Question

The experimental data for the reaction
$2A+B_2⟶2AB$ is
The rate equation for the above data is

Exp.	[A]	$[B_2]$	Rate $(M{s}^{-1})$
1	0.50	0.50	$1.6\times 10^{-4}$
2.	0.05	1.00	$3.2\times 10^{-4}$
3.	1.00	1.00	$3.2\times 10^{-4}$

• A. rate = k $[B_2]$
• B. rate = k $[B_2{]}^2$
• C. rate = k $[A{]}^2[B{]}^2$
• D. rate = k $[A{]}^2[B]$

Answer 1

Answer: (A)

Consider the following rate law equation,
$\frac{dx}{dt}=k[A{]}^m[B_2{]}^n$
$1.6\times 10^{-4}=k[0.50{]}^m[0.50{]}^n...(i)$
$3.2\times 10^{-4}=k[0.50{]}^m[1.0{]}^n...(ii)$
$3.2\times 10^{-4}=k[1.00{]}^m[1.0{]}^n...(iii)$
From Eqs. (ii) and (iii)
$\frac{3.2\times 10^{-4}}{3.2\times 10^{-4}}=\frac{k[1.00{]}^m[1.0{]}^n}{k[0.50{]}^m[1.0{]}^n}$
$1=2^{m}or{2}^0=2^m$
$\therefore m=0$
From Eqs. (i) and (ii)
$\frac{3.2\times 10^{-4}}{1.6\times 10^{-4}}=\frac{k[0.50{]}^m[1.0{]}^n}{k[0.50{]}^m[0.50{]}^n}$
$2=2^n$
or $2^1=2^n$
$\therefore n=1$
Hence, rate
$\left(\frac{dx}{dt}\right)=k[A{]}^0[B_2{]}^1$
$=k[B_2]$