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

SECTION 1

Historical Context & Motivation

The concept of chemical equilibrium arose from a fundamental puzzle: why do many reactions appear to stop before all reactants are consumed? By the mid-nineteenth century, chemists had observed that certain reactions seemed to reach a stationary state in which both reactants and products coexisted indefinitely. This observation defied the prevailing assumption that all reactions proceed to completion, and it demanded a quantitative framework capable of predicting the final composition of a reacting mixture. The development of the equilibrium constant provided exactly that framework, transforming equilibrium from a qualitative curiosity into a predictive, calculable quantity.

1864 • Guldberg & Waage — Law of Mass Action

Norwegian chemists Cato Guldberg and Peter Waage proposed that the rate of a reaction is proportional to the product of the concentrations of the reactants, each raised to a power. Their law of mass action laid the groundwork for expressing equilibrium as a ratio of product to reactant concentrations.

1884 • Le Châtelier's Principle

Henri Le Châtelier articulated that a system at equilibrium, when subjected to a perturbation, will shift to partially counteract that change. This principle provided a qualitative tool for predicting equilibrium shifts, complementing the quantitative constant.

1886 • Van 't Hoff's Thermodynamic Treatment

Jacobus Henricus van 't Hoff connected the equilibrium constant to thermodynamic quantities, demonstrating that K depends on temperature through the relationship involving the standard Gibbs free energy change. This linked the constant to the energetic favorability of a reaction.

1901 • Nobel Prize to Van 't Hoff

Van 't Hoff received the first Nobel Prize in Chemistry, in part for his work on chemical equilibrium and reaction kinetics, cementing the equilibrium constant as a central concept in physical chemistry.

1923 • Lewis & Randall — Fugacity and Activity

Gilbert N. Lewis and Merle Randall formalized the use of activities and fugacities in equilibrium expressions, extending the equilibrium constant framework to non-ideal systems encountered in real-world chemistry.

The central question that motivated this entire body of work can be stated simply: given a balanced chemical equation, how can we write a single expression whose numerical value tells us precisely where the equilibrium lies—whether products or reactants dominate the mixture at a given temperature? The answer is the equilibrium constant expression, and mastering how to construct it from a balanced equation is the focus of this lesson.

SECTION 2

Core Principles & Definitions

Before writing any equilibrium constant expression, it is essential to internalize several foundational ideas. A reversible reaction is one in which both the forward and reverse processes occur simultaneously. At dynamic equilibrium, the rates of the forward and reverse reactions are equal, meaning the macroscopic concentrations of all species remain constant even though molecular-level transformations continue unceasingly. The equilibrium constant is a dimensionless number (in the thermodynamic sense, when activities are used) that encapsulates the composition of the system at equilibrium for a particular reaction at a specific temperature.

Products Over Reactants

Equilibrium constant expressions are always written as the concentrations (or partial pressures) of products in the numerator divided by reactants in the denominator. This convention is universal and non-negotiable.

Stoichiometric Coefficients as Exponents

Each concentration or partial pressure term is raised to the power equal to its stoichiometric coefficient in the balanced equation. Doubling all coefficients squares the value of K.

Only Gases and Aqueous Species Appear

Pure solids and pure liquids have activities of 1 and are omitted from the expression. Only species in the gas phase (g) or aqueous solution (aq) contribute variable terms.

Kc vs. Kp

K_c uses molar concentrations (mol/L), while K_p uses partial pressures (typically in atm or bar). K_p applies only to reactions involving gases.

Temperature Dependence

The numerical value of K changes with temperature but is independent of initial concentrations, catalysts, or pressure changes. A catalyst speeds attainment of equilibrium without altering K.

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KEY TAKEAWAY

Think of the equilibrium constant as a scorecard for a reaction: if K is large (≫ 1), products dominate the mixture at equilibrium, much like a lopsided scoreboard favoring the home team. If K is small (≪ 1), reactants dominate, and the reaction barely proceeds. The expression itself is the formula for computing that score, and the stoichiometric coefficients determine how heavily each species is weighted—analogous to how different plays in a game might carry different point values.

SECTION 3

Visual Explanation — Constructing K from a Balanced Equation

The following diagram illustrates the systematic procedure for translating a balanced chemical equation into both the K_c and K_p expressions. It uses a generic reaction aA + bB ⇌ cC + dD to show how coefficients map to exponents and how the products-over-reactants convention is applied.

The diagram shows a balanced gas-phase equation at the top, with arrows leading to the K_c expression (left, cyan border) and the K_p expression (right, violet border). The golden box lists the three cardinal rules, and the green box at the bottom shows how K_p and K_c are interconverted.

Notice that the structure of both expressions is identical: a ratio of product terms to reactant terms, each raised to the appropriate stoichiometric power. The only difference is what quantity is substituted—molar concentrations in square brackets for K_c, or partial pressures for K_p. If a reaction involves species in different phases (heterogeneous equilibrium), remember to exclude any pure solid or pure liquid from the expression, since their activities are defined as unity.

SECTION 4

Mathematical Framework

The mathematical formalism behind equilibrium constants draws directly from the law of mass action and thermodynamics. For a general balanced equation, we define two constants and the equation that relates them. Understanding these equations allows you to convert between the two forms and to recognize how changes in stoichiometry affect the value of K.

EQUILIBRIUM CONSTANT (CONCENTRATION)

Kc = [C]ᶜ [D]ᵈ / ([A]ᵃ [B]ᵇ)

For the reaction aA + bB ⇌ cC + dD. Square brackets denote equilibrium molar concentrations in mol/L. Each concentration is raised to the power of its stoichiometric coefficient in the balanced equation.

EQUILIBRIUM CONSTANT (PRESSURE)

Kp = (P_C)ᶜ (P_D)ᵈ / ((P_A)ᵃ (P_B)ᵇ)

P denotes the equilibrium partial pressure of each gaseous species, typically in atmospheres (atm). This expression is used exclusively for gas-phase or mixed-phase reactions involving at least one gas.

Kp – Kc RELATIONSHIP

Kp = Kc × (RT)^Δn

R = 0.08206 L·atm/(mol·K), T = temperature in Kelvin, and Δn = (sum of moles of gaseous products) − (sum of moles of gaseous reactants). When Δn = 0, K_p = K_c.

THERMODYNAMIC CONNECTION

ΔG° = −RT ln K

The standard Gibbs free energy change is related to the equilibrium constant. A negative ΔG° corresponds to K > 1 (products favored), while a positive ΔG° corresponds to K < 1 (reactants favored). This equation bridges thermodynamics and equilibrium.

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Important Convention

If the reaction is reversed, the new K is the reciprocal (1/K) of the original. If the equation is multiplied by a factor n, the new K equals the original K raised to the nth power: K' = Kⁿ. These rules follow directly from the algebraic structure of the expression.

SECTION 5

Heterogeneous Equilibria & Special Cases

Many important equilibria involve species in more than one phase. Such reactions are called heterogeneous equilibria. The critical rule is that pure solids and pure liquids have a constant concentration (their density divided by molar mass, which does not change during the reaction) and are assigned an activity of 1. Consequently, they do not appear in the equilibrium expression. Only species whose concentrations can vary—gases and dissolved solutes—are included.

Four examples of equilibrium constant expressions demonstrating the omission rule. Pure solids (s) and pure liquids (l) are tagged red and excluded, while gases (g) and aqueous ions (aq) are tagged green and included. Each box shows the resulting K expression.

The four examples above cover the most commonly encountered scenarios. Example 1 (CaCO₃ decomposition) and Example 3 (AgCl dissolution) are classic heterogeneous cases where the K expression is remarkably simple because most species are pure solids. Example 2 (water decomposition) shows that even when the liquid is the solvent, it is excluded. Example 4 (ammonia synthesis) represents a homogeneous gas-phase equilibrium where all species appear in the expression. Carefully reading the phase labels in parentheses after each formula is the single most important step when writing K expressions for heterogeneous reactions.

SECTION 6

Worked Example

Let us apply the principles developed above to a concrete problem involving both K_c and K_p for the synthesis of sulfur trioxide.

Writing Kc and Kp, then Converting Between Them

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Step 1 — Write the Balanced Equation

The balanced equation for the contact process is: 2 SO₂(g) + O₂(g) ⇌ 2 SO₃(g). All species are gases, so this is a homogeneous equilibrium and every species appears in the expression.
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Step 2 — Write the Kc Expression

Place products in the numerator and reactants in the denominator, with stoichiometric coefficients as exponents.

K_c = [SO₃]² / ([SO₂]² [O₂])
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Step 3 — Write the Kp Expression

Replace each molar concentration with the corresponding partial pressure.

K_p = (P_SO₃)² / ((P_SO₂)² × P_O₂)
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Step 4 — Calculate Δn

Δn = (moles of gaseous products) − (moles of gaseous reactants) = 2 − (2 + 1) = 2 − 3 = −1.

Δn = −1
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Step 5 — Convert Kc to Kp

Using K_p = K_c × (RT)^Δn. Suppose K_c = 4.32 × 10² at 600 K. Then K_p = 4.32 × 10² × (0.08206 × 600)⁻¹ = 4.32 × 10² / 49.24 = 8.78.

K_p = 8.78 at 600 K

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Why Kp < Kc Here

Because Δn is negative (fewer moles of gas on the product side), multiplying by (RT)⁻¹ reduces the numerical value. When Δn > 0, K_p > K_c. When Δn = 0, the two are equal.

SECTION 7

Comparing Kc and Kp — When to Use Each

A common source of confusion for students is deciding whether to write K_c or K_p for a given problem. The choice depends on the information available and the phases of the species involved. The table below provides a concise comparison to guide your decision-making.

Feature	Kc	Kp
Quantity used	Molar concentration [mol/L]	Partial pressure [atm or bar]
Applicable phases	Aqueous and/or gas-phase species	Gas-phase species only
Best used when	Solution-phase data are given; reactions in aqueous solution	Gas-phase data (pressures) are given; all species are gases
Relationship	K_p = K_c × (RT)^Δn	K_c = K_p / (RT)^Δn
When Δn = 0	K_c = K_p	K_p = K_c
Units (in practice)	Often reported as dimensionless (when using standard state of 1 M)	Often reported as dimensionless (when using standard state of 1 atm or 1 bar)

Comparison of Kc and Kp

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KEY TAKEAWAY

K_c and K_p are two dialects of the same language. Choosing between them is like choosing between Celsius and Fahrenheit: both describe the same physical reality (the equilibrium position), but one may be more convenient given the data at hand. If a problem gives you partial pressures for a gas-phase reaction, write K_p directly. If it gives molarities or if aqueous species are involved, write K_c. And if you need to convert, the (RT)^Δn factor bridges the two.

SECTION 8

Connection to Thermodynamics & Advanced Equilibrium

The equilibrium constant expressions you have learned here serve as the entry point to a deeper thermodynamic treatment. In advanced coursework, you will encounter the thermodynamic equilibrium constant (K), which is defined in terms of activities rather than concentrations or pressures. For ideal solutions and ideal gases, activities reduce to molar concentrations and partial pressures divided by their standard-state values (1 M and 1 atm, respectively), making K_c and K_p excellent approximations. In non-ideal systems, however, activity coefficients must be applied.

Concept	General Chemistry Level	Physical Chemistry / Advanced Level
Equilibrium constant	K_c or K_p using concentrations or pressures	K defined via activities (a_i = γ_i × [i]/c°)
Non-ideal behavior	Ignored; ideal solutions and ideal gases assumed	Accounted for via fugacity (gases) and activity coefficients (solutions)
Temperature dependence	Acknowledged but not derived	Quantified via the van 't Hoff equation: d(ln K)/dT = ΔH°/RT²
Reaction quotient	Q compared to K to predict direction of shift	ΔG = ΔG° + RT ln Q; equilibrium when ΔG = 0 (Q = K)

General Chemistry vs. Advanced Treatment

Looking ahead, the equilibrium constant also connects to electrochemistry through the Nernst equation (E = E° − (RT/nF) ln Q), to acid-base chemistry through K_a and K_b, and to solubility through K_sp. All of these are special cases of the general equilibrium constant expression you are learning to write. Mastering K_c and K_p now provides the scaffolding on which these more specialized constants are built.

SECTION 9

Practice Problems

Question 1: CONCEPTUAL

Explain why pure liquid water (H₂O(l)) does not appear in the equilibrium constant expression for the reaction: CO₂(g) + H₂O(l) ⇌ H₂CO₃(aq). What would change if water were in the gas phase?

Show answer

Pure liquid water has an activity of 1 because its concentration is essentially constant (~55.5 M) and is incorporated into the value of the equilibrium constant itself. Therefore, it is omitted from the K expression, giving K_c = [H₂CO₃] / [CO₂]. If water were in the gas phase, H₂O(g), its partial pressure or molar concentration would be variable, and it would need to be included as a term in the denominator: K_c = [H₂CO₃] / ([CO₂][H₂O]). The key principle is that only species with variable activities appear in the expression.

Question 2: BASIC CALCULATION

Write the K_c and K_p expressions for the reaction: 4 NH₃(g) + 5 O₂(g) ⇌ 4 NO(g) + 6 H₂O(g). Then calculate Δn for this reaction.

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K_c = [NO]⁴ [H₂O]⁶ / ([NH₃]⁴ [O₂]⁵). K_p = (P_NO)⁴ (P_H₂O)⁶ / ((P_NH₃)⁴ (P_O₂)⁵). Δn = (4 + 6) − (4 + 5) = 10 − 9 = +1. Since all species are gaseous, all terms appear in both expressions. The positive Δn indicates that K_p = K_c × RT, making K_p numerically larger than K_c at any temperature above 0 K.

Question 3: INTERMEDIATE

For the reaction: 2 Fe(s) + 3 CO₂(g) ⇌ Fe₂O₃(s) + 3 CO(g), write the K_c expression. If K_c = 12.0 for this reaction, what is K_c for the reverse reaction? What is K_c if the equation is multiplied by 2?

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Fe(s) and Fe₂O₃(s) are pure solids and are omitted. K_c = [CO]³ / [CO₂]³. For the reverse reaction, K'_c = 1/K_c = 1/12.0 = 0.0833. When the equation is multiplied by 2, all coefficients double, so K''_c = K_c² = (12.0)² = 144. Notice that Δn = 3 − 3 = 0 for the original equation, so K_p would equal K_c = 12.0 for this system.

Question 4: APPLIED

In the Haber process, N₂(g) + 3 H₂(g) ⇌ 2 NH₃(g), K_c = 0.500 at 400 °C. Calculate K_p at this temperature. Use R = 0.08206 L·atm/(mol·K).

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First, convert temperature: T = 400 + 273 = 673 K. Next, calculate Δn = 2 − (1 + 3) = −2. Apply the conversion: K_p = K_c × (RT)^Δn = 0.500 × (0.08206 × 673)⁻² = 0.500 × (55.23)⁻² = 0.500 × (1/3050) = 0.500 × 3.279 × 10⁻⁴ = 1.64 × 10⁻⁴. The much smaller K_p value makes physical sense: because the reaction decreases the number of gas moles (Δn < 0), K_p is much smaller than K_c.

Question 5: CRITICAL THINKING

Consider two reactions at the same temperature: (I) A(g) ⇌ 2B(g), K_c = 4.0; (II) A(g) ⇌ 2B(g) + C(s), K_c = 4.0. A student claims the equilibrium concentrations of B must be identical in both systems because the K_c values are the same. Evaluate this claim.

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The student's claim is correct, but only if the initial concentration of A is the same in both experiments and no other equilibrium constraints differ. For reaction (I), K_c = [B]²/[A] = 4.0. For reaction (II), C is a pure solid and omitted, so the expression is also K_c = [B]²/[A] = 4.0—mathematically identical. Given the same initial conditions and the same algebraic form for K, the equilibrium concentrations solved from the ICE table would indeed be the same. However, this analysis assumes C(s) remains present at equilibrium (its activity stays 1). If all solid C were consumed, the system would no longer be at equilibrium as described by expression (II), and the reaction would be forced to completion. This subtlety—the requirement that a pure solid or liquid be physically present for the equilibrium to hold—is often overlooked but critical in heterogeneous equilibria.

SUMMARY

Summary — Equilibrium Constant Expressions

Summary

The equilibrium constant expression is constructed by placing the product concentrations (or partial pressures) in the numerator and the reactant concentrations (or partial pressures) in the denominator, each raised to the power of its stoichiometric coefficient from the balanced equation. Pure solids and pure liquids are omitted because their activities are unity. Two forms exist: Kc (using molar concentrations) and Kp (using partial pressures), related by Kp = Kc × (RT)^Δn.

Key manipulations include: reversing the equation inverts K, and multiplying coefficients by n raises K to the nth power. The value of K is temperature-dependent but independent of initial concentrations or the presence of a catalyst. A large K (≫ 1) indicates product-favored equilibrium, while a small K (≪ 1) indicates reactant-favored equilibrium. These expressions form the quantitative foundation for all equilibrium calculations in chemistry.