Question

The experimental data for the reaction
$2 A + B_2 \longrightarrow 2 AB$ 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
$ \big(\frac{dx}{dt}\big)=k [A]^0 [B_2]^1$
$ =k [B_2]$