Advertisements
Advertisements
प्रश्न
For a first order reaction, show that time required for 99% completion is twice the time required for the completion of 90% of reaction.
Show that for a first order reaction, time required for 99% completion is twice the time required for 90% completion of the reaction.
Advertisements
उत्तर १
99% completion means that,
x = 99% of [R]0
or, [R] = [R]0 − 0.99 [R]0 = 0.01[R]0
For a first-order reaction,
t = `2.303/k log [R]_0/[[R]]`
∴ t99% = `2.303/k log [R]_0/(0.01[R]_0)`
= `2.303/k log 10^2`
= `2 xx 2.303/k`
90% completion means that,
[R] = [R]0 − 0.99[R]0
= 0.1[R]0
∴ t90% = `2.303/k log [R]_0/(0.1[R]_0)`
= `2.303/k log 10`
= `2.303/k`
∴ `t_(99%)/t_(90%) = ((2 xx 2.303/k)/((2.303/k)))`
= 2
or, t99% = 2 × t90%
उत्तर २
If the initial concentration is [A]0, for 99% completion,
[A] = `[A]_0 - [A]_0 xx 99/100`
= `[A]_0 xx 1/100`
The time required for 99% completion is given by:
t99% = `2.303/k log_10 ([A]_0)/([A]_0 xx 1/100)`
= `2.303/k log_10 100/1` ...(i)
For 90% completion,
[A] = `[A]_0 - [A]_0 xx 90/100`
= `[A]_0 xx 10/100`
Time required for 90% completion is given by:
t90% = `2.303/k log_10 ([A]_0)/([A]_0 xx 10/100)`
= `2.303/k log_10 100/10` ...(ii)
Dividing eq. (i) by eq. (ii), we get,
`t_(99%)/t_(90%) = (log_10 100/1)/(log_10 100/10)`
`t_(99%)/t_(90%)` = 2
∴ t99% = 2 × t90%
उत्तर ३
For a first-order reaction,
t = `2.303/k log_10 [A]_0/([A])`
For 99% completion,
[A]0 = 100
[A] = 100 − 99 = 1
∴ t99 = `2.303/k log_10 100/1`
∴ t99 = = `(2 xx 2.303)/k` ...(i)
For 90% completion,
[A]0 = 100,
[A] = 100 − 90 = 10
∴ t90 = `2.303/k log_10 100/10`
∴ t90 = = `(1 xx 2.303)/k` ...(ii)
From equations (i) and (ii),
t99 = 2 × t90.
Thus, the time required for 99% completion for a first-order reaction is twice the time required for 90% completion.
APPEARS IN
संबंधित प्रश्न
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)
A first order reaction takes 30 minutes for 50% completion. Calculate the time required for 90% completion of this reaction.
(log 2 = 0.3010)
Calculate the half-life of a first order reaction from the rate constant given below:
200 s−1
The half-life for radioactive decay of 14C is 5730 years. An archaeological artifact containing wood had only 80% of the 14C found in a living tree. Estimate the age of the sample.
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).
The unit of rate constant for zero order reaction is _______.
(A) t–1
(B) mol dm–3 t–1
(C) mol–1 dm3 t–1
(D) mol–2 dm6 t–1
The half-life period of zero order reaction A → product is given by
(a) `([A]_0)/k`
(b) `0.693/k`
(c) `[A]_0/(2k)`
(d) `(2[A]_0)/k`
Show that the time required for 99% completion is double of the time required for the completion of 90% reaction.
In a first-order reaction A → product, 80% of the given sample of compound decomposes in 40 min. What is the half-life period of the reaction?
The half life period of a radioactive element is 140 days. After 560 days, 1 g of element will be reduced to
Show that for a first order reaction half life is independent of initial concentration.
Read the passage given below and answer the following question.
|
Are there nuclear reactions going on in our bodies? There are nuclear reactions constantly occurring in our bodies, but there are very few of them compared to the chemical reactions, and they do not affect our bodies much. All of the physical processes that take place to keep a human body running are chemical processes. Nuclear reactions can lead to chemical damage, which the body may notice and try to fix. The nuclear reaction occurring in our bodies is radioactive decay. This is the change of a less stable nucleus to a more stable nucleus. Every atom has either a stable nucleus or an unstable nucleus, depending on how big it is and on the ratio of protons to neutrons. The ratio of neutrons to protons in a stable nucleus is thus around 1 : 1 for small nuclei (Z < 20). Nuclei with too many neutrons, too few neutrons, or that are simply too big are unstable. They eventually transform to a stable form through radioactive decay. Wherever there are atoms with unstable nuclei (radioactive atoms), there are nuclear reactions occurring naturally. The interesting thing is that there are small amounts of radioactive atoms everywhere: in your chair, in the ground, in the food you eat, and yes, in your body. The most common natural radioactive isotopes in humans are carbon-14 and potassium-40. Chemically, these isotopes behave exactly like stable carbon and potassium. For this reason, the body uses carbon-14 and potassium-40 just like it does normal carbon and potassium; building them into the different parts of the cells, without knowing that they are radioactive. In time, carbon-14 atoms decay to stable nitrogen atoms and potassium-40 atoms decay to stable calcium atoms. Chemicals in the body that relied on having a carbon-14 atom or potassium-40 atom in a certain spot will suddenly have a nitrogen or calcium atom. Such a change damages the chemical. Normally, such changes are so rare, that the body can repair the damage or filter away the damaged chemicals. The natural occurrence of carbon-14 decay in the body is the core principle behind carbon dating. As long as a person is alive and still eating, every carbon-14 atom that decays into a nitrogen atom is replaced on average with a new carbon-14 atom. But once a person dies, he stops replacing the decaying carbon-14 atoms. Slowly the carbon-14 atoms decay to nitrogen without being replaced, so that there is less and less carbon-14 in a dead body. The rate at which carbon-14 decays is constant and follows first order kinetics. It has a half-life of nearly 6000 years, so by measuring the relative amount of carbon-14 in a bone, archeologists can calculate when the person died. All living organisms consume carbon, so carbon dating can be used to date any living organism, and any object made from a living organism. Bones, wood, leather, and even paper can be accurately dated, as long as they first existed within the last 60,000 years. This is all because of the fact that nuclear reactions naturally occur in living organisms. |
Approximately how old is a fossil with 12 g of Carbon-14 if it initially possessed 32 g of Carbon-14? (Given log 2.667 = 0.4260)
Which radioactive isotope would have the longer half-life 15O or 19O? (Given rate constants for 15O and 19O are 5.63 × 10–3 s–1 and k = 2.38 × 10–2 s–1 respectively.)
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 rate of a first order reaction is 0.04 mol litre-1 s-1 at 10 minutes and 0.03 mol litre-1 sec-1 at 20 minutes after initiation. The half-life of the reaction is ______ min.
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)`.
Calculate the half-life of a first order reaction from the rate constant given below:
2 min−1
