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:

$A/mol{L}^{-1}$	$0.20$	$0.20$	$0.40$
$B/\text{mol}{L}^{-1}$	$0.30$	$0.10$	$0.05$
$r_0/mol{L}^{-1}{s}^{-1}$	$5.07\times 10^{-5}$	$5.07\times 10^{-5}$	$1.43\times 10^{-4}$

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

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

$\Rightarrow x=1.5$