The rate of a reaction: $\displaystyle \mathrm{A}+\mathrm{B} \longrightarrow$ product is given below as a function of different initial concentrations of A and B.

Experiment	[A]$\displaystyle / \mathrm{mol} \mathrm{L}^{-1}$	[B]/mol L $\displaystyle { }^{-1}$	Initial Rate $\displaystyle / \mathrm{mol} \mathrm{L}^{-1} \mathrm{~min}^{-1}$
$\displaystyle 1$	$\displaystyle 0.01$	$\displaystyle 0.01$	$\displaystyle 5 \times 10^{-3}$
$\displaystyle 2$	$\displaystyle 0.02$	$\displaystyle 0.01$	$\displaystyle 1 \times 10^{-2}$
$\displaystyle 3$	$\displaystyle 0.01$	$\displaystyle 0.02$	$\displaystyle 5 \times 10^{-3}$

Calculate the order of the reaction with respect to A and B. Determine the rate constant of the reaction.

Rate = k [ A]p [B]q $\displaystyle 5$·$\displaystyle 0$ × $\displaystyle 10$–$\displaystyle 3$= k [ $\displaystyle 0.01$]p [$\displaystyle 0.01$]q………Eq $\displaystyle 1$ $\displaystyle 1$·$\displaystyle 0$ × $\displaystyle 10$–$\displaystyle 2$= k [ $\displaystyle 0.02$]p [$\displaystyle 0.01$]q………Eq $\displaystyle 2$ $\displaystyle 5$·$\displaystyle 0$ × $\displaystyle 10$–$\displaystyle 3$= k [ $\displaystyle 0.01$]p [$\displaystyle 0.02$]q………Eq $\displaystyle 3$ On Comparing (eq1) and (eq3) $\displaystyle 1$= ($\displaystyle 2$)q q = $\displaystyle 0$ On Comparing (eq1) and (eq2) ($\displaystyle 2$)$\displaystyle 1$= ($\displaystyle 2$)p p = $\displaystyle 1$ Order w·r·t A = $\displaystyle 1$ Order w·r·t B= $\displaystyle 0$ From eq $\displaystyle 1$ $\displaystyle 5$·$\displaystyle 0$ × $\displaystyle 10$–$\displaystyle 3$= k [ $\displaystyle 0.01$]$\displaystyle 1$ [$\displaystyle 0.01$]$\displaystyle 0$ k =$\displaystyle 0.5$ min–$\displaystyle 1$