Question

In a first order reaction $x\to y$, the concentration of ‘$x$’ changes from 0.1 M to 0.025 M in 40 minutes. The rate of the reaction (mol/L min) when the concentration of $x$ is 0.01 M would be:

• A. $3.47\times {10}^{-5}$
• B.
  $3.47\times {10}^{-4}$
• C. $5.8\times {10}^{-4}$
• D. $5.8\times {10}^{-6}$

Answer 1

The correct option is B

$3.47\times {10}^{-4}$
We know that:
$ln{A}_0-lnA=kt$
$ln\frac{A_0}{A}=kt$
$ln\frac{0.1}{0.025}=k\times 40$
$ln4=k\times 40$
$k=\frac{ln4}{40}\left(1\right)$
Rate of reaction when concentration $0.01$
Rate$=k\left[0.01\right]$
On substituting value of $k$ from (1) , we get
$Rate=3.47\times {10}^{-4}mol/Lmin$