Question

During the kinetic study of the reaction, $2A + B\longrightarrow C + D$, following results were obtained . Based on the above data which one of the following is correct?

Run	$A/mol\,L^{-1}$	$B/mol\,L^{-1}$	Initial rate of formation of $D/mol\,L^{-1}\,min^{-1}$
I	0.1	0.1	$6.0 \times 10^{-3}$
II	0.3	0.2	$7.2 \times 10^{-2}$
III	0.3	0.4	$2.88 \times 10^{-1}$
IV	0.4	0.1	$2.40 \times 10^{-2}$

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

Answer 1

The Correct Option is D

Let the order of reaction with respect to A is x
and with respect to B is y. Thus,
rate=k$[A]^x [B]^y$
($x$ and $y$ are stoichiometric coefficient)
For the given cases,
I. rate = $k (0.1)^x \, (0.1)^y =6.0 \times 10^{-3}$
II. rate = $k (0.3)^x \, (0.2)^y =7.2 \times 10^{-2}$
III. rate = $k (0.3)^x \, (0.40)^y =2.88 \times 10^{-1}$
IV. rate = $k (0.34)^x \, (0.1)^y =2.40 \times 10^{-2}$
Dividing E (I) by E (IV), we get
$ \big(\frac{0.1}{0.4}\big)^x \big(\frac{0.1}{0.1}\big)^y =\frac{6.0 \times 10^{-3}}{2.4 \times 10^{-2}}$
or $ \big(\frac{1}{4} \big)^x = \big(\frac{1}{4} \big)^1$
$\therefore \hspace25mm x=1$
On dividing E (II) by E (Ill), we get
$ \big(\frac{0.3}{0.3}\big)^x \big(\frac{0.2}{0.4}\big)^y =\frac{7.2 \times 10^{-2}}{2.88 \times 10^{-1}}$
or $\big(\frac{1}{2} \big)^y = \frac{1}{4}$
or $ \big(\frac{1}{2} \big)^y = \big(\frac{1}{2} \big)^2$
$\therefore \hspace25mm y=2$
Thus, rate law is,
$ rate = k[A]^1 [B]^2$
$ = k[A][B]^2$