AP Chemistry equation sheet: Every formula, worked example, and how to use it in the exam

The College Board hands you a three-page reference packet for every AP (Advanced Placement) Chemistry exam. It has the periodic table, a page of equations, and a page of constants. You get it in both the multiple-choice and the free-response sections, so you never have to memorize the ideal gas constant or Henderson-Hasselbalch. What you do have to know is when to reach for each formula and what to do with it once it's in front of you.

This guide walks the equation sheet from top to bottom the way it's laid out in the official 2026 exam reference. For each equation you'll get a short read on what it means, when it shows up on the exam, and a worked example. If you'd rather see the equations taught inside full topic lessons with practice questions, Cognito's AP Chemistry course has a lesson attached to every one of them.

What's on the sheet. Three pages: A periodic table, a page of equations grouped into atomic structure, equilibrium, kinetics, gases and solutions, thermodynamics, and electrochemistry, and a page of constants (Planck's constant, speed of light, Faraday constant, gas constant, Rydberg constant, standard atmospheric pressure). Everything else, including standard reduction potentials for specific half-cells or heats of formation for specific compounds, is either given in the question or expected knowledge.

The first block of equations covers how energy, frequency, and wavelength relate to one another. These come up in questions about photoelectron spectroscopy, atomic emission spectra, and the hydrogen atom.

E = hν links the energy of a photon to its frequency, where h is Planck's constant (6.626 x 10 to the minus 34 joule-seconds). c = λν connects wavelength and frequency through the speed of light (2.998 x 10 to the 8 meters per second). Combine them and you get E = hc/λ, which is the version you actually use most because wavelength is usually what the question gives you.

Worked example: A photon has a wavelength of 656 nanometers (the red line in the hydrogen spectrum). Its energy is E = (6.626 x 10 to the minus 34)(2.998 x 10 to the 8) / (656 x 10 to the minus 9) = 3.03 x 10 to the minus 19 joules. That's the energy released when an electron in hydrogen drops from n = 3 to n = 2.

The equilibrium block gives you the general forms for Kc, Kp, Ka, Kb, and Kw, plus pH, pOH, pKa, pKb, and the Henderson-Hasselbalch equation. The exam expects you to write the specific expression for whatever reaction it hands you, then plug in either concentrations (Kc) or partial pressures (Kp).

For a reaction aA + bB reacts to form cC + dD, Kc = [C]^c [D]^d / [A]^a [B]^b, with pure solids and pure liquids left out. Kp swaps concentrations for partial pressures in atmospheres. The Kp = Kc(RT)^Δn conversion isn't printed on the sheet, so if a question asks you to move between them you'll need to remember it (Δn is moles of gas products minus moles of gas reactants).

Worked example: For N2 + 3H2 reacts to form 2NH3 at 500 K, if [N2] = 0.10 M, [H2] = 0.20 M, and [NH3] = 0.15 M at equilibrium, Kc = (0.15)^2 / (0.10)(0.20)^3 = 0.0225 / 0.0008 = 28. Notice how the coefficients become exponents, not multipliers. That's the single most common place students lose the point.

The acid-base equations sit inside this block too. Kw = [H+][OH-] = 1.0 x 10 to the minus 14 at 25 degrees Celsius, pH = -log[H+], pH + pOH = 14, Ka x Kb = Kw for a conjugate acid-base pair, and pH = pKa + log ([A-]/[HA]) (the Henderson-Hasselbalch equation, printed directly on the sheet). If you're given a Ka of 1.8 x 10 to the minus 5 for acetic acid, the Kb of its conjugate base (acetate) is 1.0 x 10 to the minus 14 divided by 1.8 x 10 to the minus 5, which is 5.6 x 10 to the minus 10.

The kinetics block has the general rate law, the two integrated rate laws that appear most, and the first-order half-life.

Rate = k[A]^m [B]^n is the general rate law, where m and n are experimental orders you find from data, not from stoichiometry. The mistake to avoid is reading the balanced equation and assuming the orders match the coefficients. They usually don't.

Worked example: If doubling [A] doubles the rate while [B] is held constant, then m = 1. If doubling [B] quadruples the rate with [A] held constant, then n = 2. The rate law is Rate = k[A][B]^2, third order overall.

For first-order reactions, ln[A] = ln[A]0 - kt is the integrated rate law. Rearranged, [A] = [A]0 x e^(-kt). The half-life is t1/2 = (ln 2)/k = 0.693/k, and crucially it doesn't depend on starting concentration. If a first-order reaction has a rate constant of 0.030 per second, its half-life is 0.693 / 0.030 = 23 seconds, whether you start with 1 M or 0.001 M.

The second-order integrated form (1/[A]t - 1/[A]0 = kt) is also on the sheet. Which one to use is determined by which straight-line plot the data give you. A plot of ln[A] versus time is linear for first order, 1/[A] versus time is linear for second order, and [A] versus time is linear for zero order.

The Arrhenius equation (k = A e^(-Ea/RT)) isn't on the sheet. If a question asks you to extract activation energy from rate constants at two temperatures, you'll need it from memory.

The gas block gives you the ideal gas law and Dalton's law of partial pressures. The solutions block gives you molarity and the dilution equation.

PV = nRT with R = 0.08206 liter-atmospheres per mole-kelvin (or 8.314 joules per mole-kelvin if you're working in SI units). Pick R based on the units of pressure the question uses. Atmospheres and liters means 0.08206; pascals and cubic meters means 8.314.

Worked example: 2.0 grams of helium at 25 degrees Celsius in a 5.0 liter flask. Moles of He = 2.0 / 4.0 = 0.50. P = nRT/V = (0.50)(0.08206)(298) / 5.0 = 2.4 atmospheres.

P_total = P1 + P2 + ... is Dalton's law, useful whenever you collect a gas over water or mix gases in a container. The partial pressure of each gas is its mole fraction times the total pressure.

M1V1 = M2V2 handles any dilution problem. Take 25 mL of 6.0 M HCl and dilute to 500 mL and the new concentration is (6.0)(25) / 500 = 0.30 M.

Six equations sit in this block, and they show up in almost every long free-response question one way or another.

q = mcΔT is the calorimetry equation. Given the mass of water, its specific heat (4.18 joules per gram per degree Celsius), and the temperature change, you can find the heat absorbed or released. If 100 grams of water heats from 22.0 to 28.5 degrees Celsius, q = (100)(4.18)(6.5) = 2,700 joules absorbed.

ΔH_rxn = Σ ΔHf(products) - Σ ΔHf(reactants) lets you calculate the enthalpy of a reaction from standard heats of formation, which the question will provide. Careful with signs: Elements in their standard state have ΔHf = 0, so O2, N2, and graphite drop out.

ΔS_rxn = Σ S(products) - Σ S(reactants) works the same way, using absolute entropies (which are always positive, unlike enthalpies of formation).

ΔG = ΔH - TΔS ties enthalpy and entropy together into free energy. T must be in kelvin, and if ΔH is in kilojoules while ΔS is in joules per kelvin, convert one of them before plugging in. Missing the units conversion is the most common wrong answer on this equation.

ΔG = -RT ln K links free energy to the equilibrium constant. A negative ΔG means K is greater than 1 (products favored); a positive ΔG means K is less than 1 (reactants favored); ΔG = 0 means K = 1.

Worked example: For a reaction where ΔG = -22.9 kilojoules per mole at 298 K, K = e^(-ΔG/RT) = e^(22,900 / (8.314 x 298)) = e^(9.24) = 1.0 x 10 to the 4. A modestly negative free energy corresponds to a fairly large equilibrium constant.

The electrochemistry block on the sheet is short: it links free energy to cell potential and gives you the definition of current.

ΔG° = -nFE°, where n is moles of electrons transferred and F is the Faraday constant (96,485 coulombs per mole). A positive E° gives a negative ΔG°, which is spontaneous, which is a galvanic (voltaic) cell. Combined with ΔG° = -RT ln K (also on the sheet), this lets you move between voltage, free energy, and the equilibrium constant.

I = q/t relates current, charge, and time – the workhorse for any electrolysis calculation. Combined with the Faraday constant, it converts a measured current over a given time into moles of electrons transferred, which then gives moles of product deposited.

What isn't printed: the Nernst equation (E = E° - (0.0592/n) log Q at 25 degrees Celsius) is not on the sheet, so if a question moves away from standard conditions you'll need to remember it. Q is the reaction quotient, calculated the same way you'd calculate K but with current (non-equilibrium) concentrations.

Worked example: A copper-zinc voltaic cell has E° = 1.10 volts and n = 2. If [Zn2+] = 1.0 M and [Cu2+] = 0.010 M, then Q = [Zn2+]/[Cu2+] = 100, log Q = 2, and E = 1.10 - (0.0592/2)(2) = 1.10 - 0.0592 = 1.04 volts. Lower copper concentration on the cathode side reduces the cell voltage, as Le Chatelier would predict.

Standard reduction potentials for specific half-cells (Cu2+/Cu, Zn2+/Zn, and so on) also aren't on the sheet. The question stem or an appended table provides any E° values you need.

The sheet is a safety net, not a shortcut. Three habits separate students who use it well from those who waste time on it.

First, know what's on it and what isn't. Standard reduction potentials, heats of formation, Ka values for specific acids, Ksp values for specific salts – all of those are given in the question or expected knowledge, not on the sheet. Don't scan the sheet for them; scan the question stem.

Second, know which equation each unit uses. Kinetics questions almost always need a rate law or an integrated rate law (and the Arrhenius equation from memory if two-temperature data appears). Electrochemistry usually needs ΔG° = -nFE° (on the sheet) and, for non-standard conditions, the Nernst equation from memory. Thermochemistry needs q = mcΔT for calorimetry and Hess-law style summing for anything involving standard values. Building a mental map from unit to formula is faster than a page-turn search.

Third, plug in with units. Every AP Chemistry FRQ (free-response question) that asks for a numerical answer wants the units. Setting up the expression with units attached – moles per liter, kelvin, joules – catches unit errors before you hit the calculator. The full unit-by-unit walkthrough on Cognito's AP Chemistry course has practice questions grouped by which equation you need, which is a good way to build that muscle memory.

One final note. The equation sheet you'll see in May 2026 is the same one used in May 2025, and the College Board doesn't change it often. Any commercial reference sheet dated 2025 or 2026 will match. If you find a version that's missing the second-order integrated rate law or has different constants, it's either older or third-party. Stick to the official version from apcentral.collegeboard.org. AP prep sits alongside all the other AP courses Cognito supports, so if you're taking two or three APs together the same study rhythm carries across.