The following results have been obtained during the kinetic studies of the reaction:
$2 A + B \rightarrow C + D$

Experiment	$[A] / mol \, L^{-1}$	$[B] / mol \, L^{-1}$	Initial rate of formation of $D / mol \, L^{-1} \, min^{-1}$
$I$	$0.1$	$0.1$	$6.0 \times 10^{-3}$
$II$	$0.3$	$0.2$	$7.2 \times 10^{-2}$
$III$	$0.3$	$0.4$	$2.88 \times 10^{-1}$
$IV$	$0.4$	$0.1$	$2.40 \times 10^{-2}$

Determine the rate law and the rate constant for the reaction.

(N/A) Let the order of the reaction with respect to $A$ be $x$ and with respect to $B$ be $y$.
Therefore,the rate of the reaction is given by:
Rate $= k [A]^x [B]^y$
According to the experimental data:
$6.0 \times 10^{-3} = k [0.1]^x [0.1]^y$ $(i)$
$7.2 \times 10^{-2} = k [0.3]^x [0.2]^y$ $(ii)$
$2.88 \times 10^{-1} = k [0.3]^x [0.4]^y$ $(iii)$
$2.40 \times 10^{-2} = k [0.4]^x [0.1]^y$ $(iv)$
Dividing equation $(iv)$ by $(i)$:
$\frac{2.40 \times 10^{-2}}{6.0 \times 10^{-3}} = \frac{k [0.4]^x [0.1]^y}{k [0.1]^x [0.1]^y}$
$4 = (\frac{0.4}{0.1})^x$ $\Rightarrow 4 = 4^x$ $\Rightarrow x = 1$
Dividing equation $(iii)$ by $(ii)$:
$\frac{2.88 \times 10^{-1}}{7.2 \times 10^{-2}} = \frac{k [0.3]^x [0.4]^y}{k [0.3]^x [0.2]^y}$
$4 = (\frac{0.4}{0.2})^y$ $\Rightarrow 4 = 2^y$ $\Rightarrow 2^2 = 2^y$ $\Rightarrow y = 2$
Thus,the rate law is: Rate $= k [A] [B]^2$
Calculating the rate constant $k$ using experiment $I$:
$k = \frac{6.0 \times 10^{-3}}{(0.1) (0.1)^2} = \frac{6.0 \times 10^{-3}}{0.001} = 6.0 \, L^2 \, mol^{-2} \, min^{-1}$