We have given intial concentration 2x10^-3and 1x10^-3. 
  the intial concentration equal to  2.40 x 10^-4 Ms^-1 and 0.60 x10^-4 MS^-1

Let the rate be  = $\mathrm{k}{\left[\mathrm{A}\right]}^{\mathrm{x}}$

From the trial (i) and (ii), we get

                      $\frac{2.40 \times {10}^{-4} \mathrm{M} {\mathrm{s}}^{-1}}{0.6 \times {10}^{-4} \mathrm{M} {\mathrm{s}}^{-1}} = {\left[\frac{2\times {10}^{-3}}{1\times {10}^{-3}}\right]}^{\mathrm{x}}$
or                                          $4 = (2{)}^{\mathrm{x}} \mathrm{or} \mathrm{x} = 2.$
Thus reaction is of second order

                       $$\begin{gathered} \mathrm{Rate} = \mathrm{k}{\left[\mathrm{A}\right]}^2 \\ \mathrm{k} = \frac{\mathrm{Rate}}{{\left[\mathrm{A}\right]}^2} = \frac{2.4\times {10}^{-4} \mathrm{mol} {\mathrm{L}}^{-1} {\mathrm{s}}^{-1}}{(2\times {10}^{-3} \mathrm{mol} {\mathrm{L}}^{-1}{)}^2} \\ = \frac{2.4 \times {10}^{-4} \mathrm{mol} {\mathrm{L}}^{-1} {\mathrm{s}}^{-1}}{4 \times {10}^{-6} {\mathrm{mol}}^2 {\mathrm{L}}^{-2}} = 0.6 \times {10}^2 {\mathrm{mol}}^{-1} \mathrm{L} {\mathrm{s}}^{-1} \end{gathered}$$

We have to assume that law canbe written as  $\mathrm{Rate} = \mathrm{k}{\left[\mathrm{A}\right]}^{\mathrm{\alpha }} {\left[\mathrm{B}\right]}^{\mathrm{\beta }}$

Thus 

$\frac{{\mathrm{r}}_2}{{\mathrm{r}}_1} = \frac{\mathrm{k}({\mathrm{A}}_2{)}^{\mathrm{\alpha }} ({\mathrm{B}}_2{)}^{\mathrm{\beta }}}{\mathrm{k}{\left[{\mathrm{A}}_1\right]}^{\mathrm{\alpha }} {\left[{\mathrm{B}}_1\right]}^{\mathrm{\beta }}}$

or      $\frac{4.8 \times {10}^{-2}}{1.2 \times {10}^{-2}} = \frac{(1{)}^{\mathrm{\alpha }} (2.0{)}^{\mathrm{\beta }}}{(1{)}^{\mathrm{\alpha }} {\left[{\mathrm{B}}_1\right]}^{\mathrm{\beta }}} \mathrm{or} \frac{4}{1} = {\left(\frac{2}{1}\right)}^{\mathrm{\beta }} \mathrm{or} \mathrm{\beta } = 2.$

From reaction no. (iv) and (i)    $\frac{{\mathrm{r}}_4}{{\mathrm{r}}_1} = \frac{\mathrm{k}({\mathrm{A}}_4{)}^{\mathrm{\alpha }} ({\mathrm{B}}_4{)}^{\mathrm{\beta }}}{\mathrm{k}{\left[{\mathrm{A}}_1\right]}^{\mathrm{\alpha }} {\left[{\mathrm{B}}_1\right]}^{\mathrm{\beta }}}$

or             $\frac{4.9 \times {10}^{-2}}{1.2 \times {10}^{-2}} = \frac{(4{)}^{\mathrm{\alpha }}}{(1{)}^{\mathrm{\alpha }}}\times \frac{(1{)}^{\mathrm{\beta }}}{(1{)}^{\mathrm{\beta }}} \mathrm{or} \mathrm{\alpha } = 1.$

The order of the reaction is first order with respect to A and second order with respect to B. Overall order of the reaction is 3.

Rate = k[H₂] [NO]²
The order of each reactant
Order with respect to H₂ = 1
Order with respect to NO = 2.

Overall order = 1 + 2 = 3.

For the first order reaction can given by
 
$\mathrm{k} = \frac{2.303}{\mathrm{t}}\log \frac{{\left[\mathrm{A}\right]}_0}{{\left[\mathrm{A}\right]}_{\mathrm{t}}}$

At t = 900 seconds, 

${\left[\mathrm{A}\right]}_0 = 50.8 \mathrm{and} {\left[\mathrm{A}\right]}_{\mathrm{t}} = 19.7 \mathrm{at} \mathrm{t} = 1800 \mathrm{seconds}, {\left[\mathrm{A}\right]}_0 = 50.8 \mathrm{and} {\left[\mathrm{A}\right]}_{\mathrm{t}} = 7.62$
                                                                 $\mathrm{k} = \frac{2.303}{\mathrm{t}}\log \frac{{\left[\mathrm{A}\right]}_0}{{\left[\mathrm{A}\right]}_{\mathrm{t}}}$
(i)                         $\mathrm{k}=\frac{2.303}{900 \mathrm{s}}\log \frac{50.8}{19.7} = \frac{2.303}{900 \mathrm{s}}\times 0.4114 = 1.048 \times {10}^{-3} {\mathrm{s}}^{-1}$

(ii)                        $\mathrm{k} = \frac{2.303}{1800 \mathrm{s}}\log \frac{50.8}{7.62} = \frac{2.303}{1800 \mathrm{s}}\times 0.839 = 1.052 \times {10}^{-3} {\mathrm{s}}^{-1}$

Since the value of k is both case is almost same thus it is first order reaction.