For the reaction. A + 2 B ⟶ C , the reaction rate is doubled if the concentration of A is doubled.…

Rate of the given reaction is doubled when concentration of $\text{'}A\text{'}$ is doubled and it is quadrupled when concentrations of $\text{'}A\text{'}$ and  $\text{'}B\text{'}$ are raised four times.

Let the rate,  $\mathrm{R}=\mathrm{k} [\mathrm{A}{]}^{\mathrm{x}} [\mathrm{B}{]}^{\mathrm{y}} ...\left(\mathrm{i}\right)$

$\therefore$ When concentration of $A$ is doubled, the new

rate $\left(R_1\right)$will be

${\mathrm{R}}_1=\mathrm{k}\left[2\mathrm{A}{]}^{\mathrm{x}}\right[\mathrm{B}{]}^{\mathrm{y}} ...\left(ii\right)$

But ${\mathrm{R}}_1=2\mathrm{R}$

$\therefore k\left[2A{]}^x\right[B{]}^y=2k\left[A{]}^x\right[B{]}^y$.....[From (i) and (ii) ]

$\therefore k{2}^x\left[A{]}^x\right[B{]}^y=2k\left[A{]}^x\right[B{]}^y$

$\therefore 2^x=2$ and thus $x=1$

Similarly, when concentrations of both $A$ and $B$are quadrupled, the new rate $\left(R_2\right)$ will be

${\mathrm{R}}_2=\mathrm{k}\left[4\mathrm{A}{]}^{\mathrm{x}}\right[4\mathrm{B}{]}^{\mathrm{y}} ....\left(\mathrm{iii}\right)$

But ${\mathrm{R}}_2=4\mathrm{R}$

$\therefore k\left[4A{]}^x\right[4B{]}^y=4k\left[A{]}^x\right[B{]}^y$

.....[From(i)and (iii)]

$\cancel{k} 4^x\left[\cancel{A}{]}^x{4}^y\right[\cancel{B}{]}^y=4\cancel{k}\left[\cancel{A}{]}^x\right[\cancel{B}{]}^y$

$\therefore 4^x·4^y=4$

$\therefore 4.4^y=4 (\because \mathrm{x}=1)$

$\therefore 4^y=1$ and thus $y=0$

Substituting the values of $x$ and $y$ in$\left(i\right)$, the rate expression for the givcn rcaction is

rate,  $\mathrm{R}=\mathrm{k}\left[\mathrm{A}{]}^1\right[\mathrm{B}{]}^0 \therefore \mathrm{R}=\mathrm{k}\left[\mathrm{A}\right]$

Thus, the overall order of the given reaction is  $1.$