# Module 3 Kinetics Lecture note

Pages: 19 (2867 words) Published: January 11, 2015
Module 3:
Kinetics of Chemical Reactions
Introduction to Kinetics
Chemical thermodynamics has answered the question “can a specified chemical reaction take place?” (i.e. is the reaction thermodynamically favourable?). • However, chemical thermodynamics hasn‟t answered the question “HOW FAST will a specified chemical reaction occur?”

o Many thermodynamically favourable reactions are so slow (ex: metamorphic transformation of rocks, corrosion of marble sculptures owing to weathering) that they can be considered not to happen!
Derivation of equilibrium constant (gases)
Consider the following reaction, where reactants (A, B) and products (C, D) are all gases: 𝑎𝐴 𝑔 + 𝑏𝐵 𝑔 → 𝑐𝐶 𝑔 + 𝑑𝐷 𝑔 (where 𝑎, 𝑏, 𝑐, 𝑑 are stoichiometric coefficients) The change in molar Gibbs free energy of that reaction:

∆𝐺 = 𝑐𝜇𝐶 + 𝑑𝜇𝐷 − 𝑎𝜇𝐴 + 𝑏𝜇𝐵 Which can be rewritten as:
∆𝐺 = ∆𝐺 0 + 𝑅𝑇 ln 𝐾
For an ideal gas, Chemical potential 𝜇𝑖 is defined as
𝜇𝑖 𝑝𝑖 = 𝜇𝑖0 + 𝑅𝑇 𝑙𝑛

𝑝𝑖
𝑝0

Therefore,
∆𝐺 = 𝑐 𝜇𝐶0 + 𝑅𝑇 𝑙𝑛

𝑝𝐶
𝑝𝐷
𝑝𝐴
𝑝𝐵
+ 𝑑 𝜇𝐷0 + 𝑅𝑇 𝑙𝑛 0 − 𝑎 𝜇𝐴0 + 𝑅𝑇 𝑙𝑛 0 − 𝑏 𝜇𝐵0 + 𝑅𝑇 𝑙𝑛 0 0
𝑝
𝑝
𝑝
𝑝

At equilibrium ∆𝐺 = 0 and therefore:

0 = ∆𝐺 0 + 𝑅𝑇 𝑙𝑛

𝑝𝐶
𝑝0
𝑝𝐴
𝑝0

𝑐

𝑎

𝑝𝐷
𝑝0
𝑝𝐵
𝑝0

𝑑

𝑏

We recognise the term in brackets to be the dimensionless equilibrium constant as defined above. It follows that:
• ∆𝐺 0 = −𝑅𝑇𝑙𝑛𝐾
Page | 1

Comparing these two expressions, we can write
𝑝𝐶
𝑝0
𝐾=
𝑝𝐴
𝑝0

𝑐

𝑎

𝑝𝐷
𝑝0
𝑝𝐵
𝑝0

𝑑

𝑏

 Using ΔG0 = –RT ln K we can predict the value of K, at a given T, from ΔG0 (which we can get using the thermodynamic relationships discussed earlier) Ask yourself: What is the significance K? Why is the above expression relevant? • This formula will hold in a wide range of cases, as long as the values of activity are appropriately defined. For pure substances (100% pure), activity is considered to be 1. Consider the following reaction:

2 𝐶𝑟 𝑠 + 3 𝑂2 𝑔 → 𝐶𝑟2 𝑂3 (𝑠)
The chemical equilibrium constant of this reaction is given by 𝐾=

𝑎𝐶𝑟2 𝑂3
2 3
𝑎𝐶𝑟
𝑝𝑂2

Assuming 𝐶𝑟 and its oxide (𝐶𝑟2 𝑂3 ) 100% pure, their activity will be 1. Mathematically the activity can be expressed as: 𝑎𝐶𝑟 = 𝑎𝐶𝑟2 𝑂3 = 1 1
𝐾= 3
𝑝𝑂2
𝑝𝑂2 =

3

𝐾

Ask yourself: What is the significance of this derivation? What will happen if the reactant and products are not 100% pure?
Problem Solving Methodology
A common question is “what conversion of a particular reaction will be achieved at temperature T, pressure p, if equilibrium is attained”?
Step 1: Write down the chemical evaluation, and look up the values of ΔH f0, S0 or ΔGf0 at 298K from standard thermodynamic data table (in the appendix section of most of the standard books in this field). Also look up expressions for CP of the species involved. Following methods can be used for solving problems.

Method 1:
o Evaluate ΔH0(298 K), ΔG0 (298 K) and thus get K(298 K) = exp(–ΔG0/RT) o Evaluate ΔCP0, and use this to get an expression for ΔH0 as a function of T. o Put this expression into the van‟t Hoff equation and integrate to find the value of K(T). Method 2:

o Evaluate ΔH0(298 K) and ΔS0 (298 K)
o Evaluate ΔCP0, and perform the integrals to get ΔH0(T) and ΔS0(T) Page | 2

o Hence calculate ΔG0(T) = ΔH0(T) – T ΔS0(T), and then get K(T) = exp(–ΔG0/RT). In both cases, K can be used to predict the conversion of reactant if equilibrium is achieved at the stated T and p.

Rate of Reaction
Chemical engineers define rA, the rate of reaction of species A, as the number of moles of A created per second per unit volume.
1 𝑑𝑁 𝐴
,
𝑑𝑡

o For a batch system: 𝑟𝐴 = 𝑉
moles of A in the vessel

where NA is the number of

o For the special case of a batch system at constant volume, this gives:
𝑑𝐶𝐴
𝑟𝐴 =
𝑑𝑡

Fig.1: Change in concentration
with respect to time upon reaction

For a continuous system: rate can be defined as the number of moles of product formed (in this case, A) in the universe.
𝑢𝑛𝑖𝑣𝑒𝑟𝑠𝑒
1 𝑑𝑁𝐴
.
𝑑𝑡

𝑟𝐴 = 𝑉

(Note that a continuous system at...