Question

For a reaction $A+2 B \longrightarrow C+D$, the following data were obtained :

Expt. Initial concentration $\left( mol \,L ^{-1}\right)$			Initial rate of formation of $D\left(mol\ , L^{-1}\right.\,min\,\left.^{-1}\right)$
	[A]	[B]
1	0.1	0.1	$6.0 \times 10^{-3}$
2	0.3	0.2	$7.2 \times 10^{-2}$
3	0.3	0.4	$2.88 \times 10^{-1}$
4	0.4	0.1	$2.4 \times 10^{-2}$

The correct rate law expression will be

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

Answer 1

Answer: (B)

Let the order of reaction with respect to $A$ is $x$ and $B$ is $y$.
Rate $=k[A]^{x}[B]^{y} 1$.
1.Rate $=6.0 \times 10^{-3}=(0.1)^{x}(0.1)^{y} \ldots$...(i)
2. $7.2 \times 10^{-2}=(0.3)^{x}(0.2)^{y} \ldots$..(ii)
3. $2.88 \times 10^{1}=(0.3)^{x}(0.4)^{y} \ldots$..(ii)
4. $2.4 \times 10^{2}=(0.4)^{x}(0.1)^{y} \ldots$...(iv)
On dividing Eq. (i) by (iv), we get
$\frac{6.0 \times 10^{-3}}{2.4 \times 10^{-2}}=\left(\frac{0.1}{0.4}\right)^{x}\left(\frac{0.1}{0.1}\right)^{y}$
$\therefore x=1$
On dividing Eq. (ii) by (iii), we get
$\frac{7.2 \times 10^{-2}}{28.8 \times 10^{-2}}=\left(\frac{0.3}{0.3}\right)^{x}\left(\frac{0.2}{0.4}\right)^{y}$
$\frac{1}{4}=\left(\frac{1}{2}\right)^{y}$
$y=2$
Thus, the rate law is rate $=k[ A B ]^{2}$