Reversible Reaction Rates

Introduction

Many reactions are reversible. Reactant becomes product while product can become reactant, and the measured concentration change is the difference between these fluxes. At equilibrium the net rate is zero because the opposing rates match, not because molecular motion or reaction stops.

Core explanation

For simple elementary A ⇌ B, let forward rate be kf[A] and reverse rate kr[B]. Then d[B]/dt=kf[A]−kr[B] in a fixed-volume one-to-one system. Early in a run starting with pure A, [B]≈0 and reverse rate is small, so B initially rises. As B accumulates, reverse rate grows and forward rate may fall as A declines. Eventually kf[A]eq=kr[B]eq.

For this simple elementary pair, equilibrium ratio [B]eq/[A]eq=kf/kr when ideal concentration approximations apply. This relates kinetic constants to the equilibrium composition. It does not mean any arbitrary overall reversible reaction has K equal to a ratio of one forward and one reverse apparent rate constant; mechanisms and stoichiometric powers matter.

The approach to equilibrium can be exponential in this simple A ⇌ B model. With total concentration C=[A]+[B] constant, d[B]/dt=kf(C−[B])−kr[B]=kfC−(kf+kr)[B]. Thus the relaxation rate constant is kf+kr. A plateau in [B] does not prove B formation has stopped; forward and reverse molecular conversions continue at equal rates.

A catalyst can speed both directions and shorten the time to equilibrium without changing the equilibrium ratio at fixed temperature. Temperature can change kf and kr and therefore K. This distinction prevents the false statement that adding catalyst moves the equilibrium to products just because product appears faster initially.

If a product is continuously removed, the system may never settle into the same closed-vessel equilibrium. The reverse rate can remain small because [B] stays low, allowing continued net forward conversion. This is a process condition, not a change in the intrinsic equilibrium constant. Likewise, adding more A changes current rates and the reaction quotient, after which the system moves toward a new equilibrium composition at the same K.

In complex reversible systems, a measured net rate can include several forward and reverse pathways. A simple graph plateau might also reflect reactant exhaustion or instrument drift rather than equilibrium. Testing reversibility can involve starting from both directions or perturbing a settled mixture and observing relaxation.

Step-by-step reasoning

1. Write forward and reverse rate expressions for a justified model. 2. Subtract reverse from forward for net product formation. 3. Set net rate to zero to derive equilibrium concentration relation. 4. Distinguish zero net change from zero individual flux. 5. Check whether the system is closed and conditions remain fixed.

Visual explanation

Draw A ⇌ B with two opposing arrows that grow and shrink during approach. At equilibrium, make arrows equal thickness and show flat concentration-time curves. Write kf[A]eq=kr[B]eq beneath the diagram.

Real-world analogy

People can enter and leave a room at equal rates while the number inside remains constant. The stable headcount is a dynamic balance, not evidence that no one moves.

Real-world example

An esterification mixture can approach a stable composition while ester forms and hydrolyzes in opposite directions. Changing conditions or removing a product can drive further net formation, while a catalyst mainly changes approach speed.

Why?

Why does reverse rate increase as product accumulates? In a simple mass-action model, reverse rate is proportional to product concentration, so more B provides more reverse-reacting molecules.

Common misconception

“At equilibrium, all reaction rates are zero.” Forward and reverse rates are equal and can both be nonzero; only their difference, the net rate, is zero.

Worked example

For A ⇌ B, kf=0.30 min⁻¹ and kr=0.10 min⁻¹. At equilibrium, [B]/[A]=kf/kr=3 under the simple model. If total concentration is 0.80 M, then [A]+[B]=0.80 and [B]=3[A], so [A]=0.20 M and [B]=0.60 M. Forward rate is 0.30×0.20=0.060 M min⁻¹; reverse rate is 0.10×0.60=0.060 M min⁻¹. Net rate is zero although both fluxes continue.

Quick check

1. What condition defines dynamic equilibrium for simple A ⇌ B? Answer: kf[A]eq=kr[B]eq, equal forward and reverse rates.

Exam focus

Write net rate as forward minus reverse, solve simple equilibrium ratios and explicitly state that molecular conversions continue. Qualify the kf/kr relation to a justified elementary model.

Advanced insight

The relaxation time for ideal first-order A ⇌ B is approximately 1/(kf+kr). Measuring how quickly a perturbed mixture returns to equilibrium can therefore reveal kinetic constants beyond the equilibrium ratio alone.

Summary

Reversible reactions have opposing forward and reverse fluxes. Product concentration levels off when those rates become equal; it does not mean reactions stop. In simple elementary A ⇌ B, the equilibrium ratio is kf/kr.

Practice questions

1. If forward and reverse rates are each 0.05 M s⁻¹, what is net rate? Answer: Zero, although both processes continue at 0.05 M s⁻¹. 2. For simple A ⇌ B, what is [B]eq/[A]eq if kf=2kr? Answer: Two. 3. What does a catalyst change at fixed temperature in this model? Answer: It can speed approach by increasing both opposing rates without changing the equilibrium ratio.