(rate constant K is equal to collision frequency A if E_a = 0)
In zero order reactions, increase in the concentration of reactant does not alter the rate.
So statement (i) is wrong.
\(k=A{ e }^{ -\left( \frac { { E }_{ a } }{ RT } \right) }\)
if E_a = 0 so, statement (ii) is correct, and statement (iii) is wrong
k = Ae⁰
k = A
In k = In A - \({ \left( \frac { { E }_{ a } }{ R } \right) }\) \(\left( \frac { 1 }{ T } \right) \)
This equation is of the form of a straight line y = mx+c
A plot of In k Vs \(\left( \frac { 1 }{ T } \right) \) gives a straight line with a negative slope
So statement (iv) and (v) are wrong.

A catalyst provides a new path to the reaction with low activation energy. i.e., it lowers the activation energy.

\(k=\frac { 2.303 }{ t } log\frac { \left[ { A }_{ 0 } \right] }{ \left[ A \right] } \)
\(k=\frac { 1 }{ t } ln \frac { \left[ { A }_{ 0 } \right] }{ \left[ A \right] } \)
\(e^{kt}=\frac { \left[ { A }_{ 0 } \right] }{ \left[ A \right] }\)
[A] = [A₀] e^kt
In this case
k = x min^-1 and [A₀]  = 0.01 M = 1 x 10^-2M
t = 1 hour = 60 min
[A] = (1 x 10^-2)e^-60x

Let [A₀] = 100;
When t = t_99.9%; [A] = (100-99.9) = 0.1
\(k=\frac { 2.303 }{ t } \log\left( \frac { [{ A }_{ 0 }] }{ [A] } \right) \)
\({ t }_{ 99.9\% }=\frac { 2.303 }{ K } \log\left( \frac { 100 }{ 0.1 } \right) \)
\({ t }_{ 99.9\% }=\frac { 2.303 }{ K } \log1000\)
\({ t }_{ 99.9\% }=\frac { 2.303 }{ K } (3)\)
\({ t }_{ 99.9\% }=\frac { 6.909 }{ K } \)
\({ t }_{ 99.9\% }=10\times \frac { 0.69 }{ K } \)
\({ t }_{ 99.9\% }={ 10 } t_{ 1/2 }\)

Given data: Rate constant of a first order reaction (k) = 0.45 sec^-1
Formula: \({ t }_{ \frac { 1 }{ 2 } }=\frac { 0.693 }{ k } \)
Solution: \({ t }_{ \frac { 1 }{ 2 } }=\frac { 0.693 }{ k } =\frac { 0.693 }{ 0.45 } \)
Half-life period = 1.54 sec.

(i) The differential rate equation will be 
\(\frac { -d\left[ R \right] }{ dt } \) = k[A][B]²
(ii) if the concentration of B is increased three times, then
\(\frac { -d\left[ R \right] }{ dt } \) = k[A] [3B]²
= 9.k [A] [3B]²
∴ The rate of reaction will increase 9 times.
(iii) When the concentrations of both A and B are doubled
\(\frac { -d\left[ R \right] }{ dt } \) =k [A] [B]²
= k [2A] [2B]²
= 8.k [A] [B]²
∴ The rate of reaction will increase 8 times.

Let the concentration of the reactant be [A] = a
Rate of reaction R = k [A]²
R=ka²
(i) R = k (2a)²
= 4 ka²
= 4R
∴ The rate of the reaction would increase by 4 times.
(ii) If the concentration of the reactant is reduced to half [A] = \(\frac { 1 }{ 2 } \) a than the rate of reaction would be
\(R=k\left( \frac { 1 }{ 2 } a \right) 2\)
\(=\frac { 1 }{ 4 } k{ a }^{ 2 }\)
\(=\frac { 1 }{ 4 } kR\)
∴ The rate of the reaction would reduced by \({ \frac { 1 }{ 4 } }^{ th }\)

For first order reaction
\(k=\frac { 2.303 }{ t } \log { \frac { a }{ a-x } } \)
For t = t_1/2, \(x=\frac { a }{ 2 } \)
\({ t }_{ \frac { 1 }{ 2 } }=\frac { 2.303 }{ k } \log { \frac { a }{ a-\frac { a }{ 2 } } } \)
\(\frac { 2.303 }{ k } \log2=\frac { 2.303\times 0.3010 }{ 7.39\times { 10 }^{ -5 } } \)
= 9.38 x 10³ s.

Formula:
k₁ = A.e^-E_a/RT
Given:
Rate constant, k₁ - 3.0 x 10^-4 s^-1
Frequency factor, A = 6.0 x 10¹⁴ s^-1
Activation energy, E_a = 104.4 kJ
= 104400 J
Temperatures, T₁ = 25^o C; T₂ =∞
Solution:
\({ k }_{ 2 }=A.{ e }^{ -\frac { { E }_{ a } }{ RT } }\)
k₂ = 6.0 x 10¹⁴ x \(\frac { 104400 }{ { e }^{ 8.314\times \infty } } \)
k₂ = 6.0 x 10¹⁴ x 1   \(\left( \therefore e\frac { -{ { E }_{ a } } }{ R\infty } =1 \right) \)
∴ The rate constant at T ⟶ ∞ is 6.0 x 10¹⁴ s^-1

(i) (a) Molecularity of the reaction is the number of molecules taking part in an elementary step.
(b) For this we require at least a single molecule leading to the value of minimum molecularity of one.
(c) Hence, molecularity of any reaction can never be equal to zero.
(ii) For elementary reaction, (i.e) the reaction which proceeds in a single step, order, and molecularity have the same value.

(i) Reaction Rate = \(k{ \left[ A \right] }^{ 2 }{ \left[ B \right] }{ \left[ L \right] }^{ 3/2 }\) ...(1)
When [L] = [4L]
Rate = \(k{ \left[ A \right] }^{ 2 }{ \left[ B \right] }{ \left[ 4L \right] }^{ 3/2 }\)
The Reaction Rate = \(8(k{ \left[ A \right] }^{ 2 }{ \left[ B \right] }{ \left[ L \right] }^{ 3/2 })\) ...(2)
Comparing (1) and (2) rate is increased by 8 times
(ii)  [A] = [2A] and [B] = [2B]
Reaction Rate = \(k{ \left[ 2A \right] }^{ 2 }\left[ 2B \right] { \left[ L \right] }^{ 3/2 }\)
Reaction Rate = \(\\ 8(k{ \left[ A \right] }^{ 2 }{ \left[ B \right] }{ \left[ L \right] }^{ 3/2 })\)  ...(3)
Comparing (1) and (3) rate is increased by 8 times
(iii) \(\left[ A \right] =\left[ \frac { A }{ 2 } \right] \)
Reaction Rate = \(k{ \left[ \frac { A }{ 2 } \right] }^{ 2 }{ \left[ B \right] }{ \left[ L \right] }^{ 3/2 }\)
Reaction Rate = \(\frac { 1 }{ 4 } \left( k\left[ { A }^{ 2 } \right] { \left[ B \right] }{ \left[ L \right] }^{ 3/2 } \right) \) ...(4)
Comparing (1) and (4); rate is reduced to (1/4) times.
(iv) \(\left[ A \right] =\left[ \frac { 1 }{ 3 }A \right] and\left[ L \right] =\left[ 4L \right] \)
Rate = \(k{ \left[ \frac { 1 }{ 3 }A \right] }^{ 2 }{ \left[ B \right] }{ \left[ 4L \right] }^{ 3/2 }\)
Rate = \(\left( \frac { 8 }{ 9 } \right) \left( k{ \left[ A \right] }^{ 2 }{ \left[ B \right] }{ \left[ L \right] }^{ 3/2 } \right)\) ...(5)
Comparing (1) and (5); rate is reduced to \(\frac { 8 }{ 9 } \) times.

A Reaction whose rate depends on the reactant concentration raised to the first power is called a first order reaction. For a first order reaction,
A ⟶ product
Rate law can be expressed as
Rate = k[A]^-1
\(\frac { -d\left[ A \right] }{ \left[ A \right] } =k dt\)    ...(1)
Integrate the above equation between the limits of time t = 0 and time equal to t, while the concentration varies from the initial: concentration [A_o] to [A] at the later time.
\(\int _{ { [A }_{ 0 }] }^{ [A] }{ \frac { -d\left[ A \right] }{ \left[ A \right] } } =k\int _{ 0 }^{ t }{ dt } \)
\({ \left( -In\left[ A \right] \right) }_{ \left[ { A }_{ 0 } \right] }^{ \left[ A \right] }=k{ \left( t \right) }_{ 0 }^{ t }\)
\(-In\left[ A \right] -\left( In\left[ { A }_{ 0 } \right] \right) =k\left( t-0 \right) \)
\(-In\left[ A \right] -\left( In\left[ { A }_{ 0 } \right] \right) =kt\)
\(In\left( \frac { \left[ { A }_{ 0 } \right] }{ \left[ A \right] } \right) =kt\) ...(2)
This equation is in natural logarithm. To convert it into usual logarithm with base 10, we have to multiply the term by 2.303.

\(2.303\quad log\left( \frac { \left[ { A }_{ 0 } \right] }{ \left[ A \right] } \right) =kt\)
\(k=\frac { 2.303 }{ t } log\left( \frac { \left[ { A }_{ 0 } \right] }{ \left[ A \right] } \right) \)    ...(3)
Equation (2) can be written in the form
y = mx+c as below
In [A₀]-In[A] = kt
In[A] = In[A₀]-kt
⇒ y = c + mx
If we follow the reaction by measuring the concentration of the reactants at regular time interval 't' a plot of In[A] against 't' yields a straight line with a negative slope. From this, the rate constant is calculated.

Average rate of reaction:
The average rate is defined as the ratio of change in the final concentration of reactants and the initial concentration of reactants over the entire time period of reaction.
Average rate=\(\frac{-[Final \space concentration \space of \space reactants - Initial \space concentration \space of \space reactants}{(Change \space in \space time)}\)
R \(=\frac{-\left(\left[A_{2}\right]-\left[A_{1}\right]\right)}{\left(t_{2}-t_{1}\right)}=-\left(\frac{\Delta[A]}{\Delta t}\right)\)
[A]₁ = Concentration of reactant A₁ at time t₁
[A]₂ = Concentration of reactant A₂ at time t₂
Instantaneous rate of reaction:
The rate of reaction at any particular instant during the course of reaction is called as instantaneous rate.
Instantaneous rate \(=(\text { Average rate })_{\Delta t \rightarrow 0}\)
Rate of the reaction \(=\left(\frac{-\Delta \mathrm{A}}{\Delta \mathrm{t}}\right)\)

Given data:
k₁ = 2.45 x 10^-5 sec^-1; T₁ = 273 K
k₂ = 16.2 x 10^-4 sec^-1; T₂ = 303 K
R = 8.314 JK^-1 mol^-1
Formula: \(\log { \frac { { k }_{ 2 } }{ { k }_{ 1 } } } =\frac { { E }_{ a } }{ 2.303R } \left[ \frac { { T }_{ 2 }-{ T }_{ 1 } }{ { T }_{ 1 }{ T }_{ 2 } } \right] \)
Solution:
\(\log { \frac { 16.2\times { 10 }^{ -4 } }{ 2.45\times { 10 }^{ -5 } } } =\frac { { E }_{ a } }{ 2.303\times 8.314 } \left( \frac { 303-273 }{ 273\times 303 } \right) \)
\(1.8203=\frac { { E }_{ a } }{ 2.303\times 8.314 } \times \left[ \frac { 30 }{ 273\times 303 } \right] \)
∴ E_a = 52802.3 x 1.8203
= 96116 J / mol
E_a = 96.116 kJ mol^-1.

Rate = k [x]² = ka²
[x] = a mol^-1
If the concentration of x is in cross three time, then
(x) = 3a mol L^-1
Rate = R(3a)² = 9 ka²
Hence, the rate of formation will increase by 9 times.

Arrhenius suggested that the rates of most reactions vary with temperature in such a way that the rate constant is directly proportional to \({ e }^{ -\left( \frac { { E }_{ a } }{ RT } \right) }\) and he proposed a relation between the rate constant and temperature.
\(k=A{ e }^{ -\left( \frac { { E }_{ a } }{ RT } \right) }\) ....(1)
Where A the frequency factor,
R the gas constant,
E_a the activation energy of the reaction and,
T the absolute temperature (in K)
(ii) The frequency factor (A) is related to the frequency of collisions (number of collisions per second) between the reactant molecules. The factor A does not vary significantly with temperature and hence it may be taken as a constant.
(iii) E_a is the activation energy of the reaction, which Arrhenius considered as the minimum energy that a molecule must have to posses to react.
(iv) Taking logarithm on both side of the equation (1)
In k = In A + In \({ e }^{ -\left( \frac { { E }_{ a } }{ RT } \right) }\)
In k = In A -\(\left( \frac { { E }_{ a } }{ RT } \right) \)       (∴ In e = 1)
In k = In A - \({ \left( \frac { { E }_{ a } }{ R } \right) }\) \(\left( \frac { 1 }{ T } \right) \)....(2)
y = c = m x
The above equation is of the form of a straight line y = mx+c
(v) A plot of In k Vs \(\left( \frac { 1 }{ T } \right) \) gives a straight line with a negative slope -\(\frac { { E }_{ a } }{ R } \) If the rate constant for a reaction at two different temperatures is known, we can calculate the activation energy as follows.
At temperature T = T¹; the rate constant k = k₁
In k₁ = In A - \(^{ \left( \frac { { E }_{ a } }{ { RT }_{ 1 } } \right) }\)      ....(3)
At temperature T = T₂; the rate constant k = k₂
In k₂ = In A - \(\left( \frac { { E }_{ a } }{ { RT }_{ 2 } } \right) \)    ......(4)
(4) - (3)
In k₂ - In k₁ = - \(\left( \frac { { E }_{ a } }{ { RT }_{ 2 } } \right) \) +  \(^{ \left( \frac { { E }_{ a } }{ { RT }_{ 1 } } \right) }\)
In \(\left( \frac { { k }_{ 2 } }{ { k }_{ 1 } } \right) \) = \(\frac { { E }_{ a } }{ R } \) \(\left( \frac { 1 }{ T_{ 1 } } -\frac { 1 }{ { T }_{ 2 } } \right) \)
2.303 log \(\left( \frac { { k }_{ 2 } }{ { k }_{ 1 } } \right) \)= \(\frac { { E }_{ a } }{ R } \) \(\left( \frac { { T }_{ 2 }-{ T }_{ 1 } }{ { T }_{ 1 }{ T }_{ 2 } } \right) \)
log \(\left( \frac { { k }_{ 2 } }{ { k }_{ 1 } } \right) \) = \(\frac { { E }_{ a } }{ 2.303R } \) \(\left( \frac { { T }_{ 2 }-{ T }_{ 1 } }{ { T }_{ 1 }{ T }_{ 2 } } \right) \)
In k₂ - k₁ = - \(\left( \frac { { E }_{ a } }{ R{ T }_{ 2 } } \right) \) + \(\left( \frac { { E }_{ a } }{ R{ T }_{ 1 } } \right) \)
This equation can be used to calculate E_a from rate constants k₁ and k₂ at temperatures T₁ and T₂.

(i) Average rate of reaction between the time interval, 30 to 60 seconds
\(=\frac { d\left[ Ester \right] }{ dt } \)
\(=\frac { 0.31-0.17 }{ 60-30 } =\frac { 0.14 }{ 30 } \)
= 4.67 x 10^-3 mol L^-1 s^-1.
(ii) For a pseudo first order reaction,
\(k=\frac { 2.303 }{ t } \log { \frac { { \left[ R \right] }_{ 0 } }{ \left[ R \right] } } \)
For, t = 303
\({ k }_{ 1 }=\frac { 2.303 }{ t } \log { \frac { 0.55 }{ 0.31 } } \)
For, t = 60 s
\({ k }_{ 2 }=\frac { 2.303 }{ 60 } \log { \frac { 0.55 }{ 0.17 } } \)
For, t = 90 s
\({ k }_{ 3 }=\frac { 2.303 }{ 90 } \log { \frac { 0.55 }{ 0.085 } } \)
= 2.075 x 10^-2 s^-1
The average rate constant,
\(k=\frac { { k }_{ 1 }+{ k }_{ 2 }+{ k }_{ 3 } }{ 3 } \)
\(=\frac { \left( 1.911\times { 10 }^{ -2 } \right) +\left( 1.957\times { 10 }^{ -2 } \right) +\left( 2.075\times { 10 }^{ -2 } \right) }{ 3 } \)
= 1.98 x 10^-2 s^-1.

(i) Collision theory is based on the kinetic theory of gases. According to this theory, a chemical reaction occurs as a result of collisions between the reacting molecules.
(ii) Let us understand this theory by considering the following reaction.
A₂(g) + B₂(g) ⟶ 2AB(g)
(iii) If we consider that, the reaction between A₂ and B₂ molecules proceeds through collisions between them, then the rate would be proportional to the number of collisions per second.
(iv) Rate ∝ number of molecules colliding per litre per second (collision rate).
(v) The number of collisions is directly proportional to the concentration of both A₂ and B₂.
Collison rate ∝ [A₂][B₂]
Collision rate = Z [A₂][B₂]
(vi) Where, Z is a constant
(vii) A fraction of effective collisions (f) is given by the following expression
\(f={ e }^{ \frac { { -E }_{ a } }{ RT } }\)
(xiv) This fraction of collisions is further reduced due to orientation factor i.e., even if the reactant collides with sufficient energy, they will not react unless the orientation of the reactant molecules is suitable for the formation of the transition state.
(viii) The diagram illustrates the importance of proper alignment of molecules which leads to reaction.

(ix) The fraction of effective collisions (f) having proper orientation is given by the steric factor p.
⇒ Rate = p x f x collision rate
\(\Rightarrow Rate=p\times { e }^{ \frac { -Ea }{ RT } }\times Z\left[ { A }_{ 2 } \right] \left[ { B }_{ 2 } \right] \quad ...(1)\)
As per the rate law,
Rate = \(k=\left[ { A }_{ 2 } \right] \left[ { B }_{ 2 } \right] \quad ...(2)\)
Where k is the rate constant
On comparing equation (1) and (2), the rate constant k is
\(k=pZ{ e }^{ \frac { -Ea }{ RT } }\)

(a) Elementary reaction
Each and Every single step in a reaction mechanism is called an elementary reaction.
Rate = k[A] [B]
(b)

Order of reaction	Molecularity of a reaction
Order of reaction is the sum of the powers of concentration terms involved in the experimentally determined rate law.	Molecularity of a reaction is the total number of reactant species that are involved in an elementary step.
It can be zero (or) fractional (or) integer	It is always a whole number, cannot be zero or a fractional number.
It is assigned for a overall reaction.	It is assigned for each elementary step of the mechanism.

Rate = K[X]ⁿ[y]^m
4 x 10^-3 mol L^-1s^-1= 2 x 10^-2s^-1(0.2mol L^-1)ⁿ(0.2mol L^-1)^m
\(\frac { 4\times { 10 }^{ -3 }mol\quad { L }^{ -1 }{ s }^{ -1 } }{ 2\times { 10 }^{ -2 }{ s }^{ -1 } } =\) (0.2)^n+m(mol L^-1)^n+m
0.2(mol L^-1) = (0.2)^n+m(mol L^-1)^n+m
Comparing the powers on both sides
The overall order of the reaction n + m = 1

a) \(Rate=\frac { -1 }{ 2 } \frac { d[NO] }{ dt } =\frac { -d[{ O }_{ 2 }] }{ dt } =\frac { 1 }{ 2 } \frac { d\left[ { NO }_{ 2 } \right] }{ dt } \)
b) \(\frac { -d\left[ { O }_{ 2 } \right] }{ dt } =\frac { 1 }{ 2 } \frac { d\left[ { NO }_{ 2 } \right] }{ dt } \)
\(\frac { d[{ NO }_{ 2 }] }{ dt } =2 \times\left( \frac { -d\left[ { O }_{ 2 } \right] }{ dt } \right) =2\times 0.2 \ { mol \ L }^{ -1 }{ s }^{ -1 }\)
= 0.4 mol L^-1s^-1

For n^th reaction
\(K\alpha \frac { 1 }{ { C }^{ n-1 } } \)

The order of the reaction is zero.

Rate decreases with time as the reaction proceeds and the average rate cannot be used to predict the rate of the reaction at any instant. The rate of the reaction, at a particular instant during the reaction is called the instantaneous rate. So instantaneous rate in prepared over average rate.

The study of chemical kinetics helps us
(i) To determine the rate of a chemical reaction.
(ii) In optimizing the process conditions of industrial manufacturing processes.

(i) Let A = 100M, [A₀] - [A] = 20M,
For the zero order reaction
\(k=\left( \frac { \left[ { A }_{ 0 } \right] -\left[ A \right] }{ t } \right) \)
(i) 20% completion  \(k=\left( \frac { 20M }{ 20min } \right) \) = 1 mol L^-1 min^-1 
(ii) 80% completion
\(\mathrm{K}=1 \mathrm{~mol} \mathrm{~L}{ }^{-1} \mathrm{~s}^{-1} ;\left[\mathrm{A}_{0}\right]=100 \mathrm{M} ;\left[\mathrm{A}_{0}\right]-[\mathrm{A}]=80 \mathrm{M} ; \mathrm{t}=?\)
\(\therefore t=\left(\frac{\left[A_{O}\right]-[A]}{K}\right)=\frac{80}{1}=80 \mathrm{mins}\)

K = 1.54 \(\times\) 10^-3s^-1; t_1/2 = ?
\({ t }_{ \frac { 1 }{ 2 } }=\frac { 0.693 }{ k } \)
\(=\frac { 0.693 }{ 1.54\times { 10 }^{ -3 } } \)
= 450 s

(i) First order reaction
(ii) First order reaction
(iii) \(\frac{1}{2}+2=2 \frac{1}{2}\); Pseudo first order reaction

For the first order reaction
\(\mathrm{k}=\frac{2.303}{\mathrm{t}} \log \frac{\mathrm{A}_{\mathrm{o}}}{\mathrm{A}}\) (or)
\(\mathrm{k} =\frac{2.303}{t} \log \frac{P_o}{P_t} \)
Here, P_o=0.08 atm (P_t)=0.04 atm; t = 60 s
\(\mathrm{k}=\frac{2.303}{60} \log \frac{[0.08]}{[0.04]}\)
\(=0.0383 \times \log 2  \)
\(=0.0383 \times 0.3010  \)
\(=1.152 \times 10^{-2} \mathrm{~s}^{-1}\)

Reaction rate = k[x]^3/2[y]^1/2
(i) Over all order of reaction = (3/2 + 1/2)=2 
i.e., second order reaction.
(ii) Since the rate expression does not contain the concentration of z, the reaction is zero order with respect to z.