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$(mol\, L^{-1} s^{-1})$
$1$.	$2.5\times10^{-4}$	$3\times10^{-5}$	$5\times10^{-4}$
$2$.	$5\times10^{-4}$	$6\times10^{-5}$	$4\times10^{-3}$
$3$.	$1\times10^{-3}$	$6\times10^{-5}$	$1.6\times10^{-2}$

• 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. $\left(1\right), 5\times10^{-4} = k \left(2.5\times 10^{-4}\right)^{x} \left(3\times10^{-5}\right)^{y} \ldots\left(i\right)$
From exp. $\left(2\right), 4\times10^{-3}=k\left(5\times10^{-4}\right)^{x} \left(6\times10^{-5}\right)^{y} \ldots\left(ii\right)$
Dividing $\left(ii\right)$ by $\left(i\right)$, $\frac{4\times10^{-3}}{5\times10^{-4}}=2^{x}\cdot2^{y}=8$
From exp. $\left(3\right), 1.6\times10^{-2}=k\left(1\times10^{-3}\right)^{x} \left(6\times10^{-5}\right)^{y} \ldots\left(iii\right)$
Dividing $\left(iii\right)$ by $\left(ii\right)$, $\frac{1.6\times10^{-2}}{4\times10^{-3}}=2^{x}=4$
or $x = 2, y = 1$
Hence, order with respect to $A$ is $2$ and with respect to $B$ is $1$.