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%

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

\(k=\frac { 2.303 }{ t } \log\frac { \left( { V }_{ \infty }-{ V }_{ 0 } \right) }{ \left( { V }_{ \infty }-{ V }_{ 0 } \right) } \)
\(k=\frac { 2.303 }{ 20 } \log\frac { 50.4-20.2 }{ 50.4-25.6 } \)
\(=0.1151 \log\frac { 30.2 }{ 24.8 } \)
= 0.1151 log 1.2479
= 0.1151 x 0.0959
= 11.03 x10^-3 min^-1
When t = 40 mts
\(k=\frac { 2.303 }{ 40 } \log\frac { 50.4-20.2 }{ 50.4-25.6 } \)
\(=0.0576\times \log\frac { 30.2 }{ 20.9 } \)
= 9.19 x 10^-3 min^-1
When t = 60 mts
\(k=\frac { 2.303 }{ 60 } \log\frac { 50.4-20.2 }{ 50.4-25.6 } \)
\(=0.03838\times \log\frac { 30.2 }{ 17.6 } \)
= 0.03838 x 0.2343
= 8.99 x 10^-3 min^-1
The constant values of K show that the reaction is of first order.

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

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)

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.

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