Unimolecular Mass-Action Kinetics

From a Single Reaction Event to a Concentration Trajectory

A unimolecular conversion describes a reaction step in which one molecule of a species \(A\) converts to product. When the step follows mass-action kinetics, the rate of disappearance of \(A\) is proportional to its concentration. This simple relation leads to a specific prediction over time: the concentration declines exponentially, not by the same fixed amount in every time interval.

The result is useful in QSP models whenever a single species is removed through one first-order conversion under fixed conditions. The introductory material on elementary reaction steps and rate laws establishes the mass-action idea; here the focus is deriving the concentration trajectory and interpreting the rate constant and half-life.

Definition: For the single conversion \(A \rightarrow P\), unimolecular mass-action kinetics is the model in which the disappearance rate of \(A\) is \(k[A]\), where \([A]\) is the concentration of \(A\) and \(k\) is a constant, nonnegative rate constant. With no production of \(A\) and no other removal process, the concentration satisfies \(d[A]/dt=-k[A]\).

This equation describes the concentration of \(A\), not the concentration of product \(P\). The model treats product formation as the consequence of \(A\)'s conversion but does not require tracking product concentration. Its assumptions are important: the modeled volume is well mixed and constant, temperature and other conditions are fixed so that \(k\) does not change, and the specified conversion is the only process affecting \([A]\). The rate law is first order in \(A\): doubling \([A]\) at a given time doubles the instantaneous disappearance rate.

Concentration might be measured in \(\mathrm{nmol/L}\), \(\mathrm{mg/L}\), or another concentration unit. Since \(d[A]/dt\) has concentration-per-time units and \([A]\) has concentration units, \(k\) must have units of inverse time, such as \(\mathrm{h}^{-1}\). A rate constant is not itself a rate of concentration change; the rate at a particular moment is \(k[A]\).

Deriving the Exponential Law

Write \(C(t)=[A](t)\) and let \(C_0=[A](0)\). For \(C_0>0\), separation of variables gives the concentration trajectory directly. The derivation also shows why the ratio of concentrations, rather than their difference, is central to this model.

\frac{dC}{dt}=-kC \quad\Longrightarrow\quad \frac{dC}{C}=-k\,dt \quad\Longrightarrow\quad \int_{C_0}^{C(t)}\frac{dC}{C}=-k\int_0^t dt \quad\Longrightarrow\quad \ln\!\left(\frac{C(t)}{C_0}\right)=-kt \quad\Longrightarrow\quad C(t)=C_0 e^{-kt}.

The last expression is nonnegative for a nonnegative initial concentration. If \(C_0=0\), the differential equation also gives \(C(t)=0\) for all \(t\). The exponential solution applies for \(k\geq 0\); the interpretation of half-life and target times requires care at the boundary \(k=0\).

Result: Under the stated constant first-order model, \([A](t)=[A]_0e^{-kt}\). If \(k>0\), the concentration decreases toward zero but does not reach zero at any finite time when \([A]_0>0\). If \(k=0\), it remains equal to \([A]_0\) for all time.

The equation can be checked by substitution: at \(t=0\), it gives \(C(0)=C_0\), and its derivative is \(dC/dt=-kC_0e^{-kt}=-kC(t)\), as required. This verifies both the initial condition and the rate law. The decreasing curve is steepest in absolute concentration change at the start, when \(C\) is largest; later, the rate becomes smaller as less \(A\) remains.

Worked Example: Predicting Concentration After a Fixed Time

Suppose a hypothetical species has an initial concentration of \(80\ \mathrm{nmol/L}\) and converts with \(k=0.200\ \mathrm{h}^{-1}\). Calculate its concentration after 6.00 h, assuming the single first-order model and fixed conditions hold.

The model is \([A](t)=[A]_0e^{-kt}\). Substituting the given values gives

[A](6.00\ \mathrm{h}) =(80\ \mathrm{nmol/L})e^{-(0.200\ \mathrm{h}^{-1})(6.00\ \mathrm{h})} =80e^{-1.20}\ \mathrm{nmol/L} =24.1\ \mathrm{nmol/L}.

The exponent is dimensionless because \(\mathrm{h}^{-1}\) is multiplied by hours. The predicted concentration is \(24.1\ \mathrm{nmol/L}\), about 30.1% of the initial concentration. It is not correct to subtract the same concentration amount in each hour; the instantaneous loss depends on the concentration remaining.

Half-Life and Time to a Target

For \(k>0\) and \([A]_0>0\), the half-life \(t_{1/2}\) is the time required for the concentration to fall to one-half its initial value. Substituting \([A](t_{1/2})=[A]_0/2\) into the exponential law and solving for time gives the relationship between half-life and rate constant. If \([A]_0=0\), the concentration remains zero, so the time to reach half the initial concentration is not uniquely defined.

\frac{[A]_0}{2}=[A]_0e^{-kt_{1/2}} \quad\Longrightarrow\quad \frac{1}{2}=e^{-kt_{1/2}} \quad\Longrightarrow\quad t_{1/2}=\frac{\ln(2)}{k}, \qquad k>0,\quad [A]_0>0.

Because the initial concentration cancels, the half-life is independent of \([A]_0\) under this model. After each additional half-life, the concentration halves again: after \(m\) half-lives it is \([A]_0(1/2)^m\). The half-life has units of time, as it should, because \(1/k\) has time units.

More generally, for \(k>0\), the time to reach a positive target concentration \([A]_{\mathrm{target}}\) with \(0<[A]_{\mathrm{target}}\leq[A]_0\) follows by solving the same trajectory equation. A target equal to the initial concentration is reached at time zero. A lower positive target is reached at a positive finite time. Zero is approached asymptotically, not attained at finite time.

[A]_{\mathrm{target}}=[A]_0e^{-kt} \quad\Longrightarrow\quad t=\frac{\ln\!\left([A]_0/[A]_{\mathrm{target}}\right)}{k}, \qquad k>0,\quad 0<[A]_{\mathrm{target}}\leq[A]_0.

Worked Example: Calculating Half-Life and Checking the Trajectory

For a hypothetical conversion with \(k=0.200\ \mathrm{h}^{-1}\), calculate the half-life and predict the concentration after one half-life if the initial concentration is \(80\ \mathrm{nmol/L}\).

Since \(k>0\), the half-life formula applies:

t_{1/2}=\frac{\ln(2)}{k} =\frac{0.6931}{0.200\ \mathrm{h}^{-1}} =3.47\ \mathrm{h}.

At one half-life, the concentration is \(80/2=40.0\ \mathrm{nmol/L}\). Substituting the unrounded half-life into the trajectory gives the same result:

[A](t_{1/2})=(80\ \mathrm{nmol/L})e^{-(0.200\ \mathrm{h}^{-1})(3.4657\ \mathrm{h})} =40.0\ \mathrm{nmol/L}.

This agreement checks the calculation. For this model, changing the initial concentration would change the concentration at a given time but not the \(3.47\ \mathrm{h}\) half-life.

Worked Example: Time to Reach a Lower Positive Concentration

A hypothetical species starts at \(50.0\ \mathrm{mg/L}\) and has \(k=0.150\ \mathrm{h}^{-1}\). How long does the model predict it will take to reach \(10.0\ \mathrm{mg/L}\)?

The target is positive and is below the initial concentration, and \(k>0\), so the target-time formula applies:

t=\frac{\ln([A]_0/[A]_{\mathrm{target}})}{k} =\frac{\ln(50.0/10.0)}{0.150\ \mathrm{h}^{-1}} =\frac{\ln(5)}{0.150\ \mathrm{h}^{-1}} =10.7\ \mathrm{h}.

The ratio \(50.0/10.0\) is dimensionless, leaving the result in hours. Substitution confirms the prediction: \(50.0e^{-(0.150)(10.7)}\ \mathrm{mg/L}\) is approximately \(10.0\ \mathrm{mg/L}\), with small differences possible from rounding the displayed time.

The Zero-Rate Case and Other Boundary Checks

The value \(k=0\) is allowed by the differential equation, but it is a special case for time calculations. When \(k=0\), the equation becomes \(d[A]/dt=0\), so the concentration stays at its initial value. There is no finite half-life. No lower positive target is reached; only the unchanged target \([A]_{\mathrm{target}}=[A]_0\) occurs, at time zero. Thus, the formulas involving division by \(k\) must be stated only for \(k>0\).

Worked Example: Handling a Zero Rate Constant

A hypothetical model specifies \([A]_0=12.0\ \mathrm{\mu mol/L}\) and \(k=0\ \mathrm{h}^{-1}\). Determine the concentration at 5.00 h and whether the species has a finite half-life.

Using the trajectory, \([A](5.00\ \mathrm{h})=(12.0\ \mathrm{\mu mol/L})e^{-(0\ \mathrm{h}^{-1})(5.00\ \mathrm{h})}=12.0\ \mathrm{\mu mol/L}\). Since the concentration never falls to \(6.00\ \mathrm{\mu mol/L}\), no finite half-life exists. Substituting \(k=0\) into \(\ln(2)/k\) would be undefined and would not be a valid calculation.

Checking the Model Against Concentration Data

With \(k>0\) and positive concentrations, taking the logarithm of the concentration ratio gives a linear relationship with time:

\ln\!\left(\frac{[A](t)}{[A]_0}\right)=-kt.

Thus, if measurements follow the model closely, the logarithm of the fraction remaining should decline approximately linearly with time, with slope \(-k\). This is a diagnostic, not proof that the mechanism is unimolecular: measurement error, limited observation times, and other mechanisms can also produce data that appear approximately linear on this scale. Concentrations used in the logarithm must be positive; zero or below-detection measurements cannot be logged as ordinary positive values.

A practical check is to compare observed concentrations with the curve predicted from the proposed \(k\) and initial value. Systematic curvature in the log-ratio plot, changing apparent slopes across time, or residuals that repeatedly fall above and below the curve in a structured way can signal that constant first-order loss is inadequate. Before changing the model, check units, time origin, concentration measurements, and whether conditions plausibly stayed fixed. Random scatter around the curve is different from a consistent pattern.

If all assumptions are credible and the rate constant is known, the exponential formula gives an exact model solution. Numerical integration is unnecessary for this one-equation case. Numerical methods become useful when a model has multiple interacting states or rates that vary with time, but the simple exponential solution should still be a reference check when a subsystem has constant first-order loss.

The main alternative is not a different way to calculate the same answer, but a different model for how concentration changes. A constant-amount loss model, for example, predicts a fixed absolute decrease per unit time rather than a fixed fractional decrease. If data show a nonconstant log-ratio slope or other systematic departure, forcing an exponential trajectory can produce biased predictions, especially when extrapolating to long times. More complicated or changing conditions require a model suited to those features; this tutorial's formula alone cannot identify which alternative is correct.

Common Mistakes and Limitations

Using the half-life formula at \(k=0\). The expression \(\ln(2)/k\) is defined here only for \(k>0\). At \(k=0\), concentration is constant and there is no finite half-life.
Applying a target-time formula to zero concentration. For \(k>0\), \(\ln([A]_0/[A]_{\mathrm{target}})/k\) requires a positive target. The model approaches zero asymptotically and does not reach it in finite time.
Giving \(k\) concentration-per-time units. \(k\) has inverse-time units; \(k[A]\) has concentration-per-time units.
Assuming the exponential shape proves the mechanism. A good fit is consistent with the model but does not establish that a single unimolecular conversion is the true mechanism.
Ignoring changing conditions or other processes. A changing rate constant, production of \(A\), or additional removal routes can invalidate the single-process equation as written.

Key takeaway: For a single-species conversion with constant first-order mass-action kinetics, \([A](t)=[A]_0e^{-kt}\). If \(k>0\), \(t_{1/2}=\ln(2)/k\), independent of the initial concentration. If \(k=0\), concentration is constant and neither a finite half-life nor a lower positive target time exists.

Check Your Understanding

Work each question using the model assumptions stated above. Solutions are included after the questions.

A species starts at \(30.0\ \mathrm{nmol/L}\) and \(k=0.100\ \mathrm{h}^{-1}\). What concentration does the model predict after 10.0 h?
For \(k=0.400\ \mathrm{day}^{-1}\), calculate the half-life and state its units.
A concentration falls from \(64.0\ \mathrm{mg/L}\) to \(16.0\ \mathrm{mg/L}\) in 6.00 h under the model. Calculate \(k\) and the half-life.
For \(k=0\), can the model reach a target concentration below its initial value? Does it have a finite half-life?
Why does a straight-line pattern in \(\ln([A](t)/[A]_0)\) versus time support, but not prove, the constant first-order model?

Solutions. (1) \([A](10.0)=30.0e^{-(0.100)(10.0)}=11.0\ \mathrm{nmol/L}\). (2) \(t_{1/2}=0.6931/(0.400\ \mathrm{day}^{-1})=1.73\ \mathrm{days}\). (3) The remaining fraction is \(16.0/64.0=0.250\), so \(k=\ln(64.0/16.0)/(6.00\ \mathrm{h})=\ln(4)/6.00=0.231\ \mathrm{h}^{-1}\); hence \(t_{1/2}=0.6931/0.231=3.00\ \mathrm{h}\). (4) No. At \(k=0\), concentration remains at its initial value, so no lower target is reached and no finite half-life exists. (5) The model predicts exactly a line with slope \(-k\) under its assumptions, so such a pattern is consistent with it. Other mechanisms or limited, noisy data may also look approximately linear, so the pattern alone does not identify the mechanism.