An organic compound undergoes first-order decomposition. If the time taken for the $60 \%$ decomposition is $540 \mathrm{~s}$, then the time required for $90 \%$ decomposition will be ________ s. (Nearest integer).

Given: $\ln 10=2.3 ; \log 2=0.3$

For the first order reaction, <br/><br/>$k=\frac{1}{t} \ln \frac{a}{a-x}$ <br/><br/>$\Rightarrow$ $t=\frac{1}{k} \ln \frac{a}{a-x}$ <br/><br/>When reaction is $60 \%$ completed, <br/><br/>$x=\frac{60}{100} a=0.6 a, t=540 \text { seconds } ;$ <br/><br/>$k= \frac{1}{t_1} \ln \frac{a}{a-0.6 a}$ <br/><br/>$\therefore$ $t_1=\frac{1}{k} \ln \frac{a}{0.4 a}$ <br/><br/>When reaction is $90 \%$ completed, i.e., $x=0.9 a$ <br/><br/>$$ \begin{aligned} k= \frac{1}{t_2} \ln \frac{a}{a-0.9 a} \\\\ \Rightarrow t_2=\frac{1}{k} \ln \frac{a}{0.1 a} \\\\ \therefore \frac{t_1}{t_2} =\frac{\frac{1}{k} \ln \frac{a}{0.4 a}}{\frac{1}{k} \ln \frac{a}{0.1 a}} \\\\ \Rightarrow \frac{540}{t_2} =\frac{\ln \frac{10}{4}}{\ln 10} \end{aligned} $$ <br/><br/>$\Rightarrow$ $\frac{540}{t_2} =\frac{2.3\log 10-2.3\log 4}{2.3\log 10}$ <br/><br/>$\Rightarrow$ $\frac{540}{t_2} =\frac{2.3-2.3(0.6)}{2.3}$ <br/><br/>$\Rightarrow$ $\frac{540}{t_2} =0.4$ <br/><br/>$\Rightarrow$ $t_2=\frac{540}{0.4}=1350$ sec