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Midterm & Final Reference · Ultra-Dense A4
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CHEMICAL EQUILIBRIUM ↗ TAP
Equilibrium constant K
aA + bB ⇌ cC + dD K = [C]ᶜ[D]ᵈ / ([A]ᵃ[B]ᵇ)

K depends only on temperature, not concentrations. K_c uses concentrations; K_p uses partial pressures (gases). Pure solids and liquids are excluded from K expressions.

Q vs K — direction
CompareDirectionWhat happens
Q < K→ forwardmore products form
Q = Kat equilibriumno net change
Q > K← reverseproducts → reactants
Le Chatelier
Stress shifts equilibrium to oppose the change. Add reactant → forward. Remove product → forward. Increase pressure → fewer-mole side. Heat exo → reverse; heat endo → forward.
K conventions
Reverse rxn: K_rev = 1/K. Multiply rxn by n: K^n. Add rxns: multiply Ks. Pure liquids/solids: omit from expression.
K_p = K_c (RT)^Δn Δn = (moles gas products) − (moles gas reactants)

ICE table: set up Initial, Change, Equilibrium concentrations. Solve quadratic for x. Approximation x ≪ initial valid if K is small (typically K < 10⁻³).

⚡ EXAM TRAP — K DEPENDS ON T ONLY

Adding more reactant doesn't change K — only shifts position toward products. Adding inert gas at constant V doesn't change anything. Only temperature changes K. Many students think pressure changes the constant; it doesn't.

THERMODYNAMICS II — GIBBS & SPONTANEITY ↗ TAP
Three sister functions
QuantitySymbolWhat it captures
EnthalpyΔHheat at const P
EntropyΔSdisorder change
Gibbs free energyΔGspontaneity, max usable work
ΔG = ΔH − T·ΔS spontaneous ⇔ ΔG < 0
Sign analysis
4 cases
(i) ΔH < 0, ΔS > 0: always spontaneous. (ii) ΔH > 0, ΔS < 0: never spontaneous. (iii) ΔH < 0, ΔS < 0: at low T. (iv) ΔH > 0, ΔS > 0: at high T.
Standard ΔG vs ΔG
ΔG°: standard conditions, related to K. ΔG: actual conditions: ΔG = ΔG° + RT ln Q.
ΔG° = −RT ln K large K (favors products) ⇔ very negative ΔG°

Entropy of universe: ΔS_univ = ΔS_sys + ΔS_surr ≥ 0 (2nd law). Spontaneous = ΔS_univ > 0. Notice the universe, not just system.

⚡ EXAM TRAP — SPONTANEOUS ≠ FAST

Diamond → graphite is spontaneous (ΔG < 0) but takes geological time (kinetics). Spontaneity says nothing about rate. ΔG only addresses thermodynamic favorability; rate needs activation energy analysis.

ACID-BASE EQUILIBRIA ↗ TAP
pH and Ka basics
pH = −log[H⁺] pOH = −log[OH⁻] pH + pOH = 14 (at 25°C)Ka = [H⁺][A⁻]/[HA] pKa = −log Ka pKa + pKb = 14 (conjugate pair)
Acid strength comparison
TypeExamplesBehavior
Strong acidsHCl, HBr, HI, HNO₃, H₂SO₄ (1st), HClO₄fully dissociate
Weak acidsCH₃COOH (pKa 4.76), HCN (9.21), HF (3.17)partial dissociation
Strong basesNaOH, KOH, Ca(OH)₂, Ba(OH)₂fully dissociate
Weak basesNH₃ (pKb 4.75), aminespartial
Conjugate pairs
Strong acid → very weak conjugate base. Weak acid → conjugate base of moderate strength (basic salt). NH₄⁺ slightly acidic; CH₃COO⁻ slightly basic.
Polyprotic
H₂SO₄, H₃PO₄, H₂CO₃: stepwise dissociation. Ka1 ≫ Ka2 ≫ Ka3 each successive H harder to remove.

Salt hydrolysis: NH₄Cl in water → acidic (NH₄⁺ donates H⁺). NaC₂H₃O₂ → basic (acetate accepts H⁺). NaCl → neutral.

⚡ EXAM TRAP — pH FROM WEAK ACID

For weak acid HA at concentration C, [H⁺] ≠ C. Use ICE: x² ≈ Ka·C when x ≪ C (5% rule). pH = −log(√(Ka·C)). Common error: assuming weak acid dissociates fully like a strong one.

KINETICS ↗ TAP
Rate law
rate = k [A]^m [B]^n m, n = orders, found EXPERIMENTALLY (not from coefficients)

Overall order = m + n. Units of k depend on order. Most reactions have small integer orders (0, 1, 2).

Integrated rate laws
OrderIntegrated formLinear plotHalf-life
0[A] = [A]₀ − kt[A] vs t[A]₀/(2k)
1ln[A] = ln[A]₀ − ktln[A] vs t0.693/k (const)
21/[A] = 1/[A]₀ + kt1/[A] vs t1/(k[A]₀)
Arrhenius
k = A · e^(−Ea/RT). Higher T → higher k (exponential). Plot ln k vs 1/T → slope = −Ea/R.
Catalyst
Lowers Ea → speeds both forward + reverse equally → no change in K. Doesn't shift equilibrium, just gets there faster.

Mechanism: overall reaction = sum of elementary steps. Slowest step = rate-determining. Order in reactants matches the rate law of the rate-limiting step (only).

⚡ EXAM TRAP — ORDERS ≠ COEFFICIENTS

For 2NO + O₂ → 2NO₂, you might guess rate = k[NO]²[O₂]. But experiment may show rate = k[NO][O₂]² depending on the mechanism. Always determine orders empirically from data, not from balanced equation coefficients.

ELECTROCHEMISTRY ↗ TAP
Galvanic cell
Spontaneous redox; chemical → electrical energy. Cathode = reduction (+), anode = oxidation (−)

Memorize: Red Cat An Ox. Salt bridge maintains charge balance. e⁻ flow from anode to cathode through external wire.

QuantityFormula
Cell EMF (standard)E°_cell = E°_cathode − E°_anode
ΔG° from EMFΔG° = −nFE°_cell
K from EMFln K = nFE°/(RT)
Nernst eqE = E° − (RT/nF) ln Q
F = 96485 C/mol e⁻ n = mol e⁻ transferred Q = reaction quotient
Standard reduction potentials
Higher E° = stronger oxidizing agent. F₂/F⁻ = +2.87 V (best oxidizer). Li⁺/Li = −3.04 V (best reducer).
Electrolysis
Non-spontaneous: external power forces redox. Time × current = total charge. Q = It; mol = Q/(nF). Faraday's laws.

Sign check: spontaneous galvanic E°_cell > 0, ΔG < 0, K > 1. Electrolysis is the reverse — input energy required.

⚡ EXAM TRAP — SIGNS DON'T FLIP IN E° TABLE

When you reverse a half-reaction (oxidation instead of reduction), flip the sign of E°. But when adding two half-cells to get cell EMF: don't multiply E° by stoichiometric coefficient — E° is intensive. Multiply ΔG, not E°.

BUFFERS & TITRATION ↗ TAP
Buffer = weak acid + conjugate base
Henderson–Hasselbalch: pH = pKa + log([A⁻] / [HA])

A buffer resists pH changes. Best buffering when [A⁻] ≈ [HA] (pH ≈ pKa). Buffer capacity scales with absolute concentrations.

Titration regions
RegionWhat dominatespH calc
Initialweak acid onlyx² ≈ Ka·C
Pre-equivbuffer (HA + A⁻)HH equation
Half-equiv[HA] = [A⁻]pH = pKa
Equivalenceonly A⁻pH from Kb hydrolysis
Post-equivexcess strong basejust OH⁻
Pick the right buffer
For pH X, choose acid with pKa within ±1 of X. CH₃COOH/CH₃COO⁻ buffers near pH 4.76. NH₄⁺/NH₃ buffers near pH 9.25.
Indicator selection
Pick indicator whose pKa ≈ pH at equivalence point. Phenolphthalein (8-10) for weak-acid + strong-base. Methyl red (4-6) for strong-acid + weak-base.

Equivalence ≠ neutralization to pH 7. For weak acid + strong base, equivalence pH > 7 (basic salt left over). Strong + strong: pH = 7. Strong acid + weak base: pH < 7.

⚡ EXAM TRAP — TITRATING WEAK + STRONG ≠ pH 7

Titrating CH₃COOH with NaOH gives equivalence at pH ≈ 8.7 (acetate is basic). Indicator must turn around there, not at pH 7. Universal-indicator pH 7 endpoint loses you the experiment.

SOLUBILITY EQUILIBRIA — Ksp ↗ TAP
Ksp expression
AgCl(s) ⇌ Ag⁺(aq) + Cl⁻(aq) Ksp = [Ag⁺][Cl⁻]

Solid AgCl excluded from K. Smaller Ksp = less soluble. Ksp values vary across many orders of magnitude (10⁻¹⁰ to 10⁻³⁵).

Molar solubility from Ksp
1:1 salt (AgCl)
Let s = molar solubility. [Ag⁺] = s, [Cl⁻] = s. Ksp = s². s = √Ksp.
1:2 or 2:1 (PbI₂, Ag₂CrO₄)
PbI₂: Ksp = s · (2s)² = 4s³. s = ∛(Ksp/4). Be careful with the stoichiometric factors.
Common Ksp valuesKsp
AgCl1.8 × 10⁻¹⁰
BaSO₄1.1 × 10⁻¹⁰
CaCO₃4.5 × 10⁻⁹
Mg(OH)₂5.6 × 10⁻¹²

Common-ion effect: adding a shared ion decreases solubility. Add Cl⁻ to AgCl saturated solution → [Ag⁺] drops to maintain Ksp. Le Chatelier in action.

Q vs Ksp: Q < Ksp = unsaturated (more dissolves). Q > Ksp = supersaturated (precipitates). Q = Ksp = saturated.

⚡ EXAM TRAP — STOICHIOMETRY IN Ksp

For Ag₂CrO₄: 2 Ag⁺ + 1 CrO₄²⁻. Solubility s gives [Ag⁺] = 2s, [CrO₄²⁻] = s. Ksp = (2s)²s = 4s³. Forgetting the (2s)² costs the entire problem. Always write the dissolution equation first.

DECISION BOX — WHICH TECHNIQUE? ↗ TAP
Match question to method
Question says…Use § fromApproach
'find K' or 'eq concentrations'§ ①ICE table; small-x approx if K<10⁻³
'shift in equilibrium'§ ①Le Chatelier: oppose the stress
'pressure / volume change'§ ①Δn(gas) tells direction
'pH of strong acid/base'§ ②full dissociation, direct log
'pH of weak acid'§ ②x² ≈ Ka·C; check 5% rule
polyprotic acid§ ②treat each Ka step; usually Ka1 dominates
salt acidity (NH₄Cl, NaOAc)§ ②hydrolysis: cation acid, anion base
'buffer pH' / Henderson§ ③pH = pKa + log([A⁻]/[HA])
'titration midpoint / equiv'§ ③identify region: HH or weak hydrolysis
'will it precipitate?'§ ④compute Q, compare to Ksp
solubility from Ksp§ ④watch stoichiometry: 4s³ for 1:2 etc.
'find rate law from data'§ ⑤compare runs, find orders, then k
'half-life of 1st order'§ ⑤t½ = 0.693/k (constant only for 1st order)
'find Ea'§ ⑤Arrhenius: ln(k₂/k₁) = (Ea/R)(1/T₁ − 1/T₂)
'mechanism + RDS'§ ⑤slow step rate law = overall rate law
'is rxn spontaneous?'§ ⑥ΔG = ΔH − TΔS; sign analysis
'find ΔG° from K'§ ⑥ΔG° = −RT ln K
'galvanic cell EMF'§ ⑦E°_cell = E°_cath − E°_anode
'electrolysis: how much deposited'§ ⑦Q = It; mol = Q/(nF)
'Nernst non-standard E'§ ⑦E = E° − (RT/nF) ln Q
Sanity checklist
Units? Sig figs? Sign of ΔG / E? Q vs K direction? Reasonable magnitude (compare to nearby pKa, Ksp values)?
FRQ template
State the equation (balanced!). Set up ICE. Identify approximations. Solve. Verify approximations valid. Box answer.
⚡ EXAM TRAP — UNIT MISMATCH

R = 8.314 J/(mol·K) — match J or convert to kJ for ΔH. F = 96485 C/mol e⁻. Always confirm units when mixing equations. Off-by-1000 errors cascade through the whole problem.

⚡ FINAL EXAM TRAP — Q vs K REASONING

'If Q < K, reaction shifts toward products' — this is the universal direction-check. Use it for: equilibrium shifts, solubility (Q vs Ksp), galvanic cell direction (ΔG = ΔG° + RT ln Q). One framework, many applications.

GEN CHEM II · Comprehensive Cram Sheet · Ultra-Dense A4
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