1 = 7.7

14.2

The equilibrium constant Kp for the reaction

is 158 at 1000K. What is the equilibrium pressure of O2 if the PNO = 0.400 atm and PNO = 0.270 atm?2

2NO2 (g) 2NO (g) + O2 (g)

14.2

Kp = 2PNO PO

2

PNO2

2

PO2 = Kp

PNO2

2

PNO2

PO2 = 158 x (0.400)2/(0.270)2 = 347 atm

Calculating Equilibrium ConstantsCalculating Equilibrium Constants• Proceed as follows:

– Tabulate initial and equilibrium concentrations (or partial pressures) given.

– If an initial and equilibrium concentration is given for a species, calculate the change in concentration.

– Use stoichiometry on the change in concentration line only to calculate the changes in concentration of all species.

– Deduce the equilibrium concentrations of all species.

• Usually, the initial concentration of products is zero. (This is not always the case.)

• When in doubt, assign the change in concentration a variable.

PROBLEM: Place 1.00 mol each of HPROBLEM: Place 1.00 mol each of H22 and I and I22 in a 1.00 L in a 1.00 L

flask. Calc. equilibrium concentrations.flask. Calc. equilibrium concentrations.

HH22(g) + I(g) + I22(g) 2 HI(g)(g) 2 HI(g)

Kc = [HI]2

[H2 ][I2 ] = 55.3Kc =

[HI]2

[H2 ][I2 ] = 55.3

Typical CalculationsTypical Calculations

HH22(g) + I(g) + I22(g) (g) 2 HI(g), K2 HI(g), Kcc = 55.3 = 55.3

Step 1. Set up table to define EQUILIBRIUM concStep 1. Set up table to define EQUILIBRIUM concentrations.entrations.

[H[H22]] [I[I22]] [HI][HI]

Initial Initial 1.001.00 1.001.00 00

ChangeChange

EquilibEquilib

Step 1. Set up table to define EQUILIBRIUM concentStep 1. Set up table to define EQUILIBRIUM concentrations.rations.

[H[H22]] [I[I22]] [HI][HI]

Initial Initial 1.001.00 1.001.00 00

ChangeChange -x-x -x-x +2x+2x

EquilibEquilib 1.00-x1.00-x 1.00-x1.00-x 2x2x

where where xx is defined as amount of H is defined as amount of H22 and I and I22 consumed on consumed on

approaching equilibrium.approaching equilibrium.


Step 2. Put equilibrium concentrations into KStep 2. Put equilibrium concentrations into Kcc e e

xpression.xpression.

Kc = [2x]2

[1.00 - x][1.00 - x] = 55.3Kc =

[2x]2

[1.00 - x][1.00 - x] = 55.3


Step 3. Solve KStep 3. Solve Kcc expression - take square root expression - take square root

of both sides.of both sides.

x = 0.79x = 0.79

7.44 = 2x

1.00 - x7.44 =

2x1.00 - x




x = 0.79x = 0.79

Therefore, at equilibriumTherefore, at equilibrium

[H[H22] = [I] = [I22] = 1.00 - x = 0.21 M] = 1.00 - x = 0.21 M

7.44 = 2x

1.00 - x7.44 =

2x1.00 - x




x = 0.79x = 0.79

Therefore, at equilibriumTherefore, at equilibrium

[H[H22] = [I] = [I22] = 1.00 - x = 0.21 M] = 1.00 - x = 0.21 M

[HI] = 2x = 1.58 M[HI] = 2x = 1.58 M

7.44 = 2x

1.00 - x7.44 =

2x1.00 - x


Nitrogen Dioxide Nitrogen Dioxide EquilibriumEquilibrium

NN22OO44(g) (g) 2 2

NONO22(g)(g)

Nitrogen Dioxide Nitrogen Dioxide EquilibriumEquilibrium

NN22OO44(g) (g) 2 2

NONO22(g)(g)

Nitrogen Dioxide EquilibriumNitrogen Dioxide EquilibriumNN22OO44(g) (g) 2 NO 2 NO22(g)(g)

If initial concentration of NIf initial concentration of N22OO44 is 0.50 M, what are the equilibrium is 0.50 M, what are the equilibrium

concentrations?concentrations?

Step 1. Set up an equilibrium tableStep 1. Set up an equilibrium table

[N[N22OO44]] [NO[NO22]]

InitialInitial 0.500.50 00

ChangeChange

EquilibEquilib

Kc = [NO2 ]2

[N2O4 ] = 0.0059 at 298 KKc =

[NO2 ]2

[N2O4 ] = 0.0059 at 298 K


If initial concentration of NIf initial concentration of N22OO44 is 0.50 M, what are the equilibrium is 0.50 M, what are the equilibrium

concentrations?concentrations?

Step 1. Set up an equilibrium tableStep 1. Set up an equilibrium table

[N[N22OO44]] [NO[NO22]]

InitialInitial 0.500.50 00

ChangeChange -x-x +2x+2x

EquilibEquilib 0.50 - x0.50 - x 2x2x

Kc = [NO2 ]2

[N2O4 ] = 0.0059 at 298 KKc =

[NO2 ]2

[N2O4 ] = 0.0059 at 298 K

Nitrogen Dioxide EquilibriumNitrogen Dioxide EquilibriumNN22OO44(g) (g) 2 NO2 NO22(g) (g)

Step 2. Substitute into KStep 2. Substitute into Kcc expression and solve. expression and solve.

Rearrange: Rearrange: 0.0059 (0.50 - x) = 4x0.0059 (0.50 - x) = 4x22

0.0029 - 0.0059x = 4x0.0029 - 0.0059x = 4x22

4x4x22 + 0.0059x - 0.0029 = 0 + 0.0059x - 0.0029 = 0

This is a This is a QUADRATIC EQUATIONQUADRATIC EQUATION

axax22 + bx + c = 0 + bx + c = 0

a = 4a = 4 b = 0.0059 b = 0.0059 c = -0.0029c = -0.0029

Kc = 0.0059 = [NO2 ]2

[N2O4 ]=

(2x)2

(0.50 - x) Kc = 0.0059 =

[NO2 ]2

[N2O4 ]=

(2x)2

(0.50 - x)

Solve the quadratic equation for x.Solve the quadratic equation for x.

axax22 + bx + c = 0 + bx + c = 0

a = 4a = 4 b = 0.0059 b = 0.0059 c = -0.0029c = -0.0029

x = -b b2 - 4ac

2ax =

-b b2 - 4ac2a

Nitrogen Dioxide EquilibriumNitrogen Dioxide EquilibriumNN22OO44(g) (g) 2 NO2 NO22(g) (g)



axax22 + bx + c = 0 + bx + c = 0

a = 4a = 4 b = 0.0059 b = 0.0059 c = -0.0029c = -0.0029

x = -b b2 - 4ac

2ax =

-b b2 - 4ac2a

x = -0.0059 (0.0059)2 - 4(4)(-0.0029)

2(4)x =

-0.0059 (0.0059)2 - 4(4)(-0.0029)2(4)



axax22 + bx + c = 0 + bx + c = 0

a = 4a = 4 b = 0.0059 b = 0.0059 c = -0.0029c = -0.0029

x = -0.00074 ± 1/8(0.046)x = -0.00074 ± 1/8(0.046)1/21/2 = -0.00074 ± 0.027 = -0.00074 ± 0.027

x = -b b2 - 4ac

2ax =

-b b2 - 4ac2a

x = -0.0059 (0.0059)2 - 4(4)(-0.0029)

2(4)x =

-0.0059 (0.0059)2 - 4(4)(-0.0029)2(4)


x = -0.00074 ± 1/8(0.046)x = -0.00074 ± 1/8(0.046)1/21/2 = -0.00074 ± 0.027 = -0.00074 ± 0.027

x = 0.026 or -0.028x = 0.026 or -0.028

But a negative value is not reasonable.But a negative value is not reasonable.

ConclusionConclusion

[N[N22OO44] = 0.050 - x = 0.47 M] = 0.050 - x = 0.47 M

[NO[NO22] = 2x = 0.052 M] = 2x = 0.052 M

x = -0.0059 (0.0059)2 - 4(4)(-0.0029)

2(4)x =

-0.0059 (0.0059)2 - 4(4)(-0.0029)2(4)

Le Châtelier’s Principle

1. System starts at equilibrium.2. A change/stress is then made to system at

equilibrium.• Change in concentration• Change in volume• Change in pressure• Change in Temperature• Add Catalyst

3. System responds by shifting to reactant or product side to restore equilibrium.

• Consider the production of ammonia

• As the pressure increases, the amount of ammonia present at equilibrium increases.

• As the temperature decreases, the amount of ammonia at equilibrium increases.

• Can this be predicted?• Le Châtelier’s Principle: if a system at equilibrium is

disturbed, the system will move in such a way as to counteract the disturbance.

N2(g) + 3H2(g) 2NH3(g)

14-5 Using Le Chatelier’s Principle to Predict Shifts in Chemical Equilibria

Le Châtelier’s PrincipleLe Châtelier’s Principle

Le Châtelier’s PrincipleLe Châtelier’s Principle

Change in Reactant or Product ConcentrationsChange in Reactant or Product Concentrations

• Consider the Haber process

• If H2 is added while the system is at equilibrium, the syste

m must respond to counteract the added H2 (by Le Châtel

ier).

• That is, the system must consume the H2 and produce pro

ducts until a new equilibrium is established.

• Therefore, [H2] and [N2] will decrease and [NH3] increases.

N2(g) + 3H2(g) 2NH3(g)

Le Châtelier’s PrincipleLe Châtelier’s PrincipleChange in Reactant or Product ConcentrationsChange in Reactant or Product Concentrations

Le Châtelier’s Principle

• Changes in Concentration continued

Change Shifts the Equilibrium

Increase concentration of product(s) left

Decrease concentration of product(s) right

Decrease concentration of reactant(s)

Increase concentration of reactant(s) right

left14.5

aA + bB cC + dD

AddAddRemove Remove

Le Châtelier’s PrincipleLe Châtelier’s PrincipleChange in Reactant or Product ConcentrationsChange in Reactant or Product Concentrations• Adding a reactant or product shifts the equilibrium a

way from the increase.• Removing a reactant or product shifts the equilibrium

towards the decrease.• To optimize the amount of product at equilibrium, we

need to flood the reaction vessel with reactant and continuously remove product (Le Châtelier).

• We illustrate the concept with the industrial preparation of ammonia

N2(g) + 3H2(g) 2NH3(g)

Le Châtelier’s PrincipleLe Châtelier’s PrincipleChange in Reactant or Product ConcentrationsChange in Reactant or Product Concentrations

Le Châtelier’s PrincipleLe Châtelier’s PrincipleChange in Reactant or Product ConcentrationsChange in Reactant or Product Concentrations• N2 and H2 are pumped into a chamber.

• The pre-heated gases are passed through a heating coil to the catalyst bed.

• The catalyst bed is kept at 460 - 550 C under high pressure.

• The product gas stream (containing N2, H2 and NH3) is passed over a cooler to a refrigeration unit.

• In the refrigeration unit, ammonia liquefies but not N2 or H2.

Le Châtelier’s PrincipleLe Châtelier’s PrincipleChange in Reactant or Product ConcentrationsChange in Reactant or Product Concentrations• The unreacted nitrogen and hydrogen are recycled wi

th the new N2 and H2 feed gas.

• The equilibrium amount of ammonia is optimized because the product (NH3) is continually removed and the reactants (N2 and H2) are continually being added.

Effects of Volume and PressureEffects of Volume and Pressure• As volume is decreased pressure increases.• Le Châtelier’s Principle: if pressure is increased the s

ystem will shift to counteract the increase.

Le Châtelier’s PrincipleLe Châtelier’s PrincipleEffects of Volume and PressureEffects of Volume and Pressure• That is, the system shifts to remove gases and

decrease pressure.• An increase in pressure favors the direction that has

fewer moles of gas.• In a reaction with the same number of product and

reactant moles of gas, pressure has no effect.• Consider

N2O4(g) 2NO2(g)

Le Châtelier’s PrincipleLe Châtelier’s PrincipleEffects of Volume and PressureEffects of Volume and Pressure• An increase in pressure (by decreasing the volume)

favors the formation of colorless N2O4.

• The instant the pressure increases, the system is not at equilibrium and the concentration of both gases has increased.

• The system moves to reduce the number moles of gas (i.e. the forward reaction is favored).

• A new equilibrium is established in which the mixture is lighter because colorless N2O4 is favored.

Le Châtelier’s PrincipleLe Châtelier’s PrincipleEffect of Temperature ChangesEffect of Temperature Changes• The equilibrium constant is temperature dependent.• For an endothermic reaction, H > 0 and heat can be

considered as a reactant.• For an exothermic reaction, H < 0 and heat can be

considered as a product.• Adding heat (i.e. heating the vessel) favors away from

the increase:– if H > 0, adding heat favors the forward reaction,

– if H < 0, adding heat favors the reverse reaction.

Le Châtelier’s PrincipleLe Châtelier’s PrincipleEffect of Temperature ChangesEffect of Temperature Changes• Removing heat (i.e. cooling the vessel), favors towards

the decrease:– if H > 0, cooling favors the reverse reaction,

– if H < 0, cooling favors the forward reaction.

• Consider

for which H > 0.– Co(H2O)6

2+ is pale pink and CoCl42- is blue.

Cr(H2O)6(aq) + 4Cl-(aq) CoCl42-(aq) + 6H2O(l)

Le Châtelier’s PrincipleLe Châtelier’s PrincipleEffect of Temperature ChangesEffect of Temperature Changes

Le Châtelier’s PrincipleLe Châtelier’s PrincipleEffect of Temperature ChangesEffect of Temperature Changes

– If a light purple room temperature equilibrium mixture is placed in a beaker of warm water, the mixture turns deep blue.

– Since H > 0 (endothermic), adding heat favors the forward reaction, i.e. the formation of blue CoCl4

2-.

– If the room temperature equilibrium mixture is placed in a beaker of ice water, the mixture turns bright pink.

– Since H > 0, removing heat favors the reverse reaction which is the formation of pink Co(H2O)6

2+.

Cr(H2O)6(aq) + 4Cl-(aq) CoCl42-(aq) + 6H2O(l)

Le Châtelier’s PrincipleLe Châtelier’s PrincipleThe Effect of CatalystsThe Effect of Catalysts• A catalyst lowers the activation energy barrier for the

reaction.• Therefore, a catalyst will decrease the time taken to

reach equilibrium.• A catalyst does not effect the composition of the

equilibrium mixture.

Calculating Equilibrium Concentrationsfrom K and Initial Concentrations

1. Balance chemical equation.

2. Determine Q to see which way reaction will proceed to get to equilibrium

3. Define changes needed to get to equilibrium. Use ‘x’ to represent the change in concentration to get to equilibrium. Determine expression for equilibrium concentration.

4. Write the expression for equilibrium constant, K.

5. Substitute expressions defined in step 3 into the equilibrium constant (step 4).

6. Solve for x and plug value back to table made in step 3.

14.4

At 12800C the equilibrium constant (Kc) for the reaction

Is 1.1 x 10-3. If the initial concentrations are [Br2] = 0.063 M and [Br] = 0.012 M, calculate the concentrations of these species at equilibrium.

Br2 (g) 2Br (g)

Br2 (g) 2Br (g)

Let x be the change in concentration of Br2

Initial (M)

Change (M)

Equilibrium (M)

0.063 0.012

-x +2x

0.063 - x 0.012 + 2x

[Br]2

[Br2]Kc = Kc =

(0.012 + 2x)2

0.063 - x= 1.1 x 10-3 Solve for x

14.4

Kc = (0.012 + 2x)2

0.063 - x= 1.1 x 10-3

4x2 + 0.048x + 0.000144 = 0.0000693 – 0.0011x

4x2 + 0.0491x + 0.0000747 = 0

ax2 + bx + c =0-b ± b2 – 4ac

2ax =

Br2 (g) 2Br (g)

Initial (M)

Change (M)

Equilibrium (M)

0.063 0.012

-x +2x

0.063 - x 0.012 + 2x

x = -0.00178x = -0.0105

At equilibrium, [Br] = 0.012 + 2x = -0.009 M or 0.00844 M

At equilibrium, [Br2] = 0.062 – x = 0.0648 M14.4

14-6 Some Industrially Important Chemical Equilibria

A brief summary of the Contact Process for Sulfuric Acid:

The Contact Process:

•makes sulphur dioxide;

•converts the sulphur dioxide into sulphur trioxide (the reversi

ble reaction at the heart of the process);

•converts the sulphur trioxide into concentrated sulphuric acid.

Making the sulphur dioxide

This can either be made by burning sulphur in an excess of air:

. . . or by heating sulphide ores like pyrite in an excess of air:

In either case, an excess of air is used so that the sulphur dioxide produced is already mixed with oxygen for the next stage.

Converting the sulphur dioxide into sulphur trioxide

This is a reversible reaction, and the formation of the sulphur trioxide is exothermic.

A flow scheme for this part of the process looks like this:

Converting the sulphur trioxide into sulphuric acid

This can't be done by simply adding water to the sulphur trioxide - the reaction is so uncontrollable that it creates a fog of sulphuric acid. Instead, the sulphur trioxide is first dissolved in concentrated sulphuric acid:

The product is known as fuming sulphuric acid or oleum.This can then be reacted safely with water to produce concentrated sulphuric acid - twice as much as you originally used to make the fuming sulphuric acid.

14 chemical equilibrium contents 14-1 introduction to chemical equilibria 14-2 equilibrium constants...

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