Question

The rate constant for a first order reaction is $60s^{-1}$. 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, $k=60s^{-1},[A]=[A{]}_0/10$
Now, $t=\frac{2.303}{60}\log \frac{[A{]}_0}{[A{]}_0/10}$
$=\frac{2.303}{60}\log 10$
$=0.038s$
or $3.8\times 10^{-2}s$