Reversible Reactions and Net Flux

Why Reversible Reactions Need Two Rates

A reversible reaction can proceed in both directions at the same time. The forward and reverse processes each have a nonnegative rate, while the observed change in concentration depends on their difference. Keeping these quantities distinct prevents a common modeling error: treating the forward rate alone as the rate of change when reverse conversion is also occurring.

For a reversible elementary reaction, the forward and reverse directions have their own rate constants and mass-action rates. In this tutorial, a reaction flux is expressed in concentration per unit time. Concentrations are assumed to be measured in a fixed volume, so changes in concentration correspond directly to reaction progress.

Forward and Reverse Mass-Action Rates

Consider the elementary reaction \(aA+bB\rightleftharpoons cC\). If the forward direction is \(aA+bB\to cC\), the mass-action rates are

Here \([A]\), \([B]\), and \([C]\) denote concentrations, and \(k_f\) and \(k_r\) are the rate constants for the forward and reverse elementary steps. The exponents follow the molecularities of those elementary steps. This does not mean that a reaction's stoichiometric coefficients always determine its rate law: that mass-action form is justified here because each direction is specified as an elementary step.

The units of each rate constant depend on the order of its step, so that both \(v_f\) and \(v_r\) have units of concentration per time. For example, a unimolecular rate constant has units of inverse time, while a bimolecular rate constant has units of inverse concentration per time. The sign convention for \(J\) is a modeling choice; once chosen, it must be used consistently when writing the concentration equations.

From Net Flux to Concentration Equations

Stoichiometry determines how one unit of net reaction progress changes each species. For \(aA+bB\rightleftharpoons cC\), using the forward direction as positive gives

The coefficients multiply the net flux because a forward event consumes \(a\) units of \(A\) and \(b\) units of \(B\), and produces \(c\) units of \(C\); reverse events are already included through the subtraction in \(J\). In a model with other reactions or inputs, their contributions must be added to the corresponding concentration equations.

Solving a Reversible Unimolecular Conversion

Now consider the closed, fixed-volume conversion \(A\rightleftharpoons B\). Assume both directions are elementary and unimolecular, the rate constants are positive and constant, and no other process produces or removes \(A\) or \(B\). Write \(x=[A]\), \(y=[B]\), and use \(k_f\) and \(k_r\) for the forward and reverse rate constants. The rates and net flux are

Because each forward conversion consumes one \(A\) and produces one \(B\), the equations are

Adding the equations gives \(d(x+y)/dt=0\), so the total concentration \(C=x+y\) is constant under the stated assumptions. Substituting \(y=C-x\) into the equation for \(x\) yields a single linear ODE:

The constant concentration at which this derivative is zero is \(x_*=\frac{k_rC}{k_f+k_r}\). Define \(\lambda=k_f+k_r\), which has units of inverse time. Subtract \(x_*\) from both sides by setting \(z=x-x_*\). Since \(k_rC-\lambda x_*=0\), \(z\) obeys \(dz/dt=-\lambda z\). An integrating factor gives the solution without dividing by \(z\): multiply by \(e^{\lambda t}\) to obtain \(d(e^{\lambda t}z)/dt=0\), and then use the initial value \(z(0)=x(0)-x_*\).

The second expression follows from conservation. Equivalently, \(y(t)=\frac{k_fC}{k_f+k_r}+\bigl(y(0)-\frac{k_fC}{k_f+k_r}\bigr)e^{-\lambda t}\). The exponential factor describes the rate of approach: the characteristic time is \(1/\lambda\), and the deviation from the stationary composition is multiplied by \(e^{-1}\) over that time. These formulas assume a single isolated reversible step; additional reactions generally change the equations and their solution.

Direction Can Change During the Time Course

The sign of net flux depends on the current concentrations, not just on which direction is called forward. If \(A\) begins below \(x_*\), reverse conversion initially dominates; as \(A\) approaches \(x_*\), the net flux tends toward zero. The next example checks that behavior directly.

Checks, Alternatives, and Common Pitfalls

Several checks make a reversible-step calculation easier to trust. First, verify that \(v_f\) and \(v_r\) have the same units before subtracting them. Second, confirm that the stoichiometric equations give the expected signs. For \(A\rightleftharpoons B\), \(dx/dt=-J\) and \(dy/dt=J\), so the sum must remain constant. Third, with positive rate constants and nonnegative initial concentrations, the analytic solution should remain nonnegative and approach the stationary values without overshooting them. A numerical trajectory that loses mass or becomes negative may signal a coding, solver, or time-step problem, though the model assumptions should also be checked.

A common mistake is to call \(v_f\) the rate of change of \(A\). That ignores reverse conversion: the correct equation is \(dx/dt=-v_f+v_r=-J\). Another mistake is to infer a mass-action exponent from stoichiometry for a reaction that has not been specified as elementary. In that situation, the rate law requires separate justification; the stoichiometric equation alone does not establish it.

Writing two directional rates and then subtracting them is especially useful when the contribution of each direction matters. Writing a single net flux is a compact alternative once both rates are defined. For this unimolecular step, the analytic solution is exact and exposes conservation and the relaxation rate. For a larger system of coupled reactions, one typically writes all flux contributions in the ODE system and uses numerical integration; the same mass-action assumptions and sign checks still apply. A one-way model can be a useful approximation only when reverse flux is negligible over the conditions and time period of interest, not merely because the forward direction is the one of interest.

When deriving the exponential solution by separation of variables, dividing by a deviation \(z\) requires care. It is valid only on an interval where \(z\ne0\). On such an interval, separation gives \(z(t)=z(0)e^{-\lambda t}\), which cannot reach zero at a finite time if \(z(0)\ne0\). Continuity rules out a first finite zero, and uniqueness for this linear ODE also ensures the solution cannot depart from zero if it starts there. The integrating-factor derivation above avoids this division altogether.