Equilibrium Constant Expressions — Write equilibrium constant expressions (Kc, Kp) from balanced equations - Chemistry 2

SECTION 1

Historical Context & Motivation

For centuries, chemists assumed that every reaction ran to completion — that reactants would be entirely consumed and only products would remain. But careful experiments in the 1800s revealed something puzzling: many reactions seemed to stop before finishing. Reactants and products coexisted in the same container, and no matter how long you waited, the mixture didn't change any further. Understanding why this happened — and how to predict the final mixture — became one of the most important questions in chemistry.

1803 • Berthollet's Observation

French chemist Claude Louis Berthollet observed that some reactions in Egypt's soda lakes appeared to run in reverse. He proposed that the amounts of reactants and products could influence the direction of a reaction — an early hint of equilibrium.

1864 • Guldberg & Waage's Law of Mass Action

Norwegian scientists Cato Guldberg and Peter Waage formulated the law of mass action, stating that the rate of a reaction is proportional to the concentrations of the reactants raised to their stoichiometric coefficients. This became the foundation for equilibrium constant expressions.

1884 • Le Châtelier's Principle

Henry Le Châtelier published his famous principle describing how equilibrium systems respond to disturbances. This work reinforced the idea that equilibrium is a dynamic, predictable state.

1901 • van 't Hoff's Nobel Prize

Jacobus van 't Hoff won the first Nobel Prize in Chemistry partly for his work connecting thermodynamics to chemical equilibrium. He showed how temperature affects equilibrium constants, linking K to the energy of reactions.

The central question that emerged from this history was straightforward: if a reaction reaches a balance point, can we write a single number that tells us how far the reaction proceeds toward products? That number is the equilibrium constant, and learning to write its expression from a balanced equation is the skill you'll master in this lesson.

SECTION 2

Core Principles & Definitions

Before writing any expression, you need a clear grasp of four key ideas that form the backbone of equilibrium constant expressions. Each principle builds on the last, so take them in order.

Dynamic Equilibrium

At equilibrium, the forward and reverse reactions continue at equal rates. The concentrations of reactants and products stay constant — not because nothing is happening, but because changes cancel out.

Products Over Reactants

The equilibrium constant expression always places product concentrations in the numerator and reactant concentrations in the denominator. This convention is universal and must be followed exactly.

Coefficients Become Exponents

The stoichiometric coefficients from the balanced equation become the exponents (powers) on each concentration or pressure term. A coefficient of 2 means that term is squared.

Only Gases and Aqueous Species

Pure solids and pure liquids are excluded from equilibrium expressions. Their concentrations are effectively constant and are already absorbed into the value of K.

✦

KEY TAKEAWAY

Think of an equilibrium expression like a recipe review score. The products are what you made (numerator), and the reactants are what you started with (denominator). A large K means you produced a lot relative to what you used — the reaction favors products. A small K means most of your starting materials are still sitting there — the reaction favors reactants. The balanced equation is your recipe, and the coefficients tell you how heavily to weight each ingredient.

ℹ

Kc vs. Kp

K_c uses molar concentrations (mol/L, shown in brackets [ ]) and applies to reactions in solution or any system where concentrations are convenient. K_p uses partial pressures (in atm) and is used for gas-phase reactions. Both follow the same products-over-reactants structure.

SECTION 3

Visual Explanation

The following diagram illustrates exactly how a balanced chemical equation maps onto an equilibrium constant expression. Follow the color-coded arrows to see how each part of the equation translates into the K expression.

The balanced equation's reactants (pink) go in the denominator, products (cyan) go in the numerator, and coefficients (amber) become exponents on each term.

Notice how the diagram traces each element of the balanced equation to its position in the K_c expression. The species on the right side of the equilibrium arrow (products C and D) are always placed in the numerator, while the species on the left side (reactants A and B) are placed in the denominator. The dashed amber arrows show how each coefficient in the balanced equation becomes the exponent on that species' concentration. This same pattern applies every single time, no matter how complex the equation.

SECTION 4

Mathematical Framework

Now let's formalize the expressions you'll write. There are two forms of the equilibrium constant — K_c for concentrations and K_p for partial pressures — and an equation that connects them.

EQUILIBRIUM CONSTANT (Kc)

Kc = [C]ᶜ × [D]ᵈ / [A]ᵃ × [B]ᵇ

For the reaction aA + bB ⇌ cC + dD, where [X] = molar concentration of species X (mol/L), and the superscript letters are the stoichiometric coefficients from the balanced equation. Only gaseous (g) and aqueous (aq) species are included.

EQUILIBRIUM CONSTANT (Kp)

Kp = (P_C)ᶜ × (P_D)ᵈ / (P_A)ᵃ × (P_B)ᵇ

For gas-phase reactions, where P_X = partial pressure of gas X (usually in atm). The structure is identical to K_c — products over reactants, coefficients as exponents — but pressures replace concentrations.

RELATING Kp AND Kc

Kp = Kc × (RT)^Δn

R = ideal gas constant (0.0821 L·atm/mol·K), T = temperature in Kelvin, and Δn = (moles of gaseous products) − (moles of gaseous reactants). When Δn = 0, K_p = K_c.

⚠

Remember: Solids & Liquids Are Omitted

Pure solids (s) and pure liquids (l) do not appear in equilibrium expressions. For example, in CaCO₃(s) ⇌ CaO(s) + CO₂(g), the K_c expression is simply K_c = [CO₂]. Both CaCO₃ and CaO are pure solids and are excluded.

SECTION 5

Homogeneous vs. Heterogeneous Equilibria

Equilibrium systems are classified as either homogeneous or heterogeneous depending on the phases of the species involved. In a homogeneous equilibrium, all reactants and products are in the same phase — for example, all gases or all dissolved in solution. In a heterogeneous equilibrium, the reactants and products exist in two or more different phases, such as a solid reacting with a gas. The distinction matters because it determines which species appear in your K expression.

In homogeneous equilibria (left), every species appears in the expression. In heterogeneous equilibria (right), pure solids and liquids are omitted, which can dramatically simplify the K expression.

As you can see, the heterogeneous example simplifies dramatically. Because calcium carbonate and calcium oxide are both pure solids, their concentrations are essentially fixed — they don't change as the reaction proceeds. Chemists account for this by folding those constant values into K itself. The result is a much cleaner expression that only includes the species whose concentrations actually change: gases and dissolved (aqueous) species.

💡

Quick Check: Which Species to Include

Look at the state symbol after each species: (g) = gas → include it. (aq) = aqueous → include it. (s) = solid → leave it out. (l) = liquid → leave it out. If water is the solvent (l), omit it. If water appears as steam (g), include it.

SECTION 6

Worked Example

Let's walk through writing K_c and K_p expressions for a real reaction, step by step. We'll use the industrial synthesis of sulfur trioxide.

Writing Kc and Kp for the Synthesis of SO₃

1
Step 1 — Write the Balanced Equation

The balanced equation for the contact process step is: 2SO₂(g) + O₂(g) ⇌ 2SO₃(g). All species are gases, so this is a homogeneous equilibrium.

2SO₂(g) + O₂(g) ⇌ 2SO₃(g) — all gases, all included
2
Step 2 — Identify Products and Reactants

Products (right side of ⇌): SO₃ with a coefficient of 2. Reactants (left side of ⇌): SO₂ with a coefficient of 2, and O₂ with a coefficient of 1.

Product: SO₃ (coeff. 2) | Reactants: SO₂ (coeff. 2), O₂ (coeff. 1)
3
Step 3 — Write the Kc Expression

Place products in the numerator and reactants in the denominator. Each concentration is raised to the power of its coefficient. K_c = [SO₃]² / ([SO₂]² × [O₂])

Kc = [SO₃]² / ([SO₂]² × [O₂])
4
Step 4 — Write the Kp Expression

Since all species are gases, we can also write K_p by replacing each concentration with the corresponding partial pressure. K_p = (P_SO₃)² / ((P_SO₂)² × P_O₂)

Kp = (P_SO₃)² / ((P_SO₂)² × $P_O$ ₂)
5
Step 5 — Determine Δn for the Kp–Kc Relationship

Δn = (moles of gaseous products) − (moles of gaseous reactants) = 2 − (2 + 1) = 2 − 3 = −1. This means K_p = K_c × (RT)⁻¹, so K_p ≠ K_c for this reaction.

Δn = −1 → Kp = Kc / (RT)

SECTION 7

Common Mistakes & Pitfalls

Writing equilibrium expressions is a systematic process, but there are several common mistakes that can trip you up. The table below highlights the most frequent errors, explains why they're wrong, and shows the correct approach.

Common Mistake	Why It's Wrong	Correct Approach
Putting reactants in the numerator	By convention, products always go on top and reactants on the bottom. Reversing this gives you 1/K, not K.	Always write products over reactants — think "P over R".
Using coefficients as multipliers instead of exponents	Writing 2[NH₃] instead of [NH₃]² gives a completely different value.	Coefficients become exponents (powers), never multipliers in front of brackets.
Including solids or pure liquids	Pure solids and liquids have constant concentrations that are already built into K.	Check the state symbol: only (g) and (aq) species go into K.
Using an unbalanced equation	The coefficients determine the exponents, so wrong coefficients produce a wrong expression.	Always balance the equation first before writing K.
Confusing Kc and Kp units	Kc uses mol/L concentrations; Kp uses atm pressures. Mixing them gives meaningless results.	Use brackets [ ] for Kc and P for Kp. Convert between them using Kp = Kc(RT)^Δn.

Five common pitfalls when writing equilibrium constant expressions

✦

KEY TAKEAWAY

Think of writing a K expression like filling out a form. The balanced equation is your source document, and there's a specific place for each piece of information. Products go in the numerator, reactants go in the denominator, and coefficients become exponents. If you follow the form exactly, you'll never make an error — just like filling out a tax form correctly if you read the instructions.

SECTION 8

Connecting K to Thermodynamics & Advanced Theory

The equilibrium constant you're writing today is actually a doorway to much deeper chemistry. In more advanced courses, you'll discover that K is directly linked to the Gibbs free energy of a reaction through the equation ΔG° = −RT ln K. A large K (favoring products) corresponds to a negative ΔG° (spontaneous reaction), while a small K (favoring reactants) corresponds to a positive ΔG°. This connection reveals that the equilibrium constant isn't just an empirical ratio — it's rooted in the fundamental thermodynamics of the system.

Concept	What You Know Now	What Comes Next
K expression	Write Kc and Kp from balanced equations using products over reactants	Use K with ICE tables to calculate equilibrium concentrations
Magnitude of K	Large K → products favored; Small K → reactants favored	Relate K to ΔG° to predict spontaneity quantitatively
Reaction quotient Q	K describes the system at equilibrium	Q uses the same expression but with non-equilibrium concentrations; compare Q to K to predict shift direction
Temperature dependence	K is constant at a given temperature	van 't Hoff equation relates K at different temperatures to ΔH°

How today's lesson connects to advanced equilibrium topics

For now, the critical skill is writing the expression correctly. Every future calculation — whether finding unknown concentrations, predicting which direction a reaction shifts, or connecting equilibrium to energy — starts with a correctly written K expression. Master this step, and the rest of equilibrium chemistry becomes much more manageable.

SECTION 9

Practice Problems

Test your understanding with these five problems, arranged from conceptual to challenging. Try each one before checking the answer.

Question 1: CONCEPTUAL

For the reaction H₂(g) + I₂(g) ⇌ 2HI(g), a student writes K_c = [H₂][I₂] / [HI]². Identify the mistake and write the correct expression.

Show answer

The student placed reactants in the numerator and the product in the denominator, which is the inverse of the correct arrangement. In an equilibrium expression, products always go on top. The correct expression is K_c = [HI]² / ([H₂] × [I₂]). What the student actually wrote is the expression for the reverse reaction, which equals 1/K_c.

Question 2: BASIC CALCULATION

Write both the K_c and K_p expressions for the following reaction: PCl₅(g) ⇌ PCl₃(g) + Cl₂(g). Then determine Δn.

Show answer

Products: PCl₃ (coefficient 1) and Cl₂ (coefficient 1). Reactant: PCl₅ (coefficient 1). K_c = [PCl₃][Cl₂] / [PCl₅]. K_p = (P_PCl₃)(P_Cl₂) / P_PCl₅. Δn = (1 + 1) − 1 = +1. Since Δn ≠ 0, K_p ≠ K_c for this reaction.

Question 3: INTERMEDIATE

Write the K_c expression for the heterogeneous reaction: Fe₂O₃(s) + 3CO(g) ⇌ 2Fe(s) + 3CO₂(g). Explain which species are included and which are excluded.

Show answer

Fe₂O₃(s) is a solid — exclude it. CO(g) is a gas — include it. Fe(s) is a solid — exclude it. CO₂(g) is a gas — include it. The expression uses only the gas species: K_c = [CO₂]³ / [CO]³. Notice that the coefficient 3 on CO becomes the exponent 3 in both the numerator and denominator. You could also simplify this as K_c = ([CO₂]/[CO])³, though both forms are correct.

Question 4: APPLIED

The Haber process synthesizes ammonia: N₂(g) + 3H₂(g) ⇌ 2NH₃(g). Write K_p for this reaction. If someone gives you the K_p for the reverse reaction (decomposition of NH₃), how would the two K_p values be related?

Show answer

K_p = (P_NH₃)² / (P_N₂ × (P_H₂)³). The reverse reaction is 2NH₃(g) ⇌ N₂(g) + 3H₂(g). Its K_p expression is the reciprocal: K_p,reverse = (P_N₂ × (P_H₂)³) / (P_NH₃)² = 1/K_p,forward. Reversing a reaction inverts the equilibrium constant.

Question 5: CRITICAL THINKING

Consider the reaction N₂(g) + 3H₂(g) ⇌ 2NH₃(g) with K_c = 0.50 at a certain temperature. If the equation is rewritten as ½N₂(g) + ³⁄₂H₂(g) ⇌ NH₃(g), what is the new K_c'? Explain the relationship.

Show answer

The rewritten equation is exactly half of the original equation — every coefficient has been divided by 2. When you multiply all coefficients in an equation by a factor n, the new K equals the original K raised to the power n. Here, n = ½, so K_c' = (K_c)^(1/2) = √(0.50) ≈ 0.71. The new expression is K_c' = [NH₃] / ([N₂]^(1/2) × [H₂]^(3/2)). This shows that K depends on how you write the balanced equation, which is why you must always specify which equation your K refers to.

SUMMARY

Lesson Summary

Summary

The equilibrium constant expression is built directly from a balanced chemical equation by placing product concentrations in the numerator and reactant concentrations in the denominator. Each term is raised to the power of its stoichiometric coefficient. K_c uses molar concentrations ([ ]) while K_p uses partial pressures, but both follow the same products-over-reactants structure.

In heterogeneous equilibria, pure solids and pure liquids are excluded from the expression because their concentrations are constant. The two constants are related by Kp = Kc(RT)^Δn, where Δn is the change in moles of gas. A large K indicates that products are favored at equilibrium, while a small K indicates that reactants are favored. Reversing a reaction inverts K, and multiplying coefficients by a factor raises K to that power.