Arrhenius Equation and Activation Energy

Complete guide to Arrhenius equation, activation energy, temperature dependence of rate constant, and energy profiles for JEE

12 min read

Arrhenius Equation and Activation Energy

Real-Life Connection: Why Refrigeration Preserves Food

Why does milk last for days in the refrigerator but spoils in hours at room temperature? Why do chemical hand warmers activate when you break an inner pouch? Why do fireflies glow only when warm?

The answer: Activation energy and the Arrhenius equation!

Food preservation:

  • At 25°C: Bacterial growth rapid (k large)
  • At 4°C: Bacterial growth slow (k small)
  • Temperature decrease → Exponential decrease in rate

Chemical hand warmers:

  • Breaking pouch exposes iron to air
  • Fe + O₂ reaction has high activation energy
  • Needs initial heat to overcome Ea barrier
  • Once started, releases more heat (exothermic)

Biological processes:

  • Enzymes optimized for 37°C (body temperature)
  • Fever (↑T) → faster metabolism
  • Hypothermia (↓T) → slower metabolism

The Arrhenius equation quantifies how temperature affects reaction rates through activation energy!

What is Activation Energy?

Definition

Activation energy (Ea) is the minimum energy that colliding molecules must possess for a reaction to occur.

$$\boxed{E_a = \text{Energy barrier that must be overcome}}$$

Energy Profile Diagram

Energy
  |
  |      ╱╲  ← Transition State
  |     ╱  ╲
  |    ╱    ╲
Ea→   |      ╲
  |   |       ╲_____ Products
  |___Reactants
  |
  |_________________
     Reaction Progress

Key points:

  • Reactants must climb energy barrier
  • Transition state (activated complex) at the top
  • Ea = height of barrier
  • Products may be higher or lower energy (ΔH)

Why Activation Energy Exists

Not all collisions lead to reaction!

For reaction to occur, molecules need:

  1. Sufficient energy (≥ Ea)
  2. Proper orientation
  3. Collision to occur

Ea represents:

  • Energy to break old bonds (partially)
  • Energy to rearrange atoms
  • Energy to form transition state

Relationship to Rate

$$\boxed{\text{Higher } E_a \rightarrow \text{Slower reaction}}$$ $$\boxed{\text{Lower } E_a \rightarrow \text{Faster reaction}}$$

Examples:

  • Combustion reactions: Low Ea → Fast (can explode)
  • Rusting of iron: High Ea → Slow (years)
  • Diamond → Graphite: Very high Ea → No observable change (thermodynamically favorable but kinetically impossible!)

The Arrhenius Equation

Basic Form

$$\boxed{k = Ae^{-E_a/RT}}$$

where:

  • k = rate constant
  • A = pre-exponential factor (frequency factor)
  • Ea = activation energy (J/mol or kJ/mol)
  • R = gas constant = 8.314 J mol⁻¹ K⁻¹
  • T = absolute temperature (Kelvin)

Physical Meaning

A (pre-exponential factor):

  • Represents collision frequency and orientation factor
  • How often molecules collide with proper orientation
  • Units: same as k (depends on order)
  • Typically 10⁸ to 10¹⁴ for most reactions

e^(-Ea/RT) (Boltzmann factor):

  • Fraction of molecules with energy ≥ Ea
  • Temperature-dependent term
  • At higher T, more molecules have E ≥ Ea

Logarithmic Form

Taking natural logarithm of both sides:

$$\boxed{\ln k = \ln A - \frac{E_a}{RT}}$$

This is equation of straight line: y = mx + c

  • y = ln k
  • x = 1/T
  • m = -Ea/R (slope)
  • c = ln A (intercept)

Common logarithm form:

$$\boxed{\log k = \log A - \frac{E_a}{2.303RT}}$$

Two-Temperature Form

Most useful for JEE problems!

For two different temperatures T₁ and T₂:

$$\ln k_1 = \ln A - \frac{E_a}{RT_1}$$ $$\ln k_2 = \ln A - \frac{E_a}{RT_2}$$

Subtracting:

$$\boxed{\ln\frac{k_2}{k_1} = \frac{E_a}{R}\left(\frac{1}{T_1} - \frac{1}{T_2}\right)}$$

or

$$\boxed{\log\frac{k_2}{k_1} = \frac{E_a}{2.303R}\left(\frac{1}{T_1} - \frac{1}{T_2}\right)}$$

Interactive Demo: Explore Activation Energy and Temperature Effects

See how temperature affects reaction rates through the Arrhenius equation. Adjust activation energy and temperature to observe changes in rate constant.

This is the most frequently used form in JEE!

Arrhenius Plot

Graph: ln k vs 1/T

Plotting ln k against 1/T gives a straight line:

ln k
  |
  |        ╲
  |         ╲
  |          ╲
  |           ╲
  |            ╲
  |_____________╲____
       1/T

From the plot:

  • Slope = -Ea/R
  • Ea = -slope × R
  • Intercept = ln A

Determining Ea Graphically

Method:

  1. Perform reaction at different temperatures
  2. Measure k at each temperature
  3. Plot ln k vs 1/T
  4. Find slope (should be negative)
  5. Calculate: Ea = -slope × R

Example: If slope = -10,000 K

$$E_a = -(-10000) \times 8.314 = 83,140 \text{ J/mol} = 83.14 \text{ kJ/mol}$$

Effect of Temperature on Rate

Temperature Coefficient (Q₁₀)

Rule of thumb: Rate approximately doubles for every 10°C rise

$$\boxed{Q_{10} = \frac{k_{T+10}}{k_T} \approx 2 \text{ to } 3}$$

Deriving Temperature Effect

From Arrhenius equation:

$$\ln\frac{k_2}{k_1} = \frac{E_a}{R}\left(\frac{T_2 - T_1}{T_1T_2}\right)$$

For small temperature change:

$$\frac{k_2}{k_1} \approx e^{E_a\Delta T/(RT^2)}$$

Key insight:

  • High Ea → strong temperature dependence
  • Low Ea → weak temperature dependence

Why Small T Change Causes Large k Change?

Exponential relationship!

e^(-Ea/RT) is extremely sensitive to T

Example: For Ea = 50 kJ/mol

  • At 300 K: e^(-50000/(8.314×300)) = e^(-20.05) = 1.94 × 10⁻⁹
  • At 310 K: e^(-50000/(8.314×310)) = e^(-19.41) = 3.63 × 10⁻⁹

87% increase in k for just 10 K rise!

Collision Theory

Why Arrhenius Equation Works

Collision theory provides theoretical basis:

$$\text{Rate} = Z \times f \times p$$

where:

  • Z = collision frequency
  • f = fraction with E ≥ Ea = e^(-Ea/RT)
  • p = steric factor (proper orientation)
$$k = Z \times p \times e^{-E_a/RT}$$

A = Z × p (pre-exponential factor)

Maxwell-Boltzmann Distribution

At any temperature, molecules have range of kinetic energies:

Number
  |     T₁
  | ╱╲
  |╱  ╲___  T₂ (higher)
  |     ╲  ╱╲
  |      ╲╱  ╲___
  |           |
  |___________|_______ Energy
              Ea

Key points:

  • At higher T, distribution shifts right and flattens
  • More molecules have E ≥ Ea at higher T
  • Area under curve beyond Ea increases exponentially with T

Catalyst and Activation Energy

How Catalysts Work

Catalyst provides alternative pathway with lower Ea

Without Catalyst:        With Catalyst:
Energy                   Energy
  |                        |
  |   ╱╲                   |  ╱╲
  |  ╱  ╲                  | ╱  ╲╱╲
  | ╱    ╲                 |╱      ╲
  |╱      ╲___             |        ╲___
  |_____________           |_____________
     Progress                 Progress

  Ea(uncat) = 100 kJ/mol   Ea(cat) = 50 kJ/mol

Effect on rate:

$$\frac{k_{cat}}{k_{uncat}} = e^{-(E_{a,cat} - E_{a,uncat})/RT}$$

Example: At 300 K, reducing Ea from 100 to 50 kJ/mol:

$$\frac{k_{cat}}{k_{uncat}} = e^{50000/(8.314 \times 300)} = e^{20.05} = 5.1 \times 10^8$$

Rate increases by ~500 million times!

Important Points

  1. Catalyst lowers Ea for both forward and reverse reactions
  2. No effect on ΔH (thermodynamics unchanged)
  3. No effect on equilibrium constant K
  4. Only affects rate, not equilibrium position
  5. Catalyst is not consumed (regenerated)

Energy Profiles: Detailed Analysis

Exothermic Reaction

Energy
  |
  |      ╱╲  ← Transition State
  |     ╱  ╲
  |    ╱    ╲
  |___╱ Ea,f ╲
  |           ╲_____ Products
  |                Ea,r
  |
  |↑ΔH<0 (negative)
  |_________________
     Reaction Progress

Characteristics:

  • Products lower energy than reactants
  • ΔH < 0 (exothermic)
  • Ea(forward) < Ea(reverse)
  • Releases energy

Relationship:

$$\boxed{\Delta H = E_{a,f} - E_{a,r}}$$

For exothermic: Ea,f < Ea,r

Endothermic Reaction

Energy
  |
  |      ╱╲  ← Transition State
  |     ╱  ╲
  |    ╱    ╲
  |   ╱ Ea,f ╲_____ Products
  |__╱            ↑
  |               |Ea,r
  |               |
  |↑ΔH>0 (positive)
  |_________________
     Reaction Progress

Characteristics:

  • Products higher energy than reactants
  • ΔH > 0 (endothermic)
  • Ea(forward) > Ea(reverse)
  • Requires energy input

Relationship:

$$\boxed{\Delta H = E_{a,f} - E_{a,r}}$$

For endothermic: Ea,f > Ea,r

Key Relationships

$$E_{a,forward} - E_{a,reverse} = \Delta H$$ $$E_{a,forward} = \Delta H + E_{a,reverse}$$

(endothermic)

$$E_{a,reverse} = E_{a,forward} - \Delta H$$

(general)

Memory Tricks

“ARRHENIUS” Mnemonic

Activation energy in exponent (negative) Rate constant k on left RT in denominator Higher T means higher k Exponential relationship Needs absolute temperature (Kelvin) Intercept gives ln A Useful for temperature dependence Slope gives -Ea/R

“CAT” for Catalyst

Cuts activation energy (lowers Ea) Accelerates both directions (forward and reverse) Thermodynamics unchanged (ΔH and K same)

“HEAT” for Temperature Effect

Higher T increases k Exponential increase Activation energy determines sensitivity Ten degrees doubles rate (approximately)

Common JEE Mistakes

Mistake 1: Temperature Units

Wrong: Using temperature in °C in Arrhenius equation

Correct: Always use Kelvin!

  • T(K) = T(°C) + 273.15
  • Never use Celsius in exponential term

Mistake 2: Ea Units

Wrong: Mixing kJ/mol with R = 8.314 J mol⁻¹ K⁻¹

Correct: Units must match!

  • If Ea in kJ/mol, use R = 8.314 × 10⁻³ kJ mol⁻¹ K⁻¹
  • OR convert Ea to J/mol

Mistake 3: Sign of Slope

Wrong: “Slope of ln k vs 1/T is positive”

Correct: Slope is -Ea/R (always negative!)

  • Ea is always positive
  • Therefore slope is negative

Mistake 4: Catalyst and Equilibrium

Wrong: “Catalyst increases product yield”

Correct: Catalyst has no effect on equilibrium

  • Only increases rate to reach equilibrium
  • K unchanged

Mistake 5: ΔH vs Ea Confusion

Wrong: “ΔH = Ea”

Correct: ΔH = Ea(forward) - Ea(reverse)

  • These are different quantities!
  • Ea always positive, ΔH can be positive or negative

Practice Problems

Level 1: JEE Main Foundation

Problem 1: A reaction has Ea = 50 kJ/mol. At 300 K, k = 10⁻⁵ s⁻¹. Calculate k at 310 K.

Solution:

$$\ln\frac{k_2}{k_1} = \frac{E_a}{R}\left(\frac{1}{T_1} - \frac{1}{T_2}\right)$$ $$\ln\frac{k_2}{10^{-5}} = \frac{50000}{8.314}\left(\frac{1}{300} - \frac{1}{310}\right)$$ $$\ln\frac{k_2}{10^{-5}} = 6014.4 \times (0.003333 - 0.003226)$$ $$\ln\frac{k_2}{10^{-5}} = 6014.4 \times 0.000107 = 0.644$$ $$\frac{k_2}{10^{-5}} = e^{0.644} = 1.90$$ $$k_2 = 1.90 \times 10^{-5} \text{ s}^{-1}$$

90% increase for 10 K rise!

Level 2: JEE Main/Advanced

Problem 2: For a reaction, rate constant increases by factor of 4 when temperature increases from 300 K to 320 K. Calculate activation energy.

Solution:

$$\ln\frac{k_2}{k_1} = \frac{E_a}{R}\left(\frac{1}{T_1} - \frac{1}{T_2}\right)$$

Given: k₂/k₁ = 4

$$\ln 4 = \frac{E_a}{8.314}\left(\frac{1}{300} - \frac{1}{320}\right)$$ $$1.386 = \frac{E_a}{8.314} \times (0.003333 - 0.003125)$$ $$1.386 = \frac{E_a}{8.314} \times 0.000208$$ $$E_a = \frac{1.386 \times 8.314}{0.000208} = 55,360 \text{ J/mol}$$ $$E_a = 55.4 \text{ kJ/mol}$$

Problem 3: The following data is obtained:

T (K)k (s⁻¹)1/T (K⁻¹)ln k
3001.0×10⁻⁵0.00333-11.51
3204.5×10⁻⁵0.00313-10.01
3401.8×10⁻⁴0.00294-8.62

Calculate: (a) Ea from graph (b) Pre-exponential factor A

Solution:

(a) Graphical method:

Slope = Δ(ln k) / Δ(1/T)

Using first and last points:

$$\text{Slope} = \frac{-8.62 - (-11.51)}{0.00294 - 0.00333} = \frac{2.89}{-0.00039} = -7410 \text{ K}$$ $$E_a = -\text{slope} \times R = 7410 \times 8.314 = 61,607 \text{ J/mol} = 61.6 \text{ kJ/mol}$$

Algebraic method:

Using two-temperature formula with T₁ = 300 K, T₂ = 340 K:

$$\ln\frac{1.8 \times 10^{-4}}{1.0 \times 10^{-5}} = \frac{E_a}{8.314}\left(\frac{1}{300} - \frac{1}{340}\right)$$ $$\ln 18 = \frac{E_a}{8.314} \times 0.000392$$ $$2.89 = \frac{E_a}{8.314} \times 0.000392$$ $$E_a = 61,300 \text{ J/mol} = 61.3 \text{ kJ/mol}$$

(b) From any data point, using k = Ae^(-Ea/RT):

At 300 K:

$$1.0 \times 10^{-5} = A \times e^{-61300/(8.314 \times 300)}$$ $$1.0 \times 10^{-5} = A \times e^{-24.58}$$ $$1.0 \times 10^{-5} = A \times 2.09 \times 10^{-11}$$ $$A = \frac{1.0 \times 10^{-5}}{2.09 \times 10^{-11}} = 4.78 \times 10^5 \text{ s}^{-1}$$

Level 3: JEE Advanced

Problem 4: For a reaction, the rate doubles when temperature increases from 27°C to 37°C. Calculate: (a) Activation energy (b) Temperature at which rate will be four times the rate at 27°C

Solution:

(a) T₁ = 27°C = 300 K, T₂ = 37°C = 310 K

k₂ = 2k₁

$$\ln 2 = \frac{E_a}{8.314}\left(\frac{1}{300} - \frac{1}{310}\right)$$ $$0.693 = \frac{E_a}{8.314} \times 0.000107$$ $$E_a = \frac{0.693 \times 8.314}{0.000107} = 53,800 \text{ J/mol} = 53.8 \text{ kJ/mol}$$

(b) For k₃ = 4k₁:

$$\ln 4 = \frac{53800}{8.314}\left(\frac{1}{300} - \frac{1}{T_3}\right)$$ $$1.386 = 6469.8 \times \left(0.003333 - \frac{1}{T_3}\right)$$ $$2.14 \times 10^{-4} = 0.003333 - \frac{1}{T_3}$$ $$\frac{1}{T_3} = 0.003333 - 0.000214 = 0.003119$$ $$T_3 = 320.5 \text{ K} = 47.5°\text{C}$$

Problem 5: A reaction has ΔH = -80 kJ/mol and Ea(forward) = 50 kJ/mol. Calculate: (a) Ea(reverse) (b) Energy profile diagram values

Solution:

(a)

$$\Delta H = E_{a,f} - E_{a,r}$$ $$-80 = 50 - E_{a,r}$$ $$E_{a,r} = 50 - (-80) = 130 \text{ kJ/mol}$$

(b) Energy profile:

  • Reactants at 0 kJ (reference)
  • Transition state at +50 kJ (height of forward barrier)
  • Products at -80 kJ (below reactants)
  • Reverse barrier from products = 130 kJ

Verification: 50 - 130 = -80 ✓

Problem 6:** In presence of catalyst, activation energy reduces from 100 kJ/mol to 60 kJ/mol. By what factor does rate increase at 300 K?

Solution:

$$\frac{k_{cat}}{k_{uncat}} = \frac{Ae^{-E_{a,cat}/RT}}{Ae^{-E_{a,uncat}/RT}}$$ $$= e^{-(E_{a,cat} - E_{a,uncat})/RT}$$ $$= e^{-(60000 - 100000)/(8.314 \times 300)}$$ $$= e^{40000/2494.2} = e^{16.04}$$ $$= 9.2 \times 10^6$$

Rate increases by factor of ~10⁷!

This shows the dramatic effect of catalysts!

Relationship to Equilibrium

Connection to Equilibrium Constant

At equilibrium:

$$K = \frac{k_f}{k_r}$$

From Arrhenius equation:

$$K = \frac{A_f e^{-E_{a,f}/RT}}{A_r e^{-E_{a,r}/RT}}$$ $$K = \frac{A_f}{A_r} \times e^{-(E_{a,f} - E_{a,r})/RT}$$

Since Ea,f - Ea,r = ΔH:

$$K = \frac{A_f}{A_r} \times e^{-\Delta H/RT}$$

This connects kinetics to thermodynamics!

van’t Hoff Equation

Taking ln:

$$\ln K = \ln\frac{A_f}{A_r} - \frac{\Delta H}{RT}$$

This is the van’t Hoff equation for temperature dependence of K!

See: Chemical Equilibrium

Connection to Other Topics

Within Chemical Kinetics

Thermodynamics Connections

Organic Chemistry Applications

Physics Connections

JEE Previous Year Questions

JEE Main 2021: Rate constant of a reaction at 200 K is 10⁴ times the rate constant at 300 K. What is the activation energy?

Solution:

$$\ln\frac{k_1}{k_2} = \frac{E_a}{R}\left(\frac{1}{T_2} - \frac{1}{T_1}\right)$$ $$\ln\frac{10^4 k}{k} = \frac{E_a}{8.314}\left(\frac{1}{300} - \frac{1}{200}\right)$$ $$\ln 10^4 = \frac{E_a}{8.314} \times (-0.00167)$$ $$9.21 = -\frac{E_a}{8.314} \times 0.00167$$

This gives negative Ea, which is impossible!

Correction: The question likely meant k₂₀₀ = k₃₀₀/10⁴ (rate decreases at lower T)

$$\ln\frac{k_{300}}{k_{200}} = \ln 10^4 = 9.21$$ $$9.21 = \frac{E_a}{8.314} \times 0.00167$$ $$E_a = 45,900 \text{ J/mol} = 45.9 \text{ kJ/mol}$$

JEE Advanced 2020: For an exothermic reaction with Ea = 60 kJ/mol and ΔH = -40 kJ/mol, what is Ea for reverse reaction?

Solution:

$$\Delta H = E_{a,f} - E_{a,r}$$ $$-40 = 60 - E_{a,r}$$ $$E_{a,r} = 100 \text{ kJ/mol}$$

Quick Revision Points

  1. Arrhenius equation: k = Ae^(-Ea/RT)
  2. Ea = activation energy (always positive)
  3. Higher T → higher k (exponential)
  4. ln k vs 1/T gives straight line (slope = -Ea/R)
  5. Two-temperature form most useful for JEE
  6. Always use Kelvin for temperature
  7. Catalyst lowers Ea, increases k dramatically
  8. ΔH = Ea(forward) - Ea(reverse)
  9. Higher Ea → slower reaction, stronger T-dependence
  10. Q₁₀ ≈ 2-3 (rate doubles per 10°C rise)

Summary

The Arrhenius equation is fundamental to understanding:

  • Temperature dependence of reaction rates
  • Activation energy concept
  • Catalyst mechanisms
  • Connection between kinetics and thermodynamics

Key takeaways:

  • Exponential relationship between k and T
  • Small temperature changes cause large rate changes
  • Catalysts work by lowering Ea
  • Graphical methods for determining Ea
  • Energy profiles visualize reaction pathways

Mastering the Arrhenius equation is essential for:

  • Solving JEE kinetics problems
  • Understanding industrial process optimization
  • Predicting reaction behavior at different temperatures
  • Connecting kinetics to thermodynamics

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