Rate laws and reversible reactions

Every reaction you met in school was drawn as a one-way arrow running to completion. Real reactions are not like that: products can react back to re-form reactants, and it is the tug-of-war between the forward and reverse rates that decides where a system actually ends up. A reaction does not "finish" — it relaxes toward a moving balance point where the two opposing rates become equal. This article treats the simplest reversible elementary reaction, $A \rightleftharpoons B$ , as a pair of coupled rate processes. You already know rate laws, order, and the integrated first-order law from Class 11–12; what is new here is combining a forward step and a reverse step into one differential equation, solving it, and watching the kinetic result reproduce the equilibrium constant $K_{eq}$ you know from thermodynamics.

The core insight

For a reversible elementary reaction $A \rightleftharpoons B$ , the system relaxes exponentially toward equilibrium at a rate set by $k_1+k_{-1}$ , and the equilibrium constant is exactly the ratio of the two rate constants, $K_{eq}=k_1/k_{-1}$ .

The core idea

Setting up the net rate. Consider the elementary reaction

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

where $k_1$ is the forward rate constant and $k_{-1}$ is the reverse rate constant. Because $B$ is continuously reacting back to $A$ , the net rate of disappearance of $A$ is the forward rate minus the reverse rate, not the forward rate alone:

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

where: $[A]$ and $[B]$ are the instantaneous concentrations of $A$ and $B$ , and $k_1$ , $k_{-1}$ carry units appropriate to a first-order (unimolecular) process here, $\text{s}^{-1}$ .

Rewriting in terms of extent of reaction. Starting from pure $A$ at concentration $a_0$ , let $x$ be the amount converted to $B$ at time $t$ , so $[A]=a_0-x$ and $[B]=x$ . Substituting turns the rate law into a single linear ODE in $x$ :

\frac{dx}{dt} = k_1(a_0-x) - k_{-1}x = k_1a_0 - (k_1+k_{-1})x

This one equation carries all the physics of the approach to equilibrium; the rest of the article is about reading it correctly and solving it.

Two things to notice immediately. First, as $t\to\infty$ the rate $dx/dt$ must go to zero — not because the reaction has "stopped," but because the forward and reverse rates have become equal. Second, the combination $k_1+k_{-1}$ , not either constant alone, is what will turn out to control how fast the system gets there. The figure below sketches the shape this produces: concentration climbing quickly at first, then flattening out as it nears a plateau, never overshooting it.

Figure

A concentration versus time curve for a reversible reaction rising smoothly from zero and flattening toward a plateau equilibrium value without overshoot

Reading the plateau. The plateau value the curve settles at is the equilibrium extent $x_{eq}$ , found by setting $dx/dt=0$ in the ODE above: $x_{eq}=k_1a_0/(k_1+k_{-1})$ . At that point the forward and reverse rates are equal — dynamic balance, not a frozen system — and this equality is exactly the kinetic route to the equilibrium constant, worked out below.

Derivation ASolving the reversible first-order ODE for the approach to equilibrium

The goal is to solve $\dfrac{dx}{dt} = k_1a_0 - (k_1+k_{-1})x$ for $x(t)$ , starting from $x(0)=0$ , and to show that the result is an exponential climb toward $x_{eq}=k_1a_0/(k_1+k_{-1})$ with a rate constant equal to the sum $k_1+k_{-1}$ , which is what ties the shape of the approach curve to both rate constants at once rather than to either one alone.

Step 1. Identify the equilibrium value directly from the ODE. At equilibrium the net rate is zero, $dx/dt=0$ , so $k_1a_0=(k_1+k_{-1})x_{eq}$ , giving $x_{eq}=\dfrac{k_1a_0}{k_1+k_{-1}}$ . This is the plateau the curve must approach, found without integrating anything.

Step 2. Rewrite the ODE using $x_{eq}$ . Since $k_1a_0=(k_1+k_{-1})x_{eq}$ , substitute back: $\dfrac{dx}{dt} = (k_1+k_{-1})x_{eq} - (k_1+k_{-1})x = (k_1+k_{-1})(x_{eq}-x)$ . The equation now says the rate of approach is proportional to how far $x$ still is from its final value — the classic signature of an exponential relaxation.

Step 3. Separate variables and integrate. Write $\dfrac{dx}{x_{eq}-x} = (k_1+k_{-1})\,dt$ and integrate both sides from $0$ to $t$ (with $x=0$ at $t=0$ ): $-\ln\!\left(\dfrac{x_{eq}-x}{x_{eq}}\right) = (k_1+k_{-1})t$ .

Step 4. Solve for $x(t)$ . Exponentiating and rearranging gives $x_{eq}-x = x_{eq}\,e^{-(k_1+k_{-1})t}$ , so

x(t) = x_{eq}\left[1-e^{-(k_1+k_{-1})t}\right]

Checking the limits confirms the picture: at $t=0$ , $x=0$ as required; as $t\to\infty$ , $x\to x_{eq}$ , the plateau from Step 1; and the rate at which the exponential decays is $k_1+k_{-1}$ — the sum, so even a reaction with a sluggish forward step can settle quickly if its reverse step is fast.

Connecting to $K_{eq}$ . At true equilibrium the forward and reverse rates are equal (detailed balance for this elementary step): $k_1[A]_{eq}=k_{-1}[B]_{eq}$ , so

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

This holds rigorously for a single elementary (microscopically reversible) step; for a multi-step mechanism the observed $K_{eq}$ is still the ratio of the overall effective forward and reverse rate constants, but the individual step constants combine according to the mechanism rather than appearing one-for-one. The same equilibrium argument extends to a bimolecular pair $A+B\rightleftharpoons C+D$ : setting $dx/dt=0$ in $\dfrac{dx}{dt}=k_1(a_0-x)^2-k_{-1}x^2$ gives $\dfrac{[C]_{eq}[D]_{eq}}{[A]_{eq}[B]_{eq}}=k_1/k_{-1}=K_{eq}$ — the familiar mass-action law, now derived from rate constants rather than assumed from thermodynamics.

Worked example

The isomerisation $A\rightleftharpoons B$ starts from pure $A$ at $a_0=1.00\ \text{mol L}^{-1}$ . At equilibrium, $[B]_{eq}=0.75\ \text{mol L}^{-1}$ , and a relaxation measurement shows the system has covered $63\%$ of its total approach to equilibrium (i.e. $x=0.63\,x_{eq}$ ) after $t=20\ \text{s}$ . Find $k_1$ , $k_{-1}$ , and the time needed to reach $90\%$ of the approach.

Step 1. Get $K_{eq}$ from the equilibrium concentrations. $[A]_{eq}=1.00-0.75=0.25\ \text{mol L}^{-1}$ , so $K_{eq}=k_1/k_{-1}=0.75/0.25=3.0$ .

Step 2. Get $k_1+k_{-1}$ from the relaxation data. From $x(t)=x_{eq}[1-e^{-(k_1+k_{-1})t}]$ , the fraction achieved is $1-e^{-(k_1+k_{-1})t}$ . Setting this to $0.63\approx1-e^{-1}$ means $(k_1+k_{-1})t=1$ , so $k_1+k_{-1}=1/20\ \text{s}=0.050\ \text{s}^{-1}$ .

Step 3. Solve the two equations simultaneously. Substituting $k_1=3.0k_{-1}$ into $k_1+k_{-1}=0.050$ gives $4.0k_{-1}=0.050$ , so $k_{-1}=0.0125\ \text{s}^{-1}$ and $k_1=3.0\times0.0125=0.0375\ \text{s}^{-1}$ . Check: $k_1/k_{-1}=0.0375/0.0125=3.0=K_{eq}$ , consistent.

Step 4 (stretch). Time for $90\%$ approach. Set $1-e^{-(k_1+k_{-1})t}=0.90$ , so $e^{-(k_1+k_{-1})t}=0.10$ , giving $(k_1+k_{-1})t=\ln(10)=2.303$ . With $k_1+k_{-1}=0.050\ \text{s}^{-1}$ , $t=2.303/0.050=46.1\ \text{s}$ — about $2.3\times$ the earlier $63\%$ time, as expected for an exponential approach.

Final answer: $k_1=0.0375\ \text{s}^{-1}$ , $k_{-1}=0.0125\ \text{s}^{-1}$ , and the system reaches $90\%$ of equilibrium after about $46\ \text{s}$ .

Common traps

Treating "reversible" as "instantly at equilibrium." The approach is exponential in time with rate constant $k_1+k_{-1}$ ; how fast the system gets there is separate information from where it ends up.
Assuming $K_{eq}=k_1/k_{-1}$ holds step-by-step for any multi-step mechanism. It is exact for a single elementary step only; a composite mechanism's $K_{eq}$ is the ratio of overall effective constants, not a simple product of individual steps.
Believing equilibrium means the reaction has stopped. At equilibrium the net rate is zero because forward and reverse rates are equal and nonzero — this is dynamic balance, not a frozen system.
Applying the first-order exponential formula to a bimolecular pair. $x(t)=x_{eq}[1-e^{-(k_1+k_{-1})t}]$ is derived for $A\rightleftharpoons B$ only; the $A+B\rightleftharpoons C+D$ case needs partial fractions for an explicit $x(t)$ .

Ask the tutor: if you'd like to work through the full partial-fractions integration of the reversible bimolecular rate equation, or practise extracting $k_1$ and $k_{-1}$ from a different set of relaxation data.

Why does the relaxation rate depend on the sum k1 plus k-1 instead of just the forward constant k1?

Because both processes are pulling the system toward the same balance point at once. Even if the forward step is slow, a fast reverse step still shuttles material back and forth quickly, so the net concentration profile settles rapidly — it just settles at a value skewed toward reactants because $k_1$ is small. The math bears this out directly: solving the coupled rate law gives an exponential with decay constant $k_1+k_{-1}$ , never $k_1$ or $k_{-1}$ in isolation. Only the equilibrium position ( $x_{eq}$ , hence $K_{eq}$ ) depends on the ratio; the speed of getting there depends on the sum.

If K_eq = k1/k-1, does that mean I can find k1 and k-1 separately just from the equilibrium concentrations?

No — equilibrium concentrations alone only ever give you the ratio $K_{eq}=k_1/k_{-1}$ , one equation in two unknowns. To pin down $k_1$ and $k_{-1}$ individually you need a second, independent piece of information: a kinetic (time-dependent) measurement such as the relaxation time, which fixes the sum $k_1+k_{-1}$ . Combining the ratio from equilibrium with the sum from kinetics is exactly the two-equation, two-unknown system solved in the worked example above.

Does this K_eq = k1/k-1 relation work for reactions that are not single elementary steps, like most reactions you meet in a lab?

Not directly, and this is a common overreach. The derivation here relies on detailed balance for one elementary step, where the forward and reverse processes are genuinely the reverse of each other at the molecular level. A real multi-step mechanism has its own intermediate steps, each with its own forward and reverse rate constant; the overall observed $K_{eq}$ is still the ratio of the overall effective forward and reverse rate constants of the composite process, but you cannot read off $K_{eq}$ from any single step's $k_1$ and $k_{-1}$ without accounting for the full mechanism.

Rate laws and reversible reactions

The core idea

Worked example

Common traps

Before this

Where this leads