Question

For the given reaction $A+B⟶$ Products, following data were obtained:

	$\left[A_0\right]$	$\left[B_0\right]$	$R_0\left(mol{litre}^{-1}{sec}^{-1}\right)$
$1.$	$0.1$ $M$	$0.2$ $M$	$0.05$
$2.$	$0.2$ $M$	$0.2$ $M$	$0.10$
$3.$	$0.1$ $M$	$0.1$ $M$	$0.05$

(i) Write the rate law expression.
(ii) Find the rate constant.

Answer 1

Let order with respect to $A$ and $B$ are $\alpha$ and $\beta$ respectively. Rate law expression may be given as,
Rate $=k{\left[A\right]}^{\alpha }{\left[B\right]}^{\beta }$
$0.05=k{\left[0.1\right]}^{\alpha }{\left[0.2\right]}^{\beta }$ ...(i)
$0.10=k{\left[0.2\right]}^{\alpha }{\left[0.2\right]}^{\beta }$ ...(ii)
$0.05=k{\left[0.1\right]}^{\alpha }{\left[0.1\right]}^{\beta }$ ...(iii)
Dividing equation (i) by equation (ii), we get
$\frac{1}{2}={\left[\frac{1}{2}\right]}^{\alpha }$, i.e., $\alpha =1$
Dividing equation (i) by equation (iii), we get
$1={\left[2\right]}^{\beta }$, i.e., $\beta =0$
$\therefore$ Rate $=k{\left[A\right]}^1{\left[B\right]}^0$
Substituting the value of $\alpha$ and $\beta$ in equation (i), we get
$0.05=k{\left[0.1\right]}^1{\left[0.2\right]}^0$
$k=0.5{sec}^{-1}$