Chemical Kinetics
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Chemical Kinetics
The branch of physical chemistry which deals with study of rate of chemical reaction and mechanism by which the reaction proceed is known as chemical kinetics.
Rate of chemical reaction (r)
Rate of chemical reaction is defined as the change in concentration of reactant or product with respect to time. It is denoted by (r) and given by:
where Δx = change in concentration
and Δt = change in time.
It’s unit is mol L−1 s−1.
Types of rate of chemical reaction
1) Average rate of reaction
The rate of chemical reaction due to change in concn of reactant or product over specified interval of time is called average rate of reaction. It is denoted by rav.
Let us consider a chemical reaction:
The average rate of reaction is:
where −ve sign indicate decrease in concn and +ve sign indicate increase in concentration.
2) Instantaneous rate of reaction (r)
The rate of chemical reaction due to change in concn of reactant or product at a particular interval of time is called instantaneous rate of reaction. It is denoted by r and given by:
It is equal to the slope when a plot is drawn bet’n time and concentration.
3) Equivalent rate of reaction (req)
The rate of chemical reaction due to change in concn of reactant or product for one mole each of reactant and product is called equivalent rate of reaction. It is denoted by (req).
Let us consider a chemical reaction:
The equivalent rate of reaction is:
Questions
1*
For a reaction 2 O3 → 3 O2, the rate of decomposition of O3 is 1.2 × 10−4 mol L−1 s−1. What will be the rate of formation of O2 in soln?
The rate of given chemical reaction is:
Rate of decomposition of O3 = −Δ[O3]/Δt = 1.2 × 10−4 mol L−1 s−1.
Rate of formation of O2 = Δ[O2]/Δt = ?
rate of decomposition of O3 = rate of formation of O2
−(1/2) Δ[O3]/Δt = +(1/3) Δ[O2]/Δt
(1/2) × 1.2 × 10−4 = (1/3) Δ[O2]/Δt
3 × (1/2) × 1.2 × 10−4 = Δ[O2]/Δt
1.8 × 10−4 mol L−1 s−1 = Δ[O2]/Δt
Hence, the rate of formation of O2 is 1.8 × 10−4 mol L−1 s−1.
2)
For a reaction 2HI → H2 + I2, if the formation of iodine is 9.1 × 10−6 mol L−1 s−1, what will be the disappearance of HI?
The rate of given chemical reaction is:
rate of formation of iodine Δ[I2]/Δt = 9.1 × 10−6
rate of decomposition of HI [−Δ[HI]/Δt] = ?
rate of decomposition of HI = rate of formation of iodine.
−(1/2) Δ[HI]/Δt = Δ[I2]/Δt
(1/2) Δ[HI]/Δt = 9.1 × 10−6 mol L−1 s−1
Δ[HI]/Δt = 9.1 × 10−6 × 2 mol L−1 s−1
Δ[HI]/Δt = 1.82 × 10−5 mol L−1 s−1
Hence, the rate of disappearance of HI = 1.82 × 10−5 mol L−1 s−1.
3)
For a reaction 2N2O5 → 4NO2 + O2, the rate of disappearance of N2O5 is 4 × 10−6 mol L−1 s−1. What will be the rate of formation of nitrogen dioxide (NO2)?
The rate of given chemical reaction is:
rate of disappearance of N2O5 = −Δ[N2O5]/Δt = 4 × 10−6
rate of formation of NO2 = Δ[NO2]/Δt = ?
rate of disappearance of N2O5 = rate of formation of NO2
R = Δ[N2O5]/Δt = (1/4) Δ[NO2]/Δt
R = (1/2) × 4 × 10−6 = (1/4) Δ[NO2]/Δt
R = (4/2) × 4 × 10−6 = Δ[NO2]/Δt
R = 8 × 10−6 mol L−1 s−1 = Δ[NO2]/Δt
Hence, the rate of formation of NO2 is 8 × 10−6 mol L−1 s−1.
4)
For a reaction 2N2O5 → 4NO2 + O2, calculate the rate of each component in mol s−1 when 2.24 litre of O2 gas that NTP are produce in 30 min. Calculate the:
Rate of disappearance [−ΔN2O5/Δt] = ?
Rate of formation of Δ[NO2]/Δt = ?
Rate of formation of Δ[O2]/Δt = ?
No. of mole of O2 = given vol / 22.4
= 2.24 / 22.4
= 0.1 mol of O2
No of concentration of [O2] = 0.1 mol.
Rate of formation of O2 = +Δ[O2]/Δt
= 0.1 / (30 × 60)
= 5.55 × 10−5 mol s−1
Rate of formation of NO2 = rate of formation of O2
(1/4) Δ[NO2]/Δt = Δ[O2]/Δt
(1/4) Δ[NO2]/Δt = 5.55 × 10−5
Δ[NO2]/Δt = 4 × 5.55 × 10−5
Δ[NO2]/Δt = 2.22 × 10−4
Rate of formation of N2O5 = Rate of formation of NO2
−(1/2) Δ[N2O5]/Δt = (1/4) × 2.22 × 10−4
(1/2) Δ[N2O5]/Δt = (1/4) × 2.22 × 10−4
Δ[N2O5]/Δt = 2 × (1/4) × 2.22 × 10−4
Δ[N2O5]/Δt = 1.11 × 10−4
Factors affecting on rate of chemical reaction
- Concn of reactant
- Surface area of reactant
- Temperature
- Catalyst
- Nature of reactant
a) Concn of reactant
With increase in concn of reactant increases rate of chemical reaction due to increase in probability of effective collision to give product.
b) Surface area of reactant
Higher the surface area of reactant expose for rxn higher will be the right contact for effective collision of reactant and faster will be the rate of chemical reaction.
c) Temperature
With increase in temp increases the K.E of reactant and contact molecule and hence rate of rxn also increases. The effect of temp in rate of reaction is express in terms of temp coefficient (T.C).
What is temp coefficient?
It is defined as the ratio of rate constant of a reaction at 2 different temperature differing by 10°C.
d) Catalyst
Use of +ve catalyst increases the rate of reaction by decreasing activation energy and −ve catalyst decreases the rate of reaction by increasing activation energy.
e) Nature of reactant
The chemical reaction involves by using similar substance is fast due to proton transfer but the reaction by non-polar and covalent substance is slow.
Rate law / rate law equation / rate law expression
Rate law expression is defined as the equation which shows how rate of a rxn is related with concentration of reactant with product order.
Let us consider a chemical rxn:
It’s rate law is:
where r = rate of chemical rxn
K = rate constant
x = order of rxn w.r.t A
y = order of rxn w.r.t B
z = order of rxn w.r.t C
When molar concn [A] = [B] = [C] = 1 mol L−1, then eqn (1) becomes:
Hence, rate constant (K) is defined as the rate of chemical rxn when concentration of each reactant is 1 mol L−1.
Again, if x = y = z = 0 then eqn can be written as r = K.
Therefore rate constant K can be written as also be defined as the rate of chemical rxn when power of each reactant is zero.
Order of reaction
The actual concentration of reactant which affect the rate of chemical rxn is called order of rxn. Mathematically it is the sum of power of concentration of reactant in rate law.
Let us consider a chemical rxn:
x + y + z = overall order of reaction.
i) If x + y + z = 0, then the rxn is zero order.
ii) If x + y + z = 1, then the rxn is first order.
iii) If x + y + z = 2, then the rxn is second order.
iv) If x + y + z = 3, then the rxn is third order.
Molecularity of rxn
The no of molecule involve in chemical rxn for effective collision to give product at a particular condition of temp and pressure is called molecularity of rxn. Depending upon the molecule rxn is divided into uni, bi and tri-molecular rxn.
If a balanced chemical rxn involves only one mole of reactant then the rxn is called uni-molecular rxn.
Example
Difference bet’n Order and Molecularity
| Order | Molecularity |
|---|---|
| i) It is sum of the power of concn terms raised in the rate law equation. | i) It is number of atoms or ions or molecules that collide simultaneously to give products in elementary reaction. |
| ii) It may be zero, small −ve, +ve or fractional number. | ii) It has only a small whole number of positive integral value except zero and more than three. |
| iii) It is determined for overall reaction. | iii) Each reaction of a multistep rxn have their molecularity. Molecularity of overall multistep reaction is molecularity of the slowest step. |
| iv) Order is same for whole rxn whether it is simple or complex. | iv) Molecularity of complex reaction has no meaning and is expressed for each elementary step. |
Zero order Rxn
The chemical rxn whose rate is independent on the initial concn of reactant is called zero order reaction.
Let us consider a chemical reaction:
It’s rate law is:
where K0 is rate constant of zero order rxn.
Unit of rate constant of zero order rxn is mol L−1 s−1.
Example
Decomposition of ammonia on Pt surface:
Decomposition of HI on gold surface:
Rxn bet’n H2 and Cl2 in presence of sunlight:
Integrated rate law for zero order rxn
Let us consider a chemical rxn:
It’s rate law is:
Let us consider initial concn of reactant (A) is a mol L−1. After sometime x mol L−1 product is obtain then concn of reactant (A) at time t is (a−x) mol L−1.
| A | Product | |
|---|---|---|
| t = 0 | a mol L−1 | 0 |
| time = t | (a−x) mol L−1 | x mol L−1 |
At time t, [A] = (a−x) mol L−1.
r = K0[a−x]0
dx/dt = K0 × 1
dx = K0dt
on integration:
∫dx = K0∫dt
x = K0t + C …(2)
Initially t = 0 and x = 0
0 = K0 × 0 + C
C = 0
putting the value of C in eqn (2):
x = K0t + 0
x = K0t
K0 = x/t …(3)
Which is required expression for integrated rate constant of zero order rxn.
Half-life period
The time required to complete half of the reactant converted into product is called half life period.
It is denoted by t1/2 or t0.5.
At half-life period:
then eqn (3) becomes:
t1/2 = a / 2K0 …(4)
t1/2 ∝ a
Hence, half-life period of zero order rxn is directly proportional to initial concn of reactant.
First Order Reaction
The chemical rxn whose rate depends on first power of initial concn of reactant is called first order reaction. It’s unit is [Unreadable in source].
Integrated rate law of first order Rxn
Let us consider a chemical rxn:
It’s rate law is:
Let us consider initial concn of reactant (A) is a mol L−1. After sometime x mol L−1 product is obtained then concn of reactant (A) at time t is (a−x) mol L−1.
| A | Product | |
|---|---|---|
| t = 0 | a mol L−1 | 0 |
| t = t | (a−x) mol L−1 | x mol L−1 |
At time t, [A] = (a−x) mol L−1.
r = K1[A]1
dx/dt = K1(a−x)
dx/(a−x) = K1dt
on integration:
∫ dx/(a−x) = K1 ∫dt
−ln(a−x) = K1t + C …(2)
Initially t = 0 and x = 0
putting t = 0 and x = 0 in eqn (2):
−ln(a−0) = K1 × 0 + C
−ln(a) = C
∴ C = −ln(a)
putting ‘C’ in eqn (2):
−ln(a−x) = K1t − ln(a)
ln(a) − ln(a−x) = K1t
ln [a/(a−x)] = K1t
2.303 log [a/(a−x)] = K1t
K1 = (2.303/t) log [a/(a−x)] …(3)
which is required expression for integrated rate constant of first order rxn.
Half life period
The time required to complete half of the reactant converted into product is called half life period.
t = t1/2 and x = a/2
the eqn (3) becomes:
K1 = (2.303/t) log [a/(a−x)]
K1t1/2 = 2.303 log [a/(a−a/2)]
K1t1/2 = 2.303 log [a/(a/2)]
t1/2 = (2.303/K1) log [2a/a]
t1/2 = (2.303/K1) log(2)
t1/2 = 0.693/K1 …(4)
Numerical
2x + y → product
The rate law of given chemical rxn is:
From 1st exp:
from 2nd exp:
from 3rd exp:
from 4th exp:
Dividing eqn (2) by (5):
(7 × 10−3) / (2.8 × 10−2) = [K(0.1)a(0.1)b] / [K(0.4)a(0.1)b]
0.25 = (0.1/0.4)a
(0.25)1 = (0.25)a
a = 1
Similarly, dividing eqn (3) by (4):
(8.4 × 10−2) / (3.36 × 10−1) = [K(0.3)a(0.2)b] / [K(0.3)a(0.4)b]
0.25 = (0.2/0.4)b
0.25 = (0.5)b
(0.5)2 = (0.5)b
b = 2
i) order w.r.t x is first and y is second order and overall order of rxn is a+b = 1+2 = 3rd order.
ii) From eqn (2):
7 × 10−3 = K[0.1]1[0.1]2
7 × 10−3 = K[0.1 × 0.01]
7 × 10−3 = K[1 × 10−3]
7 = K
K3 = 7 L2 mol−2 s−1
iii) Write its rate law.
iv) half life of rxn with respect to x:
t1/2 = 0.693/k
= 0.693/7
t1/2 = 0.099 sec.
v) Rate formation of product when [x] = 0.6 mol L−1 and y = 0.3 mol L−1.
(r) = K3[x]1[y]2
= 7 × (0.6)1 × (0.3)2
r = 0.378 mol L−1 s−1
Kinetic plot of first order rxn
We know:
log [a/(a−x)] = (K1/2.303) × t …(1)
eqn (1) is in the form eqn of straight line with slope K1/2.303.
Again,
log a − (K1/2.303) × t = log(a−x) …(2)
eqn (2) is in the form eqn of straight line with slope −K1/2.303.
Second order rxn
The chemical rxn whose rate depends on second power of initial concn of reactant is called second order reaction.
Integrated rate law of second order rxn
Let us consider a chemical rxn:
It’s rate law is:
where K2 → rate constant of second order rxn and it’s unit is L mol−1 s−1.
Let us consider initial concn of reactant (A) is a mol L−1. After sometime x mol L−1 product is obtained. Then concentration of reactant (A) at time t is a−x mol L−1.
| A | Product | |
|---|---|---|
| t = 0 | a mol L−1 | 0 |
| time = t | (a−x) mol L−1 | x mol L−1 |
At time t, [A] = (a−x) mol L−1.
dx/dt = K2(a−x)2
dx/(a−x)2 = K2dt
on solving:
K2 = x / [t a(a−x)]
Half-time period (t1/2 or t0.5)
The time required to complete half of the reactant convert into product is called half-life period.
It is denoted by t1/2 or t0.5.
At half-life period:
∴ t1/2 = 1 / K2a
Third order rxn
The chemical rxn whose rate depends on third power of initial concn of reactant is called third order reaction.
Integrated rate law of third order rxn
Let us consider a chemical rxn:
It’s rate law is:
where K3 → rate constant of third order rxn and it’s unit is L mol−1 s−1.
Let us consider initial concn of reactant (A) is a mol L−1. After sometime x mol L−1 product is obtained. Then concn of reactant (A) at time t is a−x mol L−1.
| A | Product | |
|---|---|---|
| t = 0 | a mol L−1 | 0 |
| time = t | (a−x) mol L−1 | x mol L−1 |
At time t, [A] = (a−x) mol L−1.
r = K3[a−x]3
dx/dt = K3(a−x)3
dx/(a−x)3 = K3dt
on solving:
K3 = x(2a−x) / [2ta2(a−x)2]
Half-life period (t1/2 or t0.5)
The time required to complete half of the reactant convert into product is called half-life period.
It is denoted by t1/2 or t0.5.
At half life period:
∴ t1/2 = 3 / (2K3a2)
Pseudo order rxn
The chemical rxn which appears to be higher order but actually follow lower order kinetics is called pseudo order rxn.
Example: Acid catalysed esterification (hydrolysis of ester)
r = K[R′−COOR][H2O]
r = K′[R′−COOR]
Collision theory / Activation Energy and activated complex
According to this theory, a chemical reaction takes place only due to collision bet’n reactant molecules having certain fraction of total collision are capable to give product. Such collision are called effective collision.
Condition for effective collision
- The colliding molecules must possess sufficient energy in excess than average energy.
- The orientation of molecule energy must be presented enough to produce product. The minimum amount of additional energy is called activation energy.
The conversion of reactant to product through the formation of product at the top of the energy barrier is known as activated complex or transition state (T.S).
Catalysis
A substance which alters the rate of chemical rxn but itself remains chemically unchanged at the end of the rxn is known as catalyst or catalytic agent and the phenomenon is known as catalysis. It may be classified as:
Positive catalyst
The catalyst which increase the rate of a chemical rxn but itself remains chemically unchanged at the end of the rxn is called positive catalyst.
eg:
Negative catalyst
The catalyst which decrease the rate of chemical rxn is called negative catalyst and the phenomenon is known as negative catalysis. A negative catalyst is also known as an inhibitor and process is known as inhibition.
Autocatalyst
When one of the product of a rxn itself act as a catalyst for the rxn then such a product is known as autocatalyst and phenomenon is known as autocatalysis.
Induced catalyst
When one rxn catalyse the other rxn and both rxn are occurs in the same vessel such rxn is called induced catalyst and phenomenon is known as induced catalysis. eg: oxidn of sodium sulphite catalysed the oxidn of sodium arsenite.
Different types of catalysis
- Homogeneous catalysis
- Heterogeneous catalysis
1) Homogeneous catalysis
A catalytic reaction in which the catalyst, the reactant and the products are present in the same phase are called homogeneous catalysis.
2) Heterogeneous catalysis
The catalytic rxn in which the catalyst present in a different physical phase form the reactant is termed as heterogeneous catalysis or contact catalysis.
Characteristics of catalysis
- A catalyst remains unchanged in mass and chemical composition at the end of a chemical rxn.
- A small amount of catalyst is sufficient to catalysed the rate of chemical rxn.
- A catalyst is highly effective when it is finely divided.
- A catalyst is specific in nature.
- Activity of catalyst is increased in the presence of promoters.
- A catalyst does not affect the equilibrium position.
Acid-base catalysis
Homogeneous catalytic rxn which are catalyzed by acids or base are known as acid-base catalysis.
Acid catalysis
The rxn in which are catalysed by H+ ions, dissociated acid or cations of weak base are known as acid catalysis.
Base catalysis
The reaction which are catalysed by OH− ion, dissociated base or anion of weak base are known as base catalysis.
Enzyme Catalysis
Enzyme catalysts are the protein molecule present in the living system which catalyse the rxn taking place in the living cell and also help in controlling their rxn rates. The rxn occurs in a specific site on the protein molecule called active site. The reactant in an enzyme rxn are reformed to an substrate. Enzyme catalyst are highly efficient biological catalysts.
Mechanism of enzyme catalysed rxn on the basis of lock and key theory (mode)
By Fischer in 1898 and Paul Erld and D-D Woods.
According to this theory a particular lock can be opened by a particular key. In the same way particular enzyme act on a particular substrate. The molecule of substrate which have complementary shape fit into these cavities just as a key fit into lock. Then the enzyme formed an activated complex with the substrate and its decomposition result into product.
Enzyme
They are long chain of protein molecule in which molecules are [Unreadable in source] each other to form a rigid colloidal particles with cavities on surface. These cavities have characteristics shape and have many active groups.
Discussion
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