Bimolecular Association Rates

Association Between Two Distinct Reactants

When two molecular species associate, the rate depends on how often molecules of each species encounter one another. For an elementary step involving one molecule of \(A\) and one molecule of \(B\), the mass-action model makes that dependence explicit: the rate is proportional to the product of their concentrations. This gives a second-order rate law, even though each reactant is a distinct species.

The concentration-time behavior is not generally a single exponential. As the reaction proceeds, both reactants are depleted, and their concentrations remain linked because each association event consumes one molecule of each. The resulting equations can be solved directly. The form of the solution depends on whether the initial concentrations are equal, unequal, or whether one reactant is absent.

Definition: For the elementary association \(A+B\rightarrow C\), the mass-action event rate is \(r=k[A][B]\). Here \([A]\), \([B]\), and \([C]\) denote concentrations, \(r\) is the number of association events per unit volume per unit time, and \(k\) is the association rate constant. For concentrations expressed in \(\mathrm{M}\) and time in seconds, \(k\) has units \(\mathrm{M}^{-1}\mathrm{s}^{-1}\).

This rate law assumes a well-mixed, constant-volume system in which the modeled step is elementary, concentrations represent the available reactants, and the rate constant is constant over the interval considered. The mass-action rate is proportional to the frequency of encounters between \(A\) and \(B\) under those conditions. It is not a claim that every reaction’s rate law can be inferred just by looking at its overall chemical equation.

From Encounter Rate to Concentration Equations

For \(A+B\rightarrow C\), each event removes one \(A\), removes one \(B\), and forms one \(C\). With no other sources, sinks, or reactions, the event rate therefore gives the concentration changes:

\frac{d[A]}{dt}=-k[A][B],\qquad \frac{d[B]}{dt}=-k[A][B],\qquad \frac{d[C]}{dt}=k[A][B].

The negative signs indicate depletion of reactants; the positive sign indicates product formation. The two reactants have the same instantaneous rate of concentration loss, but their concentrations need not be equal: their difference stays constant.

\frac{d}{dt}\bigl([B]-[A]\bigr) =\frac{d[B]}{dt}-\frac{d[A]}{dt} =0.

Let the initial concentrations be \(A_0=[A](0)\) and \(B_0=[B](0)\), and define the constant difference \(D=B_0-A_0\). Then \([B](t)-[A](t)=D\) throughout the reaction. This conservation relation is useful both for solving the equations and for checking a numerical or analytic result.

Equal Initial Concentrations

If \(A_0=B_0>0\), the concentration difference is zero, so the two concentrations remain equal at all times. Write their common value as \(x(t)\). The coupled equations reduce to \(dx/dt=-kx^2\). Separating variables and integrating from \(0\) to \(t\) gives the integrated second-order association equation:

\int_{A_0}^{x(t)}\frac{dx}{x^2} =-k\int_0^t dt \quad\Longrightarrow\quad -\frac{1}{x(t)}+\frac{1}{A_0}=-kt,

and therefore

[A](t)=[B](t)=\frac{A_0}{1+kA_0t}.

The product concentration, if initially \(C_0\), follows from the one-to-one stoichiometry and the absence of other processes: \([C](t)=C_0+A_0-[A](t)\). The reactant concentrations approach zero as time increases, but this idealized solution does not reach zero at any finite time.

Worked Example: Equal Starting Concentrations

Suppose a hypothetical well-mixed system has \(A_0=B_0=10.0\ \mathrm{\mu M}\), \(k=0.0200\ \mathrm{\mu M}^{-1}\mathrm{min}^{-1}\), and \(C_0=0\). Find the concentrations after \(5.00\) minutes.

Because the starting concentrations are equal, they remain equal and the equal-concentration solution applies:

[A](5.00)=[B](5.00) =\frac{10.0\ \mathrm{\mu M}} {1+(0.0200\ \mathrm{\mu M}^{-1}\mathrm{min}^{-1})(10.0\ \mathrm{\mu M})(5.00\ \mathrm{min})} =\frac{10.0}{2.00}\ \mathrm{\mu M} =5.00\ \mathrm{\mu M}.

The product concentration is \(C_0+A_0-[A](5.00)=0+10.0-5.00=5.00\ \mathrm{\mu M}\). The result is consistent with the one-to-one association: \(5.00\ \mathrm{\mu M}\) of each reactant has been consumed and \(5.00\ \mathrm{\mu M}\) of product has formed.

Unequal Positive Initial Concentrations

Now suppose \(A_0>0\), \(B_0>0\), and \(D=B_0-A_0\ne 0\). Since \([B]=[A]+D\), either reactant can be used to reduce the two equations to one. A convenient derivation uses the concentration ratio. It requires both initial concentrations to be positive, so that the logarithms and ratio below are defined. Dividing each concentration equation by its concentration and subtracting gives:

\frac{d}{dt}\ln\!\left(\frac{[A]}{[B]}\right) =\frac{1}{[A]}\frac{d[A]}{dt} -\frac{1}{[B]}\frac{d[B]}{dt} =-k[B]+k[A] =-kD.

Integration, with \([A](0)=A_0\) and \([B](0)=B_0\), yields \([A](t)/[B](t)=(A_0/B_0)e^{-kDt}\). Combining this ratio with \([B]-[A]=D\) gives the unequal-concentration solution:

[A](t)=\frac{A_0D}{B_0e^{kDt}-A_0}, \qquad [B](t)=\frac{B_0D e^{kDt}}{B_0e^{kDt}-A_0}.

Although \(D\) can be positive or negative, both concentrations in these expressions remain positive for finite \(t\ge 0\) when \(A_0,B_0>0\) and \(k\ge0\). The denominator is nonzero: if \(D>0\), then \(B_0e^{kDt}\ge B_0>A_0\); if \(D<0\), then \(B_0e^{kDt}\le B_0<A_0\). In either case the numerator and denominator have matching signs. When \(k=0\), the formulas reduce to the constant initial concentrations, as required.

Worked Example: Unequal Positive Starting Concentrations

Consider a hypothetical association with \(A_0=2.00\ \mathrm{\mu M}\), \(B_0=5.00\ \mathrm{\mu M}\), \(k=0.100\ \mathrm{\mu M}^{-1}\mathrm{min}^{-1}\), and \(C_0=0\). Calculate the concentrations after \(2.00\) minutes.

The difference is \(D=B_0-A_0=3.00\ \mathrm{\mu M}\), and the dimensionless exponent is \(kDt=(0.100)(3.00)(2.00)=0.600\). Substitution into the unequal-concentration solution gives:

[A](2.00) =\frac{(2.00)(3.00)} {(5.00)e^{0.600}-2.00}\ \mathrm{\mu M} =\frac{6.00}{9.1106-2.00}\ \mathrm{\mu M} =0.8438\ \mathrm{\mu M},

[B](2.00) =\frac{(5.00)(3.00)e^{0.600}} {(5.00)e^{0.600}-2.00}\ \mathrm{\mu M} =3.8438\ \mathrm{\mu M}.

The difference check gives \(3.8438-0.8438=3.0000\ \mathrm{\mu M}=D\). The amount consumed from each reactant is \(2.00-0.8438=1.1562\ \mathrm{\mu M}\) and \(5.00-3.8438=1.1562\ \mathrm{\mu M}\), respectively. Thus \([C](2.00)=1.1562\ \mathrm{\mu M}\), consistent with the stated reaction and initial product concentration.

Zero Initial Concentrations and Limiting Cases

The ratio derivation does not apply if either initial concentration is zero. In a closed system with no source term, however, the equations make this case straightforward: if \(A_0=0\) or \(B_0=0\), then \([A][B]=0\) initially and remains zero, so neither reactant changes and no new product forms. This includes the case in which both reactants start at zero. The unequal-concentration formulas are not needed to establish this result.

Worked Example: One Reactant Is Initially Absent

In a hypothetical system, let \(A_0=0\ \mathrm{\mu M}\), \(B_0=4.00\ \mathrm{\mu M}\), \(C_0=0\), and \(k=0.0500\ \mathrm{\mu M}^{-1}\mathrm{min}^{-1}\). What are the concentrations at \(t=10.0\) minutes?

At the initial time, the reaction rate is \(r(0)=kA_0B_0=(0.0500)(0)(4.00)=0\ \mathrm{\mu M}\mathrm{min}^{-1}\). Since there is no source of \(A\), it remains absent, and the rate remains zero. Therefore \([A](10.0)=0\ \mathrm{\mu M}\), \([B](10.0)=4.00\ \mathrm{\mu M}\), and \([C](10.0)=0\ \mathrm{\mu M}\). The unequal-positive-concentration formula must not be used here because its logarithmic-ratio derivation requires \(A_0>0\).

Checks, Alternatives, and Limitations

Before interpreting a calculated concentration trajectory, check that its assumptions match the modeled setting. In particular, the closed-system formulas do not include synthesis, degradation, transport, dilution, or additional reaction steps. If any of these processes are important, the concentration difference need not remain constant, and the formulas above no longer describe the full system.

Several checks help identify algebraic or implementation errors:

Initial condition: setting \(t=0\) in a solution must return \(A_0\) and \(B_0\).
Difference invariant: for the stated one-to-one association without other processes, \([B](t)-[A](t)=B_0-A_0\).
Direction of change: for \(k\ge0\) and nonnegative concentrations, both reactants cannot increase under these equations.
Units: \(k[A][B]\) has units concentration per time; the exponent \(kDt\) is dimensionless.
Boundary cases: when \(k=0\), no concentration changes; when either initial reactant is absent, the association rate stays zero.

The equal-concentration formula is simpler and avoids a difference term in the denominator. The unequal-concentration formula handles two distinct positive starting amounts directly. A numerical solution of the coupled differential equations is an alternative when the system includes additional processes or lacks a convenient closed form; for this isolated association step, it should agree with the analytic solution and satisfy the same difference check. Numerical integration does not remove the need to justify the model assumptions.

A frequent mistake is to set \([A](t)=[B](t)\) merely because both reactants are consumed at the same rate. Equal rates of change do not imply equal concentrations: their initial difference is preserved. A second pitfall is to apply the logarithmic ratio formula when one starting concentration is zero. The correction is to use the equal-concentration solution only when \(A_0=B_0>0\), the unequal solution only when both are positive and unequal, and the zero-concentration argument when either is absent.

Key takeaway: For an elementary \(A+B\) association under constant-volume, closed-system conditions, \(r=k[A][B]\). Equal positive initial concentrations follow a simple reciprocal-time solution; unequal positive concentrations follow a ratio-based solution; and if either reactant is initially absent, no association occurs without a source.

Check Your Understanding

Try each question, then compare your reasoning with the solutions below.

For \(A+B\rightarrow C\), what units must \(k\) have if concentration is measured in \(\mathrm{mM}\) and time in hours?
If \(A_0=4.00\ \mathrm{\mu M}\) and \(B_0=7.00\ \mathrm{\mu M}\), what is \([B](t)-[A](t)\) under the stated closed-system model?
For equal positive initial concentrations \(A_0=B_0=6.00\ \mathrm{\mu M}\) and \(k=0.0250\ \mathrm{\mu M}^{-1}\mathrm{min}^{-1}\), calculate \([A](4.00\ \mathrm{min})\).
Why is the ratio derivation invalid when \(A_0=0\), and what happens to the concentrations if \(B_0>0\) and there is no source of \(A\)?
Name two checks that can reveal an incorrect concentration-time calculation.

Solutions. (1) Since \(r=k[A][B]\) must have units \(\mathrm{mM}\,\mathrm{h}^{-1}\), \(k\) has units \(\mathrm{mM}^{-1}\mathrm{h}^{-1}\). (2) The difference is invariant, so \([B](t)-[A](t)=7.00-4.00=3.00\ \mathrm{\mu M}\). (3) Use the equal-concentration formula: \([A](4.00)=6.00/[1+(0.0250)(6.00)(4.00)]\ \mathrm{\mu M}=6.00/1.600=3.75\ \mathrm{\mu M}\). (4) The ratio and its logarithm require positive concentrations, so they are undefined at \(A_0=0\). Without a source of \(A\), the rate remains zero; \(A\) stays zero and \(B\) stays at \(B_0\). (5) For example, check that the solution returns the initial conditions at \(t=0\), preserves the concentration difference, has dimensionless exponents, and does not predict increasing reactant concentrations under this model.

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