Question

Sr.No.	$[A{]}_0/M$(lonic conc.)	$[B{]}_0/M$(lonic conc.)	rate/$\left(Mse{c}^{-1}\right)$
1	$1.6\times {10}^{-3}$	$5\times {10}^{-2}$	${10}^{-3}$
2	$3.2\times {10}^{-3}$	$5\times {10}^{-2}$	$4\times {10}^{-3}$
3	$1.6\times {10}^{-3}$	${10}^{-1}$	$2\times {10}^{-3}$
4	$3.2\times {10}^{-3}$	${10}^{-1}$	$8\times {10}^{-3}$

For the reaction $A+B⟶C$; starting with different initial concentration of A and B, initial rate of reaction were determined graphically in four experiments :
Rate law for reaction from above data is :

• A. $r=k\left[A{]}^2\right[B{]}^2$
• B. $r=k\left[A{]}^2\right[B]$
• C. $r=k\left[A\right][B{]}^2$
• D. $r=k\left[A\right]\left[B\right]$

Answer 1

The correct option is B $r=k\left[A{]}^2\right[B]$

The rate is given by,

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

From experiments $1$ and $2$,

${10}^{-3}=[1.6\times {10}^{-3}{]}^x[5\times {10}^{-2}{]}^y$ ...... (Exp 1)

$4\times {10}^{-3}=[3.2\times {10}^{-3}{]}^x[5\times {10}^{-2}{]}^y$ ......(Exp 2)

When the concentration of B is kept constant and the concentration of $A$ is doubled, the rate of the reaction becomes four times. Hence, the order of the reaction with respect to $A$ is $2.$

Similarly, from experiments $1$ and $3$, when the concentration of $A$ is kept constant and the concentration of $B$ is doubled, the rate of the reaction is doubled. Hence, the order of the reaction with respect to $B$ is $1.$

Hence, the rate law for reaction from above data is $r=k\left[A{]}^2\right[B]$.