Question

Half-life period of a first order reaction is $10\, min$. Starting with initial concentration $12\, M$, the rate after $20\, min$ is

• A. $0.0693 \,M \,min^{-1}$
• B. $0.693 \times 3\,M \, min ^{-1}$
• C. $0.0693 \times \,3\,M \, min^{-1}$
• D. $0.0693 \times \,4\,M \,min^{-1}$

Answer 1

Answer: (C)

$0.0693 \times \,3\,M \, min^{-1}$

Answer 2

Key concept As we know, rate constant remains same throughout the reaction. For first order reaction, half-life of the reactant does not depend upon the its concentration. Hence, we can easily calculate the value of rate constant with the help of provided half-life value by using the formula:

$t_{1 / 2}=\frac{0.693}{k}$. Now, we can calculate the

concentration of reactant after $20 min$ and then rate of reaction.

Given, $t_{1 / 2}=10\, min$
$Sin\, t_{1 / 2} =\frac{0.693}{k}$ (k = rate constant) ;

$k =\frac{0.693}{10}=0.0693$

Stage
Time
Concentration

Intial
t=0
12 M

$t_{1/2}=10\,min$
6 M

$t=20\,min$
3 M

Since, it is a first order reaction.

Rate $= k[A] =0.0693 \times 3M \,min^{-1}$