Question

For the non - stoichiometry reaction: $2A+B\to C+D$, the following kinetic data were obtained in three separate experiments, all at $298K$

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

The rate law for the formation of C is :

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

Answer 1

Answer: (D)

$2A+B⟶C+D$

Rate of Reaction $=\frac{-1}{2}\frac{d[A]}{dt}=-\frac{d[B]}{dt}$

$=\frac{d[C]}{dt}=\frac{d[D]}{dt}$

Let rate of Reaction $=k[A{]}^x[B{]}^y$

Or, $\frac{d[C]}{dt}=k[A{]}^x[B{]}^y$

Now from table,

$1.2\times 10^{-3}=k[0.1{]}^x[0.1{]}^y\dots (i)$

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

$2.4\times 10^{-3}=k[0.2{]}^x[0.1{]}^y\dots (iii)$

Dividing equation (i) by (ii)

$\Rightarrow \frac{1.2\times 10^{-3}}{1.2\times 10^{-3}}=\frac{k[0.1{]}^x[0.1{]}^y}{k[0.1{]}^x[0.2{]}^y}$

$\Rightarrow 1={\left[\frac{1}{2}\right]}^y$

$\Rightarrow y=0$

Now Dividing equation (i) by (iii)

$\Rightarrow \frac{1.2\times 10^{-3}}{2.4\times 10^{-3}}=\frac{k[0.1{]}^x[0.1{]}^y}{k[0.2{]}^x[0.1{]}^y}$

$\Rightarrow {\left[\frac{1}{2}\right]}^1={\left[\frac{1}{2}\right]}^x$

$\Rightarrow x=1$

Hence $\frac{d[C]}{dt}=k[A{]}^1[B{]}^0$