Question

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

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

• 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

Answer: (B)

Rate $=k{\left[A\right]}^x{\left[B\right]}^y$
From exp. $\left(1\right),5\times 10^{-4}=k{\left(2.5\times 10^{-4}\right)}^x{\left(3\times 10^{-5}\right)}^y\dots \left(i\right)$
From exp. $\left(2\right),4\times 10^{-3}=k{\left(5\times 10^{-4}\right)}^x{\left(6\times 10^{-5}\right)}^y\dots \left(ii\right)$
Dividing $\left(ii\right)$ by $\left(i\right)$, $\frac{4\times 10^{-3}}{5\times 10^{-4}}=2^x\cdot 2^y=8$
From exp. $\left(3\right),1.6\times 10^{-2}=k{\left(1\times 10^{-3}\right)}^x{\left(6\times 10^{-5}\right)}^y\dots \left(iii\right)$
Dividing $\left(iii\right)$ by $\left(ii\right)$, $\frac{1.6\times 10^{-2}}{4\times 10^{-3}}=2^x=4$
or $x=2,y=1$
Hence, order with respect to $A$ is $2$ and with respect to $B$ is $1$.