Pseudo First Order Reaction
The order of a reaction is sometimes altered by conditions. Consider a chemical reaction between two substances when one reactant is present in large excess. During the hydrolysis of 0.01 mol of ethyl acetate with 10 mol of water, amounts of the various constituents at the beginning (t = 0) and completion (t) of the reaction are given as under.
The concentration of water does not get altered much during the course of the reaction. So, in the rate equation Rate = k'[CH3COOC2H5] [H2O] the term [H2O] can be taken as constant. he equation, thus, becomes Rate = k[CH3COOC2H5] where k = k'[H2O] and the reaction behaves as first order reaction. The molecularity of acidic hydrolysis of sucrose and esters is 2, whereas their order is 1. In both the reactions water is in excess so that its concentration remains constant throughout the reaction. The rate of reaction therefore depends only on the concentration of sucrose and ester in two reactions respectively. So the reactions in which the molecularity is 2 or 3 but they conform to the first order kinetics are known as pseudo first order reactions OR pseudo uni-molecular reactions.
C12H22O11 + H2O + H+ → C6H12O6(glucose) + C6H12O6(fructose)
CH3COOC2H5(ester) + H2O + H+ → CH3COOH + C2H5OH
(In both the reactions, H ion acts as a catalyst)
nth Order kinetics
Half-life (t1/2) : at t = t1/2 , [A]t =
⇒ k =
⇒ t 1/2 = ⇒
Problem : 5
For the non-equilibrium process, A + B → Products, the rate is first order w.r.t A and second order w.r.t. B. If 1.0 mole each of A and B are introduced into a 1 litre vessel and the initial rate were 1.0 x 10-2 mol/litre-sec, calculate the rate when half of the reactants have been used.
Sol. Rate = K[A] [B]2
Therefore, 10-2 = K 2
or K = 10-2 litre2 mol-2 sec-1
Now ratell = 10-2 x 0.5 x (0.5)2
or New rate = 1.2 x 10-3 mol/L-sec
Analysis of some important first-order reactions
Decomposition of Hydrogen peroxide (H2O2)
H2O2(g) → H2O(g) + 1/2O2(g)
The rate of this first order reaction is measured by titrating a fixed volume of H2O2 (undecomposed) against a standard solution of KMnO4. Here KMnO4 acts as oxidising agent and H2O2 as reducing agent. The volumes of KMnO4 used for H2O2 after regular intervals of time are as follows.
t = 0
Vol. of KMnO4
Volume of KMnO4 at t = 0 corresponds to volume of H2O2 initially present.
⇒ A0 ∝ V0
Volume of KMnO4 at time instants t1, t2, t3, .................... corresponds to volume of H2O2 remaining after t1, t2, t3, .................
⇒ A ∝ Vt
Now it being a first order reaction, follows first order kinetics, so
k t = 2.303 log10
Now using the above expression, if we calculate the values of k for different time intervals t1, t2, ........... (for actual numerical data), the values of k should be same if the reaction follows first order kinetics.
Decomposition of ammonium nitrite (NH4NO2)
and benzene diazonium chloride (C6H5N = NCl)
NH4NO2(g) → 2 H2O(g) + N2(g)
C6H5 - N = N - Cl(g) → C6H5 - Cl(g) + N2(g)
The rate of both the reaction is studied (measured) in similar manner. The volume of nitrogen (N2) is collected after a regular interval of time as follows
t = 0
Vol. of N2
At t = 0, clearly the volume of N2 = 0
Time instant t = ∞ means the end of a reaction i.e., when whole of NH4NO2 or C6H5 - N = N - Cl is decomposed.
⇒ At t = ∞, V∞ corresponds to the initial volume of NH4NO2 or C6H5 - N = N - Cl
(Note that the ratio of stoichiometric coefficient for both N2 : NH4NO2 or N2 : C6H5N = NCl is 1 : 1)
⇒ A0 ∝ V∞
At t = t1, t2, t3................ the volume of N2 corresponds to concentration of product formed i.e., equal to x.
⇒ x ∝ Vt
⇒ A0 - x ∝ V∞ - Vt
⇒ kt = 2.303 log10
Hydrolysis of Esters (CH3COOC2H5 )
CH3COOC2H5 (ester) + H2O + HCl(H+ ) → CH3COOH + C2H5OH
The reaction rate is measured by titrating the acid (CH3COOH) produced against a standard alkali solution. Note that when a test sample is prepared from the reacting mixture, there are two acids : one is mineral acid H (HCl or any other) and second is CH3COOH produced. So volume of alkali used gives titration value for both acids. The data is collected in the following manner.
t = 0
Vol. of NaOH
At t = 0, V0 is the volume NaOH used to neutralise the mineral acid present (H ) being used as catalyst.
(At t = 0, no CH3COOH is yet produced)
At t = ∞ (i.e., at the end of hydrolysis), V∞, is the volume of NaOH used to neutralise whole of CH3COOH plus vol. of HCl present At t = ∞, volume of CH3COOH corresponds to volume of ester taken initially
⇒ A0 ∝ V∞ - V0 (as V0 = vol. of HCl)
At t = t1, t2, t3............ V1, V2, V3, ................corresponds to vol. of HCl plus vol. of CH3COOH being produced.
⇒ x ∝ Vt - V0
⇒ A0 - x ∝ (V∞ - V0) - (Vt - V0)
⇒ A0 - x ∝ V∞ - Vt
⇒ kt = 2.303 log10
Inversion of Cane Sugar (C12H22O11)
C12H22O11 + H2O + H+ → C6H12O6(glucose) + C6H12O6 (fructose)
The rate is measured by measuring the change in the angle of rotation (optical activity) by a polarimeter. Sucrose is dextro-rotatory, glucose is dextro-rotatory and fructose is leavo-rotatory. The change produced in rotatory power in time t gives a measure of x, the quantity of sucrose decomposed in that time. The total change in the rotatory power produced at the end of the reaction gives the measure of A0, the initial concentration of sucrose.
If r0, r1 and r∞ represent rotations at the start of reaction, after time t and at the end of reaction respectively,
⇒ A0 ∝ r0 - r∞ and x ∝ r0 - rt
DECOMPOSITION OF AsH3(g)
In first-order reactions involving gases, sometime measuring the pressure of the reaction mixture is very good method for measuring reaction rates.
For example consider decomposition of arsine gas (AsH3)
AsH3(g) → As(s) +
The rate of reaction is measured as the increase in pressure of the reaction mixture. Note that there is an increase in number of moles of the gaseous products to the right, so as the reaction proceeds, there will be an increase in pressure of contents (P ∝ n).
Let the initial pressure of AsH3(g) is P0, if x is the decrease in pressure of AsH3(g) after time t.
Arsenic is solid, so P(AS) = 0
After time t, let Pt be the total pressure, then
Pt = P(AsH3) + P(H2) = (P0 - x) +
⇒ Pt = P0 + ⇒ x = 2(Pt - P0)
Now A0 ∝ P0
and A ∝ P0 - x ≡ P0 - 2 (Pt - P0) ≡ 3P0 - 2Pt
⇒ kt = 2.303 log10
On similar pattern, please try to write the expression for Ist order rate law for following first-order reactions. (in terms of P0 and Pt)
1. N2O(g) → N2(g) + O2(g)
2. (CH3)3C - O - O - C(CH3)(g) → 2(CH3)2C = O(g) + C2H6(g)
Complex (First order) kinetics
(A) Parallel Kinetics
Rate of change of A = [rate of change of A]I + [rate of change of A]II
- = K1 [A] + K2 [A], ,
% of B in the mix of A & B =
% of C in the =
Problem : 6
An organic compound A decomposes following two parallel first order mechanisms :
; and k1 = 1.3 x 10-5 sec-1.
Calculate the concentration ratio of C to A, if an experiment is allowed to start with only A for one hour.
But k1 = 1.3 x 10-5 sec-1 ; k2 = 9 x 1.3 x 10-5 sec-1 = 117 x 10-5 sec-1
(k1 + k2) = (1.3 x 10-5) + (11.7 x 10-5) sec-1 = 13 x 10-5 sec-1 ....(1)
Also ⇒ [B]t = ...(2)
⇒ ; = (k1 + k2)t
⇒ [from eq. (2)] ⇒
⇒ = 13 x 10-5 x 60x 60 = 0.468 [from eq. (1)]
⇒ 1 + = 1.5968 ;
(2) Series Kinetics
Graph of [A], [B], [C] Vs t:
Time when [B] is maximum
O < t < teq Chemical kinetics, teq t < Chemical Equilibrium
At equilibrium, rf = rb , k1 [A] equilibrium = k2 [B] equilibrium ,
= - k1 [A] + k2 [B], = - k1 [A] + k2 [ [A]o - [A] ] = - (k1 + k2) [A] + k2 [A]o
= (k1 + k2)
By substituting the value from equation (I)
Problem : 7 For a reversible first order reaction,
A B ; Kf = 10-2 sec-1
and = 4 ; If A0 = 0.01 ML-1 and B0 = 0, what will be concentration of B after 30 sec ?
Therefore, Kb = 0.25 x 10-2 and xeq = = 0.008
30 = log
Therefore, x = 2.50 x 10-3
(1) Initial Rate Method
A + B product
rate = k [A]m [B]n ; Order = m + n
2 x 10-3
4 x 10-3
32 x 10-3
Experiment (1) and Experiment (2)
n = 1
Experiment (1) and Experiment (3)
m = 4
Order (m + n) = 4 + 1 = 5
(2) Half - life method
2 hr 0.2
n = 0
(3) Integrated Rate law Method
t = 0 1000 M
t = 60 sec 100 M
t = 120 sec 10 M
n = 0 k = = = 15
n = 1
(4) Ostwald Isolation method
rate = k[A]m [B]n [C]o [D]p - - - - - - - -
Experiment 1 : [A] = In small quantity ; [B], [C], [D] - - - - - - - in excess
The rate equation reduces to
rate = k' [A]m r1 = k' [A]1m
r2 = k' [A]2m
Experiment 2: [B] = In small quantity . & [A], [C], [D] - - - - - - in excess.
rate = k' [B]n repeated
Order of reaction = m + n + o + p + - - - - - - - -
Activation Energy (Ea)
A mixture of magnesium and oxygen does not react at room temperature. But if a burning splinter is introduced to the mixture, it burns vigorously. Similarly a mixture of methane and oxygen does not react at room temperature, but if a burning match-stick is put in the mixture, it burns rapidly. Why it happen like this, that some external agents has to be introduced in order to initiate the reaction ?
According to the theory of reaction rates "for a chemical reaction to take place, reactant molecules must make collisions among themselves". Now in actual, only a fraction of collisions are responsible for the formation of products, i.e., not all collisions are effective enough to give products. So the collisions among reactant molecules are divided into two categories :-
Effective collisions and In-effective collisions
Effective collisions are collisions between the molecules which have energies equal to or above a certain minimum value. This minimum energy which must be possessed by the molecules in order to make an effective collision (i.e., to give a molecule of products) is called as threshold energy. So it is the effective collisions which bring about the occurrence of a chemical reaction.
Ineffective collisions are the collisions between the molecules which does not posses the threshold energy. These can not result in a chemical reaction.
Now most of the times, the molecules of reactants do not possess the threshold energy. So in order to make effective collisions (i.e., to bring about the chemical reaction), an additional energy is needed to be absorbed by the reactant molecules. This additional energy which is absorbed by the molecules so that they achieve the threshold energy is called as energy of activation or simply activation energy. It is represented as Ea.
A reaction which needs higher activation energy is slow at a given temperature.
For example : is faster at ordinary temperature whereas the following reaction :
CO(g) + → CO2(g) is slower at the same temperature as the value of Ea for the second reaction is much higher.
Factors Affecting Rate of Reaction
(a) Catalyst : The rate of reaction increased by addition of catalyst, because catalyst lowers, the activation energy and increased the rate of reaction. A catalyst is a substance which increases the rate of a reaction without itself undergoing any permanent chemical change. For example, MnO2 catalyses the following reaction so as to increase its rate considerably.
2KClO3 2 KCl + 3O2
The word catalyst should not be used when the added substance reduces the rate of reaction. The substance is then called inhibitor. The action of the catalyst can be explained by intermediate complex theory. According to this theory, a catalyst participates in a chemical reaction by forming temporary bonds with the reactants resulting in an intermediate complex. This has a transitory existence and decomposes to yield products and the catalyst.
It is believed that the catalyst provides an alternate pathway or reaction mechanism by reducing the activation energy between reactants and products and hence lowering the potential energy barrier as shown in Fig. It is clear from Arrhenius equation that lower the value of activation energy faster will be the rate of a reaction. A small amount of the catalyst can catalyse a large amount of reactants. A catalyst does not alter Gibbs energy, DG of a reaction. It catalyses the spontaneous reactions but does not catalyse non-spontaneous reactions. It is also found that a catalyst does not change the equilibrium constant of a reaction rather, it helps in attaining the equilibrium faster, that is, it catalyses the forward as well as the backward reactions to the same extent so that the equilibrium state remains same but is reached earlier.
(b) Temperature : With increase in temperature the rate of reaction increases. It is generally found for every 10o increase in temperature. The rate constant double.
The ratio of rate constants with 10o difference in their temperature is called temperature coefficient.
= Q = Temperature coefficient of reaction 2
(c) Concentration :
Rate = A e-Ea/RT [A]m [B]n - - - - - - - - -
With increase in concentration of reactants the rate of the reaction increases because number of, collision (effective collisions) increases.
(d) Nature of Reactants :
Ionic Reactants :
Generally ionic reactions in aq. media are fast than the reaction involving covalent reactants.
As covalent reactants involving breaking of bond then formation of bond where as ionic reaction involve in single step.
(e) Surface Area : Increase in surface area increases the number of collisions and hence rate increases
(f) Radiation : Some reactions exposes to sunlight also increases the rate of reaction.