A → B Reactant Product If formation of compound B follows the first order of kinetics and after 70…

$\begin{array}{ccc}\left[\mathrm{A}\right]& \to & \left[\mathrm{B}\right]\\ \mathrm{Reactant}& & \mathrm{Product}\end{array}$

Since the reaction follows the first order kinetics and after $70$ minutes the concentration of $\left[\mathrm{A}\right]$ was found to be half of its initial concentration. That means 70 minutes is the half life time for the reaction. And we know that for a first order reaction half life time is given as.

$$\begin{gathered} {\mathrm{t}}_{\frac{1}{2}} = \frac{0.693}{\mathrm{K}} \\ \mathrm{K}=\frac{0.693}{{\mathrm{t}}_{{\displaystyle \frac{1}{2}}}}=\frac{0.693}{70\times 60 \sec .} \end{gathered}$$

$$\begin{gathered} =\frac{6930}{7\times 6}\times {10}^{-6} {\sec }^{-1} \\ = 165 \times {10}^{-6} {\sec }^{-1} \\ \mathrm{So}, \mathrm{x} = 165 \end{gathered}$$