In a first order reaction if concentartion of reactant drops from $0.8 \mathrm{~mol} \mathrm{~L}^{-1}$ to $0…
In a first order reaction if concentartion of reactant drops from $0.8 \mathrm{~mol} \mathrm{~L}^{-1}$ to $0.4 \mathrm{~mol} \mathrm{~L}^{-1}$ in 15 minute. What is the time required to drop concentration from $0.1 \mathrm{~mol} \mathrm{~L}^{-1}$ to 0.025 mol $\mathrm{L}^{-1}$.
7.5 minute
15 minute
30 minute
60 minute
Solution
As the concentration of reactant is reduced to half i.e., from $0.8 \mathrm{~mol} \mathrm{~L}^{-1}$ to $0.4 \mathrm{~mol} \mathrm{~L}^{-1}$ in 15 minutes,
$\therefore \quad \mathrm{t}_{1 / 2}=15$ minutes
For a first order reaction,
$\therefore \quad$ Time required to drop the concentration from
$\begin{aligned}
0.1 \mathrm{~mol} \mathrm{~L}^{-1} \text { to } 0.025 \mathrm{~mol} \mathrm{~L}^{-1} & =2 \times \mathrm{t}_{1 / 2} \\
& =2 \times 15 \text { minutes } \\
& =30 \text { minutes }
\end{aligned}$
