JEE MAIN - Chemistry (2023 - 30th January Evening Shift - No. 18)
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: $\ln 10=2.3 ; \log 2=0.3$
Given: $\ln 10=2.3 ; \log 2=0.3$
Answer
1350
Explanation
For the first order reaction,
$$ k=\frac{1}{t} \ln \frac{a}{a-x} $$
$$ \Rightarrow $$ $$ t=\frac{1}{k} \ln \frac{a}{a-x} $$
When reaction is $60 \%$ completed,
$$ x=\frac{60}{100} a=0.6 a, t=540 \text { seconds } ; $$
$k= \frac{1}{t_1} \ln \frac{a}{a-0.6 a}$
$$ \therefore $$ $$t_1=\frac{1}{k} \ln \frac{a}{0.4 a} $$
When reaction is $90 \%$ completed, i.e., $x=0.9 a$
$$ \begin{aligned} k= \frac{1}{t_2} \ln \frac{a}{a-0.9 a} \\\\ \Rightarrow t_2=\frac{1}{k} \ln \frac{a}{0.1 a} \\\\ \therefore \frac{t_1}{t_2} =\frac{\frac{1}{k} \ln \frac{a}{0.4 a}}{\frac{1}{k} \ln \frac{a}{0.1 a}} \\\\ \Rightarrow \frac{540}{t_2} =\frac{\ln \frac{10}{4}}{\ln 10} \end{aligned} $$
$$ \Rightarrow $$ $\frac{540}{t_2} =\frac{2.3\log 10-2.3\log 4}{2.3\log 10}$
$$ \Rightarrow $$ $\frac{540}{t_2} =\frac{2.3-2.3(0.6)}{2.3}$
$$ \Rightarrow $$ $\frac{540}{t_2} =0.4$
$$ \Rightarrow $$ $$ t_2=\frac{540}{0.4}=1350 $$ sec
$$ k=\frac{1}{t} \ln \frac{a}{a-x} $$
$$ \Rightarrow $$ $$ t=\frac{1}{k} \ln \frac{a}{a-x} $$
When reaction is $60 \%$ completed,
$$ x=\frac{60}{100} a=0.6 a, t=540 \text { seconds } ; $$
$k= \frac{1}{t_1} \ln \frac{a}{a-0.6 a}$
$$ \therefore $$ $$t_1=\frac{1}{k} \ln \frac{a}{0.4 a} $$
When reaction is $90 \%$ completed, i.e., $x=0.9 a$
$$ \begin{aligned} k= \frac{1}{t_2} \ln \frac{a}{a-0.9 a} \\\\ \Rightarrow t_2=\frac{1}{k} \ln \frac{a}{0.1 a} \\\\ \therefore \frac{t_1}{t_2} =\frac{\frac{1}{k} \ln \frac{a}{0.4 a}}{\frac{1}{k} \ln \frac{a}{0.1 a}} \\\\ \Rightarrow \frac{540}{t_2} =\frac{\ln \frac{10}{4}}{\ln 10} \end{aligned} $$
$$ \Rightarrow $$ $\frac{540}{t_2} =\frac{2.3\log 10-2.3\log 4}{2.3\log 10}$
$$ \Rightarrow $$ $\frac{540}{t_2} =\frac{2.3-2.3(0.6)}{2.3}$
$$ \Rightarrow $$ $\frac{540}{t_2} =0.4$
$$ \Rightarrow $$ $$ t_2=\frac{540}{0.4}=1350 $$ sec
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