13 equilibrium

7/28/2019 13 Equilibrium

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CHEMICAL

EQUILIBRIUM

Cato Maximilian Guldberg andhis brother-in-law Peter Waagedeveloped the Law of MassAction


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Chemical Equilibrium

Reversible Reactions:

A chemical reaction in which the productscan react to re-form the reactants

Chemical Equilibrium:

When the rate of the forward reactionequals the rate of the reverse reactionand the concentration of products andreactants remains unchanged

2HgO(s) 2Hg(l) + O2(g)Arrows going both directions ( ) indicates equilibrium in a chemical equation


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

Remember this fromChapter 12?

Why was it so important tomeasure reaction rate at

the start of the reaction(method of initial rates?)


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


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Law of Mass ActionFor the reaction:

WhereKis the equilibrium constant,and is unitless

jA + kB lC + mD

kj

ml

BADCK

][][][][


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Product Favored EquilibriumLarge values for Ksignify the reaction is

product favored

When equilibrium is achieved, mostreactant has been converted to product


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Reactant Favored EquilibriumSmall values for Ksignify the reaction is

reactant favored

When equilibrium is achieved, very littlereactant has been converted to product


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Writing an Equilibrium Expression

2NO2(g) 2NO(g) + O2(g)K =???

Write the equilibrium expression for thereaction:

2

2

2

2

][

][][

NO

ONOK


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Conclusions about Equilibrium Expressions

The equilibrium expression for a reaction

is the reciprocal for a reaction written inreverse

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

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

2

2

22

][

][][

NO

ONOK

][][

][1

2

2

2

2'

ONO

NO

K

K


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Conclusions about Equilibrium Expressions

When the balanced equation for a reaction

is multiplied by a factor n, the equilibriumexpression for the new reaction is theoriginal expression, raised to the nth power.

2NO2(g)

2NO(g) + O2(g)

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

2

2

2

][

][][

NO

ONOK

][

]][[

2

2

1

22

1

1

NO

ONOKK


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Equilibrium Expressions Involving Pressure

For the gas phase reaction:

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

))(( 3

2

22

3

HN

NH

P PP

PK

pressurespartialmequilibriuarePPP HNNH 223 ,,

n

p RTKK

)(


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Heterogeneous Equilibria

The position of a heterogeneous equilibriumdoes not depend on the amounts of pure solidsor liquids present

Write the equilibrium expression for

the reaction:PCl5(s) PCl3(l) + Cl2(g)

Puresolid Pureliquid ][ 2ClK

2ClpPK


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The Reaction Quotient

For some time, t, when the system is not atequilibrium, the reaction quotient, Qtakes theplace of K, the equilibrium constant, in thelaw of mass action.

jA + kB lC + mD

kj

ml

BADCQ

][][][][


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Significance of the Reaction Quotient

If Q= K, the system is at equilibrium

If Q > K, the system shifts to the left,

consuming products and forming reactantsuntil equilibrium is achieved

If Q < K, the system shifts to the right,

consuming reactants and forming productsuntil equilibrium is achieved


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Solving for Equilibrium Concentration

Consider this reaction at some temperature:H2O(g) + CO(g) H2(g) + CO2(g) K= 2.0Assume you start with 8 molecules of H2O

and 6 molecules of CO. How many moleculesof H2O, CO, H2, and CO2 are present atequilibrium?

Here, we learn about ICE the mostimportant problem solving technique in thesecond semester. You will use it for thenext 4 chapters!


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H2O(g) + CO(g) H

2(g) + CO

2(g) K= 2.0

Step #1:We write the law of mass actionfor the reaction:

]][[

]][[0.2

2

22

COOH

COH


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H2O(g) + CO(g) H2(g) + CO2(g)Initial:

Change:

Equilibrium:

Step #2:We ICE the problem, beginningwith the Initial concentrations

8 6 0 0

-x -x +x +x

8-x 6-x x x


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Equilibrium: 8-x 6-x x x

Step #3:We plug equilibrium concentrationsinto our equilibrium expression, and solve for x

H2O(g) + CO(g) H2(g) + CO2(g)

))(8(

))((0.2

COx

xx

x = 4


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Solving for Equilibrium ConcentrationStep #4:Substitute x into our equilibriumconcentrations to find the actual concentrations

H2O(g) + CO(g) H2(g) + CO2(g)

Equilibrium: 8-x 6-x x x

Equilibrium: 8-4=4 6-4=2 4 4

x = 4

13 equilibrium

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