Reaction Eqb

Physical Chemistry I

Instr: Dr. Jayadevan. K. P., B309

Chapter 6: Reaction Equilibrium in Ideal Gas Mixtures 6.1-6.4

Review of concepts about Eqb.

Material Eqb.: is maximized at Eqb.

Phase Eqb.: 0

Reaction Eqb.: 0

Consider (hypothetical) reaction of an ideal gas mixture:

Syst Univ

i ii

i ii

S S

dnα

αµ

ν µ

+

=

=

∑∑

∑

BITS Pilani, K K Birla Goa Campus

Consider (hypothetical) reaction of an ideal gas mixture:

Eqb. Constant

c dC D

o o

A

aA bB cC dD

P PP P

P

+ +

=

⇌

0 where 1 bar.

ln

a bB

o o

o oeq

PP

P P

G RT K

=

∆ = −

Chemical Potential of an (pure) Ideal Gas

• Chemical potentialis an intensive property andfor ideal gasesdepends on T and P


• Variation of chemical potentialµ of a pure ideal gas with pressureat constant temperature. µo is thestandard chemical potential

Chemical Potential of an (pure) Ideal Gas

d i v i d i n g b y n o . o f m o l e s :

F o r c o n s t a n t T :

m m m

m

d G S d T V d P

d G d S d T V d P

G

R T

µµ

= − +

= = − +=

∵


22 1

1

1 1

c o n s t . p u r e i d e a l g a s .

i f 1 b a r = >

( , ) ( , ) l n

( , ) ( )

( ) l n

m

o o

oo

R Td V d P d P

P

T

PT P T P R T

P

P P T P T

PT R T

P

µ

µ µ

µ µ

µ µ

= =

− =

= = =

= +

∵

Chemical Potentials in an Ideal Gas Mixture

1 : all

2 Pure gas separated from

mixture through a

membrane permeable to gas

Eqb. Partial Pressure:

. , ,

.totPV n RT T P n

i

i

=


Eqb. Partial Pressure:

for pure substance

Phase Eqb. between mixture and pure

Mixt

*

*

:i i

i i

P x P P

i

µ µ

≡ =

=

1 2

ure at

at Eqb.

: ideal gas mixture

*

* *

, ,

( , , , , ...) ( , ) ( , )

i

i i

i i i i i

T P x

P x P P

T P x x T x P T Pµ µ µ

≡ =

= =

Chemical Potentials in an Ideal Gas Mixture

:

F u n d am en ta l E q n . fo r

Id ea l g as m ix tu re , P 1 b ar

lno ii i

o

o

PR T

Pµ µ

= +

=


d ep en d s o n ly o n tem p era tu reoµ

Each of U, H, S, G and CP for an ideal gas mixtureis the sum of the correspondingthermodynamic functions for the pure gases calculatedfor each pure gas occupying a volume equal tomixture’s volume at a pressure equal to itspartial pressure in the mixture and at a temperature

equal to its temperature in the mixture.

Ideal-Gas Reaction Eqb.

Consider a general ideal gas reaction:

At Eqb.: 0i i

aA bB cC dD

ν µ+ +

=∑⇌


in terms of

or 0

:

i ii

A B C D

C D A B

a b c d

c d a b

µµ µ µ µ

µ µ µ µ+ = +

+ − − =

∑


lno ii i o

PRT

P

P P

µ µ = +


0

ln ln

ln ln

o oC DC Do o

o A BA Bo o

P Pc cRT d dRT

P P

P Pa aRT b bRT

P P

µ µ

µ µ

+ + +

− − − − =


( ) ( ) ( ) ( )-------------(A)

ln ln ln ln

o o o oC D A B

o o o oC D A B

c d a b

RT c P P d P P a P P b P P

µ µ µ µ+ − −

=− + − −


-------------(A)

L.H.S of (A)

, , ( )o o oT i mT i i i

i i

o o o oC D A B

G G T

c d a b

ν ν µ

µ µ µ µ

∆ = =

= + − −

∑ ∑∵


( ) ( )( ) ( )

From and R.H.S of Eqn. (A):

, ,

, ,

( )

ln

o

c do oC eq D eqo

a bo oA eq B eq

G T

P P P PG RT

P P P P

∆

∆ = −


( ) ( )( ) ( )( ) ( )

where P 1 bar

, ,

, ,

, ,

ln

A eq B eq

c do oC eq D eqo o

P a bo oA eq B eq

o oP

P P P P

P P P PK

P P P P

G RT K

= ≡

∆ = −


( )0

0

,

,

ln

( ) ln

o oi i i i i eqb

i i

o oi i i i eqb

i i

RT P P

T RT P P

ν µ ν µ

ν µ ν

= + =

+ =

∑ ∑

∑ ∑


and

and since , ,

, ,

( )

( )

( )

i i

i i i i i ii i i i i

o oi m T i

o o oT i m T i i i

i i

a b a b ca c a

T G

G G T

µ

ν ν µ

+ = + =

=

∆ = =

∑ ∑ ∑ ∑ ∑

∑ ∑

∵


( )

( )1 2 1 2

,

,

( ) ln

ln

ln ln ln .... ln ln( ... )

i

o oi i eq

i

oi eq

i

i n ni

G T RT P P

RT P P

a a a a a a a

ν

ν∆ = −

= −

= + + + =

∑

∑

∑∵


( )

( )

1 21

Ideal gas reaction Eqb.

=>

,

,

ln ...

lni

i

o

n

i ni

o oT i eq

i

o oP i eq

i

o G RTP

a a a a

G RT P P

K P P

K e

ν

ν

=

− ∆

= =

=> ∆ = −

≡

=

∏

∏

∏

Summary: Ideal Gas Reaction Eqb.Ideal gas reaction:

0 or 0

( ) ln

ln

i i i ii i

oi i o

o o

A

PT RT

P

G RT K

ν ν µ

µ µ = +

∆ = −

∑ ∑⇌ ⇌


are negative for reactants and positive products

:(Std. Eqb.constant) products in the numerator &

reactants in the deno

lno oP

i

oP

G RT K

K

ν∆ = −∵

( )

23

minator, dimensionless

If only partial pressure is considered

will have a dimension (e.g. NH formation, pressure )

:

i

P

P

P ii

K

K

K Pν

−

= ∏

A mixture of 11.02 mmol of H 2S and 5.48 mmol of CH 4was placed in an empty container along with a Pt catalyst and the eqb.

2 H2S (g) + CH4 (g) 4 H2 (g) + CS2 (g) was established at 700 oC and 762 torr . The reaction

Ideal Gas Reaction Eqb.: Kpo

and ∆Go

⇌


was established at 700 C and 762 torr . The reaction mixture was removed from the catalyst and rapidly cooled to room temperature, where the rates of forward and reverse reactions are negligible. Analysis of Eqb. Mixture found 0.711 mmol of CS2.Find Kp

o and ∆Go


and ∆Go

2

2

Eqb. Composition calculation:

Products:

0 711 mmol

=> 4 0 711 mmol 2 84 mmol H at Eqb.

.

. .CSn =

× =


4

4

2

Reactants:

0 711 mmol reacted

=> At Eqb.: 5 48 0 711 4 77 mmol

2 0 711 1 42 mmol reacted

=> At Eqb.:

.

. . .

. .

CH

CH

H S

n

n

n

=

= − =

= × =

2 11 02 mmol 1 42 9 6 mmol. . .H Sn = − =


and ∆Go

C a lc u la t i o n o f p a r t i a l p r e s s u r e s :

w h e r e 7 6 2 t o r r .

M o le f r a c t i o n s :

9 6= = 0 .5 3 6 , = 0 .2 6 6 ,

1 7 9 2

.

.

i i

H S C H

P x P P

x x

= =


2 4

2 2

2

4 2

= = 0 .5 3 6 , = 0 .2 6 6 , 1 7 9 2

= 0 .1 5 8 , = 0 .0 3 9 7

P a r t i a l p r e s s u r e s :

0 5 3 6 7 6 2 4 0 8 t o r r

2 0 3 t o r r , 1 2 0 t o r r

.

.

H S C H

H C S

H S

C H H

C S

x x

x x

P

P P

P

= × =

= =

23 0 3 t o r r.=


and ∆Go

( ) ( )( ) ( )

( ) ( )

2 2

2 4

4

2

4120 750 30 3 750.

o oH CSo

Po o

H S CH

P P P PK

P P P P=

= =


( ) ( )( ) ( )2

973

-1 -1

120 750 30 3 7500 000331

408 750 203 750

at 700 C or 973 K

8 314 Jmol K 973 K 0 000331

64 8 kJ/mol

..

ln

. ln .

.

o o oPG RT K

= =

∆ = −

= − × ×

=

Concentration and Mole-Fraction Eqb. Constants

Expressing partial pressures in terms of concn.:

Ideal gas mixture.

ii

ii i

nc

Vn RT

P c RTV

=

= =


( ) ( )( ) ( )

( ) ( )( ) ( )

, ,

, ,

, ,

, ,

f do oF eq D eqo

P a bo oA eq B eq

f d f d a bo o oF eq D eq

a b oo oA eq B eq

VaA bB fF dD

c RT P c RT PK

c RT P c RT P

c c c c c RT

Pc c c c

+ − −

+ +

×=

×

× = ×

×

⇌

Concentration and Mole -Fraction Eqb. Constants

2 2 3

d im ensionally eq . to

N 3 H 2 N H

2 1 3 2

/

( ) ( ) ( )

/

o oc R T P

n m o l f d a b

g g g

n m o l

∆ = + − −+

∆ = − − = −⇌


( )3

S td .C oncn .E qb .C onst:

w here 1 m o l/L 1 m o ldm

,

/

io oC i eq

i

o

n m o loo oP C o

K c c

c

R T cK K

P

ν

−

∆

=

= ≡

=

∏


like is dimensionless

depends only on

and are constants shows that

,

o oC P

oP

o o

K K

K T

c P

∵


( )

and are constants shows that

only

Mole-fraction Eqb. Const.:

= ,

( )

i

oC

x i eqi

c P

K f T

K xν

=

∏


Except for reactions with:

0 is a function of and

/

, .

n mol

oP x o

PK K

P

n K T P

∆ = ×

∆ =


0 is a function of and

Note: Any ideal gas Eqb.

can be solved using only at 1 bar

Only indirectly related to and

, .

ln

.

x

o oP

o oP

oC x

n K T P

K P

G RT K

K K

∆ =

=

∆ = −

Qualitative discussion of Chemical Eqb.

1

0 => very large,

favors reactants

=> is very small

o

o

oP G RT

o G RT

oP

Ke

G e

K

∆

∆

=

∆ >>


=> is very small

and vice versa

Exponential reln. between

and unless range:

12 12 will be very large/small

if is l

,

- ,

P

o o oP

o oP

K

G K G

RT G RT K

T

∆ ∆

< ∆ <

∵

ess, varies rapidly with

At high :

o oP

o o

K G

T G T S

∆

∆ ≈ − ∆

Qualitative discussion of Chemical Eqb.: Kp

o vs T

At low temperatures,low value of Kp

o impliesLarge positive


Large positive Std. Gibbs energy.(∆Go)

Temperature dependence of Eqb. constant

2

1

ln

ln ( )

ooP

o o oP

GK

RT

d K G d G

dT RT RT dT

∆= −

∆ ∆= −


=>

,,

,,

om io o

i m i ii i

m m m

om i o

m i

dT RT RT dT

dGG G

dT

dG S dT V dP

dGS

dT

ν ν∆ = =

= − +

= −

∑ ∑∵

∵

Temperature dependence of Eqb. constant

2

2

V an 't H off E qn.

,

ln

ln

oo o

i m ii

o o oP

o oP

d GS S

dT

d K G S

dT R T R T

d K H

dT R T

ν∆ = − = − ∆

∆ ∆= > = +

∆= > =

∑


2

1

2

2

22

1

2

1 1 2

V an 't H off E qn.

1 1

N eglecting dependence o f

ln

( )ln

( )

( )ln

( )

ooP

o oTPo TP

o oPoP

o

dT R T

Hd K dT

R T

K T HdT

K T R T

K T H

K T R T T

T H

= > =

∆=

∆=

∆= −

∆

∫

Find Kpo at 600 K for the reaction

N2O4 (g) 2 NO2 (g) (a) using the approximation that ∆Ho is independent of T(b) using approximation that ∆CP

o is independent of T

Temperature dependence of Eqb. Constant: Example

⇌

(a) 572 kJ/mol 4730 J/mol. ,o oH G∆ = ∆ =


298 298

298

4600

(a) 572 kJ/mol 4730 J/mol

0 148

57200 1 1

0 148 8314 J/mol-K 298 15 K 600 K

11 6

163 10

,

,

. ,

.

ln. . .

.

.

o oK K

oP K

oP

oP

H G

K

K

K

∆ = ∆ =

=

≈ −

=

≈ ×

Temperature dependence of Eqb. Constant: Example

1 1 1

298

(b)

Substituting this in Eqn for

temp. dependence of and 2 88 J/mol-K

1 1

,

( ) ( ) ( )( )

.

( ) ( )

o o oP

o oP P K

o o

H T H T C T T T

K C

K T H T

∆ = ∆ + ∆ −

∆ = −

∆


2 1

1 1 2

1 2 1

1 2

4600

1 1

1

1 52 10,

( ) ( )ln

( )

( )ln

.

o oPoP

oP

oP K

K T H T

K T R T T

C T T T

R T T

K

∆≈ −

∆+ + −

= ×

Temperature dependence of Eqb. Constant


Vant Hoff Eqn. in differentials1

1 vs is linear and from the slope

can be calculated.

, and can be calculated

ln

ln

( ) ,

o oP

oP

o

o o o oP

d K H

Rd

T

KT

H

K f T G H S

∆= −

∆= ∆ ∆ ∆

Ideal-Gas Eqb. Calculations:Steps summary

1 from a table of values.

2.

3. Eqb. mole no.:

, ,.

ln

o o oT i f T i f T i

i

o oT P

G G G

G RT K

ν∆ = ∆ ∆

∆ = −

∑


3. Eqb. mole no.:

+

4. const.T&P

At fixed T and V: if V is known.

,i i o i eq

i i i ii

ii

n n

P x P n n P

n RTP

V

ν ξ=

= =

=

∑

Ideal-Gas Eqb. Calculations:Steps summary

( )5 as a fn. of

solve for

6 Eqb. mole numbers from and in step 3

.

.

i

i

o oP i eq

i

P x

K P P

n

νξ

ξ

= ∏


6 Eqb. mole numbers from and in step 3. eq inξN2 H2 NH3

INITIAL

MOLES

1.0 2.0 0.5

CHANGE -z -3z 2z

EQB.

MOLES

1.0-z 2.0-3z 0.5 + 2z

6.12: For the reaction N 2O4 (g) 2 NO2 (g), measurements of compositions of the eqb. Mixtures gave Kp

o= 0.144 at 25oC and Kpo= 0.321 at 35oC. Find

∆Ho, ∆Go , ∆So at 25oC for this reaction.6.25 Calculation of Eqb. amounts

Ideal-Gas Eqb. Calculations:Example

⇌


6.25 Calculation of Eqb. amounts

• Calculation of ∆Go, KPo : 6.3, 6.4, 6.5, 6.9

• Ideal Gas Reaction (True/False): 6.10, 6.11• Calculation of ∆Go, ∆Ho, ∆So : 6.12, 6.13, 6.14• Temperature dependence of KP

o : 6.15, 6.16, 6.17, 6.19, 6.33• Calculation of Eqb. Composition and KP

o : 6.24, 6.25, 6.26, 6.33

Example Problem List


Reaction Eqb

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