Arrhenius Equation Calculator
Inputs
| Pre-exponential Factor | 1e13 |
|---|---|
| Activation Energy | 50 kJ/mol |
| Temperature | 298.2 K |
Arrhenius Equation Calculator
Calculate the rate constant k = A·exp(−Ea/RT) from the pre-exponential factor, activation energy, and temperature, plus the Boltzmann fraction of activated collisions.
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Details
Understanding the Arrhenius equation
The Arrhenius equation relates a reaction's rate constant to temperature and activation energy:
k=A⋅e−Ea/(RT)| Symbol | Quantity | Unit |
|---|---|---|
| k | Rate constant | depends on reaction order |
| A | Pre-exponential factor | same as k |
| E_a | Activation energy | J/mol (enter in kJ/mol) |
| R | Gas constant | 8.314 J/(mol·K) |
| T | Absolute temperature | K |
The exponential term is the Boltzmann fraction — the fraction of molecular collisions that have at least the activation energy. This calculator reports both k and f.
Worked example
What is the rate constant for a reaction with A = 1 × 10¹³, Ea = 50 kJ/mol at 25 °C (298.15 K)?
f=e−50,000/(8.314×298.15)=e−20.165≈1.74×10−9 k=1×1013×1.74×10−9≈1.74×104Only about 1.7 in every billion collisions has enough energy to react at room temperature.
Temperature dependence
The Arrhenius equation predicts that k increases exponentially with temperature. For reactions near room temperature, a rough rule of thumb is that the rate roughly doubles for every 10 °C rise, although the exact factor depends on Ea.
| Temperature | Boltzmann fraction (Ea = 50 kJ/mol) |
|---|---|
| 0 °C (273 K) | 4.1 × 10⁻¹⁰ |
| 25 °C (298 K) | 1.7 × 10⁻⁹ |
| 50 °C (323 K) | 6.2 × 10⁻⁹ |
| 100 °C (373 K) | 5.2 × 10⁻⁸ |
The pre-exponential factor A
The pre-exponential factor A (also called the frequency factor) represents the collision frequency corrected for geometric orientation. Its numerical value and units depend on the reaction order:
- First-order reactions: A has units of s⁻¹
- Second-order reactions: A has units of L/(mol·s)
Because rate-constant units vary by order, this calculator treats A and k as plain numbers and focuses on computing their ratio at a given temperature.
Finding Ea from two temperatures
If the rate constant is measured at two temperatures, the activation energy can be extracted without knowing A. Taking the ratio of the two Arrhenius expressions and applying the natural log:
ln(k1k2)=−REa(T21−T11)Solving for Ea:
Ea=−R⋅1/T2−1/T1ln(k2/k1)Limitations
The Arrhenius equation is an empirical model that works well for many elementary reactions over moderate temperature ranges. It assumes A and Ea are constant with temperature, which is only an approximation. At very high temperatures or for reactions with quantum tunneling, more elaborate treatments are needed.
Frequently Asked Questions (FAQ)
What is the Arrhenius equation?
The Arrhenius equation is k = A × exp(−Ea/(R × T)), where k is the rate constant, A is the pre-exponential (frequency) factor, Ea is the activation energy in J/mol, R = 8.314 J/(mol·K) is the gas constant, and T is the absolute temperature in kelvin. It describes how the rate constant of a chemical reaction depends on temperature: higher temperature increases k exponentially because a larger fraction of collisions has enough energy to overcome the activation barrier.
What is activation energy?
Activation energy (Ea) is the minimum kinetic energy that colliding reactant molecules must have for a reaction to occur. Molecules with energy below Ea bounce off each other unchanged. Ea is typically given in kJ/mol. A reaction with high activation energy is strongly temperature-dependent — raising the temperature dramatically increases the fraction of molecules that exceed the threshold, speeding up the reaction.
What is the pre-exponential factor A?
The pre-exponential factor A (also called the frequency factor or attempt frequency) represents the rate at which collisions occur with the correct geometric orientation, regardless of energy. It sets an upper limit on the rate constant: even if every collision had infinite energy, k could not exceed A. In practice A is determined experimentally by measuring k at several temperatures and extrapolating the Arrhenius plot to 1/T = 0.
How do I find Ea from two rate constants at two temperatures?
If you know k₁ at T₁ and k₂ at T₂, divide the two Arrhenius expressions to cancel A. Taking the natural log gives ln(k₂/k₁) = −(Ea/R) × (1/T₂ − 1/T₁). Rearranging: Ea = −R × ln(k₂/k₁) / (1/T₂ − 1/T₁). For example, if k doubles from 300 K to 310 K, ln(2) ≈ 0.693, and 1/310 − 1/300 ≈ −1.075 × 10⁻⁴ K⁻¹, giving Ea ≈ 8.314 × 0.693 / 1.075 × 10⁻⁴ ≈ 53.6 kJ/mol.