Writing and Reading Keq
Products over reactants, each raised to its coefficient
For a general reversible reaction aA + bB ⇌ cC + dD, the equilibrium constant expression is written as Keq = [C]^c[D]^d / [A]^a[B]^b — the concentrations of the products (raised to their respective coefficients) divided by the concentrations of the reactants (raised to their respective coefficients). This single number, Keq, is fixed for a given reaction at a given temperature, regardless of the specific starting concentrations used to reach equilibrium.
When the expression uses molar concentrations, the constant is specifically called Kc. When it's written instead using partial pressures (common for reactions involving gases), it's called Kp. Both describe the same underlying equilibrium, just expressed in different units, and they're related to each other through a formula involving temperature and the change in moles of gas.
Not every species in a balanced equation appears in the Keq expression. Pure solids and pure liquids are excluded entirely, because their concentration (technically, their activity) doesn't meaningfully change and is treated as a constant already folded into K itself. Only aqueous species (dissolved in solution) and gaseous species actually appear in the expression, since only their concentrations can meaningfully vary.
💡 What the Size of K Actually Tells You
The numerical value of Keq is a direct indicator of where a reaction's equilibrium lies. A large K (much greater than 1) means that at equilibrium, products dominate — the reaction proceeds essentially to completion, favoring the forward direction. A small K (much less than 1) means reactants dominate at equilibrium — very little product actually forms, and the reaction strongly favors the reverse direction. A K value close to 1 means significant amounts of both reactants and products are present at equilibrium, with neither side strongly favored.
Critically, K says nothing about how FAST a reaction reaches equilibrium — that's a separate question, governed by reaction kinetics and rate constants, not by the equilibrium constant.
Write
Writing a Keq expression from a balanced equation
Given any balanced reversible equation, the process is mechanical: place all product concentrations in the numerator (each raised to its coefficient), place all reactant concentrations in the denominator (each raised to its coefficient), and exclude any pure solids or pure liquids from the expression entirely — leaving only aqueous and gaseous species.
For N₂(g) + 3H₂(g) ⇌ 2NH₃(g), the expression is Keq = [NH₃]² / ([N₂][H₂]³) — note the coefficients (1, 3, 2) become the exponents on each corresponding concentration.
Exclude
Recognizing which species to leave out
A pure solid or pure liquid participating in a reaction still matters chemically, but its concentration doesn't change in any meaningful way as the reaction proceeds — because of this, its "activity" is defined as exactly 1 and it's simply omitted from the K expression, rather than written in with an exponent of zero.
For CaCO₃(s) ⇌ CaO(s) + CO₂(g), both solids are excluded entirely, leaving simply Keq = [CO₂] — the equilibrium in this reaction depends only on the concentration of the one gaseous species involved.
Q vs K
The reaction quotient Q — predicting which direction a reaction will shift
The reaction quotient Q uses the exact same mathematical expression as Keq, but calculated using whatever concentrations exist at any given moment — not necessarily the equilibrium concentrations. Comparing Q to K tells you which direction a reaction not yet at equilibrium will proceed: if Q < K, there's proportionally too little product yet, so the reaction proceeds forward (toward products) to reach equilibrium. If Q > K, there's proportionally too much product already, so the reaction proceeds in reverse (toward reactants). If Q = K, the reaction is already at equilibrium, and there's no net shift in either direction.
Mixing reactants together for the very first time, before any reaction has occurred, gives Q = 0 (since product concentration is zero) — since 0 is necessarily less than any positive K, the reaction always proceeds forward initially, exactly as expected.
🔬 Applied Scenario — Using Q to Predict Reaction Direction
Given a reaction's known Keq value and a snapshot of current concentrations, calculating Q and comparing it to K is exactly how chemists predict which way a system not yet at equilibrium will shift.
A
A reaction has a known Keq of 4.0 at a given temperature. At some moment before equilibrium is reached, the current concentrations are measured and used to calculate Q, following the identical expression used for K, just with these non-equilibrium concentrations plugged in.
B
The calculation yields Q = 1.5. Comparing this to the known Keq of 4.0 shows Q < K.
C
Since Q < K, the reaction must shift forward, producing more product — there's proportionally too little product relative to reactant right now compared to where the true equilibrium sits, so the forward reaction will proceed until Q rises to meet K, at which point the system reaches true equilibrium.
D
If the calculated Q had instead come out greater than 4.0, the correct conclusion would flip — the reaction would need to shift in reverse, converting some product back into reactant, until Q decreased down to match K.
⚠️ Most Common Equilibrium Constant Mistakes
Including pure solids or liquids in the K expression. Students frequently forget to exclude a pure solid or liquid species, writing an extra term into the expression that shouldn't be there — remembering that only aqueous and gaseous species vary in concentration is the key to avoiding this.
Forgetting to raise each concentration to its coefficient. Writing [C][D] instead of the correct [C]^c[D]^d is a very common shortcut error — every coefficient in the balanced equation becomes an exponent in the K expression, without exception.
Confusing Q < K with "the reaction is too slow" rather than "not enough product yet." Q and K describe concentration ratios, not reaction speed — a reaction can have a very fast rate but still show Q < K simply because it hasn't yet had time to build up enough product, and K itself carries no information about how fast that buildup happens.
✓ Quick Self-Test
1. Write the general Keq expression for aA + bB ⇌ cC + dD.
2. Which two types of species are always excluded from a Keq expression, and why?
3. What does a very large K value (much greater than 1) tell you about a reaction at equilibrium?
4. If Q is calculated to be greater than K, which direction will the reaction shift to reach equilibrium?
5. Does the value of Keq tell you anything about how fast a reaction reaches equilibrium?
Answers:
1. Keq = [C]^c[D]^d / [A]^a[B]^b.
2. Pure solids and pure liquids — their concentration (activity) doesn't meaningfully change as the reaction proceeds, so they're treated as a constant already folded into K and left out of the expression entirely.
3. That the reaction strongly favors products at equilibrium — the forward reaction proceeds essentially to completion, with very little reactant remaining.
4. In reverse (toward reactants) — Q > K means there's proportionally too much product already relative to true equilibrium, so the reaction shifts backward until Q decreases to match K.
5. No — K describes only the position of equilibrium (the ratio of products to reactants once equilibrium is reached), not the speed at which that equilibrium is reached, which is governed separately by reaction kinetics.