Half-life of a reversible first-order reaction of the form A ⇄ B

I am wondering if the half-life of a reversible first-order reaction of the form $\ce{A <--> B}$ is defined as


1. the time in which half of the amount of reactant which should react up to equilibrium, has reacted, or,
2. the time in which half of the total amount of reactant has reacted.


I only found [this](https://www.toppr.com/ask/question/halflife-of-a-reversible-reaction-may-be-defined-as-the-time-in-which-half-of) source.




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**Example:** Species $\ce{A}$ reacts in first-order kinetics. The initial concentration is $1 \,\pu{mol l^{-1}}$. Its equilibrium concentration is $0.3 \,\pu{mol l^{-1}}$. After $40\,\pu{s}$, its concentration is $0.65 \,\pu{mol l^{-1}}$. After $60 \,\pu{s}$. its concentration is $0.5 \,\pu{mol l^{-1}}$. Is the half-life $40 \, \pu{s}$ (because $0.35 \,\pu{mol l^{-1}}$ of total $0.7 \,\pu{mol l^-1}$ that react, have reacted), or $60 \,\pu{s}$?




The [Wikipedia article](https://en.wikipedia.org/wiki/Half-life#In_non-exponential_decay) on half-life states:



> 
> The term "half-life" is almost exclusively used for decay processes that are exponential (such as radioactive decay or the other examples above), or approximately exponential (such as biological half-life discussed below). In a decay process that is not even close to exponential, the half-life will change dramatically while the decay is happening. In this situation it is generally uncommon to talk about half-life in the first place, but sometimes people will describe the decay in terms of its "first half-life", "second half-life", etc., where the first half-life is defined as the time required for decay from the initial value to 50%, the second half-life is from 50% to 25%, and so on.
> 
> 
> 


Concentrations of a reaction approaching equilibrium do not decay exponentially (for one, they don't approach zero as time passes).


The kinetics of a reaction approaching equilibrium can be complicated. If you start without product, kinetics are governed by the rate law of the forward reaction in the beginning. Then, as product is being made, the rate law of the reverse reaction plays into the kinetics as well. Surprisingly, as you get very close to equilibrium, the net forward rate decays exponentially (i.e. apparent first-order reaction) no matter what the rate laws for the forward and reverse reactions are. This is used in [temperature-jump](https://www.nobelprize.org/uploads/2018/06/eigen-lecture.pdf) and other relaxation methods to study fast reactions.



> 
> [OP] I was wondering if the half-life of a reversible reaction is defined as [...]
> 
> 
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For processes that don't follow a pure exponential decay, the half-life is ill-defined. There is no good answer to this multiple-choice question. For example, what do you do when the reaction reaches equilibrium before half of the reactant is used up?


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In the reaction $A\rightleftharpoons B$ the relaxation rate constant, i.e. the rate of approach to equilibrium is exponential and given by $k=k\_1+k\_{-1}$. The forwards and reverse reactions have rate constants $k\_1,k\_{-1}$ respectively. The half-life, if you want to define it is, $\ln(2)/k$.


If the reaction is $A+B\rightleftharpoons C+D$ then the relaxation time is still exponential but now depends on the equilibrium concentrations.


You can solve the kinetics $A\rightleftharpoons B$ using $x=[A\_0]-[A]=[B]$, write down the rate equation $d[A]/dt$, substitute for $x$ and then integrate $dx/dt$ from $0\to t$. Next represent the equilibrium constant as $K\_e=k\_1/k\_{-1}=x\_e/([A\_0]-x\_e)$ where subscript $e$ is the equilibrium amount. You should find that


$$\displaystyle x=x\_e(1-e^{-(k\_1+k\_{-1})t})$$


which is a rising exponential as $x$ is the amount of B at time $t$. When $t=0$ this is zero and at long times $x\to x\_e$ as expected. Species A falls as B increases (from $1$ to $0.3$ in your example) and with the same rate constant $k$.


The initial amount of A is $1$ and equilibrium amount $0.3$ (so $x\_e=0.7$) and the half life should be the time to reach half of the range from $1$ to $0.3$ as this is the range that A covers.


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