Question

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

Expt. Initial concentration $\left(mol{L}^{-1}\right)$			Initial rate of formation of $D(mol,L^{-1}min{}^{-1})$
	[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 $=6.0\times 10^{-3}=(0.1{)}^x(0.1{)}^y\dots$...(i)
2. $7.2\times 10^{-2}=(0.3{)}^x(0.2{)}^y\dots$..(ii)
3. $2.88\times 10^1=(0.3{)}^x(0.4{)}^y\dots$..(ii)
4. $2.4\times 10^2=(0.4{)}^x(0.1{)}^y\dots$...(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 $=k[AB{]}^2$