Midterm Exam 2 Chemistry Study (Ole Miss, Dargon, Chem 106)

Catalysts, intermediates & energy diagrams

• A catalyst lowers activation energy (Eₐ) — it does not change ΔH or the overall thermodynamics of the reaction.

• An intermediate is formed in one step and consumed in another — it appears in the mechanism but never in the overall equation.

• On an energy diagram: Eₐ = height of each transition-state peak above its reactant plateau; ΔH = difference between reactant and product energy levels.

Reaction mechanisms & rate law

• The rate-determining step (slow step) dictates the rate law — never the fast step.

• If an intermediate appears in the slow-step rate law, substitute it out using the equilibrium expression from the fast step before it.

• Overall reaction order = sum of all exponents in the rate law.

Rate = k[A]ᵐ[B]ⁿ  →  exponents come from the slow step only

Arrhenius equation

• Always convert temperature from °C → K by adding 273.

• Two-temperature form lets you find Eₐ or compare rate constants without knowing A.

k = A·e^(−Eₐ/RT)    ln(k₂/k₁) = (Eₐ/R)(1/T₁ − 1/T₂)

R = 8.314 J/(mol·K). Eₐ will be in J/mol — watch units carefully.

Interpreting K values

• K ≫ 1: equilibrium favors products.

• K ≪ 1: equilibrium favors reactants.

• K ≈ 1: significant amounts of both present.

• Only temperature changes K. Adding/removing species, changing pressure/volume, or adding a catalyst do NOT change K — they only shift the reaction position.

K manipulation rules

• Reverse the reaction → K becomes 1/K.

• Multiply all coefficients by n → K becomes Kⁿ.

• Add two reactions → K_total = K₁ × K₂.

Q vs K — predicting shift direction

• Q < K → reaction shifts right (toward products) to reach equilibrium.

• Q > K → reaction shifts left (toward reactants).

• Q = K → system is at equilibrium, no net shift.

Le Châtelier's principle

• Add reactant or remove product → shifts right.

• Add product or remove reactant → shifts left.

• Increase pressure (decrease volume) → shifts toward side with fewer gas moles.

• Solids and pure liquids: excluded from K — adding/removing them has no effect on equilibrium.

• Increasing temperature: shifts toward the endothermic direction (also changes K).

Writing Kc & ICE tables

• Kc = [products]^coeff / [reactants]^coeff — exclude pure solids and liquids.

• ICE: Initial → Change (+x or −x) → Equilibrium. Substitute equilibrium expressions into K = ... and solve for x.

Kp = Kc(RT)^Δn    Δn = moles gaseous products − moles gaseous reactants

Δn counts gas-phase species only — ignore solids and liquids in the Δn count.

Strong acids & bases — memorize these

7 strong acids

• HCl, HBr, HI (hydrogen halides — NOT HF)

• HNO₃, H₂SO₄, HClO₄, HClO₃

HF is a weak acid — don't confuse it with HCl/HBr/HI.

Strong base groups

• Group 1 hydroxides: LiOH, NaOH, KOH, RbOH, CsOH

• Group 2 hydroxides: Ca(OH)₂, Sr(OH)₂, Ba(OH)₂

Ba(OH)₂ gives 2 OH⁻ per formula unit — [OH⁻] = 2 × [Ba(OH)₂]!

Acid-base definitions

• Arrhenius: acid produces H⁺ in water; base produces OH⁻.

• Brønsted–Lowry: acid = H⁺ donor; base = H⁺ acceptor.

• Lewis: acid = electron-pair acceptor; base = electron-pair donor.

• Amphiprotic species can act as either acid or base (e.g., HCO₃⁻, H₂O, HSO₄⁻).

Conjugate pairs, pKa, Ka ranking

• Conjugate base of a strong acid → very weak base (negligible Kb).

• Stronger acid → weaker conjugate base, and vice versa.

• Lower pKa = stronger acid. Higher pKa = weaker acid.

• Ka × Kb = Kw = 1.0 × 10⁻¹⁴ (at 25°C) for a conjugate acid-base pair.

To find Kb for the conjugate base when given Ka: Kb = Kw / Ka.

pH calculations

• Strong acid: [H⁺] = concentration of acid (fully dissociates) → pH = −log[H⁺].

• Strong base: find [OH⁻] (remember ×2 for Group 2!) → pOH = −log[OH⁻] → pH = 14 − pOH.

• Weak acid: [H⁺] = √(Ka × C) (shortcut when x ≪ C) → pH = −log[H⁺].

• Weak base: [OH⁻] = √(Kb × C) → pOH = −log[OH⁻] → pH = 14 − pOH. Don't stop at pOH!

• Percent dissociation: % = ([H⁺] / C₀) × 100.

Kw, temperature & salt classification

• Neutral pH = 7 only at 25°C. As temperature increases, Kw increases, so neutral pH decreases below 7.

• Salt classification: strong acid + strong base → neutral salt. Weak acid + strong base → basic salt. Strong acid + weak base → acidic salt.

• Polyprotic acids: Ka₁ ≫ Ka₂ ≫ Ka₃. First dissociation dominates pH calculation.