Reversible reactions, dynamic equilibrium, equilibrium expressions, Le Châtelier's principle, and the science of balance in chemical systems
You have spent the last several chapters learning how reactions happen and how fast they go. You studied collision theory, activation energy, and all the factors that speed reactions up or slow them down. Now a new question arises: what happens when a reaction does not simply run to completion and stop?
Many chemical reactions behave exactly this way. They do not convert all reactants into products and then finish. Instead, they reach a state where both the forward reaction and the reverse reaction are happening simultaneously, and the amounts of reactants and products stop changing. This condition is called chemical equilibrium.
Equilibrium is one of the most important ideas in all of chemistry. It explains why some reactions seem to "get stuck" partway through. It explains how the human body maintains stable concentrations of oxygen, carbon dioxide, and acids in the blood. It explains why industrial chemists must carefully adjust temperature and pressure to get the most product out of a reaction. It explains how caves form, how oceans absorb carbon dioxide, and why certain medicines work the way they do.
This chapter builds equilibrium understanding from the ground up. You will start with the idea of reversible reactions, move to what equilibrium actually means at the particle level, learn how to write simple equilibrium expressions, and then master one of the most powerful tools in chemistry — Le Châtelier's principle — which predicts how equilibrium systems respond to being disturbed. The approach throughout is conceptual first and mathematical second.
Nitrogen and hydrogen gas are sealed in a container at high temperature. Ammonia begins to form — but the ammonia also starts breaking back down into nitrogen and hydrogen. Does the reaction just cancel itself out? What determines how much ammonia ends up in the flask? By the end of this chapter, you will be able to answer these questions precisely, and you will understand why industrial chemists have worked for over a century to get the most ammonia out of this exact reaction.
In earlier chapters, when you balanced a chemical equation, you imagined the reaction going from left to right. Reactants combine and are converted into products. Once the reactants are used up, the reaction stops. For many reactions this picture is accurate enough — when wood burns, the products do not reassemble into wood and oxygen. Those reactions are said to go to completion.
But many other reactions do not behave this way. Consider what happens when nitrogen gas and hydrogen gas are combined in a sealed container at high temperature:
N₂(g) + 3 H₂(g) → 2 NH₃(g)
This reaction produces ammonia gas. But at the same time, the ammonia molecules that have just formed can decompose back into nitrogen and hydrogen:
2 NH₃(g) → N₂(g) + 3 H₂(g)
Both reactions are occurring at the same time. The first, running left to right, is the forward reaction. The second, running right to left, is the reverse reaction. When both directions are possible, the reaction is reversible. Chemists use a double arrow to show this:
N₂(g) + 3 H₂(g) ⇌ 2 NH₃(g)
Reversibility is especially common for gas-phase reactions in sealed containers (products cannot escape and begin reacting in reverse), for ion-exchange reactions in solution, for the dissolution of slightly soluble substances, and for acid-base proton transfer reactions. Reactions where a gas escapes, a solid settles out, or very large amounts of energy are released tend to be much less reversible under ordinary conditions.
| Type of Reaction | Why Reversibility Is Common | Example |
|---|---|---|
| Gas-phase in sealed container | Products accumulate and can react with each other in reverse | N₂ + 3 H₂ ⇌ 2 NH₃ |
| Dissolution equilibrium | Dissolved ions can return to solid crystal surface | AgCl(s) ⇌ Ag⁺(aq) + Cl⁻(aq) |
| Acid-base in water | Proton transfer is reversible; conjugate acid and base coexist | HA ⇌ H⁺ + A⁻ |
| Gas dissolving in liquid | Gas molecules re-enter solution; dissolved gas molecules escape | CO₂(g) + H₂O ⇌ H₂CO₃(aq) |
Hemoglobin in your blood reversibly binds oxygen — picking it up in the lungs and releasing it in your tissues. Carbonated drinks go flat because CO₂ dissolved in water can escape back into the gas phase. In both cases, the double arrow ⇌ is the correct symbol.
Imagine you put nitrogen and hydrogen into a sealed container at high temperature. At the very beginning, no ammonia is present. Only the forward reaction can occur. As ammonia builds up, the reverse reaction begins — but slowly, because there is little ammonia to break apart. The reverse reaction speeds up as ammonia concentration grows. Meanwhile, the forward reaction slows down because N₂ and H₂ are being consumed.
Eventually a moment arrives when the rate of the forward reaction exactly equals the rate of the reverse reaction. Ammonia is forming at the same speed it is breaking down. The concentrations of all substances stop changing. This condition is chemical equilibrium.
Dynamic equilibrium is the state in which the forward and reverse reactions occur at equal rates. Both reactions continue — the system is active — but concentrations do not change because every molecule formed is balanced by a molecule breaking apart.
Imagine a busy staircase between two floors. People move constantly up and down. If the number going up equals the number going down at every moment, the number of people on each floor stays constant — even though movement never stops. That is dynamic equilibrium at the molecular level.
In practice, you know a reaction has reached equilibrium when the concentrations of all reactants and products stop changing over time, when the color of a colored solution or gas mixture stops changing, and when properties like pressure, pH, or conductivity become constant. These observations do not mean the reaction has stopped — they mean forward and reverse rates have matched each other.
You can reach the same equilibrium starting from either side. Begin with only N₂ and H₂, or begin with only NH₃ — at the same temperature and total amount of material, you arrive at the same equilibrium mixture. The equilibrium state belongs to the reaction and the conditions, not to the starting point.
Students often hear the word "equilibrium" and assume reactants and products are present in equal amounts. This is one of the most common misconceptions in chemistry — and it is incorrect.
Equilibrium means the rates of the forward and reverse reactions are equal — not the concentrations of reactants and products.
At equilibrium, a reaction mixture might contain far more product than reactant. It might contain far more reactant than product. The exact ratio depends on the specific reaction and the temperature.
| Situation | Example | What It Means |
|---|---|---|
| Products favored | H₂ + I₂ ⇌ 2 HI at 425°C | Mostly HI at equilibrium; small amounts of H₂ and I₂ remain |
| Reactants favored | N₂ + O₂ ⇌ 2 NO at room temperature | Almost no NO at equilibrium; N₂ and O₂ barely affected |
| Mixed amounts | Many reactions at intermediate temperatures | Significant amounts of both reactants and products present |
Equal rates ≠ equal concentrations. Never confuse the two. The ratio of products to reactants at equilibrium is set by the chemistry and temperature, not by any requirement for equal amounts.
Chemists have developed a mathematical tool to summarize the equilibrium position of a reaction. It is called the equilibrium expression, and it gives a number — the equilibrium constant K — that describes the ratio of products to reactants at equilibrium.
For the general reaction: a A + b B ⇌ c C + d D
The equilibrium expression is: K = [C]ᶜ × [D]ᵈ / ([A]ᵃ × [B]ᵇ)
Square brackets [ ] represent molar concentration (mol/L). Products go in the numerator; reactants go in the denominator. Each concentration is raised to the power of its stoichiometric coefficient.
K = [NH₃]² / ([N₂] × [H₂]³)
K = [HI]² / ([H₂] × [I₂])
Pure solids and pure liquids are not included. Their concentrations do not change meaningfully as a reaction proceeds. Only gases and dissolved aqueous species appear in the K expression.
K = [CO₂]| Value of K | Interpretation |
|---|---|
K much greater than 10³ | Products strongly favored; reaction proceeds nearly to completion |
K between 10⁻³ and 10³ | Significant amounts of both reactants and products at equilibrium |
K much less than 10⁻³ | Reactants strongly favored; very little product at equilibrium |
The value of K changes only with temperature. Changes in concentration, pressure, or the addition of a catalyst shift the equilibrium position but do not change K. Temperature is the only factor that changes the equilibrium constant itself.
Once a system reaches equilibrium, concentrations remain constant — unless something disturbs the system. When a change is imposed, the system responds by shifting in a direction that partially counteracts the change and eventually reaches a new equilibrium.
This behavior was stated clearly in 1884 by the French chemist Henri-Louis Le Châtelier. His principle, Le Châtelier's principle, can be stated as:
If a change is imposed on a system at equilibrium, the equilibrium will shift in the direction that tends to reduce the effect of that change.
This principle is not a formula — it is a reasoning strategy. It lets you predict, without calculation, how an equilibrium system will respond to disturbances including changes in concentration, pressure, volume, temperature, and the presence of a catalyst.
When a system is disturbed, one reaction temporarily becomes faster than the other. The system is no longer at equilibrium. Reactions continue at unequal rates until concentrations shift enough to restore the balance between forward and reverse rates. The system "shifts" toward whichever direction absorbs or undoes the disturbance — but only partially. The new equilibrium will always reflect that the disturbance occurred.
One of the most common ways to disturb an equilibrium is to add more of one substance. The system responds by consuming the added substance — the equilibrium shifts toward the side that uses it up.
Consider the ammonia reaction: N₂(g) + 3 H₂(g) ⇌ 2 NH₃(g)
If more N₂ is added to a system already at equilibrium, the forward reaction rate increases. More collisions between N₂ and H₂ produce more NH₃. The system shifts to the right — toward products — until a new equilibrium is reached with higher [NH₃] and lower [H₂].
If NH₃ is continuously removed — for example, by cooling it so it liquefies and settles out — the equilibrium constantly shifts to the right, producing more ammonia. This is actually one strategy industrial chemists use to increase ammonia yield.
| Change Made | Direction of Shift | Reasoning |
|---|---|---|
| Add reactant | Shifts right → toward products | System consumes the added reactant |
| Remove reactant | Shifts left ← toward reactants | System replenishes what was removed |
| Add product | Shifts left ← toward reactants | System consumes the added product |
| Remove product | Shifts right → toward products | System replenishes what was removed |
Changing concentration shifts the equilibrium position but does not change the value of K. Think of K as the target ratio. When concentrations are disturbed, the system rearranges itself until the product-to-reactant ratio returns to that same target value at the same temperature.
For equilibrium reactions involving gases, changes in pressure or volume can shift the equilibrium position. This does not apply to reactions involving only liquids or solids.
Increasing pressure by compressing a gas-phase system means more molecules per unit volume — higher concentrations of all gases. The system responds by shifting toward the side of the equation with fewer moles of gas, which partially reduces the pressure. Decreasing pressure by expanding the volume shifts the system toward the side with more moles of gas.
The key is counting the total gas molecules on each side of the balanced equation.
Left side: 1 mol N₂ + 3 mol H₂ = 4 mol gas
Right side: 2 mol NH₃ = 2 mol gas
| Change Made | Direction of Shift |
|---|---|
| Increase pressure (decrease volume) | Shifts toward side with fewer moles of gas |
| Decrease pressure (increase volume) | Shifts toward side with more moles of gas |
| Equal moles of gas on both sides | No shift |
| Add inert gas at constant volume | No shift (partial pressures of reactants and products unchanged) |
Changes in concentration and pressure shift the equilibrium position but do not change the value of K. Temperature is different. Changing the temperature actually changes the value of K.
To understand why, think about whether a reaction is exothermic or endothermic. For equilibrium purposes, you can treat heat as either a product or a reactant:
Exothermic reaction: heat is released → treat heat as a product on the right side.
N₂(g) + 3 H₂(g) ⇌ 2 NH₃(g) + heat
Endothermic reaction: heat is absorbed → treat heat as a reactant on the left side.
heat + N₂(g) + O₂(g) ⇌ 2 NO(g)
If you increase the temperature, you are adding heat. The system responds by absorbing the added heat — shifting in the endothermic direction. If you decrease the temperature, you are removing heat. The system responds by producing heat — shifting in the exothermic direction.
| Change Made | Direction of Shift | Effect on K |
|---|---|---|
| Increase temperature | Shifts in endothermic direction | K changes (increases if endothermic reaction, decreases if exothermic) |
| Decrease temperature | Shifts in exothermic direction | K changes (decreases if endothermic reaction, increases if exothermic) |
For exothermic reactions like ammonia synthesis, lower temperatures favor more product at equilibrium — but also slow the reaction dramatically. Industrial chemists use a compromise temperature balanced with a catalyst. This tension between thermodynamic yield and kinetic speed is one of the central challenges of chemical engineering.
In Chapter 21, you learned that a catalyst increases the rate of a chemical reaction by providing an alternative pathway with lower activation energy. The question now is: how does a catalyst affect equilibrium?
The answer may surprise you: a catalyst does not shift the equilibrium position, and it does not change the value of K.
A catalyst lowers the activation energy of both the forward reaction and the reverse reaction equally. If both reactions are sped up by the same factor, their rate ratio stays the same. The equilibrium condition — equal forward and reverse rates — is reached more quickly, but the equilibrium position is unchanged. A catalyst is a tool for reaching equilibrium faster. It is not a tool for shifting where equilibrium lies.
| Effect of Catalyst | Result |
|---|---|
| Rate of forward reaction | Increases |
| Rate of reverse reaction | Increases (equally) |
| Equilibrium position | Unchanged |
| Value of K | Unchanged |
| Time to reach equilibrium | Decreases |
Even though a catalyst does not change yield at equilibrium, it is extraordinarily valuable because it allows industries to operate at lower temperatures. Lower temperatures can be safer, cheaper, and — for exothermic reactions — more favorable for yield. In the Haber-Bosch process, an iron-based catalyst enables acceptable reaction speed at the lower temperature that favors ammonia formation.
The concepts built in this chapter operate in industrial plants, in the oceans, in the human bloodstream, and in every living cell. Real-world equilibrium chemistry is not an abstraction — it feeds the world, keeps blood pH stable, and controls how much CO₂ the ocean can absorb.
Ammonia, NH₃, is the starting material for nitrogen-based fertilizers that support food production for billions of people. Its industrial synthesis is one of the most consequential applications of equilibrium chemistry:
N₂(g) + 3 H₂(g) ⇌ 2 NH₃(g) ΔH = −92 kJ/mol (exothermic)
| Le Châtelier Factor | Industrial Choice | Reasoning |
|---|---|---|
| Temperature | ~400–500°C (compromise) | Low temperatures favor NH₃ (exothermic), but slow the reaction; high temperature needed for practical speed |
| Pressure | 150–300 atm | High pressure shifts to right (fewer moles of gas on product side: 2 vs. 4); dramatically increases yield |
| Catalyst | Iron catalyst with promoters | Compensates for slowness at the compromise temperature; does not shift K |
| Remove product | NH₃ continuously condensed out | Removing product shifts equilibrium right; drives more ammonia formation |
Your blood carries oxygen using hemoglobin (Hb), a protein that reversibly binds O₂:
Hb(aq) + O₂(aq) ⇌ HbO₂(aq)
In the lungs, where oxygen concentration is high, the equilibrium shifts to the right — hemoglobin picks up oxygen. In the body's tissues, where oxygen is being consumed and its concentration is lower, the equilibrium shifts to the left — oxyhemoglobin releases oxygen. The same chemical equilibrium runs in two different directions in two different locations in your body — Le Châtelier's principle operating in a living system.
When carbon dioxide from the atmosphere dissolves in seawater, a series of equilibria are established:
CO₂(g) ⇌ CO₂(aq)
CO₂(aq) + H₂O(l) ⇌ H₂CO₃(aq)
H₂CO₃(aq) ⇌ H⁺(aq) + HCO₃⁻(aq)
As human activity adds more CO₂ to the atmosphere, Le Châtelier's principle predicts that more CO₂ dissolves into the ocean, forming more carbonic acid, releasing more H⁺ ions, and lowering the ocean's pH. This is the chemical basis of ocean acidification — a significant environmental concern studied by marine chemists and ecologists worldwide.
Your blood maintains a nearly constant pH of about 7.4 through buffer systems — equilibrium reactions that absorb added acid or base and prevent drastic pH changes. The carbonate buffer system is the most important:
H⁺(aq) + HCO₃⁻(aq) ⇌ H₂CO₃(aq) ⇌ H₂O(l) + CO₂(g)
When blood becomes too acidic, the equilibrium shifts right, consuming H⁺. When blood becomes too basic, it shifts left, releasing H⁺. You will study buffer systems in full in Chapter 23.
Answer each question in one or two sentences. These questions test understanding before detailed practice begins.
For each situation, predict whether the equilibrium shifts left, shifts right, or does not shift. Explain briefly.
PCl₅(g) ⇌ PCl₃(g) + Cl₂(g) ΔH = +88 kJ/mol (endothermic)
2 CO(g) + O₂(g) ⇌ 2 CO₂(g) ΔH = −566 kJ/mol (exothermic)
Study each graph description carefully before answering.
Graph 1: A graph shows concentration of reactants and products over time for a reversible reaction. At time zero, only reactants are present. Reactant concentration decreases rapidly at first, then levels off and becomes constant. Product concentration increases rapidly at first, then levels off and becomes constant.
Graph 2: A graph shows the rates of the forward and reverse reactions over time. The forward rate starts high and decreases. The reverse rate starts at zero and increases. Eventually the two lines meet and remain equal.
Write the equilibrium expression K for each reaction. Remember to omit pure solids and liquids.
Both forward and reverse reactions continue at equilibrium. Concentrations stop changing because both reactions run at equal rates. Equilibrium is a dynamic state, not a frozen one.
Equilibrium means equal rates, not equal concentrations. The value of K — which is almost never exactly 1 — determines the actual ratio of products to reactants.
A catalyst lowers activation energy equally for both forward and reverse reactions. It speeds up reaching equilibrium but does not change where equilibrium lies. K is unaffected by a catalyst.
K depends only on temperature. Adding more reactant shifts the equilibrium position — more product forms until the ratio returns to K — but K itself is unchanged at constant temperature.
Increasing pressure shifts equilibrium toward the side with fewer moles of gas — which may be either the reactant or product side. If gas moles are equal on both sides, no shift occurs at all.
These questions draw on concepts from earlier chapters in Part IV and Part V.
Chapter 22 has given you a powerful framework for understanding how chemical systems reach balance and respond to change. The concept of dynamic equilibrium will appear again immediately in Chapter 23, which is dedicated to acids and bases. Acid-base chemistry is fundamentally an equilibrium story. Strong acids ionize nearly completely — a reaction lying very far to the right. Weak acids establish a true equilibrium between the molecular acid and its ions, described by an acid dissociation constant Ka. The pH scale is built on equilibrium concentrations of H⁺ ions. You will also encounter buffers — solutions that resist pH change because of carefully balanced equilibria between weak acids and their conjugate bases — and Le Châtelier's principle will explain exactly how they work. Everything from this chapter carries forward directly into Chapter 23 and every chapter that follows.