t_1/2 α \(\frac{1}{[A_{0}]^{n-1}}\)...(1)
If [A₀ = 2[A₀]; then t_1/2 = 2t_1/2
2t_1/2 α \(\frac{1}{[2A_{0}]^{n-1}}\)...(2)
(2)/(1) = \(2= \frac{1}{[2A_{0}]^{n-1}} \times \frac{1}{[A_{0}]^{n-1}}\)
\(2= \frac {[2A_{0}]^{n-1}} {[A_{0}]^{n-1}}\)
\(2= \frac {1} {2}^{n-1}\)
2 = (2^-1)^n-1
2¹ = (2^-n+1)
n = 0

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

c) Assertion is true but reason is false

Average rate = \(-\frac { \triangle \left( R \right) }{ \triangle t } =-\frac { { \left[ R \right] }_{ 2 }-{ \left[ R \right] }_{ 1 } }{ { t }_{ 2 }-{ t }_{ 1 } } \)
\(=-\frac { 0.02M-0.03M }{ 25min } =\frac { -0.01M }{ 25min } \)
= 4 x 10^-4 M min^-1 and 
= \(-\frac { -0.01m }{ 25\times 60 } \) = 6.66 x 10^-6 Ms^-1

(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.

Given t_1/2= 0.6392 \(\times\)10⁴ s
To solve: when t = 100 min,
\(\frac { [{ A }_{ 0 }]-[A] }{ [{ A }_{ 0 }] } \times 100=?\)
We know that
For a first order reaction, \({ t }_{ 1/2 }=\frac { 0.6932 }{ k } \)
\(k=\frac { 0.6932 }{ 6.932\times { 10 }^{ 4 } } \)
\(k={ 10 }^{ -5 }{ s }^{ -1 }\)
\(k=\left( \frac { 1 }{ t } \right) In\left( \frac { [{ A }_{ 0 }] }{ [{ A }] } \right) \)
\({ 10 }^{ -5 }{ s }^{ -1 }\times 100\times 60s=In\left( \frac { [{ A }_{ 0 }] }{ [{ A }] } \right) \)
\(0.06-In\ \left( \frac { [{ A }_{ 0 }] }{ [{ A }] } \right) \)
\(\frac { [{ A }_{ 0 }] }{ [{ A }] } ={ e }^{ 0.06 }\)   ?(given : e^0.06 = 1.06)
\(\frac { [{ A }_{ 0 }] }{ [{ A }] } =1.06\)
\(\therefore \frac { [{ A }_{ 0 }]-[{ A }] }{ [{ A }_{ 0 }] } \times 100\%\)
\(=\left( 1-\frac { [{ A }] }{ [{ A }_{ 0 }] } \right) \times 100\%\)
=\(\left( 1-\frac { 1 }{ 1.06 } \times 100\% \right) \)  
= 5.66%

Pseudo first order reaction:
A second order reaction can be altered to a first order reaction by taking one of the reactants in large excess, such reaction is called pseudo first order reaction.
Example: Acid Hydrolysis of an ester,
\(C{ H }_{ 3 }COOC{ H }_{ 3(aq) }+{ H }_{ 2 }{ O }_{( l) }\overset { { H }^{ + } }{ \longrightarrow } C{ H }_{ 3 }COO{ H }_{ (aq) }+C{ H }_{ 3 }O{ H }_{ (aq) }\)
Rate = \(k\left[ { CH }_{ 3 }COOC{ H }_{ 3 } \right] \left[ { H }_{ 2 }O \right] \)
If the reaction is carried out with a large excess of water, there is no significant change in the concentration of water during hydrolysis. i.e., the concentration of water remains almost constant.
Now, we can define k [H₂O] = k'; Therefore the above rate equation becomes
Rate = \(k'\left[ { CH }_{ 3 }COOC{ H }_{ 3 } \right] \)
Thus it follows first order kinetics.

(i) A catalyst is substance which alters the rate of a reaction without itself undergoing any permanent chemical change. They may participate in the reaction, but again regenerated and the end of the reaction. In the presence of a catalyst, the energy of activation is lowered and hence, greater number of molecules can cross the energy barrier and change over to products, thereby increasing the rate of the reaction.
(ii) The reaction between KMnO₄ and H₂SO₄ and oxalic acid is catalysed by MnSO₄ and increases the rate of oxidation of C₂O₄^2- by MnO₄^-.

a) First order with respect to Br⁻, first order with respect to BrO₃⁻ and second order with respect to H⁺. Hence the overall order of the reaction is equal to 1 + 1 + 2 = 4
b) Order of the reaction with respect to acetaldehyde is \(\frac{3}{2}\) and overall order is also \(\frac{3}{2}\)

\(\mathrm{k}=\mathrm{Ae}^{-\mathrm{E}_{\mathrm{a}} / \mathrm{RT}}\)
\(\log \mathrm{k}=\frac{-\mathrm{E}_{\mathrm{a}}}{2.303 \mathrm{RT}}+\log \mathrm{A} \text { (or) } \log \mathrm{A}=\log \mathrm{k}+\frac{\mathrm{E}_{\mathrm{a}}}{2.303 \mathrm{RT}}\)
\(\mathrm{k}=1.8 \times 10^{-5} \mathrm{~s}^{-1} ;\)
\(\mathrm{Ea}=22.5 \mathrm{k} \mathrm{Cal} \mathrm{mol}^{-1}=22500 \mathrm{Cal} \mathrm{mol}^{-1} \)
\(\log A=\log \left(1.8 \times 10^{-5}\right)+\frac{22500}{2.303 \times 1.987 \times 313} \)
\(=\log 1.8-5 \log {10}+15.71 \)
\(=0.2553-5+15.71 \)
\(\log A=10.9653 \)
\(A=\text { Antilog } 10.9653 \)
\(=9.232 \times 10^{10} \text { collisions } \mathrm{s}^{-1} \text {. }\)

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