Question

During the kinetic study of the reaction, $2A+B\to C+D$, following results were obtained:

Run	[A] ${\text{mol L}}^{-1}$	[B] ${\text{mol L}}^{-1}$	Initial rate of formation of D ${\text{mol L}}^{-1}{\text{min}}^{-1}$
I.	0.1	0.1	$6.0\times {10}^{-3}$
II.	0.3	0.2	$7.2\times {10}^{-2}$
III.	0.3	0.4	$2.88\times {10}^{-1}$
IV.	0.4	0.1	$2.40\times {10}^{-2}$

(Where, $\left[A\right]$ and $\left[B\right]$ are inital concentration of $A$ and $B$ respectively.)

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

Answer 1

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

Let, the rate of reaction be given by:
$\text{Rate}=k\left[A{]}^a\right[B{]}^b$
Now, consider $II$ and $III$, where $\left[A\right]$ is constant
$\frac{7.2\times {10}^{-2}}{2.88\times {10}^{-1}}=\frac{\left[0.3{]}^a\right[0.2{]}^b}{\left[0.3{]}^a\right[0.4{]}^b}$
$\frac{1}{4}=(\frac{1}{2}{)}^b\Rightarrow b=2$
Now consider, $I$ and $IV$
$\frac{6.0\times {10}^{-3}}{2.4\times {10}^{-2}}=\frac{\left[0.1{]}^a\right[0.1{]}^b}{\left[0.4{]}^a\right[0.1{]}^b}$
$\frac{1}{4}=(\frac{1}{4}{)}^a$
$\Rightarrow a=1$
$\therefore Rate=k\left[A\right][B{]}^2$