Question

$t_{1 / 2}$ for a first order reaction is $10\, \min$. Starting with $10\, M$, the rate after $20\, \min$ is

• A. $ 0.0693\, M\ \min^{-1}$
• B. $ 0.0693\times 5\,M\, \min^{-1}$
• C. $ 0.0693\times 2.5\,M\,\min^{-1}$
• D. $ 0.0693\times 10\,M\, \min^{-1}$

Answer 1

Answer: (C)

Given, $t_{1 / 2}=10\, \min$
$t=20\, \min$
$N_{0}=10\, M$
$\Rightarrow t=n t_{1 / 2}$
$20=n \times 10$
$\therefore n=2$
$\Rightarrow \frac{N}{N_{0}}=\left(\frac{1}{2}\right)^{n}$
$\Rightarrow \frac{N}{10}=\left(\frac{1}{2}\right)^{2}$
or $\frac{N}{10} =\frac{1}{4}$
$\therefore N =\frac{10}{4}=2.5\, M$
$\therefore $ Rate $=k[A]$
$=0.0693 \times 2.5\, M\, \min ^{-1}$