Advertisements
Advertisements
प्रश्न
The half life period of a radioactive element is 140 days. After 560 days, 1 g of element will be reduced to
पर्याय
`(1/2) "g"`
`(1/4) "g"`
`(1/8) "g"`
`(1/16) "g"`
Advertisements
उत्तर
`(1/16) "g"`
APPEARS IN
संबंधित प्रश्न
A first order reaction takes 40 minutes for 30% decomposition. Calculate t1/2 for this reaction. (Given log 1.428 = 0.1548)
Calculate the half-life of a first order reaction from the rate constant given below:
200 s−1
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)
Show that the time required for 99% completion is double of the time required for the completion of 90% reaction.
The rate constant for a first order reaction is 1.54 × 10−3 s−1. Calculate its half life time.
For the given first order reaction A → B the half life of the reaction is 0.3010 min. The ratio of the initial concentration of reactant to the concentration of reactant at time 2.0 min will be equal to ______. (Nearest integer)
Assertion (A): The half-life of a reaction is the time in which the concentration of the reactant is reduced to one-half of its initial concentration.
Reason (R): In first-order kinetics, when the concentration of reactant is doubled, its half-life is doubled.
Calculate the half-life of a first order reaction from the rate constant given below:
2 min−1
A first order reaction takes 40 min for 30% decomposition. Calculate `"t"_(1/2)`.
