ch 6_equilibrium-1 (1)

7/25/2019 CH 6_Equilibrium-1 (1)

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Chapter 6: Chemical Equilibrium

6.1 The Equilibrium Condition

6.2 The E uilibrium Constant

6.3 Equilibrium Expressions Involving Pressures

6.4 The Concept of Activity

6.5 Heterogeneous Equilibria

6.6 Applications of the Equilibrium Constant

. o v ng qu r um ro ems

6.8 Le Chtelier's Principle6.9 Equilibria Involving Real Gasses

Stoichiometry vs. Equilibrium Previous stoichiometric calculations have assumed

complete reactions

Limitin rea ent is used u

Maximum amount of product is formed

2NO2 N2O4

Most reactions, however, do not

reddish-brown colorless

reach completion.

They reach chemical equilibrium NO2

NO2andN2O4


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Chemical Equilibrium

Chemical reactions are not a one-way street

2 NO2 (g) N2O4 (g)

Every reaction has a forward and a reverse!

reaction.

Equilibrium calculations look at the forward andreverse reactions.

Reaching Equilibrium on theMacroscopic and Molecular Levels

DECOMPOSITION

N2O4 (g) 2 NO2 (g)Colorless Brown


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2 NO2 (g) N2O4 (g)

Two reactions are ha enin :


Chemical equilibrium is the state at which theconcentrations of all reactants and products remain

Dimerization of nitrogen dioxideDecomposition of dinitrogen tetroxide

e same.

At chemical equilibrium the rate of formation and therate of decomposition are equal.

EQUILIBRIUMOver time, the reactionlooks like this:

N2O4(g)

2 NO2(g)

e amoun o 2 4decreases

The amount of NO2increases

Eventually theconcentrations of

changing.

Equilibrium, however,is a dynamicprocess.

TIME


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Characteristics of Chemical

Equilibrium States Reaching equilibrium requires that reactionsoccur.

Chemical reactions occur via collisions of reactingmolecules. The energy released during collisions breaksbonds in reactant molecules, allowing the atoms torearrange to form a product.

,macroscopic evidence of further change.

Reached through dynamic balance offorward and reverse reaction rates.

Chemical Equilibrium Equilibrium is reached when the forward and reverse

reactions are balanced, so there is no net change in theconcentrations of reactants or roducts.

Where this balancing point occurs can be different for eachreaction:

2H2 (g)+ O2 (g) 2H2O (g) (leaves few reactants)

Equilibrium lies far to the right

2CaO (s) 2Ca (s)+ O2 (g) (~no solid Ca forms)

Equilibrium lies far to the left


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SO, how do we know if the equilibrium is

to the left?

in the middle?

Where will equilibrium be?

WE DO EXPERIMENTS!!

And Later Thermodynamics!!

Initial and Equilibrium Concentrations for the

N2O4-NO2 System at 100C

Initial Equilibrium Ratio

[N2O4] [N2O4] [N2O4][NO2] [NO2] [NO2]2

0.1000 0.0000 0.0491 0.1018 0.211

0.0000 0.1000 0.0185 0.0627 0.212

0.0500 0.0500 0.0332 0.0837 0.211=K

0.0750 0.0250 0.0411 0.0930 0.210

Experiments carried out at a certain temperature, but withdifferent initial concentrations, yield the same value for the

EQUILIBRIUM CONSTANT, K.


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The Equilibrium Constant

This is a constant at agiven temperature the equilibrium

constant



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The Equilibrium Constant:N

2

+ 3 H2

2 NH3

The Equilibrium Constant:

N2 + 3 H2 2 NH3

Varying the starting concentrations gives differentequilibrium concentrations, but K is the same in each case

(6.02 x 10-2)


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Equilibrium Constant - UNITS

We actuallyexpress K in terms of activity of reactants and

products:

Activity of

This definition of the e uilibrium constant re laces the

product E

definition provided by Zumdahl in Section 6.2.

This definition means Kis unitless, as it should be! (Zumdahlonly uses units in Ch. 6 and calls them apparent units.)

Activities To relate concentration (or pressure) of a reactant or

product to its activity, we divide by a reference

For concentration, the reference is 1 M

aHCl(aq) = [HCl]/1M

For pressure, the reference is 1 atm

aHe(g) = PHe/1 atm

ure so s an pure qu s ave an ac v y oaHg(l) = 1 aFe(s) = 1


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The Equilibrium Constant (contd.)

Example from Zumdahl (done with activities)

At equilibrium [NH3] = 3.1 x 10-2 M, [N2] = 8.5 x 10

-1 M, and[H2] = 3.1 x 10

-3 M. What is the equilibrium constant?

We write Kc to indicate that we are using concentrations.

The Equilibrium Constant (contd.)

Notice that dividing by a reference state makes Kunitless.


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Reaching Equilibrium from DifferentStarting Points

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

[CO][H2]2

Different initialconcentrations

eren naconcentrations

SAME ratio ofproducts / reactants

The Equilibrium Constant A reaction can have manyequilibrium constants

ou need to s ecif the tem erature of the reaction

At that temperature, the equilibrium constant isconstant, regardless of initial concentrations.

The set of equilibrium concentrations (based on yourinitial concentration) is called the equilibriumposition.

nce ere are an n n e num er o n aconcentrations, there are also an infinite numberof equilibriumpositions for each constant!


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2H2 (g)+ O2 (g) 2H2O (g) (leaves few reactants)

Equilibrium lies far to the right

What can Ktell us about this reaction?Equilibrium concentration of reactants is lowEquilibrium concentration of products is highSince Kis t e ratio o pro ucts over reactants

K must be large

The Equilibrium Constant2CaO (s) 2Ca (s)+ O2 (g) (~no solid Ca forms)

Equilibrium lies far to the left

What can Ktell us about this reaction?

Equilibrium concentration of reactants is high

Since Kis the ratio of products over reactants

K must be small


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Equilibrium Constant:

Equilibrium:

Constant vs. Position ra o o pro uc s o reac an s

depends only on TEMPERATURE

versus

Chemical equilibrium (the equilibrium position)

a dynamic situation

can be affected by temperature, pressure, and/orinitial concentration of reactants and products.

Solving Equilibrium Problems At this point, you should be able to do these things

involvin the e uilibrium constant:

Write an equilibrium expression

Use the magnitude of Kto predict if theequilibrium lies to the left or right (favors productsor reactants)

Given equilibrium concentrations, calculate K


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Sample ProblemsProblem: Write the equilibrium constant expression for

the following reactions:a. N2 (g) + O2 (g) 2NO (g)

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

c. N2O4 (g) 2NO2 (g)

Sample ProblemsProblem: Predict where the equilibrium lies for thesereactions

a. N2 (g) + O2 (g) 2NO (g); K= 1 x 10-30

b. 2CO (g) + O2 (g) 2CO2 (g); K= 2.2 x 1022

c. N2O4 (g) 2NO2 (g); K= 0.211

Plan: Remember that large Kfavors products, small K

favors reactants.


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Sample Problems

Problem: Predict where the equilibrium lies for thesereactions

-30. 2 2

K is very small, the reaction favors reactants

b. 2CO (g) + O2 (g) 2CO2 (g); K= 2.2 x 1022

K is very large, the reaction favors products

c. N2O4 (g) 2NO2 (g); K= 0.211

K is intermediate (somewhat close to one), thereaction has significant amounts of reactants andproducts at equilibrium

Sample ProblemsProblem: At 360C, the reaction

COCl CO + Cl

has equilibrium concentrations of:

[COCl2] = 0.480 M, [CO] = 2.0 x 10-2 M, [Cl2] = 2.0 x 10

-2 M

What is the value of Kat 360C for this reaction?

Plan: Write the equilibrium constant expression for thereaction and then substitute in the equilibrium concentrationsto solve for K.


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Sample Problems

Problem: At 360C, the reaction

COCl CO + Cl

has equilibrium concentrations of:

[COCl2] = 0.480 M, [CO] = 2.0 x 10-2 M, [Cl2] = 2.0 x 10

-2 M

What is the value of Kat 360C for this reaction?

K= [CO][Cl2] = [2.0 x 10-2][2.0 x 10-2]

[COCl2] [0.480]

K= 8.3 x 10-4

Determining Equilibrium Concentrations from K

Methane can be made by reacting carbon disulfide withhydrogen gas:

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

o . .reaction mixture in a 5.0 L flask is: 0.250 mol CS2, 1.10mol of H2, and 0.45 mol of H2S. How many moles ofmethane are formed?

Plan:

Calculate theequilibrium

concentrations

Use moles and V

Concentration ofmethane

Use K and equilibriumexpression

Convertmolarity to

moles


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Determining Equilibrium Concentrations from K

[CS2] =

0.250 mol

5.0 L= 0.05 M

H S =0.450 mol

= 0.09 M

[H2] =

1.10 mol

5.0 L= 0.22 M

[CH4] = ?

[CH4][H2S]2

[CS2][H2]4

K = = 27.8

.

(27.8)(0.05)(0.22)4= 0.40 M

K[CS2][H2]4

2[CH4] = =

moles CH4 = 0.40 M x 5.0 L = 2.0 moles

.2

CS2(g) + 3 O2(g) CO2(g) + 2 SO2(g)

The equilibrium expression for a reaction written in reverse isthe reciprocal of that for the original expression:

Using and Rearranging K

[CO2][SO2]2

3K =

If the original reaction is multiplied by a factor n, the newequilibrium constant is the original raised to the power n.

CO2(g) + 2 SO2(g) CS2(g) + 3 O2(g)

2 2

[CS2][O2]3

[CO2][SO2]2

K = =1

K

2 CS2(g) + 6 O2(g) 2 CO2(g) + 4 SO2(g)

[CO2]2[SO2]

4

[CS2]2[O2]

6K = = K2


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The following equilibrium concentrations were observed for thereaction between CO and H2 to form CH4 and H2O at 927

oC.

CO (g) + 3 H2 (g) CH4 (g) + H2O (g)

Example

[CO] = 0.613 mol/L [CH4] = 0.387 mol/L[H2] = 1.839 mol/L [H2O] = 0.387 mol/L

a) Calculate the value of K at 927oC for this reaction.

b) Calculate the value of the equilibrium constant at 927oC for:H2O (g) + CH4 (g) CO (g) + 3 H2 (g)

c) Calculate the value of the equilibrium constant at 927oC for:1/3 CO (g) + H2 (g) 1/3 CH4 (g) + 1/3 H2O (g)

a) Calculate the value of K at 927oC for this reaction.

Example

K = =[CO] [H2]

3

[CH4] [H2O] (0.387) (0.387)

(0.613)(1.839)3= 0.0393


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Example contd.

b) Calculate the value of the equilibrium constant at 927oC for:

H2O (g) + CH4 (g) CO (g) + 3 H2 (g)

[CO][H2]3 (0.613) (1.839)3

.[H2O][CH4] (0.387) (0.387)

1K

K = = = 25.451

0.0393

c) Calculate the value of the equilibrium constant at 927oC for:

Shortcut if K is known: K is just the reciprocal of K from part a:

1/3 CO (g) + H2 (g) 1/3 CH4 (g) + 1/3 H2O (g)

K = =[CO]1/3 [H2]

[H2O]1/3[CH4]1/3 (0.387)1/3 (0.387)1/3

(0.613)1/3 (1.839)

Shortcut if K is known: K =(K)1/3 = (0.0393)1/3 = 0.3399

SummarySome Characteristics of the Equilibrium Expression

The e uilibrium ex ression for a reaction written in reverse isthe reciprocal of that for the original reaction.

Knew = 1/Koriginal

When the balanced equation for a reaction is multiplied by afactor n, the equilibrium expression for the new reaction is

the original expression raised to the nth power.

Knew = (Koriginal)n


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Types of Equilibrium Constants

, , .

Solubility product (precipitation reactions; Ch. 8)

Stability constant (complexation reactions; Ch. 8)

Standard potentials (redox reactions; Ch. 11)

Equilibrium arises through dynamic balancebetween forward and reverse reactions

Forward: 2NO2 (g) NO3 (g) + NO (g)k1

Reverse: NO3 (g) + NO (g) 2NO2 (g)k-1

Forward rate = k1[NO2][NO2] = k1[NO2]2

Reverse rate = k-1[NO3][NO]

k1 and k-1 reflect probabilities that one collision leads toreaction


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Kinetic Approach to Equilibrium

k1

NO2

2 k1 NO NO3

1

k1

3NO

2 2

K

The RATE of forward reaction decreases as the reactionprogresses since the initial concentration of reactantsdecreases over time.

A value of K >> 1 means that at equilibrium the system willconsist mainly of products

Extent of the Reaction

A value of K


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The Reaction Quotient, QOR

How do we know the reaction is ate uilibrium?

Consider the reaction:

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

The reaction quotient, Q is defined as:

c d

Qhas same form as K, but the concentrations are the actualconcentrations at any time (t) rather than the concentrationsafter equilibrium is reached.

[A]t

a [B]t

bQ =

The equilibrium constant describes equilibrium

Equilibrium Constant, K

vs.Reaction Quotient, Q

concentrations The reaction quotient uses initial concentrations to

determine in which direction a reaction will proceed.

if QK , the system will consume products(reverse reaction; shifts to the left)


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Reaction Direction and theRelative Sizes of Q and K

Excess reactantsinitially

Excess productsinitially

N2O4 (g)

2NO2 (g)

Q = [NO2]2

[N2O4]

TIME


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Ways of Expressing Q and K values

Form of Chemical Equation Form of Q Value of K

Forward reaction: A B Qfwd= Kfwd =[B]

A

[B]eq

Reversed reaction: B A Q = = K=

Reaction as sum of two steps:

eq

1 [A]

Qfwd [B]1

Kfwd

(1) A CQoverall = Q1 x Q2 Koverall = K1 x K2

[A]

[C]Q1 =

(2) C B

Reaction with pure solid or liquid component:A(s) B Q = [B] K = [B]

[C]

[B]Q2 =

Predicting Reaction Direction using Q

Consider the following reaction:

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

1.00 mol CH4, 1.00 mol CS2, 2.00 mol H2S and 2.00 mol H2are mixed in a 250 mL vessel at 960oC. If K = 0.036 at thistemperature, calculate Q and figure out in which directionwill the reaction proceed in order to reach equilibrium?

Plan:

Calculate actualconcentrations

Use moles andvessel volume

Calculate Q Compare to K


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[CH4]0= 1.00 mol/0.250 L = 4.00 M

[H2S]0 = 2.00 mol/0.250 L = 8.00 M

= =K = 0.036

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

2 0 . . .

[H2]0 = 2.00 mol/0.250 L = 8.00 M

Calculate the value of Q:

[CS2]o[H2]o4

2

(4.00 M)(8.00 M)4

2Q = = = 64

Comparing Q and K: Q > K, so the reaction goes to the left.

[reactants] increase and [products] decrease

4 o 2 o . .

N2 (g) + 3 H2 (g) 2 NH3 (g)For the synthesis of ammonia at 500oC, the equilibrium constant is6.0 x 10-2. Predict the direction in which the system willshift to reach equilibrium in each of the following cases.a) [NH3]0 = 1.0 x 10

-3 M [N2]0= 1.0 x 10-5 M [H2]0=2.0 x 10

-3 M

= -4 = -5 = -13 0 . 2 0 . 2 0 .

c) [NH3]0 = 1.0 x 10-4 M [N2]0= 5.0 M [H2]0= 1.0 x 10

-2 M

Solution:

a) First we calculate Q:

Q = =[NH3]o

2

[N2]o[H2]o3

(1.0 x 10-3 )2

(1.0 x 10-5 )(2.0 x 10-3 )3

= . x

Since K = 6.0 x 10-2 L2/mol2, Qis much greater than K. For thesystem to attain equilibrium, [products] must decrease and[reactants] must increase.

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


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Q = =

= -2

Example contd.b) We calculate the value of Q:

[NH3]o2

[N2]o[H2]o3

(2.00 x 10-4

)2

(1.50 x 10-5 ) (3.54 x 10-1 )3)

K = 6.0 x 10-2 L2/mol2

Q = =

.

In this case Q = K, so the system is at equilibrium..

c) The value of Q is:

[NH3]o2

N H 3

(1.0 x 10-4 )2

5.0 1.0 x 10-2 3

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

= 2.0 x 10-3

Here Q is less than K, so the system will shift to the right, attainingequilibrium by increasing [product] and decreasing [reactants]:

More ammonia!

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

Solving Equilibrium ProblemsGiven the reaction:

2 2

And initial concentrations:

[H2]0 = 1.00 M

[F2]0 = 2.00 M

What will the equilibrium concentrations be forreactants and products?


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First: Make sure the reaction is balanced!

H ( ) + F ( ) 2 HF ( )

Solving Equilibrium Problems

Second: Construct the equilibrium expression:

2

2

HF

1 M

2 2.

H F

1 M 1 M

C

Third: Using the stoichiometry of the reaction, deriveexpressions for concentrations at equilibrium.


ou can n roug s process y cons er ng einitial, change in, and equilibrium concentrations. Easyto do using an ICE table:

Conc. (M) H2 (g) F2 (g) 2 HF (g)

H2 (g) + F2 (g) 2 HF (g)

Initial 1.00 2.00 0

Change -x -x +2x

Equilibrium 1.00 - x 2.00 - x 2x


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-x

What is this x of which you speak?

PCl5(g) PCl3(g) + Cl2(g)

x x

Initially only PCl5 present.

As reaction proceeds, for each PCl5 that decomposes,

one PCl3 and one Cl2 is formed (stoichiometry).

x represents the change in composition between theinitial conditions and equilibrium.

Third: Using the stoichiometry of the reaction, deriveexpressions for concentrations at equilibrium.


ou can n roug s process y cons er ng einitial, change in, and equilibrium concentrations. Easyto do using an ICE table:

Conc. (M) H2 (g) F2 (g) 2 HF (g)

H2 (g) + F2 (g) 2 HF (g)

Initial 1.00 2.00 0

Change -x -x +2x



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Fourth: Substitute equilibrium concentrations into theequilibrium expression and solve.

2


2

2 2

HF

1 M1.15 10

H F

1 M 1 M

CK

Species Equilib. Conc. (M)

H2 1.00 - xF2 2.00 - x

HF 2x

By solve we mean solvefor x which can use todetermine equilib.concentrations!

Continuing with the example:

2

2

2 Mx


2 1 M

1.15 101.00 M 2.00 M 1.00 2.00

1 M 1 M

x

x x x x

22

1.15 10 1.00 2.00 2x x x

We need to solve this last equation forx.

2 2 2 21.11 10 3.45 10 2.30 10 0x x


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The last expression is a quadratic equation of the form:

2 + + =


The solution to this equation can be determined usingthe quadratic formula:

2 2 2 21.11 10 3.45 10 2.30 10 0x x

24

2

b b ac

x a

Applying the quadratic formula to our example:

a b c


2 2 2 21.11 10 3.45 10 2.30 10 0x x

22 2 2 2

2

2

3.45 10 3.45 10 4 1.11 10 2.30 104

2 2 1.11 10

2.14,0.968

b b acx

a

x

Which root is correct? The answer is the correct root isthe one that makes physical sense. Recall that since[H2]eq = 1.00 x. Only x = 0.968 makes physical sense.


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Last: Using expression forx, determine the equilibriumconcentrations.0.968x


Species Equilib. Conc. (M)

H2 1.00 - x 3.2 x 10-2 M

F2 2.00 - x 1.03 M

HF 2x 1.94 M

Which species are present to the greatest extent atequilibrium? What does that tell us about the extent towhich the reaction proceeds towards products?

Zumdahls Steps:

Write the balanced equation for the reaction.


Write the equilibrium expression.

List the initial concentrations.

Define the change in concentrations using the stoichiometry ofthe reaction.

Using the initial concentrations and change, write down theexpressions for equilibrium concentrations.

,and solve.

Using the physically-reasonable solution, determine theequilibrium concentrations.


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Calculating Equilibrium Partial PressuresUsing Quadratic Equation

The reaction between nitrogen and oxygen to form nitrogenmonoxide proceeds according to the following reaction:

N2 (g) + O2 (g) 2 NO(g) K = 4.1x10-4 at 2000K

Initially 0.500 moles of N2 and 0.860 moles of O2 are put into a2.00 L vessel. Calculate the concentrations of all the species atequilibrium.

n a concen ra on o 2 = . mo . = .

Initial concentration of O2

= 0.860 mol/2.00 L = 0.430 M

Initial concentration of NO = 0

Q = 0, therefore the reaction proceeds to the right

Construct the reaction table.

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


Conc. (M) N2 (g) O2 (g) 2NO (g)

Initial 0.250 0.430 0

Change -x -x +2x



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Substituting the equilibrium concentrations from the I.C.E. table intothe equilibrium expression:


[N2][O2]K = = = 4.10 x 10-4

x

(0.250-x)(0.430-x)

This expression simplifies to:

(4.00)x2 + (2.79 x 10-4)x (4.41 x 10-5) = 0

x = -3.35 x 10-3

and x = 3.28 x 10-3

Since only the positive root leads to all positive concentrations, weignore the negative root.

Calculating Equilibrium Partial Pressures

Using Quadratic Equation

Calculating equilibrium concentrations:[NO] = 2x = 6.56x10-3 M[N2] = 0.250 M x = 0.247 M[O2] = 0.430 M x = 0.427 M

ec :

K = = = 4.10x10-4[NO]2

[N2][O2]

(6.56x10-3)2

(0.247)(0.427)


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Solving equilibrium problems withsimplifying assumptions

Phosgene decomposes into CO and Cl2 when heated according.

Calculate the concentration of all species at equilibrium if

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

K = 8.3x10-4 at 360oC

5.00 moles of phosgene are placed into a 10.0 L flask.

[COCl2] = = 0.500 M5.00 mol

10.0 L

COCl2 (g) CO (g) + Cl2 (g)-4

Solving equilibrium problems with

simplifying assumptions

.

Conc. (M) COCl2 (g) CO (g) Cl2 (g)

Initial 0.500 0 0

Change -x +x +x

Equilibrium 0.500 - x x x


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K = = = 8.3 x 10-4[CO][Cl2]

[COCl2]

x2

0.500 - x

Solving equilibrium problems withsimplifying assumptions

For K to be so small, (0.500 x) >>x2, so 0.500-x 0.500

= 8.3 x 10-4x2

0.500 - x

x2

0.500

x = 2.037 x 10-2 2.04 x 10-2

This ASSUMPTION can ONLY be made for SMALL VALUES of K!!Use the 5% rule if the error is less than 5%, the assumption is valid.

(if x < 5% of[ ]o, then [ ] < 5%...assumption is valid)

Check to see if this assumption works 0.500 2.04 x 10-2 = 0.4796

Approximation gives error of ~4% (0.500/0.4796 = 95.92%).

Calculating K from Concentration DataProblem: Hydrogen iodide decomposes at moderate temperatures by thereaction:

When 4.00 mol HI were placed in a 5.00 L vessel at 458oC, the equilibriummixture was found to contain 0.442 mol I . What is the value of K ?

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

Plan: First we calculate the molar concentrations of HI and I2, and thensubstitute them into the equilibrium expression to find the value of Kc.

Solution:

Starting conc. of HI = = 0.800 M4.00 mol5.00 L

0.442 molqu r um conc. o 2 = = .5.00 L

Conc. (M) 2HI (g) H2 (g) I2 (g)

Starting 0.800 0 0Change - 2x +x +xEquilibrium 0.800 - 2x x x = 0.0884


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Calculating K from Concentration Data

[HI] = (0.800 (2 * 0.0884)) M = 0.623 M

[I2] = x = [H2] = 0.0884 M

Kc = = = 0.0201[H2] [I2]

[HI]2

(0.0884)(0.0884)

(0.623)2

The e uilibrium constant for the decom osition of h dro en iodide at458oC is only 0.0201, meaning that the decomposition does notproceed very far under these temperature conditions.

Equilibrium We talked about Q, the equilibrium quotient, as a

sna shot of where the reaction is at an time t.

Three possibilities: Q > K, Q = K, Q < K

[C]tc[D]t

d

[A]ta[B]t

bQ =[C]eq

c[D]eq

d

[A]eqa[B]eq

bK =

Today well explore what happens if we push areaction out of equilibrium by adding or removingreactants or products.


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Factors that control the position of chemical equilibriaare:

Equilibrium Position

Concentration

Pressure

Temperature

: ca a ys s n uence ra e, no pos on o equ r um

The Effect of a Change in Concentration

Given an equilibrium equation such as :

CH4 (g) + NH3 (g) HCN (g) + 3 H2 (g)

Add NH3 Forces equilibrium toproduce more product.

What happens to Q when we add NH3?

3 [NH3] increases

[CH4]t [NH3]tQ = Q < K, so more

products must form toreturn to equilibrium


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The Effect of a Change inConcentration

CH4 (g) + NH3 (g) HCN (g) + 3 H2 (g)

Forces the reaction equilibrium to go backto the left and produce more of the reactants.

Remove NH3

What happens to Q when we remove NH3?

3 [NH ] decreasest 2 t

[CH4]t [NH3]tQ = Q then increases

Q > K, so morereactants must form toreturn to equilibrium


CH4 (g) + NH3 (g) HCN (g) + 3 H2 (g)

Add H2Forces equilibrium to gotoward the reactants.

CH ( ) + NH ( ) HCN ( ) + 3 H ( )

Remove HCNForces equilibrium to make moreproduct and replace the lost HCN.


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Le Chteliers Principle

If a change in conditions (a stress) is imposed on a systemat equilibrium, the equilibrium position will shift in a directionthat tends to reduce that change in conditions.

For example, in the reaction above:

Add A or B reaction will make more product

Remove A or B reaction will form more reactants

A + B C + D + Energy

Add C or D reaction will form more reactants

Remove C or D reaction will shift to form more products

If the reaction is heated you will get more reactants

If the reaction is cooled you will get more products

The Effect of a Change inConcentration

a gaseous reac an or pro uc s a e o asystem at equilibrium, the system will shift in adirection to reduce the concentration of theadded component.

If a gaseous reactant or product is removed from a

system at equi i rium, t e system wi s i t in adirection to increase the concentration of theremoved component.


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Consider the following reaction:

2 H S + O 2 S s + 2 H O


What happens to:

(a) [H2O] if O2 is added?

The reaction proceeds to the right so [H2O] increases.

(b) [H2S] if O2 is added?Some H2S reacts with the added O2 to move the reaction to

the right, so [H2S] decreases.

2 H2S (g) + O2 (g) 2 S (s) + 2 H2O (g)


(c) [O2] if H2S is removed?

The reaction proceeds to the left to re-form H2S, more O2is formed as well, [O2] increases.

(d) [H2S] if S(s) is added?

S is a solid, so its activity does not change. Thus, [H2S] isunchanged.


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The Effect of a Change inPressure (Volume)

are nearly incompressible). For gases, pressure changes canoccur in three ways: Change the concentration of a gaseous component Add an inert gas (one that does not take part in the

reaction) Chan e the volume of the reaction vessel

When you change the volume of a system ate uilibrium the e uilibrium osition shifts to reduce the

The Effect of a Change in Pressure(Volume)

effect of the change. If the volume decreases (increased partial

pressure), the total number of gas moleculesdecreases (reaction shifts to side with fewer molesof gas).

If the volume increases (decreased partial

pressure), the total number of gas moleculesincreases (reaction shifts to side with more moles ofgas).


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Pressure Change

2 NO2 (g)N2O4 (g)

Brown Colorless

We can summarizepressure changes asfollows:

Increase pressureequilibrium shiftstoward side with fewermoles of as

Reduce pressure

equilibrium shiftstoward side with moremoles of gas

At equilibriumPressure is increased

rapidly (concentration ofboth gases increases)

Equilibrium is re-

established, shifts tofewer moles of gas

We can add or remove a gaseous reactant or product!

but this is just the same as changing the concentration ofa s ecies.

Cant we change P in other ways?

We can add a gas that is not involved in the reaction!

adding a chemically inert gas to the reaction mixture willchange the TOTAL pressure, but will have no effect on thechemistry and, therefore, no effect on the equilibrium.

So the onl P chan e that has an effect on e uilibriumposition is a change in external P, which you get bychanging the volume of the reaction system.


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How would you change the total pressure or volume in thefollowing reactions to increase the yield of the products:

a CaCO s CaO s + CO

The Effect of a Change in Pressure (Volume)

The only gas is the product CO2. To move the reaction to the right,increase the volume.

(b) S(s) + 3 F2 (g) SF6 (g)

With 3 moles of gas on the left and only one on the right, weincrease the pressure (decrease volume) to form more SF6.

(c) Cl2(g) + I2(g) 2 ICl (g)

The number of moles of gas is the same on both sides of theequation, so a change in pressure or volume will have noeffect on the equilibrium.

The Effect of a Change in Temperature

Only temperature changes will alter the equilibrium constant, and thatis why we always specify the temperature when giving the value of K.

The best way to look at temperature effects is to realize thattemperature is a component of the equation, the same as a reactant orproduct.

O2 (g) + 2 H2 (g) 2 H2O (g) + Energy = Exothermic

Energy + 2 H2O (g) 2 H2 (g) + O2 (g) = Endothermic

A temperature increase favors the endothermic direction and atemperature decrease favors the exothermic direction.


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Why would a reaction be exothermic (release heat)? Chemical bonds of the products are more stable than

those of the reactants.

Changes in Temperature (contd.)

During solvation (Ch. 4), the potential energy of the ionsdissolved in solution is less than in the solid form.

Why would a reaction be endothermic (absorb heat)?

em ca on s o pro uc s are ess s a e an oseof the reactants.

Changes in Temperature (contd.)Heat + N2O4 (g) 2 NO2 (g)

o or ess 2 4 gasdecomposes to formreddish-brown NO2.

This reaction involvesthe breaking of achemical bond.

The breaking of the chemical bond required energy (in theform of heat) to be added to the system.

Therefore, this reaction is endothermic (with heat on thereactant side).


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Changes in Temperature (contd.)

What if we increase thetemperature at which the

Heat + N2O4 (g) 2 NO2 (g)

reaction occurs.

Increase in T means anincrease in energy available tobreak the chemical bond.

Equilibrium shifts to products!

We can summarize temperature effects as follows:

For an endothermic reaction, an increase in T shifts equilibriumtowards products

For an exothermic reaction, an increase in T shifts equilibriumtowards reactants

Exothermic vs. Endothermic ReactionsExothermic Endothermic

Role of energy in the Released as a Absorbed as areaction: product reactant

Effect of increasingtemperature:

Equilibrium shiftstoward thereactants

Equilibrium shiftstoward theproducts

Change in K because Decreases Increases

Change in concentration at constant T changes theequilibrium position, but not the value of K (ratio of

products to reactants).


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How does an increase in temperature affect the equilibriumconcentration ofthe substance in bold and Kfor the followingreactions:

Example

(a) CaO(s) + H2O (l) Ca(OH)2 (aq) + energy

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

(c) SO2 (g) + energy S(s) + O2(g)

How does an increase in temperature affect the equilibriumconcentration ofthe substance in bold and Kfor the followingreactions:

Example

(a) CaO(s) + H2O (l) Ca(OH)2 (aq) + energy

Increasing T shifts the system to the left, where it absorbs energy.[Ca(OH)2] and K decrease.

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

Increasing T shifts the system to the right, where it absorbs

energy. [CO2] an K ncrease.(c) SO2 (g) + energy S(s) + O2(g)

Increasing T shifts the system to the right, where it absorbsenergy. [SO2] will decrease and K increases.


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Effect of Various Stresses on anEquilibrium System

Stress Net Direction of ReactionEffect on Value of

K

Concentration: Increase [reactant] Decrease [reactant]

Toward formation of products Toward formation of reactants

None None

Pressure (constant T): Decrease P (raise V) Increase P (lower V)

Toward larger amount of gas Toward smaller amount of gas

None None

Temperature: ncrease owar pro uc s energy s a

reactant (endothermic) Toward reactants if energy is a

product (exothermic)

ncreases wincreasing T

K decreases withincreasing T

Catalyst added: None Rates of forward and reverse

reactions increase equally None

ch 6_equilibrium-1 (1)

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