What is the formula for calculating the half-life of a third-order reaction?

Question

I have searched and found nothing! anybody who knows, please help!

Answer 1

Dear Melomane,
Only 1st order equations have the exponential decay which leads to a natural time constant.
A 3rd order reaction has time dependence given by:
d[A]/dt = -k[A]^3

This integrates up to give: T = 1/( 2k[A]^2)¦ evaluated at the two times in question

You can see that the time to go to half concentration depends on the concentration, unlike the first order case where the time to half value is independent of the concentration.

If the 3rd order reaction has time dependence given by:
d[A]/dt = -k[A] * [B]^2

or d[A]/dt = -k[A][B][C]

And [B] and [C] are only marginally altered then you have a pseudo-1st order reaction which then has a definable half life.

The reason that the 1st order reaction gives a half life is that when the equation Integrates up it produces a Logarithmic function which when subtracted produces a quotient and the starting amounts just cancel by division!

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