Question

The experimental data for the reaction, $2A+B_2⟶2AB$, is as follows:

Expt. No.	$\left[A\right]$ $\left(mol{L}^{-1}\right)$	$\left[B_2\right]$ $\left(mol{L}^{-1}\right)$	Rate $\left(mol{L}^{-1}{s}^{-1}\right)$
$1.$	$0.50$	$0.50$	$1.6\times {10}^{-4}$
$2.$	$0.50$	$1.00$	$3.2\times {10}^{-4}$
$3.$	$1.00$	$1.00$	$3.2\times {10}^{-4}$

Write the most probable equation for the rate of reaction giving reason for your answer.

Answer 1

From an examination of above data, it is clear that when the concentration of $B_2$ is doubled, the rate is doubled. Hence, the order of reaction with respect to $B_2$ is one.
Further, when concentration of $A$ is doubled, the rate remains unaltered. So, order of reaction with respect to $A$ is zero.
The probable rate law for the reaction will be
$-\frac{dx}{dt}=k\left[B_2\right]{\left[A\right]}^0=k\left[B_2\right]$
Alternatively, Rate $=k{\left[B_2\right]}^{\alpha }$
$1.6\times {10}^{-4}=k{\left[0.5\right]}^{\alpha }$
$3.2\times {10}^{-4}=k{\left[1\right]}^{\alpha }$
On dividing we get, $\alpha =1$
$\therefore$ Rate $=k{\left[A\right]}^0{\left[B_2\right]}^1=k\left[B_2\right]$