Question

The rate constant for a first order reaction is $60 \,s^{-1}$. How much time it will take to reduce the concentration of the reactant to $1 / 10^\text{ th}$ of its initial value?

• A. $ 3.8\times 10^{-2}\,s $
• B. $ 1.26\times 10^{13}\,s $
• C. $ 2.01\times 10^{13}\,s $
• D. $ 1.097\times 10^{3} $

Answer 1

Answer: (A)

For a first order reaction,
$t=\frac{2.303}{k} \log \frac{[A]_{0}}{[A]}$
Here, $A_{0}=$ initial concentration
$A=$ final concentration
Given, $k=60 s^{-1},[A]=[A]_{0} / 10$
Now, $t=\frac{2.303}{60} \log \frac{[A]_{0}}{[A]_{0} / 10}$
$=\frac{2.303}{60} \log 10$
$=0.038 s$
or $3.8 \times 10^{-2} s$