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

Rate = k[A]ⁿ
Rate = \(\frac{-\mathrm{d}[\mathrm{A}]}{\mathrm{dt}}\)
unit of rate = \(\frac{mol L^{-1}}{s}\)=mol L^-1/s^-1
unit of rate constant 
\(=\frac{ (mol{ L }^{ -1 }{ S }^{ -1 }) }{ ({ mol }{ L }^{ -1 })^n } \)
= mol^1-nL^n-1s^-1
in the case
rate = k [Acetone]^3/2
n = 3/2
= mol^1-(3/2)L^(3/2)-1s^-1
(mol^-(1/2) L^(1/2) s^-1).

It is know that,
\(t=\frac { 2.303 }{ k } \log { \frac { { \left[ R \right] }_{ 0 } }{ \left[ R \right] } } \)
= \(\frac { 2.303 }{ 60{ s }^{ -1 } } \log\frac { 1 }{ \frac { 1 }{ 16 } } \)
= \(\frac { 2.303 }{ 60{ s }^{ -1 } } \log16\)
= 4.6 x 10^-2 s (approximately)
Hence the required time is 4.6 x 10^-2 s

\(2{ NH }_{ { 3 }_{ (g) } }\overset { pt }{ \longrightarrow } { N }_{ { 2 }_{ (g) } }+3{ H }_{ { 2 }_{ (g) } }\)
Rate = \(-\frac { 1 }{ 2 } \frac { d\left[ { NH }_{ 3 } \right] }{ dt } =\frac { d\left[ { N }_{ 2 } \right] }{ dt } =\frac { 1 }{ 3 } \frac { d\left[ { N }_{ 2 } \right] }{ dt } \)
However, it is given that the reaction is of zero order. Therefore
\(-\frac { 1 }{ 2 } \frac { d\left[ { NH }_{ 3 } \right] }{ dt } =\frac { d\left[ { N }_{ 2 } \right] }{ dt } =\frac { 1 }{ 3 } \frac { d\left[ { N }_{ 2 } \right] }{ dt } k\)
= 2.5 x 10^-4 mol L^-1 S^-1
The rate of production of N₂ is
\(\frac { d\left[ { N }_{ 2 } \right] }{ dt } \)=  2.5 x 10^-4 mol L^-1 S^-1
The rate of production of H₂ is
\(\frac { d\left[ { N }_{ 2 } \right] }{ dt } \)= 3 x 2.5 x 10^-4 mol L^-1 s^-1
= 7.5 x 10^-4 mol L^-1 s^-1.

(i) All radioactive transformations follow first order kinetics. For example,
₉₂U²³⁸ ⟶ ₉₀U²³⁴ +₂He⁴
(ii) Decomposition of sulphuryl chloride in the gas phase proceeds by first order kinetics.
SO₂Cl_2(g) ⟶ SO_2(g) + Cl_2(g)
(iii) Inversion of sucrose in acidic aqueous medium follows first order reaction.
C₁₂H₂₂O₁₁ +H₂O \(\overset { H+ }{ \longrightarrow } \) C₆H₁₂O₆ +C₆H₁₂O₆

(i) The step which has the lowest rate value among the other steps of the reaction is called as the rate determining step (or) rate limiting step: (or)
(ii) The overall rate of a reaction is controlled by the slowest step in a reaction called the rate determining step.
Example:
\(2 \mathrm{~A}+\mathrm{B} \rightarrow \mathrm{C}+\mathrm{D}\) going by two steps like,
\( \mathrm{A}+\mathrm{B} \stackrel{\mathrm{k}_{1}}{\longrightarrow} \mathrm{C}+\mathrm{Z}-(1) \text { Step }(\text { slow }) \)
\(Z+A \stackrel{k_{2}}{\longrightarrow} D-(2) \text { Step }(\text { fast }) \)
Over all reaction:  \(2 \mathrm{~A}+\mathrm{B} \rightarrow \mathrm{C}+\mathrm{D}\)
Here \(A+B \underset{\text { Slow }}{\stackrel{K_{1}}{\longrightarrow}} C+Z\), step is the rate determining step. For the decomposition of hydrogen peroxide catalysed by I^-.
2H₂O_2(aq)\(\rightarrow\) 2H₂O_(I) + O₂(g)
It is experimentally found that the reaction is first order with respect to both H₂,O₂, and I^-, which indicates that I^- is also involved in the reaction. The mechanism involves the following steps.
Step: 1
H₂O_2(aq)+I^-1_(aq) \(\rightarrow\) H₂O_(l)+OI^-1_(aq)
Step: 2
H₂O_2(aq)+OI^-1_(aq)\(\rightarrow\) H₂O + I^-_(aq) + O_(g)
Overall reaction is
2H₂O_2(aq) \(\rightarrow\) 2H₂O_(l) + O_2(g)
These two reactions are elementary reactions. Adding equation (1), and (2) gives the overall reaction. Step 1 is the rate determining step, since it involves both H₂,O₂ and I^-, the overall reaction is bimolecular.

(i) The half life of a reaction is defined as the time required for the reactant concentration to reach one half its initial value.
(ii)   The rate constant for a first order reaction is given by
\(k=\frac { 2.303 }{ t } log\frac { \left[ { A }_{ 0 } \right] }{ \left[ A \right] } \)
\(at\quad t={ t }_{ \frac { 1 }{ 2 } };\left[ A \right] =\frac { \left[ { A }_{ 0 } \right] }{ 2 } \)
\(k=\frac { 2.303 }{ t_{ 1/2 } } log\frac { \left[ { A }_{ 0 } \right] }{ \frac { \left[ { A }_{ 0 } \right] }{ 2 } } \)
\(k=\frac { 2.303 }{ { t }_{ 1/2 } } log2\)
\(k=\frac { 2.303\times 0.3010 }{ { t }_{ 1/2 } } =\frac { 0.6932 }{ { t }_{ 1/2 } } \)
\({ t }_{ 1/2 }=\frac { 0.6932 }{ k } \)
This equation has no concentration term So, the half life of a first order reaction is independent of initial concentration.

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

(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
\(k=pZ{ e }^{ \frac { -Ea }{ RT } }\)

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 plot of log k against  \(\frac { 1 }{ T } \) values gives a straight line with slope value equal to - E_a/ 2.303R and intercept value equals to log A.

(ii) The plot gives a negative slope straight line also.
(iii) From the slope of straight line, E_a can be calculated.

The reaction is a zero order reaction whose rate is independent on the concentration of reactants.

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

Ea = 200 kJ mol^-1=200 \(\times\) 10³J mol^-1
A = 1.6 \(\times\) 10¹³s^-1; T = 600 K; R = 8.314 JK mol^-1
\(k=A{ e }^{ -\left( \frac { Ea }{ RT } \right) }\)
\(k=1.6\times { 10 }^{ 13 }{ s }^{ -1 }{ e }^{ -\left( \frac { 200\times 10^3}{ 8.314 \times 600 } \right) }\)
\(k=1.6\times { 10 }^{ 13 }{ s }^{ -1 }{ e }^{ -\left( 40.09 \right) }\)
\(k=1.6\times { 10 }^{ 13 }\times 3.8\times { 10 }^{ -18 }{ s }^{ -1 }\)
\(k=6.08\times { 10 }^{ -5 }{ s }^{ -1 }\)