My textbook (not an international standard) states that for the first order reactions only

The quantity of reactant remaining after $n$ half lives is

$$A_n = \frac{A_0}{2^n},$$

where $A_0$ is the initial concentration and $A_n$ is the concentration after $n$ half lives.

Shouldn't this statement be true for all orders of reaction?

Upon calculation (for 0 order reaction), I observed that half life time reduced with progress of reaction.

Doing same calculation for First order reaction, I observed that half life time remained constant with progress of reaction.

So, is it this(inconsistency of half life time period) due to which Half life concept is not extended to other order of reactions?

I only wanted to confirm regarding concentration after n half lives, as in a single half life of any order reaction, concentration of reactant reduces to half of its initial value.

  • 6
    $\begingroup$ Does this answer your question? Half-life equation for 2nd order kinetics $\endgroup$ Nov 4, 2020 at 15:51
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    $\begingroup$ Reactions of all other orders do not have half-lives. $\endgroup$ Nov 4, 2020 at 16:30
  • $\begingroup$ @IvanNeretin can you please give one example? I am unaware of it. $\endgroup$ Nov 5, 2020 at 6:13
  • 1
    $\begingroup$ @SujayGhosh Well, if for the 2nd order reaction $\ce{A + B -> C}$ is $[A]=2 [B]$, how is defined the half concentration ? $[A]/2$ or $[B]/2$ or $([A] + [B])/2$ or $[A][B]/2$ ? Half-time is very unuseful concept for these rate orders, even if I agree we can formally define it. What if $[A]=10000 [B]$ ? $\endgroup$
    – Poutnik
    Nov 5, 2020 at 7:48
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    $\begingroup$ @SujayGhosh The half life of a reaction has meaning only in special cases: For a first-order reaction, the half life of the reactant may be called the half life of the reaction. For a reaction involving more than one reactant, with the concentrations of the reactants in their stoichiometric ratios, the half life of each reactant is the same, and may be called the half life of the reaction. As I have said previously, it may be formally defined, but it is not practically used nor called so. $\endgroup$
    – Poutnik
    Nov 5, 2020 at 9:33