Activation Energy Calculator
Calculate activation energy using the Arrhenius equation from rate constants at two different temperatures. Activation energy (Ea) represents the minimum energy required for a chemical reaction to occur. Essential for chemical kinetics, reaction rates, and understanding temperature dependence of reactions.
Calculate Activation Energy
Temperature 1 (T₁)
Temperature 2 (T₂)
Load Example Problem:
Activation Energy Formulas
Arrhenius Equation
1. Standard Arrhenius Equation:
\( k = A e^{-E_a/RT} \)
Where \(k\) is rate constant, \(A\) is pre-exponential factor (frequency factor), \(E_a\) is activation energy, \(R\) is gas constant (8.314 J/mol·K), and \(T\) is absolute temperature in Kelvin. Shows exponential temperature dependence of reaction rates.
2. Linearized Form (Natural Logarithm):
\( \ln k = \ln A - \frac{E_a}{RT} \)
Linear form: \(y = mx + b\) where \(y = \ln k\), \(x = 1/T\), slope \(m = -E_a/R\), intercept \(b = \ln A\). Plot ln k vs 1/T gives straight line with slope -Ea/R. Useful for graphical determination of activation energy.
3. Two-Temperature Form (Most Practical):
\( \ln\left(\frac{k_2}{k_1}\right) = \frac{E_a}{R}\left(\frac{1}{T_1} - \frac{1}{T_2}\right) \)
Calculates Ea from two rate constants at different temperatures. Rearrange: \(E_a = \frac{R \ln(k_2/k_1)}{(1/T_1 - 1/T_2)}\). If rate doubles (k₂ = 2k₁), ln(2) = 0.693. Most common form for Ea calculations in practice.
4. Solving for Activation Energy:
\( E_a = \frac{R \ln(k_2/k_1)}{(1/T_1 - 1/T_2)} = \frac{-R \ln(k_2/k_1)}{(1/T_2 - 1/T_1)} \)
Both forms equivalent. Use \(R = 8.314\) J/mol·K for Ea in J/mol, or \(R = 1.987\) cal/mol·K for cal/mol. Convert: 1 cal = 4.184 J. Temperatures must be in Kelvin. \(T(K) = T(°C) + 273.15\).
Activation Energy Practice Problems
Problem 1: Rate Constant Doubles with Temperature
Question: Calculate activation energy for a reaction whose rate constant doubles when temperature increases from 18°C to 28°C.
Solution:
Given: T₁ = 18°C = 291 K, T₂ = 28°C = 301 K, k₂/k₁ = 2
Formula: \(E_a = \frac{R \ln(k_2/k_1)}{(1/T_1 - 1/T_2)}\)
Calculate: \(1/T_1 - 1/T_2 = 1/291 - 1/301 = 0.003436 - 0.003322 = 0.000114\) K⁻¹
\(E_a = \frac{8.314 \times \ln(2)}{0.000114} = \frac{8.314 \times 0.693}{0.000114} = 50,500\) J/mol
Answer: 50.5 kJ/mol or 12.1 kcal/mol
Problem 2: First Order Reaction with Half-Lives
Question: First order reaction has t½ = 30 min at 27°C and t½ = 10 min at 47°C. Find Ea.
Solution:
T₁ = 27°C = 300 K, T₂ = 47°C = 320 K
For first order: \(k = 0.693/t_{1/2}\)
k₁ = 0.693/30 = 0.0231 min⁻¹, k₂ = 0.693/10 = 0.0693 min⁻¹
\(\ln(k_2/k_1) = \ln(0.0693/0.0231) = \ln(3) = 1.099\)
\(1/T_1 - 1/T_2 = 1/300 - 1/320 = 0.00333 - 0.00313 = 0.00020\) K⁻¹
\(E_a = \frac{8.314 \times 1.099}{0.00020} = 45,700\) J/mol
Answer: 45.7 kJ/mol or 10.9 kcal/mol
Problem 3: Given Rate Constants
Question: Rate constant k = 3.7 × 10⁻⁵ s⁻¹ at 500 K and k = 1.7 × 10⁻³ s⁻¹ at 600 K. Calculate Ea.
Solution:
\(\ln(k_2/k_1) = \ln(1.7 \times 10^{-3} / 3.7 \times 10^{-5}) = \ln(45.95) = 3.827\)
\(1/T_1 - 1/T_2 = 1/500 - 1/600 = 0.002 - 0.00167 = 0.000333\) K⁻¹
\(E_a = \frac{8.314 \times 3.827}{0.000333} = 95,500\) J/mol
Answer: 95.5 kJ/mol or 22.8 kcal/mol
Frequently Asked Questions
What is activation energy?
Activation energy (Ea) is the minimum energy required for reactant molecules to overcome the energy barrier and form products in a chemical reaction. It represents the height of the energy barrier between reactants and products in an energy diagram. Higher activation energy means slower reaction at given temperature - molecules need more kinetic energy to react. Ea is measured in kJ/mol or kcal/mol. Typical values: 50-100 kJ/mol for many reactions. Catalysts work by lowering activation energy, allowing faster reactions without changing equilibrium. Temperature increases reaction rate by giving more molecules energy above Ea threshold.
How do you calculate activation energy from rate constants?
To calculate activation energy from two rate constants at different temperatures, use the Arrhenius equation: Ea = R × ln(k₂/k₁) ÷ (1/T₁ - 1/T₂). Steps: (1) Convert temperatures to Kelvin (K = °C + 273.15). (2) Calculate ln(k₂/k₁). (3) Calculate (1/T₁ - 1/T₂). (4) Multiply ln(k₂/k₁) by gas constant R (8.314 J/mol·K). (5) Divide by (1/T₁ - 1/T₂). Example: If k doubles from 300K to 310K, ln(2) = 0.693, (1/300 - 1/310) = 0.000108, Ea = 8.314 × 0.693/0.000108 = 53.3 kJ/mol.
What is the Arrhenius equation?
The Arrhenius equation is k = A × e^(-Ea/RT), relating reaction rate constant (k) to temperature (T) and activation energy (Ea). Components: k = rate constant, A = pre-exponential/frequency factor (collision frequency), Ea = activation energy, R = gas constant (8.314 J/mol·K), T = absolute temperature (Kelvin). Shows exponential temperature dependence - small temperature increase causes large rate increase for high Ea reactions. Linearized form: ln k = ln A - Ea/RT. Plot ln k vs 1/T gives straight line with slope -Ea/R. Named after Svante Arrhenius (1889). Fundamental equation in chemical kinetics.
Why must temperatures be in Kelvin for Arrhenius equation?
Kelvin is required because the Arrhenius equation involves absolute temperature in the exponential term e^(-Ea/RT). Using Celsius or Fahrenheit would give incorrect results because they have arbitrary zero points. Kelvin scale starts at absolute zero (0 K = -273.15°C), the lowest possible temperature. The gas constant R is calibrated for Kelvin (8.314 J/mol·K). Temperature ratios and differences in 1/T calculations only make physical sense with absolute scale. Conversion: K = °C + 273.15. Always convert before calculations. Example: 25°C = 298.15 K (not just 25). Using °C in formula would dramatically underestimate activation energy.
What is a typical activation energy value?
Typical activation energies range from 40-400 kJ/mol (10-100 kcal/mol) depending on reaction type. Low Ea (< 40 kJ/mol): Fast reactions like radical reactions, ion combinations. Medium Ea (40-150 kJ/mol): Most organic reactions, enzymatic reactions. High Ea (> 150 kJ/mol): Combustion reactions, bond-breaking steps. Very high Ea (> 250 kJ/mol): Requires harsh conditions or catalysts. For comparison: C-H bond energy ~400 kJ/mol, but Ea for breaking it ~160 kJ/mol (not full bond energy). Zero activation energy: diffusion-controlled reactions. Negative "apparent" Ea: rare, indicates complex mechanisms or pre-equilibria.
How does activation energy affect reaction rate?
Activation energy inversely affects reaction rate - higher Ea means slower reaction. According to Arrhenius equation, rate constant k ∝ e^(-Ea/RT). The exponential relationship means even small Ea changes dramatically impact rate. At room temperature (298 K), increasing Ea by 5.7 kJ/mol decreases rate by factor of 10. Rule of thumb: for many reactions, rate doubles for every 10°C increase (Ea ~50 kJ/mol). Physical interpretation: Ea determines fraction of molecular collisions with sufficient energy to react. Higher Ea = smaller fraction = slower reaction. Temperature dependence is stronger for high-Ea reactions. Catalysts lower Ea, increasing rate without changing thermodynamics.