Question

During the kinetic study of the reaction, $2A+B⟶C+D$, 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 $D/mol{L}^{-1}mi{n}^{-1}$
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

Answer: (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 Eq. (I) by Eq. (IV), we get
$\left(\frac{0.1}{0.4}{)}^x\right(\frac{0.1}{0.1}{)}^y=\frac{6.0\times 10^{-3}}{2.4\times 10^{-2}}$
or $(\frac{1}{4}{)}^x=(\frac{1}{4}{)}^1$
$\therefore x=1$
On dividing Eq. (II) by Eq. (Ill), we get
$\left(\frac{0.3}{0.3}{)}^x\right(\frac{0.2}{0.4}{)}^y=\frac{7.2\times 10^{-2}}{2.88\times 10^{-1}}$
or $(\frac{1}{2}{)}^y=\frac{1}{4}$
or $(\frac{1}{2}{)}^y=(\frac{1}{2}{)}^2$
$\therefore y=2$
Thus, rate law is,
$rate=k[A{]}^1[B{]}^2$
$=k[A][B{]}^2$