Faraday's laws, Kohlrausch's law, conductance, standard electrode potentials, Gibbs free energy & cell potential, complex rate laws, Arrhenius deep-dive, and catalysis for NEET.
1. Faraday's Laws of Electrolysis
💡 Why this matters in real life
Electrolysis is used to refine metals (like copper purification), electroplate jewelry, produce chlorine and aluminum, and recharge batteries. Faraday's laws tell us exactly how much product forms for a given amount of electricity — essential for industrial electrochemical processes. Every time you charge your phone, Faraday's laws are at work.
Fig 1.1: Electrolytic cell for copper purification. Cu²⁺ ions migrate to the cathode, where they deposit as copper metal.
🧠 The Big Idea — In Plain Words
Faraday's First Law: The mass (m) of a substance deposited at an electrode is directly proportional to the charge (Q) passed: m = ZQ, where Z is the electrochemical equivalent. Since Q = It (current × time): m = ZIt.
Faraday's Second Law: When the same quantity of electricity passes through different electrolytes, the masses deposited are proportional to their equivalent weights: m₁/m₂ = E₁/E₂.
🔍 Step-by-Step: Deriving m = ZIt
1
The charge of one mole of electrons is the Faraday constant.F = e × N_A = 1.602×10⁻¹⁹ × 6.022×10²³ = 96485 C/mol. This is the charge of 1 mole of electrons.
2
For an ion with charge n, the charge required to deposit 1 mole is nF. For Cu²⁺: n = 2, so charge = 2F. For Ag⁺: n = 1, so charge = F. For Al³⁺: n = 3, so charge = 3F.
3
Mass deposited = (Molar mass × Charge passed) / (n × F).m = (M × Q) / (n × F) = (M × I × t) / (n × F). The electrochemical equivalent Z = M/(nF). For Cu (M = 63.5, n = 2): Z = 63.5/(2×96485) = 3.29×10⁻⁴ g/C.
m = M·I·t / (n·F) F = 96485 C/mol Z = M/(nF) For same Q: m₁/m₂ = E₁/E₂
🔍 Common Mistake
Common Mistake — n-Value
The n in Faraday's law is the number of electrons involved in the half-reaction, NOT the oxidation state alone. For Al³⁺ → Al, n = 3. For Cu²⁺ → Cu, n = 2. For O₂ evolution (2H₂O → O₂ + 4H⁺ + 4e⁻), n = 4. Also: make sure current is in amperes and time in seconds.
🎯 NEET Pattern
🎯 NEET Pattern — Faraday's Laws
• "How much Cu (M = 63.5, n = 2) is deposited by 2 A for 965 s?" → Q = 2×965 = 1930 C. m = (63.5×1930)/(2×96485) = 0.635 g.
• "Same current passed through CuSO₄ and AgNO₃. Ratio of masses?" → m_Cu/m_Ag = E_Cu/E_Ag = (63.5/2)/(108/1) = 31.75/108 = 0.294.
• "How much time for 1.08 g Ag (M = 108, n = 1) with 1 A?" → t = m·n·F/(M·I) = 1.08×1×96485/(108×1) = 965 s.
✍️ Worked Example
Example 1: Faraday's Law
A current of 5 A is passed through a CuSO₄ solution for 30 minutes. Calculate the mass of Cu deposited. (Cu = 63.5 g/mol, n = 2, F = 96485 C/mol)
Solution: Q = 5 × 30×60 = 9000 C. m = (63.5 × 9000)/(2 × 96485) = 571500/192970 = 2.96 g. Key insight: Always check that time is in seconds, not minutes.
2. Kohlrausch's Law
💡 Why this matters in real life
Kohlrausch's law allows us to determine the conductivity of weak electrolytes at infinite dilution — something we cannot measure directly because weak electrolytes don't fully dissociate even at high dilution. This is essential for calculating the degree of dissociation and the dissociation constant of weak acids and bases.
🧠 The Big Idea — In Plain Words
At infinite dilution (Λₘ°), each ion contributes a fixed amount to the total molar conductivity, regardless of the other ion it's paired with. Λₘ° = ν⁺λ⁺° + ν⁻λ⁻°, where ν is the number of ions and λ° is the ionic conductivity at infinite dilution. This law is used to find Λₘ° for weak electrolytes by adding the Λₘ° of strong electrolytes.
🔍 Step-by-Step: Applying Kohlrausch's Law
1
At infinite dilution, ions behave independently.Λₘ° = λ⁺° + λ⁻° for a binary electrolyte. For example, NaCl: Λₘ°(NaCl) = λ°(Na⁺) + λ°(Cl⁻). These ionic conductivities are tabulated constants.
2
For weak electrolytes like CH₃COOH, Λₘ° can be found using strong electrolytes.Λₘ°(CH₃COOH) = Λₘ°(HCl) + Λₘ°(CH₃COONa) − Λₘ°(NaCl). This works because: (H⁺+Cl⁻) + (Na⁺+CH₃COO⁻) − (Na⁺+Cl⁻) = H⁺ + CH₃COO⁻
3
Degree of dissociation α = Λₘ/Λₘ°. For a weak electrolyte at concentration C, α can be found. Then the dissociation constant: Kₐ = Cα²/(1−α). For very weak acids (α ≪ 1): Kₐ ≈ Cα².
Λₘ° = ν⁺λ⁺° + ν⁻λ⁻° α = Λₘ/Λₘ° Kₐ = Cα²/(1−α)
🔍 Common Mistake
Common Mistake — Units
Molar conductivity (Λₘ) has units of S·cm²/mol or S·m²/mol. Specific conductivity (κ) has units S/cm or S/m. The relationship: Λₘ = κ × 1000/C (if C is in mol/L and κ in S/cm). Always check which unit system the problem uses.
🎯 NEET Pattern
🎯 NEET Pattern — Kohlrausch's Law
• "Λₘ°(HCl) = 426 S·cm²/mol, Λₘ°(NaCl) = 126, Λₘ°(CH₃COONa) = 91. Find Λₘ°(CH₃COOH)." → 426 + 91 − 126 = 391 S·cm²/mol.
• "Λₘ for 0.01 M CH₃COOH is 16.2 S·cm²/mol. Find α if Λₘ° = 391." → α = 16.2/391 = 0.0414 (4.14%).
• "Which has highest Λₘ°: NaCl, MgCl₂, AlCl₃?" → AlCl₃ (more ions: 4 per formula unit).
✍️ Worked Example
Example 2: Kohlrausch's Law
Λₘ for 0.00 5 M CH₃COOH is 19.2 S·cm²/mol. Λₘ°(HCl)=426, Λₘ°(NaCl)=126, Λₘ°(CH₃COONa)=91. Find Kₐ for CH₃COOH.
Solution: Λₘ°(CH₃COOH) = 426 + 91 − 126 = 391. α = 19.2/391 = 0.0491. Kₐ = Cα²/(1−α) = 0.005×(0.0491)²/(0.951) = 1.27×10⁻⁵. Key insight: This matches the standard Kₐ of acetic acid (1.8×10⁻⁵ approximately).
3. Conductance & Cell Constant
💡 Why this matters in real life
Conductivity measurements are used to check water purity (pure water has very low conductivity), monitor industrial processes, test battery electrolytes, and even measure salinity in environmental monitoring. A simple conductivity meter can detect dissolved ions in water.
🧠 The Big Idea — In Plain Words
Resistance (R) of a solution follows R = ρ·l/A, where ρ is resistivity, l is distance between electrodes, and A is electrode area. Conductance G = 1/R. Specific conductivity (κ) = 1/ρ = G × (l/A) = G × cell constant. Molar conductivity Λₘ = κ/C. As concentration decreases, Λₘ increases because interionic attractions weaken.
🔍 Step-by-Step: Variation of Conductivity with Concentration
1
For strong electrolytes, Λₘ follows Kohlrausch's empirical law.Λₘ = Λₘ° − A√C. A plot of Λₘ vs √C is linear. The intercept gives Λₘ°, and the slope gives the constant A (related to ion-ion interactions).
2
For weak electrolytes, Λₘ increases sharply at very low concentrations. This is because the degree of dissociation increases rapidly as concentration decreases. The Λₘ vs √C plot is not linear — it curves upward at low concentrations.
3
The cell constant G* = l/A must be determined experimentally. This is done by measuring the resistance of a standard KCl solution of known conductivity. G* = κ(KCl) × R(KCl). Once known, the same cell constant is used for unknown solutions.
G = 1/R κ = G × G* Λₘ = κ/C Strong: Λₘ = Λₘ° − A√C
🔍 Common Mistake
Common Mistake — Specific vs Molar Conductivity
Specific conductivity (κ) decreases with dilution (fewer ions per unit volume). Molar conductivity (Λₘ) increases with dilution (each mole of electrolyte contributes more to conductivity). This distinction is frequently tested in NEET.
🎯 NEET Pattern
🎯 NEET Pattern — Conductance
• "Resistance of 0.1 M KCl in a cell is 100 Ω. κ(KCl) = 0.0129 S/cm. Find cell constant." → G* = κ × R = 0.0 129 × 100 = 1.29 cm⁻¹.
• "Λₘ of 0.01 M solution is 120 S·cm²/mol. κ = Λₘ × C/1000 = 120 × 0.01/1000 = 0.0 012 S/cm."
• "Which has higher Λₘ: 0.1 M NaCl or 0.001 M NaCl?" → 0.001 M (more dilute → less ion-pairing → higher Λₘ).
✍️ Worked Example
Example 3: Conductance
A conductivity cell has electrodes of area 2 cm² separated by 1 cm. The resistance of 0.01 M KCl solution is 200 Ω. Find κ and Λₘ. (κ(KCl) = 0.00141 S/cm)
Solution: Cell constant = l/A = 1/2 = 0.5 cm⁻¹. G = 1/200 = 0.005 S. κ = 0.005 × 0.5 = 0.0025 S/cm. Λₘ = κ×1000/C = 0.0025×1000/0.01 = 250 S·cm²/mol. Key insight: Factor of 1000 converts L to cm³.
4. Standard Electrode Potentials
💡 Why this matters in real life
The electrochemical series tells us which metals will corrode, which batteries produce higher voltage, and which reactions are feasible. It's why zinc protects iron from rusting (galvanization), why lithium batteries have high voltage, and why some metals displace others from solution.
🧠 The Big Idea — In Plain Words
The standard hydrogen electrode (SHE) is assigned E° = 0 V. All other electrode potentials are measured relative to SHE under standard conditions. A more positive E° means the species has a greater tendency to be reduced (gain electrons). A more negative E° means it tends to be oxidized (lose electrons). The electrochemical series ranks elements by E°: F₂ (+2.87 V) is the strongest oxidizing agent, Li⁺ (−3.04 V) is the strongest reducing agent.
🔍 Step-by-Step: Calculating Cell Potential
1
Cell potential (EMF) = E°(cathode) − E°(anode). The cathode is where reduction occurs (higher E°), and the anode is where oxidation occurs (lower E°). For the Daniel cell: Cu²⁺/Cu (E° = +0.34 V), Zn²⁺/Zn (E° = −0.76 V). E°_cell = 0.34 − (−0.76) = +1.10 V.
2
A positive E°_cell indicates a spontaneous reaction. ΔG° = −nFE°_cell. If E°_cell > 0, ΔG° < 0 → spontaneous. If E°_cell < 0, the reaction is non-spontaneous (requires external voltage).
3
The electrochemical series helps predict displacement reactions. A metal with more negative E° can displace a metal with more positive E° from its salt solution. Zn (E° = −0.76 V) displaces Cu²⁺ (E° = +0.34 V) because Zn is more easily oxidized.
Standard reduction potentials are always written as reduction reactions. When a half-cell undergoes oxidation, the sign is reversed. But the easiest method: keep all values as reduction potentials, identify cathode (higher E°) and anode (lower E°), and use E°_cell = E°_cathode − E°_anode. Don't flip signs manually.
🎯 NEET Pattern
🎯 NEET Pattern — Electrode Potentials
• "E° for Zn²⁺/Zn = −0.76 V, Cu²⁺/Cu = +0.34 V. Find E°_cell." → 0.34 − (−0.76) = 1.10 V.
• "Can Fe (E° = −0.44 V) displace Cu from CuSO₄?" → Yes, Fe is more negative → Fe is oxidized, Cu²⁺ is reduced.
• "Arrange in increasing reducing power: Li (−3.04), Mg (−2.36), Cu (+0.34)." → Cu < Mg < Li (more negative = stronger reducing agent).
✍️ Worked Example
Example 4: Standard Electrode Potentials
Calculate the standard EMF of the cell: Mg|Mg²⁺||Ag⁺|Ag. E°(Mg²⁺/Mg) = −2.36 V, E°(Ag⁺/Ag) = +0.80 V.
Solution: Anode (oxidation): Mg. Cathode (reduction): Ag. E°_cell = 0.80 − (−2.36) = +3.16 V. ΔG° = −nFE° = −2 × 96485 × 3.16 = −610 kJ/mol. Highly spontaneous. Key insight: This is why Mg is used as a sacrificial anode to protect steel.
5. Gibbs Free Energy & Cell Potential
💡 Why this matters in real life
The relationship between Gibbs free energy, cell potential, and equilibrium constant connects three fundamental concepts in chemistry. It allows us to calculate battery voltage from thermodynamic data, predict if a reaction is feasible, and determine equilibrium constants from electrochemical measurements.
🧠 The Big Idea — In Plain Words
The maximum electrical work from an electrochemical cell equals −nFE. This work equals the change in Gibbs free energy: ΔG = −nFE. Under standard conditions: ΔG° = −nFE°. Since ΔG° = −RT ln K, we get E° = (RT/nF) ln K. At 298 K: E° = (0.0591/n) log K. These equations form the bridge between thermodynamics and electrochemistry.
🔍 Step-by-Step: Deriving E° = (0.0591/n) log K
1
Start with three fundamental equations.ΔG° = −nFE° (from electrochemistry), ΔG° = −RT ln K (from thermodynamics). Equate: −nFE° = −RT ln K → E° = RT ln K / nF
2
Convert ln to log₁₀ and substitute constants.E° = (2.303RT/nF) × log K. At 298 K: R = 8.314 J/(mol·K), F = 96485 C/mol. 2.303 × 8.314 × 298 / 96485 = 0.0591. So: E° = (0.0591/n) × log K
3
Application to equilibrium constant calculation. For a cell with E° = 1.10 V (Daniel cell, n = 2): log K = nE°/0.0591 = 2 × 1.10/0.0591 = 37.2 → K = 1.6×10³⁷. This enormous K means the reaction goes virtually to completion.
ΔG° = −nFE° ΔG° = −RT ln K E° = (0.0591/n) log K (at 298 K)
🔍 Common Mistake
Common Mistake — Sign of ΔG°
A positive E° gives negative ΔG° (spontaneous). A negative E° gives positive ΔG° (non-spontaneous). Students sometimes confuse: E° > 0 means the reaction as written (reduction at cathode, oxidation at anode) is spontaneous. Also: E° is intensive, but ΔG° is extensive — doubling the reaction doubles ΔG° but doesn't change E°.
🎯 NEET Pattern
🎯 NEET Pattern — Gibbs & Cell
• "E° = 0.46 V for a 2-electron cell. Find K." → log K = 2×0.46/0.0591 = 15.56 → K = 3.6×10¹⁵.
• "For a cell, K = 10¹⁰ at 298 K. Find E° for n = 1." → E° = 0.0591 × 10/1 = 0.591 V.
• "ΔG° = −100 kJ/mol for a 2-electron reaction. Find E°." → E° = −ΔG°/(nF) = 100000/(2×96485) = 0.518 V.
✍️ Worked Example
Example 5: Gibbs & Cell
For the reaction Zn + Cu²⁺ → Zn²⁺ + Cu, E° = 1.10 V, n = 2. Find ΔG° and K at 298 K.
Solution: ΔG° = −2 × 96485 × 1.10 = −212267 J = −212.3 kJ/mol. log K = 2 × 1.10/0.0591 = 37.2. K = 10³⁷·² ≈ 1.6×10³⁷. Key insight: E° = 1.10 V corresponds to a very large equilibrium constant — the reaction is essentially complete.
6. Complex Rate Laws & Reaction Mechanisms
💡 Why this matters in real life
Most real-world reactions don't follow simple zero/first/second order kinetics. Understanding complex rate laws is essential for modeling atmospheric chemistry, biochemical pathways, and industrial processes. The concept of rate-determining step explains why some reactions are faster or slower than their stoichiometry suggests.
🧠 The Big Idea — In Plain Words
The rate-determining step (RDS) is the slowest step in a multistep reaction mechanism — it determines the overall reaction rate. Molecularity is the number of molecules that collide in an elementary step (unimolecular, bimolecular, termolecular). Pseudo-first-order reactions occur when one reactant is in large excess, making its concentration effectively constant. The Arrhenius equation k = Ae^(−Eₐ/RT) describes temperature dependence.
🔍 Step-by-Step: Rate-Determining Step
1
Consider a two-step mechanism. Step 1 (slow): A + B → C, Step 2 (fast): C → D + E. The overall rate is determined by Step 1: Rate = k[A][B]. The intermediate C is consumed as fast as it's formed.
2
If the first step is fast and reversible, followed by a slow step. Step 1 (fast): A ⇌ B + C. Step 2 (slow): B + D → E. The rate depends on [B], which is in equilibrium: [B] = K[A] (assuming C is constant). So: Rate = k₂[B][D] = k₂K[A][D] = k_obs[A][D]
3
Pseudo-first-order approximation. For reaction A + B → products: If [B]₀ ≫ [A]₀, [B] stays nearly constant. Rate = k[A][B] ≈ k'[A] where k' = k[B]₀. The reaction appears to be first order. This is common in hydrolysis reactions where water is the solvent.
Rate determined by slowest step Pseudo-1st-order: k' = k[B]₀ Molecularity: number of reacting species
🔍 Common Mistake
Common Mistake — Order vs Molecularity
Order is experimental (determined from data) and can be fractional, zero, or negative. Molecularity is theoretical (from mechanism) and must be a positive integer ≤ 3. For elementary reactions only, order equals molecularity. For complex reactions, order may differ from the stoichiometric coefficients.
🎯 NEET Pattern
🎯 NEET Pattern — Complex Kinetics
• "A 2-step mechanism with slow first step: overall rate = k[A][B]."
• "CH₃COOCH₃ + H₂O → CH₃COOH + CH₃OH in large excess water: pseudo-first-order."
• "Rate = k[A]²[B]⁰. The order is 2 (w.r.t. A) + 0 (w.r.t. B) = 2 overall. Fractional order possible."
✍️ Worked Example
Example 6: Rate-Determining Step
A proposed mechanism: (1) NO₂ + NO₂ → NO₃ + NO (slow), (2) NO₃ + CO → NO₂ + CO₂ (fast). Write the overall reaction and rate law.
Solution: Overall: NO₂ + CO → NO + CO₂ (NO₃ is intermediate, NO₂ is catalyst). Rate law: Rate = k[NO₂]² (from the slow step). Key insight: The rate law depends only on the slow step, not on the overall stoichiometry.
7. Arrhenius Equation — Temperature Dependence
💡 Why this matters in real life
The Arrhenius equation explains why reactions speed up when heated — the rule of thumb that a 10°C rise doubles the reaction rate. It's used to determine activation energy, predict reaction rates at different temperatures, and understand why some reactions are temperature-sensitive while others aren't.
🧠 The Big Idea — In Plain Words
The Arrhenius equation k = Ae^(−Eₐ/RT) has two factors: A (frequency factor — how often molecules collide with correct orientation) and e^(−Eₐ/RT) (Boltzmann factor — fraction of molecules with energy ≥ Eₐ). The linear form ln k = ln A − Eₐ/RT allows us to determine Eₐ from experimental data by plotting ln k vs 1/T.
🔍 Step-by-Step: Determining Activation Energy
1
The logarithmic form of the Arrhenius equation.ln k = ln A − Eₐ/(RT). Rearranged: ln k = (−Eₐ/R)(1/T) + ln A. This is y = mx + c: ln k vs 1/T gives a straight line with slope = −Eₐ/R.
2
Two-point form for calculating Eₐ.ln(k₂/k₁) = (Eₐ/R)(1/T₁ − 1/T₂). This is useful when you have rate constants at two temperatures. For example, if k doubles when T increases from 300 K to 310 K: Eₐ = R·ln(2)/(1/300−1/310).
3
The pre-exponential factor A represents collision frequency and steric effects. According to collision theory: k = P·Z·e^(−Eₐ/RT), where Z is the collision frequency and P is the steric factor (orientation requirement). For gas-phase reactions, Z ≈ 10¹⁰ L·mol⁻¹·s⁻¹. P ranges from 1 (favorable orientation) to 10⁻⁶ (stringent orientation).
ln(k₂/k₁) = (Eₐ/R)(1/T₁ − 1/T₂) | Slope of ln k vs 1/T = −Eₐ/R
🔍 Common Mistake
Common Mistake — Temperature in K
Always use Kelvin, never Celsius. A 10°C change = 10 K change, but the ratio 1/T₁ − 1/T₂ is not the same as 1/(T₁−T₂). For example, 300 K to 310 K: 1/300 − 1/310 = 0.00333 − 0.00323 = 1.08×10⁻⁴ K⁻¹. Don't use the difference directly.
🎯 NEET Pattern
🎯 NEET Pattern — Arrhenius Deep-Dive
• "Rate doubles when T rises from 300 K to 310 K. Find Eₐ." → Eₐ = R·ln2/(1/300−1/310) = 8.314×0.693/1.08×10⁻⁴ = 53.6 kJ/mol.
• "A plot of ln k vs 1/T has slope −5000 K. Find Eₐ." → Eₐ = 5000 × 8.314 = 41570 J/mol = 41.6 kJ/mol.
• "Two reactions have Eₐ = 20 kJ and 80 kJ. Which is more temperature sensitive?" → The one with higher Eₐ (80 kJ) is more temperature sensitive — a given T increase causes a larger rate increase.
✍️ Worked Example
Example 7: Arrhenius Equation
A reaction has k = 0.02 s⁻¹ at 300 K and k = 0.08 s⁻¹ at 320 K. Find Eₐ and the pre-exponential factor A.
Solution: ln(0.08/0.02) = ln(4) = 1.386. Eₐ = 8.314 × 1.386/(1/300−1/320) = 11.52/2.08×10⁻⁴ = 55385 J/mol ≈ 55.4 kJ/mol. For A: ln A = ln k + Eₐ/RT = ln(0.02) + 55385/(8.314×300) = −3.912 + 22.2 = 18.29. A = 8.8×10⁷ s⁻¹. Key insight: A high A means many collisions per second.
8. Catalysis — Lowering Activation Energy
💡 Why this matters in real life
Catalysts are essential in modern life — catalytic converters in cars reduce pollution, enzymes in your body digest food, the Haber process makes fertilizer using iron catalyst, and zeolites crack petroleum into gasoline. A catalyst can increase reaction rates by factors of millions or more, making otherwise impractical processes commercially viable.
🧠 The Big Idea — In Plain Words
A catalyst provides an alternative reaction pathway with a lower activation energy. It participates in the reaction but is regenerated unchanged. The catalyst does NOT affect ΔG, ΔH, or the equilibrium constant — it only speeds up the rate at which equilibrium is reached. Homogeneous catalysts are in the same phase as reactants (e.g., acid catalysts). Heterogeneous catalysts are in a different phase (e.g., solid catalysts for gas reactions).
🔍 Step-by-Step: How Catalysts Work
1
A catalyst lowers Eₐ without changing ΔH. The Arrhenius equation: k = Ae^(−Eₐ/RT). If Eₐ decreases, k increases exponentially. Lowering Eₐ by just 10 kJ/mol can increase the rate by a factor of 50 at 300 K (e^(10000/8.314×300) = e^4 = 55).
2
Catalyst is not consumed. In the catalyzed mechanism: Step 1: A + Cat → A−Cat. Step 2: A−Cat + B → AB + Cat. The catalyst is regenerated in Step 2. Net: A + B → AB. The catalyst does not appear in the overall balanced equation.
3
Enzyme catalysis — the lock-and-key model. Enzymes are highly specific biological catalysts. The substrate binds to the enzyme's active site (lock-and-key or induced fit). The Michaelis-Menten equation describes enzyme kinetics: Rate = V_max[S]/(K_m + [S]). At low [S], rate ∝ [S]. At high [S], rate = V_max (saturation).
Catalyst: lowers Eₐ, unchanged at end, no effect on ΔG or K_eq Enzyme kinetics: Rate = V_max[S]/(K_m + [S])
🔍 Common Mistake
Common Mistake — Catalyst Effect on Equilibrium
A catalyst speeds up both forward and reverse reactions equally. It does NOT shift the equilibrium position. The equilibrium constant K is unchanged. The catalyst only helps equilibrium be reached faster. Also: catalysts are specific — a catalyst for one reaction may not work for another.
🎯 NEET Pattern
🎯 NEET Pattern — Catalysis
• "Which is true: catalyst changes ΔG, ΔH, or Eₐ?" → Only Eₐ (activation energy) is changed. ΔG and ΔH remain the same.
• "Example of homogeneous catalysis?" → Acid-catalyzed hydrolysis of esters. Catalyst (H⁺) is in the same aqueous phase.
• "Example of heterogeneous catalysis?" → Haber process: Fe(s) catalyzes N₂(g) + 3H₂(g) ⇌ 2NH₃(g). Different phases.
✍️ Worked Example
Example 8: Catalysis
An uncatalyzed reaction has Eₐ = 120 kJ/mol. A catalyst lowers Eₐ to 80 kJ/mol. How much faster is the catalyzed reaction at 300 K?
Solution: k_cat/k_uncat = e^{(Eₐ_uncat − Eₐ_cat)/RT} = e^{(40000)/(8.314×300)} = e^{16.04} = 9.2×10⁶. The reaction is about 9.2 million times faster. Key insight: A 40 kJ/mol reduction in Eₐ causes an enormous rate increase due to the exponential relationship.
End of Session 4 — Electrochemistry & Kinetics Deep-Dive. All four chemistry derivation sessions are now complete.