Question

The following results have been obtained during the kinetic studies of the reaction:
$2A+B\to C+D$

Experiment	$\left[A\right]/mol{L}^{-1}$	$\left[B\right]/mol{L}^{-1}$	Initial rate of formation of $D/mol{L}^{-1}mi{n}^{-1}$
I	$0.1$	$0.1$	$6.0\times {10}^{-3}$
II	$0.3$	$0.2$	$7.2\times {10}^{-2}$
III	$0.3$	$0.4$	$2.88\times {10}^{-1}$
IV	$0.4$	$0.1$	$2.40\times {10}^{-2}$

Determine the rate law and the rate constant for the reaction.

Answer 1

The rate law expression is, $rate=k\left[A{]}_0^p\right[B{]}_0^q$.

From experiments $I$ and $IV$, we get ,
$rat{e}_1=k\left(0.1{)}^p\right(0.1{)}^q=6.0\times {10}^{-3}$ ......(i)
$rat{e}_4=k\left(0.4{)}^p\right(0.1{)}^q=2.40\times {10}^{-2}$ ......(ii)

Divide equation (ii) by equation (i)
$\frac{(0.4{)}^p}{(0.1{)}^p}=\frac{2.40\times {10}^{-2}}{6.0\times {10}^{-3}}=4$

$4^p=4$
$p=1$

From experiments $II$ and $III$, we get,
$rat{e}_2=k\left(0.3{)}^p\right(0.2{)}^q=7.2\times {10}^{-2}$ ......(iii)
$rat{e}_3=k\left(0.3{)}^p\right(0.4{)}^q=2.88\times {10}^{-1}$ ......(iv)

Divide equation (iv) by equation (iii)
$\frac{(0.4{)}^q}{(0.2{)}^q}=\frac{2.88\times {10}^{-1}}{7.2\times {10}^{-2}}=4$

$2^q=4$

$q=2$

The rate law expression is, $rate=k\left[A\right][B{]}^2$.

Substitute values in equation (i)
$6.0\times {10}^{-3}=k\left(0.1\right)(0.1{)}^2$

$k=6.0L^{2}mo{l}^{-2}mi{n}^{-1}$