### Basis concepts of Thermochemistry

```Basis concepts of Thermochemistry
Goal: Using the general concept of minimal Gibbs free energy for chemical reactions
at constant T& P to derive law of mass action
To apply the concept of minimal Gibbs free energy in chemical reactions we need
generalization of single component systems (such as water only)
G   N and dG  SdT  VdP  dN
to
Multi-component systems
For now restriction to single phase systems (think of chemical reaction of gases into a gaseous product
such as 2 + 2 ⇄ 22 )
The Gibbs free energy of a multi-component system with Nj particles in each
G = G(N1 , … , Nj , … , T, P)
Using homogeneity of G according to
1 , … ,  , … , ,  =  1 , … ,  , … , ,

+⋯
+⋯=
1 1

and especially for  = 1

G =  1 + ⋯   + ⋯ =
1

as generalization of G   N
Likewise dG  SdT  VdP  dN is generalized into
dG = −SdT + VdP +



On the way towards the law of mass action let’s derive a useful relation
(Gibbs-Duham) between change of the chemical potentials
as a result of particle exchange
With
dG =

=

+

dG = −SdT + VdP +
and
dG − dG = 0 = −SdT + VdP +
-
−

Gibbs-Duhem equation
−SdT + VdP-
When T and P are constant

= 0
=0
telling us that the chemical potentials
of the different components are related
For example: in a 2 component mixture when knowing 1 we can calculate
2 from 1 1 + N2 d2 = 0
Law of mass action
Let’s consider a chemical reaction among  species (components)
For example the reaction 32 + 2 ⇌ 23
Dh=-92,5kJ/mol
Let’s write this in the form -32 − 2 + 23 =0
Or in a general notation
1 1 + 2 2 + ⋯ +   = 0
stoichiometric coefficient
For example: -32 − 2 + 23 =0 with
Aj is a product if  > 0
3 = 2 and 3 product
Aj is a reactant if  < 0  2 = −3 and 2 reactant ,  2 = −1 and 2 reactant
Reaction goes in both directions: where is the equilibrium
At constant T and P: equilibrium determined by
G = G(N1 , … , Nj , … , T, P)
minimum
dG = 0
dG = −SdT + VdP +
With

=0

and T,P constant
in equilibrium
If we look, e.g., at the forward direction of the reaction we identify

=
=0
Consequences of this relation for ideal gases
The chemical potential of an ideal gas can be derived from G 
implying
(

1
V k T
) = ( ) = = B

N
N
P
dG  SdT  VdP  dN
μ=
N

( )   + f(T)

μ=  / +  Φ()
For each component (reactant and product) of the chemical reaction we can write

μ =
+  Φj ()

With
=0
(under the assumption that the components can be described by ideal gases)
and
partial pressure of component j

=  = P [Aj ]

concentration of component j
total pressure  =

(
[

[ ]

+ [ ] + Φj ()] =0
=

[
+ Φj () ]

−   [   +Φj () ]

Law of mass action
[ ] = (, )

according Dalton’s law
+ Φj () ) =0
[ ] = −

[ ]

Example:
32 + 2 ⇌ 23
3 2
=
2 3 2
where (, )
Equilibrium constant
Consequences from law of mass action in equilibrium
[ ] = (, )

If (, ) large
requires
large with  > 0 meaning product concentration large
small with  < 0 meaning reactant concentration low
If (, ) small
requires
small with  > 0 meaning product concentration low
large with  < 0 meaning reactant concentration large
What can we say about P and T dependence of
Can we use P to shift the equilibrium to the side of reactants or products
=

−   [   +Φj () ]

=

If

−
=

−

−   Φ()
= 0 no P dependence of

−   Φj ()
If   > 0 increase in P decreases
Equilibrium shifts towards reactants
If   < 0 increase in P increases
Equilibrium shifts towards products
Can we use T to shift the equilibrium to the side of reactants or products
−

=

From
μ =

−

=−

ln  =

Φ ()
Φ

+  Φj ()

μ

−

Φ ()
Φ
Next we replace
by a quantity closely related to the

experiment such as the heat released in an exothermal reaction
μ

= μ +    2
−

Φ
=   +  Φj  +

Φ

A
From
=  +
μ

μ = ℎ +
=

using
μ
μ = ℎ +

dG  SdT  VdP  dN
A
B

& B
Φ
hj = −

2
From

Φ

=−

=

and
hj = −   2
Φ

ℎ
2

Δℎ
=
2
where

ℎ = Δℎ
is the enthalpy of the reaction
For a better understanding let’s integrate the equation
2
ln
=
1
2
1
Δℎ

2
(2 ) =
−Δℎ 1 1
−
1   2 T1
If Δℎ < 0 heat leaves the system, the reaction is exothermic
decreases with increasing T and the yield goes down
(consider the reaction of hydrogen and oxygen into water. At high temperature (about 6000K ) there is
virtually no yield
for water and at even higher T water dissociates into hydrogen and oxygen.)
If Δℎ > 0 the reaction is endotherm
increases with increasing T and the yield goes up
The Haber-Bosch process
An example to utilize the full potential of the law of mass action
Industrial route to ammonia (Nobel Prize in Chemistry 1918, Fritz Haber)
Let’s recall:
32 + 2 ⇌ 23
Dh=-92,5kJ/mol
http://en.wikipedia.org/wiki/File:Fritz_Haber.png
Fritz Haber, 1868-1934
At room temperature the reaction is slow (note: our equilibrium considerations say
Is it a good idea to increase T to increase reaction speed
Equilibrium consideration from law of mass action limits this possibility
Since Δℎ < 0
reaction is exothermic  decreases with increasing T and the
yield goes down
Compromise:
T not higher than needed
for catalyst to work (400 C)
http://en.wikipedia.org/wiki/Haber_process
Can pressure be utilized to favor the forward direction
Law of mass action provides the answer: Yes
Let’s recall -32 − 2 + 23 =0
1 1 + 2 2 + ⋯ +   = 0

=

−
−

Means here
3 = 2
2 = −3
2 = −1
Φ ()
= −3 − 1 + 2 = −2 < 0
high-pressure reactor (1913) in the ammonia plant
of the Badische Anilin und Soda Fabrik (BASF)

increases with increasing P
Expenses/technical reasons of high P equipment bring limitations to this approach
Compromise: 6–18 MPa (59–178 atm)
Finally: we have seen from the discussion of stability conditions
A perturbation away from equilibrium creates a force driving the system back
(Le Chatelier’s principle)
removing NH3 will increase yield of production
3 2
=
2 3 2
(In the Haber-Bosch process ammonia is
removed as a liquid)
```