First-order reaction

A first-order reaction has a reaction rate proportional to one reactant concentration in the simple single-species case. For an irreversible $\mathrm{A}⟶\mathrm{B}$ in a closed system at fixed volume, take $[\mathrm{A}](0)=A_0$, $[\mathrm{B}](0)=0$, and constant rate constant $k>0$. Its rate equation and solution areA short check is enough: differentiating $A_0{e}^{-kt}$ gives $-k{A}_0{e}^{-kt}$, and the two concentrations sum to $A_0$, as required by the one-to-one stoichiometry. Thus the same exponential decay describes both remaining reactant and decreasing reaction rate. Fixed $k$ means a fixed fractional loss per unit time, not a fixed absolute consumption rate.

Choose illustrative values $A_0=1\mathrm{mol}{\mathrm{L}}^{-1}$ and $k=0.2{\mathrm{s}}^{-1}$, with $t$ measured in seconds. Then $[\mathrm{A}]=e^{-0.2t}\mathrm{mol}{\mathrm{L}}^{-1}$, $[\mathrm{B}]=(1-e^{-0.2t})\mathrm{mol}{\mathrm{L}}^{-1}$, and $v=0.2e^{-0.2t}\mathrm{mol}{\mathrm{L}}^{-1}{\mathrm{s}}^{-1}$. The half-life isAt each successive half-life, half the remaining reactant is consumed; $t_{1/2}$ is independent of $A_0$ for this first-order reaction. At $t=10\mathrm{s}$, the calculated values are $[\mathrm{A}]\approx 0.135335\mathrm{mol}{\mathrm{L}}^{-1}$, $[\mathrm{B}]\approx 0.864665\mathrm{mol}{\mathrm{L}}^{-1}$, and $v\approx 0.027067\mathrm{mol}{\mathrm{L}}^{-1}{\mathrm{s}}^{-1}$. These are model predictions for chosen parameters, rather than experimental measurements.

The left panel tracks the two concentrations; the right tracks their instantaneous conversion reaction rate. Their different vertical units matter: $v=k[\mathrm{A}]$, so the numerical curves need not have the same height. This gives a practical way to translate a measured concentration into a forecast of both conversion and waiting time.

The coefficient $k$ of a first-order concentration term in a rate law. Its units are inverse time; two reversible first-order directions contribute additively to the relaxation rate.