Question

For the non-stoichiometric reaction 2A + B ⟶ C + D. The following kinetic data were obtained in three separate experiments, all at 298 K.

Initial concentration (A)	Initial concentration (B)	Initial rate of formation of C $\left(mol{L}^1{S}^{-1}\right)$
0.1M	0.1M	$1.2\times {10}^{-3}$
0.1M	0.2M	$1.2\times {10}^{-3}$
0.2M	0.1M	$2.4\times {10}^{-3}$

The rate law for the formation of C is :
(IIT-JEE-2014)

• A. $\frac{dc}{dt}=k\left[A\right][B{]}^2$
• B. $\frac{dc}{dt}=k\left[A\right]$
• C. $\frac{dc}{dt}=k\left[A\right]\left[B\right]$
• D. $\frac{dc}{dt}=k\left[A{]}^2\right[B]$

Answer 1

The correct option is

A

$\frac{dc}{dt}=k\left[A\right]$

(i) $R=k\left[A{]}^x\right[B{]}^y$

$1.2x{10}^{-3}=k\left[0.1{]}^x\right[0.1{]}^y$ …(i)

$1.2x{10}^{-3}=k\left[0.1{]}^x\right[0.2{]}^y$ …(ii)

Divide Eq. (ii) by Eq. (i)

$1=(2{)}^y$

∴ $(2{)}^0=(2{)}^y\therefore y=0$

(ii) $1.2x{10}^{-3}=k\left[0.1{]}^x\right[0.1{]}^y$

= $k\left[0.1{]}^x\right(\because y=0)$ …(iii)

$2.4x{10}^{-3}=k\left[0.2{]}^x\right[0.1{]}^0$ …(iv)

Divide Eq. (iv) by Eq. (iii)

$(2{)}^1=(2{)}^x\therefore x=1.$

R = k[A]

Hence answer (a) is correct.