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 𝑂2 + 2𝐶𝑂 ⇄ 2𝐶𝑂2 )
The Gibbs free energy of a multi-component system with Nj particles in each
component reads
G = G(N1 , … , Nj , … , T, P)
Using homogeneity of G according to
𝐺 𝜆𝑁1 , … , 𝜆𝑁𝑗 , … , 𝑇, 𝑃 = 𝜆𝐺 𝑁1 , … , 𝑁𝑗 , … , 𝑇, 𝑃
𝜕𝐺
𝜕𝐺
𝑁 +⋯
𝑁 +⋯=𝐺
𝜕𝜆𝑁1 1
𝜕𝜆𝑁𝑗 𝑗
and especially for 𝜆 = 1
𝜕𝐺
𝜕𝐺
G = 𝜕𝑁 𝑁1 + ⋯ 𝜕𝑁 𝑁𝑗 + ⋯ =
1
𝑗
𝑗 𝜇𝑗 𝑁𝑗
as generalization of G   N
Likewise dG  SdT  VdP  dN is generalized into
dG = −SdT + VdP +
𝑗
𝜇𝑗 𝑑𝑁𝑗


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 = 𝑑
𝑗 𝜇𝑗 𝑁𝑗
=
𝑗 𝑑𝜇𝑗 𝑁𝑗
+
𝑗 𝜇𝑗 𝑑𝑁𝑗
dG = −SdT + VdP +
and
dG − dG = 0 = −SdT + VdP +
𝑗 𝜇𝑗 𝑑𝑁𝑗 - 𝑗 𝑑𝜇𝑗 𝑁𝑗
−
𝑗 𝜇𝑗 𝑑𝑁𝑗
𝑗 𝜇𝑗 𝑑𝑁𝑗
Gibbs-Duhem equation
−SdT + VdP-
When T and P are constant
𝑗
𝑗 𝑑𝜇𝑗 𝑁𝑗
𝑑𝜇𝑗 𝑁𝑗 = 0
=0
telling us that the chemical potentials
of the different components are related
For example: in a 2 component mixture when knowing 𝜇1 we can calculate
𝜇2 from 𝑁1 𝑑𝜇1 + N2 d𝜇2 = 0
Law of mass action
Let’s consider a chemical reaction among  species (components)
For example the reaction 3𝐻2 + 𝑁2 ⇌ 2𝑁𝐻3
Dh=-92,5kJ/mol
Let’s write this in the form -3𝐻2 − 𝑁2 + 2𝑁𝐻3 =0
Or in a general notation
𝜁1 𝐴1 + 𝜁2 𝐴2 + ⋯ + 𝜁𝜎 𝐴𝜎 = 0
stoichiometric coefficient
For example: -3𝐻2 − 𝑁2 + 2𝑁𝐻3 =0 with
Aj is a product if 𝜁𝑗 > 0
𝜁 𝑁𝐻3 = 2 and 𝑁𝐻3 product
Aj is a reactant if 𝜁𝑗 < 0 𝜁 𝐻2 = −3 and 𝐻2 reactant , 𝜁 𝑁2 = −1 and 𝑁2 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
𝑗 𝜇𝑗 𝑑𝑁𝑗
=0
𝑗 𝜇𝑗 𝑑𝑁𝑗
and T,P constant
in equilibrium
If we look, e.g., at the forward direction of the reaction we identify
𝑗 𝜇𝑗 𝜁𝑗
𝑑𝑁𝑗 = 𝜁𝑗
=0
Consequences of this relation for ideal gases
The chemical potential of an ideal gas can be derived from G 
implying
(
𝜕𝜇
1 𝜕𝐺
V k T
)𝑇 = ( )𝑇 = = B
𝜕𝑃
N 𝜕𝑃
N
P
dG  SdT  VdP  dN
μ=
N
𝜕𝜇
( ) 𝑇 𝑑𝑃 + f(T)
𝜕𝑃
μ=𝑘𝐵 𝑇𝑙𝑛 𝑃/𝑃𝑟 + 𝑘𝐵 𝑇Φ(𝑇)
For each component (reactant and product) of the chemical reaction we can write
𝑃𝑗
μ𝑗 =𝑘𝐵 𝑇𝑙𝑛
+ 𝑘𝐵 𝑇Φj (𝑇)
𝑃𝑟
𝑗 𝜇𝑗 𝜁𝑗
With
=0
(under the assumption that the components can be described by ideal gases)
and
partial pressure of component j
𝑁𝑗
𝑃𝑗 = 𝑃 = P [Aj ]
𝑁
concentration of component j
total pressure 𝑃 =
𝑘𝐵 𝑇
𝑗 𝜁𝑗 (𝑙𝑛 𝑃
𝑗 𝜁𝑗 [ 𝑙𝑛
𝑗
𝑒
𝑃
𝑃𝑟
[𝐴𝑗 ]
𝑃𝑟
+ 𝑙𝑛[𝐴𝑗 ] + Φj (𝑇)] =0
=𝑒
𝑃
𝜁𝑗 [ 𝑙𝑛
+ Φj (𝑇) ]
𝑃
𝑗
𝑟
𝑃
− 𝑗 𝜁𝑗 [ 𝑙𝑛 𝑃 +Φj (𝑇) ]
𝑟
Law of mass action
[𝐴𝑗 ]𝜁𝑗 = 𝜅(𝑃, 𝑇)
𝑗
according Dalton’s law
+ Φj (𝑇) ) =0
𝑙𝑛[𝐴𝑗 ]𝜁𝑗 = −
𝜁𝑗
𝑗 𝑙𝑛[𝐴𝑗 ]
𝑗 𝑃𝑗
Example:
3𝐻2 + 𝑁2 ⇌ 2𝑁𝐻3
𝑁𝐻3 2
=𝜅
𝐻2 3 𝑁2
where 𝜅(𝑃, 𝑇)
Equilibrium constant
Consequences from law of mass action in equilibrium
[𝐴𝑗 ]𝜁𝑗 = 𝜅(𝑃, 𝑇)
𝑗
If 𝜅(𝑃, 𝑇) large
requires
𝐴𝑗 large with 𝜁𝑗 > 0 meaning product concentration large
𝐴𝑗 small with 𝜁𝑗 < 0 meaning reactant concentration low
If 𝜅(𝑃, 𝑇) small
requires
𝐴𝑗 small with 𝜁𝑗 > 0 meaning product concentration low
𝐴𝑗 large with 𝜁𝑗 < 0 meaning reactant concentration large
What can we say about P and T dependence of 𝜿
Can we use P to shift the equilibrium to the side of reactants or products
𝜅= 𝑒
𝑃
− 𝑗 𝜁𝑗 [ 𝑙𝑛 𝑃 +Φj (𝑇) ]
𝑟
𝑃
𝜅=
𝑃𝑟
If
𝑗 𝜁𝑗
− 𝑗 𝜁𝑗
= 𝑒
𝑃
−𝑙𝑛 𝑃
𝑟
𝑒 − 𝑗 𝜁𝑗 Φ𝑗(𝑇)
= 0 no P dependence of 𝜅
𝑗 𝜁𝑗
− 𝑗 𝜁𝑗 Φj (𝑇)
If 𝑗 𝜁𝑗 > 0 increase in P decreases 𝜅
Equilibrium shifts towards reactants
If 𝑗 𝜁𝑗 < 0 increase in P increases 𝜅
Equilibrium shifts towards products
Can we use T to shift the equilibrium to the side of reactants or products
− 𝑗 𝜁𝑗
𝑃
𝜅=
𝑃𝑟
𝜕𝑙𝑛𝜅
𝜕𝑇
From
μ𝑗 =𝑘𝐵 𝑇 𝑙𝑛
𝑇
𝑒−
𝑗 𝜁𝑗
=−
𝑃
𝑃
ln 𝜅 = 𝑙𝑛
𝑃𝑟
Φ𝑗 (𝑇)
𝑑Φ𝑗
𝜁𝑗
𝑑𝑇
𝑗
𝑃𝑗
+ 𝑘𝐵 𝑇Φj (𝑇)
𝑃𝑟
𝜕μ𝑗
𝜕𝑇
𝑃
−
𝑗
𝜁𝑗 Φ𝑗 (𝑇)
𝑑Φ
Next we replace
by a quantity closely related to the
𝑑𝑇
experiment such as the heat released in an exothermal reaction
𝜕μ𝑗
𝜕𝑇
= μ𝑗 + 𝑘 𝐵 𝑇 2
− 𝑗 𝜁𝑗
𝑃
𝑃𝑗
𝑑Φ𝑗
= 𝑘𝐵 𝑙𝑛 + 𝑘𝐵 Φj 𝑇 + 𝑘𝐵 𝑇
𝑃𝑟
𝑑𝑇
𝑑Φ𝑗
𝑑𝑇
A
From
𝐺 = 𝐻 + 𝑇𝑆
𝜕μ𝑗
𝜕𝑇
μ𝑗 = ℎ𝑗 + 𝑇𝑠𝑗
= 𝑠𝑗
𝑃
using
𝜕μ𝑗
μ𝑗 = ℎ𝑗 + 𝑇
𝜕𝑇
dG  SdT  VdP  dN
A
B
𝑃
& B
𝑑Φ𝑗
hj = − 𝑘𝐵 𝑇
𝑑𝑇
2
From
𝜕𝑙𝑛𝜅
𝜕𝑇
𝑑Φ𝑗
𝜁𝑗
𝑑𝑇
𝑗
=−
𝑃
𝜕𝑙𝑛𝜅
𝜕𝑇
=
𝑃
𝜕𝑙𝑛𝜅
𝜕𝑇
and
hj = − 𝑘𝐵 𝑇 2
𝑑Φ𝑗
𝑑𝑇
𝑗 𝜁𝑗 ℎ𝑗
𝑘𝐵 𝑇 2
𝑃
Δℎ
=
𝑘𝐵 𝑇 2
where
𝑗
𝜁𝑗 ℎ𝑗 = Δℎ
is the enthalpy of the reaction
For a better understanding let’s integrate the equation
𝜅 𝑇2
ln
=
𝜅 𝑇1
𝑇2
𝑇1
Δℎ
𝑑𝑇
𝑘𝐵 𝑇 2
𝜅(𝑇2 ) = 𝜅
−Δℎ 1 1
−
𝑇1 𝑒 𝑘𝐵 𝑇2 T1
If Δℎ < 0 heat leaves the system, the reaction is exothermic
𝜅 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
𝜅 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:
3𝐻2 + 𝑁2 ⇌ 2𝑁𝐻3
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
nothing about kinetics ! )
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 𝜅 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 -3𝐻2 − 𝑁2 + 2𝑁𝐻3 =0
𝜁1 𝐴1 + 𝜁2 𝐴2 + ⋯ + 𝜁𝜎 𝐴𝜎 = 0
𝑃
𝜅=
𝑃𝑟
𝑗
− 𝑗 𝜁𝑗
𝑒−
𝑗 𝜁𝑗
Means here
𝜁 𝑁𝐻3 = 2
𝜁 𝐻2 = −3
𝜁 𝑁2 = −1
Φ𝑗 (𝑇)
𝜁𝑗 = −3 − 1 + 2 = −2 < 0
high-pressure reactor (1913) in the ammonia plant
of the Badische Anilin und Soda Fabrik (BASF)
𝜅
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
𝑁𝐻3 2
=𝜅
𝐻2 3 𝑁2
(In the Haber-Bosch process ammonia is
removed as a liquid)