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

(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^-.

Volume of KMnO₄ used is proportional to the amount of H₂O₂ present. If the reaction is of first order, it must obey the equation.
\(k=\frac { 2.303 }{ t } log\frac { \left[ { A } \right] _{ 0 }}{ \left[ A \right] } \)
(or) \(k=\left( \frac { 2.303 }{ t } \right) log\left( \frac { { V }_{ 0 } }{ { V }_{ 1 } } \right) \)
(i) \( \mathrm{V}_{\mathrm{o}}=46.1 \mathrm{~mL} ; \mathrm{t}=10 \mathrm{mins} ; \mathrm{V}_{\mathrm{t}}=29.8 \mathrm{~mL}\)
\(k =\frac{2.303}{10} \log \left(\frac{46.1}{29.8}\right)  \)
\(=0.2303 \log 1.5469  \)
\(=0.2303 \times 0.1894=0.0436 \mathrm{~min}^{-1}\)
\(\mathrm{t}=20 \mathrm{mins} ; \mathrm{V}_{\mathrm{t}}=19.3 ; \mathrm{V}_{\mathrm{o}}=46.1 \mathrm{~mL} \)
\(k =\frac{2.303}{20} \log \left(\frac{46.1}{19.3}\right)  \)
\(=0.11515 \times \log 2.388  \)
\(=0.11515 \times 0.3780  \)
\(=0.0435 \mathrm{~min}^{-1}\)
Since the value of k comes out to be almost constant for the reaction, it is of first order. The mean value of k = 0.04355 min^-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
\(k=pZ{ e }^{ \frac { -Ea }{ RT } }\)

\(\log K=\log A-\frac { { E }_{ a } }{ 2.303R } \left( \frac { 1 }{ T } \right) \)
y = c + mx
\(m=-\frac { { E }_{ a } }{ 2.303R } \)
E_a = -2.303 Rm
E_a = -2.303 x 8.314 x (-4000)
E_a = 76,589J  mol^-1
E_a = 76.589 KJ mol^-1

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) 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 \(\frac{3}{2}\)

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

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

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

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
\(k=A{ e }^{ -\left( \frac { Ea }{ RT } \right) }\)
\(k=1.6\times { 10 }^{ 13 }{ s }^{ -1 }{ e }^{ -\left( \frac { 200\times 10^3}{ 8.314 \times 600 } \right) }\)
\(k=1.6\times { 10 }^{ 13 }{ s }^{ -1 }{ e }^{ -\left( 40.09 \right) }\)
\(k=1.6\times { 10 }^{ 13 }\times 3.8\times { 10 }^{ -18 }{ s }^{ -1 }\)
\(k=6.08\times { 10 }^{ -5 }{ s }^{ -1 }\)

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

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)