Module 3 Kinetics Lecture note

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:
?? ? + ?? ? → ?? ? + ?? ? (where ?, ?, ?, ? are stoichiometric coefficients)
The change in molar Gibbs free energy of that reaction:
∆? = ??? + ??? − ??? + ???
Which can be rewritten as:
∆? = ∆? 0 + ?? ln ?
For an ideal gas, Chemical potential ?? is defined as
?? ?? = ??0 + ?? ??

??
?0

Therefore,
∆? = ? ??0 + ?? ??

??
??
??
??
+ ? ??0 + ?? ?? 0 − ? ??0 + ?? ?? 0 − ? ??0 + ?? ?? 0
0
?
?
?
?

At equilibrium ∆? = 0 and therefore:

0 = ∆? 0 + ?? ??

??
?0
??
?0

?

?

??
?0
??
?0

?

?

We recognise the term in brackets to be the dimensionless equilibrium constant as defined above.
It follows that:
• ∆? 0 = −?????
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Comparing these two expressions, we can write
??
?0
?=
??
?0

?

?

??
?0
??
?0

?

?

 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 ?? ? + 3 ?2 ? → ??2 ?3 (?)
The chemical equilibrium constant of this reaction is given by
?=

???2 ?3
2 3
???
??2

Assuming ?? and its oxide (??2 ?3 ) 100% pure, their activity will be 1.
Mathematically the activity can be expressed as: ??? = ???2 ?3 = 1
1
?= 3
??2
??2 =

3

?

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)
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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 ?? ?
,
??

o For a batch system: ?? = ? 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:
???
?? =
??

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.
????????
1 ???
.
??

?? = ?

(Note that a continuous system at steady state has

?? ?
??

= 0 and

?? ?
??

= 0).

o Units of rA are mol dm-3 s-1 or of similar nature.
Rate Law Expressions
• We obtain expressions for rates of reaction in terms of the concentrations of the species present.
These are termed “rate laws”.
• General rate law expression:
?? = ?(?? , ?? , ?? , … ) o Note that chemists often use [A] to denote the concentration of reagent A.
• In some cases, the rate law is a simple function: o For example, the reaction: A + B → products, may have rate law:
?? = −??? ?? o This example is 1st order with respect to A and 1st order with respect to B. The overall order of reaction is 2. o k is called the rate coefficient or rate constant.

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o Note that the units of k change depending on the overall order of the reaction. For first order it has units of time-1.
For second order it has units of conc.-1 time-1
In other cases, the rate law may not be such a simple expression:
? ? ?

1 ? ?. o For example: ?? = − 1+?
?
2 ?

o This example is 1st order with respect to A, but the order with respect to B varies between 0 and
1 depending on the magnitude of ?2 ?? .
The stoichiometric coefficients ? i in a reaction need to be kept in mind when considering the rates of reaction of different species: o For example, consider the reaction: A + 2B → 3C o In this case, every mole of A reacting consumes two moles of B and produces 3 moles of C o Mathematically: −?? = −

??
2

=

??
3

?

o In general, ? ? is a constant for a chemical reaction.
?

• The stoichiometric coefficient is often taken into account when defining the rate constant of simple-order reactions.
For instance, the rate constant k for the second-order reaction 2A → products is conventionally defined through ?? = −2???2 , because each of the reaction as written removes 2 molecules of A.
Rate law expressions cannot normally be written down by inspection. This is partly because the overall reaction may actually depend on a sequence of different reaction steps (the reaction mechanism). Measuring rate constants
In many cases, the change in concentration of reagents and/or products can be followed over time
(e.g. spectroscopically by following a change in the absorbance of a particular light frequency, or by following a change in conductivity or some other physical parameter). o This enables the order of reaction to be checked, and the rate constant obtained.
• Example: 1st order reaction, batch system, constant volume:

Integrating, we get

???
= −?1 ??
??
?? = ??0 exp −?1 ?

Graph of ln(CA) against t will be linear in this case with gradient –k1
•Example: 2nd order reaction, batch system, constant volume:
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Integrating, we get

???
= −?2 ??2
??
1
1
−
= −?2 ?
??0 ??

• Graph of 1/CA against t will be linear in this case with gradient k2.
In some cases, the concentration of one reagent may be greatly in excess so that it has negligible change during the reaction with another component. o For instance, a reaction A + 2B → products with rate law
?? = −??? ??2 may be treated as having approximate rate law
?? = −?′?? if the concentration of B is far greater than the concentration of A (?. ?. ? ≫ [?]). In this special case, experimental rate data could be analysed assuming pseudo-first-order behaviour.
• Rate constants can also be obtained from initial value of the rate of reaction at time t=0 if the form of the rate law is known.
• Note that rate constants have units (e.g. s–1 for a first-order reaction; [concentration]–1s–1 for a second-order reaction).
Reaction Mechanisms
An overall reaction may actually depend on a sequence of different reaction steps.
• “Elementary reactions” are the simplest reaction steps possible. They represent the actual molecular collisions occurring when chemical bonds are made and/or broken. o This means that the rate expression for an elementary reaction can be written down by inspection (unlike the overall rate law expression).
• Example: the reaction between NO and O2 is believed to take place according to the following reaction mechanism:
Reaction
No.
1
2
3

Reaction
2NO → N2O2
N2O2 → 2NO
N2O2 + O2 → 2NO2

Elementary reaction rate
(by inspection) k1[NO]2 k2[N2O2] k3[N2O2][O2] The overall reaction is 2NO + O2 → 2NO2.
The overall rate expression can‟t be written down by simple inspection.
In chemical kinetics, when two or more steps occur in series, then slowest step becomes the rate determining/ rate limiting step.

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• In some cases, the “rate-determining step” in the reaction mechanism can be identified and this might allow the overall form of rate law to be predicted. For instance in the mechanism for reaction between NO and O2, o If reaction 1 is the rate-determining step (and reaction 3 is rapid), we would anticipate the overall rate law to be rate = k1[NO]2 (and be independent of oxygen concentration). o On the other hand, if reaction 3 is the rate-determining step:
 We anticipate overall rate law to be rate = k3[N2O2][O2]
 But when reaction 3 is rate limiting (slowest), we know that reactions 1 and 2 are rapid.
This implies that 2NO ⇌ N2O2 will be at thermodynamic equilibrium.
Hence [N2O2] = K[NO]2 where K is an equilibrium constant.
 Hence in this case we expect overall rate = k3K[NO]2[O2]
In real life it is not possible to prove a reaction mechanism is correct from experimentally observed rate laws – all one can do is show that a proposed reaction mechanism is consistent with the experimental rate law.
• For instance, the reaction between NO and O2 is observed experimentally to be 2nd order with respect to NO, and 1st order with respect to O2. o This supports the mechanism proposed above with step 3 being the rate-determining step. o The evidence doesn‟t, however, prove that the proposed mechanism is correct – there may be another mechanism that gives a similar overall rate law. o More evidence, such as the detection of the N 2O2 intermediate by spectroscopic means, is required for one to be able to say the proposed mechanism is correct.
Steady State Hypothesis
The steady state hypothesis (SSH) is commonly used to predict overall rate laws from a reaction mechanism. • This states that the concentration of any reactive intermediate species involved may be taken as constant. i.e., they are being removed by reaction at the same rate that they are being formed.
At steady state, for intermediate compound [?]
?[?]
=0
??
Which can be rephrased as: ???? ?? ????????? ?? ? − ???? ?? ??????? ?? ? = 0
• For instance, in the mechanism for reaction between NO and O 2, we can treat N2O2 as a reactive intermediate and assume that it is at steady state.
• In that case:
?[? ? ]
SSH states that ??2 2 = 0 o But the elementary reactions imply:
?[?2 ?2 ]
??

= k1 [NO]2 – ?2 N2 O2 – ?3 N2 O2 O2 = 0
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o This gives
N2 O2

k1 [NO]2
=
?2 + ?3 O2

o Overall rate of product formation = ?3 N2 O2 O2 =

k 1 k 3 [NO ]2 O 2
? 2 +? 3 O 2

Ck [NO ]2 O 2
3 /? 2 ) O 2

3 o Overall rate of product formation = 1+(?

(where ? = ?1 /?2 )

Note that this derivation hasn‟t made an assumption about what is the rate-determining step. o If

?3
?2

O2 ≪ 1, then overall rate = Ck 3 [NO]2 O2 which corresponds to the equation envisaged if

step 3 was the rate determining step. o If

?3
?2

O2 ≫ 1, then overall rate = k1 [NO]2 which corresponds to the equation envisaged if step

1 was the rate determining step. o If

?3
?2

O2 ~1, then we need the full rate expression, as reaction steps 1 and 3 are both

influencing the overall rate.
• Note that in more complicated reaction mechanisms, it may be necessary to assume more than one intermediate species is at steady state.
Temperature dependence of Rate Constants
• Rate constants are often found to obey the Arrhenius equation:
? = ? ??? −

??
??

where ? is a constant and is called the pre-exponential factor, and ?? is termed the activation energy for the reaction.
• Examples:
Reaction
2N2O5(g) → 4NO2(g)+O2(g)
Cl(g)+O3(g) → ClO(g) +O2(g)
CH(g)+CH4(g) → C2H4(g)+H(g)

Activation Energy (?? ) (kJ mol-1)
160
2.1
−1.7

• If rate constants are measured at more than one temperature, then the Arrhenius parameters A and ?? may be obtained from a graph of ln k against 1/T. (Arrhenius plot).

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Fig. 2: The dependence of „k‟ on T

Fig. 3: The dependence of „ln k‟ on 1/T

Knowledge of the Arrhenius equation parameters allows us to predict reaction rates at other
(higher or lower) temperatures.
• The value of A corresponds to the rate constant predicted at infinite temperature. For first-order reactions it is usually of the order of 1013 s–1. This corresponds to the frequency of molecular vibrations. Bond breaking can‟t take place faster than this.
Ask yourself: What is the justification for the form of the Arrhenius equation?
Can we predict the parameters A and ?? without measuring them directly? We shall describe briefly two models that help answer these questions: o kinetic theory of gases o transition state theory
Kinetic theory of gases
Kinetic theory of gases is used for predicting rates constants of gas-phase reactions.
• Consider the elementary gas-phase reaction A + B → products
• The number of collisions per second per unit volume between gas molecule A and gas molecule
B depends on their concentration:
ZAB = α CACB
• For ideal gases α may be predicted from the kinetic theory of gases. This assumes that molecules in gases behave as solid spheres moving in random directions with a distribution of velocities. Consequently, number of collisions per second depends on the size of the molecules, their mass, and it has weak temperature dependence.
• However, only a certain fraction of collisions will result in reaction. The molecules need sufficient energy to break the bonds necessary for the reaction to take place: the probability that they have
?
sufficient energy is given by ??? − ??? where ?? is an activation barrier that needs to be overcome. o There will also be an orientation dependence; for instance, the molecules may need to collide
“head-on” or “side-on” to react. The fraction of molecules that collide in the correct orientation is φ.
• This leads to an overall expression:
?
?
?
rate = [∅? ??? − ??? ]?? ?? = ∅??? ?? ??? − ??? = ? ??? − ???
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from which the Arrhenius parameters A and ?? can be identified.
Transition State Theory
During a chemical reaction, there is an energy barrier that needs to be overcome (otherwise the reaction would take place instantaneously).
• This can be represented diagrammatically in a reaction profile diagram:

Fig. 4: Reaction profile diagram
• The reaction coordinate axis is a description of the course of the reaction:
e.g. for H + F2 → HF + F it could be related to the H–F distance.
• The parameters A and ?? can be predicted theoretically using a science called statistical thermodynamics. The result (not proven here) is that: the parameter A includes, among other terms, the entropy change ΔS‡ to get to the transition state. o ?? is the enthalpy change (ΔH‡) to get to the transition state. For this reason, reaction profile diagrams are often drawn with enthalpy rather than energy on the vertical axis:

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Fig. 5: Reaction profile diagram
• Do not confuse the transition state with a reaction intermediate. o The transition state occurs at an energy maximum: such species can’t be detected though theoreticians infer that they “exist” for ~10–13 s (the duration of a molecular vibration). o A reaction intermediate occurs at a local energy minimum; such species exist for finite times and can be detected (e.g. spectroscopically) in some cases.

Fig. 6: Reaction profile diagram exhibiting reaction intermediates
Catalysts
A catalyst is a substance that increases the rate of a chemical reaction without being consumed in the process.

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• Catalysts function by altering the reaction mechanism to provide a more facile route for the reaction (e.g. one that has lower activation energy).
Note that catalysts can’t change the thermodynamics of a chemical reaction. o a thermodynamically unfavourable reaction will never “go”, no matter what catalyst is present. o You can‟t get more conversion than that predicted by thermodynamic equilibrium constants, no matter what catalyst is present.
• In homogeneous catalysis, the catalyst and reagents are all in the same phase. o Example: dissolving a few drops of Br2 in aqueous H2O2 causes decomposition of hydrogen peroxide to take place in a few minutes (rather than several weeks in the absence of Br2).
Probable mechanism:
?2 ?2 ?? + ??2 ?? → 2?? − ?? + 2? + ?? + ?2 (?)
?2 ?2 ?? + 2?? − ?? + 2? + ?? → 2?2 ? ? + ??2 ??
Overall reaction

2?2 ?2 ?? → 2?2 ? ? + ?2 (?)

• In heterogeneous catalysis, the catalyst and reagents are in a different phase.
Example: N2 + H2 do not normally react together, but in presence of a solid metal catalyst (usually
Fe) at 250°C they form ammonia (NH3).
Case Study
Oxidation of pure graphite, plasma sprayed substrate and spark plasma sintered pellets can be performed in Differential Scanning Calorimeter (DSC)/Thermo-Gravimetric Analyzer (TGA) (model:
Perkin Elmer, maximum temperature 1000˚C, maximum weight of sample = 100 ?, equipped with oxygen & nitrogen supply facility).
Since, Al2O3 remains chemically inert; Al2O3 powder can be taken as reference material into the sample pan in order to avoid contamination of pan from the sample or vice versa. Perkin Elmer
Thermal Analyzer software can be used for thermal analysis of the sample. The heating and cooling cycle can be divided in four steps:
a) Holding at 50˚C for 2 min in N2 atmosphere (flow rate: 20 mL/min)
b) Heating from 50˚C to T˚C (oxidation temperature) @10˚C/min in N2 atmosphere (flow rate:
10 mL/min)
c) Isothermal holding at T˚C for t hrs (holding time) in O2 atmosphere (flow rate: 10 mL/min)
d) Cooling from T˚C to 30˚C @ 10˚C/min in N2 atmosphere (flow rate: 10 mL/min)
Interpretation:
i) % weight loss can be plotted as a function of temperature. ii) Activation energy of oxidation is determined from rate of oxidation using Arrhenius relationship. The following model was used for activation energy determination.
Rate of oxidation (?? ) at temperature (?) can be defined as
?? =

?????? ?? ???? ?? ?????? ???? ?????? ??????? ?? ?????? ?????? ???? ?? ??? ???? ??????? ????
???? ?????? ??????? ???? ? ?? ??????????? ??

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Let ?1 and ?2 be the rates of oxidation at absolute temperatures T1 and T2 respectively, ?? be the activation energy necessary for oxidation. The reaction rates can be expressed as:
??

?1 = ? ??? − ??

1

??

?2 = ???? − ??

2

?1
?2

= ???⁡ −

??

1

?

?1

1

−?

2

Taking „??‟ on both sides
??

?1
?2

=−
?? ?

??

1

?

?1

1 2
?? = ? −?
??
1

2

1

−?

2

?1
?2

Here, R = Universal gas constant (8.314 J/mol.K)
The equation of linear fit of the straight line is
? = ?? + ? where, m is slope of the line, c is intercept.
Comparing above equations, it can be written that
??
?=−?
Using above relationships, activation energy can be calculated as
?? = ?. ?
(if ?? is expressed in kilo Joule)

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