Question

The rate of a certain reaction depends on concentration according to equation:
$\frac{-d\left[A\right]}{dt}=\frac{k_1\left[A\right]}{1+k_2\left[A\right]}$
What will be the order of reaction when the concentration is very low?

Answer 1

Given: $\frac{-d\left[A\right]}{dt}=\frac{k_1\left[A\right]}{1+k_2\left[A\right]}$
When [A] is very low
$⟹[1+k_2[A\left]\right]=k^{{}^{\prime }}$
$\therefore \frac{-d\left[A\right]}{dt}=\frac{k_1\left[A\right]}{k^{{}^{\prime }}}=k^{\prime \prime }\left[A\right]$
Thus, the order of reaction is one.