Equilibrium in Chemical
Reactions

Chemical Reaction Equilibrium
Reaction rate and % conversion are both important.
Do not necessarily vary with conditions in same way.
Conversion depends on the yield at equilibrium.
High conversion/low rate - large reactor, little purification equipment

Low conversion/high rate - small reactor, large purification system
Conversion is determined by chemical reaction eqm.
Rate is determine by chemical kinetics.

Chemical Reaction Equilibrium

Stoichiometric Numbers &
Reaction Co-ordinate
Mass is conserved in a chemical reaction.
Not necessarily number of moles.
e.g.
N2 + 3H2
2NH3
4 moles react to give 2 moles of product.
In the above reaction product moles - reactant moles = -2 but in an atom balance
2NH3 - N2 - 3H2 = 0

Stoichiometric Numbers Vi defined:
Vi =overall change in moles during reaction v1.N2 + v2.H2 + v3.NH3 = 0 v1 = -1; v2 = -3; v3 = +2 vi = v1 + v2 +v3 = -2
i.e. product have positive numbers and reactants negative.

In general viFi = 0 where Fi are the chemical species present.
NOTE.
NOT A MATHEMATICAL EQUATION BUT A
CHEMICAL REACTION EQUATION.

In the above example
F1 = N2; F2 = H2 and F3 = NH3
Thus
v1.F1 + v2.F2 + v3.F3 + v4.F4 + ……. = 0 or viFi = 0 ni are changes in number of moles as reaction n n proceeds. ni

i 1

vi

i 1

= extent to which the reaction has proceeded
(Reaction Co-ordinate)

Since Vi are constant for the reaction:

dn1

v1d

Extent of reaction can be used to determine the number of moles of each reaction component after the reaction has gone some way.

Example 12
Assume 10 moles of H2, reacts with 5 moles N2 then: dn N 2 vN 2

dn H 2 vH 2

dn NH 3 v NH 3

d

dn N 2
1

or

dn H 2
3

dn NH 3
2

For N2 dnN2 = -d nN 2

d

nN 2

3 d

nH 2 10

dn N 2
5

5

nN 2

0

5

0

For H2 dnH2 = -3d nH 2

dn H 2
10

0

3(

0)

nH 2

10 3

d

For NH3 dnNH3 = +2d
Thus
n NH 3

dn NH 3
0

2 d

n NH 3

2(

0)

0

n NH 3

2

In tabulated form:
Component
N2
H2
NH3
Total

Moles in
5
10
0
15

Moles out
510 - 3
2
15 - 2

Mole fraction out
(5 - )/(15 - 2 )
(10 - 3 )/(15 - 2 )
2 /(15 - 2 )
1

Example 12b
A+B
C
A + 2D
E
Reaction co-ordinates: for first reaction for second reaction
Initially 10 moles A, 10 moles D and 5 moles B.
Find the expressions for mole fractions out:
Component
A
B
C
D
E
Total

Moles in
10
5
0
10
0
25

Moles out
10 - 510 - 2
25 - - 2

Mole fraction out
(10 - - )/(25 - - 2 )
(5 - )/(25 - - 2 )
/(25 - - 2 )
(10 - 2 )/(25 - - 2 )
/(25 - - 2 )
1

Criteria for equilibrium
Same as before for phase behaviour.
Irreversible dS > O equilibrium dS = 0
Irreversible (dG)T,P< 0 equilibrium (dG)T,P = 0 is the extent of reaction variable. defines composition of the system.
Eqm composition exists when G is at a min.
Eqm may be approached from either side.
For several simultaneous reactions, the composition that minimises G is eqm.
This corresponds to = E

Equilibrium Constant
For an open system, for a chemically reacting system, with a change in the number of moles of reactants and products n dG Vdp SdT

g i dni i 1

But,
Now
So

ˆi

gi dni vi

the partial chemical potential.

d n n

g i dni i 1

ˆ i .vi d

i 1

Thus at constant T & p

n

G
T,p

This is the gradient of the G vs.

ˆ i .vi

i 1

curve.

At equilibrium
At any T

Thus

gi g i0 gi ˆi

gi

n

G
T,p

RTlnf i
RTlnf i 0
RTlnfˆi
g i0

ˆ i .vi

0

i 1

gi

g

0 i fˆi
RTln 0 fi RTlnaˆ i

RTlnaˆ i

Note: standard state used is pure substance at T of system and at one fixed P;
THIS IS DIFFERENT from Lewis Randall standard state.

n

ˆ i vi

0

i 1

n

RTlnaˆ i

vi g i0

0

n

i 1 n n

vi g

0 i i 1

vi RTlnaˆ i n vi g i0 i 1

i 1 n RTlnaˆ ivi n RTln

i 1

aˆ

is given the symbol

i 1

i 1

vi g i0

0

vi g i

n

i 1

0

i 1

n

n

K is defined as

aˆ ivi

g 0 = Standard Gibbs Free Energy Change

aˆ ivi

for the Reaction.

i 1

vi i aˆ .aˆ .aˆ ......aˆ v1 1

v2
2

v3
3

g0

vn n g0

RT lnK

For chemical reactions standard state: pure component at 1 atm and reaction T.
For gaseous reactions, the ideal gas