Michael Farabaugh 2026 AP Chemistry Practice Exam Study Guide

General Information and Exam Conditions

  • Total Content: Michael Farabaugh's 2026 AP Chemistry Practice Exam (Multiple Choice Section).

  • Total Questions: 60 Multiple Choice Questions.

  • Time Limit: 1 hour and 30 minutes.

  • Standard Conditions: Unless otherwise specified, calculations assume a temperature of 298K298\,K, a pressure of 1.0atm1.0\,atm, and aqueous solutions.

  • Formula References: All calculations and justifications utilize standard chemical principles, including the Ideal Gas Law (PV=nRTPV = nRT), thermodynamics (ΔG=ΔHTΔS\Delta G = \Delta H - T\Delta S), and stoichiometry.

Gases and Kinetic Molecular Theory

  • Partial Pressure in Mixtures:     * In a mixture of Ar(g)Ar(g) and O2(g)O_2(g), the partial pressure of a gas is proportional to its mole fraction (XiX_i).     * Formula: Pi=Xi×PtotalP_i = X_i \times P_{total}.     * Example: A particle diagram shows 4 ArAr atoms and 2 O2O_2 molecules (total 6 particles). With a total pressure of 3.0atm3.0\,atm, the partial pressure of O2O_2 is PO2=26×3.0atm=1.0atmP_{O2} = \frac{2}{6} \times 3.0\,atm = 1.0\,atm.

  • Addition of Non-Reactive Gases:     * Adding an inert gas (like NeNe) to a rigid container at constant temperature increases the total pressure (PtotalP_{total}) but does not change the partial pressures of the existing gases (PiP_i) because the moles, temperature, and volume for those specific gases remain constant.

  • Deviations from Ideal Behavior:     * Gases deviate most from ideal behavior at high pressures and low temperatures.     * Molecular factors: High intermolecular forces (IMF) and large molecular volumes increase deviations. CF4CF_4 is more likely to deviate than H2H_2, O2O_2, or CH4CH_4 due to having a larger, more polarizable electron cloud resulting in stronger London dispersion forces.

  • Maxwell-Boltzmann Distribution:     * Particle speed distributions depend on temperature and molar mass.     * At the same temperature, heavier gases have a lower average speed and a narrower distribution (Curve 1) compared to lighter gases (Curve 2).     * Example: Ar(g)Ar(g) (MW40g/molMW \approx 40\,g/mol) at 300K300\,K would be represented by a curve shifted left compared to Ne(g)Ne(g) (MW20g/molMW \approx 20\,g/mol) at 300K300\,K.

Kinetics and Reaction Rates

  • Factors Affecting Rate:     * Concentration: Increasing the concentration of reactants (e.g., HCl(aq)HCl(aq)) generally increases the rate and decreases the time to completion.     * Surface Area: Fine powders react faster than single chunks due to increased contact area.     * Temperature: Higher temperatures increase kinetic energy and collision frequency.

  • Rate Laws:     * For a second-order reaction (e.g., 2NO2(g)2NO(g)+O2(g)2NO_2(g) \rightarrow 2NO(g) + O_2(g)), the rate is proportional to the square of the concentration: Rate=k[NO2]2\text{Rate} = k[NO_2]^2.     * If concentration triples (0.020M0.060M0.020\,M \rightarrow 0.060\,M), the rate increases by a factor of 32=93^2 = 9.

  • Reaction Mechanisms:     * The rate-determining step (RDS) is the slowest step, represented by the highest activation energy peak on an energy profile.     * Thermodynamics of the overall reaction depends on the energy difference between reactants and products (ΔHobs=EproductsEreactants\Delta H_{obs} = E_{products} - E_{reactants}).

  • Half-life (t1/2t_{1/2}):     * A constant half-life is a signature of first-order kinetics.     * Rate constant (kk) for first-order reactions has units of s1s^{-1}.

Atomic Structure and Periodic Trends

  • Ionization Energy:     * First ionization energy generally increases across a period (left to right) and decreases down a group.     * Example: SrSr has a larger first ionization energy than RbRb, CsCs, or BaBa.

  • Atomic and Ionic Radii:     * Radii increase down a group and decrease across a period due to effective nuclear charge (ZeffZ_{eff}).     * Anions (e.g., S2S^{2-}) are larger than their neutral atoms; cations are smaller.     * Order: Cl < S < S^{2-}.

  • Photoelectron Spectroscopy (PES):     * Peak position (binding energy) corresponds to the attraction of electrons to the nucleus.     * 1s1s electrons are closest to the nucleus and have the highest binding energy (peaks appearing further left on the x-axis).

  • Mass Spectrometry:     * Determines relative abundance of isotopes.     * Example: If 85Rb^{85}Rb has a 72%72\% abundance and 87Rb^{87}Rb has a 28%28\% abundance, the peak at 85 will be roughly 2.5 times higher than the peak at 87.

Chemical Bonding and Intermolecular Forces

  • VSEPR and Geometry:     * T-shaped Geometry: Results from 5 electron domains (AX3E2AX_3E_2), such as in ClF3ClF_3.

  • Lattice Enthalpy:     * Strength of ionic bonds in a solid crystal. Calculated via Coulomb's Law: Fq1q2r2F \propto \frac{q_1 q_2}{r^2}.     * Higher charges (Mg2+Mg^{2+} vs Na+Na^+) and smaller radii (FF^- vs BrBr^-) result in stronger attractions. MgF2MgF_2 has stronger attractions than NaFNaF, NaBrNaBr, or MgBr2MgBr_2.

  • Resonance and Bond Order:     * Molecules with resonance (like NO2NO_2^-) have intermediate bond orders. If there is one single bond and one double bond in two resonance structures, the bond order is 1.51.5.

  • Intermolecular Forces (IMFs) and Boiling Points:     * London Dispersion Forces: Present in all molecules; strength increases with polarizability (larger electron clouds). CH3ICH_3I (BP 316K316\,K) has a higher boiling point than CH3ClCH_3Cl (BP 195K195\,K) because Iodine has a larger, more polarizable electron cloud.     * Dipole-Dipole: Present in polar molecules (e.g., CH2F2CH_2F_2).     * Hydrogen Bonding: Occurs when H is bonded to N, O, or F. This significantly raises boiling points (e.g., alcohols vs ethers).

  • Chromatography:     * Separation based on differential affinity for stationary and mobile phases.     * In a polar stationary phase with a nonpolar solvent, a nonpolar molecule (like benzene) will travel faster/further than a polar molecule (like phenol).

Thermochemistry and Thermodynamics

  • Enthalpy (\Delta H):     * Bond breaking is endothermic (\Delta H > 0); bond formation is exothermic (\Delta H < 0).

  • Standard Enthalpy of Formation (\Delta H_f^o):     * ΔHrxno=ΣΔHfo(products)ΣΔHfo(reactants)\Delta H_{rxn}^o = \Sigma \Delta H_f^o (products) - \Sigma \Delta H_f^o (reactants).     * For CCl4(g)+4HF(g)CF4(g)+4HCl(g)CCl_4(g) + 4HF(g) \rightarrow CF_4(g) + 4HCl(g), with ΔHo=106kJ/molrxn\Delta H^o = -106\,kJ/mol_{rxn}, ΔHfo\Delta H_f^o for CCl4(g)CCl_4(g) is calculated to be 103kJ/mol-103\,kJ/mol.

  • Entropy (\Delta S):     * Related to disorder. Processes increasing moles of gas have \Delta S > 0.     * If moles of gas remain constant (CH4(g)+2O2(g)CO2(g)+2H2O(g)CH_4(g) + 2O_2(g) \rightarrow CO_2(g) + 2H_2O(g)), ΔS\Delta S is close to zero.

  • Gibbs Free Energy and Favorability:     * ΔGo=ΔHoTΔSo\Delta G^o = \Delta H^o - T\Delta S^o.     * Thermodynamically favorable reactions have \Delta G^o < 0.     * If \Delta H^o < 0 and \Delta S^o < 0, the reaction is favorable at low temperatures but unfavorable at high temperatures (e.g., NH3(g)+HCl(g)NH4Cl(s)NH_3(g) + HCl(g) \rightarrow NH_4Cl(s)).

  • Specific Heat Calculations:     * q=mcΔTq = mc\Delta T.     * Example: A 71.0g71.0\,g sample of AlAl absorbing 6.24kJ6.24\,kJ: ΔT=6240J71.0g×0.896J/goC98.1oC\Delta T = \frac{6240\,J}{71.0\,g \times 0.896\,J/g\cdot^oC} \approx 98.1^oC. Final temperature: 22.0+98.1=120.1oC22.0 + 98.1 = 120.1^oC.

Chemical Equilibrium

  • Equilibrium Constants (KpK_p and KcK_c):     * Expression excludes solids and liquids. For CH3NH2(aq)+H2O(l)CH3NH3+(aq)+OH(aq)CH_3NH_2(aq) + H_2O(l) \rightleftharpoons CH_3NH_3^+(aq) + OH^-(aq), K=[CH3NH3+][OH][CH3NH2]K = \frac{[CH_3NH_3^+][OH^-]}{[CH_3NH_2]}.

  • The Reaction Quotient (QQ):     * If Q < K, the reaction shifts right (toward products).     * If Q > K, the reaction shifts left (toward reactants).

  • Le Chatelier’s Principle:     * Increasing temperature for an exothermic reaction (\Delta H < 0) shifts equilibrium to the left, increasing reactants and decreasing the value of KpK_p.

  • Thermodynamics and Equilibrium:     * ΔGo=RTln(K)\Delta G^o = -RT\ln(K).     * Large KK values (K > 1) correspond to positive EcelloE^o_{cell} and negative ΔGo\Delta G^o.

Acids, Bases, and Solubility

  • Weak vs. Strong Acids:     * For a strong acid, pH=log[Acid]pH = -\log[Acid]. If observed pH > -\log[Acid], the acid is weak.     * Sample calculation: For a 0.20M0.20\,M acid with pH=2.7pH = 2.7, [H+]=102.72×103M[H^+] = 10^{-2.7} \approx 2 \times 10^{-3} M. Ka=(2×103)20.20=2×105K_a = \frac{(2 \times 10^{-3})^2}{0.20} = 2 \times 10^{-5}.

  • Titrations:     * The equivalence point volume depends on the total moles of acid (n=M×Vn = M \times V).     * At the half-equivalence point, pH=pKapH = pK_a.     * Doubling the titrant concentration (e.g., 0.0750M0.150MNaOH0.0750\,M \rightarrow 0.150\,M\,NaOH) halves the volume required to reach the equivalence point.

  • Buffers:     * Buffer capacity depends on the concentration of the buffer components. A 1.0M1.0\,M buffer has a higher capacity and smaller pHpH change than a 0.10M0.10\,M buffer when a base is added.

  • Salt pH:     * Salts of weak acids (like NaFNaF) are basic because the conjugate base (FF^-) hydrolyzes water to produce OHOH^-: F+H2OHF+OHF^- + H_2O \rightleftharpoons HF + OH^-.

  • Solubility Product (KspK_{sp}):     * Solubility (molarity of a saturated solution) is an intensive property and does not change with volume. However, the total number of moles of dissolved ions increases with volume in a saturated solution.

Stoichiometry and Solutions

  • Limiting Reactants:     * Determined by the mole ratio in the balanced equation.     * Example: 4Cr+3O22Cr2O34Cr + 3O_2 \rightarrow 2Cr_2O_3. Given 52gCr52\,g\,Cr (1.0mol1.0\,mol) and 32gO232\,g\,O_2 (1.0mol1.0\,mol). 1.0molCr1.0\,mol\,Cr requires 0.75molO20.75\,mol\,O_2. Since we have excess O2O_2, CrCr is the limiting reactant. Maximum Cr2O3Cr_2O_3 produced is 0.5mol0.5\,mol, or 76g76\,g.

  • Particulate Models of Solutions:     * Ion-dipole interactions: Positive ions (K+K^+) are surrounded by the oxygen ends (partial negative) of water; negative ions (ClCl^-) are surrounded by the hydrogen ends (partial positive).

  • Percent Composition:     * Mass %=Mass of elementTotal mass of compound×100\text{Mass \%} = \frac{\text{Mass of element}}{\text{Total mass of compound}} \times 100.     * NH3NH_3 has a higher N %\text{N \%} (82%\approx 82\%) than NO2NO_2, N2ON_2O, or HNO3HNO_3.