Propanone reacts with iodine in acid solution as shown in this equation.

${\mathrm{CH}}_3\mathrm{C}\left(\mathrm{O}\right){\mathrm{CH}}_3+{\mathrm{I}}_2\stackrel{{\mathrm{H}}^{+}}{⟶}{\mathrm{CH}}_3\mathrm{C}\left(\mathrm{O}\right){\mathrm{CH}}_2\mathrm{I}+\mathrm{HI}$

These data were obtained when the reaction was studied

$\left[{\mathrm{CH}}_8\mathrm{C}\left(\mathrm{O}\right){\mathrm{CH}}_8\right],\mathbf{M}$	$\left[{\mathrm{a}}_2\right],\mathrm{M}$	$\left[{\mathrm{B}}^{+}\right],\mathrm{M}$	Relative Rate
0.010	0.010	0.010	1
0.020	0.010	0.010	2
0.020	0.020	0.010	3
0.020	0.010	0.020	4

What is the rate equation for the reaction?

$n=0m=1$

$\begin{array}{l}\frac{4}{2}={\left(\frac{0.01}{0.02}\right)}^n{\left(\frac{0.02}{0.01}\right)}^P\\ \frac{4}{2}=2\\ 2=2\\ p=1\end{array}$

Rate  $=k{\left[{\mathrm{CH}}_3-{\mathrm{C}}^{\mathrm{C}}-{\mathrm{CH}}_3\right]}^{\mathrm{\prime }}{\left[{\mathrm{H}}^{+}\right]}^{\mathrm{\prime }}$

option D.

$\begin{array}{r}\text{ rate }{\left.=K{\left[C{n}_3\right]}_{-in}\right]}^m{\left(I_L\right]}^n{\left(x^{+}\right]}^P\\ \text{ (1) }1=k\left(0.01{)}^m\right(0.01{)}^n(0.01{)}^p\\ \text{ (2) }z=k\left(0.02{)}^m\right(0.01{)}^n(0.01{)}^p\\ \text{ (3) }2=k\left(0.02{)}^m\right(0.02{)}^n(0.01{)}^p\\ 4=k\left(0.02{)}^m\right(0.01{)}^n(0.02{)}^p\\ \text{ taking iohto (2)/(1) }\\ \frac{2}{1}={\left(\frac{0.02}{0.01}\right)}^m{\left(\frac{0.01}{0.01}\right)}^n\left(\frac{0.01}{0.01}\right)\\ 2=2^m\\ m=1\\ \text{ taking satio (3) (2) }\\ \frac{2}{2}={\left(\frac{0.02}{0.01}\right)}^n\\ 1=2^n\\ n=0\end{array}$