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(A) The rate law is given by $r = k[A]^x[B]^y$.Comparing experiments $I$ and $II$:$[A]$ is doubled,$[B]$ is constant,and the rate $r$ is doubled.There

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$a, b$ and $c$

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(A) The rate law is given by $r = k[A]^x[B]^y$.Comparing experiments $I$ and $II$:$[A]$ is doubled,$[B]$ is constant,and the rate $r$ is doubled.Therefore,$x = 1$.Comparing experiments $II$ and $III$:$[A]$ is constant,$[B]$ is doubled,and the rate $r$ is doubled.Therefore,$y = 1$.The rate law is $r = k[A][B]$. Statement $(b)$ is correct.Using experiment $I$ to find $k$:$2 	imes 10^{-4} = k 	imes (1 	imes 10^{-2}) 	imes (2 	imes 10^{-2})$$2 	imes 10^{-4} = k 	imes (2 	imes 10^{-4})$$k = 1 \ mol^{-1} \ L \ sec^{-1}$. Statement $(a)$ is correct.If both $[A]$ and $[B]$ are doubled:$r' = k(2[A])(2[B]) = 4k[A][B] = 4r$.Thus,the rate increases four times. Statement $(c)$ is correct.Therefore,all statements $(a), (b),$ and $(c)$ are correct.

$S.NO.$	$[A] \ mol \ L^{-1}$	$[B] \ mol \ L^{-1}$	$Rate \ mol \ L^{-1} \ sec^{-1}$
$I$	$1 \times 10^{-2}$	$2 \times 10^{-2}$	$2 \times 10^{-4}$
$II$	$2 \times 10^{-2}$	$2 \times 10^{-2}$	$4 \times 10^{-4}$
$III$	$2 \times 10^{-2}$	$4 \times 10^{-2}$	$8 \times 10^{-4}$

$Exp.$	$[A] \ (mol \ L^{-1})$	$[B] \ (mol \ L^{-1})$	Initial rate $(mol \ L^{-1} \ s^{-1})$
$1$	$2.5 \times 10^{-4}$	$3 \times 10^{-5}$	$5 \times 10^{-4}$
$2$	$5 \times 10^{-4}$	$6 \times 10^{-5}$	$4 \times 10^{-3}$
$3$	$1 \times 10^{-3}$	$6 \times 10^{-5}$	$1.6 \times 10^{-2}$

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