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प्रश्न
After 2 hours, a radioactive substance becomes `(1/16)^"th"` of original amount. Then the half life ( in min) is
विकल्प
60 minutes
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उत्तर
30 minutes
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संबंधित प्रश्न
Sucrose decomposes in acid solution to give glucose and fructose according to the first order rate law. The half life of the reaction is 3 hours. Calculate fraction of sucrose which will remain after 8 hours.
A first order reaction takes 23.1 minutes for 50% completion. Calculate the time required for 75% completion of this reaction.
(log 2 = 0.301, log 3 = 0.4771, log 4 = 0.6021)
The experimental data for decomposition of N2O5
\[\ce{2N2O5 -> 4NO2 + O2}\]
in gas phase at 318 K are given below:
| t/s | 0 | 400 | 800 | 1200 | 1600 | 2000 | 2400 | 2800 | 3200 |
| 102 × [N2O5]/mol L−1 | 1.63 | 1.36 | 1.14 | 0.93 | 0.78 | 0.64 | 0.53 | 0.43 | 0.35 |
- Plot [N2O5] against t.
- Find the half-life period for the reaction.
- Draw a graph between log [N2O5] and t.
- What is the rate law?
- Calculate the rate constant.
- Calculate the half-life period from k and compare it with (ii).
A first order reaction takes 10 minutes for 25% decomposition. Calculate t1/2 for the reaction.
(Given : log 2 = 0.3010, log 3 = 0.4771, log 4 = 0.6021)
Define half life of a reaction.
A first-order reaction takes 69.3 min for 50% completion. What is the time needed for 80% of the reaction to get completed? (Given: log 5 = 0.6990, log 8 = 0.9030, log 2 = 0.3010)
The amount of C-14 isotope in a piece of wood is found to be 1/16th of its amount present in a fresh piece of wood. The age of wood, half-life period of C-14 is 5770 years, is ______ years.
A reaction has a half-life of 1 min. The time required for 99.9% completion of the reaction is ______ min.
(Round off to the nearest integer).
[Use: In 2 = 0.69; In 10 = 2.3]
Show that the half-life of zero order reaction is `t_(1/2) = ([A]_0)/(2k)`.
A first order reaction takes 40 min for 30% decomposition. Calculate t1/2.
