Arrhenius Equation, Activation Energy and Catalysis

• Apply the Arrhenius equation k = A·e^(−Ea/RT) to find k at any temperature
• Calculate activation energy Ea from two rate constants at different temperatures
• Interpret the graph of ln k vs 1/T and extract Ea from slope
• Explain how catalysts increase reaction rate without shifting equilibrium
• Distinguish homogeneous, heterogeneous, and enzyme catalysis

Level 1 — Foundations

Arrhenius Equation
$k = A \cdot e^{-E_a/RT}$

• A = frequency factor (pre-exponential factor) — collision frequency × steric factor
• Ea = activation energy in J/mol (energy barrier molecules must overcome)
• R = 8.314 J/mol·K (gas constant)
• T = absolute temperature in Kelvin
• Higher T → exponent less negative → k increases exponentially

Physical meaning: Only those collisions with energy ≥ Ea and correct orientation lead to products. Fraction of molecules with E ≥ Ea = e^(−Ea/RT) (Maxwell-Boltzmann).

Catalyst: A substance that provides an alternative reaction pathway with lower Ea, increasing k without being consumed. Does NOT change ΔH, ΔG, or equilibrium constant K.

Level 2 — JEE Depth

Logarithmic Forms
$\ln k = \ln A - \frac{E_a}{RT}$ $\log k = \log A - \frac{E_a}{2.303RT}$

Graph: ln k vs 1/T → straight line, slope = −Ea/R, y-intercept = ln A

Two-Temperature Form (Most Used in JEE)
$\log\frac{k_2}{k_1} = \frac{E_a}{2.303R}\left(\frac{1}{T_1} - \frac{1}{T_2}\right)$

Sign convention: if T₂ > T₁, then (1/T₁ − 1/T₂) > 0, so k₂ > k₁ ✓

Rule of Thumb (Van't Hoff)
A 10°C rise approximately doubles the rate for Ea ≈ 50 kJ/mol.
More precisely: k at (T+10)/k at T ≈ 2 for typical organic reactions at room temperature.

Effect of Catalyst on Energy Profile

• Catalyst lowers Ea for both forward and reverse reactions equally
• Rate of forward reaction increases: k_f increases
• Rate of reverse reaction increases by same factor: k_r increases
• K = k_f/k_r remains unchanged → equilibrium position unchanged
• This is a critical JEE conceptual point: catalyst speeds approach to equilibrium but does NOT shift it

Types of Catalysis

Enzyme Catalysis — Lock and Key Model

• Active site of enzyme is complementary in shape to specific substrate
• Enzyme-substrate (ES) complex forms → lowers Ea dramatically
• Highly specific: one enzyme for one substrate type
• Affected by pH, temperature, inhibitors (unlike inorganic catalysts)

Negative Catalysts / Inhibitors
Substances that decrease reaction rate by blocking active sites or reacting with catalyst — not technically "negative catalysts" in IUPAC terms but tested in JEE MCQs.

JEE Traps

• Ea must be in J/mol (not kJ/mol) when using R = 8.314 J/mol·K — always convert
• In two-temperature formula, keep T in Kelvin, never Celsius
• Catalyst changes k but NOT equilibrium K — a classic MCQ trap: "catalyst increases equilibrium yield" is FALSE

Example 1: Finding Activation Energy from Two Rate Constants

Given: k₁ = 1.5×10⁻³ s⁻¹ at T₁ = 300 K, k₂ = 4.5×10⁻³ s⁻¹ at T₂ = 320 K
Find: Ea

Step 1: Apply two-temperature Arrhenius formula
  log(k₂/k₁) = (Ea/2.303R)(1/T₁ − 1/T₂)

Step 2: Calculate log(k₂/k₁)
  k₂/k₁ = 4.5×10⁻³ / 1.5×10⁻³ = 3
  log(3) = 0.4771

Step 3: Calculate (1/T₁ − 1/T₂)
  1/300 − 1/320 = (320 − 300)/(300 × 320) = 20/96000 = 2.083×10⁻⁴ K⁻¹

Step 4: Solve for Ea
  0.4771 = Ea / (2.303 × 8.314) × 2.083×10⁻⁴
  0.4771 = Ea × (2.083×10⁻⁴ / 19.147)
  0.4771 = Ea × 1.088×10⁻⁵
  Ea = 0.4771 / (1.088×10⁻⁵)
  Ea = 43,850 J/mol ≈ 43.85 kJ/mol

Answer: Ea ≈ 43.85 kJ/mol

Example 2: Calculating k from Arrhenius Equation

Given: Ea = 75,000 J/mol, A = 1×10¹³ s⁻¹, T = 500 K, R = 8.314 J/mol·K
Find: k

Step 1: Calculate Ea/RT
  Ea/RT = 75000 / (8.314 × 500) = 75000 / 4157 = 18.04

Step 2: Calculate e^(−Ea/RT)
  e^(−18.04) = ?
  Use: ln(e^(−18.04)) = −18.04
  Convert to log₁₀: log₁₀ = −18.04/2.303 = −7.832
  So e^(−18.04) = 10^(−7.832) = 1.472×10⁻⁸

Step 3: k = A × e^(−Ea/RT)
  k = 1×10¹³ × 1.472×10⁻⁸
  k = 1.472×10⁵ s⁻¹

Answer: k ≈ 1.47×10⁵ s⁻¹

Note: In JEE, you'll usually be given log tables or asked to express log k.
  log k = log A − Ea/(2.303RT) = 13 − 18.04/2.303 = 13 − 7.832 = 5.168
  k = 10^5.168 = 1.47×10⁵ s⁻¹ ✓

• Q1: Ea = 100 kJ/mol, A = 10¹² s⁻¹. Write the expression for log k at temperature T.
• Q2: Rate constant doubles when temperature rises from 300 K to 310 K. Find Ea.
• Q3: A catalyst reduces Ea from 80 kJ/mol to 50 kJ/mol. Does the equilibrium constant change?
• Q4: A graph of log k vs 1/T has slope −5000 K. Find Ea.
• Stretch: Q5: Two reactions have the same A but Ea₁ = 40 kJ/mol and Ea₂ = 80 kJ/mol. At 300 K, what is the ratio k₁/k₂? What does this reveal about the sensitivity of high-Ea reactions to temperature?

NCERT Chapter 3 link: Section 3.4 covers the Arrhenius equation, effect of temperature on rate, activation energy, and catalysis. The energy profile diagram (potential energy vs reaction coordinate) is explicitly in NCERT — understand the shape for both catalysed and uncatalysed reactions.

Exam connections: JEE Mains: one or two numerical on two-temperature Arrhenius formula to find Ea or k. JEE Advanced: energy profile diagrams, multi-step mechanism identifying rate-determining step and its Ea, enzyme catalysis and inhibition. "Effect of catalyst on K" is a repeated conceptual MCQ.

Study strategy: Memorise the two-temperature formula as your primary tool — it avoids needing A. Practise the log calculation path (not the exponential path) since JEE provides log tables. Sketch the energy profile diagram (reactants → transition state → products) for both catalysed and uncatalysed paths until it is automatic.

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