Question

Rate For the reaction $A+B\to products$, what will be the order of reaction with respect of $A$ and $B$?

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

• A. $1$ with respect to $A$ and $2$ with respect to $B$
• B. $2$ with respect to $A$ and $1$ with respect to $B$
• C. $1$ with respect to $A$ and $1$ with respect to $B$
• D. $2$ with respect to $A$ and $2$ with respect to $B$

Answer 1

The correct option is B $2$ with respect to $A$ and $1$ with respect to $B$

$Rate=k\left[A{]}^x\right[B{]}^y$

From exp. $\left(1\right),5\times {10}^{-4}=k(2.5\times {10}^{-4}{)}^x(3\times {10}^{-5})y.....(i)$

From exp. $\left(2\right),4\times {10}^{-3}=k(5\times {10}^{-4}{)}^x(6\times {10}^{-5}{)}^y....\left(ii\right)$

Dividing (ii) by (i), $\frac{4\times {10}^{-3}}{5\times {10}^{-4}}=2^x\cdot 2^y=8$
From exp. $\left(3\right),1.6\times {10}^{-2}=k(1\times {10}^{-3}{)}^x(6\times {10}^{-5}{)}^y....\left(iii\right)$

Dividing (iii) by (ii), $\frac{1.6\times {10}^{-2}}{4\times {10}^{-3}}=2^x=4$
or $x=2,y=1$
Hence, order with respect to $A$ is $2$ and with respect to $B$ is $1$.