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Using the rate expression for the first order kinetics and expressing the half-life time, it can be proven the half-life time $t_\frac 1 2$ of the first order reaction is independent of its initial concentration:

$$t_\frac 1 2 = \frac{\ln 2}{k}$$

Can we intuitively explain that the first order half-life time is independent of its starting concentration and remains constant without the aid of the derived expression for the half-life time said above?

For the zeroth order reaction, it can intuitively be explained, but for the first order reaction it seems a little hard.

andselisk
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Sudhagar
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  • It is ok following this argument, I can say when the particle count goes on decreasing half life too can go on decreasing. but how comes that the half life is "exactly" equal to initial concentration. – Sudhagar Jul 11 '22 at 06:26
  • That is not about kinetics of the 1st order and looks rather confused. Half life cannot be equal to concentration even for the simple reason of different dimension. – Poutnik Jul 11 '22 at 06:28
  • I m sorry for that typo and wanted to say successive half lives are exactly equal and independent of initial concentrations. – Sudhagar Jul 11 '22 at 08:22
  • Sure. If 10e23 units decay to 1/2 and another 10e23 units decay to 1/2, than in summary 20e23 units decay to 1/2. They do not decay faster nor slower just because there is more or less of them. – Poutnik Jul 11 '22 at 08:26
  • Thanks.. got your point, thinking over your answers. Treating the first order kinetics in terms of "reacting units" as you have mentioned earlier and the fraction of units that undergoes reaction (proportionality constant) is intuitive for me to rationalize that the half life period is constant. It makes me think that molecules of first order kinetics comes in half life packets. – Sudhagar Jul 11 '22 at 11:46
  • My professor said that for first order kinetics,between 2 successive half life times amount of reactant to be consumed is halved. Similarly, the rate from one period to the next one is also halved due to first order kinetics . Therefore, half life time doesn't change. Though I understood his point, I wasn't convinced & looked for better explanation – Sudhagar Jul 11 '22 at 11:57
  • You may want to mark Porphyrin's answer as accepted. It is nicer and he has more cool ID. – Poutnik Jul 11 '22 at 12:04

2 Answers2

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The basic assumption is that the probability of reaction of a given molecule (or disintegration of a radioactive atom) in a given time interval is independent of past history, and depends only on the time interval. If it is assumed that the time interval we choose is sufficiently small the chance of reaction is proportional to this time interval and the exponential decay law can be derived from this. The constant of proportionality is called the rate constant $k$. These assumptions mean that we cannot predict exactly when a given molecule will react but because of the proportionality (the rate constant) we can after many measurements know the form of the decay. Of course we assume that all the molecules of a given type are identical and behave in the same way and independently of one another.

We can measure the average time our type of molecule takes to decay by making making many repeated measurements and this average is the reciprocal of the rate constant and is related to the half life as $\ln(2)/k$.

porphyrin
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  • This one essentially means that the infinestimal time interval dt is proportional to reaction probability dN/ N: going by this, same time intervals mean same reaction probability. By extended argument, for same half life times we have same extent of reaction that is to say half life times are constant for first order kinetics. – Sudhagar Jul 11 '22 at 16:36
  • I didn't get the point " the average time" – Sudhagar Jul 11 '22 at 16:47
  • @Sudhagar It is probabilistic behaviour, that has clean, exponential outcome for very large set of molecules. – Poutnik Jul 12 '22 at 03:18
  • I only used average time as an example as this is mathematically straightforward to calculate and can be measured experimentally of course. – porphyrin Jul 12 '22 at 08:36
  • A fuller answer is here https://chemistry.stackexchange.com/questions/159111/fluorescence-what-mechanism-regulates-the-rate-of-discharge-of-energy/159171#159171 – porphyrin Jul 12 '22 at 08:39
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A particle has some probability to decay (Radioactive decays are ideal illustrative examples.) or react during some time interval. With more particles, the count of particles is directly proportional to particle count that will do so during that interval.

Therefore, the relative portion of particles reacting during that period does not depend on the amount nor concentration (assuming very high particle numbers converging to matter continuity). When the time interval leads to a portion being the half, it is the half life time.

Poutnik
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