Given: Half life period (t_1/2) = 10 mins
Formula: \(k=\frac { 2.303 }{ t } \log { \frac { a }{ a-x } } \)
Solution:
\(k=\frac { 0.693 }{ { t }_{ \frac { 1 }{ 2 } } } =\frac { 0.693 }{ 10 } \)
= 0.693 min^-1 = 6.93 x 10^-2 min^-1
Time taken = 1 hour = 60 minutes
a = 100%
x = ?
\(k=\frac { 2.303 }{ t } log\frac { a }{ a-x } \)
\(=\frac { 2.303 }{ 60 } \) [log 100 - log(a - x)]
log 100 - log(a-x) = \(\frac { 6.93\times { 10 }^{ -2 }\times 60 }{ 2.303 } \)
log 100 - log(a-x) = 1.8060
log (a - x) = log 100 - 1.8060
= 2 - 1.8060
log (a - x) = 0.1940
a - x = Antilog of 0.1949
a - x = 1.563%

Given data: Time required for 99.9% completion is about 10 times its half life period. Prove that for a first order reaction the time required for 99.9% completion is about 10 times its half life period.
Formula: \(k=\frac { 2.303 }{ t } \log { \frac { a }{ a-x } } \)
\({ t }_{ \frac { 1 }{ 2 } }=\frac { 0.693 }{ k } \)
Solution:
\({ t }_{ 99.9 }=\frac { 2.303 }{ k } \log\frac { 100 }{ 100-99.9 } \)
\(=\frac { 2.303 }{ k } \log\frac { 100 }{ 0.1 } \)
\(=\frac { 2.303 }{ k } \log1000\)
\({ t }_{ 99.9 }=\frac { 2.303\times 3 }{ k } \)
\({ t }_{ 99.9 }=\frac { 6.909 }{ k } ;\)
\({ t }_{ \frac { 1 }{ 2 } }=\frac { 0.693 }{ k } \)  \(\therefore k=\frac { 0.693 }{ { t }_{ \frac { 1 }{ 2 } } } \)
\(\frac { { t }_{ 99.9\% } }{ { t }_{ \frac { 1 }{ 2 } } } =\frac { 6.909 }{ k } \times \frac { k }{ 0.693 }\)
\({ t }_{ 99.9 }=\frac { 6.909 }{ 0.693 } \times { t }_{ \frac { 1 }{ 2 } }\)
t_99.9 = 10 x t_1/2
Time required for 99.9% completion of the reaction = 10t_1/2

Given data: Time taken for completion of 30% of the reaction = 12 mins.
a = 100; x = 30; a - x = 70 and t =12 min
Formula: \(k=\frac { 2.303 }{ t } \log { \frac { a }{ a-x } } \)
Solution:
For a first order reaction,
\(\frac { 2.303 }{ 12 } \log\frac { 100 }{ 70 } \)
\(=\frac { 2.303 }{ 12 } \times 0.1549\) = 0.02972 min^-1
k = 2.97 x 10^-2 min^-1
If t = 65.33 minutes, x = ?
\(k=\frac { 2.303 }{ 65.33 } \log\frac { 100 }{ 100-x } \)
\(0.02972=\frac { 2.303 }{ 65.33 } \log\frac { 100 }{ 100-x } \)
\(\log\frac { 100 }{ 100-x } \)= \(\frac { 0.02972\times 65.33 }{ 2.303 } =0.8430\)
\(\frac { 100 }{ 100-x }\)  =Antilog of 0.8430
100 = 6.966 (100 - x)
100 = 696.6 - 6.966x
-6.966x = 100-696.6 = 596.6
\(x=\frac { 596.6 }{ 6.966 } =85.62%\)
The reaction completed in 65.33 minutes

Given data: Time taken for 25% completion of the reaction = 100 mins
a = 100; x = 25 and a - x = 75; t = 100min
Formula: \(k=\frac { 2.303 }{ t } \log { \frac { a }{ a-x } } ;{ t }_{ \frac { 1 }{ 2 } }=\frac { 0.693 }{ k } \)
Solution:
\(k=\frac { 2.303 }{ 100 } \log\frac { 100 }{ 100-25 } \)
\(=\frac { 2.303 }{ 100 } \log\frac { 100 }{ 75 } =\frac { 2.303 }{ 100 } \log\frac { 4 }{ 3 } \)
\(=\frac { 2.303 }{ 100 } \times 0.1249\)
Rate constant(k) = 2.8773 x 10^-3 min^-1
\({ t }_{ \frac { 1 }{ 2 } }=\frac { 0.693 }{ k } =\frac { 0.693 }{ 2.8773\times { 10 }^{ -3 }{ ,min }^{ -1 } } \)
\(=\frac { 693 }{ 2.8773 } =240.85\)
Half-life period (t_1/2) = 240.85 minutes

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.

(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] [B]²
Therefore, the rate of reaction will increase 9 times.
(iii) When the concentration of both A and B are doubled.
\(-\frac { d\left[ R \right] }{ dt } \) = k[A] [B]² 
= k [2A] [2B]²
Therefore, the rate of reaction will increase 8 times.

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)

Rate = k [x]² = ka²
[x] = a mol^-1
If the concentration of x is in cross three time, then
(x) = 3a mol L^-1
Rate = R(3a)² = 9 ka²
Hence, the rate of formation will increase by 9 times.

Average rate = \(-\frac { \triangle \left( R \right) }{ \triangle t } =-\frac { { \left[ R \right] }_{ 2 }-{ \left[ R \right] }_{ 1 } }{ { t }_{ 2 }-{ t }_{ 1 } } \)
\(=-\frac { 0.02M-0.03M }{ 25min } =\frac { -0.01M }{ 25min } \)
= 4 x 10^-4 M min^-1 and 
= \(-\frac { -0.01m }{ 25\times 60 } \) = 6.66 x 10^-6 Ms^-1

(i) The magnitude of order of a reaction may be zero or fractional or integral values. Order is never fractional for elementary reaction.
(ii) It should be determined only by experiments.
(iii) Simple reactions possess low values of order like n = 0, 1, 2 reactions with order greater than or equal to 3.0 are called complex reactions.
(iv) Some reactions show fractional order depending on rate.
(v) Higher order reactions may be experimentally converted into simpler order (pseudo) reactions by using excess concentrations of one or more reactants.

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) Average rate of reaction between the time interval, 30 to 60 seconds
\(=\frac { d\left[ Ester \right] }{ dt } \)
\(=\frac { 0.31-0.17 }{ 60-30 } =\frac { 0.14 }{ 30 } \)
= 4.67 x 10^-3 mol L^-1 s^-1.
(ii) For a pseudo first order reaction,
\(k=\frac { 2.303 }{ t } \log { \frac { { \left[ R \right] }_{ 0 } }{ \left[ R \right] } } \)
For, t = 303
\({ k }_{ 1 }=\frac { 2.303 }{ t } \log { \frac { 0.55 }{ 0.31 } } \)
For, t = 60 s
\({ k }_{ 2 }=\frac { 2.303 }{ 60 } \log { \frac { 0.55 }{ 0.17 } } \)
For, t = 90 s
\({ k }_{ 3 }=\frac { 2.303 }{ 90 } \log { \frac { 0.55 }{ 0.085 } } \)
= 2.075 x 10^-2 s^-1
The average rate constant,
\(k=\frac { { k }_{ 1 }+{ k }_{ 2 }+{ k }_{ 3 } }{ 3 } \)
\(=\frac { \left( 1.911\times { 10 }^{ -2 } \right) +\left( 1.957\times { 10 }^{ -2 } \right) +\left( 2.075\times { 10 }^{ -2 } \right) }{ 3 } \)
= 1.98 x 10^-2 s^-1.

The initial rate of the reaction is
Rate = k [A] [B] 2
= (2.0 x 10^-6 mol^-2 L² S^-1) (0.1  mol L^-1)²
= 8.0 x 10^-9 mol^-2 L² s^-1
When [A] is reduced from 0.1 mol L^-1 to 0.06 mol^-1, the concentration of A reacted = (0.1 - 0.06) mol L^-1 = 0.04 mol L^-1.
∴ The concentration of B reaction 
= \(\frac { 1 }{ 2 } \) x 0.04 mol L^-1
= 0.02 mol L^-1
Then, concentration of B available [B] = (0.2 - 0.02) mol L^-1
= 0.18 mol L^-1.
After [A] is reduced to 0.06 mol L^-1, the rate of the reaction is given by,
Rate = k [A] [B] 2
= (2.0 x 10^-6 mol^-2 L² s^-1) (0.06 mol L^-1) (0.18 mol L^-1)²
= 3.89 mol L^-1 s^-1.

Given data: Rate constant of a first order reaction (k) = 0.45 sec^-1
Formula: \({ t }_{ \frac { 1 }{ 2 } }=\frac { 0.693 }{ k } \)
Solution: \({ t }_{ \frac { 1 }{ 2 } }=\frac { 0.693 }{ k } =\frac { 0.693 }{ 0.45 } \)
Half-life period = 1.54 sec.