Reversible (Opposing) Reactions

Drop a spoonful of sugar into hot water and stir. The dissolution seems to finish in seconds and then simply stop. Nothing appears to happen afterward, yet on a molecular level sugar molecules are still leaving and rejoining the crystal surface — the two flows have merely become equal. Reversible, or opposing, reactions are the kinetic bridge between rate laws and equilibrium constants. They explain why concentrations settle at fixed but nonzero values rather than racing to completion. This topic matters because almost every real reaction is reversible to some degree. Even reactions that look "complete," like strong acid neutralization, are formally opposed reactions with an equilibrium constant so large that reverse flow is negligible. Understanding the opposing-reaction framework lets you predict how fast a system reaches equilibrium, not just where it ends up. This article builds the simplest case, a reversible first-order reaction, from the two elementary rate laws. You will derive the exponential approach to equilibrium. You will connect the relaxation time to both rate constants. You will show that $K = k_1/k_{-1}$ falls directly out of the kinetics. Two worked examples anchor the algebra to real numbers.

Key takeaways

A reversible reaction A ⇌ B has a forward rate $k_1[A]$ and a reverse rate $k_{-1}[B]$ acting simultaneously.
Net rate of change is the difference of the two one-way rates, not just the forward rate alone.
Concentrations approach equilibrium exponentially, with a relaxation time $\tau = 1/(k_1+k_{-1})$ .
At equilibrium the forward and reverse rates are equal, giving $K = k_1/k_{-1}$ .
This single relation links kinetics (rate constants) to thermodynamics (equilibrium constant) for any elementary opposing reaction.
Key idea: equilibrium is a dynamic balance of equal and opposite rates, and the sum $k_1+k_{-1}$ (not either constant alone) sets how fast that balance is reached.

The core insight

In a reversible reaction the observed concentration change is the small difference between two competing one-way rates, so the system relaxes exponentially toward equilibrium with a rate constant equal to the sum $k_1+k_{-1}$ , while the equilibrium position itself is set by their ratio $K=k_1/k_{-1}$ .

The physical picture

Consider the simplest nontrivial kinetic system: a unimolecular interconversion,

\text{A} \underset{k_{-1}}{\overset{k_1}{\rightleftharpoons}} \text{B}

Molecules of A convert to B with rate constant $k_1$ , and molecules of B convert back to A with rate constant $k_{-1}$ . Both processes run continuously and simultaneously — the reaction never "switches off" the reverse channel once B appears. Early on, when little B exists, the reverse reaction is negligible and A disappears almost as if the reaction were irreversible and first order. As B accumulates, the reverse rate grows, clawing back some of A. Eventually the forward consumption of A exactly balances the reverse regeneration of A, and the net rate becomes zero — not because molecules stop moving, but because the two flows cancel. This is fundamentally different from an irreversible first-order decay (see Integrated Rate Law: First Order), where $[A]$ falls to zero at infinite time. Here $[A]$ falls only to its equilibrium value, $[A]_{eq} > 0$ , and does so along a modified exponential curve whose time constant depends on both rate constants together. A useful mental model: think of two people transferring coins between two jars at rates proportional to how many coins are currently in the jar they're taking from. Even though coins never stop moving, the number in each jar becomes fixed once the transfer rates balance. That steady-state distribution is the equilibrium; the speed of approach to it is the kinetics.

Figure

Concentration profiles for a reversible first-order reaction approaching equilibrium

Figure

Rate–versus–time picture: forward and reverse rates converge to a common value at equilibrium

The core relations

-\frac{d[A]}{dt} = k_1[A] - k_{-1}[B]

where: net rate of loss of A is the forward rate minus the reverse rate.

[A]_t = [A]_{eq} + \big([A]_0 - [A]_{eq}\big)\, e^{-t/\tau}

where: $[A]_t$ is the concentration at time $t$ , $[A]_{eq}$ the equilibrium value, and $\tau$ the relaxation time.

\tau = \frac{1}{k_1+k_{-1}}

where: $\tau$ is the characteristic time for approach to equilibrium, set by the sum of both rate constants.

K = \frac{[B]_{eq}}{[A]_{eq}} = \frac{k_1}{k_{-1}}

where: $K$ is the equilibrium constant, equal to the ratio of forward to reverse rate constants.

The core relations

Derivation AApproach to equilibrium for A ⇌ B

Goal. Starting from the two elementary rate laws, derive how $[A]$ evolves in time and show that it relaxes exponentially to $[A]_{eq}$ with time constant $\tau = 1/(k_1+k_{-1})$ , and that $K=k_1/k_{-1}$ .

Step 1. Write the net rate law. Let $[A]_0$ be the initial concentration of A, with no B present initially, so mass balance gives

[A] + [B] = [A]_0

The net rate of disappearance of A is the forward rate minus the reverse rate:

-\frac{d[A]}{dt} = k_1[A] - k_{-1}[B]

Step 2. Eliminate B using mass balance. Substitute $[B] = [A]_0 - [A]$ :

-\frac{d[A]}{dt} = k_1[A] - k_{-1}\big([A]_0-[A]\big) = (k_1+k_{-1})[A] - k_{-1}[A]_0

Step 3. Identify the equilibrium concentration. At equilibrium the net rate is zero, so setting $d[A]/dt = 0$ and calling that concentration $[A]_{eq}$ :

(k_1+k_{-1})[A]_{eq} = k_{-1}[A]_0 \quad\Rightarrow\quad [A]_{eq} = \frac{k_{-1}}{k_1+k_{-1}}[A]_0

Step 4. Recast the rate law in terms of the displacement from equilibrium. Define $x = [A] - [A]_{eq}$ . Subtracting the equilibrium condition from the general rate law of Step 2 gives

-\frac{dx}{dt} = (k_1+k_{-1})\,x

This is now a simple first-order decay in the displacement from equilibrium, exactly analogous to the integrated first-order law but with rate constant $(k_1+k_{-1})$ instead of a single $k$ .

Step 5. Integrate. Separating variables and integrating from $t=0$ (where $x_0 = [A]_0-[A]_{eq}$ ) to time $t$ :

x(t) = x_0\, e^{-(k_1+k_{-1})t}

[A]_t = [A]_{eq} + \big([A]_0-[A]_{eq}\big)e^{-(k_1+k_{-1})t}

Step 6. Read off the relaxation time and the equilibrium constant. The exponent shows that the natural time constant is

\tau = \frac{1}{k_1+k_{-1}}

Finally, since equilibrium is defined by equal forward and reverse rates, $k_1[A]_{eq} = k_{-1}[B]_{eq}$ , rearranging gives the thermodynamic link:

\boxed{K = \frac{[B]_{eq}}{[A]_{eq}} = \frac{k_1}{k_{-1}}, \qquad \tau = \frac{1}{k_1+k_{-1}}}

Worked examples

Example 1 — Foundational: finding both rate constants from equilibrium and relaxation data. A cis–trans isomerization, cis ⇌ trans, is monitored spectroscopically starting from pure cis at $[A]_0 = 0.500$ M. At equilibrium, $[trans]_{eq} = 0.320$ M. Independently, a small perturbation experiment shows the system relaxes toward equilibrium with $\tau = 25.0$ s. Find $k_1$ and $k_{-1}$ . Step 1: get the equilibrium constant. Mass balance gives $[A]_{eq} = 0.500 - 0.320 = 0.180$ M.

K = \frac{[B]_{eq}}{[A]_{eq}} = \frac{0.320}{0.180} = 1.78

Step 2: get the sum of rate constants from $\tau$ .

k_1 + k_{-1} = \frac{1}{\tau} = \frac{1}{25.0\ \text{s}} = 0.0400\ \text{s}^{-1}

Step 3: solve the two equations simultaneously. From $K = k_1/k_{-1} = 1.78$ , write $k_1 = 1.78\,k_{-1}$ . Substituting:

1.78\,k_{-1} + k_{-1} = 0.0400\ \text{s}^{-1} \quad\Rightarrow\quad 2.78\,k_{-1} = 0.0400\ \text{s}^{-1}

k_{-1} = 0.0144\ \text{s}^{-1}, \qquad k_1 = 1.78 \times 0.0144 = 0.0256\ \text{s}^{-1}

Sanity check: $k_1+k_{-1} = 0.0256+0.0144 = 0.0400\ \text{s}^{-1}$ ✓, matching $1/\tau$ . Also $k_1/k_{-1} = 0.0256/0.0144 = 1.78 = K$ ✓. Both constants are positive and of plausible magnitude for a thermal isomerization with a modest barrier.

Example 2 — Applied: predicting the composition at a specific time, and comparing to the naive irreversible estimate. For the same isomerization ( $k_1 = 0.0256\ \text{s}^{-1}$ , $k_{-1} = 0.0144\ \text{s}^{-1}$ , $[A]_0 = 0.500$ M, $[A]_{eq}=0.180$ M), find $[A]$ at $t = 40.0$ s, and compare with the (incorrect) prediction obtained by treating the reaction as simple irreversible first-order decay with rate constant $k_1$ alone. Correct treatment (reversible kinetics).

[A]_t = [A]_{eq} + ([A]_0-[A]_{eq})e^{-(k_1+k_{-1})t}

[A]_{40} = 0.180 + (0.500-0.180)\,e^{-(0.0400)(40.0)}

= 0.180 + 0.320\, e^{-1.60}

Since $e^{-1.60} = 0.2019$ ,

[A]_{40} = 0.180 + 0.320(0.2019) = 0.180 + 0.0646 = 0.245\ \text{M}

So $[B]_{40} = 0.500 - 0.245 = 0.255$ M. Naive irreversible estimate (common error). If one mistakenly ignored the reverse reaction and used $[A]_t = [A]_0 e^{-k_1 t}$ with $k_1 = 0.0256\ \text{s}^{-1}$ :

[A]_{40} = 0.500\,e^{-(0.0256)(40.0)} = 0.500\,e^{-1.024} = 0.500(0.359) = 0.180\ \text{M}

Comparison and sanity check. The naive model predicts $[A]$ has almost reached its true equilibrium value (0.180 M) already at 40 s — a substantial underestimate of how much A remains, because it ignores B converting back to A. The correct reversible treatment gives 0.245 M, noticeably higher. Both models correctly reduce to the same limit at $t\to\infty$ , since the naive one asymptotes to zero while the true equilibrium concentration is 0.180 M — this discrepancy is exactly the effect the reversible framework is built to capture. As $t\to\infty$ , $[A]_t \to [A]_{eq} = 0.180$ M in the correct treatment, while the naive treatment erroneously drives $[A]\to 0$ .

Limiting cases & common traps

Situation	What happens	Key relation
$k_{-1} \ll k_1$ (K very large)	Reaction looks irreversible; $[A]_{eq}\approx 0$	$\tau \approx 1/k_1$ , recovers ordinary first-order kinetics
$k_1 \ll k_{-1}$ (K very small)	Little product forms; A barely changes	$[A]_{eq}\approx [A]_0$
$k_1 = k_{-1}$	Equal populations at equilibrium, $K=1$	$\tau = 1/(2k_1)$
Early times, $t \ll \tau$	Reverse reaction negligible; looks like simple first-order decay	$[A]\approx [A]_0 e^{-k_1 t}$
Long times, $t \gg \tau$	System has fully relaxed; concentrations essentially constant	$[A]\to [A]_{eq}$

The chemist behind it

The dynamic-equilibrium picture — reactions running both ways at equal rates rather than simply halting — was central to the work of Henri Louis Le Chatelier and, more mechanistically, to the physical chemists who built kinetic theory in the late 19th century. But the person most responsible for turning "opposing rates" into quantitative kinetics of fast reversible reactions was Manfred Eigen, who pushed beyond classical mixing experiments into the microsecond and nanosecond regime using rapid perturbations. His relaxation methods, developed in the 1950s, measured $\tau = 1/(k_1+k_{-1})$ directly for reactions previously considered "instantaneous," such as proton transfer in water. This work sits at the doorway to the next topic in this sequence: relaxation methods and fast reactions.

Origins · 1 minManfred Eigen and how this was discovered

Manfred Eigen

1927–2019 · Pioneered relaxation techniques (temperature-jump, pressure-jump) to measure rate constants of extremely fast reversible reactions, sharing the 1967 Nobel Prize in Chemistry

Ask the tutor: How does this exponential relaxation picture change for a reversible reaction that is second order in one direction, like A + B ⇌ C?

Why isn't the net rate law for A ⇌ B just a simple first-order equation like the irreversible case?

Because two processes compete simultaneously: A converting to B, and B converting back to A. The net rate is their difference, $k_1[A]-k_{-1}[B]$ , not a single term. Once you substitute the mass-balance relation and define the displacement from equilibrium, the algebra collapses back into first-order form — but with rate constant $(k_1+k_{-1})$ instead of just $k_1$ . That is why the mathematics looks similar to Integrated Rate Law: First Order despite the extra physics.

Does the relation K = k1/k−1 apply to any reversible reaction, not just A ⇌ B?

It applies rigorously to elementary reversible steps, where forward and reverse rate laws come directly from reaction molecularity. For an elementary equilibrium $a\text{A} \rightleftharpoons b\text{B}$ , the same equal-rates argument at equilibrium gives $K = k_1/k_{-1}$ with appropriate concentration powers. For overall reactions made of multiple elementary steps, the relation still holds thermodynamically, but the "rate constants" involved are usually composite quantities from a full mechanism, not simple elementary rate constants.

How can I tell experimentally whether a reaction is reversible or effectively irreversible?

Follow the concentration profile to long times. If it plateaus at a nonzero reactant concentration rather than going to zero, the reaction is measurably reversible. Quantitatively, if $K = k_1/k_{-1}$ is very large (say, greater than about 100), the equilibrium concentration of reactant becomes negligible and the system behaves indistinguishably from an irreversible first-order decay within experimental error.

What is the difference between τ here and the half-life used for irreversible first-order reactions?

The irreversible half-life is $t_{1/2}=\ln2/k$ , describing decay all the way to zero. The relaxation time $\tau = 1/(k_1+k_{-1})$ describes decay of the displacement from equilibrium, which never reaches zero concentration but rather a finite equilibrium value. Both are exponential time constants, but $\tau$ combines both rate constants, whereas an irreversible half-life involves only one.

How does this topic connect to the pre-equilibrium approximation used in more complex mechanisms?

The pre-equilibrium approximation assumes a fast reversible step reaches its own local equilibrium quickly, characterized by exactly this $K=k_1/k_{-1}$ relation, before a slower subsequent step consumes the product. Recognizing that a reversible step relaxes on a timescale $\tau=1/(k_1+k_{-1})$ tells you when that assumption is valid — namely when $\tau$ for the fast step is much shorter than the timescale of the slow step.

Reversible (Opposing) Reactions

The physical picture

The core relations

The core relations

Worked examples

Limiting cases & common traps

The chemist behind it

Before this

Where this leads