Question

The rate of the reaction starting with initial concentrations $2\times {10}^{3}M$ and $1\times {10}^{3}M$ are equal to $2.40\times {10}^{4}M{s}^1$ and $0.60\times {10}^{4}M{s}^1$ respectively. Calculate the order of reaction with respect to reactant:

Answer 1

$r_0=k[A_0{]}^a$
$\frac{\left(r_0{)}_1\right)}{\left(r_0{)}_2\right)}={\left\{\frac{[A_0{]}_1}{[A_0{]}_2}\right\}}^a$
a = $\frac{log\left[\right(r_0{)}_1\left)/\right(r_0{)}_2\left)\right]}{log\left[A_0{]}_1/\right[A_0{]}_2}$
=
$\frac{log(2.40\times 104/0.60\times 104)}{log(2\times 103/1\times {10}^3)}=\frac{log4}{log2}=2$
Thus, reaction is of second order.
k = $\frac{Rate}{[A{]}^2}$
=
$\frac{2.4\times 104mol{L}^1{s}^1}{(2\times {10}^{3}mol{L}^{1{)}^2}}$ = $0.6\times {10}^{4}mo{l}^{-1}{L}^{-1}{s}^1$