Chapter 31 Chemical Equilibrium - ctlsfasu · 4 Chemical Equilibrium • Therefore, many reactions do not go to completion but rather form a mixture of products and unreacted reactants,

1

Chapter 31

Chemical

Equilibrium

Copyright (c) 2011 by Michael A. Janusa, PhD. All rights reserved.

2

Many chemical reactions do not completely convert

reactants to products. Stop somewhere between no rxn and

complete rxn. A + B C + D

some left some formed

reversible (both directions)

Chemical Equilibrium

Previously we have assumed that chemical

reactions results in complete conversion of

reactants to products:

A + B C + D

No A or B remaining or possibly an excess of A or B but

not both and eventually reaction stops.

exchange, constant conc.,

Ratef = Rater “equilibrium”

A + B C + D

A + B C + D

4

Chemical Equilibrium

• Therefore, many reactions do not go to

completion but rather form a mixture of

products and unreacted reactants, in a

dynamic equilibrium.

– A dynamic equilibrium consists of a forward

reaction, in which substances react to give products,

and a reverse reaction, in which products react to

give the original reactants.

– Chemical equilibrium is the state reached by a

reaction mixture when the rates of the forward

and reverse reactions have become equal.

5

Graphically we can represent this A + B C + D

The concentrations and reaction rate (less collisions, less component)

of A and B decreases over time as the concentrations and reaction rate

of C and D increases (more collisions, more component) over time

until the rates are equal and the concentrations of each components

reaches a constant. This occurs at what we call equilibrium -- Rf = Rr.

If the rates are equal, then there must be a relationship to show this.

3H2 + CO CH4 + H2O

3H2 + CO CH4 + H2O rate decreases over time

CH4 + H2O 3H2 + CO rate increases over time

Rf=Rr

Ebbing, D. D.; Gammon, S. D. General

Chemistry, 8th ed., Houghton Mifflin, New

York, NY, 2005.

7

For the reverse reaction we have,

C + D A + B Rr = kr [C][D]

We know at equil that Rf = Rr ;therefore, we can set these two

expressions as equal

kf [A][B] = kr [C][D]

Rearrange to put constants on one side we get

]][[

]][[

BA

DC

k

kK

r

f

If we assume these reactions are elementary rxns (based on

collisions), we can write the rate laws directly from the

reaction:

A + B C + D Rf = kf [A][B]

8

Constant divided by constant just call a new constant K. This ratio is

given a special name and symbol called equilibrium constant K

relating to the equilibrium condition at a certain temperature (temp

dependent) for a particular reaction relating conc of each component.

This is basically a comparison between forward and reverse reaction

rates. At equilibrium, the ratio of conc of species must satisfy K.

]][[

]][[

BA

DC

k

kK

r

f

9

The Equilibrium Constant

• Every reversible system has its own “position of

equilibrium”- K- under any given set of conditions.

– The ratio of products produced to unreacted reactants

for any given reversible reaction remains constant under

constant conditions of pressure and temperature. If the

system is disturbed, the system will shift and all the

concentrations of the components will change until

equilibrium is re-established which occurs when the ratio of

the new concentrations equals "K". Different constant conc

but ratio same as before.

– The numerical value of this ratio is called the equilibrium

constant for the given reaction, K.

10

The Equilibrium Constant • The equilibrium-constant expression for a reaction is

obtained by multiplying the equil concentrations ( or partial

pressures) of products, dividing by the equil concentrations (or

partial pressures) of reactants, and raising each concentration to

a power equal to its coefficient in the balanced chemical

equation.

dDcC bBaA

ba

dc

BA

DCK

][][

][][b

B

a

A

d

D

c

Cp

PP

PPK

)()(

)()(

c

The molar concentration of a substance is denoted by writing its formula in

square brackets for aq solutions. For gases can put Pa - atm. As long as use M

or atm, K is unitless; liquids and solids = 1; setup same for all K’s (Ka, Kb, etc.)

Temp dependent; any changes, ratio will still equal K when equil established

11

31.1 The Equilibrium Constant, K

• The law of mass action states that the value of the

equilibrium constant expression K is constant for a

particular reaction at a given temperature, whenever

equilibrium concentrations are substituted.

• When at equil, conc ratio equals K and concs are

constant but if disturbed, values change until equil

reached, different constant conc but same ratio of K).

12

Obtaining Equilibrium Constants

for Reactions

• Equilibrium concentrations for a reaction

must be obtained experimentally and

then substituted into the equilibrium-

constant expression in order to

calculate Kc.

13


for Reactions

• Consider the reaction below

O(g)H (g)CH (g)H 3 )g(CO 242

Suppose we started with initial concentrations of

CO and H2 of 0.100 M and 0.300 M, respectively.

0.100 M 0.300 M 0 0

Obviously shift to right, decrease CO and H2 and

increase CH4 and water until equil established.

14

– When the system finally settled into equilibrium we

determined the equilibrium concentrations to be as

follows.

Reactants Products

[CO]eq = 0.0613 M

[H2] eq = 0.1839 M

[CH4] eq = 0.0387 M

[H2O] eq = 0.0387 M


for Reactions

O(g)H (g)CH (g)H 3 )g(CO 242

15

– If we substitute the equilibrium concentrations,

we obtain:

93.3)1839.0)(0613.0(

)0387.0)(0387.0(3cK


for Reactions

O(g)H (g)CH (g)H 3 )g(CO 242

eqeq

eqeq

cHCO

OHCHK

3

2

24

][][

][][

16

– Regardless of the initial concentrations (whether

they are reactants or products or mixture), the law

of mass action dictates that the reaction will

always settle into an equilibrium where the

equilibrium-constant expression equals Kc.


for Reactions

O(g)H (g)CH (g)H 3 )g(CO 242

17

As an example, let’s repeat the previous

experiment, only this time starting with initial

concentrations of products (note: if only products

to start shift left until equil established):

[CH4]initial = 0.2000 M and [H2O]initial = 0.2000 M

Obviously shift to left decrease CH4 and H2O and

increase CO and H2 until equil established.


for Reactions

O(g)H (g)CH (g)H 3 )g(CO 242

0 0 0.2000 M 0.2000 M

18

– We find that these initial concentrations result in

the following equilibrium concentrations.

Reactants Products

[CO] eq = 0.0990 M

[H2] eq = 0.2970 M

[CH4] eq = 0.1010 M

[H2O] eq = 0.1010 M


for Reactions

O(g)H (g)CH (g)H 3 )g(CO 242

19

– Substituting these values into the equilibrium-

constant expression, we obtain the same result.

93.3)2970.0)(0990.0(

)1010.0)(1010.0(3cK

– Whether we start with reactants or products at any

initial conc, the system establishes the same ratio

at same temperature.


for Reactions

O(g)H (g)CH (g)H 3 )g(CO 242

eqeq

eqeq

cHCO

OHCHK

3

2

24

][][

][][

20


• The equilibrium constant, K, is the value obtained

for the equilibrium-constant expression when

equilibrium concentrations (not just any conc but

equil conc) are substituted.

– A large K, K>1, indicates large concentrations of

products at equilibrium.

– A small K, K<1, indicates large concentrations of

unreacted reactants at equilibrium. HW 14

21


– Do same set up for all equil equations (future chapters), just

different subscript describing the reaction in question: i.e.

acid hydrolysis - acid + water - Ka, Kb, Kc, Kp, Ksp

– Kc is based on conc (M) and Kp is based on pressures (atm).

There is a difference between Kc and Kp and to jump

between these two, there is an equation to use. Let's show

how to jump between.

22

31.1.1 The Equilibrium Constant in

Terms of Pressure, Kp

• In discussing gas-phase equilibria, it is often

more convenient to express concentrations in

terms of partial pressures rather than

molarities.

– It can be seen from the ideal gas equation (PV =nRT) that

the partial pressure of a gas is proportional to its molarity

– P proportional to M - n/V ; therefore handled the same way.

MRTRTV

nP )(

23

The Equilibrium Constant, Kp

– Consider the reaction below.

O(g)H (g)CH (g)H 3 )g(CO 242

– The equilibrium-constant expression in terms of partial

pressures becomes (same way but partial pressures instead

of M):

3)(

2

24

HCO

OHCH

pKPP

PP3

2

24

][H ][

O][ ][

CO

HCHKc

24

The Equilibrium Constant, Kp

• In general if you need to jump between

K's, the numerical value of Kp differs

from that of Kc.

where ng is the sum of the moles of gaseous

products in a reaction minus the sum of the moles

of gaseous reactants.

ng

cp RTKK )(

25

A Problem to Consider

• Consider the reaction

– Kc for the reaction is 280 at 1000. K . Calculate Kp

for the reaction at this temperature.

(g)SO 2 )g(O)g(SO2 322

26


– We know that

From the equation we see that ng = 2 – 3 = -1. We

can simply substitute the given reaction temperature

and the value of R (0.0821 L.atm/mol.K) to obtain Kp.

• Consider the reaction at 1000. K and Kc

= 280

(g)SO 2 )g(O)g(SO2 322

ng

cp RTKK )(

27


– Since

3.4 K) 1000 0821.0( 280 -1

KmolatmL

pK

• Consider the reaction

(g)SO 2 )g(O)g(SO2 322

ng

cp RTKK )(

28


• Applying Stoichiometry to an Equilibrium Mixture

(basic setup for future problems).

– What is the composition of the equilibrium mixture

if it contains 0.080 mol NH3 at equilibrium?

– Suppose we place 1.000 mol N2 and 3.000 mol H2

in a reaction vessel at 450 oC and 10.0

atmospheres of pressure. The reaction is

(g)2NH )g(H3)g(N 322

322

23

c]H][N[

]NH[K

29

– This procedure for many types of problems, however in

this problem given equil quantity of NH3 ;therefore, figure

out rest.

• Using the information given, set up the

following table. (ratio works for atm, M, mols,

etc.) (g)2NH )g(H3 )g(N 322

Initial, no

Change,

n

Equil, neq

1.000 3.000 0

- - +

– The equilibrium amount of NH3 was given as 0.080 mol.

Therefore, 2x = 0.080 mol NH3 (x = 0.040 mol).

x 3x 2x

1.000 - x 3.000 - 3x 2x = 0.080 mol

30


• Using the information given, set up the

following table.

Equilibrium amount of N2 = 1.000 - 0.040 = 0.960 mol N2

Equilibrium amount of H2 = 3.000 - (3 x 0.040) = 2.880 mol H2

Equilibrium amount of NH3 = 2x = 0.080 mol NH3

(g)2NH )g(H3 )g(N 322

Starting 1.000 3.000 0 Change -x -3x +2x

Equilibrium 1.000 - x 3.000 - 3x 2x = 0.080 mol

HW 15

x = 0.040 mol

31

31.2 Changing the Chemical

Equation

• Similar to the method of combining reactions

that we saw using Hess’s law in

thermochemistry, we can combine equilibrium

reactions whose K values are known to

obtain K for the overall reaction.

– With Hess’s law, when we reversed reactions

(change sign) or multiplied them prior to adding them

together (mult by factor). We had to manipulate the

H’s values to reflect what we had done.

– The rules are a bit different for manipulating K.

32

2. If you multiply/divide each of the coefficients in an

equation by the same factor (2, 1/2, …), raise Kc to the

same power (2, 1/2, …). (known as coefficient rule, Kn)

3. When you finally combine (that is, add) the individual

equations together, take the product of the equilibrium

constants to obtain the overall K.(rule of multiple

equilibria, K1 x K2 x K3… = KT)

1. If you reverse a reaction, invert the value of K.

(reciprocal rule, 1/K)

Equilibrium Constant for the Sum

of Reactions

33

• For example, nitrogen and oxygen can

combine to form either NO(g) or N2O (g)

according to the following equilibria.

NO(g) )g(O2

1)g(N

2

122

O(g)N )g(O)g(N 2221

2

Kc = 6.4 x 10-16

Kc = 2.4 x 10-18

(1)

(2)

Kc = ?

– Using these two equations, we can obtain K for

the formation of NO(g) from N2O(g):

NO(g) 2 )g(O)g(ON 221

2overall

Equilibrium Constant for the Sum

of Reactions

reverse and multiply by factor of 1

same and multiply by factor of 2

34

NO(g) 2 )g(O)g(ON 221

2

NO(g) 2 )g(O)g(N 22Kc = (6.4 x 10-16)2 =

4.1 x 10-31

(1)

)g(O (g)N O(g)N 221

22 Kc = (2)

overall

18-10 4.2

1

131731 107.1)102.4()101.4()(overallKc

HW 16

NO(g) )g(O2

1)g(N

2

122

O(g)N )g(O)g(N 2221

2

Kc = 6.4 x 10-16

Kc = 2.4 x 10-18

= 4.2 x 1017

35

31.1.2 Heterogeneous Equilibria

• A heterogeneous equilibrium is an equilibrium that

involves reactants and products in more than one

phase. Up to now all our reactions have been

homogeneous - all gases or aqueous solutions.

– The equilibrium of a heterogeneous system is

unaffected by the amounts of pure solids or liquids

present, as long as some of each is present.

– The concentrations of pure solids and liquids are

always considered to be “1 activity” and therefore, do

not appear in the equilibrium expression. Solids and

pure liquids have no effect on conc or pressure.

36

Heterogeneous Equilibria

• Consider the reaction below.

(g)H CO(g) )()( 22 lOHsC

2

2 ]][[

HCOp

c

PPK

HCOK

HW 17

37

31.3 Direction of Reaction

• How could we predict the direction in which a

reaction at non-equilibrium conditions will shift to

reestablish equilibrium? Remember did example

with only reactants, obviously had to go right and if

only products obviously must go left but what if have

some of R and P?

– To answer this question, substitute the current

concentrations into the reaction quotient

expression and compare it to Kc.

– The reaction quotient, Qc, is an expression that

has the same form as the equilibrium-constant

expression but whose concentrations are not

necessarily at equilibrium.

38

Predicting the Direction of Reaction

• For the general reaction

dDcC bBaA

the Qc expresssion would be (i=initial):

ba

dc

c]B[]A[

]D[]C[Q

ii

ii

ba

dc

cBA

DCK

eqeq

eqeq

][][

][][

39

Predicting the Direction of Reaction

• For the general reaction

dDcC bBaA

– If Qc = Kc, then the reaction is at equilibrium.

– If Qc > Kc, the reaction will shift left toward reactants until equil

reached.

– If Qc < Kc, the reaction will shift right toward products until equil

reached.

ba

dc

c]B[]A[

]D[]C[Q

ii

ii

ba

dc

c]B[]A[

]D[]C[Q

ii

ii

40


– Consider the following equilibrium.

– A 50.0 L vessel contains 1.00 mol N2, 3.00 mol

H2, and 0.500 mol NH3. Is the sytem at equil? If

not, in which direction (toward reactants or

toward products) will the system shift to

reestablish equilibrium at 400 oC?

– Kc for the reaction at 400 oC is 0.500.

(g)2NH )g(H3 )g(N 322

41


First, calculate concentrations from moles

of substances.

(g)2NH )g(H3 )g(N 322

3

22

2

3

][][

][

ii

ic

HN

NHQ

Next, plug into Q expression

42


(g)2NH )g(H3 )g(N 322

– Substituting these concentrations into the reaction

quotient gives:

1.23)0600.0)(0200.0(

)0100.0(Q

3

2

c

Note: you cannot just look at number of mols or molarity to decide direction of reaction

because must account for the size of K.

43

0.0100 M 0.0600 M 0.0200 M


(g)2NH )g(H3 )g(N 322

– Because Qc = 23.1 is greater than Kc = 0.500,

the reaction will go to the left (toward

reactants - consume products, produce

reactants) as it approaches equilibrium.

HW 18

44

31.4 Calculating K and Equilibrium

Quantities

• Once you have determined the

equilibrium constant, K, for a reaction,

you can use it to calculate the

concentrations of substances in the

equilibrium mixture.

45

Calculating Equilibrium

Concentrations

– Suppose a gaseous mixture contained 0.30 mol CO,

0.10 mol H2, 0.020 mol H2O, and an unknown

amount of CH4 per liter at equilibrium.

– What is the concentration of CH4 at equilibrium in this

mixture? The equilibrium constant Kc equals 3.92.

– Note: the amounts given are equil amounts; therefore

can plug into K eq.

O(g)H (g)CH (g)H 3 )g(CO 242

– For example, consider the following equilibrium

mixture.

46


Concentrations

– First, calculate concentrations from moles of substances

then plug into K expression.

O(g)H (g)CH (g)H 3 )( 242gCO

eqeq

eqeq

cHCO

OHCHK

3

2

24

][][

][][

47


Concentrations

– Substituting the known concentrations and the value

of Kc gives:

3

4

)10.0)(30.0(

)020.0]([92.3

CH

?? 0.30 M 0.10 M 0.020 M

O(g)H (g)CH (g)H 3 )g(CO 242

48


Concentrations

– You can now solve for [CH4].

MCH 059.0)020.0(

)10.0)(30.0)(92.3(][

3

4

– The concentration of CH4 in the equil mixture is

0.059 mol/L.

?? 0.30 M 0.10 M 0.020 M

O(g)H (g)CH (g)H 3 )g(CO 242

49

31.4.1 Calculating Equilibrium Quantities

from Initial Values (Perfect Square)

• Suppose we begin a reaction with

known amounts of starting materials

and want to calculate the quantities at

equilibrium.

50


Concentrations – Consider the following equilibrium.

• Suppose you start with 1.000 mol each of carbon

monoxide and water in a 50.0 L container.

Calculate the molarity of each substance in the

equilibrium mixture at 1000 oC.

• Kc for the reaction is 0.58 at 1000 oC.

(g)H(g)CO )g(OH)g(CO 2220 0

- +

Which way will reaction shift?

51


Concentrations

– First, calculate the initial molarities of CO and H2O.

(g)H(g)CO )g(OH)g(CO 222

0 0

52


Concentrations

• The starting concentrations of the products are 0.

• We must now set up a table of concentrations (starting, change,

and equilibrium expressions in x, ICE table).

0.0200 M 0.0200 M 0 M 0 M


53


Concentrations

– Let x be the moles per liter of product formed.

[ ]o 0.0200 0.0200 0 0

[ ]

[ ]eq

– The equilibrium-constant expression is:

]OH][CO[

]H][CO[K

2

22c


- - + + x x x x

0.0200-x 0.0200-x x x

54

– Solving for x.

Starting 0.0200 0.0200 0 0 Change -x -x +x +x

Equilibrium 0.0200-x 0.0200-x x x

– Substituting the values for equilibrium concentrations, we

get:

)x0200.0)(x0200.0(

)x)(x(58.0


]OH][CO[

]H][CO[K

2

22c

55


Concentrations

– Solving for x.



2

2

)x0200.0(

x58.0


56


Concentrations – Solving for x.



– Taking the square root of both sides

because perfect square, we get:

x

xx

xx

x

x

76.10152.0

76.00152.0

)0200.0(76.0

)0200.0(76.0


2

2

)0200.0(58.0

x

x

Mx 0086.076.1

0152.0

57


Concentrations – Solving for equilibrium concentrations.



– If you substitute for x in the last line of the table you obtain the

following equilibrium concentrations. If plug into Keq, should

equal K or close to it because of sign fig for a check


HW 19

[CO]eq = 0.0200 – x = 0.0200 – 0.0086 = 0.0114 M

[H2O]eq = 0.0200 – x = 0.0200 – 0.0086 = 0.0114 M

[CO2]eq = x = 0.0086 M

[H2]eq = x = 0.0086 M

58


Concentrations

• The preceding example illustrates the

three steps in solving for equilibrium

concentrations.

1. Set up a table of concentrations (starting, change, and

equilibrium expressions in x – ICE table).

2. Substitute the expressions in x for the equilibrium

concentrations into the equilibrium-constant equation.

3. Solve the equilibrium-constant equation for the values

of the equilibrium concentrations.

59

Another example: If the initial pressure of C is 1.0 atm, what

would be the partial pressures of each species at equil.

Initial, Po 0 0 1.0

Change, P

Equil, Peq

HW 20

A + B 2C Kp = 9.0

+ + - x x 2x

x x 1.0-2x

60

Another example: If the initial pressure of C is 0.10 atm and A

and B are 1.00 atm, what would be the partial pressures of each

species at equil.

Initial, Po 0.10 1.00 1.00

Change, P

Equil, Peq

HW 21

2C A + B

Kp = 0.016

Q > K therefore, shift left

+ - - 2x x x

0.10 + 2x 1.00 – x 1.00 - x

61

31.4.2 Calculating Equilibrium Quantities

from Initial Values (Quadratic Formula)

• In some cases it is necessary to solve a

quadratic equation to obtain equilibrium

concentrations - not a perfect square.

• The next example illustrates how to

solve such an equation.

62


Concentrations

– Consider the following equilibrium.

• Suppose 1.00 atm H2 and 2.00 atm I2 are placed in

a 1.00-L vessel. What are the partial pressures of

all species when it comes to equilibrium at 458 oC?

• Kp at this temperature is 49.7.

HI(g)2 )g(I)g(H 22

63


Concentrations

– The concentrations of substances are as follows.

HI(g)2 )g(I)g(H 22

Initial, Po 1.00 2.00 0

Change, P

Equil, Peq

- - + x x 2x

1.00 - x 2.00 – x 2x

64


Concentrations

– The concentrations of substances are as follows.

Po 1.00 2.00 0

P -x -x +2x

Peq 1.00-x 2.00-x 2x

– Substituting our equilibrium concentration expressions

gives:

7.49)x00.2)(x00.1(

)x2(K

2

p

HI(g)2 )g(I)g(H 22

65


Concentrations

– Solving for x.

– Because the right side of this equation is not a perfect square (sq over

sq), you must solve the quadratic equation.

HI(g)2 )g(I)g(H 22

Starting 1.00 2.00 0

Change -x -x +2x

Equilibrium 1.00-x 2.00-x 2x

a

acbbx

cbxax

2

4

0

2

2

66

49.7(1.00-x)(2.00-x) = (2x)2

49.7(2.00 - 3.00x + x2) = 4x2

99.4 - 149.1x + 49.7x2 = 4x2

45.7x2 - 149.1x + 99.4 = 0

7.49)x00.2)(x00.1(

)x2(K

2

p

a b c

a

acbbx

cbxax

2

4

0

2

2

)7.45(2

)4.99)(7.45(4)1.149()1.149( 2

x

4.91

32.18170)81.22230()1.149(x

4.91

49.4060)1.149(x

4.91

72.631.149x

934.04.91

38.85x

33.24.91

82.212x

67


Concentrations

– However, x = 2.33 gives a negative value to 1.00 - x (the

equilibrium concentration of H2), which is not possible.

possible 0.934Only xIf you do the shift correctly, you may get a negative and

positive number, take the positive number and if two positive

numbers, take the smaller number. However, if you neglect the

shift and select it incorrectly, you will end up with either a +

and - # or two -#'s. In this case the negative number will be

correct or the smaller of the two negative numbers will be

correct. Bottom line examine numbers carefully when

selecting answer and selecting proper shift makes life easier

(use Q when necessary).

68


Concentrations

• Solving for equilibrium pressures.

– If you substitute 0.934 for x in the last line of the table

you obtain the following equilibrium concentrations.

HI(g)2 )g(I)g(H 22

Starting 1.00 2.00 0

Change -x -x +2x

Equilibrium 1.00-x 2.00-x 2x

HW 22

69

31.5 Effect of Changing the Reaction

Conditions Upon Equilibrium

• Obtaining the maximum amount of

product from a reaction depends on the

proper set of reaction conditions which

gets us to:

– Le Chatelier’s principle states that when a

system in a chemical equilibrium is disturbed by a

change of temperature, pressure, or

concentration, the equilibrium will shift in a way

that tends to counteract this change until

equilibrium is established again.

70

31.5.1 Adding or Removing a

Species

• If a chemical system at equilibrium is disturbed by adding a gaseous or aqueous species (not solid or liquid R or P), the reaction will proceed in such a direction as to consume part of the added species. Conversely, if a gaseous or aqueous species is removed (complex or escape gas), the system shifts to restore part of that species. This shift will occur until equilibrium is re-established.

• A + B C + D add - shift to opposite side,

remove - shift towards that side.

71

31.5.2 Compression or Expansion • A pressure change caused by changing the volume of the

reaction vessel can affect the yield of products in a gaseous reaction; only if the reaction involves a change in the total moles of gas present

• Ex. N2O4 (g) 2NO2 (g)

• Suppose system is compressed by pushing down a piston (decrease volume of space), which way would the shift be that would benefit and use the available space wisely?

• Shift to smaller number of gas molecules to pack more efficiently and relieve the increase in pressure due to piston coming down.

72

Effects of Pressure Change

• Basically the reactants require less volume

(that is, fewer moles of gaseous reactant) and

by decreasing the volume of the reaction vessel

by increasing the pressure, the rxn would shift

the equilibrium to the left (toward reactants)

until equil is established.

73

Effects of Pressure Change

• Literally “squeezing” the reaction (increase P) will cause a shift

in the equilibrium toward the fewer moles of gas.

• Reducing the pressure in the reaction vessel by increasing its

volume would have the opposite effect.

• Decrease P, increase V, shift larger mols of gas (L &S

not compressible)

• Increase P, decrease V, shift to smaller mols of gas

• In the event that the number of moles of gaseous

product equals the number of moles of gaseous

reactant, vessel volume/pressure will have no effect

on the position of the equilibrium; no advantage to

shift one way over the other.

74

SO2 (g) + 1/2 O2 (g) SO3 (g)

Increase P?

Shift to right toward smaller mols gas

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

Decrease P?

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

Decrease P?

C (s) + H2O(g) CO (g) + H2 (g)

Increase P?

Shift to left toward larger mols gas

No shift

Shift to left toward smaller mols gas; note

that C is a solid.

75

31.5.3 Temperature

• Temperature has a significant effect on

most reactions. – Reaction rates generally increase with an increase

in temperature. Consequently, equilibrium is

established sooner.

– However, when you add or remove

reactants/products or change pressure, the result

is that the system establishes new conc of species

but ratio still equal to same K at that temperature.

For temp changes, the numerical value of the

equilibrium constant Kc varies with

temperature. K temp dependent.

76

Effect of Temperature Change

• Let’s look at “heat” as if it were a

product in exothermic reactions and

a reactant in endothermic reactions.

• We see that increasing the temperature is

related to adding more product (in the case of

exothermic reactions) or adding more

reactant (in the case of endothermic

reactions).

• This ultimately has the same effect as if heat

were a physical entity.

A + B C + D + heat exo

A + B + heat C + D endo

77


• For example, consider the following generic

exothermic reaction.

• Increasing temperature would be like adding

more product, causing the equilibrium to

shift left.

• Since “heat” does not appear in the equilibrium-

constant expression, this change would result

in a smaller numerical value for Kc

(numerator smaller and den larger)

negative) is H( heat""products reactants

78


• For an endothermic reaction, the opposite is

true.

• Increasing temperature would be analogous to

adding more reactant, causing the

equilibrium to shift right.

• This change results in more product at

equilibrium, and a larger numerical value for

Kc (larger numerator, smaller den)

positive) is H( products reactantsheat""

79


• In summary:

– For an exothermic reaction ( H is negative) the

amounts of reactants are increased (shift left) at

equilibrium by an increase in temperature (Kc is

smaller at higher temperatures).

– For an endothermic reaction ( H positive) the

amounts of products are increased (shift right) at

equilibrium by an increase in temperature (Kc is

larger at higher temperatures).

Note: opposite occurs when lower temperature.

80

Effect of a Catalyst

• A catalyst is a substance that increases

the rate of a reaction but is not

consumed by it.

– It is important to understand that a catalyst has

no effect on the equilibrium composition of a

reaction mixture.

– A catalyst merely speeds up the attainment of

equilibrium but does not cause it shift one way or

the other, just get to direction is was going faster.

81

Consider the system I2 (g) 2I (g) H = 151 kJ

Suppose the system is at equilibrium at 1000oC. In which direction will rxn

occur if

a.) I atoms are added?

b.) the system is compressed?

c.)the temp is increased?

d.)effect increase temp has on K?

e.) add catalyst?

HW 23

Add gas products, shift left until equil established.

Increase P, decrease V, shift towards smaller mols of gas; therefore to the left until equil.

Endo reaction, therefore similar to adding reactants and rxn shifts to the right

Producing more products (shift right) and less reactants, therefore larger K

No effect, if reaction wasn’t at equil just getting there faster.

Documents

Chapter 31 Chemical Equilibrium - ctlsfasu · 4 Chemical Equilibrium • Therefore, many reactions do not go to completion but rather form a mixture of products and unreacted reactants,