What does the reaction quotient Q represent?
Answer: The ratio of product to reactant concentrations/pressures at any point in time, calculated like Keq but using current (non-equilibrium) values.
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MCAT General Chemistry: Equilibrium, Acids, Bases and Thermodynamics (University)
Equilibrium BasicsLe Chatelier's PrincipleAcid-Base EquilibriaBuffersTitrationsThermodynamicsKineticsElectrochemistryGasesSolubilityColligative Properties
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Equilibrium Basics
If Q < Keq, which way does the reaction shift?
Answer: Forward, toward products, until Q rises to equal Keq.
If Q > Keq, which way does the reaction shift?
Answer: Reverse, toward reactants, until Q falls to equal Keq.
Write the equilibrium constant expression for aA + bB ⇌ cC + dD.
Answer: Keq = [C]^c[D]^d / ([A]^a[B]^b)
Explanation
Pure solids and liquids are omitted. Example: for N2 + 3H2 ⇌ 2NH3, Keq = [NH3]^2 / ([N2][H2]^3).
Le Chatelier's Principle
Effect of increasing pressure (decreasing volume) on a gas-phase equilibrium?
Answer: Shifts toward the side with fewer moles of gas.
Effect of increasing temperature on an exothermic reaction at equilibrium?
Answer: Shifts reverse (toward reactants); Keq decreases.
Explanation
Treat heat as a product: adding heat pushes the reaction backward.
Effect of increasing temperature on an endothermic reaction at equilibrium?
Answer: Shifts forward (toward products); Keq increases.
Explanation
Treat heat as a reactant: adding heat pushes the reaction forward.
Effect of adding more reactant to a system at equilibrium?
Answer: Shifts forward, consuming added reactant and forming more product.
Effect of removing product from a system at equilibrium?
Answer: Shifts forward to replace the removed product.
Acid-Base Equilibria
Relationship between Ka, Kb, and Kw for a conjugate acid-base pair?
Answer: Ka × Kb = Kw = 1.0×10⁻¹⁴ at 25°C.
Formula for pKa?
Answer: pKa = −log(Ka); a smaller pKa means a stronger acid.
Explanation
Example: Ka = 1.8×10⁻⁵ (acetic acid) → pKa = 4.74.
pH of a strong acid solution of concentration C?
Answer: pH = −log[H+] = −log(C), assuming complete dissociation.
Explanation
Example: 0.01 M HCl → pH = −log(0.01) = 2.
Approximate pH of a weak monoprotic acid at concentration C?
Answer: [H+] ≈ √(Ka × C); pH = −log[H+].
Explanation
Example: Ka = 1.8×10⁻⁵, C = 0.1 M → [H+] ≈ √(1.8×10⁻⁶) ≈ 1.34×10⁻³ M → pH ≈ 2.87.
Relationship between pH and pOH at 25°C?
Answer: pH + pOH = 14 (since Kw = 10⁻¹⁴).
Buffers
Henderson-Hasselbalch equation?
Answer: pH = pKa + log([A⁻]/[HA])
Explanation
Example: pKa = 4.76, [A⁻]/[HA] = 10 → pH = 4.76 + log(10) = 5.76.
When is a buffer's capacity greatest?
Answer: When [A⁻] = [HA], i.e., pH = pKa; the buffer best resists pH change near this ratio.
What happens to a buffer's pH when a small amount of strong acid is added?
Answer: H+ reacts with A⁻, converting it to HA; pH drops only slightly instead of sharply.
Titrations
On a titration curve of a weak acid with strong base, identify the equivalence point and half-equivalence point.
Answer: Equivalence point: sharp pH jump where moles acid = moles base added. Half-equivalence point: midpoint before that, where pH = pKa.
Explanation
At half-equivalence, [HA] = [A⁻], so by Henderson-Hasselbalch pH = pKa.
pH at the half-equivalence point of a weak acid titration?
Answer: pH = pKa, because [HA] = [A⁻] at that point.
At the equivalence point of a weak acid–strong base titration, is the solution acidic, basic, or neutral?
Answer: Basic (pH > 7), due to hydrolysis of the conjugate base formed.
pH at the equivalence point of a strong acid–strong base titration?
Answer: 7 (neutral); the salt formed does not hydrolyze.
Thermodynamics
Gibbs free energy equation?
Answer: ΔG = ΔH − TΔS
Explanation
ΔH = enthalpy change, T = temperature (K), ΔS = entropy change. Example: ΔH=−50 kJ, ΔS=−100 J/K, T=298K → ΔG=−50,000−298(−100)=−20,200 J (spontaneous).
Sign of ΔG for a spontaneous process?
Answer: ΔG < 0 (negative); the process is exergonic and spontaneous as written.
Spontaneity when ΔH < 0 and ΔS > 0?
Answer: Always spontaneous at all temperatures (ΔG is always negative).
Spontaneity when ΔH > 0 and ΔS < 0?
Answer: Never spontaneous at any temperature (ΔG is always positive).
Relationship between ΔG° and the equilibrium constant K?
Answer: ΔG° = −RT ln K
Explanation
R=8.314 J/mol·K, T in Kelvin. Example: K=100, T=298K → ΔG°=−8.314×298×ln(100)≈−11.4 kJ/mol.
Kinetics
General form of a rate law and how are orders determined?
Answer: Rate = k[A]^m[B]^n; m and n are found experimentally by seeing how rate changes when one concentration is varied while others are held constant (not from stoichiometry).
Explanation
Example: doubling [A] quadruples rate → order in A is 2.
Arrhenius equation?
Answer: k = A·e^(−Ea/RT)
Explanation
A = pre-exponential factor, Ea = activation energy, R = gas constant, T = temperature (K). Raising T increases k exponentially.
How does a catalyst affect reaction rate and activation energy?
Answer: Lowers Ea by providing an alternate pathway, increasing rate in both directions; does not change ΔG or Keq.
Electrochemistry
In a galvanic (voltaic) cell, where is oxidation and what are the electrode signs?
Answer: Oxidation occurs at the anode; anode is negative, cathode is positive. The cell generates electrical energy spontaneously.
In an electrolytic cell, where is oxidation and what are the electrode signs?
Answer: Oxidation still occurs at the anode, but the anode is positive and cathode negative; an external power source drives the nonspontaneous reaction.
Nernst equation?
Answer: E = E° − (RT/nF) ln Q, or at 25°C: E = E° − (0.0592/n) log Q
Explanation
n = moles of electrons transferred, F = Faraday's constant. Example: Daniell cell E°=1.10V; if Q<1, E>E°.
Relationship between E°cell and ΔG°?
Answer: ΔG° = −nFE°cell
Explanation
Example: n=2, F=96,485 C/mol, E°=1.10V → ΔG°=−2(96,485)(1.10)≈−212 kJ/mol.
Gases
Ideal gas law?
Answer: PV = nRT
Explanation
P=pressure, V=volume, n=moles, R=gas constant, T=temperature (K). Example: 1 mol at STP (1 atm, 273K) → V=22.4 L.
Dalton's law of partial pressures?
Answer: Ptotal = P1 + P2 + ... + Pn, the sum of each gas's partial pressure in a mixture.
Explanation
Example: if Ptotal=760 mmHg with PO2=160 mmHg, then PN2 (and others)=600 mmHg.
Solubility
Ksp expression for a salt AxBy?
Answer: Ksp = [A]^x[B]^y, ion concentrations at saturation (pure solid excluded).
Explanation
Example: AgCl, Ksp=1.8×10⁻¹⁰ → solubility = √Ksp ≈ 1.34×10⁻⁵ M.
What is the common ion effect?
Answer: Adding an ion already present in the equilibrium shifts the dissolution equilibrium backward, decreasing solubility of the sparingly soluble salt.
Explanation
Example: adding NaCl to a saturated AgCl solution lowers AgCl solubility.
Colligative Properties
What are colligative properties and what do they depend on?
Answer: Properties that depend only on the number of solute particles, not their identity: boiling point elevation, freezing point depression, vapor pressure lowering, osmotic pressure.
Freezing point depression formula?
Answer: ΔTf = i·Kf·m
Explanation
i = van't Hoff factor, Kf = cryoscopic constant, m = molality. Example: 1 m NaCl (i=2), Kf=1.86 → ΔTf=3.72°C.
Boiling point elevation formula?
Answer: ΔTb = i·Kb·m
Explanation
i = van't Hoff factor, Kb = ebullioscopic constant, m = molality.
Osmotic pressure formula?
Answer: π = iMRT
Explanation
M=molarity, R=gas constant, T=temperature (K), i=van't Hoff factor. Example: 0.1 M NaCl (i=2), T=298K → π=2(0.1)(0.0821)(298)≈4.89 atm.
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