### Chemical Kinetics Question 9

#### Question 9 - 30 January - Shift 2

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

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

## Show Answer

#### Answer: (1350)

#### Solution:

$\frac{t_1}{t_2}=\frac{\frac{1}{K} \ln \frac{a_0}{0.4 a_0}}{\frac{1}{K} \ln \frac{a_0}{0.1 a_0}}$

$\frac{540}{t_2}=\frac{\ln \frac{10}{4}}{\ln 10}$

$\frac{540}{t_2}=\frac{\log 10-\log 4}{\log 10}$

$\frac{540}{t_2}=\frac{1-0.6}{1}$

$\Rightarrow \frac{540}{t_2}=0.4$

$\Rightarrow t_2=\frac{540}{0.4}=1350 sec$