An organic compound undergoes first order decomposition. If the...

For the First order reaction, integrated rate equation can be written as follows,

$t = \dfrac{1}{k}ln\dfrac{{a}_{0}}{{a}_{t}}t = timek = rateconstant{a}_{0} = intialconcentration{a}_{t} = concentrationattimet$

$\dfrac{{t}_{1}}{{t}_{2}} = \dfrac{\dfrac{1}{k}ln\dfrac{{a}_{0}}{0 . 4{a}_{0}}}{\dfrac{1}{k}ln\dfrac{{a}_{0}}{0 . 1{a}_{0}}}$

$\dfrac{540}{{t}_{2}} = \dfrac{ln\dfrac{10}{4}}{\ln 10}$

$\dfrac{540}{{t}_{2}} = \dfrac{\log 10 - \log 4}{\log 10}$

$\dfrac{540}{{t}_{2}} = \dfrac{1 - 0 . 6}{1}$

$⇒ \dfrac{540}{{t}_{2}} = 0 . 4$

$⇒ {t}_{2} = \dfrac{540}{0 . 4} = 1350sec$