[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)

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

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.