Download - 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
+
=
=
∑∑
∑
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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
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• 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
= − +
= = − +=
∵
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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
=
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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
= +
=
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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+ +
=∑⇌
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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+ = +
+ − − =
∑
Ideal-Gas Reaction Eqb.
lno ii i o
PRT
P
P P
µ µ = +
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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
+ + +
− − − − =
Ideal-Gas Reaction Eqb.
( ) ( ) ( ) ( )-------------(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+ − −
=− + − −
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-------------(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
∆ = =
= + − −
∑ ∑∵
Ideal-Gas Reaction Eqb.
( ) ( )( ) ( )
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
∆
∆ = −
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( ) ( )( ) ( )( ) ( )
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
= ≡
∆ = −
Ideal-Gas Reaction Eqb.
( )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
= + =
+ =
∑ ∑
∑ ∑
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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
+ = + =
=
∆ = =
∑ ∑ ∑ ∑ ∑
∑ ∑
∵
Ideal-Gas Reaction Eqb.
( )
( )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∆ = −
= −
= + + + =
∑
∑
∑∵
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( )
( )
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 = +
∆ = −
∑ ∑⇌ ⇌
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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
⇌
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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
Ideal Gas Reaction Eqb.: Kpo
and ∆Go
2
2
Eqb. Composition calculation:
Products:
0 711 mmol
=> 4 0 711 mmol 2 84 mmol H at Eqb.
.
. .CSn =
× =
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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 = − =
Ideal Gas Reaction Eqb.: Kpo
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
= =
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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.=
Ideal Gas Reaction Eqb.: Kpo
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=
= =
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( ) ( )( ) ( )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
=
= =
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( ) ( )( ) ( )
( ) ( )( ) ( )
, ,
, ,
, ,
, ,
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
∆ = + − −+
∆ = − − = −⇌
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( )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
−
∆
=
= ≡
=
∏
Concentration and Mole -Fraction Eqb. Constants
like is dimensionless
depends only on
and are constants shows that
,
o oC P
oP
o o
K K
K T
c P
∵
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( )
and are constants shows that
only
Mole-fraction Eqb. Const.:
= ,
( )
i
oC
x i eqi
c P
K f T
K xν
=
∏
Concentration and Mole -Fraction Eqb. Constants
Except for reactions with:
0 is a function of and
/
, .
n mol
oP x o
PK K
P
n K T P
∆ = ×
∆ =
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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
∆
∆
=
∆ >>
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=> 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
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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
∆= −
∆ ∆= −
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=>
,,
,,
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∆ = − = − ∆
∆ ∆= > = +
∆= > =
∑
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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∆ = ∆ =
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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
∆ = ∆ + ∆ −
∆ = −
∆
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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
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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∆ = ∆ ∆
∆ = −
∑
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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
= ∏
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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
⇌
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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
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