Question

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

Initial Concentration $\left[A\right]$	Initial Concentration $\left[B\right]$	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:

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

Answer 1

The correct option is B $\frac{d\left[C\right]}{dt}=k\left[A\right]$

Let order with respect to $A$ and $B$ are $\alpha$ and $\beta$ respectively.
$\therefore \frac{dc}{dt}=k{\left[A\right]}^{\alpha }{\left[B\right]}^{\beta }$
$1.2\times {10}^{-3}=k{\left[0.1\right]}^{\alpha }{\left[0.1\right]}^{\beta }$ ...(i)
$1.2\times {10}^{-3}=k{\left[0.1\right]}^{\alpha }{\left[0.2\right]}^{\beta }$ ...(ii)
$2.4\times {10}^{-3}=k{\left[0.2\right]}^{\alpha }{\left[0.1\right]}^{\beta }$ ...(iii)
Dividing (ii) by (i): $2^{\beta }=1$ $\therefore \beta =0$
Dividing (iii) by (i): $2^{\alpha }=2$ $\therefore \alpha =1$
Rate law will be:
$\frac{dc}{dt}=k\left[A\right]$