According to Arrhenius equation
\(\log\left( \frac { { k }_{ 2 } }{ { k }_{ 1 } } \right) =\frac { { E }_{ a } }{ 2.303R } \left( \frac { { T }_{ 2 }-{ T }_{ 1 } }{ { T }_{ 1 }{ T }_{ 2 } } \right) \)
T₂ = 400K ; k₂ = 0.04 s^-1
T₁ = 200K ; k₁ = 0.02 s^-1
\(\log\left( \frac { 0.04}{ 0.02} \right) =\frac { { E }_{ a } }{ 2.303\times 8.314} =\left( \frac { 400-200 }{ 200\times 400 } \right) \)
\(\log(2)=\frac { { E }_{ a } }{ 2.303\times 8.314} =\left( \frac { 1 }{ 400 } \right) \)
E_a = log(2) \(\times\) 2.303 \(\times\) 8.314 \(\times\) 400
= 0.3010 \(\times\) 2.303 \(\times\) 8.314 \(\times\) 400
E_a = 2305 J mol^-1 = 2.305 kJ mol^-1

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 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%

For a first order reaction
\(\\ \\ k=\frac { 2.303 }{ t } log\left( \frac { [{ A }_{ 0 }] }{ [A] } \right) \\ \)   ..(1)
Let[A₀] =100M
When 
t = t_90%; [A] = 10M (given that t_90% = 8hours)
t = t_80%; [A ] = 20M
\(k=\frac { 2.303 }{ { t }_{ 80\% } } \log\left( \frac { 100 }{ 20 } \right) \)
\({ t }_{ 80\% }=\frac { 2.303 }{ K } \log(5)\) ....(2)
Find the value of k using the given data
\(k=\frac { 2.303 }{ { t }_{ 90\% } } \log\left( \frac { 100 }{ 10 } \right) \)
\(k=\frac { 2.303 }{ 8 } \log10\)
\(k=\frac { 2.303 }{ 8 } ...(3)\)
Substitute the value of k in equation (2)
\({ t }_{ 80\% }\frac { 2.303 }{ 2.303/8hours } \log(5)\)
t_80%= 8 hours x 0.6989
t_80%= 5.59 hours

Let \(\left[A_{0}\right]=100 \%\), t = 50 minutes
Then [A]=100 - 40 = 60 %
(1) \(\mathrm{k}=\frac{2.303}{\mathrm{t}} \log \frac{\left[\mathrm{A}_{\mathrm{o}}\right]}{[\mathrm{A}]}\)
\(=\frac{2.303}{50} \log \left(\frac{100}{60}\right) \)
\(=\frac{2.303}{50} \log 1.667  \)
\(=\frac{2.303}{50} \times 0.2219 \)
\(\mathrm{k}=0.010216 \mathrm{~min}^{-1}  \)
\(\mathrm{k}=1 \times 10^{-2} \mathrm{~min}^{-1}\)
(2)  \( t =\frac{2.303}{0.010216} \log \left(\frac{100}{20}\right) \)
\(t =\frac{2.303}{{0.010216}}\times 0.6990\)
= 225.43 \(\times \) 0.6990
t = 157.58 min.
The time at which the reaction will be 80% complete is 157.58 min.

For a first order reaction
\(k=\frac { 2.303 }{ t } \log\frac { \left[ { A }_{ 0 } \right] }{ \left[ A \right] } \)
\(k=\frac { 2.303 }{ t } \log\frac { { V }_{ \infty } }{ { V }_{ \infty }-{ V }_{ 1 } } \)
V_∞= 58.3 ml.

As the rate constants are constant through out it is a first order reaction. 

\(\mathrm{k}=0.693 / \mathrm{t}_{1 / 2}\)
\(\mathrm{k}=\frac{0.693}{8.0}=0.08 \mathrm{~min}^{-1}\)
For the first order reaction:
\(t =\frac{2.303}{k} \log \frac{\left[A_{0}\right]}{[A]}  \)
\(t =\frac{2.303}{0.087} \log \left(\frac{100}{1}\right)=26.47 \log 10^{2}  \)
\(t =2 \times 26.47 \log 10  \)
\(t =52.94 \mathrm{~min}\)

(i) The chemical reaction involves breaking of certain existing bonds of the reactant and forming new bonds which lead to the product.
(ii) The net energy involved in this process is dependent on the nature of the reactant and hence the rates are different for different reactants.
Example:
Let us compare the following two reactions that you carried out in volumetric analysis.
1) Redox reaction between ferrous Ammonium Sulphate (FAS) and KMnO₄.
2) Redox reaction between oxalic acid and KMnO₄.
(i) The oxidation of oxalate ion by KMnO₄ is relatively slow compared to the reaction between KMnO₄ and Fe²⁺. In fact heating is required for the reaction between KMnO₄ and Oxalate ion and is carried out at around 60^oC.
(ii) The physical state of the reactant also plays an important role to influence the rate of reactions.
(iii) Gas phase reactions are faster as compared to the reactions involving solid or liquid reactants.
Ex : Na_(s) + I₂(vap)   [Faster]
Na_(s) + I₂(s)   [Slower]
KI_(aq) + Pb(NO3)_2(aq) → PbI₂ (yellow) [Faster]
KI_(s) + Pb(NO3)_2(s) → PbI₂ (yellow)  [Slower]

(i) The rate of a reaction increases with the increase in the concentration of the reactants.
(ii) The effect of concentration is explained on the basis of collision theory of reaction rates.
(iii) According to this theory, the rate of a reaction depends upon the number of collisions between the reacting molecules.
(iv) Higher the concentration, greater is the possibility for collision and hence increase the rate.

If the monomer is represented by X. Then
\(2 \mathrm{M} \rightarrow(\mathrm{M})_{2}\)
Since the reaction is of second order, the rate of reaction will be given by,
\(\text { Rate }=\mathrm{k}[\mathrm{M}]^{n}  \)
\(7.5 \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}=\mathrm{k}\left(0.05 \mathrm{~mol} \mathrm{} \mathrm{~L}^{-1}\right)^{2} \)
\(\mathrm{k} =\frac{Rate}{[M]^{n}}  \)
\(\mathrm{k} =\frac{7.5 \times 10^{-3}}{(0.05)^{2}}  \)
\(=3 \mathrm{~mol}^{-1} \mathrm{~L} \mathrm{~s}^{-1}\)

(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.

(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.

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

1. Photochemical reaction between H₂ and Cl₂ ;  \(\mathrm{H}_{2(\mathrm{g})}+\mathrm{Cl}_{2(\mathrm{g})} \stackrel{h v}{\longrightarrow} 2 \mathrm{HCl}_{(\mathrm{g})}\)
2. Decomposition of N₂O on hot platinum surface \(\mathrm{N}_{2} \mathrm{O}_{(\mathrm{g})} \rightleftharpoons \mathrm{N}_{2(\mathrm{~g})}+\frac{1}{2} \mathrm{O}_{2(\mathrm{~g})}\)

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

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.

(a) Rate = \(k{ \left[ x \right] }^{ 3/2 }{ \left[ y \right] }^{ 0 }=k[x]^{3/2}\)
(b) 2NO + Br₂ ⟶ 2NOBr
Rate = k[NO]²[Br₂]

(i) Order of the reaction =1; \(\mathrm{t}_{1 / 2}=60 \mathrm{~s} ; \mathrm{k}=?\)
\(\mathrm{k}=\frac{0.6932}{\mathrm{t}_{\frac{1}{2}}} \)
\(=\frac{0.6932}{60}  \)
\(k =1.155 \times 10^{-2} \mathrm{~s}^{-1}\)
(ii) \(\left[\mathrm{A}_{0}\right]=100 \% ; \mathrm{t}=180 \mathrm{~s} ;[\mathrm{A}]=? ; \mathrm{k}=1.155 \times 10^{-2} \mathrm{~s}^{-1}\)
For first order reaction
\(\mathrm{k}=\frac{2.303}{\mathrm{t}} \log \frac{\left[\mathrm{A}_{0}\right]}{[\mathrm{A}]} \)
\(1.155 \times 10^{-2} =\frac{2.303}{180} \log \left(\frac{100}{[A]}\right)  \)
\(\frac{0.01155 \times 180}{2.303} =\log \left(\frac{100}{[A]}\right)  \)
\(0.9027 =\log 100-\log [\mathrm{A}]  \)
\(\log [\mathrm{A}] =\log 100-0.9027 \)
\(\log [A]=2-0.9027 \)
\(\log [A]=1.0972 \)
[A] = antilog  of (1.0972)
[A] =12.51 %