EquilibriumEquilibrium

L. SchefflerL. Scheffler

Lincoln High SchoolLincoln High School

20092009

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Equilibrium SystemsEquilibrium Systems

Many chemical reactions are reversible. Many chemical reactions are reversible. Such reactions do not go to completion. Such reactions do not go to completion. There is a state of balance between the There is a state of balance between the

products and the reactantsproducts and the reactants When the concentration of neither the When the concentration of neither the

reactants nor the products is changing, reactants nor the products is changing, the system is in equilibrium.the system is in equilibrium.

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Chemical equilibrium occurs in chemical Chemical equilibrium occurs in chemical reactions that are reversible. In a reaction reactions that are reversible. In a reaction such as: such as:

CHCH44(g) + H(g) + H22O(g) O(g) CO(g) + 3H CO(g) + 3H22 (g) (g)

The reaction can proceed in both directionsThe reaction can proceed in both directions

CO(g) + 3HCO(g) + 3H22 (g) (g) CH CH44(g) + H(g) + H22O(g) O(g)

Chemical EquilibriumChemical Equilibrium

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Equilibrium Equilibrium ConditionsConditions

At equilibrium the rate of reaction in the At equilibrium the rate of reaction in the forward direction and the rate in the forward direction and the rate in the reverse direction are equalreverse direction are equal

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An Equilibrium SystemAn Equilibrium System

CHCH44(g) + H(g) + H22O(g) O(g) CO(g) + 3H CO(g) + 3H22 (g) (g)

After some of the products are created, After some of the products are created, the products begin to react to form the the products begin to react to form the reactants.reactants.

At equilibrium there is no net change in At equilibrium there is no net change in the concentrations of the reactants and the concentrations of the reactants and products.products.

The concentrations do not change but The concentrations do not change but they are not necessarily equal.they are not necessarily equal.

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At equilibrium two opposing At equilibrium two opposing processes are taking place at processes are taking place at equal rates.equal rates.

In other words the rate in the In other words the rate in the forward direction = the rate in forward direction = the rate in the reverse directionthe reverse direction

ExamplesExamples

H2O (l) H2O (g)

NaCl (s) NaCl (aq)H2O

CO (g) + 2 H2 (g) CH3OH (g)6

Dynamic Dynamic EquilibriumEquilibrium

Equilibrium Equilibrium ConditionsConditions

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HH22O + CO O + CO H H22 + CO + CO22

Law of Mass ActionLaw of Mass Action

Given the reactionGiven the reaction

aA + bB aA + bB cC + dDcC + dD

The rate in the forward direction isThe rate in the forward direction is

rate forward = krate forward = kf f [A][A]aa [B] [B]bb

The rate in the reverse direction isThe rate in the reverse direction is

rate reverse = krate reverse = krr [C] [C]cc [D] [D]dd

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Law of Mass ActionLaw of Mass Action

At equilibrium these rates are equalAt equilibrium these rates are equal

rate forward = rate reverserate forward = rate reverse

kkf f [A][A]aa [B] [B]bb = k = krr [C] [C]cc [D] [D]dd

The ratio of the rate constants isThe ratio of the rate constants is

KeqKeq == [C][C]cc [D] [D]dd

[A][A]aa [B] [B]bb

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Writing Equilibrium Writing Equilibrium ExpressionsExpressions

NN22 (g)(g) + 3 H + 3 H2 2 (g)(g) 2NH2NH3 3 (g)(g)

2 SO2 SO22 (g)(g) + O + O2 2 (g) (g) 2SO 2SO3 3 (g)(g)

HH22 (g)(g) + Br + Br2 2 (g)(g) 2 HBr 2 HBr(g)(g)

2N2N22O O (g)(g) 2 N 2 N2 2 (g)(g) + O + O2 2 (g)(g)

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AnswersAnswers Check your work against the following:Check your work against the following:

4. 4. KKeqeq ==[N[N2 2 ]]2 2 [O[O22]]

[N[N22O]O]22

3. 3. KKeqeq ==[HBr][HBr]22

[H[H22] ] [Br[Br22]]

2. 2. KKeqeq ==[SO[SO33]]22

[SO[SO22]]22 [O [O22]]

1. 1. K Keqeq ==[NH[NH33]]22

[N[N22]] [H [H22]]33

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Reaction QuotientReaction Quotient

The equilibrium constant is a constant ratio The equilibrium constant is a constant ratio only when the system is in equilibrium.only when the system is in equilibrium.

If the system it not at equilibrium the ratio is If the system it not at equilibrium the ratio is known as a Reaction Quotientknown as a Reaction Quotient

If the reaction quotient is equal to the If the reaction quotient is equal to the equilibrium constant then the system is at equilibrium constant then the system is at equilibriumequilibrium

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K>>1: K>>1: The reaction is product-favored; The reaction is product-favored; equilibrium concentrations of products are equilibrium concentrations of products are greater than equilibrium concentrations of greater than equilibrium concentrations of reactants.reactants.

K<<1: K<<1: TheThe reaction is reactant-favored; reaction is reactant-favored; equilibrium concentrations of reactants are equilibrium concentrations of reactants are greater than equilibrium concentrations of greater than equilibrium concentrations of products.products.

The Meaning of the The Meaning of the Equilibrium ConstantEquilibrium Constant

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Calculating Calculating Equilibrium ConstantsEquilibrium Constants

Nitrogen dioxide decomposes at high temperatures Nitrogen dioxide decomposes at high temperatures according to this equation:according to this equation:

2 NO2 NO22 (g) (g) 2 NO (g) + O 2 NO (g) + O22 (g) (g)

If the equilibrium concentrations are as follows: If the equilibrium concentrations are as follows: [NO[NO22]= 1.20 M, [NO] = 0.160, and [O]= 1.20 M, [NO] = 0.160, and [O22] = 0.080 M;] = 0.080 M; calculate the equilibrium constant.calculate the equilibrium constant.

KKeq eq ==[NO][NO]22 [O [O22]]

[NO[NO22]] 22

So KSo Keqeq = =(0.160)(0.160)22 (0.080)(0.080)

= 0.00142= 0.00142(1.20)(1.20)22

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CalculatingCalculating Equilibrium Constants Equilibrium Constants

The equilibrium equation for the oxidation of The equilibrium equation for the oxidation of sulfur dioxide is as follows:sulfur dioxide is as follows:

2SO2SO22 (g)(g) + O + O2 2 (g)(g) 2 SO 2 SO3 3 (g)(g)

If the equilibrium concentrations are as follows: If the equilibrium concentrations are as follows: [SO[SO22 ]= 0.44 M, [O ]= 0.44 M, [O22] = 0.22, and [SO] = 0.22, and [SO33] = 0.78 M] = 0.78 M , , Calculate the equilibrium constantCalculate the equilibrium constant

SoSo K Keqeq = =(0.78)(0.78)22

= 14.3= 14.3(0.44)(0.44)22 (0.22) (0.22)

KKeqeq==[SO[SO33]]22

[SO[SO22]]22 [O [O22]]

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Practice ProblemPractice Problem 1 1

The equilibrium equation for the carbon monoxide with The equilibrium equation for the carbon monoxide with steam to produce hydrogen gas is as follows:steam to produce hydrogen gas is as follows:

COCO22 (g) + H(g) + H 22 (g) (g) CO (g) + H CO (g) + H22O (g) O (g)

If the equilibrium concentrations are as follows: If the equilibrium concentrations are as follows: [CO] = 1.00 [CO] = 1.00 M, [HM, [H22O] = 0.025, [COO] = 0.025, [CO22] = 0.075 M and [H] = 0.075 M and [H22] = 0.060 M] = 0.060 M, ,

calculate the equilibrium constant.calculate the equilibrium constant.

KKeq eq ==[CO] [H[CO] [H22O]O]

[CO[CO22]] [H [H22]]

So KSo Keqeq = =(1.00)(1.00) (0.025)(0.025)

= 0.139= 0.139(0.075) (0.060)(0.075) (0.060)

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Calculating Equilibrium Calculating Equilibrium ConcentrationsConcentrations

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Equilibrium Equilibrium Calculations –Using I. Calculations –Using I.

C. E. ModelsC. E. Models Equilibrium constants and concentrations Equilibrium constants and concentrations

can often be deduced by carefully can often be deduced by carefully examining data about the initial and examining data about the initial and equilibrium concentrationsequilibrium concentrations

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IInitial nitial CChange hange EEquilbriumquilbrium

Equilibrium Equilibrium CalculationsCalculations

ICE Model problem 1ICE Model problem 1 Hydrogen and iodine are in equilibrium with Hydrogen Hydrogen and iodine are in equilibrium with Hydrogen

iodide to this reaction: iodide to this reaction:

HH2 2 + I+ I22

2HI 2HI

Suppose that 1.5 mole of HSuppose that 1.5 mole of H22 and 1.2 mole of I and 1.2 mole of I22 are are

placed in a 1.0 dmplaced in a 1.0 dm33 container. At equilibrium it was container. At equilibrium it was found that there were 0.4 mole of HI. Calculate the found that there were 0.4 mole of HI. Calculate the equilibrium concentrations of [Hequilibrium concentrations of [H22] and [I] and [I22] and the ] and the

equilibrium constant.equilibrium constant.

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Equilibrium Equilibrium CalculationsCalculations

ICE Model Problem 1 -- ICE Model Problem 1 -- SolutionSolutionHydrogen and iodine are in equilibrium with Hydrogen Hydrogen and iodine are in equilibrium with Hydrogen

iodide to this reaction: iodide to this reaction: HH2 2 + I+ I22

2HI 2HI

Suppose that 1.5 mole of HSuppose that 1.5 mole of H22 and 1.2 mole of I and 1.2 mole of I22 are placed are placed in a 1.0 dmin a 1.0 dm33 container. At equilibrium it was found that container. At equilibrium it was found that there were 0.4 mole of HI. Calculate the equilibrium there were 0.4 mole of HI. Calculate the equilibrium concentrations of [Hconcentrations of [H22] and [I] and [I22] and the equilibrium ] and the equilibrium constant.constant.

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II CC EE Since 2x = 0.4, x = 0.2Since 2x = 0.4, x = 0.2[H[H22 ] 1.5 - x 1.5- x ] 1.5 - x 1.5- x [H2 ] = 1.5 – 0.2 = 1.3[H2 ] = 1.5 – 0.2 = 1.3

[ I[ I22 ] 1.2 - x 1.2 –x ] 1.2 - x 1.2 –x [I2 ] = 1.2 – 0.2 = 1.0[I2 ] = 1.2 – 0.2 = 1.0

[HI ] 0 +2x 0.4[HI ] 0 +2x 0.4

Keq = Keq = [HI][HI]22 = = (0.4) (0.4)2 2 = 0.123 = 0.123 [H[H22 ] [ I ] [ I22 ] (1.3) (1.0) ] (1.3) (1.0)

Equilibrium Equilibrium CalculationsCalculations

ICE Model Problem 2ICE Model Problem 2 Sulfur dioxide reacts with oxygen to produce sulfur Sulfur dioxide reacts with oxygen to produce sulfur

trioxide according to this reaction: trioxide according to this reaction:

2 SO2 SO2 2 + O+ O22

2SO 2SO33

Suppose that 1.4 mole of SOSuppose that 1.4 mole of SO22 and 0.8 mole of O and 0.8 mole of O22 are are

placed in a 1.0 dmplaced in a 1.0 dm33 container. At equilibrium it was container. At equilibrium it was found that there were 0.6 dmfound that there were 0.6 dm33 of SO of SO33. Calculate the . Calculate the

equilibrium concentrations of [SOequilibrium concentrations of [SO22] and [O] and [O22] and the ] and the

equilibrium constant.equilibrium constant.

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Equilibrium Equilibrium CalculationsCalculations

ICE Model Problem 2 - ICE Model Problem 2 - SolutionSolutionSulfur dioxide reacts with oxygen to produce sulfur Sulfur dioxide reacts with oxygen to produce sulfur

trioxide according to this reaction: trioxide according to this reaction:

2 SO2 SO2 2 + O+ O22 2SO 2SO33

Suppose that 1.4 mole of SOSuppose that 1.4 mole of SO22 and 0.8 mole of O and 0.8 mole of O22 are are placed in a 1.0 dmplaced in a 1.0 dm33 container. At equilibrium it was found container. At equilibrium it was found that there were 0.6 dmthat there were 0.6 dm33 of SO of SO33. Calculate the equilibrium . Calculate the equilibrium concentrations of [SOconcentrations of [SO22] and [O] and [O22] and the equilibrium ] and the equilibrium constant.constant.

II CC EE Since 2x = 0.6, x = 0.3Since 2x = 0.6, x = 0.3

[SO[SO2 2 ] 1.4 -2x 1.4-2x ] 1.4 -2x 1.4-2x [SO[SO2 2 ] =1.4 – 2( 0.3) = 0.8] =1.4 – 2( 0.3) = 0.8

[O[O2 2 ] 0.8 -x 0.8 –x] 0.8 -x 0.8 –x [O[O2 2 ] = 0.8 – 0.3 = 0.5] = 0.8 – 0.3 = 0.5

[SO[SO33] 0 +2x 0.6] 0 +2x 0.6

Keq = Keq = [SO[SO33]]2 2 = = (0.6) (0.6)2 2 = 0.281= 0.281

[SO[SO2 2 ]]22 [O [O2 2 ] (0.8)] (0.8)22(0.5)(0.5) 22

Calculating Equilibrium Calculating Equilibrium Concentrations from KConcentrations from Keqeq

KKeq eq ==[[C2H4]] [H [H22]]

[[C2H6]]

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What is the concentration for each substance at equilibrium for the following gaseous reaction

C2H6    C2H4 + H2     KC = 1.01

if the initial concentration of ethene, C2H4 and that of hydrogen are both 0.300 M?

IInitialnitial CChangehange EEquilibriumquilibrium

CC22HH44 0.3000.300 - x- x 0.300-x0.300-x

HH22 0.3000.300 - x- x 0.300-x0.300-x

CC22HH66 00 + x+ x xx

1.011.01 ==[[0.300-x0.300-x][][0.300-x0.300-x]]

[[x]]1.01x = 0.09 - 0.6x +x2

x2 - 1.61x +0.09 = 0

x = 1.61+ (1.61)2-4(1)(0.09)

2(1)

x = 0.0580 and 1.552. The second root is extraneous. so [CC22HH66] = 0.0580 and [H2 ] = [CC22HH44] = 0.242

Le Chatelier’s PrincipleLe Chatelier’s Principle

– Le Chatelier's Principle states: When a Le Chatelier's Principle states: When a system in chemical equilibrium is system in chemical equilibrium is disturbed by a change of temperature, disturbed by a change of temperature, pressure, or a concentration, the pressure, or a concentration, the system shifts in equilibrium system shifts in equilibrium composition in a way that tends to composition in a way that tends to counteract this change of variable. counteract this change of variable.

– A change imposed on an equilibrium A change imposed on an equilibrium system is called a system is called a stressstress

– The equilibrium always responds in The equilibrium always responds in such a way so as to counteract the such a way so as to counteract the stressstress

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Le Chatelier’s PrincipleLe Chatelier’s Principle

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• A stress is any sudden change in conditions that drives the system out of equilibrium.

• When a stress is placed on an equilibrium system, the system will shift in such a way so as to lessen or mitigate the stress and restore equilibrium.

• A stress usually involves a change in the temperature, pressure, or in the concentration of one or more of the substances that are in equilibrium.

• Le Chatelier's principle predicts the direction of the change in the equilibrium.

Applications ofApplications ofLe Chatelier’s PrincipleLe Chatelier’s Principle

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N2 (g) +3 H2 (g) 2NH3 (g) H = -92 kJ

Haber’s process for the production of ammonia is an example of an industrial equilibrium system. We will use this equilibrium as a model to explain the how Le Chatelier’s principle operates with the following stresses:• Change in the concentration of one of the components• Changes in pressure• Changes in temperature• Use of a catalyst

Applications of Applications of Le Chatelier’s Principle Le Chatelier’s Principle

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N2 (g) +3 H2 (g) 2NH3 (g) H = - 92 kJ

The equilibrium constant for this reaction is

Any change in the concentration (or partial pressure) of any component will cause the equilibrium to shift in such a way so as to return the equilibrium constant to its original value.

KKeq eq ==[[NH3]]22

[N[N22]] [H [H22]]3 3

Le Chatelier’s Principle –Le Chatelier’s Principle –The Concentration EffectThe Concentration Effect

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N2 (g) +3 H2 (g) 2NH3 (g) H = - 92 kJ

An increase in the concentration of N2 results ina decrease H2 and an increase in NH3 in such a way to keep the equilibrium constant the same

KKeq eq ==[[NH3]]22

[N[N22]] [H [H22]]3 3

Le Chatelier’s Principle – Le Chatelier’s Principle – The Concentration Effect The Concentration Effect

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N2 (g) +3 H2 (g) 2NH3 (g) H = - 92 kJ

Likewise an increase in the concentration of H2 results in a decrease in N2 and an increase in NH3 in such a way to keep the equilibrium constant the same

KKeq eq ==[[NH3]]22

[N[N22]] [H [H22]]3 3

Le Chatelier’s Principle – Le Chatelier’s Principle – The Concentration Effect The Concentration Effect

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N2 (g) +3 H2 (g) 2NH3 (g) H = - 92 kJ

An increase in the concentration of NH3 results in a increase in N2 and an increase in H2 in such a way to keep the equilibrium constant the same.

KKeq eq ==[[NH3]]22

[N[N22]] [H [H22]]3 3

Le Chatelier’s Principle – Le Chatelier’s Principle – The Temperature Effect The Temperature Effect

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N2 (g) +3 H2 (g) 2NH3 (g) H = - 92 kJ

The reaction is exothermic in the forward direction.

An increase in the temperature would trigger a response in the heat consuming (endothermic) direction.

An increase in temperature therefore causes the reaction to shift in the reverse direction. Some NH3 decomposes to N2 and H2.

Le Chatelier’s Principle –Le Chatelier’s Principle –The Pressure Effect The Pressure Effect

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N2 (g) +3 H2 (g) 2NH3 (g) H = - 92 kJ

All molecules in the equilibrium are gases When the reaction proceeds in the forward

direction the number of moles are reduced from 4 to 2.

Pressure is proportional to the number of moles of gas

An increase in pressure therefore causes the reaction to moved in the forward direction. Some N2 and H2 combine to form more NH3

Le Chatelier’s Principle – Le Chatelier’s Principle – The Effect of Catalysts The Effect of Catalysts

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N2 (g) +3 H2 (g) 2NH3 (g) H = - 92 kJ

Catalysts lower the activation energy A catalyst affects the forward and the reverse

direction equally There is no change in the equilibrium position

from a catalyst A catalyst decreases the time required for the

system to achieve equilibrium

Heterogeneous EquilibriaHeterogeneous Equilibria

In a heterogeneous equilibrium, the components are In a heterogeneous equilibrium, the components are in two different phases.in two different phases.

A common form of a heterogeneous equilibrium is the A common form of a heterogeneous equilibrium is the equilibrium which exists between a solid and an equilibrium which exists between a solid and an aqueous solutionaqueous solution

When a substance dissolves in water there is an When a substance dissolves in water there is an equilibrium established between the solid an its equilibrium established between the solid an its dissolved ionsdissolved ions

ExampleExample AgCl (s) AgCl (s) Ag Ag++ (aq) + Cl (aq) + Cl-- (aq) (aq)

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Heterogeneous EquilibriaHeterogeneous Equilibria

AgCl (s) AgCl (s) Ag Ag++ (aq) + Cl (aq) + Cl-- (aq) (aq) In this example the concentration of the solid phase In this example the concentration of the solid phase

is essentially 1. The equilibrium constant then takes is essentially 1. The equilibrium constant then takes the formthe form

K = [AgK = [Ag++ ][Cl ][Cl--]. ]. This form is known as a solubility product and is This form is known as a solubility product and is

usually designated as Kusually designated as Kspsp

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Different designations of Different designations of equilibrium constantsequilibrium constants

KKeqeq General designation for an equilibrium constant General designation for an equilibrium constant Kc Equilibrium constant based on concentrationKc Equilibrium constant based on concentration Kp Equilibrium constant based on pressure (gases)Kp Equilibrium constant based on pressure (gases) Ksp Solubility productKsp Solubility product Ka Acid equilibrium constantKa Acid equilibrium constant Kb Base equilibrium constantKb Base equilibrium constant Kw Ion product of waterKw Ion product of water

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