(i) Let [A]₀ = 100.
Then, [A]_t at 53 minutes = (100 – 50) = 50 and [A], at 100 minutes = (100 – 73) = 27.
Substituting t and concentration values in the integrated rate equation for first-order reaction.
                                               $\mathrm{k} = \frac{2.303}{\mathrm{t}}\log \frac{{\left[\mathrm{A}\right]}_0}{{\left[\mathrm{A}\right]}_{\mathrm{t}}}$

At t = 53 min,      

               $\mathrm{k}=\frac{2.303}{53 \min }\log \frac{100}{50}=\frac{2.303}{53 \min }\log 2$

                                                $= \frac{2.303}{53 \min }\times 0.3010 = 0.013 {\min }^{-1}$

At t = 100 min,
               
                                     $$\begin{gathered} \mathrm{k}=\frac{2.303}{100 \min }\log \frac{100}{27} \\ = \frac{2.303}{53 \min }\times 0.5686 = 0.013 {\min }^{-1} \end{gathered}$$

Since the value of k is constant, the order of reaction is 1.

(ii) For a first order reaction, the time required to complete any fraction is independent of the initial concentration of reactant.
∴ 73% of N₂O will decompose when the initial concentration is 600 mm which corresponds to a pressure of $\frac{600 \mathrm{mm}}{100} = 438 \mathrm{mm}.$

For the first order reaction the Rate constant

$\mathrm{k}=\frac{0.693}{{\mathrm{t}}_{1/2}}$

${\mathrm{t}}_{1/2}$ is given 5 min, hence      

$\mathrm{k}=\frac{0.693}{5}=0.1386 {\min }^{-1}$

Let initial concentration ‘a’ of A be 100 mol L^–1. To reach 25% of initial concentration means, (a – x) = 25 mol L^–1.

$\mathrm{k} = \frac{2.303}{\mathrm{t}}\log \frac{\mathrm{a}}{\mathrm{a}-\mathrm{x}}$

Or

 $\mathrm{t}=\frac{2.303}{0.1386}\log \frac{100}{25}$


or                         

$$\begin{gathered} \mathrm{t}=\frac{2.303}{0.1386}\log 4 \\ \mathrm{t} = \frac{2.303}{0.1386}\times 0.6021 \mathrm{or} \mathrm{t} = 10.00 \min . \end{gathered}$$


Ostwald Isolation Method: In this method, the concentration of all the reactants are taken in large excess except that of one. The concentration change only for this reactant is significant as other are so much in excess that practically there is no change in their concentrations. The constant terms may be combined with the rate constant and we may write

$\mathrm{Rate} = \mathrm{k}{\left[\mathrm{A}\right]}^{\mathrm{\alpha }} {\left[\mathrm{B}\right]}^{\mathrm{\beta }} {\left[\mathrm{C}\right]}^{\mathrm{\gamma }} = {\mathrm{k}}_0{\left[\mathrm{A}\right]}^{\mathrm{\alpha }}$

The value of ‘a’, i.e., the order of reaction with respect to A can be determined by the methods given above.

0–100 s, rate = – [3.50 – 400] x 10³ Pa/100 s = 5 Pa / s
100-200 s, rate = – [3.00 – 3.50] x 10³ Pa/100 s = 5 Pa / s
200-300 s, rate = – [2.50 – 3.00] x 10³ Pa / 100 s = 5 Pa / s

We notice that the rate remains constant, therefore, reaction is of zero order.
k = rate = 5 Pa / s

t_{1 / 2} = initial concentration or pressure/2K

= 4.00 x 10³ Pa / 2 x 5 Pa s^–1 = 400 s.

Half-life of a reaction: The half-life of a reaction represented as t_1/2 is the time required for the reactant concentration to drop to one half of its initial value. Consider the zeroth order reaction.
R → Products

Integrate Zeroth order rate equation.

[R] = – kt + [R]₀

where at time t = t_1/2, the fraction of [R] that remains [R] / [R]₀, therefore above equation can be written

[R]₀/2 = – k t_1/2 + [ R]₀
k_1/2 = [R]₀ /2
                      ${\mathrm{t}}_{1/2} = {\left[\mathrm{R}\right]}_0/2\mathrm{k}$

for the first order integrated rate equation

$\mathrm{In}{\left[\mathrm{R}\right]}_{\mathrm{t}}/{\left[\mathrm{R}\right]}_0 = -\mathrm{kt}$


at time ${\mathrm{t}}_{1/2},$  ${\left[\mathrm{R}\right]}_{\mathrm{t}}/{\left[\mathrm{R}\right]}_0 = 1/2$

$$\begin{gathered} \mathrm{In} \frac{1}{2} = -\mathrm{kt} 1/2 \\ {\mathrm{kt}}_{1/2} = {\mathrm{In}}^2 \\ {\mathrm{t}}_{1/2} = 2.303 \log 2/\mathrm{k} = 0.693 \mathrm{k} \end{gathered}$$

The half life depends on reactant concentration in different order of reactions as follows.
For zero order raction t_1/2 ∝ [R]₀. For first order reaction t_1/2 is independent of R₀, for second order reaction t_1/2 ∝ 1/[R]₀.
For nth order reaction t_1/2 ∝ 1/[R]₀^n–1.

For first order reaction                

$\mathrm{t}=\frac{2.303}{\mathrm{k}}\log \frac{\mathrm{a}}{\mathrm{a}-\mathrm{x}}$

                              $$\begin{gathered} {\mathrm{t}}_{0.999} = \frac{2.303}{\mathrm{k}}\log \frac{\mathrm{a}}{(\mathrm{a}-0.999 \mathrm{a})} [\because \mathrm{x} = 0.999 \mathrm{a}] \\ = \frac{2.303}{\mathrm{k}}\log \frac{\mathrm{a}}{0.001 \mathrm{a}} \\ = \frac{2.303}{\mathrm{k}}\log 1000 = \frac{2.303}{\mathrm{k}}\times 3 ...\left(\mathrm{i}\right) \end{gathered}$$


and                             

$$\begin{gathered} {\mathrm{t}}_{1/2} = \frac{2.303}{\mathrm{k}}\log \frac{\mathrm{a}}{(\mathrm{a}-0.5 \mathrm{a})} \\ = \frac{2.303}{\mathrm{k}}\log 2 = \frac{2.303}{\mathrm{k}}\times 0.3010 \end{gathered}$$......(ii)

Dividing (i) by (ii), we get  

$\frac{{\mathrm{t}}_{0.999}}{{\mathrm{t}}_{0.5}} = \frac{2.303}{\mathrm{k}}\times 3\times \frac{\mathrm{k}}{2.303\times 0.3010}=10$


or                                         ${\mathrm{t}}_{0.999} = 10 \times {\mathrm{t}}_{0.5}$