Question

In a reaction between $A$ and $B$, the initial rate of reaction $r_{0}$ was measured for different initial concentrations of $A$ and $B$ as given below:

$A / m o l L^{-1}$	$0.20$	$0.20$	$0.40$
$B / \text{mol} L^{-1}$	$0.30$	$0.10$	$0.05$
$r_{0} / m o l L^{-1} s^{-1}$	$5.07 \times 10^{-5}$	$5.07 \times 10^{-5}$	$1.43 \times 10^{-4}$

The order of the reaction with respect to $A$ is

• A. 1.5
• B. 0.5
• C. 1
• D. 2

Answer 1

Answer: (A)

Let the rate law, rate $=k[A]^{x}[B]^{y}$

From the data given

$5.07 \times 10^{-5}=k[0.20]^{x}[0.30]^{y} \,\,\,\,\, \dots(i)$

$5.07 \times 10^{-5}=k[0.20]^{x}[0.10]^{y} \,\,\,\,\, \dots(ii)$

$1.43 \times 10^{-4}=k[0.40]^{x}[0.05]^{y} \,\,\,\,\, \dots(iii)$

On dividing eq. (i) by (ii), we get

$1=\frac{[0.30]^{y}}{[0.10]^{y}} $ i.e., y=0

On dividing eq. (iii) by eq. (ii), we get

$\frac{1.43 \times 10^{-4}}{5.07 \times 10^{-5}}=\frac{[0.40]^{x}}{[0.20]^{x}} \frac{[0.10]^{y}}{[0.05]^{y}}$

as $y=0$

Thus, $ 2.82=\left(\frac{0.40}{0.20}\right)^{x} $

$\Rightarrow (2)^{x}=2.82$

$ \Rightarrow x=1.5$