Q. Half-life period of a first order reaction is $10\, min$. Starting with initial concentration $12\, M$, the rate after $20\, min$ is
Solution:
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 $= k[A] =0.0693 \times 3M \,min^{-1}$
| Stage | Time | Concentration |
|---|---|---|
| Intial | t=0 | 12 M |
| $t_{1/2}=10\,min$ | 6 M | |
| $t=20\,min$ | 3 M |