Introduction to Kinetics and Equilibrium

Kineticsandequilibriumaretwoofthemostimportant areas in chemistry Entire books andimportantareasinchemistry.Entirebooksandcoursesattheundergraduateandgraduatelevelaredevotedtothem.

Chemicalkinetics thestudyoftheratesofchemicalprocesses

Equilibrium theconditionofasysteminwhichcompetinginfluencesarebalanced

Ch i l ilib i th t t i hi h th t ti f thChemicalequilibrium thestateinwhichtheconcentrationsofthereactantsandproductshavenonetchangeovertime

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concentrationsatequilibriumaredeterminedbythermodynamics(rG)

Physical Equilibrium

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Chemical Equilibrium

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Is Equilibrium Common?

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Demo

Equilibrium 1: H pressure balanced

We have physical equilibrium:

Equilibrium 1: H2 pressure balanced by balloon wall surface tension

Equilibrium 2: Balloon buoyancy H2 and O2

filled balloon q y y

balanced by the string tensionballoon

Do we have chemical equilibrium?? NO!Do we have chemical equilibrium?? NO!

2H2(g) +O2(g) 2 H2O (g) rG =457.2kJ/mol

Combustion to make water is favored, but extremely slow at PSCB 100 temp.

Removing constraints: heating with candle

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Removing constraints: heating with candledestroys equilibria 1&2 and (nearly) achieveschemical equilibrium simultaneously

Limiting ReagentsWas all the H2 and O2 in the balloon converted to water?

It depends on the initial amounts of the two gases Suppose:It depends on the initial amounts of the two gases. Suppose:

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

Before candle: 0.75 atm 0.75 atm

After candle: 0 0 atm 0 38 atm

0.0 atm

0 75 atmChange: - 2x - x + 2x

After candle: 0.0 atm 0.38 atm 0.75 atm

Limiting reagent:

80s Australian rock band?

?Reagent in excessLimiting reagent:molecule that is

used up, causing the reaction to stop

g

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Reactions That Don't Go to CompletionReactionseither:

gotocompletion(alllimitingreactantsareconsumed)g p ( g )

e g combustion reactions (like the balloon demo)

reactants products

e.g.combustionreactions(liketheballoon demo)

orestablishachemicalequilibrium (somereactants,someproducts)

reactants products

thesearethereactionsintroducedinthispartofthecourse

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Reactions That Don't Go to Completion

Whenanequilibriumisestablished,eachreactant(andproduct)hasai d h i i hnonzeroconcentrationandtheseconcentrationsareconstantwith

respecttotime.

Whenanequilibriumisestablished,thelimitingreagenthasanonzeroconcentration.

Thereactionsdonotstopatequilibrium.Theratesoftheforwardandreversereactionsarebalanced,sothereisnonetchangeinconcentrations.

concentrationisspecifiedasMolesperLiterandwrittenas[reactant]or[product],

or

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aspartialpressuresforgasphasereactions

A Gas Phase Equilibrium

Considerasimplegasphasereaction,theconversionofcis2butenetotrans2butene, an example of isomerization:trans 2 butene,anexampleofisomerization:

H3C CH3 H3C H

C CHH

C CCH3H

cis 2 butene trans 2 butenecis2butene trans2butene

The isusedtoindicateanequilibrium.

Here,bothspeciesaregases.

Thetotalnumberofmolesofgasremainsconstant.

Thisreactiondoesnotgotocompletion.

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g p Anequilibriumisestablished.

A Gas Phase Equilibrium

cis trans

at400CKineticregion

Equilibriumregion

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Equilibrium Constant Intheequilibriumregion,theconcentrationsofproductsandreactants

arerelatedinan equationcalledtheequilibriumconstantexpression

Inthe2butenecase,

cis trans

(at400C)

Kc iscalledtheequilibriumconstant. ForKc, speciesareexpressedasM (MolesL1) Products in numerator, reactants in denominator.Productsinnumerator,reactantsindenominator. Concentrationsraisedtotheirstoichiometriccoefficients

inthebalancedreaction(1inthiscase).

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Sample QuestionA quantity of cis-2-butene is added to a 2 Liter flask and heated to 400C for2 years. The concentration of trans-2-butene was then determined to be 0.50M. What is the concentration of cis-2-butene in the flask?M. What is the concentration of cis 2 butene in the flask?

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Generic Equilibrium Constant

a A + b B c C + d DaA+bB cC+dD

Constant for a giventemperature

Ratio of equilibrium concentrations of products over reactants Concentrations raised to the stoichiometric coefficients in the

balanced reaction equation

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Equilibrium Constant

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Origin of Equilibrium ConstantEquilibrium occurs when the rates of the forward

and reverse reactions are exactly equal

rateforward =ratereverse

Reaction rate is the number (mol) of molecules produced or d i h i l ti ti l (L) ti ( )consumed in a chemical reaction per reaction volume (L) per time (s)

rateforward

rate

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rateforward

Origin of Equilibrium ConstantFor simple reactions (like this one), reaction rate is proportional to the

concentrations of the reactants raised to their stoichiometric coefficients

rateforwardRate definition: ratereverse

rateforward = kf x [A] ratereverse = kr x [B]Rate law:

rate constants

At equilibrium: kf x [A] = kr x [B] Kc =

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f r c

ExampleThe reaction of CO with Cl2 to form COCl2 is another single-step reaction. A vessel is filled with only CO and Cl2. Describe how equilibrium is achieved and the connection between the reaction rates and the equilibrium constant.

Balanced Reaction:

connection between the reaction rates and the equilibrium constant.

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

rateforward = kf x [CO][Cl2]

Initially, we have only reactants: CO(g)+Cl2(g) COCl2(g)

forward f [ ][ 2]

Initially:rateforward >>ratereverse

As products form, the rate of the reverse reaction increases:

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

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ratereverse = kr x [COCl2]

Whentheforwardandreverseratesbecomeequal,thesystemhasq , yachievedchemicalequilibrium.

rateforward =ratereverse

Equilibriumconstant expression

Equilibriumconstant

constantexpression

=constant

Law of Mass Action Productsinnumerator,reactantsindenominator. Byconvention,theunitsofKc (ifany)areignored

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Concentrationsraisedtothestoichiometriccoefficients(1inthiscase).

Equilibrium Expressed in Partial Pressures aA+bB cC+dD

Since PV = nRT, we can write for gas phase reactions:

Kp expresses equilibrium in terms of the partial pressures

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Equilibrium Expressed in Partial Pressures In general Kc Kp , but they are related:

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Sample Problem

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Homogeneous EquilibriaRefers to reactions in which all reactants and products are in the

same phase.

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Heterogeneous EquilibriaIn heterogeneous equilibria, reactants and products are in different phases

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

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Example

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Manipulating Equilibrium Constant Expressions Eachoftheseproperchemicalequationswillhaveadifferent

equilibriumexpressionandadifferentnumericalvalueofKc:

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

1/2N2(g)+3/2H2(g) NH3(g)

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

Changing the reaction stoichiometry by a factor of x raises Kc to the power of x

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The Kcs of a forward and reverse reaction are reciprocals of each other

K of Forward and Reverse Reactions

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Multiple Equilibria

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Example

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

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

N2(g) +2O2(g) 2NO2(g)

Kc is equal to the product of the equilibrium constants for the individual reactions

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Another Example

Reactions 1, 2 and 3 are related as follows:2x{Reverse Reaction 1} + {Reaction 2} = {Reaction 3}2x{Reverse Reaction 1} + {Reaction 2} = {Reaction 3}

So:K1-1 x K1-1 x K2 = K3

Solving for K3:10-52 x 10-52 x 10129 = 1025

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Reaction Quotients Wecandeterminethedirectioninwhichareactionhastoshiftto

reachequilibriumbycalculatingthereactionquotient,Qc

TheexpressionforQc isthesameasKc.Thedifferenceisthattheinstantaneous concentration values are used when calculating Qc.instantaneousconcentration values areusedwhencalculatingQc.

Qc canvaryfromzero(noproducts)toinfinity(noreactants)

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Threepossiblesituations: Q < K Qc Kc Toomanyproducts.Theremustbeadditionalnetconversionofproductstoreactantstoachieveequilibrium.

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Sample ProblemAssumethatthereactionquotientforthefollowingreactionis1.0x108

( ) ( ) ( ) 16 ( )2NO2(g) 2NO(g)+O2(g) Kc =7.4x1016 (at25C)

From this we can conclude:Fromthiswecanconclude:

(a)Thereactionisatequilibrium.(b) E ilib i ld b h d b ddi h NO O t th t

Q>K

(b)EquilibriumcouldbereachedbyaddingenoughNOorO2 tothesystem.(c)Thereactionmustproceedfromlefttorighttoreachequilibrium.(d)Thereactionmustproceedfromrighttolefttoreachequilibrium.(e)Thereactioncanneverreachequilibrium.( ) q

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Sample ProblemIftheequilibriumconstantforthefollowingreactionisKc =1x10

2

2 C H (g) + 7 O (g) 4 CO (g) + 6 H O (g)2C2H6 (g)+7O2 (g) 4CO2 (g)+6H2O(g)

andalltheconcentrationswereinitially0.10M,y ,wecanpredictthatthereaction:

(a)isatequilibriuminitially.(b)mustshiftfromlefttorighttoreachequilibrium.(c)mustshiftfromrighttolefttoreachequilibrium.(d) h ilib i

Q

Sample Problem

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Calculating Equilibrium ConcentrationsWhen Qc Kc, we can calculate exactly how much additional reactant or product will form in order to reach equilibrium.

PCl5 (g) PCl3 (g) + Cl2 (g) Kc = 0.030

Consider this equilibrium:

A t i iti ll t i i l PCl t t ti f 0 100 M hA system initially containing only PCl5 at a concentration of 0.100 M has a Qc = 0, which is less than 0.030. What are the equilibrium concentrations of the three gases?

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Set up the problem in the following way:Setuptheprobleminthefollowingway:

PCl5 (g) PCl3 (g) + Cl2 (g) Kc = 0.030

Initial:0.1000Change:x+x+xEquilibrium:0.10 xx x

Theequationtobesolvedis:

q

Rearrange togivequadraticequation:

Solvewith quadraticformula:

Reject negative root!

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Rejectnegativeroot!x=0.042M

The concentrations at equilibrium are:

PCl5 PCl3 Cl20 10 M 0 042 M 0 042 M 0 042 M0.10 M 0.042 M

= 0.058 M0.042 M 0.042 M

Check for yourself that this gives Kc!

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Sample ProblemThe equilibrium constant for the decomposition of solid NH4Cl is Kp = 0.01

atm at a certain high temperature. Calculate the equilibrium vapor pressures of ammonia gas and HCl gas starting from pure NH4Cl (s).pressures of ammonia gas and HCl gas starting from pure NH4Cl (s).

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Sample Problem

First calculate Q, compare with K: Q = (0.012)^2/0.063 = 0.0023

Q > K, reaction goes to reactants!

Change (M) +x -2xEquilibrium (M) 0.063 + x 0.012 2x

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(0.012 2x)2 = 0.0011x + 6.93 x 10-5

4x2 - 0.048x + 0.000144 = 0.0011x + 6.93 x 10-5

4x2 - 0.0491x + 0.0000747 = 0

x = 0.0105 , x = 0.00178

Change (M) +x -2x x = 0.0105 gives nonsense;reject it

Equilibrium (M) 0.063 + x 0.012 2x reject it

At equilibrium: [Br] = 0.012 2(0.00178) = 0.00844 M

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[Br2] = 0.063 + 0.00178 = 0.0648 M

Making Approximations Wedontalwaysneedtosolvethequadratic,orhigher,equationsthat

appearinthesetypesofequilibriumproblems.

Supposewehadthisequilibrium:

2SO3 (g) 2SO2 (g)+O2 (g)Kc =1.6 1010

Initial:0.1000Ch 2 2Change:2x+2x+xEq.:0.10 2x2x x

Theequationtosolvewouldbe:

Thisisacubicequation!

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qTrulyfrightening.

Making Approximations

Notethattheequilibriumconstantisverysmall. Whenequilibriumisreached,2x willbeverysmallcomparedto0.100,sowecanignoreit.

Theresultingapproximate(butquiteaccurate)equationtosolveis

h hWhichgives:

x=7.4x105 M

2xismuchsmallerthan0.10,so0.10 2x0.10andourapproximationisvalid!

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Th t ti t ilib iTheconcentrationsatequilibriumare:

SO3 SO2 O20.10M 2(7.4x105)

0.10M

2(7.4x105)=1.5x104 M

7.4x105 M

Withthesenumbers,Kc =1.7x1010 ,veryclosetotherealvalue,

whichagainconfirmsthatourapproximationisokay.

The 5% Rule: if the change in concentration is less than 5% of the initial concentration, we can safely ignore it.5% of the initial concentration, we can safely ignore it.

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