chem alevel ocr a question

LisaPerera

how to know if something is second order from half lives
thank you

Reply 1

nxvastel

Original post

by LisaPerera

how to know if something is second order from half lives
thank you

Great question.

I'm not a very fond physical chemist; and I'd like to steer away from the maths, too, so I'll try to explain this visually and intuitively. If you want the full maths and derivation (which requires integrated rate laws + half-life equations) (beacuse it can go that far if you really wanted to), I'd suggest you check out this video: https://www.youtube.com/watch?v=QdoXxV60Ius, which seemingly just goes a teeny bit overboard for A-Level, but does help understanding. :smile:

[ZERO ORDER]

Let's just look at this without the maths. If substance A is zero order (so wouldn't appear in the rate equation), changing its concentration wouldn't affect the rate at all.
This also means in terms of half life, it doesn't matter how much or little A you have in the system - its half life would be independent of its concentration, [A].

Or in other words, particles of A are removed at a constant rate.

Suppose 10 molecules of A are removed each minute.

If you started with 100 molecules of A, it takes 5 minutes to remove 50 molecules of A.
If you started with 200 molecules of A, it takes 10 minutes to remove 100 molecules of A.

The key observation here is: [A] genuinely doesn't matter. The same, constant amount is removed everytime for a said substance. So if you have larger [A] (say 1000, would take 50 minutes to halve), the half life is longer than if you have a smaller [A] (say 500, would take 25 minutes to halve).

TL;DR: If half life depends on starting concentration, then the reactant has zero order.

[FIRST ORDER]

Now this is different. [A] now appears in the rate equation, and changing its concentration has an affect on the rate of reaction.

And this time, the half life is what you know it as. The time to halve 100 A -> 50 A is the same as the time to halve 10 A into 5 A.

In a first order reactant, half life (the time taken to halve the number of reactants in the sample) is always the same, no matter where you start.

TL;DR: A constant half life indicates first order.

[SECOND ORDER]

This is what you asked for. You can think of second order as: "two particles must collide to react". The less of A I have, the less likely it is to collide with one another, and the slower the rate.

In fact, this equation describes the relationship between half life and [A] (which is second order):

t_1/2 = 1 / k[A]
(where t_1/2 is half life and k is the rate constant.)

As shown, the half life is inversely proportional to [A]. So the smaller [A] is, the longer the half life.

Intuitively:
Many molecules -> more successful collisions -> get used up faster -> takes shorter to halve their number
Little molecules -> less successful collisions -> get used up slower -> takes longer to halve their number

So in second order, each successive half life is longer as [A] falls.

TL;DR: Each successive half life gets longer in second order.

Hope that answered your question!
Anything else, please feel free to ask.

I'm sure @TypicalNerd can explain complex physical concepts way better than me, so I'm just going to give him a nudge :smile:

Reply 2

TypicalNerd

Original post

by LisaPerera

how to know if something is second order from half lives
thank you

I think nxvastel’s answer is more than sufficient, but I’ll go into the maths for the sheer hell of it.

I do believe on another one of your threads, you asked about the shapes of the various concentration-time graphs, which I derived there.

With zero order systems, the integrated rate law is [A] = a - kt, where [A] is the current concentration and a is the initial concentration. At the first half life, [A] = 1/2 a, so 1/2 a = a - kt, hence t = 1/2 (a/k). We can repeat this calculation to find the second half life and we get the same result. Whilst this does imply a constant half life, it is dependent on a.

You find that because there is an exponential decay relationship between a first order reactant’s concentration and the time elapsed, the half life remains constant (this is just a property of exponential decay functions). We can prove this as follows:

[A] = a exp(-kt) => ln[A] = ln(a) - kt

ln[A] - ln(a) = -kt

ln([A]/a) = -kt (by the log subtraction law)

ln(a/[A]) = kt (since -ln(x) = ln(1/x) for all x)

So 1/k ln(a/[A]) = t and at the half life, [A] = 1/2 a and therefore a/[A] = 2, so the half life must be ln(2)/k. This result IS something to remember for OCR, but you do not need the proof.

For second order reactions, it’s not an exponential decay relationship - the relationship is 1/[A] = kt + 1/a, where [A] is the current concentration of A and a is the initial concentration.

Let’s consider the first half-life, where [A] = 1/2 a, so 1/[A] = 2/a. Thus, we have 2/a = kt + 1/a, hence t = 1/(ka).

Let’s now consider the time taken for the concentration to drop to 1/4 of what it was (e.g. the sum of the first and second half lives). That being the case, [A] = 1/4 a, so 1/[A] = 4/a, hence 4/a = kt + 1/a => t = 3/(ka). This is clearly more than twice the first half-life, so the half life cannot be the same each time (the second half-life is in fact 2/(ka)). We could go on and on, but this is entirely unnecessary. As nxvastel has already pointed out, the half lives get longer and longer.

In summary, this is all just what the maths dictates. [A] vs t is a curve, check the half lives to see if they are constant or not. If not, then it’s a zero order reaction.

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