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I have a doubt about the meaning of half-life of a first-order reaction. Earlier I used to think that it is the time in which concentration of reactant reduces to half of its initial but recently I got a question (I don't have that question right now) undergoing a first-order reaction with variable volume and then my teacher told that half-life is the time when moles of reactant reduces to half and as the volume is constant moles are similar as concentration.

I want to confirm this thought here.

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    $\begingroup$ Is volume constant or variable in your reaction?If it is constant, then comparing with number of moles is also correct. $\endgroup$ – Chem-Learner Jul 7 at 17:12
  • $\begingroup$ For constant volume reactions, half the reactant moles being consumed = concentration getting halved. $\endgroup$ – Eashaan Godbole Jul 7 at 17:13
  • $\begingroup$ the volume is variable and I have came across two questions where volume is variable and they used moles $\endgroup$ – user95194 Jul 7 at 17:14
  • $\begingroup$ @EashaanGodbole you are right my friend but I am confused in case of variable volume. $\endgroup$ – user95194 Jul 7 at 17:15
  • $\begingroup$ If the volume is varying with time, you will have to solve the first-order differential equation with $V(t)$ to obtain the rate equation. $\endgroup$ – Aniruddha Deb Jul 7 at 17:19
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For first order reactions (assuming the reaction $\ce{A -> B + C}$): $$-\frac{\mathrm{d}[\ce{A}]}{\mathrm{d}t} = k[\ce{A}]$$ If the volume is $V(t)$, then $[\ce{A}] = \frac{n}{V(t)}$. You will have to substitute $[\ce{A}]$ in the differential equation and solve it in order to obtain the rate equation in this case.

As for finding the half-life, Wikipedia states that it is the time required for a quantity to reduce to half its initial value. If $V(t) = V_0$, then $[\ce{A}] \propto n$ and the half life for the quantity of substance is equal to the half-life of the concentration of substance. If $V(t)$ is not constant, then you'll have different half-life times for the quantity of the substance and for the concentration of the substance.

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  • $\begingroup$ sir, I still have some problems as in the case of radioactive decay which is a first-order reaction the half-life is the time required for the decay of half of the number of nuclei. $\endgroup$ – user95194 Jul 7 at 17:38
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First order reactions are always about the total count or particles, molar amount or mass.

$$\frac {\mathrm{d}n}{\mathrm{d}t} = -k \cdot n$$ $$\frac {\mathrm{d}m}{\mathrm{d}t} = -k \cdot m$$

The point of using concentration instead is, that it is proportional to them in homogenous, constant volume scenario.

$$c = \frac nV$$ $$c = \frac m{MV}$$

so it can be expressed in terms of $c$ as well:

$$\frac {\mathrm{d}c}{\mathrm{d}t} = -k \cdot c$$

This has great advantage as $c=\mathrm{f}(t)$ can be monitored much easier than $n=\mathrm{g}(t)$ or $m=\mathrm{h}(t)$

For homogenous, variable volume scenario, concentration alone is not useful, unless multiplied by volume.

$$\frac {\mathrm{d}n}{\mathrm{d}t} = -k \cdot c \cdot V = -k \cdot n$$ $$\frac {\mathrm{d}m}{\mathrm{d}t} = -k \cdot c \cdot V \cdot M = -k \cdot m$$

For non homogenous concentration, concentration must be integrated over the volume

$$\frac {\mathrm{d}n}{\mathrm{d}t} = -k \int_V{c \cdot dV}$$ $$\frac {\mathrm{d}m}{\mathrm{d}t} = -k \cdot M \cdot \int_V{c \cdot dV}$$

or the known molar amount or mass has to be used.

With such approaches, the half-time is scenario independent:

$$t_{1/2} = \frac{ \ln{2}}k$$

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