WEEK 9_Chemical Reaction Equilibria Part 1

7/31/2019 WEEK 9_Chemical Reaction Equilibria Part 1

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THE REACTION COORDINATE

APPLICATION OF EQUILIBRIUM CRITERIA TO CHEMICAL REACTIONS THE STANDARD GIBBS-ENERGY CHANGE AND THE EQUILIBRIUM CONSTANT

EFFECT OF TEMPERATURE ON THE EQUILIBRIUM CONSTANT

EVALUATION OF EQUILIBRIUM CONSTANTS


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THE REACTION COORDINATE

General chemical reaction:

where;

By the sign convention:

Positive (+) for a product Negative (-) for a reactant

...... 44332211 AvAvAvAv

numbertricstoichiome

formulachemical

tcoefficientricstoichiome

i

i

i

v

A

v


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3

For reaction:

the stoichiometric numbers, iare

As the reaction progresses, the changes in thenumbers of moles of species present are in direct

proportion to the stoichiometric numbers. This

principle provides the equations:

The list continues to include all species.

4 2 23CH H O CO H

4 2 21 1 1 3CH H O CO H

32 1 1

2 1 3 1

etc.dndn dn dn

31 2 4

1 2 3 4

...dndn dn dn


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All terms being equal, they can be identified collectively by a

single quantity representing an amount of reaction.

A definition ofdis given by the equation:

The general relation connecting the differential change dni

with dis therefore

is called the reaction coordinate, characterizes the extent ordegree to which a reaction has taken place.

31 2 4

1 2 3 4

...dndn dn dn

d

1, 2, ..., Ni idn d i

(13.2)

(13.3)


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Integration of eq. (13.3) from an initial unreacted state where

= 0 and ni = nio to a state reached after an amount of

reaction gives

or

Summation over all species yields

or

where

Thus the mole fractions yiof the species present are related to

by

0i i ii i i

n n n

0n n

0 0i i ii i i

n n n n

0

0

i iii

nny

n n

(13.5)

(13.4)

0 0

i

i

n

i indn d

0 1, 2, ...,i i in n i N


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Example 13.1

For a system in which the following reaction occurs,

assume there are present initially 2 mol CH4, 1 mol H2O,

1 mol CO and 4 mol H2. Determine expressions for the

mole fractions yias functions of.

Solution:

For the reaction,

For the given numbers of moles of species initially

present,

4 2 23CH H O CO H

1 1 1 3 2ii

0 0 2 1 1 4 8ii

n n


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Equation (13.5) yields

The mole fractions of the species in the reacting mixture

are seen to be functions of the single variable .

4 2

2

2 1

8 2 8 2

1 4 3

8 2 8 2

CH H O

CO H

y y

y y

0

0

i iii

nnyn n


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Example 13.2Consider a vessel which initially contains only n0 mol of

water vapor. If decomposition occurs according to the

reaction,

find expression which relate the number of moles and

the mole fraction of each chemical species to the

reaction coordinate .

Solution:

For the reaction,

12 2 22

H O H O

1 1

2 21 1


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Application of eqs. (13.4) and (13.5) yields

The fractional decomposition of water vapor is

Thus, when n0 = 1, is directly related to the

fractional decomposition of the water vapor.

2 2 2

10 2

1 1 10 0 02 2 2

H O H O

n

y y yn n n

20 0 0

0 0 0

H On n n n

n n n

0

0

i iii

nny

n n

(13.5)

2 2 2

10 2H O H O

n n n n

0i i in n (13.4)


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Multireaction Stoichiometry

For multireaction, i,j designates the stoichiometric number ofspecies iin reactionj.

The general relation, eq. (13.3) becomes

Integration from ni= ni0 and j= 0 to niand jgives

Summing over all species yields

, 1, 2, ...,i i j j j

dn d i N

0 , 1, 2, ...,i i i j j j

n n i N (13.6)

0 , 0 ,i i j j i j j i i j j i

n n n


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The definition of a total stoichiometric number for

a single reaction has its counterpart here in the definition:

Combination of this last equation with eq. (13.6) gives the

mole fraction

ii

, 0whence

j i j j j

i j

n n

0 ,

0

1, 2, ...,i i j j j

i

j jj

ny i N

n

(13.7)


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Example 13.3Consider a system in which the following reactions occur:

where the numbers (1) and (2) indicate the value ofj, the

reaction index.

If there are present initially 2 mol CH4 and 3 mol H2O,

determine expressions for the yias functions of1and2.

Solution:

The stoichiometric numbers i,jcan be arrayed as follows:

4 2 2

4 2 2 2

3 1

2 4 2

CH H O CO H

CH H O CO H

i = CH4 H2O CO CO2 H2

j j

1 -1 -1 1 0 3 2

2 -1 -2 0 1 4 2


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Application of eq. (13.7) gives

The composition of the system is a function of independentvariables 1 and 2 .

0 ,

0

1, 2, ...,i i j j j

i

j jj

ny i N

n

(13.7)

4

2 2

2

1 2 1

1 2 1 2

1 2 2

1 2 1 2

1 2

1 2

2

5 2 2 5 2 2

3 25 2 2 5 2 2

3 4

5 2 2

CH CO

H O CO

H

y y

y y

y


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APPLICATION OF EQUILIBRIUM

CRITERIA TO CHEMICAL REACTIONS

Total Gibbs energy of a closed system at constant T and P

must decrease during an irreversible process and equilibrium

is reached when Gtattains its minimum value.

At this equilibrium state,

If a mixture is not in equilibrium, any reaction at constant Tand P must lead to a decrease in the total Gibbs energy of the

system.

,

0tT P

dG


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Fig. 13.1 shows the relation ofGtand . The arrows along the

curve indicate the directions of changes in (Gt)T,P. The reaction

coordinate, has its equilibrium value, eat the minimum of

the curve.

This figure indicates two distinctive

features of the equilibrium state for given

T and P,

a) Total Gibbs energy Gt is a minimum.

b) Its differential is zero.

(For a single chemical reaction)


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THE STANDARD GIBBS-ENERGY CHANGE

AND THE EQUILIBRIUM CONSTANT

Equation (11.2) provides an expression for the total

differential of the Gibbs energy:

If changes in the mole numbers ni

occur as the result of a

single chemical reaction in a closed system, then by eq. (13.3)

each dnimay be replaced by the product id.

Eq. (11.2) then becomes

Because nG is a state function, the right side of this equation

is an exact differential expression,

( ) ( ) i ii

d nG nV dP nS dT dn (11.2)

( ) ( ) i ii

d nG nV dP nS dT d

, ,

t

i i

i T P T P

GnG


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In general, the quantity represents, the rate of change

of total Gibbs energy of the system with respect to the

reaction coordinate at constant T and P. Fig. 13.1 shows that this quantity is zero at the equilibrium

state. A criterion of chemical reaction equilibrium is therefore

Recall the definition of the fugacity of a species in solution:

In addition, eq. (11.31) may be written for pure species iin its

standard state at the same temperature:

i i

i

0i ii

(13.8)

lni i iT RT f (11.46)

lno oi i iG T RT f

A standard state is a particular state of a species at temperature T and at specified

conditions of pressure, composition and physical condition e.g. gas, liquid or solid.


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The difference between these two equations is

Combining eq. (13.8) with eq. (13.9) to eliminate igives

for the equilibrium state of a chemical reaction:

or

or

whereisignifies the product over all species i.

lno ii i oi

fG RT

f

(13.9)

ln 0o oi i i i

i

G RT f f

ln 0io o

i i i i i i

G RT f f

lnoi i ii

Go

i i RTi

f f


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In exponential form, this equation becomes

where the definition ofKand its logarithm are given by

Also by definition,

Go and Kare functions of temperature.

io

i ii f f K

(13.10)

exp

ln

o

o

GK

RT

GK

RT

(13.11a)

(13.11b)

o o

i iiG G (13.12)


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The function Go iiGio in eq. (13.12) is the difference

between the Gibbs energies of the products and reactantswhen each is in its standard state as a pure substance at the

standard state pressure, but at the system temperature.

Other standard property changes of reaction are similarlydefined. For general property M:

Kis called the equilibrium constant for the reaction;

i iG

io, represented by Go, is called the standard

Gibbs-energy change of reaction.

o o

i ii

M M


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For example, the relation between the standard heat ofreaction and the standard Gibbs energy change of reaction is

developed from Eq. (6.39) written for species iin its standardstate:

Then, multiplication of both sides of this equation by vi andsummation over all species gives

Therefore, using Eq.4.14 and Eq. 13.12, relation written as

2o

od G RT

H RTdT

(13.13)

dT

RTGvdRTHv

i

o

ii

i

o

ii

)/(2

dT

RTGdRTH

o

io

i

)/(2

i

o

ii

oHvH

Eq.(4.14)


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EFFECT OF TEMPERATURE ON THE

EQUILIBRIUM CONSTANT

Because the standard state temperature is that of theequilibrium mixture, the standard property changes ofreaction, such as Go and Ho, vary with the equilibriumtemperature.

The dependence ofG

o

on T is given by eq. (13.13),

In view of eq. (13.11b), this becomes

If Ho is negative, i.e., if the reaction is exothermic, theequilibrium constant decreases as the temperature increases.Kincreases with Tfor endothermic reaction.

2

o od G RT H

dT RT

2

ln od K H

dT RT (13.14)


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IfHo is assumed independent ofT, integration of eq. (13.14)

from T to Tleads to the simple result:

A plot of ln Kvs. (1/T) the reciprocal of absolute temperature

is a straight line.

' '

1 1ln

oK H

K R T T

(13.15)


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A plot of ln K vs. (1/T) for a number of common reactions


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The rigorous development of the effect of temperature on

equilibrium constant is based on the definition of the Gibbs

energy at standard state:

Multiply by iand summation over all species gives

As a result of the definition of a standard property change of

reaction, this reduces to

The standard heat of reaction is related to temperature:

o o oG H T S (13.16)

o o o

i i iG H TS

o o o

i i i i i i

i i i

G H T S

00

oT po o

T

CH H R dT

R

(4.18)


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The temperature dependence of the standard entropy

change of reaction is:

Multiply by iand summation over all species gives

Integration gives

where So and S0o are the standard entropy changes of

reaction at temperature T and at reference temperature T0.

o o

P

dTd S C

T

00

oT

o o P

T

C dTS S R

R T

(13.17)

T

dTCdS

o

P

o

i i


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Eqs. (13.16), (4.18) and (13.17) are combined to yield

However,

Hence,

Finally, division by RT yields

0 0

0 0 0

0

o oT T

o o o o P P

T T

C CTG H H G R dT RT dT

T R R

0 00 0

o oT T

o o oP P

T T

C C

G H R dT T S RT dT R R

0 00

0

o oo H GS

T

0

0 0

0 0

0

1oo o o oo T T

P P

T T

G H H C CG dTdT

RT RT RT T R R T

(13.18)


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The first integral of eq. (13.18) is

The second integral is

where

0

2 2 3 3

0 0 0

0

11 1 12 3

oTP

T

C B C DdT A T T T R T

02

0 0 20

1

ln 12

oT

P

T

C dT D

A BT CT R T T

(13.19)

0

T

T

(4.19)


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Go/RT(= - ln K) as given in eq. (13.18) is readily calculated at

any temperature from the standard heat of reaction and the

standard Gibbs energy change of reaction at a referencetemperature (usually 298.15 K).

Factor Kmay be organized into three terms:

0 1 2K K K K

00

0

expoG

KRT

(13.21)

0 01

0

exp 1oH T

KRT T

0 02

1exp

o oT T

P P

T T

C C dTK dT

T R R T

(13.22)

(13.23)

(13.20)


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Temperature dependence of the heat capacity (eq. 4.4):

The expression for K2 becomes:

2 2PC A BT CT DT R

2 2 2

2

2 0 0 2 2

0

1 1 2 11 1 1 1exp ln

2 6 2

DK A BT CT

T

(13.24)


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EVALUATION OF EQUILIBRIUM

CONSTANTS Values ofGo for many formation reactions are tabulated in

standard references.

The reported values ofGof are not measured experimentally,but are calculated by Eq. 13.16.

Values ofGof298 and Hof298 for a limited number of chemical

compounds are listed in Table C.4 of App. C. These for atemperature of 298.15K.

Values of

G

o

for other reactions are calculated fromformation-reaction values.

In more extensive compilations of data, values of Gof andHof are given for a wide range of temperature. When data islacking, methods of estimation are available.

o o o

G H T S


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Example 13.4

Calculate the equilibrium constant for the vapor phase

hydration of ethylene at 145o

C (418.15 K) and at 320o

C(593.15 K) from data given in App. C.

Solution:

First determine the values for A, B, C and D for thereaction

= (C2H5OH) (C2H4) (H2O)

From Table C.1:

A = 3.518 1.424 3.470 = -1.376

B = (20.001 14.394 1.450) x 10-3 = 4.157 x 10-3

C = (-6.002 +4.392 0) x 10-6 = -1.610 x 10-6

D = (-0 0 0.121) x 105 = -0.121 x 105

2 4 2 2 5C H g H O g C H OH g


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Values of Ho298 and Go

298 at 25oC (298.15 K) for the

hydration reaction are found from Table C.4:

For T = 418.15 K, values of integrals in eq. (13.18) are

Substitute into eq. (13.18) for a reference temperature of

298.15 K gives:

-1298 235100 52510 241818 45792 JmoloH

-1298 168490 68460 228572 8378 JmoloG

0

0

23.121

0.0692

oT

PT

oT

P

T

C dTR

C dT

R T

418 8378 45792 45792 23.121 0.0692 1.9356

8.314 298.15 8.314 418.15 418.15

oG

RT


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For T = 593.15 K,

Finally, by eq. (13.11b)

At 418.15 K, ln K= -1.9356 and K= 1.443 x 10-1

At 593.15 K, ln K= -5.8286 and K= 2.942 x 10-3

Alternative solution by using eq. (13.21), (13.22) and (13.24).

0

0

22.632

0.0173

oT

P

T

oT

P

T

CdT

R

C dT

R T

593 8378 45792 45792 22.632 0.0173 5.8286

8.314 298.15 8.314 593.15 593.15

oG

RT

RT

GK

oln

WEEK 9_Chemical Reaction Equilibria Part 1

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