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AP Chemistry review: key concepts

AP Chemistry flashcards on atoms, bonding, stoichiometry, kinetics, thermodynamics, equilibrium, acids and bases and electrochemistry, with solved problems.

37 cards7 topicsAbout 15 minutes a roundAP Chemistry style course content (two terms of first-year university general chemistry)

Independent study material written by StudyBrix. AP and SAT are trademarks of College Board and ACT is a trademark of ACT, Inc.; neither organisation is involved with or has reviewed this deck.

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About this deck

This deck is a compact review for students taking an AP Chemistry style course or a first-year general chemistry sequence. It moves through atomic structure and periodic trends, bonding, molecular shape and intermolecular forces, gases and solutions, stoichiometry and reaction types, kinetics, thermochemistry and free energy, equilibrium, acids, bases and buffers, and electrochemistry.

Before you start, make sure you can convert units, use scientific notation, round to significant figures, and work with logarithms (pH questions need them). Keep a periodic table nearby. Most cards include a short calculation; cover the example, solve it yourself, and only then compare. If your answer differs, find the exact step where it went wrong instead of rereading the solution.

Chemistry questions often ask you to explain, not just compute. For trends, practise giving the reason in terms of Coulombic attraction (nuclear charge, distance, shielding). For equilibrium shifts, state which way Q moves relative to K. Writing one clear sentence of justification for each card is excellent preparation for free-response questions.

Common mistakes: using grams instead of moles in a balanced equation; forgetting to convert kJ and J consistently in delta G = delta H - T delta S; and saying that a catalyst shifts equilibrium (it only speeds up both directions). Students also confuse breaking bonds, which always takes energy, with forming bonds, which releases it.

Before you start: Grade 11 chemistry, scientific notation, unit conversion, basic algebra and base 10 logarithms.

All 37 cards

Cover the definitions with a sheet of paper and test yourself, or use the study modes above.

Atoms and periodic trends

Mole and molar mass: A mole is 6.022 x 10^23 particles; molar mass (g/mol) converts between grams and moles.Example: 9.0 g of water / 18.02 g/mol = 0.50 mol.
Empirical formula from percent composition: Assume 100 g, convert each element's mass to moles, divide by the smallest mole value, and round to whole number ratios.Example: 40.0% C, 6.7% H, 53.3% O: 3.33, 6.65, 3.33 mol, ratio 1:2:1, so CH2O.
Photoelectron spectroscopy (PES): A spectrum of the energies needed to remove electrons; each peak is a subshell, its position shows binding energy and its height shows how many electrons are in it.Example: Sodium shows four peaks (1s, 2s, 2p, 3s) with relative heights 2:2:6:1; the 3s peak has the lowest energy.
First ionization energy trend: Energy to remove the outermost electron rises across a period (more nuclear charge, similar shielding) and falls down a group (electron farther from the nucleus).Example: Cl has a higher first ionization energy than Na; K is lower than Na.
Atomic radius trend: Radius shrinks across a period because effective nuclear charge increases, and grows down a group as electrons occupy higher shells.Example: Mg is larger than S; Ca is larger than Mg.
Electronegativity: An atom's ability to attract shared electrons in a bond; highest near fluorine, lowest at the bottom left of the table.Example: In H-Cl the electrons sit closer to Cl, making a polar bond.

Bonding and intermolecular forces

Formal charge: Valence electrons minus nonbonding electrons minus half the bonding electrons; the best Lewis structure keeps formal charges near zero.Example: In O=C=O every atom has formal charge 0, so that structure is preferred.
VSEPR geometry: Electron domains around a central atom spread out to minimize repulsion; lone pairs push bond angles smaller.Example: NH3 has 4 domains with one lone pair: trigonal pyramidal, about 107 degrees.
Hybridization: Mixing of atomic orbitals to match the number of electron domains: 2 domains sp, 3 domains sp2, 4 domains sp3.Example: The carbon in CH4 is sp3; each carbon in C2H4 is sp2.
London dispersion forces: Temporary attractions from momentary uneven electron distribution; stronger for larger, more polarizable electron clouds.Example: I2 is a solid at room temperature while F2 is a gas.
Hydrogen bonding (intermolecular): A strong dipole attraction between H bonded to N, O or F and a lone pair on another N, O or F.Example: H2O boils at 100 C while H2S, a heavier molecule, boils near -60 C.

Stoichiometry, gases and solutions

Ideal gas law: PV = nRT relates pressure, volume, moles and kelvin temperature; R = 0.08206 L atm/(mol K).Example: P = 2.00 atm, V = 5.00 L, T = 300 K: n = 10.0/(0.08206 x 300) = 0.406 mol.
Dilution: Moles of solute stay the same when solvent is added, so M1V1 = M2V2.Example: 50.0 mL of 2.0 M diluted to 250 mL: M2 = (2.0 x 50.0)/250 = 0.40 M.
Beer-Lambert law: Absorbance is proportional to concentration: A = (molar absorptivity)(path length)(concentration), used to find unknown concentrations by spectrophotometry.Example: If 0.10 M gives A = 0.40, a sample with A = 0.20 in the same cell is about 0.050 M.
Limiting reactant: The reactant that runs out first and sets the maximum amount of product; compare moles available with the mole ratio in the balanced equation.Example: 2H2 + O2 -> 2H2O with 4 mol H2 and 1 mol O2: O2 limits, giving 2 mol H2O.
Percent yield: Actual yield divided by theoretical yield, times 100.Example: Theoretical 36.0 g, actual 27.0 g: 75.0% yield.
Net ionic equation: Shows only species that change; spectator ions present unchanged on both sides are removed.Example: AgNO3(aq) + NaCl(aq): Ag+(aq) + Cl-(aq) -> AgCl(s).
Oxidation and reduction: Oxidation is loss of electrons (oxidation number rises); reduction is gain of electrons (oxidation number falls). They always occur together.Example: Zn + Cu2+ -> Zn2+ + Cu: zinc is oxidized and Cu2+ is reduced.

Kinetics

Rate law: rate = k[A]^m[B]^n, where the orders m and n come from experiments, not from the balanced equation's coefficients.Example: Doubling [A] with [B] fixed makes the rate 4 times larger, so the reaction is second order in A.
First-order half-life: For a first-order reaction t1/2 = 0.693/k, independent of starting concentration.Example: k = 0.0231 per minute gives t1/2 = 30.0 min; after 90 min, 1/8 remains.
Integrated rate law graphs: A straight line of [A] vs t means zero order, ln[A] vs t means first order, 1/[A] vs t means second order.Example: If ln[A] vs time is linear with slope -0.05 per s, the reaction is first order with k = 0.05 per s.
Catalyst: A substance that provides a pathway with lower activation energy, speeding forward and reverse reactions equally without being used up.Example: Adding MnO2 speeds the decomposition of H2O2 but does not change the amount of O2 formed.
Rate-determining step: The slowest elementary step in a mechanism; the overall rate law matches that step's molecularity (after accounting for any fast prior equilibrium).Example: If step 1, NO2 + NO2 -> NO3 + NO, is slow, the rate law is k[NO2]^2.

Thermodynamics

Calorimetry heat equation: q = mc delta T gives the heat absorbed or released by a substance of mass m and specific heat c.Example: 100 g of water warming 5.0 C: q = 100 x 4.18 x 5.0 = 2090 J.
Hess's law: The enthalpy change of a reaction is the same however it is reached, so equations and their delta H values can be added; reversing flips the sign.Example: If A -> B is -50 kJ and B -> C is +20 kJ, then A -> C is -30 kJ.
Bond enthalpy estimate: delta H is about the sum of bond energies broken minus the sum of bond energies formed.Example: Breaking 900 kJ of bonds and forming 1100 kJ of bonds gives delta H of about -200 kJ.
Gibbs free energy: delta G = delta H - T delta S; a negative value means the process is thermodynamically favourable at that temperature.Example: delta H = -100 kJ, delta S = -0.200 kJ/K, T = 298 K: delta G = -40.4 kJ; favourable below 500 K.
Free energy and K: delta G standard = -RT ln K: a negative value means K > 1 (products favoured); a positive value means K < 1.Example: If delta G standard = 0, then ln K = 0 and K = 1.

Equilibrium, acids and bases

Reaction quotient Q: Same form as K but using current concentrations; if Q < K the reaction shifts toward products, if Q > K toward reactants.Example: K = 4.0 and Q = 1.5, so the net reaction proceeds forward.
Le Chatelier's principle: A system at equilibrium that is disturbed shifts to partly offset the disturbance.Example: Adding N2 to N2 + 3H2 <-> 2NH3 shifts the reaction right, making more NH3.
Solubility product (Ksp): Equilibrium constant for a slightly soluble salt dissolving; molar solubility comes from an ICE setup.Example: AgCl: Ksp = s^2 = 1.8 x 10^-10, so s = 1.3 x 10^-5 M.
pH and pOH: pH = -log[H3O+] and pOH = -log[OH-]; at 25 C, pH + pOH = 14.Example: [H3O+] = 1.0 x 10^-3 M gives pH 3.00 and pOH 11.00.
pH of a weak acid: Set up Ka = x^2/(C - x); if x is small compared with C, x = sqrt(Ka C) = [H3O+].Example: 0.10 M acid with Ka = 1.8 x 10^-5: x = 1.34 x 10^-3 M, pH = 2.87.
Buffer and Henderson-Hasselbalch: A buffer holds a weak acid and its conjugate base; pH = pKa + log([A-]/[HA]).Example: pKa 4.74: equal amounts give pH 4.74; ten times more base than acid gives 5.74.
Half-equivalence point: In a weak acid titration, the point where half the acid has been neutralized, so [HA] = [A-] and pH = pKa.Example: If 20.0 mL of base reaches equivalence, the pH at 10.0 mL equals the acid's pKa.

Electrochemistry

Galvanic cell potential: E standard cell = E(cathode) - E(anode) using reduction potentials; oxidation occurs at the anode and reduction at the cathode.Example: Cu2+/Cu +0.34 V and Zn2+/Zn -0.76 V: 0.34 - (-0.76) = 1.10 V.
Electrolysis and Faraday's constant: Charge q = It; moles of electrons = q/96 485 C per mol; use the half-reaction to convert to moles of product.Example: 2.00 A for 965 s: 1930 C = 0.0200 mol e-, plating 0.0100 mol Cu = 0.636 g.