In a reaction between $A$ and $B$,the initial rate of reaction $(r_0)$ was measured for different initial concentrations of $A$ and $B$ as given below:

$A / mol \ L^{-1}$	$0.20$	$0.20$	$0.40$
$B / mol \ L^{-1}$	$0.30$	$0.10$	$0.05$
$r_0 / mol \ L^{-1} \ s^{-1}$	$5.07 \times 10^{-5}$	$5.07 \times 10^{-5}$	$1.43 \times 10^{-4}$

What is the order of the reaction with respect to $A$ and $B$?

(A) Let the order of the reaction with respect to $A$ be $x$ and with respect to $B$ be $y$.
Therefore,the rate law is given by:
$r_0 = k[A]^x[B]^y$
From the given data:
$5.07 \times 10^{-5} = k[0.20]^x[0.30]^y$ $(i)$
$5.07 \times 10^{-5} = k[0.20]^x[0.10]^y$ $(ii)$
$1.43 \times 10^{-4} = k[0.40]^x[0.05]^y$ $(iii)$
Dividing equation $(i)$ by $(ii)$:
$\frac{5.07 \times 10^{-5}}{5.07 \times 10^{-5}} = \frac{k[0.20]^x[0.30]^y}{k[0.20]^x[0.10]^y}$
$1 = (3)^y$
Since $3^0 = 1$,we get $y = 0$.
Now,dividing equation $(iii)$ by $(ii)$ and substituting $y = 0$:
$\frac{1.43 \times 10^{-4}}{5.07 \times 10^{-5}} = \frac{k[0.40]^x[0.05]^0}{k[0.20]^x[0.10]^0}$
$2.82 = (2)^x$
Taking log on both sides:
$\log(2.82) = x \log(2)$
$x = \frac{0.450}{0.301} \approx 1.5$
Hence,the order of the reaction with respect to $A$ is $1.5$ and with respect to $B$ is $0$.