Q. In a first order reaction, the concentration of the reactant is reduced to $12.5\%$ in one hour. When was it half completed ?
Solution:
Key concept If initial amount of reactant is taken $100 \%$, then after three half-lifes it will remain $12.5 \%$. Hence, on dividing the given time i.e. $1 hr$ $(60 min )$ by 3, we get half-life of the reactant.
Time
Concentration
$t=0\,\,\, t=t_{1/2}$
$100]\,50]$ 1 half-life
$t=2t_{1/2}$
25] 1 half-life
$t=3t_{1/2}$
12.5] 1half-life
Total 3 half-lifes]
Total time taken for ]
3 half -lifes ] $\to \, 1hr=60\,min$
Time taken for 1 half-life $=\frac{60}{3}=20 \,min$
| Time | Concentration |
|---|---|
| $t=0\,\,\, t=t_{1/2}$ | $100]\,50]$ 1 half-life |
| $t=2t_{1/2}$ | 25] 1 half-life |
| $t=3t_{1/2}$ | 12.5] 1half-life |