\(k=\frac { 1 }{ t } ln \frac { \left[ { A }_{ 0 } \right] }{ \left[ A \right] } \)
[A₀] = 0.1: [A] = 0.05
\(k= \left[ \frac { 1 }{ t _{1/2}} \right] ln\left[ \frac { { 0.1 } }{ 0.05 }\right] \)
\(k= \left[ \frac { 1 }{ t _{1/2}} \right] ln (2)\)
t_1/2 = In(2)/K

T₁= 200 K ; k = k₁
T₂ =400 K ; k₁ = k₂ = 2k₁
log\(\frac {{ k }_{ 2 } }{ { k }_{ 1 } } =\frac { { E }_{ a } }{ 2.303R } \left[ \frac { { T }_{ 2 }-{ T }_{ 1 } }{ { T }_{ 1 }{ T }_{ 2 } } \right] \)
log \( [\frac {{2 k }_{ 1 } }{ { k }_{ 1 } }]\)
\(= \left[ \frac { E_{a} }{ 2.303\times 8.314 JK^{-1} mol^{-1} } \right] \)
\(= \left[ \frac { 400 k - 200K }{200 k \times 400 k } \right] \)
\(E_{a} = \frac{ 0.3010 \times 2.303 \times 8.314 JK^{-1}mol^{-1} \times 200K \times 400 K}{200 K}\)
E_a = 2305 J mol^-1
E_a = 2.305 kJ mol^-1

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

for n ≠ 1 t_1/2 = \(\frac{2^{n-1} -1}{(n- 1) k[A_{0}]^{-1}}\)
for n = 0; t_1/2 = \(\frac{1}{2 k[A_{0}]^{-1}}\)
t_1/2 = \(\frac{[A_{0}]}{2 k}\)
t_1/2 α [A₀] ...(1)
Given [A₀] = 0.002 M; t_1/2 = 10 min
[A₀] = 0.04M; t_1/2 = ?
Substitute in (1)
10 min α 0.02 M...(2)
t_1/2 α 0.04M ....(3)
(3)(2)
⇒ t_1/2 / 10 min
= 0.04 M/0.02 M
t_1/2 = 2 x 10 min = 20 min

\(k=\frac { 2.303 }{ t } log\frac { \left[ { A }_{ 0 } \right] }{ \left[ A \right] } \)
\(k=\frac { 1 }{ t } ln \frac { \left[ { A }_{ 0 } \right] }{ \left[ A \right] } \)
\(e^{kt}=\frac { \left[ { A }_{ 0 } \right] }{ \left[ A \right] }\)
[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

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) When the concentration of the reactant is increased by 'n' times, the rate of reaction is also increased by n times. That is, if the concentration of the reactant is doubled, the rate is doubled.
(ii) The unit of rate constant of a first order reaction is sec^-1 or time^-1.
\({ k }_{ 1 }=\frac { rate }{ (a-x) } =\frac { { mol.lit }^{ -1 }{ sec }^{ -1 } }{ { mol.lit }^{ -1 } } \)
(iii) The time required to complete a definite fraction of reaction is independent of the initial concentration, of the reactant if t_1/u is the time of one 'u' th fraction of reaction to take place then from equation.
\({ k }_{ 1 }=\frac { 2.303 }{ t } \log { \frac { a }{ a-x } } \)
\(x=\frac { a }{ u } and\quad { t }_{ \frac { 1 }{ u } }=\frac { 2.303 }{ { k }_{ 1 } } \log { \frac { a }{ a-\frac { a }{ u } } } ;\)
\({ t }_{ \frac { 1 }{ u } }=\frac { 2.303 }{ { k }_{ 1 } } \log { \frac { u }{ (u-1) } } \)
since k₁ = rate constant, t_1/u is independent of initial concentration 'a'.

(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

A reaction in which the rate is independent of the concentration of the reactant over a wide range of concentration is called a zero order reaction.
Let us consider the following by hypothetical zero order reaction.
 A⟶ product 
The rate law can be written as,
Rate = k[A]⁰ 
\(\frac { -d\left[ A \right] }{ dt } =k(1)\ \ \ \therefore \left( { \left[ A \right] }^{ 0 }=1 \right) \)
\(\Rightarrow -d\left[ A \right] =kdt\)
Integrate the above equation between the limits of [A₀] at zero time and [A] at some later time 't',
\(-\int _{ \left[ { A }_{ 0 } \right] }^{ \left[ A \right] }{ d\left[ A \right] } =k\int _{ 0 }^{ t }{ dt } \)
\(-{ \left( \left[ A \right] \right) }_{ \left[ { A }_{ 0 } \right] }^{ \left[ A \right] }=k{ \left( t \right) }_{ 0 }^{ t }\)
[A₀] - [A] = kt   
\(k=\frac { \left[ { A }_{ 0 } \right] -\left[ A \right] }{ t } \)
Straight line equation y = mx + c
ie., \(\left[ A \right] =-kt+\left[ { A }_{ 0 } \right] \)
⇒ y = c + mx
A plot [A] vs time gives a straight line with a slope of -k and y intercept of [A₀]

Rate law:
(i) Rate law or rate equation is an expression which relates the rate of a reaction with rate constant and the concentration of reactants.
(ii) For xA + yB → products
(iii) The rate law is r = k[A]^m[B]ⁿ
Rate constant:
(i) It is the rate of the reaction when the concentration of the reactants are taken unity.
In above rate law if[A] = [B] = 1, rate constant k = Rate

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 \(=\left(\frac{-\Delta \mathrm{A}}{\Delta \mathrm{t}}\right)\)

(i) The magnitude of order of a reaction may be zero or fractional or integral values. Order is never fractional for elementary reaction.
(ii) It should be determined only by experiments.
(iii) Simple reactions possess low values of order like n = 0, 1, 2 reactions with order greater than or equal to 3.0 are called complex reactions.
(iv) Some reactions show fractional order depending on rate.
(v) Higher order reactions may be experimentally converted into simpler order (pseudo) reactions by using excess concentrations of one or more reactants.

(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 of a reaction at any time depends on the concentration of the reactants which keeps on decreasing with time.

Minus sign i.e.  \(\frac { -dA }{ dt } \) or \(\frac { -dB }{ dt } \) indicates decreases in the concentration of reactants whereas + sign indicates in the concentration of products with time i.e. \(\frac { +dC }{ dt } \)

(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
\(\mathrm{K}=1 \mathrm{~mol} \mathrm{~L}{ }^{-1} \mathrm{~s}^{-1} ;\left[\mathrm{A}_{0}\right]=100 \mathrm{M} ;\left[\mathrm{A}_{0}\right]-[\mathrm{A}]=80 \mathrm{M} ; \mathrm{t}=?\)
\(\therefore t=\left(\frac{\left[A_{O}\right]-[A]}{K}\right)=\frac{80}{1}=80 \mathrm{mins}\)

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
\(\mathrm{k}=\frac{2.303}{60} \log \frac{[0.08]}{[0.04]}\)
\(=0.0383 \times \log 2  \)
\(=0.0383 \times 0.3010  \)
\(=1.152 \times 10^{-2} \mathrm{~s}^{-1}\)