Worked Problems on Reversible Reaction Kinetics

Reversible Reactions: Rates, Equilibrium, and Time

A reversible reaction can proceed in both directions at once. Its forward and reverse fluxes may each be substantial even when the concentrations are no longer changing. To analyze such a system, distinguish the two directional rates from their difference, the net rate. That distinction is essential when calculating equilibrium, interpreting a measured rate, or predicting a time course.

The examples below use closed, well-mixed systems at constant volume and temperature. Rate constants are assumed constant, and each direction follows mass-action kinetics for the reaction as written. Concentrations are nonnegative. These assumptions let us use the stated reaction rates directly; changing volume, external inputs, or additional reactions would require a different set of balance equations.

Definition: For a reversible reaction, the forward flux is the mass-action rate in the written forward direction, and the reverse flux is the rate in the opposite direction. Define the net reaction rate \(v\) as forward flux minus reverse flux. For \(A \rightleftharpoons B\), \(v=k_f[A]-k_r[B]\), so \(d[B]/dt=v\) and \(d[A]/dt=-v\). Equilibrium for this closed, single-reaction system means \(v=0\); it does not mean that both directional fluxes are zero.

The rate constants carry units that make each flux a concentration per unit time. For a first-order step, \(k_f\) or \(k_r\) has units of inverse time. For a bimolecular step such as \(A+B\to C\), its rate constant has units of concentration\(^{-1}\) time\(^{-1}\). Always write the reaction as specified: changing stoichiometric coefficients can change the mass-action expression.

Reversible First-Order Reaction: Equilibrium and Time Course

Consider \(A \rightleftharpoons B\), with forward constant \(k_f\) and reverse constant \(k_r\). In a closed constant-volume system containing only these two forms, the total concentration \(T=[A]+[B]\) remains fixed. The forward and reverse fluxes are \(k_f[A]\) and \(k_r[B]\), respectively, so the net rate in the direction \(A\to B\) is \(v=k_f[A]-k_r[B]\).

\frac{d[B]}{dt}=k_f(T-[B])-k_r[B] =k_fT-(k_f+k_r)[B].

At equilibrium, setting this derivative to zero gives the equilibrium concentrations. The same differential equation gives the full time course when the initial condition is known.

[B]_{\mathrm{eq}}=\frac{k_f}{k_f+k_r}T, \qquad [A]_{\mathrm{eq}}=\frac{k_r}{k_f+k_r}T, \qquad \lambda=k_f+k_r.

[B](t)=[B]_{\mathrm{eq}}+ \bigl([B](0)-[B]_{\mathrm{eq}}\bigr)e^{-\lambda t}, \qquad [A](t)=T-[B](t).

The parameter \(\lambda\) is the rate at which the system approaches equilibrium; its units are inverse time. The net rate decays with the same exponential factor. This follows because \(v=k_fT-\lambda[B]\), which is zero at equilibrium, and the difference \([B](t)-[B]_{\mathrm{eq}}\) is its initial difference multiplied by \(e^{-\lambda t}\). Therefore \(v(t)=v(0)e^{-\lambda t}\).

Worked Example: Equilibrium and Time Course for \(A \rightleftharpoons B\)

Suppose a hypothetical closed system has \(k_f=0.30\ \mathrm{h^{-1}}\), \(k_r=0.20\ \mathrm{h^{-1}}\), \([A](0)=8.0\ \mathrm{\mu mol\,L^{-1}}\), and \([B](0)=2.0\ \mathrm{\mu mol\,L^{-1}}\). Find the equilibrium concentrations and concentrations and net rate at \(t=2.0\ \mathrm{h}\).

The total concentration is \(T=8.0+2.0=10.0\ \mathrm{\mu mol\,L^{-1}}\). The assumptions of a closed system and one-for-one interconversion imply this total stays constant. The equilibrium values are

[B]_{\mathrm{eq}}=\frac{0.30}{0.30+0.20}(10.0) =6.00\ \mathrm{\mu mol\,L^{-1}}, \qquad [A]_{\mathrm{eq}}=10.0-6.00 =4.00\ \mathrm{\mu mol\,L^{-1}}.

Here \(\lambda=0.50\ \mathrm{h^{-1}}\), so \(\lambda t=1.00\). The time-course equations give

[B](2.0)=6.00+(2.0-6.00)e^{-1} =4.528\ \mathrm{\mu mol\,L^{-1}}, \qquad [A](2.0)=10.0-4.528483\ldots =5.472\ \mathrm{\mu mol\,L^{-1}}.

For the net rate, first calculate the initial value: \(v(0)=0.30(8.0)-0.20(2.0)=2.00\ \mathrm{\mu mol\,L^{-1}\,h^{-1}}\). Thus

v(2.0)=v(0)e^{-(0.50)(2.0)} =2.00e^{-1} =0.7358\ \mathrm{\mu mol\,L^{-1}\,h^{-1}}.

The positive sign means the net movement is still from \(A\) to \(B\). The displayed concentrations are rounded; the net rate above uses the unrounded time-course values. Substituting rounded concentrations into the rate law can produce a slightly different last digit.

Equilibrium for a Bimolecular Association

For \(A+B \rightleftharpoons C\), suppose the forward step is elementary and has rate \(k_f[A][B]\), while the reverse step has rate \(k_r[C]\). The net rate of formation of \(C\) is

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

If the initial concentration of \(C\) is zero and the reaction is the only process, let \(x\) be the concentration of \(C\) formed. One molecule of each reactant is consumed per molecule of product formed, so \([A]=[A](0)-x\), \([B]=[B](0)-x\), and \([C]=x\). The physical range is \(0\leq x\leq\min([A](0),[B](0))\). At equilibrium, the forward and reverse fluxes are equal, which gives a quadratic equation in \(x\). A mathematically valid root outside this physical range cannot describe the reaction.

Worked Example: Selecting the Physical Equilibrium Root

In a hypothetical system, \(A+B \rightleftharpoons C\) has \(k_f=0.040\ \mathrm{L\,\mu mol^{-1}\,min^{-1}}\) and \(k_r=0.20\ \mathrm{min^{-1}}\). Initially, \([A]=10.0\), \([B]=6.0\), and \([C]=0\), with concentrations in \(\mathrm{\mu mol\,L^{-1}}\). Find the equilibrium concentrations and fluxes.

Set \(x=[C]_{\mathrm{eq}}\), so \([A]_{\mathrm{eq}}=10.0-x\) and \([B]_{\mathrm{eq}}=6.0-x\), with \(0\leq x\leq6.0\). Equating fluxes gives

0.040(10.0-x)(6.0-x)=0.20x, \qquad x^2-21.0x+60.0=0.

The roots are

x=\frac{21.0\pm\sqrt{21.0^2-4(60.0)}}{2} =\frac{21.0\pm\sqrt{201}}{2}.

The roots are approximately \(3.4112766\) and \(17.5887\ \mathrm{\mu mol\,L^{-1}}\). Only the first lies in the required range \(0\leq x\leq6.0\). Using that unrounded physical root gives

[C]_{\mathrm{eq}}=3.411\ \mathrm{\mu mol\,L^{-1}}, \quad [A]_{\mathrm{eq}}=10.0-3.4112766\ldots =6.589\ \mathrm{\mu mol\,L^{-1}}, \quad [B]_{\mathrm{eq}}=6.0-3.4112766\ldots =2.589\ \mathrm{\mu mol\,L^{-1}}.

At equilibrium each directional flux is \(k_r x=0.20(3.4112766\ldots)=0.6823\ \mathrm{\mu mol\,L^{-1}\,min^{-1}}\). The net rate is zero because the two fluxes are equal, not because the reaction has stopped in both directions. For example, the rounded displayed concentrations need not reproduce equality to the final shown digit; both fluxes were calculated from the same unrounded root.

Stoichiometry Determines Species Rates

A net reaction rate is not necessarily the rate of change of every species concentration. For \(2A \rightleftharpoons B\), mass action gives \(v=k_f[A]^2-k_r[B]\), where \(v\) is the net rate of reaction events per volume per time. Each forward event consumes two \(A\) and forms one \(B\). Consequently, \(d[A]/dt=-2v\) and \(d[B]/dt=v\). The factor of two belongs in the species balance, not as an extra factor in the forward mass-action rate.

Worked Example: Net Rate and Concentration Slopes for \(2A \rightleftharpoons B\)

At a hypothetical instant, let \([A]=3.0\ \mathrm{\mu mol\,L^{-1}}\), \([B]=2.0\ \mathrm{\mu mol\,L^{-1}}\), \(k_f=0.040\ \mathrm{L\,\mu mol^{-1}\,h^{-1}}\), and \(k_r=0.15\ \mathrm{h^{-1}}\). The forward and reverse fluxes are

v_f=0.040(3.0)^2=0.360\ \mathrm{\mu mol\,L^{-1}\,h^{-1}}, \qquad v_r=0.15(2.0)=0.300\ \mathrm{\mu mol\,L^{-1}\,h^{-1}}.

Thus \(v=v_f-v_r=0.060\ \mathrm{\mu mol\,L^{-1}\,h^{-1}}\), positive in the forward direction. The species slopes are

\frac{d[A]}{dt}=-2(0.060)=-0.120\ \mathrm{\mu mol\,L^{-1}\,h^{-1}}, \qquad \frac{d[B]}{dt}=0.060\ \mathrm{\mu mol\,L^{-1}\,h^{-1}}.

These are instantaneous slopes, not predictions that the rates remain constant over an hour. As concentrations change, both mass-action fluxes generally change as well.

Choosing a Calculation and Checking It

For a simple reversible first-order reaction, the algebraic equilibrium and exponential time-course formulas are available directly. For nonlinear reactions such as bimolecular association, the equilibrium can often be found by solving the mass-action balance together with the stoichiometric concentration relationships. A time course for a nonlinear system generally requires solving its differential equations, rather than applying the first-order exponential formula. The formula for \(A\rightleftharpoons B\) does not apply merely because a reaction is reversible; it depends on first-order forward and reverse rates.

Before trusting a result, check the direction, units, and physical range. A positive net rate must point in the direction used to define \(v\). Forward and reverse fluxes have the same concentration-per-time units, while equilibrium concentrations must be nonnegative and consistent with the initial amounts available for reaction. At equilibrium, calculate both fluxes using the same unrounded concentrations and verify that they agree within rounding.

Check the rate law. Use the mass-action expression for each direction and the reaction as written.
Check units. Each flux must have units of concentration per time, and each exponential argument must be dimensionless.
Check the root. Keep only roots that satisfy nonnegative concentrations and the stoichiometric bounds.
Check the interpretation. Zero net rate means equal opposing fluxes; a nonzero gross flux may remain.

A common mistake is to set the forward flux equal to zero at equilibrium, or to report that no reaction is occurring because the net rate is zero. For the bimolecular example, that would incorrectly discard the reverse flux. The correct condition is \(k_f[A][B]=k_r[C]\); both sides can be positive even though their difference is zero.

Key takeaway: Write the directional mass-action fluxes first, define the net rate as forward minus reverse, then use stoichiometry to relate that rate to each concentration slope. At equilibrium, the opposing fluxes balance; time-course formulas require the particular rate laws and assumptions for which they were derived.

Check Your Understanding

Try each problem, then compare your work with the solutions below. All data are hypothetical.

For \(A\rightleftharpoons B\), \(k_f=0.12\ \mathrm{min^{-1}}\), \(k_r=0.08\ \mathrm{min^{-1}}\), \([A](0)=9.0\), and \([B](0)=1.0\ \mathrm{\mu mol\,L^{-1}}\). Find the equilibrium concentrations.
For the reaction in question 1, calculate \([A]\), \([B]\), and \(v\) at \(t=5.0\ \mathrm{min}\).
For \(A+B\rightleftharpoons C\), \(k_f=0.010\ \mathrm{L\,\mu mol^{-1}\,min^{-1}}\), \(k_r=0.10\ \mathrm{min^{-1}}\), and initial concentrations \(4.0,\ 3.0,\ 0\ \mathrm{\mu mol\,L^{-1}}\), respectively, determine the equilibrium extent \(x=[C]_{\mathrm{eq}}\).
For \(2A\rightleftharpoons B\), explain why a zero net rate does not imply zero forward flux.

Solutions. (1) \(T=10.0\ \mathrm{\mu mol\,L^{-1}}\), so \([B]_{\mathrm{eq}}=(0.12/0.20)(10.0)=6.00\ \mathrm{\mu mol\,L^{-1}}\) and \([A]_{\mathrm{eq}}=4.00\ \mathrm{\mu mol\,L^{-1}}\). (2) Here \(\lambda=0.20\ \mathrm{min^{-1}}\), so \(e^{-\lambda t}=e^{-1}=0.367879\). Therefore \([B](5)=6.00+(1.00-6.00)e^{-1}=4.161\ \mathrm{\mu mol\,L^{-1}}\), \([A](5)=10.0-[B](5)=5.839\ \mathrm{\mu mol\,L^{-1}}\), and \(v(5)=v(0)e^{-1}\), where \(v(0)=0.12(9.0)-0.08(1.0)=1.00\ \mathrm{\mu mol\,L^{-1}\,min^{-1}}\); hence \(v(5)=0.3679\ \mathrm{\mu mol\,L^{-1}\,min^{-1}}\). (3) Equating fluxes gives \(0.010(4-x)(3-x)=0.10x\), or \(x^2-17x+12=0\). The roots are \((17\pm\sqrt{241})/2\); only \(x=(17-\sqrt{241})/2=0.7379\ \mathrm{\mu mol\,L^{-1}}\) lies between zero and the limiting initial concentration \(3.0\). (4) At equilibrium, the condition is \(k_f[A]^2=k_r[B]\). These fluxes can be equal and positive, so their difference is zero while forward and reverse reaction events continue.