Advertisements
Advertisements
प्रश्न
Show that for a first order reaction half life is independent of initial concentration.
उत्तर
For a first order reaction, the half life is a constant i.e., it does not depend on the initial concentration.
The rate constant for a first order reaction is given by,
k = `2.303/"t" log (["A"_0])/(["A"])`
at t = `"t"_(1/2)`; [A] = `(["A"_0])/2`
k = `2.303/"t"_(1/2) log (["A"_0])/((["A"_0])/2)`
k = `2.303/"t"_(1/2) log (2)`
k = `(2.303 xx 0.3010)/"t"_(1/2)`
k = `0.6932/"t"_(1/2)`
`"t"_(1/2) = 0.6932/"k"`
APPEARS IN
संबंधित 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.
A first order reaction takes 30 minutes for 50% completion. Calculate the time required for 90% completion of this reaction.
(log 2 = 0.3010)
The rate constant for a first order reaction is 1.54 × 10−3 s−1. Calculate its half life time.
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. |
Suppose an organism has 20 g of Carbon-14 at its time of death. Approximately how much Carbon-14 remains after 10,320 years? (Given antilog 0.517 = 3.289)
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)
A sample of U238 (half-life = 4.5 × 109 years) ore is found to contain 23.8 g of U238 and 20.6 g of Pb206. The age of the ore is ______ × 109 years.
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]
Obtain a relation, `k_2/k_1 = ((t_(1/2))_2)/((t_(1/2))_1)`, where k1 and k2 are rate constants while (t1/2)1 and (t1/2)2 are half-life periods of the first order reaction at temperatures T1 and T2 respectively. Write the relation for activation energy.
A first order reaction takes 40 min for 30% decomposition. Calculate `"t"_(1/2)`.
