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

(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.

\(2{ NH }_{ { 3 }_{ (g) } }\overset { pt }{ \longrightarrow } { N }_{ { 2 }_{ (g) } }+3{ H }_{ { 2 }_{ (g) } }\)
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.

Given:
K₁ = 1.60 x 10^-5 S^-1; T₁ = 600K
K₂ = 6.36 x 10^-3 S^-1; T₂ = 700K
We know that log \(\frac { { k }_{ 2 } }{ { k }_{ 1 } } =\frac { { E }_{ a } }{ 2.303R } \left[ \frac { 1 }{ { T }_{ 1 } } -\frac { 1 }{ { T }_{ 2 } } \right] \)
Substituting the values of K₁, K_2, and R= (8.314 JK^-1 mol^-1), we get,
\(log\frac { 6.36\times { 10 }^{ -3 } }{ 1.60\times { 0 }^{ -5 } } =\frac { { E }_{ a } }{ 2.303\times 8.314 } \left[ \frac { 1 }{ 600 } -\frac { 1 }{ 700 } \right] \)
∴ E_a = 2.6 x 19.15 x 4200 J mol^-1
= 2.09 x 10⁵ J mol^-1
= 209 kJ mol^-1

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'.

A reaction is second order in A and first order in B
(i) Differential rate equation,
Rate = \(\frac { -d\left[ R \right] }{ dt } =k{ \left[ A \right] }^{ 2 }\left[ B \right] \)
(ii) When the concentration of A is increased three times, (i.e) 3A, then
Rate = k[3A]² [B]
= 9k [A]²[B] = 9 (initial rate)
This shows the rate will increase 9 times to the initial time.
(iii) When concentration of both A and B is doubled then,
Rate = k [2A]² [2B] = 8k [A]² [B] = 8 (initial rate)
This shows that rate will increase 8 times to the initial rate.

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.

(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]

Reaction Rate = k[x]ⁿ [y]ⁿ
0.15 = k[0.2]ⁿ [0.02]^m   ...(1)
0.30 = k[0.4]ⁿ [0.02]^m  ...(2)
1.20 = k[0.4]ⁿ [0.08]^m  ...(3)
eqn(3) \(\div\) eqn(2) 
\(\Rightarrow\)\(\frac { 1.2 }{ 0.3 } =\frac { k{ \left[ 0.4 \right] }^{ n }{ \left[ 0.08 \right] }^{ m } }{ k{ \left[ 0.4 \right] }^{ n }{ \left[ 0.02 \right] }^{ m } } \)
\(4={ \left( \frac { \left[ 0.08 \right] }{ \left[ 0.02 \right] } \right) }^{ m }\)
\(4^1={ \left( 4 \right) }^{ m }\)
\(m=1\)
eqn(2) \(\div\) eqn(1), \(\frac{0.30}{0.15}=\frac{\mathrm{k}[0.4]^{\mathrm{n}}[0.02]^{\mathrm{m}}}{\mathrm{k}[0.2]^{\mathrm{n}}[0.02]^{\mathrm{m}}}\)
\( 2={ \left( \frac { \left[ 0.4 \right] }{ \left[ 0.2 \right] } \right) }^{ n}\)
\(2^1= 2 ^{ n }\)
\(\therefore n=1\)
Rate = \(k{ \left[ x \right] }^{ 1 }{ \left[ y \right] }^{ 1 }\)
\(0.15=k{ \left[ 0.1 \right] }^{ 1 }{ \left[ 0.02 \right] }^{ 1 }\)
\(\frac { 0.15 }{ { \left[ 0.2 \right] }^{ 1 }{ \left[ 0.02 \right] }^{ 1 } } =k\)
k = 37.5 mol⁻¹L s⁻¹

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,
\(\text { Rate }=\mathrm{k}[\mathrm{M}]^{n}  \)
\(7.5 \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}=\mathrm{k}\left(0.05 \mathrm{~mol} \mathrm{} \mathrm{~L}^{-1}\right)^{2} \)
\(\mathrm{k} =\frac{Rate}{[M]^{n}}  \)
\(\mathrm{k} =\frac{7.5 \times 10^{-3}}{(0.05)^{2}}  \)
\(=3 \mathrm{~mol}^{-1} \mathrm{~L} \mathrm{~s}^{-1}\)

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 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) \(Rate=\frac { -1 }{ 2 } \frac { d[NO] }{ dt } =\frac { -d[{ O }_{ 2 }] }{ dt } =\frac { 1 }{ 2 } \frac { d\left[ { NO }_{ 2 } \right] }{ dt } \)
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

The rate of a reaction is affected by the following factors.
(i) Nature and state of the reactant
(ii) Concentration of the reactant
(iii) Surface area of the reactant
(iv) Temperature of the reaction
(v) Presence of a catalyst

In \(\frac { { k }_{ 2 } }{ { k }_{ 1 } } =\frac { { E }_{ a } }{ R } \left[ \frac { { T }_{ 2 }-{ T }_{ 1 } }{ { T }_{ 1 }{ T }_{ 2 } } \right] \)
In 2 = \(\frac { { E }_{ a } }{ R } \) \(\left[ \frac { 10 }{ 300\times 310 } \right] \)
E_a = 9300 R In 2
= 53.4 kJ mol^-1

\({ k }=2.303\log { \frac { \left[ { A }_{ 0 } \right] }{ \left[ A \right] } } \)
\(t=\frac { 2.303 }{ { 10 }^{ -3 } } \log { \frac { 10 }{ 2.5 } } \)
t = 2303 x 0.301 x 2 = 1386 s.

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.

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

Arrhenius equation is,
\(k=Ae^\left ({ \frac { -Ea }{ RT } } \right )\)
Here,
A \(\rightarrow\) Frequency factor
E_a \(\rightarrow\) Activation energy of the reaction
R \(\rightarrow\) Gas constant
T \(\rightarrow\) Absolute temperature (in K)

(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]

Given data: Rate = 5.2 x 10^-6 mol lit^-1 sec^-1
Initial concentration [A] = 2.6 x 10^-3
Formula: Rate = k[A]¹
Solution: Rate = 5.2 x 10^-6 mol lit^-1 sec^-1 = k x 2.6 x 10^-3 lit^-1
= \(\frac { 5.26\times { 10 }^{ -6 }mol{ \ lit }^{ -1 }{ sec }^{ -1 } }{ 2.6\times { 10 }^{ -3 }mol \ { lit }^{ -1 } } \)= 2 x 10^-3
∴ k = 2 x 10^-3 s^-1

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.

[R]₀ = 5g, [R] = 3g
K = 1.15 x 10^-3 s^-1
As the reaction is of first order
K = \(\frac { 2.303 }{ t } \log { \frac { { \left[ R \right] }_{ 0 } }{ \left[ R \right] } } \)
\(t=\frac { 2.303 }{ 1.15\times { 10 }^{ -3 }{ s }^{ -1 } } \log { \frac { 5g }{ 3g } } \)
= 200 x 10³ (log 1.667)s
= 20 x 10³ x 0.22 19 s
= 443.8 s
= 444 s (approximately)

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) 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 %