Question

For the non - stoichiometry reaction: $2A + B \rightarrow C + D$, 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 $(mol\,L^{-1}\,s^{-1})$
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)

$2 A + B \longrightarrow C + D $

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

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

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