15.1 Equilibrium and Equilibrium Constants
PRE-HEALTH PREP
Equilibrium Constants
General Chemistry 15.1
Lesson Overview
Chemical equilibrium is a dynamic state in which the forward and reverse reaction rates are equal, resulting in constant but not necessarily equal concentrations of reactants and products. The equilibrium constant K is expressed as the ratio of product concentrations to reactant concentrations, each raised to their stoichiometric coefficients, with pure solids and liquids excluded from the expression. The magnitude of K indicates whether a reaction favors products (K >> 1) or reactants (K << 1), and the relationship Kp = Kc(RT)Δn connects the concentration-based and pressure-based forms of the constant.
Table of Contents
What Is Chemical Equilibrium?
In chemistry, equilibrium refers specifically to a dynamic equilibrium — a state in which the concentrations of reactants and products are no longer changing. This is a critical concept because it overturns a simplifying assumption from first-semester general chemistry: that reactions always proceed to completion until a limiting reagent is exhausted.
In reality, most reactions do not go to completion. Instead, they reach an intermediate state — equilibrium — at which both forward and reverse reactions continue to occur, but at equal rates.
Common Misconception: "Equilibrium" does not mean the reaction has stopped. The forward and reverse reactions are both still happening; they are simply occurring at the same rate, so the net concentrations remain constant. This is what distinguishes a dynamic equilibrium from a static one.
Equilibrium Does Not Mean Equal Concentrations
One of the most persistent misconceptions students bring into this topic is the assumption that equilibrium means equal concentrations of reactants and products. This is not the case. At equilibrium, the rates are equal — not the concentrations. Depending on the reaction:
- There may be far more products than reactants.
- There may be far more reactants than products.
- There may be a significant amount of both.
The defining condition of equilibrium is that the forward rate of reaction equals the reverse rate of reaction. The resulting concentrations of reactants and products can fall anywhere on the spectrum.
The Equilibrium Constant (K)
Every equilibrium has an associated equilibrium constant, symbolized by a capital K. This is an important notational distinction: in the previous chapter on kinetics, a lowercase k represented the rate constant. Do not confuse the two.
The equilibrium constant K is affected by exactly one thing: temperature. Unlike the rate constant, K does not change when a catalyst is added, when concentrations are adjusted, or when the activation energy is altered. Only a change in temperature will produce a different K value. For this reason, equilibrium constants are always reported at a specified temperature.
Key Distinction — K vs. k: The equilibrium constant (K, uppercase) depends only on temperature. The rate constant (k, lowercase) depends on both temperature and activation energy. Do not confuse these two symbols.
How to Write an Equilibrium Constant Expression
The equilibrium constant expression — also known as the Law of Mass Action — is written as the ratio of the molar concentrations of products to the molar concentrations of reactants, with each concentration raised to the power of its stoichiometric coefficient. Square brackets denote molar concentration.
For a generic reaction of the form aA + bB ⇌ cC + dD, the equilibrium expression is:
General Form:
K = [C]ᶜ [D]ᶝ / [A]ᵃ [B]ᵇ
Products appear in the numerator; reactants appear in the denominator. Each concentration is raised to the power of its coefficient in the balanced equation.
Which Species Appear in the Equilibrium Expression?
Only aqueous (aq) and gaseous (g) species are included in the equilibrium constant expression. Pure solids and pure liquids (when acting as the solvent) are omitted.
The reason for this exclusion is rooted in thermodynamics: solids and pure liquids have an activity of exactly one. Because multiplying or dividing by one does not change the value of an expression, these terms drop out. In general chemistry, this rule is simplified to: leave out solids and liquids, keep aqueous and gaseous species.
Rule to Remember: Pure solids (s) and pure liquids acting as solvent (l) are excluded from equilibrium expressions. Only aqueous (aq) and gaseous (g) species are included.
Example: Carbon Combustion
For the reaction 2C(s) + O₂(g) ⇌ 2CO(g), carbon is a solid and is excluded. The equilibrium expression becomes:
K = [CO]² / [O₂]
Only the gaseous species appear. The coefficient of 2 on CO becomes the exponent in the expression.
Kc vs. Kp: 2 Ways to Express K for Gas-Phase Reactions
For reactions involving gases, concentrations can be expressed either as molar concentrations or as partial pressures. This leads to two distinct equilibrium constants:
- Kc — the equilibrium constant expressed in terms of molar concentrations (mol/L)
- Kp — the equilibrium constant expressed in terms of partial pressures (atm)
These two quantities are generally not numerically equal, because molar concentration and pressure are related by the ideal gas law rather than being identical. Starting from PV = nRT and solving for pressure gives P = (n/V)RT = [M]RT, where [M] is the molar concentration and RT is the extra factor that distinguishes the two.
A Note on "Keq"
In general chemistry courses and textbooks, you will often see the generic term Keq used without specifying whether molar concentrations or partial pressures are intended. In most contexts, Keq defaults to Kc. Technically, Keq is an informal designation that would not appear in a rigorous physical chemistry course. When writing equilibrium constants yourself, prefer Kc or Kp to be precise.
Important: Kp and Kc only exist for reactions involving gaseous species. If all species in a reaction are aqueous, only Kc applies — there is no Kp for purely aqueous reactions.
Concentrations Must Be Measured at Equilibrium
By definition, the concentrations appearing in the equilibrium constant expression are those measured once the system has reached equilibrium. This is implied by the notation; you do not need to add an "eq" subscript to each concentration term. The ratio equals K only when equilibrium has been established.
When a system has not yet reached equilibrium, the same ratio of concentrations and pressures is called the reaction quotient Q. Comparing Q to K is the tool used to predict which direction a reaction will shift — a topic covered in the next lesson.
What Does the Value of K Tell You?
The numerical value of an equilibrium constant is enormously informative. From a single number, you can determine whether a reaction predominantly produces products, retains mostly reactants, or yields a significant mixture of both at equilibrium.
| EQUILIBRIUM CONSTANTS | |
|---|---|
| K >> 1 | Favors Products |
| K << 1 | Favors Reactants |
| 10-3 < K < 103 | Reactants and products both present at equilibrium |
K >> 1: Reaction Favors Products
When K is much greater than 1 (for example, 109 or larger), the numerator of the expression vastly outweighs the denominator. This means there are far more products than reactants at equilibrium. We say the reaction favors the products. For industrial processes, a large K is ideal — it means the target chemical is produced in abundance.
Example: Combustion Reactions
Most combustion reactions are highly spontaneous and proceed nearly to completion. Their equilibrium constants at 298 K are on the order of 1040 to 1050. A value this large means the reaction overwhelmingly favors products — it is effectively irreversible under standard conditions.
K << 1: Reaction Favors Reactants
When K is much less than 1 (for example, 10-10), the denominator dominates. There are far more reactants than products at equilibrium. We say the reaction favors the reactants. This does not mean the product cannot be made, but forming it in large quantities will be more challenging and potentially more expensive.
K Close to 1: Significant Amounts of Both
When K falls between 10-3 and 103, neither reactants nor products are strongly preferred at equilibrium. A meaningful quantity of both will be present. We call this region "K close to one," though the boundaries are somewhat arbitrary.
Why Such Wide Boundaries? The range 10-3 to 103 accounts for the effect of stoichiometric exponents in the expression. Equal concentrations of reactants and products do not necessarily produce a K of exactly 1 when coefficients differ from 1. For example, if [CO] = [O₂] = 10 M in the expression K = [CO]² / [O₂], then K = 100/10 = 10, even though the concentrations are equal. Large exponents amplify deviations, which is why the "close to one" window is intentionally broad.
3 Possible Outcomes: Interpreting K at a Glance
- K > 103: Reaction favors products at equilibrium
- 10-3 < K < 103: Significant amounts of both reactants and products
- K < 10-3: Reaction favors reactants at equilibrium
Practice Interpretation
Given these K values, what does each tell you?
- K = 1050 — Overwhelmingly favors products; reaction goes nearly to completion.
- K = 13 — Close to one; fair amounts of both reactants and products at equilibrium.
- K = 10-10 — Heavily favors reactants; very little product forms at equilibrium.
Converting Between Kp and Kc
The relationship between Kp and Kc is derived from the ideal gas law. Since pressure equals molar concentration times RT (P = [M]RT), substituting this into the Kp expression introduces extra factors of RT that do not appear in Kc. The result is the following conversion equation:
Conversion Formula:
Kp = Kc × (RT)Δn
Where Δn is the change in the total moles of gas from reactants to products (products minus reactants). Solids, liquids, and aqueous species are excluded from this count.
When Does Kp Equal Kc?
If Δn = 0 — that is, if the total moles of gas on the product side equals the total moles of gas on the reactant side — then (RT)⁰ = 1, and Kp = Kc. This is the only scenario in which the two are numerically equal.
Exam Tip: "For which of the following reactions does Kp = Kc?" Simply count the moles of gas on each side. If they are equal (Δn = 0), then Kp = Kc. No calculation required.
Which Value of R to Use
Because the Kp/Kc conversion is derived from the ideal gas law, you must use R = 0.08206 L·atm / (mol·K) — not 8.314 J/(mol·K). Use 8.314 J/(mol·K) for energy-related calculations (such as the Arrhenius equation). Use 0.08206 L·atm/(mol·K) whenever an equation traces back to PV = nRT.
Worked Example: Finding Kp from Kc
Problem: For the reaction 2C(s) + O₂(g) ⇌ 2CO(g), the equilibrium constant Kc = 1200 at 400 K. Find Kp.
Step 1 — Identify Δn (moles of gas only):
Product side: 2 mol gas (CO). Reactant side: 1 mol gas (O₂). Carbon is a solid and is excluded.
Δn = 2 − 1 = +1
Step 2 — Apply the conversion formula:
Kp = Kc × (RT)Δn
Kp = 1200 × (0.08206 × 400)1
Kp = 1200 × 32.824
Kp ≈ 39,000 (unitless)
Interpretation: Both Kc = 1200 and Kp = 39,000 are greater than 103, confirming that this reaction strongly favors products at equilibrium.
Why Are Equilibrium Constants Unitless?
At first glance, K seems like it should have units. For the example above, Kc = [CO]² / [O₂] would have units of M² / M = M. The reason K is unitless is subtle: strictly speaking, every concentration term in the equilibrium expression is divided by a standard reference concentration of 1 M. This makes each term a dimensionless ratio, and the resulting K has no units. The reference concentration is universally implied and not written out in general chemistry, but it ensures that K is always dimensionless.
Contrast with Rate Constants: Rate constants (lowercase k) always carry units that depend on the overall reaction order. Equilibrium constants (uppercase K) are always unitless, for every reaction and at every temperature.
3 Rules for Manipulating Equilibrium Constant Expressions
A common exam question presents one reaction with a known K value and asks you to find K for a related reaction. Three fundamental algebraic manipulations cover all such problems.
Consider the following base reaction and its equilibrium constant:
Base Reaction:
2SO₂(g) + O₂(g) ⇌ 2SO₃(g) Kc = 16
Rule 1: Reversing the Reaction — Take the Reciprocal
When a reaction is written in reverse, products become reactants and vice versa. This inverts the equilibrium expression (numerator and denominator swap). Therefore, the new K equals 1 divided by the original K.
Reversed Reaction:
2SO₃(g) ⇌ 2SO₂(g) + O₂(g)
Kc = 1/16
Note: Equilibrium constants are never negative. A reversed reaction gives the reciprocal (1/K), not the negative (−K).
Rule 2: Multiplying Coefficients by a Factor — Raise K to That Power
Stoichiometric coefficients become exponents in the equilibrium expression. If all coefficients are halved, all exponents are halved, which is equivalent to taking the square root of K. If all coefficients are doubled, all exponents are doubled, which is equivalent to squaring K.
Coefficients Halved:
SO₂(g) + ½O₂(g) ⇌ SO₃(g)
Kc = √16 = 4
Do not simply divide K by 2. You must take the square root, because the exponents (not the K value itself) are being divided by 2.
General Rule for Scaling Coefficients:
- Coefficients halved → K' = K1/2 (square root)
- Coefficients multiplied by 1/3 → K' = K1/3 (cube root)
- Coefficients doubled → K' = K2 (square)
- Coefficients tripled → K' = K3 (cube)
Rule 3: Reversing and Scaling — Apply Both Rules
When a reaction is both reversed and scaled, apply both operations. The order does not matter — the result is the same either way.
Reversed and Coefficients Halved:
SO₃(g) ⇌ SO₂(g) + ½O₂(g)
Method A (invert first, then square root): 1/16 → √(1/16) = 1/4
Method B (square root first, then invert): √16 = 4 → 1/4 = 1/4
Kc = 1/4
Exam Strategy: On a timed exam, you can determine the new K by inspection once you recognize the relationship between the two reactions. But if needed, you can always write out the full equilibrium expressions for both reactions and derive the answer algebraically — it takes longer but guarantees the correct result.
Practice Problems
Given: 2SO2(g) + O2(g) ⇌ 2SO3(g) Kc = 4.0
Find Kc for each of the following related reactions:
- 2SO3(g) ⇌ 2SO2(g) + O2(g)
- SO2(g) + ½O2(g) ⇌ SO3(g)
- SO3(g) ⇌ SO2(g) + ½O2(g)
Key Takeaways
- Dynamic equilibrium means the forward and reverse reaction rates are equal, not that the reaction has stopped. Concentrations of reactants and products are constant but are not required to be equal.
- The equilibrium constant K (uppercase) is affected only by temperature — not by concentration changes, catalysts, or activation energy.
- The equilibrium constant expression places products in the numerator and reactants in the denominator, each raised to the power of its stoichiometric coefficient. Pure solids and pure liquid solvents are excluded.
- Kc uses molar concentrations; Kp uses partial pressures. Kp = Kc(RT)Δn, where Δn is the change in moles of gas. Kp = Kc only when Δn = 0.
- Use R = 0.08206 L·atm/(mol·K) in the Kp/Kc conversion; use R = 8.314 J/(mol·K) for energy-related calculations.
- Equilibrium constants are always unitless. This follows from the implicit division of each concentration term by the standard reference value of 1 M.
- A large K (>103) favors products; a small K (<10-3) favors reactants; K between 10-3 and 103 indicates significant quantities of both.
- When manipulating equilibrium expressions: reversing a reaction takes the reciprocal of K; scaling all coefficients by a factor raises K to that power (e.g., halving coefficients takes the square root).