Reversible reactions; dynamic equilibrium

Q: What is Le Châtelier's Principle?

If a system at dynamic equilibrium is subjected to a change in conditions, the equilibrium position shifts in a direction that counteracts the change.

Q: How does changing reactant or product concentration affect the position of equilibrium and egin{equation*}K_c\end{equation*}?

Adding a species shifts the equilibrium to consume it, while removing a species shifts it to replenish it; egin{equation*}K_c\end{equation*} remains unchanged.

Q: How does changing pressure affect the position of an equilibrium involving gases?

Increasing pressure shifts the equilibrium towards the side with fewer moles of gas, while decreasing pressure shifts it towards the side with more moles of gas.

Q: How does changing temperature affect the position of equilibrium and the value of egin{equation*}K_c\end{equation*}?

Increasing temperature shifts equilibrium in the endothermic direction (increasing egin{equation*}K_c\end{equation*} if endothermic, decreasing egin{equation*}K_c\end{equation*} if exothermic); temperature is the only factor that alters egin{equation*}K_c\end{equation*}.

Q: What is the effect of adding a catalyst to a system at dynamic equilibrium?

A catalyst increases the rates of both forward and reverse reactions equally, allowing equilibrium to be reached faster without changing the position of equilibrium or egin{equation*}K_c\end{equation*}.

Practice

Chemistry · Senior Secondary (HKDSE) · 9 min read · Updated: 21 Aug 2026

Topic X: Reversible Reactions & Dynamic Equilibrium

Hello! Welcome to the fascinating world of chemical equilibrium. Ever wondered why some reactions don't seem to 'finish'? Or how big chemical plants can produce huge amounts of products so efficiently? The answers lie in understanding reversible reactions and a special state called dynamic equilibrium.

In this chapter, we'll explore reactions that can go both forwards and backwards. We'll learn what it means for a reaction to be 'in balance' and discover how chemists can cleverly push a reaction in the direction they want. Don't worry if this sounds complicated; we'll break it down with simple examples and analogies. Let's get started!

Reversible vs. Irreversible Reactions

Most reactions you've learned about so far go in one direction. When you burn a piece of paper, it turns into ash. You can't easily turn the ash back into paper. This is an irreversible reaction.

Example: Burning magnesium in air. \(2\text{Mg}(s) + \text{O}_2(g) \rightarrow 2\text{MgO}(s)\)

However, many chemical reactions are two-way streets. The products can react with each other to re-form the original reactants. These are called reversible reactions. We use a special double arrow (\(\rightleftharpoons\)) to show this.

The reaction going from left to right is called the forward reaction.
The reaction going from right to left is called the reverse reaction (or backward reaction).

Example: Heating hydrated copper(II) sulphate. The blue crystals turn into a white powder when heated, but if you add water to the white powder, it turns blue again!
\(\text{CuSO}_4 \cdot 5\text{H}_2\text{O}(s) \rightleftharpoons \text{CuSO}_4(s) + 5\text{H}_2\text{O}(g)\)
(Blue hydrated solid) \(\rightleftharpoons\) (White anhydrous solid) + (Water vapour)

Key Takeaway

Irreversible reactions go in one direction (→). Reversible reactions can go in both directions (\(\rightleftharpoons\)), with a forward and a reverse reaction happening at the same time.

What is Dynamic Equilibrium?

Imagine a busy shop. People are constantly entering, and other people are constantly leaving. If the number of people entering per minute is the same as the number of people leaving per minute, the total number of people inside the shop stays constant. From the outside, it might look like nothing is changing, but inside, there is constant movement. This is a great analogy for dynamic equilibrium!

In a reversible reaction, when the rate of the forward reaction becomes equal to the rate of the reverse reaction, the system is in a state of dynamic equilibrium.

Dynamic: This means the reactions haven't stopped! Both forward and reverse reactions are still happening.
Equilibrium: This means there is no overall change. The amounts (concentrations) of reactants and products remain constant because they are being formed at the same rate they are being used up.

Characteristics of Dynamic Equilibrium

For a system to be in dynamic equilibrium, it must have these features:

The rates of the forward and reverse reactions are equal.
The concentrations of all reactants and products are constant.
It can only happen in a closed system, where no substances can enter or leave.
Macroscopic properties (like colour, pressure, concentration) do not change over time.

Quick Review: Important Distinction!

Common Mistake: Thinking that the reaction has stopped at equilibrium.
Correction: The reactions are still going! They are just perfectly balanced, like two equally matched tug-of-war teams pulling with the same force.

Key Takeaway

Dynamic equilibrium is a state in a reversible reaction where the forward and reverse reaction rates are equal, resulting in constant concentrations of reactants and products. It's a state of balance, not inactivity.

The Equilibrium Constant (Kc)

At equilibrium, the concentrations of reactants and products are constant, but that doesn't mean they are equal. Sometimes there are lots of products and very few reactants, and sometimes it's the other way around. The equilibrium constant (Kc) is a value that quantifies the position of the equilibrium.

How to Write the Kc Expression

For a general reaction:

\(a\text{A}(aq) + b\text{B}(aq) \rightleftharpoons c\text{C}(aq) + d\text{D}(aq)\)

The equilibrium constant expression is written as:

\(K_c = \frac{[\text{C}]^c[\text{D}]^d}{[\text{A}]^a[\text{B}]^b}\)

Let's break that down:

The square brackets [ ] mean "equilibrium concentration of" in \(\text{mol dm}^{-3}\).
The concentrations of the products always go on top (numerator).
The concentrations of the reactants always go on the bottom (denominator).
The stoichiometric coefficients (the numbers a, b, c, d from the balanced equation) become the powers for each concentration.

VERY IMPORTANT RULE: We DO NOT include pure solids (s) or pure liquids (l) in the \(K_c\) expression. This is because their concentration (density) is essentially constant throughout the reaction.

Example: For the reaction \(\text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g)\), the expression is:
\(K_c = \frac{[\text{NH}_3]^2}{[\text{N}_2][\text{H}_2]^3}\)
Units of \(K_c\): \(\frac{(\text{mol dm}^{-3})^2}{(\text{mol dm}^{-3})(\text{mol dm}^{-3})^3} = \text{mol}^{-2}\text{dm}^6\)

Units of Kc in HKDSE

Unlike some other constants, \(K_c\) does not have fixed units. Its unit depends on the expression:
- If total powers of products equal total powers of reactants, \(K_c\) has no unit (it is dimensionless).
- Otherwise, substitute \(\text{mol dm}^{-3}\) for each concentration term to work out the derived units.

What Does the Value of Kc Tell Us?

If Kc is large (e.g., > 1000): The numerator is much larger than the denominator. At equilibrium, products predominate. We say the equilibrium lies to the right.
If Kc is small (e.g., < 0.001): The denominator is much larger. At equilibrium, reactants predominate. We say the equilibrium lies to the left.
If Kc is around 1: There are significant amounts of both reactants and products present at equilibrium.

Remember: The numerical value of Kc for a given reaction changes ONLY when the temperature changes.

Equilibrium Calculations (ICE Table)

In HKDSE examinations, you often calculate \(K_c\) from initial amounts using an ICE table (Initial, Change, Equilibrium).

Example: In a \(1.0\text{ dm}^3\) closed vessel, \(0.50\text{ mol}\) of \(\text{H}_2(g)\) and \(0.50\text{ mol}\) of \(\text{I}_2(g)\) are heated. At equilibrium, \(0.80\text{ mol}\) of \(\text{HI}(g)\) is formed. Calculate \(K_c\).

\(\text{H}_2(g) + \text{I}_2(g) \rightleftharpoons 2\text{HI}(g)\)

Initial: \([\text{H}_2] = 0.50\text{ M}\), \([\text{I}_2] = 0.50\text{ M}\), \([\text{HI}] = 0.00\text{ M}\)
Change: \(-0.40\text{ M}\) for \(\text{H}_2\), \(-0.40\text{ M}\) for \(\text{I}_2\), \(+0.80\text{ M}\) for \(\text{HI}\)
Equilibrium: \([\text{H}_2] = 0.10\text{ M}\), \([\text{I}_2] = 0.10\text{ M}\), \([\text{HI}] = 0.80\text{ M}\)

Substitute into the expression:
\(K_c = \frac{[\text{HI}]^2}{[\text{H}_2][\text{I}_2]} = \frac{(0.80)^2}{(0.10)(0.10)} = 64\) (no unit)

Key Takeaway

Kc indicates the extent of reaction at equilibrium. Its value depends only on temperature, and its units must be derived directly from the balanced equilibrium expression.

Changing the Position of Equilibrium (Le Châtelier's Principle)

Chemical engineers often want to maximize the yield of desired products. They do this by modifying system conditions. The response of the system is governed by Le Châtelier's Principle.

Le Châtelier's Principle states: If a system at dynamic equilibrium is subjected to a change in condition (concentration, temperature, or pressure), the position of equilibrium shifts in a direction that tends to counteract the change.

1. The Effect of Changing Concentration

Changing concentration alters the position of equilibrium, but does NOT change the value of Kc.

Consider the reaction: \(\text{Fe}^{3+}(aq) + \text{SCN}^-(aq) \rightleftharpoons [\text{Fe(SCN)}]^{2+}(aq)\)
(Pale yellow) + (Colourless) \(\rightleftharpoons\) (Blood-red)

Change: Add more \(\text{Fe}^{3+}(aq)\).
System's response: Tries to consume the added \(\text{Fe}^{3+}\).
Result: Shifts to the right (forward reaction favoured; solution turns deeper red).

Change: Remove \([\text{Fe(SCN)}]^{2+}(aq)\).
System's response: Tries to replace the removed product.
Result: Shifts to the right.

Change: Add product \([\text{Fe(SCN)}]^{2+}(aq)\).
System's response: Tries to consume excess product.
Result: Shifts to the left (reverse reaction favoured).

2. The Effect of Changing Pressure (or Volume) in Gaseous Systems

Pressure changes affect systems containing gases where there is a difference in total moles of gas between reactants and products. This does NOT change \(K_c\).

Consider: \(2\text{NO}_2(g) \text{ (brown)} \rightleftharpoons \text{N}_2\text{O}_4(g) \text{ (colourless)}\)
(Reactant side = 2 moles of gas; Product side = 1 mole of gas)

Change: Increase pressure (decrease volume).
System's response: Tries to lower pressure by reducing the total number of gas molecules.
Result: Shifts to the side with fewer moles of gas (shifts to the right; colour pales).

Change: Decrease pressure (increase volume).
System's response: Tries to increase pressure by producing more gas molecules.
Result: Shifts to the side with more moles of gas (shifts to the left; colour darkens).

Note: If the number of moles of gas is the same on both sides (e.g., \(\text{H}_2(g) + \text{I}_2(g) \rightleftharpoons 2\text{HI}(g)\)), changing pressure does not shift the equilibrium position.

3. The Effect of Changing Temperature

Temperature is the ONLY factor that changes the numerical value of \(K_c\).

For an EXOTHERMIC forward reaction (\(\Delta H < 0\)):
\(\text{A} + \text{B} \rightleftharpoons \text{C} + \text{heat}\)

Increase Temperature: Equilibrium shifts to the left (endothermic direction to absorb heat). Product concentration decreases, so Kc decreases.
Decrease Temperature: Equilibrium shifts to the right (exothermic direction to release heat). Product concentration increases, so Kc increases.

For an ENDOTHERMIC forward reaction (\(\Delta H > 0\)):
\(\text{A} + \text{B} + \text{heat} \rightleftharpoons \text{C}\)

Increase Temperature: Equilibrium shifts to the right (endothermic direction). Product concentration increases, so Kc increases.
Decrease Temperature: Equilibrium shifts to the left. Product concentration decreases, so Kc decreases.

4. The Role of a Catalyst

A catalyst increases the rates of both the forward and reverse reactions equally by providing an alternative pathway with a lower activation energy. Consequently, a catalyst allows equilibrium to be reached faster, but it does not change the equilibrium position or the value of Kc.

Industrial Application: The Haber Process

In industry, chemical engineers must balance equilibrium yield and reaction rate to make processes economically viable.

\(\text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g) \quad \Delta H = -92\text{ kJ mol}^{-1}\)

Temperature: Low temperature favours a higher equilibrium yield of \(\text{NH}_3\) (exothermic reaction), but causes the reaction rate to be too slow. A compromise temperature (around 450 °C) is used to ensure a reasonable rate with acceptable yield.
Pressure: High pressure favours ammonia production (4 moles of gas \(\rightarrow\) 2 moles of gas) and increases rate. A moderate-high pressure (around 200 atm) is chosen due to equipment cost and safety constraints.
Catalyst: Finely divided iron (Fe) catalyst speeds up the reaction to operate at moderate temperatures.

Key Takeaway

Le Châtelier's Principle predicts how equilibria respond to concentration, pressure, and temperature changes. In chemical industry, operating conditions represent a compromise between equilibrium yield, reaction rate, equipment cost, and plant safety.

Quick check

Can you answer these now?

Open each question to check the key ideas from this chapter.

What is the difference between an irreversible reaction and a reversible reaction?

An irreversible reaction proceeds in only one direction, while a reversible reaction can proceed in both forward and backward directions simultaneously (egin{equation*}\rightleftharpoons\end{equation*}).

What defines a chemical system at dynamic equilibrium?

A state in a closed system where the rate of the forward reaction equals the rate of the reverse reaction, keeping all reactant and product concentrations constant.

How is the equilibrium constant (egin{equation*}K_c\end{equation*}) expression constructed for a general reaction?

egin{equation*}K_c = \frac{[\text{C}]^c[\text{D}]^d}{[\text{A}]^a[\text{B}]^b}\end{equation*}, excluding pure solids and pure liquids.

What determines the units of the equilibrium constant egin{equation*}K_c\end{equation*}?

The units depend on the stoichiometry of the expression; if the total powers of products equal those of reactants, egin{equation*}K_c\end{equation*} has no unit; otherwise, substitute egin{equation*}\text{mol dm}^{-3}\end{equation*} into each term to derive them.

What does the numerical magnitude of egin{equation*}K_c\end{equation*} indicate about the position of equilibrium?

A large egin{equation*}K_c\end{equation*} (egin{equation*}> 1000\end{equation*}) means products predominate (lies to the right), while a small egin{equation*}K_c\end{equation*} (egin{equation*}< 0.001\end{equation*}) means reactants predominate (lies to the left).

What is Le Châtelier's Principle?

If a system at dynamic equilibrium is subjected to a change in conditions, the equilibrium position shifts in a direction that counteracts the change.

How does changing reactant or product concentration affect the position of equilibrium and egin{equation*}K_c\end{equation*}?

Adding a species shifts the equilibrium to consume it, while removing a species shifts it to replenish it; egin{equation*}K_c\end{equation*} remains unchanged.

How does changing pressure affect the position of an equilibrium involving gases?

Increasing pressure shifts the equilibrium towards the side with fewer moles of gas, while decreasing pressure shifts it towards the side with more moles of gas.

How does changing temperature affect the position of equilibrium and the value of egin{equation*}K_c\end{equation*}?

Increasing temperature shifts equilibrium in the endothermic direction (increasing egin{equation*}K_c\end{equation*} if endothermic, decreasing egin{equation*}K_c\end{equation*} if exothermic); temperature is the only factor that alters egin{equation*}K_c\end{equation*}.

What is the effect of adding a catalyst to a system at dynamic equilibrium?

A catalyst increases the rates of both forward and reverse reactions equally, allowing equilibrium to be reached faster without changing the position of equilibrium or egin{equation*}K_c\end{equation*}.

Why are compromise conditions (approx. 450 °C and 200 atm) used in the industrial Haber Process?

They strike an optimal balance between acceptable equilibrium yield (egin{equation*}\text{NH}_3\end{equation*}), economical reaction rate, safety, and equipment manufacturing costs.

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