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14 - 1 Chemical Equilibrium Chemical Equilibrium Introduction to Chemical Introduction to Chemical Equilibrium Equilibrium Equilibrium Constants and Equilibrium Constants and Expressions Expressions Calculations Involving Calculations Involving Equilibrium Constants Equilibrium Constants Using Le Châtelier’s Principle Using Le Châtelier’s Principle Some Important Equilibria Some Important Equilibria

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Page 1: 14 - 1 Chemical Equilibrium Introduction to Chemical Equilibrium Equilibrium Constants and Expressions Calculations Involving Equilibrium Constants Using

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Chemical EquilibriumChemical Equilibrium

Introduction to Chemical Introduction to Chemical EquilibriumEquilibrium

Equilibrium Constants and Equilibrium Constants and ExpressionsExpressions

Calculations Involving Equilibrium Calculations Involving Equilibrium ConstantsConstants

Using Le Châtelier’s PrincipleUsing Le Châtelier’s Principle

Some Important EquilibriaSome Important Equilibria

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Chemical equilibriumChemical equilibrium

For the general chemical reaction:

aA + bB cC + dD

If A and B are brought together, a point is reached where there is no further change to the system - equilibriumequilibrium.

A chemical equilibrium is a dynamic equilibrium. Species are constantly being formed but with no net change in concentration

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Chemical equilibriumChemical equilibrium

Different types of arrows are used in chemical equations associated with equilibria.

Single arrowSingle arrow

Assumes that the reaction proceeds to completion as written.

Two single-headed arrowsTwo single-headed arrows

Used to indicate a system in equilibrium.

Two single-headed arrows of different sizes.Two single-headed arrows of different sizes.

May be used to indicate when one side of an equilibrium system is favored.

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Chemical equilibriumChemical equilibrium

Homogeneous equilibriaHomogeneous equilibriaEquilibria that involve only a single phase.

Examples.Examples.All species in the gas phase

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

All species are in solution.

HC2H2O2 (aq) H+ (aq) + C2H3O2-

(aq)

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Chemical equilibriumChemical equilibrium

Basic steps.Basic steps.For the general reaction:

A + B C

We can view the reaction as occurring in three steps.

• Initial mixing• Kinetic region• Equilibrium region

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Chemical equilibriumChemical equilibrium

Initial mixing.Initial mixing.When A and B are first brought together, there is no C present.

The reaction proceeds as

A + B C

This is just at the very start of the reaction. Things change as soon as some C is produced.

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Chemical equilibriumChemical equilibrium

Kinetic region.Kinetic region.As soon as some C has been produced, the reverse reaction is possible.

A + B C

Overall, we still see an increase in the net concentration of C.

As we approach equilibrium, the rate of the forward reaction becomes slower.

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Chemical equilibriumChemical equilibrium

Equilibrium region.Equilibrium region.A point is finally reached where the forward and reverse reactions occur at the same rate.

A + B C

There is no net change in the concentration of any of the species.

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Chemical equilibriumChemical equilibriumC

on

cen

trati

on

Time

C

B

A

EquilibriumRegion

KineticRegion

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Equilibrium constantsEquilibrium constantsand expressionsand expressions

For the general chemical reaction:

aA + bB eE + fF

The equilibrium can be expressed as:

Kc =[E]e [F]f

[A]a [B]b

Kc Equilibrium constant for a homogeneousequilibrium.

[ ]n Molar concentrations raised to the powerof their coefficients in the balancedchemical equation.

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Equilibrium constantsEquilibrium constantsand expressionsand expressions

Hetrogeneous equilibriaHetrogeneous equilibriaEquilibria that involve more than one phase.

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

Equilibrium expressions for these types of systems do notdo not include the concentrations of the pure solids (or liquids).

Kc = [CO2]

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Equilibrium constantsEquilibrium constantsand expressionsand expressions

Hetrogeneous equilibriaHetrogeneous equilibriaWe don’t include the pure solids and liquids because their concentrations do not vary. Their values end up being included in the K value.

CO2

CaO &CaCO3

As long as the temperature is constant and some solid is present, the amount of solid present has no effect on the equilibrium.

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Writing an equilibrium expressionWriting an equilibrium expression

• Write a balanced equation for the equilibrium.

• Put the products in the numerator and the reactants in the denominator.

• Omit pure solids and liquids from the expression

• Omit solvents if your solutes are dilute (<0.1M).

• The exponent of each concentration should be the same as the coefficient for the species in the equation.

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Writing an equilibrium expressionWriting an equilibrium expression

Example.Example.What would be the equilibrium expression for the following?

(NH4)2CO3 (s) 2 NH3 (g) + CO2 (g) + H2O (g)

Kc = [NH3]2 [CO2] [H2O]

(NH4)2CO3 is a pure solid so is not included Kc.

We use [NH3]2 because the coefficient for NH3(g) in the equation is 2.

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Equilibrium and rate of reactionEquilibrium and rate of reaction

Chemical reactions tend to go to equilibrium providing that reaction takes place at a significant rate.

There is no relationship between the magnitude of the equilibrium constant and the rate of a reaction.

Example.Example. 2H2 (g) + O2 (g) 2H2O (g)

Kc = 2.9 x 1031 =

However, the reaction will take years to reach equilibrium at room temperature.

[H2O][H2]2 [O2]

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Determining equilibrium constantsDetermining equilibrium constants

Equilibrium constants can be found by experiment.

If you know the initial concentrations of all of the reactants, you only need to measure the concentration of a single species at equilibrium to determine the Kc value.

Lets consider the following equilibrium:

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

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Determining equilibrium constantsDetermining equilibrium constants

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

Assume that we started with the following initial concentrations at 425.4oC.

H2 (g) 0.005 00 MI2 (g) 0.012 50 MHI (g) 0.000 00 M

At equilibrium, we determine that the concentration of iodine is 0.007 72 M

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Determining equilibrium constantsDetermining equilibrium constants

The equilibrium expression for our system is:

Kc =

Based on the chemical equation, we know the equilibrium concentrations of each species.

I2 (g) = the measured amount= 0.007 72 M

That means that 0.004 78 mol I2 reacts to produce HI in 1.00 L of solution.

[HI]2

[H2] [I2]

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Determining equilibrium constantsDetermining equilibrium constants

I2 (g) = 0.007 72 M

HI (g) = 0.004 78 M = 0.009 56 M

H2 (g) = 0.005 00 M - 0.004 78 M = 0.000 22 M

Kc = =

= 54

2 mol HI1 mol I2

[HI]2

[H2] [I2](0.009 56)2

(0.000 22)(0.007 72)

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Partial pressurePartial pressureequilibrium constantsequilibrium constants

At constant temperature, the pressure of a gas is proportional to its molarity.

Remember, for an ideal gas: PVPV = = nRTnRT

and molarity is: M = mole / liter or n/V

so: P = R T M

where R is the gas law constantT is the temperature, K.

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Partial pressurePartial pressureequilibrium constantsequilibrium constants

For equilibria that involves gases, partial pressures can be used instead of concentrations.

aA (g) + bB (g) eE (g) + fF (g)

Kp =

Kp is used when the partial pressures are expressed in units of atmospheres.

pEe pF

f

pAa pB

b

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Partial pressurePartial pressureequilibrium constantsequilibrium constants

In general, Kp = Kc, instead

Kp = Kc (RT)ng

ng is the number of moles of gaseous products minus the number of moles of gaseous reactants.

ng = (e + f) - (a + b)

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Partial pressurePartial pressureequilibrium constantsequilibrium constants

For the following equilibrium, Kc = 1.10 x 107 at 700. oC. What is the Kp?

2H2 (g) + S2 (g) 2H2S (g)

Kp = Kc (RT)ng

T = 700 + 273 = 973 K

R = 0.08206

ng = ( 2 ) - ( 2 + 1) = -1

atm Lmol K

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Partial pressurePartial pressureequilibrium constantsequilibrium constants

Kp = Kc (RT)ng

= 1.10 x 107 (0.08206 ) (973 K)

= 1.378 x105

atm Lmol K[ ]-1

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Equilibrium calculationsEquilibrium calculations

We can predict the direction of a reaction by calculating the reaction quotient.

Reaction quotient, Reaction quotient, QQ

For the reaction: aA + bB eE + fF

Q has the same form as Kc with one important difference. Q can be for any set of concentrations, not just at equilibrium.

Q =[E]e [F]f

[A]a [B]b

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Reaction quotientReaction quotient

Any set of concentrations can be given and a Q calculated. By comparing Q to the Kc value, we can predict the direction for the reaction.

QQ < < KKcc Net forward reaction will occur.

QQ = = KKcc No change, at equilibrium.

QQ > > KKcc Net reverse reaction will occur.

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Reaction quotient exampleReaction quotient example

For an earlier example

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

we determined the Kc to be 54 at 425.4 oC.

If we had a mixture that contained the following, predict the direction of the reaction.

[H2] = 4.25 x 10-3 M[I2] = 3.97 x 10-1 M[HI] = 9.83 x 10-2 M

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Reaction quotient exampleReaction quotient example

Q =

=

= 5.73

Since Q is < Kc, the system is not in equilibrium and will proceed in the forward direction.

[ HI ]2

[ H2 ] [ I2 ]

(9.83 x 10-2)2

(4.25 x 10-3)(3.97 x 10-1)

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Calculating equilibrium Calculating equilibrium concentrationsconcentrations

If the stoichiometry and Kc for a reaction is known, calculating the equilibrium concentrations of all species is possible.

• Commonly, the initial concentrations are known.

• One of the concentrations is expressed as the variable x.

• All others are then expressed in terms of x.

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Equilibrium calculation exampleEquilibrium calculation example

A sample of COCl2 is allowed to decompose. The value of Kc for the equilibrium

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

is 2.2 x 10-10 at 100 oC.

If the initial concentration of COCl2 is 0.095M, what will be the equilibrium concentrations for each of the species involved?

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Equilibrium calculation exampleEquilibrium calculation example

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

Initial conc., M 0.095 0.000 0.000

Change in conc. - X + X + Xdue to reaction

EquilibriumConcentration, M (0.095 -X) X X

Kc = =[ CO ] [ Cl2 ]

[ COCl2 ]X2

(0.095 - X)

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Equilibrium calculation exampleEquilibrium calculation example

X2

(0.095 - X)Kc = 2.2 x 10-10 =

Rearrangement gives

X2 + 2.2 x 10-10 X - 2.09 x 10-11 = 0

This is a quadratic equation. Fortunately, thereis a straightforward equation for their solution

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Quadratic equationsQuadratic equations

An equation of the form

a X2 + b X + c = 0

Can be solved by using the following

x =

Only the positive root is meaningful in equilibrium problems.

-b + b2 - 4ac

2a

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Equilibrium calculation exampleEquilibrium calculation example

-b + b2 - 4ac

2a

X2 + 2.2 x 102.2 x 10-10-10 X - 2.09 x 102.09 x 10-11-11 = 0a b b c c

X =

X = - 2.2 x 10-10 + [(2.2 x 10-10)2 - (4)(1)(- 2.09 x 10-11)]1/2

2

X = 9.1 x 10-6 M

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Equilibrium calculation exampleEquilibrium calculation example

Now that we know X, we can solve for the concentration of all of the species.

COCl2 = 0.095 - X = 0.095 M

CO = X = 9.1 x 10-6 M

Cl2 = X = 9.1 x 10-6 M

In this case, the change in the concentration of is COCl2 negligible.

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Summary of method of calculating Summary of method of calculating equilibrium concentrationsequilibrium concentrations

• Write an equation for the equilibrium.

• Write an equilibrium constant expression.

• Express all unknown concentrations in terms of a single variable such as x.

• Substitute the equilibrium concentrations in terms of the single variable in the equilibrium constant expression.

• Solve for x.

• Use the value of x to calculate equilibrium concentrations.

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Predicting shifts in equilibriaPredicting shifts in equilibria

Le Châtelier’s PrincipleLe Châtelier’s PrincipleWhen subjected to a stress, a system in equilibrium will act to relieve the stress.

The position of a chemical equilibrium will shift in a direction to relieve a stress

aA + bB cC + dD

ExampleExampleAdding A or B or removing C or D will shift equilibrium to the right.

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Predicting shifts in equilibriaPredicting shifts in equilibria

Equilibrium concentrations are based on:–The specific equilibrium–The starting concentrations–Other factors such as:

•Temperature•Pressure•Reaction specific conditions

Altering conditions will stress a system, resulting in an equilibrium shift.

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Changes in concentrationChanges in concentration

Changes in concentration do not change the value of the equilibrium constant at constant temperature.

When a material is added to a system in equilibrium, the equilibrium will shift away from that side of the equation.

When a material is removed from a system in equilibrium, the equilibrium will shift towards that sid of the equation.

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Changes in concentrationChanges in concentration

HI

H2

I2

Log

con

cen

trati

on

Time

Example.Example. I2 is added to an equilibrium mixture. The system will adjust all of the concentrations to reestablish a new equilibrium with the same Kc.

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Changes in concentrationChanges in concentration

HI

H2

I2

Log

con

cen

trati

on

Time

Example.Example. Some H2 is removed. Again, the system adjusts all of the concentrations to reestablish a new equilibrium with the same Kc.

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Changes in pressureChanges in pressure

Changing the pressure does not change the value of the equilibrium constant at constant temperature.

• Solids and liquids are not effected by pressure changes.

• Changing pressure by introducing an inert gas will not shift an equilibrium.

• Pressure changes only effect gases that are a portion of an equilibrium.

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Changes in pressureChanges in pressure

In general, increasing the pressure by decreasing volume shifts equilibria towards the side that has the smaller number of moles of gas.

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

N2O2 (g) 2NO2 (g)

Unaffected by pressureUnaffected by pressure

Increased pressure, shift to leftIncreased pressure, shift to left

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Changes in temperatureChanges in temperature

Changes in temperature usually change the value of the equilibrium constant.

• Kc can either increase or decrease with increasing temperature.

• The direction and degree of change is dependent on the specific reaction.

T, oC Kp

649 2.7 x 100

760 6.3 x 101

871 8.2 x 102

982 6.8 x 103

T, oC Kp

649 2.7 x 100

760 6.3 x 101

871 8.2 x 102

982 6.8 x 103

T, oC Kp

227 9.0 x 10-2

427 8.1 x 10-5

627 1.3 x 10-6

827 9.7 x 10-8

T, oC Kp

227 9.0 x 10-2

427 8.1 x 10-5

627 1.3 x 10-6

827 9.7 x 10-8

CH4(g) + H2O(g) 2H2(g) + CO(g) N2(g)+ 3H2(g) 2NH3(g)