\(\frac { x }{ a } =1-{ e }^{ -kt }\)

\(Step \ 1.{ O }_{ 3 }\rightleftharpoons { O }_{ 2 }+O\quad (fast)\quad \quad \quad ..(i)\)
\(Step \ 2.O+{ O }_{ 3 }\longrightarrow 2{ O }_{ 2 }\quad (slow)\quad \quad \quad...(ii)\)
\(From \ slow \ step, \ Rate \ = \ k\left[ { O }_{ 3 } \right] \left[ O \right] \ \ \ \ \ \ \ \ \ \ \ \ \ ...(iii)\)
\(From \ eqn. \ (i), \ { K }_{ eqm. }=\frac { \left[ { O }_{ 2 } \right] \left[ O \right] }{ \left[ { O }_{ 3 } \right] } or \ \left[ O \right] ={ K }_{ eqm. }\frac { \left[ { O }_{ 3 } \right] }{ \left[ { O }_{ 2 } \right] } \)
Substituting this value in eqn. (iii), we get
\(Rate=k\left[ { O }_{ 3 } \right] \times { K }_{ eqm. }{ \left[ { O }_{ 3 } \right] }/{ \left[ { O }_{ 2 } \right] }=k\prime { \left[ { O }_{ 3 } \right] }^{ 2 }{ \left[ { O }_{ 2 } \right] }^{ -1 }\) 
Thus, order w.r.t  \({ O }_{ 2 }\) = -1

For a 1st order reaction, \(k=\frac { 2.303 }{ t } \log { \frac { a }{ a-x } } \)
In terms of pressures, \(a\propto { P }_{ 0 }\)  (initial pressure)
 \(a-x\propto \)(pressure of \({ N }_{ 2 }{ O }_{ 5 }\) at time t)
\(\therefore \ k=\frac { 2.303 }{ t } \log { \frac { { P }_{ 0 } }{ P } } \) or \(\frac { kt }{ 2.303 } =\log { { P }_{ 0 } } -\log { P } \) or \(\log { P } =\frac { kt }{ 2.303 } +\log { { P }_{ 0 } } \)
Thus, plot of \(\log { P } \) vs. t will be linear if the reaction is of  1st order. Further, from the straight line plot, slope\(=-\frac { kt }{ 2.303 } \) and intercept on \(\log { P } axiz=\log { { P }_{ 0 } } \) . Thus, rate constant k and initial pressure \({ P }_{ 0 }\) can be found.

(i)\(-{d[A]\over dt}=k[A]^{1/2}\ or\ -{d[A]\over[A]^{1/2}}=k\ dt\)
Integrating both sides of the equation,- ∫ [A]112d [A] = ∫k dt or \(-{[A]^{1/2}\over 1/2}=kt+I\)
or - 2 [A]^1/2= kt + 1
When t = 0, [A] = [A_o] ∴ - 2[A_o]^1/2 = 0 + 1
Substituting this value of I in eqn. (i), we get
- 2 [A]^1/2= kt - 2 [A_o]^1/2 or kt = 2 {[A_o]^1/2-[A]^1/2 }  or  \(k={2\over t}\{[A_0]^{1/2}-[A]^{1/2}\}\)
(ii) t = t_1/2 when [A] = [A_o]/2
\(t_{1/2}={2\over k}\left\{ [A_0]^{1/2}-{[A_0]^{1/2}\over 2^{1/}}\right\}={2\over k}[A_0]^{1/2}\left\{1-{1\over \sqrt2}\right\}={2\over k}[A_0]^{1/2}{\sqrt{2}-1\over \sqrt{2}}\ or\ t_{1/2}={\sqrt{2}(\sqrt2-1)\over k}[A_0]^{1/2}\)

The given rate law equation can be written as \(-\frac { d\left[ { NH }_{ 3 } \right] }{ dt } =\frac { { k }_{ 1 }\left[ { NH }_{ 3 } \right] }{ 1/\left[ { NH }_{ 3 } \right] +{ k }_{ 2 } } \)
(i) If \(\left[ { NH }_{ 3 } \right] \) is very high, \(1/\left[ { NH }_{ 3 } \right] \) becomes negligible
 \(\therefore \)  \(-\frac { d\left[ { NH }_{ 3 } \right] }{ dt } =\frac { { k }_{ 1 } }{ { k }_{ 2 } } =k\)
i.e., rate becomes independent of concentration.Hence, it is zero order.
(ii) If \(\left[ { NH }_{ 3 } \right] \)is very small, \(1/\left[ { NH }_{ 3 } \right] \)will be very large\((>>{ k }_{ 2 })\), so that \({ k }_{ 2 }\)can be neglected in comparison to \(1/\left[ { NH }_{ 3 } \right] \). Hence,  \(-\frac { d\left[ { NH }_{ 3 } \right] }{ dt } =\frac { { k }_{ 1 } }{ 1/\left[ { NH }_{ 3 } \right] } ={ k }_{ 1 }\left[ { NH }_{ 3 } \right] \)
Thus, reaction is of 1^st order.