AP Chemistry
Atom yapısından elektrokimyaya: konu anlatımı, denklemler ve hata teşhisli sorular.
Kritik Kimya Denklemleri (Tıkla ve Kopyala)
ΔG° < 0 ise tepkime istemlidir. K > 1 ve E°_cell > 0 durumlarıyla birebir örtüşür.
Tampon çözeltilerde zayıf asit ve konjuge baz derişimleri eşitlendiğinde pH = pKa olur (Yarı-eşdeğerlik noktası).
ln[A]'ya karşı zaman (t) grafiği doğrusal çıkıyorsa tepkime 1. derecedendir ve eğim = -k olur.
A: Absorbans, ε: Molar absorpsiyon katsayısı, b: Işık yolu uzunluğu (küvet boyu), c: Molar derişim.
AP Kimya 9 Temel Ünite Haritası
AP Chemistry Theoretical Foundations & Molecular Dynamics
Thermodynamic principles, reaction mechanisms, orbital theory, and quantitative equilibrium calculations required for College Board 5/5 mastery and European chemical engineering standards:
Coulomb's Law as the Governing Principle: The force between two charges is proportional to their charges and inversely proportional to the square of the distance between them:
\(F = k \cdot |q_1 \cdot q_2| / r^2 \quad \text{and} \quad U_E = k \cdot (q_1 \cdot q_2) / r\)
\(E_{\text{photon}} = \text{Binding Energy} + \text{Kinetic Energy} \implies BE = h\nu - KE\)
Peak Height: Proportional to the relative number of electrons in that subshell (e.g., $2s^2$ is one-third the height of $2p^6$).
Peak Position: Electrons closer to the nucleus (higher effective nuclear charge $Z_{eff}$) require higher binding energy and appear further to the left on the energy axis.
Hierarchy of Intermolecular Forces (IMF):
1. Ion-Dipole: Strongest interaction in aqueous solutions (e.g., $\text{Na}^+$ surrounded by polar $\text{H}_2\text{O}$ dipoles).
2. Hydrogen Bonding: Special, unusually strong dipole-dipole interaction occurring only when $\text{H}$ is covalently bonded to highly electronegative $\text{N, O,}$ or $\text{F}$.
3. Dipole-Dipole: Electrostatic attraction between permanent molecular dipoles.
4. London Dispersion Forces (LDF): Temporary instantaneous dipoles present in all molecules. Strength increases with polarizability (greater number of electrons / more diffuse electron cloud).
Rate Laws & Reaction Orders: For $aA + bB \to \text{Products}$, $\text{Rate} = k [A]^m [B]^n$.
• Zero Order ($m=0$): $[A]_t = -kt + [A]_0$ (Linear plot: $[A]$ vs $t$; slope $= -k$). Half-life decreases as concentration drops.
• First Order ($m=1$): $\ln[A]_t = -kt + \ln[A]_0$ (Linear plot: $\ln[A]$ vs $t$; slope $= -k$). Constant half-life: $t_{1/2} = \frac{\ln 2}{k} = \frac{0.693}{k}$.
• Second Order ($m=2$): $\frac{1}{[A]_t} = kt + \frac{1}{[A]_0}$ (Linear plot: $1/[A]$ vs $t$; slope $= +k$). Half-life increases as concentration drops.
Gibbs Free Energy & Equilibrium: A reaction is thermodynamically favorable when $\Delta G^\circ < 0$:
\(\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ = -RT \ln(K) = -n F E^\circ _{\text{cell}}\)
• $\Delta H < 0$ and $\Delta S > 0 \implies \Delta G < 0$ at all temperatures.
• $\Delta H > 0$ and $\Delta S < 0 \implies \Delta G > 0$ (unfavorable) at all temperatures.
\(\text{pH} = \text{pK}_a + \log( [A^-] / [HA] )\)
Maximum Buffer Capacity: Occurs at the half-equivalence point where $[HA] = [A^-]$, so $\text{pH} = \text{pK}_a$.
Adım Adım Çözümlü AP FRQ
AP tarzı özgün soru; puanlama ölçütleriyle adım adım çözüm.
SORU (Acids, Bases & Buffer): Bir öğrenciye 25.0 mL 0.100 M zayıf asit HA (Ka = 2.0 × 10⁻⁵) çözeltisi verilmiştir. Bu çözelti 0.100 M kuvvetli baz NaOH ile titre edilmektedir.
(a) Titrasyon başlamadan önce saf HA çözeltisinin pH değerini hesaplayınız.
(b) 12.5 mL NaOH eklendiğinde oluşan çözeltinin pH değerini gerekçesiyle belirtiniz.
(c) Eşdeğerlik noktasına ulaşmak için kaç mL NaOH gerekir ve eşdeğerlik noktasındaki pH > 7 midir, = 7 midir, < 7 midir? Net iyonik denklemle açıklayınız.
• (a) Başlangıç pH [2 Puan]:
HA ⇌ H⁺ + A⁻. Ka = [H⁺][A⁻] / [HA] = x² / 0.100 = 2.0 × 10⁻⁵.x² = 2.0 × 10⁻⁶ => x = [H⁺] = 1.41 × 10⁻³ M.pH = -log(1.41 × 10⁻³) ≈ 2.85.• (b) Yarı-Eşdeğerlik Noktası [2 Puan]: Başlangıçta
0.025 L × 0.100 M = 0.0025 mol HA vardır.12.5 mL NaOH =
0.00125 mol OH⁻. HA'nın yarısı A⁻'ye dönüşür: [HA] = [A⁻].Henderson-Hasselbalch:
pH = pKa + log([A⁻]/[HA]) = pKa + log(1) = pKa.pKa = -log(2.0 × 10⁻⁵) = 5 - log(2) ≈ 4.70. Cevap: pH = 4.70.• (c) Eşdeğerlik Noktası [3 Puan]:
0.0025 mol HA için tam 0.0025 mol NaOH, yani 25.0 mL NaOH gerekir.Oluşan tuz: Na⁺ ve A⁻. A⁻ zayıf asidin konjuge bazı olduğundan hidroliz olur:
A⁻(aq) + H₂O(l) ⇌ HA(aq) + OH⁻(aq).Çözeltide OH⁻ üretildiği için pH > 7 (bazik) olur.
Unit 1: Atomic Structure and Properties
The Mole & Avogadro's Constant: One mole of any chemical entity contains precisely $N_A = 6.022 \times 10^{23}$ particles (atoms, molecules, or formula units). Molar mass $M$ ($\text{g/mol}$) provides the conversion factor between macroscopic mass and microscopic particle counts:
\(n = \frac{m}{M} \quad \text{and} \quad N = n \cdot N_A\)
\(\text{Average Atomic Mass} = \sum_{i=1}^k \left( \text{Isotopic Mass}_i \times \text{Relative Fractional Abundance}_i \right)\)
Example — Diatomic Molecule Spectra ($\text{Cl}_2$): Chlorine has two naturally occurring isotopes: $^{35}\text{Cl}$ (75%) and $^{37}\text{Cl}$ (25%). The mass spectrum of $\text{Cl}_2$ exhibits three distinct molecular ion peaks at $m/z = 70, 72,$ and $74$ with relative intensity ratios governed by the binomial distribution:
• $^{35}\text{Cl}-^{35}\text{Cl}$ ($m/z = 70$): $(0.75)^2 = 0.5625$ (Relative: $9$)
• $^{35}\text{Cl}-^{37}\text{Cl}$ ($m/z = 72$): $2(0.75)(0.25) = 0.3750$ (Relative: $6$)
• $^{37}\text{Cl}-^{37}\text{Cl}$ ($m/z = 74$): $(0.25)^2 = 0.0625$ (Relative: $1$). Ratio: $\mathbf{9:6:1}$!
Law of Definite Proportions: Pure samples of any chemical compound always contain identical proportions of elements by mass.
\(\% \text{ Mass of Element } X = \frac{\text{Mass of } X \text{ in 1 mole}}{\text{Total Molar Mass of Compound}} \times 100\%\)
1. Empirical Formula: The simplest whole-number ratio of atoms in a compound. Convert elemental mass percentages to moles, then divide by the smallest mole value.
2. Molecular Formula: $( \text{Empirical Formula} )_k$, where integer multiplier $k = \frac{\text{Experimental Molar Mass}}{\text{Empirical Formula Mass}}$.
3. Combustion Analysis: Combustion of hydrocarbon $C_x H_y O_z$ in excess $O_2$ converts all carbon into $\text{CO}_2$ ($n_C = n_{\text{CO}_2}$) and all hydrogen into $\text{H}_2\text{O}$ ($n_H = 2 \times n_{\text{H}_2\text{O}}$). The mass of oxygen is determined by difference: $m_O = m_{\text{sample}} - (m_C + m_H)$.
Quantum Electronic Architecture:
• Aufbau Principle: Electrons occupy subshells of lowest energy first ($1s < 2s < 2p < 3s < 3p < 4s < 3d$).
• Pauli Exclusion Principle: An orbital can hold a maximum of 2 electrons, which must have opposite spins ($\uparrow\downarrow$).
• Hund's Rule: Within degenerate orbitals ($p_x, p_y, p_z$), electrons occupy orbitals singly with parallel spins before pairing to minimize electron-electron repulsion.
When neutral transition metals (such as $\text{Fe}: [\text{Ar}]\,4s^2 3d^6$) are ionized, the $4s$ electrons are removed BEFORE the $3d$ electrons, because the $4s$ electrons have a higher principal quantum number ($n=4$) and reside on average further from the nucleus:
• $\text{Fe}^{2+}: [\text{Ar}]\,3d^6$ (NOT $[\text{Ar}]\,4s^2 3d^4$!)
• $\text{Fe}^{3+}: [\text{Ar}]\,3d^5$ (Stable half-filled $d$-subshell).
Based on the Photoelectric Effect: $E_{\text{photon}} = h\nu = \text{Binding Energy (BE)} + \text{Kinetic Energy (KE)}$.
1. X-axis (Binding Energy): Plotted from right-to-left (highest energy on the far left). Peaks furthest to the left correspond to core electrons closest to the nucleus ($1s$) experiencing maximal Coulombic attraction!
2. Y-axis (Relative Intensity / Peak Height): Directly proportional to the number of electrons occupying that specific subshell ($1s^2 : 2s^2 : 2p^6 \implies \text{height ratio } 2 : 2 : 6$).
3. Shifting of Peaks Across a Period: As atomic number $Z$ increases from left to right across a period, effective nuclear charge $Z_{\text{eff}}$ increases. Thus, all corresponding subshell peaks shift to higher binding energies (further to the left).
The Universal Foundation — Coulomb's Law: All periodic trends and bonding strengths are governed by the electrostatic force between charges:
\(F_e = k \cdot \frac{|q_1 \cdot q_2|}{r^2} \quad \text{and} \quad U_E = k \cdot \frac{q_1 \cdot q_2}{r}\)
Where $q_1 = Z_{\text{eff}}$ (net positive nuclear charge felt by electron), $q_2 = -1$ (electron charge), and $r$ is the average electron-nucleus distance.
• Atomic Radius: Decreases across a period because $Z_{\text{eff}} = Z - S$ increases with no change in core shielding $S$, pulling valence electrons closer to the nucleus. Increases down a group because valence electrons occupy higher principal energy levels ($n$), located further from the nucleus.
• Isoelectronic Series Radii: Ions with identical electron configurations (e.g., $\text{O}^{2-}, \text{F}^-, \text{Na}^+, \text{Mg}^{2+}$, each with 10 electrons) experience identical electron-electron repulsion. The radius is strictly determined by nuclear charge $Z$: $\text{Mg}^{2+}$ ($Z=12$) pulls electron clouds closest $\implies \text{Mg}^{2+} < \text{F}^- < \text{O}^{2-}$ ($Z=8$).
1. Group 2 vs Group 13 ($\text{Be}$ vs $\text{B}$): First IE of $\text{B}$ ($801\text{ kJ/mol}$) is lower than $\text{Be}$ ($899\text{ kJ/mol}$) because the lone $2p^1$ electron in Boron is in a higher energy subshell and is shielded by the inner $2s^2$ electrons, requiring less energy to remove.
2. Group 15 vs Group 16 ($\text{N}$ vs $\text{O}$): First IE of $\text{O}$ ($1314\text{ kJ/mol}$) is lower than $\text{N}$ ($1402\text{ kJ/mol}$) because Oxygen has a pair of electrons in one of its $2p$ orbitals ($2p_x^2 2p_y^1 2p_z^1$). Mutual electron-electron repulsion between paired electrons in the same orbital makes removing one electron easier despite Oxygen's higher nuclear charge!
Lattice Energy of Ionic Solids: $U_{\text{lattice}} \propto \frac{Q_1 Q_2}{r_0}$. The ionic charge product ($Q_1 \times Q_2$) dramatically outweighs ionic radius! For example, $\text{MgO}$ ($+2 \times -2 \implies 4$) has roughly 4 times greater lattice energy and a much higher melting point ($2852^\circ\text{C}$) than $\text{NaCl}$ ($+1 \times -1 \implies 1$, melting point $801^\circ\text{C}$).
Unit 1 Atomic Structure & Properties: İnteraktif Soru İstasyonu
Özgün AP tarzı sorular (geçmiş sınav sorusu değil). Şık seç; anında çözüm ve hata türü gelir.
Unit 2: Compound Structure and Properties (eski CED adı: Molecular & Ionic Compound Structure) — Bonding & Geometry
Chemical bonds represent thermodynamic equilibria between electrostatic attraction and nuclear repulsion. Below are the foundational potential energy curves, lattice enthalpy models, alloy structural classifications, resonance criteria, VSEPR geometries, and orbital hybridization protocols required for AP Chemistry 5/5 mastery:
Bonding exists along an unbroken spectrum dictated by the electronegativity difference ($\Delta EN$) between bonded atoms:
• Nonpolar Covalent ($\Delta EN < 0.4$): Equal or nearly equal sharing of electron density (e.g., $\text{C}-\text{H}$, $\text{Cl}_2$).
• Polar Covalent ($0.4 \le \Delta EN < 1.7$): Unequal sharing; creates a bond dipole with partial charges $\delta^+$ and $\delta^-$ (e.g., $\text{H}-\text{Cl}$, $\text{C}-\text{O}$).
• Ionic ($\Delta EN \ge 1.7$): Valence electrons transferred from metal (low ionization energy) to nonmetal (high electron affinity), forming an electrostatic lattice governed by Coulomb's Law: $F_e \propto \frac{|q_1 q_2|}{r^2}$.
Not all metal-nonmetal combinations are purely ionic! High-charge, small-radius cations (e.g., $\text{Al}^{3+}$, $\text{Be}^{2+}$) strongly polarize the electron cloud of large anions, imparting substantial covalent character (e.g., anhydrous $\text{AlCl}_3$ sublimates at low temperatures and exists as covalent $\text{Al}_2\text{Cl}_6$ dimers!).
When two neutral isolated atoms approach each other from infinite separation ($r \to \infty$), their total potential energy $U(r)$ changes according to three distinct physical regimes:
1. Large Separation ($r > r_0$): Potential energy approaches zero ($U \to 0$). Negligible electrostatic interaction.
2. Optimal Equilibrium Bond Length ($r = r_0$): Potential energy reaches an absolute minimum (the "well depth"). At this point, attractive forces (nucleus-electron) exactly balance repulsive forces (nucleus-nucleus and electron-electron). The well depth equals the Bond Dissociation Energy ($E_{\text{bond}}$).
3. Very Short Separation ($r < r_0$): Potential energy rises steeply toward positive infinity ($U \gg 0$) due to intense core electron and nuclear-nuclear Coulombic repulsion.
• Higher Bond Order (Triple vs. Double vs. Single): A triple bond ($\text{N}\equiv\text{N}$) shares 6 electrons, yielding a deeper potential energy well (greater bond energy $\sim 945\text{ kJ/mol}$) and a well minimum shifted to the left (shorter bond length $\sim 110\text{ pm}$) compared to a single bond ($\text{N}-\text{N}$, $\sim 160\text{ kJ/mol}$, $\sim 145\text{ pm}$).
• Atomic Radii Effect: Smaller bonded atoms (e.g., $\text{H}-\text{F}$ vs. $\text{H}-\text{I}$) allow nuclei to approach closer, shifting the well minimum to smaller $r_0$.
Ionic solids form a rigid three-dimensional crystal lattice of alternating cations and anions maximizing attractive contacts and minimizing repulsions.
• Brittleness & Cleavage: Applying shear stress forces layers of ions to slide. This alignment shifts like-charged ions adjacent to one another ($+ \leftrightarrow +$ and $- \leftrightarrow -$). The sudden, intense electrostatic repulsion causes the crystal to fracture catastrophically along smooth cleavage planes.
• Electrical Conductivity:
- Solid state: Non-conductive; ions are immobilized in fixed lattice positions.
- Molten liquid ($l$) or Aqueous solution ($aq$): Highly conductive; the lattice dissociates, producing freely mobile mobile cations and anions that carry electric current under an applied potential.
• High Melting/Boiling Points: Overcoming immense ionic lattice energy ($U_{\text{lattice}} \propto \frac{|q_1 q_2|}{r_0}$) requires immense thermal kinetic energy.
Electron Sea Model: Pure metals consist of an array of positive metal cations immersed in a delocalized "sea" of freely mobile valence electrons. Because metallic bonding is nondirectional, metal planes can slide past one another without shattering, providing exceptional malleability, ductility, and high thermal/electrical conductivity.
1. Substitutional Alloys ($r_{\text{solute}} \approx r_{\text{host}}$):
• Solute metal atoms have an atomic radius within $\pm 15\%$ of the host atoms and replace them directly in lattice positions.
• Examples: Brass ($\text{Cu} + \text{Zn}$), Bronze ($\text{Cu} + \text{Sn}$), Pewter ($\text{Sn} + \text{Cu}/\text{Sb}$).
• Malleability remains relatively similar; density lies between host and solute metals.
2. Interstitial Alloys ($r_{\text{solute}} \ll r_{\text{host}}$):
• Much smaller solute nonmetal atoms (e.g., Carbon, Hydrogen, Boron) occupy the small interstices (holes) between host metal atoms.
• Prime Example: Steel (small Carbon atoms occupying interstices within Iron $\text{Fe}$ crystal lattice).
• Mechanical Effect: The rigid interstitial carbon atoms severely pin and disrupt the sliding of iron metal planes. As a result, steel is much harder, stronger, and significantly less malleable than pure iron!
Formal Charge ($FC$): The apparent electrical charge assigned to an atom in a molecule assuming perfectly equal sharing of all bonding electrons: $$FC = V - N_{\text{nonbonding}} - \frac{1}{2} B_{\text{bonding}}$$ where $V$ is valence electrons of free neutral atom, $N_{\text{nonbonding}}$ is total unshared lone pair electrons, and $B_{\text{bonding}}$ is total shared bonding electrons.
1. The sum of all formal charges MUST equal the overall charge of the molecule or ion.
2. The structure with formal charges closest to zero is the most stable major contributor.
3. If non-zero formal charges are unavoidable, any negative formal charge must reside on the MORE ELECTRONEGATIVE atom (e.g., on $\text{O}$ rather than $\text{C}$ or $\text{N}$).
4. Like formal charges should not reside on adjacent connected atoms.
• Incomplete Octet: $\text{Be}$ (stable with 4 valence electrons, e.g., $\text{BeCl}_2$) and $\text{B}$ (stable with 6 valence electrons, e.g., $\text{BF}_3$).
• Odd-Electron Species (Free Radicals): $\text{NO}$, $\text{NO}_2$ (unpaired single electron on nitrogen).
• Expanded Octets: Elements in Period 3 and below ($\text{P}, \text{S}, \text{Cl}, \text{Xe}$) have energetically accessible $d$ orbitals and can accommodate 10, 12, or more valence electrons (e.g., $\text{PCl}_5, \text{SF}_6, \text{XeF}_4, \text{SO}_4^{2-}$). Period 2 elements ($\text{C}, \text{N}, \text{O}, \text{F}$) NEVER expand their octets!
VSEPR Core Principle: Electron domains (lone pairs, single, double, or triple bonds each count as ONE domain) repel one another and position themselves as far apart as possible in three-dimensional space.
$$\text{Lone Pair - Lone Pair} > \text{Lone Pair - Bonding Pair} > \text{Bonding Pair - Bonding Pair}$$ Nonbonding lone pairs are held by only one positive nucleus and occupy larger angular space, compressing adjacent bonding angles below ideal values:
• $\text{CH}_4$ (0 lone pairs): Ideal tetrahedral angle = $109.5^\circ$.
• $\text{NH}_3$ (1 lone pair): Trigonal pyramidal = $107.0^\circ$ ($< 109.5^\circ$).
• $\text{H}_2\text{O}$ (2 lone pairs): Bent = $104.5^\circ$ ($\ll 109.5^\circ$).
• Steric 2 ($sp$): Linear ($180^\circ$).
• Steric 3 ($sp^2$): Trigonal planar ($120^\circ$); with 1 lone pair = Bent ($< 120^\circ$).
• Steric 4 ($sp^3$): Tetrahedral ($109.5^\circ$); with 1 lone pair = Trigonal Pyramidal ($< 109.5^\circ$); with 2 lone pairs = Bent ($< 109.5^\circ$).
• Steric 5 ($sp^3d$ / expanded): Trigonal bipyramidal ($90^\circ, 120^\circ$); with 1 lone pair = Seesaw; with 2 lone pairs = T-shaped ($< 90^\circ$); with 3 lone pairs = Linear ($180^\circ$). (Lone pairs always occupy equatorial positions!)
• Steric 6 ($sp^3d^2$ / expanded): Octahedral ($90^\circ$); with 1 lone pair = Square Pyramidal ($< 90^\circ$); with 2 lone pairs = Square Planar ($90^\circ, 180^\circ$, trans-axial lone pairs cancel dipole $\implies$ nonpolar!).
• Every single bond $= 1\sigma$ bond (head-on overlap along internuclear axis; free rotation).
• Every double bond $= 1\sigma + 1\pi$ bond (side-by-side unhybridized $p$-orbital overlap; locks rotation).
• Every triple bond $= 1\sigma + 2\pi$ bonds.
Unit 2 Molecular & Ionic Structure: İnteraktif Soru İstasyonu
Özgün AP tarzı sorular (geçmiş sınav sorusu değil). Şık seç; anında çözüm ve hata türü gelir.
Unit 3: Properties of Substances and Mixtures (eski CED adı: Intermolecular Forces and Properties) — Kinetic & Spectroscopic Mastery
Intermolecular forces (IMFs) are attractive electrostatic interactions between separate molecules, distinct from intramolecular covalent/ionic bonds.
1. London Dispersion Forces (LDF): Present in ALL atoms and molecules without exception. Caused by momentary, unsymmetrical fluctuations in electron density creating instantaneous dipoles that induce complementary dipoles in neighboring atoms.
• The Governing Factor — Polarizability: The ease with which the electron cloud can be distorted. Species with a greater total number of electrons have larger, more diffuse electron clouds that are more polarizable, generating dramatically stronger LDFs (e.g., Boiling Points: $\text{F}_2 \, (85\text{ K}) < \text{Cl}_2 \, (239\text{ K}) < \text{Br}_2 \, (332\text{ K}) < \text{I}_2 \, (457\text{ K})$).
• Molecular Shape / Contact Area: Linear chains (e.g., n-pentane, bp $36^\circ\text{C}$) provide greater surface contact area for dispersion interactions than compact spherical isomers (e.g., neopentane / 2,2-dimethylpropane, bp $9.5^\circ\text{C}$).
2. Dipole-Dipole Attractions: Electrostatic attractions between the permanent $\delta^+$ dipole of one polar molecule and the $\delta^-$ dipole of another (e.g., $\text{HCl}, \text{CH}_3\text{Cl}$).
3. Hydrogen "Bonding": An unusually strong dipole-dipole attraction occurring ONLY when Hydrogen is covalently bonded to a small, highly electronegative atom with unshared lone pairs: strictly Nitrogen ($\text{N}$), Oxygen ($\text{O}$), or Fluorine ($\text{F}$) (e.g., $\text{H}_2\text{O}, \text{HF}, \text{NH}_3$). Generates anomalously elevated boiling points, high specific heat, and open hexagonal crystal structures upon freezing ($\text{density of ice} < \text{liquid water}$).
4. Ion-Dipole Forces: Electrostatic attraction between an ionic charge and permanent solvent dipoles (e.g., $\text{Na}^+$ or $\text{Cl}^-$ hydrated by aqueous $\text{H}_2\text{O}$). Stronger than all other IMFs and responsible for the dissolution of ionic lattices ($\Delta H_{\text{hydration}}$).
Never assume that hydrogen bonding is always stronger than London dispersion forces! A large nonpolar molecule with hundreds of electrons (e.g., heavy paraffin waxes or solid $\text{I}_2$) has vastly stronger cumulative LDFs and a much higher melting point than small hydrogen-bonding molecules like liquid $\text{H}_2\text{O}$ or volatile $\text{NH}_3$. Always compare the size of electron clouds!
The 4 Archetypes of Crystalline Solids:
• Molecular Solids: Discrete molecules held by weak IMFs (LDF, dipole-dipole, H-bonds) (e.g., dry ice $\text{CO}_2(s)$, sucrose, $\text{I}_2(s)$, ice). Low melting points ($< 300^\circ\text{C}$), soft, non-conductive.
• Ionic Solids: Cation-anion electrostatic lattices; brittle, high melting points, electrically conductive only when molten or dissolved in water.
• Metallic Solids: Cation array in delocalized electron sea; malleable, ductile, highly conductive in solid and liquid phases.
The Ideal Gas Equation & Derived Relationships:
$$PV = nRT = \left(\frac{m}{M}\right)RT$$
where $R = 0.08206 \text{ L}\cdot\text{atm}/(\text{mol}\cdot\text{K}) = 8.314 \text{ J}/(\text{mol}\cdot\text{K})$.
• Gas Density ($d$): $d = \frac{m}{V} = \frac{P \cdot M}{R \cdot T} \implies M = \frac{dRT}{P}$ (Density increases with higher pressure, lower temperature, and higher molar mass).
• Dalton's Law of Partial Pressures: $P_{\text{total}} = P_A + P_B + \dots$; $P_A = X_A \cdot P_{\text{total}}$, where mole fraction $X_A = \frac{n_A}{n_{\text{total}}}$.
• Gas Collected Over Water: $P_{\text{dry gas}} = P_{\text{barometric}} - P_{\text{H}_2\text{O}(v)}$ (Must subtract equilibrium water vapor pressure at temperature $T$!).
Kinetic Molecular Theory (KMT) Postulates:
1. Gas particles are in continuous, random, straight-line motion.
2. Individual particle volume is completely negligible compared to the total volume of the container ($V_{\text{particles}} \approx 0$).
3. Intermolecular attractions and repulsions are completely negligible.
4. Particle collisions are perfectly elastic (no kinetic energy is lost as heat).
5. Average kinetic energy is strictly proportional to absolute Kelvin temperature: $KE_{\text{avg}} = \frac{3}{2} k_B T = \frac{1}{2} m (u_{\text{rms}})^2$.
• Effect of Molar Mass at Constant $T$: Lighter gases (e.g., $\text{He}$, $M=4\text{ g/mol}$) possess a wider, flatter curve extending to much higher speeds than heavy gases (e.g., $\text{Ar}$, $M=40\text{ g/mol}$). However, at the same temperature, both gases have the EXACT same average kinetic energy ($KE_{\text{avg}}$)!
Real gases deviate from $PV = nRT$ under two conditions: HIGH PRESSURE and LOW TEMPERATURE.
$$\left( P + a \frac{n^2}{V^2} \right) (V - nb) = nRT$$ 1. Attractive Forces (Parameter $a$): Significant at low temperatures and moderate pressures. As particles move slowly, attractive IMFs pull particles together, softening their impacts against container walls. Hence, $P_{\text{real}} < P_{\text{ideal}}$ and $\frac{PV}{nRT} < 1$. Gases with strong dipoles (e.g., $\text{NH}_3, \text{H}_2\text{O}, \text{SO}_2$) show severe negative deviations.
2. Particle Volume (Parameter $b$): Dominant at extremely high pressures. Gas particles are compressed tightly; the physical volume of the molecules is no longer negligible relative to container volume. Available free space is reduced, leading to more frequent collisions: $V_{\text{real}} > V_{\text{ideal}}$ and $\frac{PV}{nRT} > 1$.
Energetics of Solution Formation:
$$\Delta H_{\text{solution}} = \Delta H_{\text{solute-solute}} \,(> 0) + \Delta H_{\text{solvent-solvent}} \,(> 0) + \Delta H_{\text{solvation}} \,(< 0)$$
• If $\Delta H_{\text{solution}} \approx 0$ or $< 0$, dissolution is thermodynamically favored by entropy ($\Delta S > 0$).
• "Like dissolves like": Polar/ionic solutes dissolve in polar solvents via dipole-dipole or ion-dipole interactions; nonpolar solutes dissolve in nonpolar solvents via London dispersion forces.
• Column Chromatography: The component with strongest affinity for the mobile phase elutes first from the bottom of the column.
Radiation-Matter Interaction Across the Electromagnetic Spectrum:
$$c = \lambda \nu \quad \text{and} \quad E_{\text{photon}} = h\nu = \frac{hc}{\lambda}$$
• Microwave Radiation: Low energy; induces transitions in molecular rotational levels.
• Infrared (IR) Radiation: Moderate energy; matches vibrational frequencies of covalent bonds, inducing molecular vibrations (stretching and bending).
• Ultraviolet/Visible (UV-Vis) Radiation: High energy; promotes valence electrons between electronic energy levels (electronic transitions).
• Wavelength Selection ($\lambda_{\text{max}}$): The spectrophotometer must be set to the wavelength of maximum absorbance, which is complementary to the solution's observed color (e.g., a blue $\text{Cu}^{2+}(aq)$ solution absorbs orange/red light $\sim 620\text{ nm}$).
1. Fingerprints or Smudges on Cuvette: Light is scattered/reflected $\implies$ measured absorbance $A$ is artificially high $\implies$ calculated concentration $c$ is OVERESTIMATED.
2. Water Droplets Left Inside Cuvette Prior to Solution Addition: Dilutes the solution $\implies$ measured $A$ is lower $\implies$ calculated concentration is UNDERESTIMATED.
3. Air Bubbles in Optical Path: Scatters transmitted light $\implies$ falsely high $A \implies$ concentration OVERESTIMATED.
4. Solution Concentration Exceeds $c > 0.1\text{ M}$ (or $A > 1.5$): Deviates from linearity; solute molecules aggregate, causing the calibration curve to plateau.
Unit 3 Intermolecular Forces & Properties: İnteraktif Soru İstasyonu
Özgün AP tarzı sorular (geçmiş sınav sorusu değil). Şık seç; anında çözüm ve hata türü gelir.
AP Chemistry — Unit 4: Chemical Reactions
A chemical reaction rearranges atoms: existing chemical bonds are broken and new bonds are formed, producing substances with new compositions and identities. A physical change alters only the state, shape or mixing of a substance; the identity of every particle is preserved (e.g., melting, boiling, sublimation, dissolving a molecular solute such as sucrose).
• Production of a gas (effervescence) that is chemically different from the reactants (e.g., $\text{H}_2$ from $\text{Mg} + \text{HCl}$).
• Formation of a precipitate (an insoluble solid) when two clear solutions are mixed.
• A color change not explained by dilution, or the emission of light.
• A temperature change in an isolated system (heat released or absorbed).
Key nuance: None of these observations is individually conclusive — boiling water produces bubbles and iodine sublimation produces a violet vapor, yet both are physical changes. Definitive proof requires evidence that a new substance with different properties has formed.
Three Equation Formats for the Same Reaction (aqueous lead(II) nitrate + potassium iodide):
• Molecular: $\text{Pb(NO}_3)_2(aq) + 2\,\text{KI}(aq) \rightarrow \text{PbI}_2(s) + 2\,\text{KNO}_3(aq)$
• Complete ionic: $\text{Pb}^{2+}(aq) + 2\,\text{NO}_3^-(aq) + 2\,\text{K}^+(aq) + 2\,\text{I}^-(aq) \rightarrow \text{PbI}_2(s) + 2\,\text{K}^+(aq) + 2\,\text{NO}_3^-(aq)$
• Net ionic: $\text{Pb}^{2+}(aq) + 2\,\text{I}^-(aq) \rightarrow \text{PbI}_2(s)$ — spectator ions ($\text{K}^+$, $\text{NO}_3^-$) are removed.
1. Write as separate aqueous ions ONLY: strong acids ($\text{HCl}, \text{HBr}, \text{HI}, \text{HNO}_3, \text{HClO}_4, \text{H}_2\text{SO}_4$ first proton), strong bases (Group 1 hydroxides, $\text{Ca(OH)}_2, \text{Sr(OH)}_2, \text{Ba(OH)}_2$) and soluble ionic salts. All $\text{Na}^+, \text{K}^+, \text{NH}_4^+$ and $\text{NO}_3^-$ salts are soluble (the only solubility rules the AP exam expects you to memorize).
2. Keep as intact formulas: solids, liquids, gases, weak acids (e.g., $\text{HC}_2\text{H}_3\text{O}_2$, $\text{HF}$), weak bases (e.g., $\text{NH}_3$) and water.
3. Cancel spectator ions, then verify BOTH mass balance and charge balance.
4. Decompose unstable products: $\text{H}_2\text{CO}_3(aq) \rightarrow \text{H}_2\text{O}(l) + \text{CO}_2(g)$.
Example (weak acid + strong base): $\text{HC}_2\text{H}_3\text{O}_2(aq) + \text{OH}^-(aq) \rightarrow \text{C}_2\text{H}_3\text{O}_2^-(aq) + \text{H}_2\text{O}(l)$
Particulate Diagrams (Topic 4.3): A balanced equation is a statement about particle counts. When a diagram shows reactant molecules in a box, (1) count every particle of each species, (2) apply the mole ratio, (3) identify the limiting reactant, and (4) draw the products together with any leftover excess reactant. Atoms are conserved: the number of each kind of atom drawn before and after must be identical.
1. Writing $\text{H}^+ + \text{OH}^- \rightarrow \text{H}_2\text{O}$ for a weak acid — weak acids are predominantly un-ionized and must remain as $\text{HA}(aq)$.
2. Splitting an insoluble reactant (e.g., $\text{CaCO}_3(s)$) into ions — solids stay intact.
3. Writing "$\text{I}_2^-$" or "$\text{Cl}_2^-$" for halide ions — the ions are monatomic: $2\,\text{I}^-$.
4. Omitting phase labels $(s), (l), (g), (aq)$ — graders expect them in net ionic answers.
The Universal Mole Bridge: Every stoichiometric conversion passes through moles; coefficients relate moles, never grams. $$\text{mass A} \xrightarrow{\div M_A} n_A \xrightarrow{\times \frac{\text{coeff. B}}{\text{coeff. A}}} n_B \xrightarrow{\times M_B} \text{mass B}$$ Other gateways into moles: solutions $n = M \times V(\text{L})$; gases $n = \frac{PV}{RT}$ (or $22.4\text{ L/mol}$ at STP, $273\text{ K}$ and $1\text{ atm}$); particles $n = N / N_A$.
1. Convert every reactant amount to moles.
2. Divide each mole value by its coefficient — the smallest quotient identifies the limiting reactant (or compute the product each reactant could form; the smaller yield wins).
3. Base ALL product calculations on the limiting reactant.
4. Excess remaining $= n_{\text{initial}} - n_{\text{consumed}}$, where $n_{\text{consumed}}$ is obtained from the limiting reactant via the mole ratio.
Percent Yield: $$\%\,\text{yield} = \frac{\text{actual yield}}{\text{theoretical yield}} \times 100\%$$ A percent yield above $100\%$ signals experimental error (e.g., a product that was not completely dried retains water mass).
1. "The reactant with the smaller mass is limiting" — FALSE. Compare moles divided by coefficients.
2. Adding reactant masses to get product mass ignores the excess reactant left over.
3. Forgetting that one formula unit can deliver several ions: $1\text{ mol CaCl}_2 \rightarrow 2\text{ mol Cl}^-$.
In a titration, a titrant of accurately known concentration is delivered from a buret into a measured volume of analyte until the equivalence point — the point at which the moles of titrant added are stoichiometrically equal to the moles of analyte present. An indicator signals the endpoint, which ideally coincides with the equivalence point.
• Diprotic acid such as $\text{H}_2\text{SO}_4$ + $\text{NaOH}$ (1:2): $n_{\text{acid}} = \frac{1}{2}\, n_{\text{NaOH}}$.
• Redox titration ($\text{MnO}_4^- + 5\,\text{Fe}^{2+}$): $n_{\text{Fe}^{2+}} = 5\, n_{\text{MnO}_4^-}$; purple $\text{MnO}_4^-$ acts as its own indicator.
• Standardization: a primary standard such as potassium hydrogen phthalate (KHP, $\text{KHC}_8\text{H}_4\text{O}_4$, $M = 204.22\text{ g/mol}$) is weighed precisely to determine the exact molarity of $\text{NaOH}$.
1. Buret rinsed with distilled water but not with titrant: titrant is diluted $\implies$ larger $V_{\text{titrant}}$ recorded $\implies$ calculated analyte concentration is OVERESTIMATED.
2. Extra distilled water added to the analyte flask: moles of analyte are unchanged $\implies$ NO EFFECT.
3. Overshooting the endpoint: excess titrant recorded $\implies$ analyte amount OVERESTIMATED.
4. Stopping before the endpoint: too little titrant recorded $\implies$ analyte amount UNDERESTIMATED.
The AP framework recognizes three major reaction classes: acid–base (proton transfer), oxidation–reduction (electron transfer), and precipitation (formation of an insoluble solid from aqueous ions). Synthesis, decomposition and combustion are descriptive labels that frequently overlap with redox (e.g., every combustion is a redox reaction).
• A base must possess a lone pair to accept the proton (e.g., the N lone pair of $\text{NH}_3$).
Oxidation is the loss of electrons (oxidation number increases); reduction is the gain of electrons (oxidation number decreases) — "OIL RIG". The species that is oxidized is the reducing agent; the species that is reduced is the oxidizing agent.
1. Free elements: $0$ (e.g., $\text{Fe}(s)$, $\text{O}_2$, $\text{Cl}_2$). Monatomic ions: equal to the ionic charge.
2. Group 1 metals $+1$; Group 2 metals $+2$; fluorine $-1$ in all compounds.
3. Hydrogen $+1$ with nonmetals ($-1$ in metal hydrides such as $\text{NaH}$).
4. Oxygen $-2$ (except peroxides such as $\text{H}_2\text{O}_2$, where it is $-1$).
5. The sum equals $0$ for a neutral compound or the overall charge for a polyatomic ion. Example: $\text{K}_2\text{Cr}_2\text{O}_7$: $2(+1) + 2x + 7(-2) = 0 \implies x = +6$.
Note: in $\text{NH}_4\text{NO}_3$ the two nitrogen atoms are in different environments: $-3$ in $\text{NH}_4^+$ and $+5$ in $\text{NO}_3^-$.
Half-Reaction Method — Acidic Solution: (1) split into oxidation and reduction half-reactions; (2) balance all atoms except O and H; (3) balance O with $\text{H}_2\text{O}$; (4) balance H with $\text{H}^+$; (5) balance charge with $e^-$; (6) multiply so electrons lost = electrons gained; (7) add, cancel, and verify mass and charge. $$\text{Cr}_2\text{O}_7^{2-}(aq) + 6\,\text{Fe}^{2+}(aq) + 14\,\text{H}^+(aq) \rightarrow 2\,\text{Cr}^{3+}(aq) + 6\,\text{Fe}^{3+}(aq) + 7\,\text{H}_2\text{O}(l)$$ Basic Solution: balance as if acidic, then add one $\text{OH}^-$ to BOTH sides for every $\text{H}^+$, combine $\text{H}^+ + \text{OH}^- \rightarrow \text{H}_2\text{O}$, and cancel water: $$2\,\text{MnO}_4^-(aq) + 6\,\text{I}^-(aq) + 4\,\text{H}_2\text{O}(l) \rightarrow 2\,\text{MnO}_2(s) + 3\,\text{I}_2(aq) + 8\,\text{OH}^-(aq)$$
1. The oxidizing agent is the species that is reduced (it takes electrons), not the one that is oxidized.
2. Oxidation numbers are per atom: $\text{Cr}$ in $\text{Cr}_2\text{O}_7^{2-}$ is $+6$, not $+12$.
3. A precipitation or neutralization reaction is NOT redox — no oxidation number changes.
4. A redox equation is balanced only if charge is balanced, not just atoms.
On the hybrid digital exam, MCQs are answered in Bluebook while FRQs are handwritten in the paper booklet. Graders award points for visible, justified reasoning — an unsupported final number often earns nothing.
1. Write the balanced equation first (with phase labels) — the mole ratio is the heart of every stoichiometry point.
2. Carry units on every number and let them cancel visibly: $25.0\text{ g Fe}_2\text{O}_3 \times \frac{1\text{ mol}}{159.70\text{ g}} \times \frac{2\text{ mol Fe}}{1\text{ mol Fe}_2\text{O}_3} \times \frac{55.85\text{ g}}{1\text{ mol Fe}} = 17.5\text{ g Fe}$.
3. Label the limiting reactant explicitly with a one-sentence justification.
4. Convert mL to L before using molarity: $18.62\text{ mL} = 0.01862\text{ L}$.
5. Significant figures: multiplication/division — keep the fewest significant figures of the measured data; addition/subtraction — keep the fewest decimal places. Exact counts and coefficients do not limit precision. Do not round intermediate values; round only the final answer.
6. Box the final answer with its unit (g, mol, L, M, %), and answer the exact quantity the prompt asks for (e.g., "mass of excess reactant remaining", not "mass consumed").
Unit 4 Chemical Reactions: İnteraktif Soru İstasyonu
Özgün AP tarzı sorular (geçmiş sınav sorusu değil). Şık seç; anında çözüm ve hata türü gelir.
AP Chemistry — Unit 5: Kinetics
The rate of a reaction is the change in concentration of a reactant or product per unit time, usually in $\text{M s}^{-1}$ ($\text{mol L}^{-1}\text{ s}^{-1}$). Rates are measured experimentally by following any property that changes with concentration: absorbance (Beer–Lambert law), pressure of a gas, pH, conductivity, or the mass of a precipitate. Factors that change the rate: reactant concentration, temperature, surface area (for heterogeneous reactions), and the presence of a catalyst.
• The instantaneous rate at time $t$ is the slope of the tangent to the $[\text{A}]$ vs. $t$ curve; the initial rate is that slope at $t = 0$.
• For most reactions the rate decreases over time because reactant concentrations fall.
A rate law expresses the rate as a function of reactant concentrations: $\text{rate} = k[\text{A}]^m[\text{B}]^n$. The exponents $m$ and $n$ are the orders with respect to each reactant; their sum is the overall order. For an overall reaction, orders are determined experimentally — they are NOT the balanced-equation coefficients. The rate constant $k$ depends on temperature (and on a catalyst), but not on concentration.
1. Choose two experiments in which only ONE reactant concentration changes.
2. Form the ratio: $\dfrac{\text{rate}_2}{\text{rate}_1} = \left(\dfrac{[\text{A}]_2}{[\text{A}]_1}\right)^m$. Concentration ×2 and rate ×1 → order 0; rate ×2 → order 1; rate ×4 → order 2. Concentration ×3 and rate ×9 → order 2.
3. Repeat for every reactant, write the rate law, then substitute ONE experiment's data to solve for $k$.
Units of $k$ (rate in $\text{M s}^{-1}$): zero order $\text{M s}^{-1}$; first order $\text{s}^{-1}$; second order $\text{M}^{-1}\text{ s}^{-1}$; third order $\text{M}^{-2}\text{ s}^{-1}$. General rule: $\text{M}^{1-n}\,\text{s}^{-1}$ for overall order $n$.
1. Copying coefficients from the overall equation into the rate law — only valid for an elementary step.
2. Taking the rate ratio itself as the order (rate ×4 means order 2, not 4).
3. Reporting $k$ without units, or with units that do not match the overall order.
4. A reactant that is zero order still appears in the balanced equation — it simply does not affect the rate.
When you have concentration-vs-time data for one reactant, graph three plots — $[\text{A}]$ vs. $t$, $\ln[\text{A}]$ vs. $t$ and $1/[\text{A}]$ vs. $t$. The plot that is a straight line reveals the order, and its slope gives $k$.
• Zero order: $[\text{A}]_t - [\text{A}]_0 = -kt$ → $[\text{A}]$ vs. $t$ is linear, slope $= -k$; $t_{1/2} = \dfrac{[\text{A}]_0}{2k}$ (depends on $[\text{A}]_0$).
• First order: $\ln[\text{A}]_t - \ln[\text{A}]_0 = -kt$ → $\ln[\text{A}]$ vs. $t$ is linear, slope $= -k$; $t_{1/2} = \dfrac{0.693}{k}$ (constant — independent of $[\text{A}]_0$; radioactive decay is first order).
• Second order: $\dfrac{1}{[\text{A}]_t} - \dfrac{1}{[\text{A}]_0} = kt$ → $1/[\text{A}]$ vs. $t$ is linear, slope $= +k$; $t_{1/2} = \dfrac{1}{k[\text{A}]_0}$ (grows as the reaction proceeds).
• After $n$ half-lives of a first-order process, the fraction remaining is $\left(\tfrac{1}{2}\right)^n$; $n$ need not be an integer.
1. Slope of $\ln[\text{A}]$ vs. $t$ is $-k$; $k$ itself is always positive.
2. Slope of $1/[\text{A}]$ vs. $t$ is $+k$ (the line rises) — do not insert a minus sign.
3. Constant successive half-lives in a data table are the fingerprint of a first-order reaction.
4. Calculate a slope from two points on the line ($\Delta y/\Delta x$), not from a single point divided by time.
An elementary reaction is a single molecular event. Its molecularity is the number of particles that collide: unimolecular ($\text{A} \rightarrow$ products, rate $= k[\text{A}]$), bimolecular ($\text{A} + \text{B}$, rate $= k[\text{A}][\text{B}]$; $2\text{A}$, rate $= k[\text{A}]^2$) and the rare termolecular step (three particles colliding simultaneously is improbable). Only for elementary steps may the rate law be written directly from the coefficients.
1. The particles collide with kinetic energy at least equal to the activation energy, $E_a$; and
2. The particles have a favorable orientation so the correct atoms can bond.
• Higher concentration → more collisions per second → faster rate (the fraction of effective collisions is unchanged).
• Higher temperature → the Maxwell–Boltzmann distribution flattens and shifts to higher energy; the fraction of collisions with energy $\geq E_a$ increases sharply. This, not the modest rise in collision frequency, is the main reason the rate increases.
• Arrhenius relationship (conceptual): $k = A\,e^{-E_a/RT}$. Here $A$ reflects collision frequency and orientation. A larger $E_a$ gives a smaller $k$, and makes $k$ more sensitive to temperature. A plot of $\ln k$ vs. $1/T$ is linear with slope $-E_a/R$.
A reaction energy profile plots potential energy against the reaction coordinate. The maximum is the transition state (activated complex) — a fleeting arrangement in which bonds are partially broken and partially formed; it cannot be isolated.
• $E_{a,\text{forward}} = E_{\text{transition state}} - E_{\text{reactants}}$
• $E_{a,\text{reverse}} = E_{\text{transition state}} - E_{\text{products}}$
• $\Delta H = E_{\text{products}} - E_{\text{reactants}} = E_{a,\text{forward}} - E_{a,\text{reverse}}$
• Exothermic ($\Delta H < 0$): products lie below reactants, so $E_{a,\text{reverse}} > E_{a,\text{forward}}$.
A reaction mechanism is a sequence of elementary steps that add up to the overall equation. An intermediate is formed in an early step and consumed in a later step. It never appears in the overall equation, and it must never appear in the final rate law. The rate-determining step (RDS) is the slowest step. The overall rate law equals the rate law of the RDS, written from its coefficients.
1. Do the elementary steps sum to the overall balanced equation? Cancel species that appear on both sides.
2. Is the rate law predicted by the slow step identical to the experimental rate law?
3. Is each step at most bimolecular? (Termolecular steps are possible but unlikely.)
Slow first step: $\text{NO}_2 + \text{NO}_2 \rightarrow \text{NO}_3 + \text{NO}$ (slow); $\text{NO}_3 + \text{CO} \rightarrow \text{NO}_2 + \text{CO}_2$ (fast). Overall: $\text{NO}_2 + \text{CO} \rightarrow \text{NO} + \text{CO}_2$, rate $= k[\text{NO}_2]^2$. $\text{CO}$ is zero order even though it is a reactant.
Fast reversible first step (Topic 5.9 pre-equilibrium): $\text{NO} + \text{Br}_2 \rightleftharpoons \text{NOBr}_2$ (fast); $\text{NOBr}_2 + \text{NO} \rightarrow 2\,\text{NOBr}$ (slow). The slow step gives rate $= k_2[\text{NOBr}_2][\text{NO}]$, but $\text{NOBr}_2$ is an intermediate. Equilibrium in step 1 gives $k_1[\text{NO}][\text{Br}_2] = k_{-1}[\text{NOBr}_2]$. Therefore $[\text{NOBr}_2] = \dfrac{k_1}{k_{-1}}[\text{NO}][\text{Br}_2]$. Substituting: $$\text{rate} = \frac{k_2 k_1}{k_{-1}}[\text{NO}]^2[\text{Br}_2] = k[\text{NO}]^2[\text{Br}_2]$$
1. Leaving an intermediate in the rate law — always eliminate it with the fast-equilibrium expression.
2. A mechanism that agrees with the rate law is consistent, not proven. Use the word "consistent" in FRQ justifications.
3. Species that are produced and later consumed are intermediates; species consumed and later regenerated are catalysts.
In a multistep energy profile, every peak is a transition state, and every trough between peaks is an intermediate. A two-step mechanism therefore shows two peaks and one trough. The step with the largest activation energy — measured from the trough or level that begins that step — is the rate-determining step.
• Lowers $E_a$ for BOTH the forward and the reverse reactions by the same amount.
• Does NOT change $\Delta H$, $\Delta G^\circ$, the equilibrium constant $K$, or the equilibrium position — equilibrium is reached faster.
• Homogeneous (same phase, e.g., $\text{Mn}^{2+}$ in solution), heterogeneous (surface catalysis on Pt, Pd or Ni — reactants adsorb, bonds weaken), acid–base catalysis, and enzymes (substrate binds the active site).
Kinetics FRQs reward a visible chain of reasoning in the paper booklet. A rate law with no supporting experiment comparison usually earns no point.
1. Name the experiments you compare: "From Experiments 1 and 2, $[\text{A}]$ doubles while $[\text{B}]$ is constant, and the rate quadruples; therefore the reaction is second order in A."
2. Write the complete rate law including $k$: $\text{rate} = k[\text{A}]^2[\text{B}]$.
3. Solve for $k$ with units that you derive on the page: $\dfrac{\text{M s}^{-1}}{\text{M}^3} = \text{M}^{-2}\text{ s}^{-1}$.
4. Graph questions: state which plot is linear and why ("$\ln[\text{A}]$ vs. $t$ is linear, so the reaction is first order; $k = -\text{slope}$"). Show the two points used for the slope.
5. Mechanism questions: (a) show that the steps add to the overall equation; (b) write the rate law of the slow step; (c) eliminate any intermediate; (d) conclude "consistent" or "not consistent".
6. Particle-level justifications must mention BOTH collision energy relative to $E_a$ AND frequency or orientation where relevant.
Unit 5 Kinetics: İnteraktif Soru İstasyonu
Özgün AP tarzı sorular (geçmiş sınav sorusu değil). Şık seç; anında çözüm ve hata türü gelir.
AP Chemistry — Unit 6: Thermochemistry
Every thermochemical statement begins by defining the system (the reacting chemicals or the substance undergoing change) and the surroundings (everything else — in the laboratory, mostly the water and the container). Energy is conserved: whatever the system loses, the surroundings gain, so $q_{\text{system}} = -q_{\text{surroundings}}$.
• Exothermic ($\Delta H < 0$, $q_{\text{sys}} < 0$): energy flows from the system to the surroundings. In a solution calorimeter, the temperature of the water rises. Examples: combustion, neutralization, freezing, condensation, dissolving $\text{CaCl}_2$ or $\text{NaOH}$.
• Endothermic ($\Delta H > 0$, $q_{\text{sys}} > 0$): energy flows from the surroundings into the system. The temperature of the water falls. Examples: melting, vaporization, sublimation, dissolving $\text{NH}_4\text{NO}_3$ (cold packs), thermal decomposition of $\text{CaCO}_3$.
• Particulate view: breaking a chemical bond or overcoming an intermolecular attraction ALWAYS requires energy (endothermic). Forming a bond or an attraction ALWAYS releases energy (exothermic). Whether the overall process is exothermic depends on which of the two contributions is larger.
An energy diagram places the enthalpy of the reactants and of the products on a vertical axis. The vertical gap between the two levels is $\Delta H$; the hump between them (when drawn) is the activation energy studied in Unit 5. $\Delta H$ depends only on the starting and finishing levels — never on the height of the peak.
• $\Delta H = H_{\text{products}} - H_{\text{reactants}}$ (final minus initial).
• Products BELOW reactants → $\Delta H < 0$ (exothermic); products ABOVE reactants → $\Delta H > 0$ (endothermic).
• For the reverse reaction, the levels swap: $\Delta H_{\text{reverse}} = -\Delta H_{\text{forward}}$.
• A catalyst lowers the peak but leaves both levels — and therefore $\Delta H$ — unchanged.
Heat ($q$) is energy transferred between two bodies because of a temperature difference. At the particulate level, faster-moving particles of the hotter body collide with slower-moving particles of the cooler body and transfer kinetic energy. The transfer continues until thermal equilibrium is reached: both bodies have the same temperature, which means their particles have the same average kinetic energy.
$q_{\text{lost by hot}} + q_{\text{gained by cold}} = 0 \quad\Longrightarrow\quad m_{\text{h}}c_{\text{h}}(T_f - T_{\text{h}}) + m_{\text{c}}c_{\text{c}}(T_f - T_{\text{c}}) = 0$ • The amount of energy transferred is equal in magnitude for both bodies, but the temperature changes are generally different: the body with the smaller $mc$ product changes temperature more.
• The final temperature always lies between the two initial temperatures, closer to the body with the larger $mc$ (usually the water).
The specific heat capacity $c$ is the energy needed to raise the temperature of 1 g of a substance by $1^\circ\text{C}$ (for water, $c = 4.18\text{ J g}^{-1}\,{}^\circ\text{C}^{-1}$, given on the AP reference sheet). The heat capacity $C$ of an object (for example, a bomb calorimeter) is the energy per degree for the whole object, in $\text{J}\,{}^\circ\text{C}^{-1}$ or $\text{kJ}\,{}^\circ\text{C}^{-1}$.
1. $q_{\text{solution}} = m\,c\,\Delta T$, with $\Delta T = T_{\text{final}} - T_{\text{initial}}$ and $m$ the TOTAL mass of solution (e.g., $50.0\text{ mL} + 50.0\text{ mL} \approx 100.0\text{ g}$).
2. $q_{\text{reaction}} = -q_{\text{solution}}$.
3. Convert J → kJ (divide by 1000).
4. $\Delta H_{\text{rxn}} = \dfrac{q_{\text{reaction}}}{n_{\text{limiting}}}$ in $\text{kJ mol}^{-1}$, with the correct sign.
Bomb calorimeter (constant volume): $q_{\text{released}} = C_{\text{cal}}\,\Delta T$; divide by the mass or moles of the sample burned.
1. Reporting $q$ in J when the answer asks for kJ (an error of $10^3$).
2. Using only one of the mixed solutions as the mass.
3. Forgetting to divide by moles: $\Delta H$ is per mole of reaction, $q$ is not.
4. Sign: a temperature rise of the solution gives a NEGATIVE $\Delta H$.
5. Heat lost to the surroundings in a poorly insulated cup makes the measured $\Delta T$ — and therefore the magnitude $|\Delta H|$ — too small.
During a phase change at constant pressure the temperature stays constant: the energy added is used to overcome intermolecular attractions (potential energy rises), not to speed up the particles (average kinetic energy is unchanged). The energy is $q = n\,\Delta H_{\text{fus}}$ or $q = n\,\Delta H_{\text{vap}}$.
• Sloped segments (one phase warming): $q = mc\Delta T$, with the $c$ of THAT phase.
• Flat segments (two phases coexisting): $q = n\,\Delta H_{\text{phase change}}$.
• Total energy = sum of all segments, every term in the same unit (kJ).
• For water: $\Delta H_{\text{fus}} = 6.01\text{ kJ mol}^{-1}$, $\Delta H_{\text{vap}} = 40.7\text{ kJ mol}^{-1}$ at $100^\circ\text{C}$. $\Delta H_{\text{vap}} \gg \Delta H_{\text{fus}}$ because vaporization separates the molecules almost completely, whereas melting only loosens the arrangement.
• Melting, vaporization and sublimation are endothermic; freezing, condensation and deposition are exothermic, with $\Delta H_{\text{condensation}} = -\Delta H_{\text{vap}}$.
The enthalpy of reaction $\Delta H_{\text{rxn}}$ is the heat exchanged at constant pressure for one mole of reaction as written, i.e., for the molar amounts given by the coefficients of the balanced equation. Enthalpy is an extensive state function, so it scales with the amount of reaction.
• Multiply the equation by a factor $n$ → multiply $\Delta H$ by $n$.
• Reverse the equation → change the sign of $\Delta H$.
• Heat for a given amount: $q = \dfrac{n_{\text{species}}}{\text{coefficient of that species}} \times \Delta H_{\text{rxn}}$.
Example: $2\,\text{H}_2(g) + \text{O}_2(g) \rightarrow 2\,\text{H}_2\text{O}(l)$, $\Delta H = -572\text{ kJ mol}_{\text{rxn}}^{-1}$, i.e., $-286\text{ kJ}$ per mole of $\text{H}_2$ burned.
• Physical states matter: forming $\text{H}_2\text{O}(l)$ releases more energy than forming $\text{H}_2\text{O}(g)$ because condensation is also exothermic.
A bond enthalpy is the energy required to break one mole of a given bond in the gas phase; it is always positive. Tabulated values are averages over many different molecules, so a $\Delta H$ calculated from them is an estimate.
2. Multiply each bond count by the coefficient of its species.
3. Broken (reactants) costs energy; formed (products) releases energy.
Example: $\text{H}_2 + \text{Cl}_2 \rightarrow 2\,\text{HCl}$: $(436 + 242) - 2(431) = -184\text{ kJ mol}^{-1}$.
1. Reversing the order (formed − broken) flips the sign. Note that the formation-enthalpy equation is products − reactants, but the bond equation is reactants − products.
2. Forgetting the coefficients ($2\,\text{HCl}$ means two $\text{H–Cl}$ bonds formed).
3. Counting a $\text{C=O}$ double bond as a single bond.
The standard enthalpy of formation $\Delta H_f^\circ$ is the enthalpy change when ONE mole of a compound forms from its elements in their standard states (most stable form at 1 atm and the stated temperature, e.g., $\text{C}(s,\text{graphite})$, $\text{O}_2(g)$, $\text{H}_2(g)$, $\text{Br}_2(l)$). By definition $\Delta H_f^\circ = 0$ for an element in its standard state.
$\Delta H^\circ = [3(-393.5) + 4(-285.8)] - [(-103.8) + 5(0)] = -2219.9\text{ kJ mol}^{-1}$.
• The same equation can be solved backwards for an unknown $\Delta H_f^\circ$ when $\Delta H^\circ_{\text{rxn}}$ is measured.
1. Reversing the order (reactants − products).
2. Dropping the coefficients $n$.
3. Subtracting a negative value incorrectly: $-(-103.8) = +103.8$.
4. A formation equation must produce exactly 1 mol of product from elements — fractional coefficients such as $\tfrac{1}{2}\text{O}_2$ are allowed.
Because enthalpy is a state function, $\Delta H$ for an overall reaction equals the sum of the $\Delta H$ values of any series of steps that add up to it. Hess's law lets you calculate $\Delta H$ for reactions that are difficult to measure directly (e.g., $\text{C}(s) + \tfrac{1}{2}\text{O}_2(g) \rightarrow \text{CO}(g)$, which always produces some $\text{CO}_2$).
1. Locate each target species in exactly ONE given equation.
2. If the species is on the wrong side, reverse that equation and change the sign of its $\Delta H$.
3. Multiply the equation so the coefficient matches the target; multiply its $\Delta H$ by the same factor.
4. Add the equations, cancel species appearing on both sides, and confirm the sum equals the target exactly.
5. Add the adjusted $\Delta H$ values.
Example: $\text{C} + \text{O}_2 \rightarrow \text{CO}_2$ ($-393.5$); reverse $\text{CO} + \tfrac{1}{2}\text{O}_2 \rightarrow \text{CO}_2$ ($-283.0$ → $+283.0$). Sum: $\text{C} + \tfrac{1}{2}\text{O}_2 \rightarrow \text{CO}$, $\Delta H = -110.5\text{ kJ mol}^{-1}$.
Thermochemistry FRQs award separate points for $q$, for the mole calculation and for the sign of $\Delta H$. Each step must be visible in the paper booklet.
1. Write $q = mc\Delta T$ with numbers and units: $q = (100.0\text{ g})(4.18\text{ J g}^{-1}\,{}^\circ\text{C}^{-1})(6.70\,{}^\circ\text{C}) = 2.80 \times 10^3\text{ J}$.
2. Convert and assign the sign with a sentence: "The temperature of the solution increased, so the reaction is exothermic and $q_{\text{rxn}} = -2.80\text{ kJ}$."
3. Divide by moles of the limiting reactant and report $\Delta H$ in $\text{kJ mol}^{-1}$ with the correct number of significant figures.
4. Bond-enthalpy and formation questions: list every term, e.g., "Broken: $4(\text{C–H}) + 2(\text{O=O})$; Formed: $2(\text{C=O}) + 4(\text{O–H})$."
5. Hess's law: rewrite each manipulated equation with its new $\Delta H$, then add.
6. Error analysis: state the direction of the change in the measured $\Delta T$ first, then the effect on the calculated $|\Delta H|$.
Unit 6 Thermochemistry: İnteraktif Soru İstasyonu
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AP Chemistry — Unit 7: Equilibrium
Many processes are reversible: evaporation and condensation, dissolution and precipitation, proton transfer between an acid and a base, electron transfer in redox couples. When a reversible process takes place in a closed system it reaches equilibrium: no further observable change occurs, reactants and products are present at the same time, and every concentration and partial pressure stays constant.
• Dynamic: the forward and reverse reactions continue at the molecular level; they simply proceed at equal rates.
• Constant composition: concentrations (or partial pressures) no longer change with time. On a concentration–time graph every curve becomes horizontal.
• Approachable from either side: starting with only reactants or only products leads to the same value of $K$ at a given temperature.
The net direction of a reversible reaction is decided by which rate is larger. If $\text{rate}_{\text{fwd}} > \text{rate}_{\text{rev}}$, reactants are converted into products; if $\text{rate}_{\text{rev}} > \text{rate}_{\text{fwd}}$, products are converted back into reactants. As the reaction proceeds, the faster rate falls (its reactants are used up) and the slower rate rises, until the two become equal.
• At $t = 0$: $\text{rate}_{\text{rev}} = 0$, because no products exist yet.
• The forward rate decreases and the reverse rate increases.
• At equilibrium the two curves meet and stay together: $\text{rate}_{\text{fwd}} = \text{rate}_{\text{rev}} \neq 0$.
For a general reaction $a\text{A} + b\text{B} \rightleftharpoons c\text{C} + d\text{D}$, the law of mass action defines the reaction quotient at ANY moment and the equilibrium constant at equilibrium. Both use the same expression; the only difference is the moment at which the concentrations are measured.
$Q_c = \dfrac{[\text{C}]^c[\text{D}]^d}{[\text{A}]^a[\text{B}]^b} \qquad Q_p = \dfrac{(P_{\text{C}})^c(P_{\text{D}})^d}{(P_{\text{A}})^a(P_{\text{B}})^b}$
• At equilibrium: $Q_c = K_c$ and $Q_p = K_p$.
• Pure solids and pure liquids are omitted, because their "concentration" does not depend on the amount present. Example: $\text{CaCO}_3(s) \rightleftharpoons \text{CaO}(s) + \text{CO}_2(g)$ gives $K_p = P_{\text{CO}_2}$.
• Water is omitted when it is the solvent (liquid), but included when it is a gas, e.g., $\text{H}_2\text{O}(g)$.
• On the AP Exam $K$ is reported without units.
For gases, $P = \dfrac{n}{V}RT = [\,\text{gas}\,]RT$, so the two constants are related by $K_p = K_c(RT)^{\Delta n}$, where $\Delta n = (\text{moles of gaseous products}) - (\text{moles of gaseous reactants})$ and $R = 0.08206\text{ L atm mol}^{-1}\text{ K}^{-1}$. When $\Delta n = 0$ (for example $\text{H}_2 + \text{I}_2 \rightleftharpoons 2\,\text{HI}$), $K_p = K_c$.
CED status: the CED (Topic 7.3, Exclusion Statement) states that conversion between $K_c$ and $K_p$ will NOT be assessed on the AP Exam. What IS assessed: noticing whether a question gives $K_c$ (use molarities) or $K_p$ (use partial pressures in atm) and never mixing the two in one expression. The conversion is useful for university courses and for the Italian TOLC-I.
1. Forgetting the exponents (coefficients).
2. Writing reactants over products (this gives $1/K$).
3. Including a solid such as $\text{C}(s)$ or a liquid such as $\text{H}_2\text{O}(l)$.
4. Substituting molarities into a $K_p$ expression.
$K$ is calculated by substituting equilibrium concentrations or partial pressures into the expression. When only the initial amounts and ONE equilibrium value are known, an ICE table (Initial, Change, Equilibrium) supplies the rest through the stoichiometry of the change row.
| atm | $\text{N}_2\text{O}_4$ | $2\,\text{NO}_2$ |
|---|---|---|
| I | 1.00 | 0 |
| C | $-x$ | $+2x$ |
| E | $1.00 - x$ | $2x$ |
$P_{\text{NO}_2} = 0.80\text{ atm}$, $P_{\text{N}_2\text{O}_4} = 0.60\text{ atm}$
$K_p = \dfrac{(0.80)^2}{0.60} = 1.1$ (two significant figures, limited by $x = 0.40$).
The size of $K$ tells you the composition of the equilibrium mixture, not how fast it is reached.
• $K \gg 1$ (e.g., $10^{10}$): products strongly favored; the reaction proceeds essentially to completion. Treat it as a limiting-reactant problem.
• $K \ll 1$ (e.g., $10^{-10}$): reactants strongly favored; very little product forms. This is the situation in which the small-$x$ approximation works.
• $K \approx 1$: comparable amounts of reactants and products; a full ICE solution is needed.
Equilibrium constants for related equations are connected by three exact rules. They are the multiplicative counterparts of the Hess's-law rules for $\Delta H$ (Topic 6.9): where $\Delta H$ is added, $K$ is multiplied.
• Reverse the equation → $K_{\text{new}} = \dfrac{1}{K}$.
• Multiply all coefficients by $n$ → $K_{\text{new}} = K^{\,n}$ (halving gives $\sqrt{K}$).
• Add two equations → $K_{\text{overall}} = K_1 \times K_2$.
• The same rules apply to $Q$.
Example: $2\,\text{A} \rightleftharpoons \text{B}$, $K = 400$. For $\tfrac{1}{2}\text{B} \rightleftharpoons \text{A}$: reverse ($1/400$), then halve: $K = (1/400)^{1/2} = 0.050$.
Given the initial conditions and $K$, first compute $Q$ to decide the direction of change, then build an ICE table and solve the resulting equation for $x$.
| M | $\text{H}_2$ | $\text{I}_2$ | $2\,\text{HI}$ |
|---|---|---|---|
| I | 0.20 | 0.20 | 0 |
| C | $-x$ | $-x$ | $+2x$ |
| E | $0.20 - x$ | $0.20 - x$ | $2x$ |
Take the square root of both sides: $\dfrac{2x}{0.20 - x} = 8 \implies x = 0.16\text{ M}$
$[\text{HI}] = 0.32\text{ M}$, $[\text{H}_2] = [\text{I}_2] = 0.04\text{ M}$. Check: $(0.32)^2/(0.04)^2 = 64$.
$\dfrac{x^2}{0.050 - x} = 2.2 \times 10^{-10}$. Because $K$ is very small, assume $0.050 - x \approx 0.050$:
$x = \sqrt{(2.2 \times 10^{-10})(0.050)} = 3.3 \times 10^{-6}\text{ M}$
5% check: $\dfrac{3.3 \times 10^{-6}}{0.050} \times 100\% = 0.0066\% < 5\%$ → the approximation is valid (the exact quadratic root agrees to three significant figures).
Approximation: $x = \sqrt{(4.0 \times 10^{-3})(0.10)} = 0.020\text{ M}$ → $0.020/0.10 = 20\% > 5\%$ → NOT valid.
Exact: $x^2 + (4.0 \times 10^{-3})x - 4.0 \times 10^{-4} = 0$
Quadratic formula, positive root: $x = 0.018\text{ M}$ (about 10% smaller than the approximate value).
Rule of thumb: the approximation is usually safe when $[\text{initial}]/K > 400$ (here $0.10/0.0040 = 25$).
1. Using moles instead of molarity (divide by the volume first).
2. Change row not proportional to the coefficients.
3. Taking the "approximation" without stating and checking the 5% criterion in an FRQ.
4. Keeping the physically impossible quadratic root (negative, or larger than the initial amount).
Particulate diagrams show the relative numbers of reactant and product particles before and at equilibrium. If the volume is fixed (or stated as 1 L), the particle counts are proportional to concentrations and can be substituted directly into $K$.
• Count each species in the box. Example for $\text{A}_2 + \text{B}_2 \rightleftharpoons 2\,\text{AB}$: 4 $\text{A}_2$, 2 $\text{B}_2$, 6 $\text{AB}$ → $K = \dfrac{6^2}{(4)(2)} = 4.5$.
• To decide whether a box is AT equilibrium, compute $Q$ from the counts and compare it with the given $K$.
• Atoms are conserved between boxes: a "next" diagram must contain the same number of A atoms and B atoms.
Le Châtelier's principle: when a system at equilibrium is disturbed, it shifts in the direction that partially counteracts the disturbance. The predicted shift can be observed through measurable properties such as color, pH and temperature.
| Stress | Shift | Does K change? |
|---|---|---|
| Add a reactant / remove a product | Toward products | No |
| Decrease volume (increase pressure) of a gas system | Toward fewer moles of gas | No |
| Add an inert gas at constant volume | No shift (partial pressures unchanged) | No |
| Dilute an aqueous system | Toward more dissolved particles | No |
| Raise T (endothermic forward reaction) | Toward products | Yes, K increases |
| Raise T (exothermic forward reaction) | Toward reactants | Yes, K decreases |
| Add a catalyst | No shift; equilibrium reached faster | No |
A disturbance makes $Q \neq K$; the system then reacts until $Q = K$ again. On FRQs, "it shifts right" earns no credit by itself. You must show why $Q$ is now smaller or larger than $K$, mathematically.
• $Q < K$: too few products relative to equilibrium → net forward reaction.
• $Q > K$: too many products → net reverse reaction.
• $Q = K$: at equilibrium; no net change.
Mathematical justification of a volume change: $\text{N}_2\text{O}_4(g) \rightleftharpoons 2\,\text{NO}_2(g)$ at equilibrium. Halving the volume instantly doubles every concentration:
$Q = \dfrac{(2[\text{NO}_2])^2}{2[\text{N}_2\text{O}_4]} = 2\,\dfrac{[\text{NO}_2]^2}{[\text{N}_2\text{O}_4]} = 2K$
Since $Q = 2K > K$, the net reaction proceeds in the reverse direction (toward fewer moles of gas) until $Q = K$. For $\text{H}_2 + \text{I}_2 \rightleftharpoons 2\,\text{HI}$, the doubling factors cancel ($2^2/2^2 = 1$), so $Q = K$ and there is no shift.
Temperature: a temperature change alters $K$ itself, not $Q$. For an exothermic forward reaction, heating decreases $K$; the old composition now gives $Q > K$, and the system shifts in reverse.
The dissolution of a sparingly soluble salt is a reversible process described by the solubility-product constant $K_{sp}$. The solid is omitted from the expression. The molar solubility $s$ is the number of moles of salt that dissolve per liter of saturated solution.
• 1:1 salt, $\text{AgCl}(s) \rightleftharpoons \text{Ag}^+ + \text{Cl}^-$: $K_{sp} = s^2$, $s = \sqrt{K_{sp}}$.
• 1:2 salt, $\text{CaF}_2(s) \rightleftharpoons \text{Ca}^{2+} + 2\,\text{F}^-$: $K_{sp} = s(2s)^2 = 4s^3$, $s = \sqrt[3]{K_{sp}/4}$.
• 2:1 salt, $\text{Ag}_2\text{CrO}_4(s) \rightleftharpoons 2\,\text{Ag}^+ + \text{CrO}_4^{2-}$: $K_{sp} = (2s)^2 s = 4s^3$.
Solved example: $\text{CaF}_2$, $K_{sp} = 3.9 \times 10^{-11}$ → $s = \sqrt[3]{(3.9 \times 10^{-11})/4} = 2.1 \times 10^{-4}\text{ M}$.
• Will a precipitate form? Compute $Q_{sp}$ from the concentrations AFTER mixing (dilution!). $Q_{sp} > K_{sp}$ → precipitate forms; $Q_{sp} \le K_{sp}$ → no precipitate.
• Salts with $K_{sp} > 1$ correspond to the "soluble" salts of the solubility rules.
A salt is less soluble in a solution that already contains one of its ions. Qualitatively, the added ion raises $Q_{sp}$ above $K_{sp}$, so the dissolution equilibrium shifts toward the solid. Quantitatively, the common ion enters the ICE table as an initial concentration.
| M | $\text{Ag}^+$ | $\text{Cl}^-$ |
|---|---|---|
| I | 0 | 0.10 |
| C | $+s$ | $+s$ |
| E | $s$ | $0.10 + s \approx 0.10$ |
• For $\text{CaF}_2$ in $0.10\text{ M NaF}$: $K_{sp} = s(0.10)^2 \implies s = 3.9 \times 10^{-9}\text{ M}$ (the fluoride concentration is SQUARED).
Two ideas that older study guides placed in Unit 7 now belong to later units in the CED effective Fall 2024. They are previewed here because they rely entirely on the Unit 7 tools of $Q$, $K$ and Le Châtelier's principle.
• A salt whose anion is the conjugate base of a weak acid ($\text{F}^-$, $\text{CO}_3^{2-}$, $\text{PO}_4^{3-}$, $\text{CH}_3\text{COO}^-$) or $\text{OH}^-$ becomes MORE soluble as the pH decreases: $\text{H}^+$ removes the anion ($\text{F}^- + \text{H}^+ \rightarrow \text{HF}$), $Q_{sp}$ drops below $K_{sp}$, and more solid dissolves.
• Salts of strong-acid anions ($\text{Cl}^-$, $\text{Br}^-$, $\text{I}^-$) are essentially unaffected by pH.
• Exclusion Statement: computations of solubility as a function of pH are not assessed on the AP Exam.
Free Energy and Equilibrium (Topic 9.5): $\Delta G^\circ = -RT\ln K$. It is developed together with entropy in Unit 9; here remember only that $K > 1$ corresponds to $\Delta G^\circ < 0$.
Equilibrium FRQs award separate points for the expression, the setup, the numerical answer and the justification. Each must be visible in the paper booklet.
1. Write the expression first, with exponents and without solids or liquids: $K_c = \dfrac{[\text{NH}_3]^2}{[\text{N}_2][\text{H}_2]^3}$. An expression point is often awarded on its own.
2. Draw the ICE table with units (M or atm) in the corner, and convert moles to molarity before the I row.
3. Show the substitution, then the answer. Report $K$ without units; report concentrations in $\text{mol L}^{-1}$ and pressures in atm.
4. Significant figures: match the least precise datum (usually 2–3 s.f.). In a value such as $3.3 \times 10^{-6}$ only the digits of the coefficient are significant. For $K$ values, keep one extra digit during the calculation and round only at the end.
5. State the approximation and check it: "Assume $x \ll 0.050$; check: $3.3 \times 10^{-6}/0.050 = 0.0066\% < 5\%$, so the assumption is valid."
6. Justify a direction with Q: "$Q = 7.0$, which is greater than $K = 0.50$; therefore the reaction proceeds in the reverse direction (toward reactants) to reach equilibrium."
• "Because $Q < K$, the rate of the forward reaction exceeds the rate of the reverse reaction until $Q = K$."
• "Adding $\text{Cl}^-$ increases $Q_{sp}$ above $K_{sp}$, so the equilibrium shifts toward the solid and the solubility of AgCl decreases."
• "Increasing the temperature of an exothermic reaction decreases the value of $K$; the reaction proceeds in reverse until $Q$ equals the new, smaller $K$."
Avoid "the reaction stops" (it is dynamic) and "the equilibrium moves" without saying in which direction and why.
Unit 7 Equilibrium: İnteraktif Soru İstasyonu
Özgün AP tarzı sorular (geçmiş sınav sorusu değil). Şık seç; anında çözüm ve hata türü gelir.
AP Chemistry — Unit 8: Acids and Bases
Unit 8 (CED weighting 11–15%, the second-largest unit on the exam) applies every equilibrium tool from Unit 7 to one family of reactions: proton transfer. You will calculate pH and pOH for strong and weak species, set up weak-acid and weak-base ICE tables, predict the products of acid–base reactions, read every region of a titration curve, explain acid strength from molecular structure, decide which form of an acid predominates at a given pH, and design and evaluate buffers with the Henderson–Hasselbalch equation. In the current CED (effective Fall 2024) Unit 8 contains Topics 8.1–8.11; Topic 8.11 (pH and Solubility, qualitative only) is previewed at the end of Unit 7 on this page. Throughout, one standard applies: whenever you claim that a system shifts, show mathematically whether $Q$ is smaller or larger than $K$. All values below assume $25^\circ\text{C}$, where $K_w = 1.0 \times 10^{-14}$, unless stated otherwise.
A Brønsted–Lowry acid is a proton ($\text{H}^+$) donor; a Brønsted–Lowry base is a proton acceptor. Every proton transfer produces two conjugate acid–base pairs that differ by exactly one $\text{H}^+$. Water is amphoteric: it can act as either an acid or a base, and it ionizes itself.
• $2\,\text{H}_2\text{O}(l) \rightleftharpoons \text{H}_3\text{O}^+(aq) + \text{OH}^-(aq) \qquad K_w = [\text{H}_3\text{O}^+][\text{OH}^-] = 1.0 \times 10^{-14}$ at $25^\circ\text{C}$.
• $\text{pH} = -\log[\text{H}_3\text{O}^+]$, $\text{pOH} = -\log[\text{OH}^-]$, and $\text{pH} + \text{pOH} = \text{p}K_w = 14.00$ at $25^\circ\text{C}$.
• A change of 1 pH unit is a tenfold change in $[\text{H}_3\text{O}^+]$; a change of 2 units is a hundredfold change.
• Neutral means $[\text{H}_3\text{O}^+] = [\text{OH}^-]$, not "pH = 7". Autoionization is endothermic, so $K_w$ increases with temperature: at $37^\circ\text{C}$, $K_w \approx 2.4 \times 10^{-14}$ and neutral water has $\text{pH} = 6.81$.
• $\text{HCO}_3^-$ donates $\text{H}^+$ (acid) and becomes its conjugate base $\text{CO}_3^{2-}$.
• $\text{NH}_3$ accepts $\text{H}^+$ (base) and becomes its conjugate acid $\text{NH}_4^+$.
Strong acids and strong bases ionize completely in water, so the ion concentrations come straight from stoichiometry; no ICE table and no $K$ are needed.
• Strong acids: $\text{HCl}$, $\text{HBr}$, $\text{HI}$, $\text{HNO}_3$, $\text{HClO}_4$, $\text{H}_2\text{SO}_4$ (first proton).
• Strong bases: Group 1 hydroxides ($\text{LiOH}$, $\text{NaOH}$, $\text{KOH}$, ...) and the heavier Group 2 hydroxides ($\text{Ca(OH)}_2$, $\text{Sr(OH)}_2$, $\text{Ba(OH)}_2$).
• Count the ions: $0.0050\text{ M Ba(OH)}_2 \rightarrow [\text{OH}^-] = 2 \times 0.0050 = 0.010\text{ M}$, $\text{pOH} = 2.00$, $\text{pH} = 12.00$.
A weak acid ionizes only partially: $\text{HA}(aq) + \text{H}_2\text{O}(l) \rightleftharpoons \text{H}_3\text{O}^+(aq) + \text{A}^-(aq)$, with $K_a = \dfrac{[\text{H}_3\text{O}^+][\text{A}^-]}{[\text{HA}]}$. A weak base accepts a proton from water: $\text{B} + \text{H}_2\text{O} \rightleftharpoons \text{BH}^+ + \text{OH}^-$, with $K_b = \dfrac{[\text{BH}^+][\text{OH}^-]}{[\text{B}]}$. For a conjugate pair, $K_a \times K_b = K_w$.
| M | HF | $\text{H}_3\text{O}^+$ | $\text{F}^-$ |
|---|---|---|---|
| I | 0.050 | 0 | 0 |
| C | $-x$ | $+x$ | $+x$ |
| E | $0.050 - x$ | $x$ | $x$ |
• Quadratic: $x^2 + (6.8 \times 10^{-4})x - 3.4 \times 10^{-5} = 0 \implies x = 5.5 \times 10^{-3}\text{ M}$, $\text{pH} = 2.26$.
Start with a weak acid at equilibrium, $K_a = \dfrac{[\text{H}_3\text{O}^+][\text{A}^-]}{[\text{HA}]}$. Dilute it tenfold: every concentration is divided by 10 at the same instant, so
$Q = \dfrac{([\text{H}_3\text{O}^+]/10)([\text{A}^-]/10)}{[\text{HA}]/10} = \dfrac{K_a}{10} < K_a$.
Because $Q < K_a$, the forward reaction is favored and a larger FRACTION of HA ionizes. Percent ionization of $0.10\text{ M}$ acetic acid is $1.3\%$; at $0.0010\text{ M}$ it is about $12.5\%$. The pH still rises (from 2.87 to 3.90), because the total $[\text{H}_3\text{O}^+]$ falls.
When an acid and a base are mixed, assume the reaction with the strong species goes to completion. Work in moles (or millimoles), find what is left over, then decide which equilibrium controls the pH.
• Strong acid + strong base: $\text{H}_3\text{O}^+ + \text{OH}^- \rightarrow 2\,\text{H}_2\text{O}$
• Weak acid + strong base: $\text{HC}_2\text{H}_3\text{O}_2 + \text{OH}^- \rightarrow \text{C}_2\text{H}_3\text{O}_2^- + \text{H}_2\text{O}$
• Weak base + strong acid: $\text{NH}_3 + \text{H}_3\text{O}^+ \rightarrow \text{NH}_4^+ + \text{H}_2\text{O}$
• Excess strong acid or strong base left $\implies$ it alone sets the pH (divide the excess moles by the TOTAL volume).
• Only a weak acid left $\implies$ $K_a$ ICE table. Only its conjugate base left $\implies$ $K_b$ ICE table.
• Weak acid AND its conjugate base both present in significant amounts $\implies$ a buffer; use Henderson–Hasselbalch (Topic 8.9). A buffer forms when a weak acid is partially neutralized by a strong base, e.g. $0.20\text{ mol HF} + 0.10\text{ mol NaOH}$.
A titration curve plots pH against the volume of titrant added. At the equivalence point, moles of added base equal the initial moles of acid (for a monoprotic acid): $M_aV_a = M_bV_b$. The half-equivalence point is where half of the acid has been converted to its conjugate base, so $[\text{HA}] = [\text{A}^-]$ and $\text{pH} = \text{p}K_a$.
| $V_{\text{NaOH}}$ (mL) | Region | Method | pH |
|---|---|---|---|
| 0 | weak acid only | $K_a$ ICE | 2.87 |
| 10.0 | buffer | $4.74 + \log(1.0/1.5)$ | 4.56 |
| 12.5 | half-equivalence | $\text{pH} = \text{p}K_a$ | 4.74 |
| 25.0 | equivalence | $K_b$ of $0.0500\text{ M A}^-$ | 8.72 |
| 30.0 | excess $\text{OH}^-$ | $0.50\text{ mmol}/55.0\text{ mL}$ | 11.96 |
• Indicator choice: pick an indicator whose $\text{p}K_{\text{In}}$ (color-change range) lies on the steep part of the curve, close to the equivalence pH.
• Polyprotic acids show one equivalence point per ionizable proton. For a diprotic acid, $\text{pH} = \text{p}K_{a1}$ at the first half-equivalence point and $\text{pH} = \text{p}K_{a2}$ halfway between the first and second equivalence points.
Acid strength depends on how easily the H–A bond breaks and on how stable the conjugate base $\text{A}^-$ is. The more stable (more delocalized) the negative charge on $\text{A}^-$, the stronger the acid.
• Oxyacids with the same central atom: more oxygen atoms $\implies$ more electron density pulled away from the O–H bond and greater delocalization of the negative charge in the conjugate base. Strength: $\text{HClO}_4 > \text{HClO}_3 > \text{HClO}_2 > \text{HClO}$.
• Electronegative substituents (inductive effect): $\text{ClCH}_2\text{COOH}$ ($\text{p}K_a \approx 2.9$) is stronger than $\text{CH}_3\text{COOH}$ ($\text{p}K_a = 4.74$), because Cl withdraws electron density and stabilizes the carboxylate anion.
• Resonance: a carboxylate ion spreads its charge over two equivalent O atoms, so carboxylic acids are far stronger than alcohols.
Comparing the pH of a solution with the $\text{p}K_a$ of an acid tells you immediately which form predominates, without any ICE table: $\dfrac{[\text{A}^-]}{[\text{HA}]} = 10^{\,\text{pH} - \text{p}K_a}$.
• $\text{pH} < \text{p}K_a$: HA (protonated form) predominates. At $\text{pH} = \text{p}K_a - 1$, $[\text{HA}]/[\text{A}^-] = 10$, so about $91\%$ is HA.
• $\text{pH} = \text{p}K_a$: $[\text{HA}] = [\text{A}^-]$ (50% each).
• $\text{pH} > \text{p}K_a$: $\text{A}^-$ predominates. Example: acetic acid ($\text{p}K_a = 4.74$) at physiological pH 7.40 has $[\text{A}^-]/[\text{HA}] = 10^{2.66} \approx 4.6 \times 10^2$.
A buffer contains a weak acid and its conjugate base (or a weak base and its conjugate acid) in comparable, significant amounts. Each component neutralizes one kind of intruder, converting a strong acid or base into a weak one.
• Added strong acid: $\text{A}^- + \text{H}_3\text{O}^+ \rightarrow \text{HA} + \text{H}_2\text{O}$ (the ratio $[\text{A}^-]/[\text{HA}]$ falls slightly, so the pH decreases slightly).
• Added strong base: $\text{HA} + \text{OH}^- \rightarrow \text{A}^- + \text{H}_2\text{O}$ (the pH increases slightly).
• Dilution with water divides $[\text{HA}]$ and $[\text{A}^-]$ by the same factor, so the ratio and therefore the pH stay essentially unchanged.
Taking $-\log$ of the $K_a$ expression gives the Henderson–Hasselbalch equation: $\text{pH} = \text{p}K_a + \log\dfrac{[\text{A}^-]}{[\text{HA}]}$. Because both species are in the same volume, mole ratios may replace concentration ratios.
1. Stoichiometry: $\text{HA} + \text{OH}^- \rightarrow \text{A}^- + \text{H}_2\text{O}$; HA: $0.100 - 0.010 = 0.090\text{ mol}$; $\text{A}^-$: $0.100 + 0.010 = 0.110\text{ mol}$.
2. Equilibrium: $\text{pH} = 4.74 + \log\dfrac{0.110}{0.090} = 4.83$.
3. Sense check: the pH rose by only 0.09 units; the same base in $1.00\text{ L}$ of pure water would give pH 12.00.
Buffer capacity is the amount of strong acid or strong base a buffer can neutralize before its pH changes significantly. It depends on the amounts of the buffer components, not on their ratio alone.
• Two buffers with the same $[\text{A}^-]/[\text{HA}]$ ratio have the same pH, but the more concentrated buffer has the larger capacity.
• Capacity against added BASE depends on the amount of HA; capacity against added ACID depends on the amount of $\text{A}^-$.
• A buffer is most effective (equal capacity in both directions) when $[\text{HA}] = [\text{A}^-]$, i.e. $\text{pH} = \text{p}K_a$; the useful range is about $\text{p}K_a \pm 1$.
Hypochlorous acid, HOCl, is a weak monoprotic acid with $K_a = 3.0 \times 10^{-8}$ at $25^\circ\text{C}$.
(a) Calculate the pH of $0.20\text{ M HOCl}$. Justify any approximation you make.
(b) A student dilutes the solution in part (a) with an equal volume of water. Using $Q$ and $K_a$, explain why the percent ionization of HOCl increases.
(c) A $50.0\text{ mL}$ sample of $0.20\text{ M HOCl}$ is mixed with $25.0\text{ mL}$ of $0.20\text{ M NaOH}$. Calculate the pH of the resulting solution.
(d) Then $1.0\text{ mmol}$ of HCl is added to the solution from part (c). Write the net ionic equation for the reaction that occurs and calculate the new pH.
Model Answer and Scoring Guide (open after attempting)
Check: $7.7 \times 10^{-5}/0.20 = 0.039\% < 5\%$, so the approximation is valid. (1 point setup, 1 point pH)
(b) Right after dilution, each concentration is halved: $Q = \dfrac{(x/2)(x/2)}{(0.20 - x)/2} = \dfrac{K_a}{2} < K_a$. Because $Q < K_a$, the forward ionization proceeds until $Q = K_a$, so a larger fraction of HOCl is ionized. (1 point for $Q < K_a$ shown mathematically, 1 point for the conclusion)
(c) HOCl: $10.0\text{ mmol}$; $\text{OH}^-$: $5.0\text{ mmol}$. After $\text{HOCl} + \text{OH}^- \rightarrow \text{OCl}^- + \text{H}_2\text{O}$: HOCl $5.0\text{ mmol}$, $\text{OCl}^-$ $5.0\text{ mmol}$. Half-neutralized, so $\text{pH} = \text{p}K_a = -\log(3.0 \times 10^{-8}) = 7.52$. (1 point stoichiometry, 1 point pH)
(d) $\text{OCl}^-(aq) + \text{H}_3\text{O}^+(aq) \rightarrow \text{HOCl}(aq) + \text{H}_2\text{O}(l)$ (1 point). $\text{OCl}^-$: $5.0 - 1.0 = 4.0\text{ mmol}$; HOCl: $5.0 + 1.0 = 6.0\text{ mmol}$. $\text{pH} = 7.52 + \log\dfrac{4.0}{6.0} = 7.35$ (1 point stoichiometry, 1 point pH). Total: 9 points.
Acid–base FRQs reward the order of your reasoning. Readers look for a visible stoichiometry step BEFORE any equilibrium step.
1. Classify first: write "strong acid", "weak acid", "buffer" or "conjugate base only" above your work.
2. Stoichiometry box in mmol (Before / Change / After) whenever a strong acid or base is added.
3. Equilibrium step: ICE table with $K_a$ or $K_b$, or Henderson–Hasselbalch with the AFTER amounts.
4. Significant figures: pH gets as many decimal places as the concentration has significant figures.
5. Check the answer's side of 7: acid solutions and weak-base titrations at equivalence must be below 7; weak-acid titrations at equivalence above 7.
• "At the half-equivalence point $[\text{HA}] = [\text{A}^-]$, so $\text{pH} = \text{p}K_a = 4.74$."
• "The pH at the equivalence point is greater than 7 because the conjugate base $\text{A}^-$ reacts with water to produce $\text{OH}^-$."
• "Because $\text{pH} < \text{p}K_a$, the protonated form HA is the predominant species."
Avoid "the buffer absorbs the acid" without naming the species that reacts and writing the equation.
Unit 8 Acids and Bases: İnteraktif Soru İstasyonu
Özgün AP tarzı sorular (geçmiş sınav sorusu değil). Şık seç; anında çözüm ve hata türü gelir.
AP Chemistry — Unit 9: Thermodynamics and Electrochemistry
Unit 9 (CED weighting 7–9%) answers the question that Units 5–8 left open: will a process happen at all, and how far will it go? You will predict and calculate entropy changes, combine $\Delta H^\circ$ and $\Delta S^\circ$ into the Gibbs free energy change $\Delta G^\circ$, separate thermodynamic favorability from reaction rate, link $\Delta G^\circ$ to the equilibrium constant $K$, and then apply the same ideas to electrochemical cells, where $\Delta G^\circ = -nFE^\circ$ and Faraday's law connects electric charge to the amount of substance. In the current CED (effective Fall 2024) Unit 9 contains Topics 9.1–9.11. Two CED notes shape this unit: the phrase "thermodynamically favored" replaces "spontaneous", and algorithmic Nernst-equation calculations are not required — the Nernst equation is used qualitatively (Topic 9.10). Throughout, one standard applies: units must match ($\Delta S^\circ$ in J/K must be converted to kJ/K before it is combined with $\Delta H^\circ$ in kJ), and every claim about favorability needs a sign or a number to back it. Constants: $R = 8.314\text{ J mol}^{-1}\text{K}^{-1}$, $F = 96\,485\text{ C mol}^{-1}$.
Entropy ($S$) measures the dispersal of matter and energy in a system. Entropy increases when:
• Matter becomes more dispersed: solid → liquid → gas; a gas expands into a larger volume at constant temperature; a solid dissolves into separate ions or molecules (usually).
• The number of moles of gas increases: for reactions involving gases, $\Delta S^\circ > 0$ when the products contain more moles of gas than the reactants.
• Energy becomes more dispersed: by kinetic molecular theory, raising the temperature broadens the distribution of particle kinetic energies, so the entropy of a substance increases with temperature.
• $\text{H}_2\text{O}(l) \rightarrow \text{H}_2\text{O}(g)$: liquid → gas, $\Delta S^\circ > 0$ (large).
• $2\,\text{SO}_2(g) + \text{O}_2(g) \rightarrow 2\,\text{SO}_3(g)$: 3 mol gas → 2 mol gas, $\Delta S^\circ < 0$.
• $\text{CaCO}_3(s) \rightarrow \text{CaO}(s) + \text{CO}_2(g)$: 0 mol gas → 1 mol gas, $\Delta S^\circ > 0$.
• $\text{Ag}^+(aq) + \text{Cl}^-(aq) \rightarrow \text{AgCl}(s)$: dissolved ions → ordered solid, $\Delta S^\circ < 0$.
Unlike enthalpy, entropy has an absolute zero point (a perfect crystal at 0 K), so tables list absolute (standard molar) entropies $S^\circ$ in $\text{J mol}^{-1}\text{K}^{-1}$. Elements in their standard states do not have $S^\circ = 0$ (contrast $\Delta H_f^\circ = 0$ for elements).
$\Delta S^\circ_{reaction} = \Sigma n S^\circ_{products} - \Sigma n S^\circ_{reactants}$, where each $S^\circ$ is multiplied by its coefficient $n$.
$S^\circ$: $\text{N}_2$ 191.6, $\text{H}_2$ 130.7, $\text{NH}_3$ 192.5 $\text{J mol}^{-1}\text{K}^{-1}$.
$\Delta S^\circ = 2(192.5) - [191.6 + 3(130.7)] = 385.0 - 583.7 = -198.7\text{ J/K}$ (per mole of reaction).
Sense check: 4 mol gas → 2 mol gas, so a negative value is expected.
$S^\circ$: $\text{CaCO}_3$ 92.9, CaO 38.2, $\text{CO}_2$ 213.7 $\text{J mol}^{-1}\text{K}^{-1}$ ⟹ $\Delta S^\circ = (38.2 + 213.7) - 92.9 = +159.0\text{ J/K}$.
$\Delta G^\circ$ is the Gibbs free energy change when all reactants and products are in their standard states (pure substances, 1 M solutions, gases at 1 atm or 1 bar). A process with $\Delta G^\circ < 0$ is thermodynamically favored; with $\Delta G^\circ > 0$ it is thermodynamically unfavored. The CED prefers "thermodynamically favored" to "spontaneous", because "spontaneous" is easily misread as "fast" or "without cause".
Two ways to obtain $\Delta G^\circ$:
• From formation data: $\Delta G^\circ_{reaction} = \Sigma n\Delta G_f^\circ(\text{products}) - \Sigma n\Delta G_f^\circ(\text{reactants})$ ($\Delta G_f^\circ = 0$ for elements in their standard states).
• From enthalpy and entropy: $\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ$, with $T$ in kelvin and $\Delta S^\circ$ converted to kJ/K.
• $\Delta H^\circ < 0$, $\Delta S^\circ > 0$ ⟹ favored at all T (no calculation needed).
• $\Delta H^\circ > 0$, $\Delta S^\circ < 0$ ⟹ favored at no T (no calculation needed).
• $\Delta H^\circ > 0$, $\Delta S^\circ > 0$ ⟹ favored at high T (entropy-driven: melting of ice, dissolving $\text{NaNO}_3$).
• $\Delta H^\circ < 0$, $\Delta S^\circ < 0$ ⟹ favored at low T (enthalpy-driven: freezing of water).
• The crossover temperature (where $\Delta G^\circ = 0$) is $T = \Delta H^\circ / \Delta S^\circ$.
$\Delta G^\circ = -92.2\text{ kJ} - (298\text{ K})(-0.1987\text{ kJ/K}) = -92.2 + 59.2 = -33.0\text{ kJ}$ ⟹ thermodynamically favored at 298 K.
For $\text{CaCO}_3(s) \rightarrow \text{CaO}(s) + \text{CO}_2(g)$, $\Delta H^\circ = +178.3\text{ kJ}$, $\Delta S^\circ = +159.0\text{ J/K}$: $\Delta G^\circ_{298} = 178.3 - 298(0.1590) = +130.9\text{ kJ}$ (unfavored), and the reaction becomes favored above $T = 178.3 / 0.1590 \approx 1121\text{ K}$ — which is why lime kilns run very hot.
Thermodynamics tells us whether a process is favored; kinetics tells us how fast it goes. Many processes with $\Delta G^\circ < 0$ occur at immeasurably slow rates. Such processes are under kinetic control, most often because their activation energy is high.
• Diamond → graphite has $\Delta G^\circ < 0$ at room conditions, yet diamonds last for billions of years.
• A mixture of $\text{H}_2$ and $\text{O}_2$ can stand for years without reacting; a spark (or a catalyst) supplies the activation energy and the reaction becomes explosive.
• Glucose in air is thermodynamically unstable with respect to $\text{CO}_2$ and $\text{H}_2\text{O}$, but it needs enzymes to react at body temperature.
"Thermodynamically favored" ($\Delta G^\circ < 0$) means that products are favored at equilibrium ($K > 1$) under standard conditions. The quantitative link is
$\Delta G^\circ = -RT\ln K \qquad K = e^{-\Delta G^\circ/RT}$
• $\Delta G^\circ < 0$ ⟺ $K > 1$ (products favored); $\Delta G^\circ > 0$ ⟺ $K < 1$ (reactants favored); $\Delta G^\circ = 0$ ⟺ $K = 1$.
• Estimation: at 298 K, $RT \approx 2.48\text{ kJ/mol}$. When $|\Delta G^\circ|$ is near zero, $K$ is close to 1; when $|\Delta G^\circ|$ is much larger than $RT$, $K$ deviates strongly from 1. Every 5.7 kJ/mol of $\Delta G^\circ$ changes $K$ by a factor of about 10 at 298 K.
1. $\Delta G^\circ = -5.70\text{ kJ/mol}$: $K = e^{5700/(8.314 \times 298)} = e^{2.30} \approx 10$.
2. $K = 1.0 \times 10^{-5}$: $\Delta G^\circ = -(8.314)(298)\ln(1.0 \times 10^{-5}) = +2.85 \times 10^4\text{ J/mol} = +28.5\text{ kJ/mol}$.
Units: $R$ is in J, so $\Delta G^\circ$ must be in J/mol inside the exponent; use the natural logarithm ln, not log.
The free energy change for dissolving a substance reflects several factors:
1. Breaking the interactions that hold the solid together (lattice energy for ionic solids): endothermic, entropy increases.
2. Reorganizing the solvent around the dissolved particles (breaking some solvent–solvent interactions, ordering solvent molecules in hydration shells): enthalpy cost, and the ordering can lower entropy.
3. Solute–solvent interactions (ion–dipole attractions): exothermic.
The signs and relative sizes of each contribution can be estimated, but predicting the overall $\Delta G^\circ$ of dissolution is difficult because the three contributions partly cancel.
A thermodynamically unfavorable process can be made to occur in two ways:
• By an external source of energy: electrical energy drives an electrolytic cell or recharges a battery; light drives the overall conversion of $\text{CO}_2$ to glucose in photosynthesis.
• By coupling it to a favorable reaction that shares a common intermediate. Because $G$ is a state function, the $\Delta G^\circ$ values of the steps add; if the sum is negative, the overall process is favored.
$\text{glutamate} + \text{NH}_3 \rightarrow \text{glutamine} + \text{H}_2\text{O}$ $\quad \Delta G^\circ = +14.2\text{ kJ/mol}$
$\text{ATP} + \text{H}_2\text{O} \rightarrow \text{ADP} + \text{P}_i$ $\quad \Delta G^\circ = -30.5\text{ kJ/mol}$
Overall: $\text{glutamate} + \text{NH}_3 + \text{ATP} \rightarrow \text{glutamine} + \text{ADP} + \text{P}_i$, $\Delta G^\circ = 14.2 + (-30.5) = -16.3\text{ kJ/mol}$ (favored). In the cell, the reactions share a phosphorylated intermediate.
Solved Example (metallurgy): $\text{Cu}_2\text{S}(s) \rightarrow 2\,\text{Cu}(s) + \text{S}(s)$, $\Delta G^\circ = +86.2\text{ kJ}$, coupled with $\text{S}(s) + \text{O}_2(g) \rightarrow \text{SO}_2(g)$, $\Delta G^\circ = -300.1\text{ kJ}$: overall $\Delta G^\circ = -213.9\text{ kJ}$.
In every electrochemical cell, oxidation occurs at the anode and reduction occurs at the cathode ("An Ox, Red Cat").
• A galvanic (voltaic) cell runs a thermodynamically favored redox reaction and produces electrical energy.
• An electrolytic cell uses an external power source to drive a thermodynamically unfavored reaction (electroplating, electrolysis of molten NaCl, recharging a battery).
Roles of the components: electrons travel through the external wire from anode to cathode; the voltmeter measures the potential difference; the salt bridge completes the circuit and keeps each half-cell electrically neutral — its anions move toward the anode compartment and its cations toward the cathode compartment.
• Anode: $\text{Zn}(s) \rightarrow \text{Zn}^{2+}(aq) + 2e^-$ — the zinc electrode loses mass; $[\text{Zn}^{2+}]$ rises.
• Cathode: $\text{Cu}^{2+}(aq) + 2e^- \rightarrow \text{Cu}(s)$ — copper plates out, the electrode gains mass, the blue color fades.
• Without a salt bridge, charge builds up in each half-cell and the current stops almost immediately.
• Electrolysis of molten NaCl: cathode $\text{Na}^+ + e^- \rightarrow \text{Na}(l)$; anode $2\,\text{Cl}^- \rightarrow \text{Cl}_2(g) + 2e^-$ — a gas evolves at the anode.
The standard cell potential is calculated from the standard reduction potentials of the two half-reactions:
$E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode}$ (both written as reductions; the half-reaction with the more positive $E^\circ$ is reduced in a galvanic cell).
• $E^\circ$ is an intensive property: multiplying a half-reaction by 2 to balance electrons does not double its $E^\circ$.
• $\Delta G^\circ = -nFE^\circ$, where $n$ is the moles of electrons transferred in the balanced equation and $F = 96\,485\text{ C/mol e}^-$. Since $1\text{ J} = 1\text{ C} \cdot \text{V}$, the result is in joules.
• $E^\circ > 0$ ⟺ $\Delta G^\circ < 0$ ⟺ $K > 1$: a thermodynamically favored (galvanic) reaction. A negative $E^\circ$ means an external potential must be applied.
$E^\circ(\text{Ag}^+/\text{Ag}) = +0.80\text{ V}$; $E^\circ(\text{Cu}^{2+}/\text{Cu}) = +0.34\text{ V}$.
1. $E^\circ_{cell} = 0.80 - 0.34 = +0.46\text{ V}$ (not $2 \times 0.80 - 0.34$).
2. $n = 2$ (two $\text{Ag}^+$ each gain one electron; one Cu loses two).
3. $\Delta G^\circ = -(2)(96\,485)(0.46) = -8.88 \times 10^4\text{ J} = -88.8\text{ kJ}$.
4. Optional link to 9.5: $\ln K = nFE^\circ/RT = 35.8 \implies K \approx 4 \times 10^{15}$.
Under nonstandard conditions, the cell potential depends on the concentrations of the active species. The cell potential is a driving force toward equilibrium: the farther the reaction is from equilibrium, the larger $|E|$.
• $E^\circ$ corresponds to $Q = 1$. As the cell operates, $Q$ rises toward $K$, $|E|$ decreases, and $E = 0$ when $Q = K$ (a "dead" battery is a cell at equilibrium).
• Changes that move the cell farther from equilibrium than $Q = 1$ (decreasing $Q$: more reactant ions, fewer product ions) increase $E$ above $E^\circ$; changes that move it closer (increasing $Q$) decrease $E$.
• Qualitative Nernst equation: $E = E^\circ - \dfrac{RT}{nF}\ln Q$. If $Q < 1$, $\ln Q < 0$ and $E > E^\circ$; if $Q > 1$, $E < E^\circ$.
• The CED notes that equilibrium arguments such as Le Châtelier's principle do not strictly apply, because an operating cell is not at equilibrium — reason with $Q$ and $K$ instead.
Faraday's laws link the amount of electric charge to the stoichiometry of the electrode reaction. One mole of electrons carries $F = 96\,485\text{ C}$.
The four-step road map (works forwards and backwards):
1. Charge: $q = I \times t$ (current in amperes, time in seconds).
2. Moles of electrons: $n_{e^-} = q / F$.
3. Moles of substance: divide by the electrons per formula unit from the half-reaction ($\text{Ag}^+$: 1, $\text{Cu}^{2+}$: 2, $\text{Al}^{3+}$: 3).
4. Mass: multiply by molar mass.
$q = 2.00 \times 1800 = 3600\text{ C}$; $n_{e^-} = 3600/96\,485 = 0.03731\text{ mol}$; $n_{\text{Cu}} = 0.03731/2 = 0.01866\text{ mol}$; mass $= 0.01866 \times 63.55 = 1.19\text{ g}$.
Solved Example 2 (finding the time): How long must 1.50 A flow to deposit 2.16 g of Ag?
$n_{\text{Ag}} = 2.16/107.87 = 0.02002\text{ mol} = n_{e^-}$; $q = 0.02002 \times 96\,485 = 1932\text{ C}$; $t = 1932/1.50 = 1288\text{ s} \approx 21.5\text{ min}$.
Cells in series carry the same charge: if 1.00 g of Cu (2 e⁻) is deposited, the silver cell in series deposits $(1.00/63.55) \times 2 \times 107.87 = 3.39\text{ g}$ of Ag.
A galvanic cell is built from a Zn electrode in 1.0 M $\text{Zn}^{2+}$ and a Ag electrode in 1.0 M $\text{Ag}^+$. $E^\circ(\text{Zn}^{2+}/\text{Zn}) = -0.76\text{ V}$, $E^\circ(\text{Ag}^+/\text{Ag}) = +0.80\text{ V}$.
(a) Write the balanced net ionic equation for the cell reaction and identify the anode.
(b) Calculate $E^\circ_{cell}$.
(c) Calculate $\Delta G^\circ$ in kJ for the reaction as written in part (a).
(d) The cell delivers a constant current of 0.150 A for 2.00 hours. Calculate the change in mass of the Ag electrode.
(e) A student replaces the 1.0 M $\text{Ag}^+$ solution with 0.10 M $\text{Ag}^+$. Predict whether the measured cell potential is greater than, less than or equal to $E^\circ_{cell}$, and justify your answer in terms of $Q$.
Model Answer and Scoring Guide (open after attempting)
(b) $E^\circ_{cell} = 0.80 - (-0.76) = +1.56\text{ V}$ (the Ag potential is not doubled). (1 point)
(c) $n = 2$: $\Delta G^\circ = -(2)(96\,485)(1.56) = -3.01 \times 10^5\text{ J} = -301\text{ kJ}$. (1 point for $n$, 1 point for value and sign)
(d) $q = 0.150 \times 7200 = 1080\text{ C}$; $n_{e^-} = 1080/96\,485 = 0.01119\text{ mol}$; Ag deposited $= 0.01119\text{ mol} \times 107.87 = 1.21\text{ g}$; the Ag electrode (cathode) gains 1.21 g. (1 point moles of electrons, 1 point mass and direction)
(e) $Q = [\text{Zn}^{2+}]/[\text{Ag}^+]^2 = 1.0/(0.10)^2 = 100 > 1$. The cell starts closer to equilibrium than under standard conditions, so $E < E^\circ_{cell}$. (1 point prediction, 1 point $Q$-based justification). Total: 8 points.
Unit 9 FRQs are scored on correct signs, correct units and explicit reasoning. A correct number with a missing sign or a J/kJ mismatch usually loses the point.
1. Predict the sign first (moles of gas for $\Delta S^\circ$; sign table for $\Delta G^\circ$), then calculate and compare.
2. Convert units on paper: "$\Delta S^\circ = -198.7\text{ J/K} = -0.1987\text{ kJ/K}$".
3. State $n$ explicitly before using $\Delta G^\circ = -nFE^\circ$ or Faraday's law.
4. Time in seconds in $q = It$.
5. Justify with $Q$ vs $K$ (or $Q$ vs 1) for nonstandard cell potentials.
• "The reaction is thermodynamically favored at 298 K because $\Delta G^\circ = -33.0\text{ kJ} < 0$."
• "Although $\Delta G^\circ < 0$, no reaction is observed because the activation energy is high; the reaction is under kinetic control."
• "Decreasing $[\text{Ag}^+]$ increases $Q$, moving the cell closer to equilibrium, so the cell potential decreases."
Avoid "the reaction is spontaneous so it happens quickly" — favorability and rate are different properties.
Unit 9 Thermodynamics and Electrochemistry: İnteraktif Soru İstasyonu
Özgün AP tarzı sorular (geçmiş sınav sorusu değil). Şık seç; anında çözüm ve hata türü gelir.