AP Chemistry Quick Review

Electron configurations fill orbitals from lowest to highest energy, and the periodic table's block structure (s, p, d, f) is a direct map of that filling order.

Aufbau principle (fill lowest energy first), Pauli exclusion principle (max 2 electrons per orbital, opposite spin), Hund's rule (fill degenerate orbitals singly before pairing). Filling order: $1s 2s 2p 3s 3p 4s 3d 4p 5s 4d 5p…$ Noble-gas core notation abbreviates the configuration up to the prior noble gas. Exceptions (e.g., Cr, Cu) arise from extra stability of half-filled/filled $d$ subshells.

An electron configuration (list of filled subshells, e.g. $1s^22s^2$) vs. an orbital diagram (boxes/arrows showing individual electron spins) --- both must obey Hund's rule, but only the diagram shows spin pairing directly.

Removing electrons from the $3d$ subshell first when forming a transition-metal cation --- despite $4s$ filling before $3d$, electrons are always removed from the HIGHEST principal quantum number ($n$) subshell first, so $Fe arrow Fe^2+$ loses its two $4s$ electrons, not $3d$ electrons.

Aufbau + Pauli + Hund fill electrons; cations lose highest-$n$ (usually $s$) electrons first, not the last ones added.

Periodic trends in size arise from the competing pull of increasing nuclear charge and the shielding provided by inner electrons, both captured by Coulomb's law.

$$F = k/q_1q_2r^2 Z_eff ≈ Z - S$$ Atomic radius decreases left$→$right across a period (rising $Z_eff$, same shielding) and increases top$→$bottom down a group (new principal energy level added). Cations are smaller than their parent atom; anions are larger.

The across-a-period radius trend (driven by increasing $Z_eff$ at constant $n$) vs. the down-a-group trend (driven by increasing $n$, which dominates despite the added protons).

Ranking isoelectronic ions (same electron count) by charge sign alone --- for a fixed number of electrons, MORE protons pulls electrons in tighter, so radius shrinks as nuclear charge rises: $Al^3+ < Mg^2+ < Na+ < F- < O^2-$.

$Z_eff ⇒$ radius $$; same electron count, more protons $⇒$ smaller ion.

Photoelectron spectroscopy directly measures the energy needed to remove electrons from each subshell, giving experimental proof of electron configuration and periodic ionization-energy trends.

$$E = h = /hc$$ First ionization energy generally increases left$→$right, decreases top$→$bottom (exceptions at Be/B and N/O due to subshell stability). On a PES spectrum: peak position (x-axis, binding energy) tells how tightly bound/close to the nucleus a subshell is (core electrons = high binding energy, far left; valence electrons = low binding energy, far right); peak height (y-axis) is proportional to the relative NUMBER of electrons in that subshell.

PES peak position (binding energy, identifies WHICH subshell) vs. peak height (relative electron count, identifies HOW MANY electrons are in that subshell).

Missing a large jump between successive ionization energies --- a big jump pinpoints exactly where core electrons begin, revealing the number of valence electrons (e.g., a huge jump after the 2nd ionization energy means 2 valence electrons).

PES peak height ratio = electron count ratio; big IE jump = boundary between valence and core electrons.

A mass spectrum reports each isotope's mass and relative abundance, which combine to give the weighted-average atomic mass listed on the periodic table.

$$Average atomic mass = (isotope mass × fractional abundance)$$ Mass spectrometer process: vaporize sample $→$ ionize $→$ accelerate in an electric field $→$ deflect by mass-to-charge ratio in a magnetic field $→$ detect.

Relative abundance (a percent that, across all isotopes of an element, must sum to 100%) vs. raw signal/peak intensity (proportional to abundance, but not itself a percentage until normalized).

Forgetting to convert percent abundances to decimal fractions before multiplying by isotope mass, or using abundances that don't sum to 1 --- always verify all fractional abundances for an element add to exactly 1 before computing the weighted average.

Avg. atomic mass $= $(isotope mass $×$ fractional abundance); fractions must total 1.

The electronegativity difference between bonded atoms determines whether the bond is ionic, covalent, or metallic, and this bonding type governs a substance's macroscopic properties.

Ionic: large EN difference, electron transfer, rigid crystal lattice, high melting point, conducts only when molten or dissolved. Covalent: small EN difference, electron sharing. Metallic: delocalized "sea" of electrons, malleable, conducts as a solid. $$Lattice Energy /Q_1Q_2r$$ (increases with greater ionic charge, decreases as ionic radius/distance increases).

Polar covalent bonds (unequal sharing, moderate EN difference) vs. ionic bonds (essentially complete electron transfer, large EN difference) --- these lie on a continuum, not a hard cutoff.

Assuming lattice energy is controlled mainly by ionic size --- charge has a far larger effect: $MgO$ ($2+/2-$ charges) has much higher lattice energy than $NaCl$ ($1+/1-$) despite similar ionic radii.

Lattice energy $$ charge product $$ distance; charge dominates over size.

Valid Lewis structures satisfy the octet rule where possible and minimize formal charge; when multiple valid electron arrangements exist, the true structure is a resonance hybrid, not a rapid flip between forms.

$$Formal Charge = (valence e^-) - (nonbonding e^-) - /12(bonding e^-)$$ The best structure minimizes formal charges (closest to zero), placing any negative formal charge on the more electronegative atom. Octet exceptions: expanded octets (period 3+, e.g. $SF6$), electron-deficient atoms ($BF3$), odd-electron radicals (e.g. $NO2$).

Resonance structures (same connectivity, only electron placement differs; the molecule is ONE hybrid, not several interconverting forms) vs. isomers (genuinely different connectivity/arrangement, distinct compounds).

Predicting unequal bond lengths from a single resonance structure --- experimentally, resonance-averaged bonds (e.g., in $NO3-$ or $O3$) are all equal length, intermediate between single and double bonds.

FC $=$ valence $-$ nonbonding $- /12$bonding; resonance $=$ one real hybrid, not alternating structures.

Molecular shape minimizes electron-domain repulsion (VSEPR), and overall molecular polarity depends on both individual bond dipoles and how symmetrically they're arranged in that shape.

Electron-domain geometry by steric number: 2 = linear, 3 = trigonal planar, 4 = tetrahedral, 5 = trigonal bipyramidal, 6 = octahedral. Lone pairs compress bond angles more than bonding pairs. A molecule with polar bonds can be NONPOLAR overall if bond dipoles cancel by symmetry (e.g. $CO2$, $CCl4$).

Electron-domain geometry (counts lone pairs, e.g. tetrahedral) vs. molecular geometry (describes only atom positions, e.g. bent or trigonal pyramidal when lone pairs are present).

Calling a molecule polar just because it contains polar bonds --- always check for symmetry-driven cancellation; $CCl4$'s four polar C--Cl bonds cancel completely due to tetrahedral symmetry, making the molecule nonpolar overall.

Steric number $→$ electron geometry; lone pairs squeeze angles; symmetric polar bonds can cancel to nonpolar.

Hybrid orbitals form a molecule's sigma-bond framework and hold lone pairs, while unhybridized p orbitals form pi bonds; the resulting bond order predicts relative bond length and strength.

$sp$ (linear), $sp^2$ (trigonal planar, allows 1 pi bond), $sp^3$ (tetrahedral, no pi bonds). Single bond $=$ 1 sigma; double bond $=$ 1 sigma $+$ 1 pi; triple bond $=$ 1 sigma $+$ 2 pi. Higher bond order $→$ shorter bond length, greater bond energy.

Sigma bonds (end-to-end overlap, allow free rotation) vs. pi bonds (side-to-side p-orbital overlap, RESTRICT rotation, the basis of cis/trans isomerism around $C=C$).

Forgetting that rotation is locked around any double or triple bond --- a molecule with a $C=C$ can have distinct cis/trans isomers, but free rotation around single (sigma-only) bonds means no such isomers exist there.

Hybridization $=$ \# sigma bonds $+$ lone pairs; pi bonds block rotation; higher bond order $→$ shorter, stronger bond.

The type and strength of the forces BETWEEN particles (not the strength of bonds within a molecule) governs bulk physical properties like boiling point, viscosity, and solubility.

Typical strength order: ionic/network $>$ hydrogen bonding $>$ dipole-dipole $>$ London dispersion. Hydrogen bonding requires H directly bonded to N, O, or F. London dispersion forces exist in ALL molecules and increase with molar mass/polarizability (surface area).

Intermolecular forces (between separate molecules; determine boiling/melting point) vs. intramolecular bonds (within a molecule; determine chemical stability) --- boiling water breaks IMFs, not the $O-H$ covalent bonds.

Assuming a heavier molecule always boils higher regardless of polarity --- always identify the STRONGEST IMF type present first (e.g., H-bonding in $HF$ or $H2O$ can outweigh a larger nonpolar molecule's dispersion forces), then compare within the same category by size/polarity.

Ion-ion/H-bond $>$ dipole-dipole $>$ London dispersion; bigger/more polarizable $→$ stronger LDFs; H-bond needs H on N/O/F.

The four solid types (ionic, metallic, covalent-network, molecular) differ in what occupies the lattice points and what holds them together, explaining their conductivity, hardness, and melting points.

Ionic: hard, brittle, high mp, conducts only molten/dissolved. Metallic: malleable, ductile, conducts as a solid (delocalized electrons). Covalent-network (e.g. diamond, $SiO2$): extremely high mp, very hard, does not conduct. Molecular: held only by IMFs, low mp, does not conduct.

Covalent-network solids (one continuous covalent bonding framework, e.g. diamond, quartz) vs. molecular solids (discrete covalent molecules held together only by weak IMFs, e.g. ice, dry ice).

Predicting a low melting point for any "covalent" solid --- the deciding question is whether it's one continuous network (very high mp) or discrete molecules (low mp); "covalent" alone doesn't determine melting point.

Ionic$→$brittle, molten conducts. Metallic$→$always conducts. Network$→$superhard/high mp. Molecular$→$soft, low mp.

Under ideal conditions, a gas's pressure, volume, temperature, and moles are linked by the ideal gas law, which follows directly from kinetic molecular theory's assumptions about particle motion.

$$PV = nRT, R = 0.08206\ L·atm/(mol·K) /P_1V_1n_1T_1 = /P_2V_2n_2T_2$$ Dalton's law: $P_total = P_i$, $P_i = X_iP_total$. Average kinetic energy depends ONLY on temperature (K): $KE_avg = /32 RT$. Graham's law: $/rate_Arate_B = sqrt/M_BM_A$.

Effusion/diffusion RATE (depends on molar mass --- lighter gas moves faster, per Graham's law) vs. average kinetic energy (equal for ALL gases at the same temperature, regardless of molar mass).

Forgetting to convert temperature to Kelvin before using $PV=nRT$ --- plugging in a Celsius value directly is the single most common numeric error on gas-law problems.

$PV=nRT$, $T$ must be Kelvin; same $T→$ same avg. KE for all gases, but lighter gas moves faster (Graham's law).

Solution concentration can be expressed several ways depending on what's held constant, and "like dissolves like" (polarity matching) predicts whether a solute will dissolve.

$$M = /mol soluteL solution m = /mol solutekg solvent M_1V_1 = M_2V_2 (dilution)$$ Polar/ionic solutes dissolve in polar solvents; nonpolar solutes dissolve in nonpolar solvents; ion-dipole forces drive dissolution of ionic solids in water.

Molarity (mol per L of SOLUTION, slightly temperature-dependent since volume expands) vs. molality (mol per kg of SOLVENT, temperature-independent --- used for colligative-property calculations for this reason).

Using total solution volume instead of solvent mass (or vice versa) when converting between molarity and molality, or forgetting that dilution changes molarity but never changes the total MOLES of solute present.

Molarity $=$ mol/L solution; Molality $=$ mol/kg solvent (temp-independent); dilution keeps moles constant.

Colligative properties depend only on the NUMBER of dissolved solute particles, not their identity, so ionic solutes that dissociate produce an amplified effect per mole dissolved.

$$ T_f = iK_fm T_b = iK_bm = iMRT$$ $i$ = van't Hoff factor (particles produced per formula unit, e.g. $i≈2$ for $NaCl$, $i≈3$ for $CaCl2$). Freezing point DEPRESSES; boiling point ELEVATES as solute is added.

Freezing point depression (solute lowers freezing point) vs. boiling point elevation (solute raises boiling point) --- both arise from the solute lowering the solvent's vapor pressure/liquid entropy, not from the solute itself freezing or boiling.

Forgetting the van't Hoff factor for ionic solutes --- treating $CaCl2$ like a non-dissociating molecular solute understates $ T$ by a factor of 3.

$ T = iKm$; ionic solutes get a bigger $i$ (more dissolved particles) $→$ bigger colligative effect.

A substance's characteristic light absorption can identify it and, via the Beer--Lambert law, determine an unknown solution's concentration.

$$A = b c$$ $A$ = absorbance, $$ = molar absorptivity (constant for a given substance/wavelength), $b$ = path length, $c$ = concentration. Absorbance is directly proportional to concentration (linear calibration curve). Observed solution color is the COMPLEMENT of the wavelength(s) most strongly absorbed.

Absorbance $A$ (directly proportional to concentration, used in Beer's-law calculations) vs. percent transmittance $%T$ (inversely/logarithmically related to concentration, NOT directly proportional).

Reading a concentration off a calibration curve for an absorbance value that falls OUTSIDE the linear range actually tested (extrapolating beyond calibrated points), or plugging $%T$ directly into Beer's law instead of converting to $A$ first.

$A= bc$: absorbance $$ concentration; stay within the calibration curve's tested range.

Classifying a reaction type reveals what governs whether it proceeds, and net ionic equations isolate only the species that actually change.

Precipitation reactions occur when combined ions form an insoluble product (apply solubility rules). Net ionic equation $=$ full equation minus spectator ions (ions unchanged on both sides). Combustion: hydrocarbon $+\ O2 → CO2 + H2O$.

Molecular (formula) equation (whole-compound formulas) vs. complete ionic equation (all soluble strong electrolytes dissociated) vs. net ionic equation (spectator ions removed, only the real reaction shown).

Leaving spectator ions in a "net ionic equation" answer, or assuming a precipitate forms without checking solubility rules --- a precipitate forms only if at least one possible product is INSOLUBLE.

Net ionic $=$ full equation $-$ spectators; precipitate only if a product is insoluble (check solubility rules).

Balanced-equation mole ratios convert between reactant and product amounts, but the limiting reactant caps how much product can actually form.

$$Percent Yield = /Actual YieldTheoretical Yield × 100$$ Limiting reactant $=$ the reactant that produces the LEAST possible product (runs out first). Molar mass converts g $≤ftrightarrow$ mol; balanced-equation coefficients convert mol A $≤ftrightarrow$ mol B.

Limiting reactant (used up, sets theoretical/maximum yield) vs. excess reactant (some remains unreacted at the end).

Identifying the limiting reactant by comparing raw grams or raw moles given, instead of comparing the moles of PRODUCT each reactant could form --- always convert through stoichiometric coefficients before comparing.

Limiting reactant $=$ least possible product; %Yield $=$ actual $$ theoretical $×100$.

A titration uses a solution of known concentration to determine an unknown concentration or amount, tracked through the stoichiometric equivalence point.

At the equivalence point: mol acid $×$ (coefficient ratio) $=$ mol base; for a monoprotic acid/base, $M_aV_a = M_bV_b$. The equivalence point is NOT the same as the endpoint.

Equivalence point (defined stoichiometrically --- moles acid equal moles base by ratio, exact) vs. endpoint (defined experimentally by indicator color change, an approximation that can introduce small error).

Applying $M_aV_a = M_bV_b$ directly to a polyprotic acid/base without adjusting for stoichiometry --- titrating $H2SO4$ with $NaOH$ requires 2 mol base per 1 mol acid, not a 1:1 ratio.

mol acid $×$ coefficient ratio $=$ mol base; endpoint (indicator) only approximates equivalence point.

In a redox reaction, electrons transfer from the species being oxidized to the species being reduced, tracked using oxidation numbers.

Oxidation $=$ loss of electrons, oxidation number increases; Reduction $=$ gain of electrons, oxidation number decreases (OIL RIG). The oxidizing agent IS REDUCED (it takes electrons from something else); the reducing agent IS OXIDIZED.

The oxidizing agent (the substance reduced --- gains electrons) vs. the reducing agent (the substance oxidized --- loses electrons) --- these labels are frequently swapped.

Labeling the oxidizing agent as "the thing that gets oxidized" --- it's the opposite: the oxidizing AGENT is itself reduced. Always assign oxidation numbers to every atom first before deciding what changed.

OIL RIG; oxidizing agent $=$ gets reduced; reducing agent $=$ gets oxidized.

A rate law, found experimentally, expresses how reaction rate depends on reactant concentrations --- it is NOT read off the balanced equation's coefficients.

$$Rate = k[A]^m[B]^n$$ $m,n$ = reaction orders (determined experimentally), overall order $=m+n$. Method of initial rates: compare trials, holding one concentration constant while another is doubled/tripled, to solve for each order.

Reaction order (empirically determined exponent in the rate law) vs. stoichiometric coefficient (from the balanced equation) --- equal only for an elementary step, never assumed equal for an overall reaction.

Assuming rate-law exponents equal the overall balanced equation's coefficients --- always derive orders from experimental initial-rate data (or a given rate-determining step) instead.

Rate $=k[A]^m[B]^n$; orders come from DATA, never from the overall balanced equation.

Integrated rate laws convert a rate expression into concentration-vs-time form, and each reaction order has a distinctive linear plot and half-life behavior.

Zero order: $[A]_t=-kt+[A]_0$ (linear: $[A]$ vs. $t$), $t_1/2=/[A]_02k$. First order: $_t=-kt+_0$ (linear: $$ vs. $t$), $t_1/2=/0.693k$ (constant). Second order: $/1[A]_t=kt+/1[A]_0$ (linear: $1/[A]$ vs. $t$), $t_1/2=/1k[A]_0$.

First-order half-life (constant, independent of starting concentration --- e.g., radioactive decay) vs. zero- and second-order half-life (both DEPEND on starting concentration).

Choosing the wrong linearized plot to identify order from a data table --- test which plot ($[A]$ vs $t$, $$ vs $t$, or $1/[A]$ vs $t$) is a straight line; only the $$ vs $t$ line confirms first order.

Straight-line test: $[A]$ vs $t→$0th; $$ vs $t→$1st; $1/[A]$ vs $t→$2nd. Only 1st-order half-life is constant.

A valid mechanism's elementary steps must sum to the overall balanced equation, and its rate law --- set by the slowest step --- must match the experimentally observed rate law.

For an elementary step, rate-law exponents DO equal that step's coefficients. The slowest step is the rate-determining step (RDS) and controls the overall rate law. Intermediates (formed then consumed) never appear in the overall equation or overall rate law; if the RDS involves one, a fast pre-equilibrium step is used to substitute it out.

A reaction intermediate (formed then consumed during the reaction, never in overall products) vs. a catalyst (present at the start, consumed then regenerated, lowers activation energy, absent from the overall net equation).

Writing a rate law that directly includes a reaction intermediate --- the valid rate law can only contain species from the OVERALL equation, so any intermediate must be eliminated algebraically using the fast pre-equilibrium step.

Steps sum to overall equation; slow step sets the rate law; intermediates must be substituted out.

A reaction occurs only when particles collide with sufficient energy and proper orientation; catalysts speed reactions by opening an alternate pathway with lower activation energy.

$$k = Ae^-E_a/RT k = -/E_aR≤ft(/1T) + A$$ Increasing temperature raises $k$ (more particles exceed $E_a$). A catalyst lowers $E_a$ (does NOT change $ H$) and is not consumed. A plot of $ k$ vs. $1/T$ has slope $-E_a/R$.

A catalyst (lowers activation energy, unchanged at the end) vs. increased temperature (raises both the fraction of particles exceeding $E_a$ AND collision frequency, without changing $E_a$ itself).

Believing a catalyst shifts a reaction's equilibrium position or changes $ H$ --- it speeds up BOTH forward and reverse reactions equally, reaching equilibrium faster without changing $K$ or the thermodynamics.

$k=Ae^-E_a/RT$; catalyst lowers $E_a$ only (not $ H$, not $K$); higher $T→$ more collisions clear $E_a$.

Heat transferred during a process can be measured via calorimetry, based on the principle that in an isolated system, heat lost by one part equals heat gained by another.

$$q = mc T, T = T_final-T_initial$$ In a coffee-cup calorimeter, $q_rxn = -q_solution$. Exothermic: $q_sys<0$, surroundings warm up. Endothermic: $q_sys>0$, surroundings cool down.

Specific heat capacity $c$ (J/(g$·^$C), intrinsic per-gram property of a substance) vs. heat capacity $C$ (J/$^$C, extrinsic, depends on the total mass of that particular object/calorimeter).

Forgetting the sign convention --- $q_rxn$ and $q_solution$ are equal in MAGNITUDE, opposite in sign; also computing $ T$ as initial $-$ final instead of final $-$ initial silently flips exothermic/endothermic conclusions.

$q=mc T$; $q_rxn=-q_solution$; exothermic $→ q<0$, surroundings warm.

Because enthalpy is a state function, $ H_rxn$ is the same regardless of pathway, letting unknown enthalpies be built by combining known reactions.

$$ H_rxn = n H_f^(products) - n H_f^(reactants)$$ Hess's law: reversing a reaction flips the sign of $ H$; multiplying a reaction by a coefficient multiplies $ H$ by that same coefficient. $ H_f^$ of any element in its standard state $=0$.

Standard enthalpy of formation $ H_f^$ (per mole of compound formed from elements in standard states) vs. enthalpy of reaction $ H_rxn$ (for the specific reaction/amounts as written).

Forgetting to flip the sign of $ H$ when reversing a given reaction in a Hess's-law combination, or forgetting to scale $ H$ when a reaction is multiplied to match coefficients --- both are silent errors that don't visibly break the algebra.

$ H_rxn= H_f^$(products)$- H_f^$(reactants); reverse $→$ flip sign; scale $→$ multiply $ H$.

Reactions involve breaking bonds in reactants (always costs energy) and forming bonds in products (always releases energy); the net balance sets whether a reaction is endo- or exothermic.

$$ H_rxn ≈ BE(bonds broken) - BE(bonds formed)$$ Breaking bonds is ALWAYS endothermic ($+$); forming bonds is ALWAYS exothermic ($-$). On an energy diagram, $ H = E_a(forward) - E_a(reverse)$.

Bond breaking (always requires energy input) vs. bond forming (always releases energy) --- never say "breaking a bond releases energy."

Reversing the sign convention (computing formed $-$ broken instead of broken $-$ formed), which flips the exothermic/endothermic conclusion; also confusing the reverse reaction's activation energy with $ H$ itself on an energy diagram.

$ H ≈ BE_broken - BE_formed$; breaking bonds costs energy, forming bonds releases it.

At equilibrium, the ratio of product to reactant concentrations (raised to their coefficients) is fixed at a given temperature; comparing $Q$ to $K$ predicts which direction a system will shift.

For $aA+bB leftharpoons cC+dD$: $$K_c = /[C]^c[D]^d[A]^a[B]^b K_p = K_c(RT)^ n$$ Pure solids/liquids are OMITTED from the expression. $Q<K→$ shifts forward; $Q>K→$ shifts reverse; $Q=K→$ at equilibrium.

$K$ (calculated with equilibrium concentrations, fixed at a given temperature) vs. $Q$ (calculated the same way but at ANY point in time, used to predict shift direction).

Including pure solids/liquids (or dilute solvent water) in the $K$ expression, or forgetting that $K$ changes ONLY with temperature --- concentration, pressure, and catalysts never change $K$'s value.

$K=$ products/reactants (solids/liquids omitted); $Q<K→$forward; $Q>K→$reverse; only $ T$ changes $K$.

An ICE (Initial, Change, Equilibrium) table tracks how concentrations change by a stoichiometrically consistent amount $x$ as a system approaches equilibrium.

Set up rows for Initial, Change ($± x$ scaled by coefficients), Equilibrium; substitute the equilibrium row into the $K$ expression and solve for $x$. The small-$x$ approximation (ignore $x$ relative to initial concentration) is valid only when $K$ is very small (rule of thumb: initial/$K>500$); always verify with the "5% rule" afterward.

Using the small-$x$ approximation (valid only for small $K$, avoids the quadratic formula) vs. needing the full quadratic formula (required when $K$ isn't small enough, or $x$ turns out $>5%$ of the initial value).

Applying the small-$x$ shortcut to a reaction with a large $K$ or high initial concentration without checking validity afterward --- this produces a plausible but numerically wrong answer; always confirm $x<5%$ of the initial concentration it was subtracted from.

ICE table $→$ solve for $x$; small-$x$ shortcut only valid if $x<5%$ of initial (check after solving!).

A system at equilibrium responds to an imposed stress (concentration, pressure/volume, or temperature change) by shifting in the direction that partially relieves it.

Add reactant/remove product $→$ shifts forward; remove reactant/add product $→$ shifts reverse. Decreasing volume (raising pressure) $→$ shifts toward FEWER moles of gas. Raising temperature: exothermic reactions shift REVERSE and $K$ decreases; endothermic reactions shift FORWARD and $K$ increases. An inert gas at constant volume has NO effect.

Temperature change (the ONLY stress that changes the numerical value of $K$) vs. concentration or pressure/volume change (shifts equilibrium POSITION, but $K$ stays the same at constant temperature).

Treating a catalyst or an inert gas addition (constant volume) as something that shifts equilibrium --- neither does; a catalyst only speeds up reaching equilibrium, and an inert gas at constant volume doesn't change any reacting species' partial pressure.

Stress $→$ shift to relieve it; only $ T$ changes $K$; catalyst/inert gas (constant V) $→$ no shift.

The solubility product $K_sp$ quantifies how far a sparingly soluble ionic solid dissociates in water, and comparing the ion product to $K_sp$ predicts whether a precipitate forms.

For $M_xA_y(s) <=> $x$ M^y+(aq) + $y$ A^x-(aq)$: $$K_sp = [M^y+]^x[A^x-]^y$$ Molar solubility $s$ relates to $K_sp$ via stoichiometry (e.g., $MA2$: $K_sp=4s^3$). Common-ion effect: adding a common ion LOWERS a salt's molar solubility. If $Q_sp>K_sp$, a precipitate forms.

$K_sp$ (a fixed equilibrium constant at a given temperature) vs. molar solubility $s$ (a concentration that varies with common ions --- NOT interchangeable, and their relationship depends on stoichiometry).

Assuming $s=sqrtK_sp$ for every salt --- that formula only holds for a 1:1 salt like $AgCl$; a salt like $PbI2$ needs $K_sp=4s^3$, not $K_sp=s^2$.

$K_sp=$[ions] to coefficients; common ion $→$ lowers solubility; $Q_sp>K_sp→$ precipitate forms.

A Bronsted--Lowry acid donates a proton to a base, forming a conjugate acid-base pair; the strength of that transfer is quantified by $K_a$/$K_b$ and reflected directly in pH.

$$K_w = K_a× K_b = [H3O+][OH-] = 1.0×10^-14 at 25^$$ $$pH = - pOH = - pH+pOH=14$$ The stronger an acid, the weaker its conjugate base (and vice versa).

Strong acid/base (100% dissociates, $K_a$ or $K_b1$) vs. concentrated acid/base (simply a large amount dissolved, independent of how completely it dissociates) --- a dilute strong acid can have lower $[H+]$ than a concentrated weak acid.

Confusing "strong" with "concentrated," or forgetting $K_w=1.0×10^-14$ holds only AT 25$^$C --- at other temperatures $K_w$ changes, so pH$=7$ is not always neutral.

$K_w=K_aK_b=10^-14$ at 25$^$C; pH$+$pOH$=14$; strong $≠$ concentrated.

Strong acids/bases dissociate completely, so $[H+]$ follows directly from stoichiometry; weak acids/bases only partially dissociate, requiring an equilibrium (ICE/$K_a$) calculation.

The 6 strong acids: $HCl, HBr, HI, HNO3, H2SO4, HClO4$; strong bases: group 1/2 hydroxides. $$Percent Ionization = /[H+]_eq[HA]_0×100$$ Percent ionization of a weak acid INCREASES upon dilution, even though $[H+]$ itself decreases.

Percent ionization rising upon dilution (a greater FRACTION ionizes) vs. actual $[H+]$/pH still moving toward neutral upon dilution (fewer total moles of acid per liter, even with a higher ionized fraction).

Solving a weak acid's pH the same way as a strong acid's (direct stoichiometry) --- a weak acid ALWAYS requires an ICE table/$K_a$ equilibrium calculation, since only a small fraction dissociates.

Strong acid: pH from stoichiometry. Weak acid: pH from ICE $+ K_a$. Dilution: % ionization up, but solution still less acidic.

A titration curve's shape (pH vs. volume of titrant) reveals whether the acid/base is strong or weak and identifies two key points: half-equivalence and equivalence.

Strong acid $+$ strong base: equivalence pH$=7$. Weak acid $+$ strong base: equivalence pH$>7$ (conjugate base hydrolyzes water). Strong acid $+$ weak base: equivalence pH$<7$. At the HALF-equivalence point of a weak-acid titration: $pH = pK_a$ (maximum buffering, flattest region). The equivalence point is the curve's steepest inflection point.

Equivalence point (moles acid $=$ moles base by stoichiometry; pH depends on what species remains, not always 7) vs. half-equivalence point (weak titrations only; pH$=$p$K_a$ exactly).

Assuming every titration's equivalence point is pH 7 --- true only for strong acid/strong base; a weak acid/strong base equivalence point is basic (pH$>7$) because the conjugate base hydrolyzes water.

Half-equivalence: pH$=$p$K_a$. Equivalence pH: $=7$ (strong/strong), $>7$ (weak acid/strong base), $<7$ (strong acid/weak base).

A buffer --- a weak acid and its conjugate base (or weak base and conjugate acid) both present in significant amounts --- resists large pH swings by consuming small additions of strong acid or base.

$$pH = pK_a + /[A^-][HA]$$ Buffer capacity is greatest when $[A^-]≈[HA]$ (pH$≈$p$K_a$). Adding strong acid converts A$^-→$HA; adding strong base converts HA$→$A$^-$; a buffer loses effectiveness once one component is nearly consumed.

A buffer solution (contains BOTH a weak acid and comparable conjugate base, resists pH change) vs. a plain weak acid solution alone (no significant conjugate base, pH changes readily with small additions).

Plugging initial (pre-reaction) mole amounts straight into Henderson--Hasselbalch instead of first reacting away any added strong acid/base stoichiometrically --- always do the neutralization step FIRST, then apply the equation to the resulting moles.

pH$=$p$K_a+([A^-]/[HA])$; best buffering when [A$^-$]$≈$[HA]; react first, THEN apply Henderson--Hasselbalch.

For related molecules, acid strength ($K_a$) can be predicted from bond polarity/strength and, for oxyacids, the number of oxygens bonded to the central atom.

Binary acids ($H-X$): strength increases as the H--X bond gets weaker/longer down a group ($HI>HBr>HCl>HF$) and with X's electronegativity across a period. Oxyacids: strength INCREASES with more oxygens on the central atom (they pull electron density from the O--H bond and stabilize the resulting conjugate base), e.g. $HClO4>HClO3>HClO2>HClO$.

The binary-acid trend (driven by H--X bond strength/length down a group) vs. the oxyacid trend (driven by the number of additional oxygens/electronegativity of the central atom, not bond length).

Applying "weaker bond $=$ stronger acid" binary-acid logic directly to oxyacids, or vice versa --- the two trends have different underlying causes and must be reasoned separately.

Binary acids: weaker H--X bond (bigger X, down a group) $→$ stronger acid. Oxyacids: more O's on central atom $→$ stronger acid.

Entropy measures the number of accessible microstates, and the second law requires that total entropy of the universe (system $+$ surroundings) increase for any spontaneous process.

$$ S_univ = S_sys + S_surr > 0 for a spontaneous process$$ Entropy generally increases with: more moles of gas produced, phase change solid$→$liquid$→$gas, higher temperature, mixing/dissolving. An exothermic reaction increases $ S_surr$ (related to $- H_sys/T$).

$ S_sys$ (entropy change of the reaction alone, can be positive OR negative) vs. $ S_univ$ (system $+$ surroundings; MUST be positive for spontaneity --- the actual second-law criterion).

Judging spontaneity from $ S_sys$ alone --- many spontaneous reactions have $ S_sys<0$ as long as they release enough heat to raise $ S_surr$ by more.

$ S_univ= S_sys+ S_surr>0$ for spontaneous; more gas moles/mixing $→$ higher entropy.

Gibbs free energy combines enthalpy and entropy into one spontaneity criterion at constant $T$ and $P$, and its sign depends on the interplay of $ H$, $ S$, and $T$.

$$ G = H - T S$$ $ G<0$: spontaneous as written; $ G>0$: nonspontaneous as written; $ G=0$: equilibrium. Sign combinations: $(- H,+ S)$ always spontaneous; $(+ H,- S)$ never spontaneous; $(- H,- S)$ spontaneous at LOW $T$; $(+ H,+ S)$ spontaneous at HIGH $T$. Also: $$ G^ = -RT K G^ = -nFE^_cell$$

$ G^$ (standard free energy, defines $K$, fixed at a given $T$) vs. $ G$ (free energy under actual/nonstandard conditions, changes as $Q$ changes, $=0$ exactly at equilibrium).

Concluding a reaction with $ G^>0$ can NEVER occur --- it only means $K<1$ (products not favored at standard conditions); the reaction can still proceed until $Q=K$, i.e. $ G=0$.

$ G= H-T S$; $ G<0→$spontaneous; $ G^=-RT K=-nFE^_cell$.

A galvanic cell harnesses a spontaneous redox reaction, physically separating oxidation and reduction to force electrons through an external circuit and generate a measurable voltage.

Oxidation occurs at the ANODE ("AN OX"); reduction occurs at the CATHODE ("RED CAT"). Electrons flow anode $→$ cathode through the external wire. $$E^_cell = E^_cathode - E^_anode$$ (using standard reduction potentials for both). A spontaneous galvanic cell has $E^_cell>0$.

Anode (oxidation, negative electrode in a galvanic cell, mass typically LOST as metal dissolves) vs. cathode (reduction, positive electrode, mass typically GAINED as metal plates out).

Subtracting $E^$ values in the wrong order (anode $-$ cathode instead of cathode $-$ anode), flipping the sign and making a spontaneous cell look nonspontaneous --- always identify the reduction half-reaction (higher/more positive $E^$) as cathode first.

AN OX (anode$=$oxidation), RED CAT (cathode$=$reduction); $E^_cell=E^_cathode-E^_anode$; spontaneous $→ E^_cell>0$.

An electrolytic cell uses an external power source to force a nonspontaneous redox reaction, and the amount of product formed is directly quantifiable from current and time applied.

Electrolytic cell: $E^_cell<0$ (nonspontaneous without outside energy). Anode $=$ oxidation and cathode $=$ reduction still, but electrode POLARITY reverses vs. a galvanic cell. $$Q = It n(mol e^-) = /QF, F=96,485\ C/mol e^-$$ Use the balanced half-reaction's electron count to convert moles of electrons to moles/grams of product.

Galvanic cell (spontaneous, $E^_cell>0$, chemical$→$electrical energy, anode is the negative terminal) vs. electrolytic cell (nonspontaneous, $E^_cell<0$, electrical$→$chemical energy, anode is now the POSITIVE terminal).

Forgetting to convert current $×$ time into moles of electrons (via Faraday's constant) before using half-reaction stoichiometry to find moles/grams of metal deposited --- a common multi-step numeric slip; also forgetting that anode/cathode labels never swap, only polarity does.

$Q=It$; mol $e^-=Q/F$; anode$=$oxidation always, but electrode polarity flips galvanic vs. electrolytic.

A table of standard reduction potentials ranks half-reactions by how readily they occur as reduction, predicting both the direction a redox reaction runs and its overall cell potential.

A more POSITIVE $E^_red$ means a stronger oxidizing agent (more easily reduced). To build a spontaneous reaction, pair the half-reaction with the more positive $E^_red$ as reduction (cathode) and reverse the other (flip its sign) as oxidation (anode). NEVER multiply $E^$ by a stoichiometric coefficient when balancing electrons --- cell potential is intensive, not extensive.

Scaling $ H$ or $ G$ by moles/coefficients when balancing an equation vs. $E^$ values, which are NEVER multiplied by coefficients when balancing electrons (potential is an intensive property).

Multiplying $E^$ by a coefficient when scaling a half-reaction to balance electrons transferred (treating it like an extensive property such as $ H$) --- one of the most common electrochemistry errors; $E^$ values are used as-is regardless of scaling.

More positive $E^_red=$ better oxidizing agent $=$ cathode; NEVER scale $E^$ by coefficients (intensive property).

One formula sheet: gas law, thermo, equilibrium, acid-base, electrochemistry --- know all cold.

When two terms sound alike, ask: fixed or variable? standard or actual? per-mole or per-mass?

Reaction type first, then the correct equation/law follows automatically.

Know the 6 strong acids and the key constants cold --- the exam reference sheet gives formulas, not these lists.

Identify the ask $→$ write the equation symbolically $→$ carry units $→$ sanity-check the sign $→$ answer in words.

60 MCQ (50%, 90 min) + 7 FRQ --- 3 long, 4 short (50%, 105 min); calculator allowed, reference sheet provided.

Balanced equation $→ K$ expression $→$ ICE table $→$ solve for $x$ $→$ check 5% rule $→$ final answer.

Identify which variable is plotted against which --- the plot type itself usually tells you the concept being tested.