The reaction $2A+B\to C+D$ follows the rate law $-\frac{d\left[B\right]}{dt}=K\left[A\right]{\left[B\right]}^2.$ Calculate the value of $(x+y)$?

S.no	${\left(A\right)}_0\times {10}^{-4}M$	${\left(B\right)}_0\times {10}^{-4}M$	Half life (sec)
1	500	2	10
2	500	4	x
3	2	250	8
4	4	500	y

 

$2A+B\to C+D$ 
$-\frac{d\left(B\right)}{dt}=K\left[A\right]{\left[B\right]}^2$ 
Experiment I and II $\to$ concentration of A is constant but concentration of B is changing.
So, we can write
$-\frac{d\left(B\right)}{dt}=K\text{'}{\left[B\right]}^{2}K\text{'}=K\left[A\right]$ 
For second order reaction $t_{1/2}\alpha \frac{1}{a}$
$a\to$ Initial concentration of [B].
$10\alpha \frac{1}{2}$ 
$x\alpha \frac{1}{4}$ 
$\frac{10}{x}=\frac{4}{2}=2$ 
$x=5$ 
If concentration of B is constant we can write
$-\frac{d\left(B\right)}{dt}=K\text{'}\left[A\right]K\text{'}=K{\left[B\right]}^2$ 
For Ist order reaction $t_{1/2}$ does not depend on initial concentration of reactant
For experiment – III 
$K\text{'}=K{\left[B\right]}^2=K{\left[250\right]}^2$ 
$\frac{0.693}{t_{1/2}}=K{\left[250\right]}^2$ 
$\frac{0.693}{8}=K{\left[250\right]}^2\mathrm{...}\left(i\right)$ 
For experiment – IV
$K\text{'}=K{\left[B\right]}^2=K{\left[500\right]}^2$ 
$\frac{0.693}{t_{1/2}}=K{\left[500\right]}^2$ 
$\frac{0.693}{y}=K{\left[500\right]}^2\mathrm{...}\left(ii\right)$ 
Form (i) and (ii)
$\frac{y}{8}=\frac{K{\left[250\right]}^2}{K{\left[500\right]}^2}=\frac{1}{4}$ 
$y=2$ 
$x+y=7$