For the reaction $A+B⟶C+2D$ experimental results were collected for three trials and the data obtained are given below:

Trial	[A],M	[B],M	Initial rate, $\mathrm{M} {\mathrm{s}}^{-}$
1 2 3	0.40 0.80 0.40	0.20 0.20 0.40	$5.5\times {10}^{-4}$ $5.5\times {10}^{-4}$ $2.2\times {10}^{-3}$

the correct rate law of the reaction is:

Let rate law $=k\left[A{]}^x\right[B{]}^y$. From experiments,

we have:

$5.5\times {10}^{-4}=k\left[0.4{]}^x\right[0.2{]}^y$                           ..(i)

$5.5\times {10}^{-4}=k[0.8{]}^x\left[0.2{\parallel }^y\right.$                       …(ii)

$2.2\times {10}^{-3}=k\left[0.4{]}^x\right[0.4{]}^y$                           …(iii)

From equations (i) and (ii) ,we have :

$\frac{5.5\times {10}^{-4}}{5.5\times {10}^{-4}}=\frac{k\left[0.4{]}^x\right[0.2{]}^y}{k\left[0.8{]}^x\right[0.2{]}^y};1={\left[\frac{0.4}{p.8}\right]}^x={\left(\frac{1}{2}\right)}^x;{\left(\frac{1}{2}\right)}^0={\left(\frac{1}{2}\right)}^x.$

So,$\mathrm{x} = 0.$

From equation(i) and (iii), we have:

$\frac{5.5\times {10}^{-4}}{2.2\times {10}^{-3}}=\frac{k\left[0.4{]}^x\right[0.2{]}^y}{k\left[0.4{]}^x\right[0.4{]}^y};\frac{1}{4}={\left(\frac{1}{2}\right)}^y;{\left(\frac{1}{2}\right)}^2={\left(\frac{1}{2}\right)}^y$

So, $\mathrm{y} = 2.$

Substituting these values in rate law, we get:

rate law $:k\left[A{]}^0\right[B{]}^2$. So, the correct answer is (1).