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).

 \(Rate=\frac { -1 }{ 2 } \frac { d[NH_3] }{ dt } \)
\(\frac { -d\left[ { N }_{ 2 } \right] }{ dt } =\frac { 1 }{ 3 } \frac { d\left[ { H}_{ 2 } \right] }{ dt } \)
\(=\frac { 1 }{ 2 } { k_1\left[ { NH}_{ 3 } \right] } = { k_2 \left[ { NH}_{ 3 } \right] }=\frac { 1 }{ 3 } { k_3\left[ { NH}_{ 3 } \right] } \)
[3/2] k₁ = 3k₂ = k₃
1.5k₁ = 3k₂ = k₃

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

(i) The rate constant is nearly doubled with a rise in temperature by 10^oC for a chemical reaction.
(ii) The temperature effect on the rate constant can be represented quantitatively by Arrhenius equation.
\(k=A{ e }^{ \frac { -{ E }_{ a } }{ RT } }\)

For first order reaction
\(k=\frac { 2.303 }{ t } \log { \frac { a }{ a-x } } \)
For t = t_1/2, \(x=\frac { a }{ 2 } \)
\({ t }_{ \frac { 1 }{ 2 } }=\frac { 2.303 }{ k } \log { \frac { a }{ a-\frac { a }{ 2 } } } \)
\(\frac { 2.303 }{ k } \log2=\frac { 2.303\times 0.3010 }{ 7.39\times { 10 }^{ -5 } } \)
= 9.38 x 10³ s.

(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^-.

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.

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

Order of the reaction = \(\frac { 1 }{ 2 } +\frac { 3 }{ 2 } =\frac { 4 }{ 2 } =2\)
The reaction is second order.

The order of the reaction is zero.

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

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

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.

The reaction is 2NH_3(g) \(\overset { Pt }{ \longrightarrow } \) N_2(g) + 3H_2(g)
Here k = 2.5 x 10^-4 mol L^-1 s^-1
The order of the reaction is zero i.e.,
Rate = k[Reactant]^o
Rate = 2.5 x 10^-4 x 1 = 2.5 x 10^-4 mol L^-1 s^-1
\(\therefore \frac { d }{ dt } \left[ { H }_{ 2 } \right] =\frac { 1 }{ 3 } \frac { d }{ dt } \left[ { H }_{ 2 } \right] \)
The rate of formation of N₂ = 2.5 x 10^-4 mol L^-1 s^-1
\(\therefore 2.5\times { 10 }^{ -4 }=\frac { 1 }{ 3 } \frac { d }{ dt } \left[ { H }_{ 2 } \right] \)
\(\therefore \frac { d }{ dt } \left[ { H }_{ 2 } \right] \) = 7.5 x 10^-4
Therefore, rate of formation of H₂ = 7.5 x 10^-4 mol L^-1 s^-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)

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