Question

Pseudo first-order rate for the reaction, $A+B⟶P$, when studied in $0.1$ $M$ of $B$ is given by $-\frac{d\left[A\right]}{dt}=k\left[A\right]$ where, $k=1.85\times {10}^4{sec}^{-1}$. Calculate the value of second order rate constant.

Answer 1

$A+B⟶P$
$-\frac{d\left[A\right]}{dt}=k\left[A\right]$
$-\frac{d\left[A\right]}{dt}=1.85\times {10}^4\times \left[A\right]$ ...(i)
Assuming the reaction to be of second order,
$-\frac{d\left[A\right]}{dt}=k^{{}^{\prime }}\left[A\right]\left[B\right]$
$-\frac{d\left[A\right]}{dt}=k^{{}^{\prime }}\left[A\right]\left[0.1\right]$ ...(ii)
Dividing equation (i) by equation (ii), we get
$1=\frac{1.85\times {10}^4}{k^{{}^{\prime }}\left[0.1\right]}$
$k^{{}^{\prime }}=1.85\times {10}^5$ $litre$ ${mol}^{-1}$ ${sec}^{-1}$