# 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 source.

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 $$60 \,\pu{s}$$?

• I've only encountered the term 'half-life' in the context of reaction kinetics when describing elementary reactions - I've not seen anyone try to define half-life for multi-step processes (such as a reversible reaction). However, in many fields, such as medicine, half-life is used commonly to measure the time it takes for some initial condition to be reduced to half, irregardless of how exactly it would happen. So I'd guess if I had to define, I'd go with the 40 s. Aug 4, 2022 at 9:19
• So basically you would go with the assumption, that because 0.7 mol/l react, we should define the half life as the time it takes for 0.35 mol/l to react, because after 0.3 mol/l there is no net change anymore? Aug 4, 2022 at 9:37
• I misunderstood the question in my earlier comment. I'd define half-life as the time it takes for the concentration to drop to half, so from 1M to 0.5 M - which I guess is 60s in your case, right? But then again I'd prefer not to define it at all. Aug 4, 2022 at 11:40
• Related Aug 4, 2022 at 14:28
• Okay yeah 60s makes sense Aug 4, 2022 at 14:55

The Wikipedia article 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 and other relaxation methods to study fast reactions.

[OP] I was wondering if the half-life of a reversible reaction is defined as [...]

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?

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.