Arrhenius Equation
The Arrhenius equation describes the temperature dependence of the rate constant of a chemical reaction.
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Formula
k = A \cdot e^{-E_A / (R \cdot T)}Variables & units – Arrhenius Equation
| Symbol | Meaning | Unit |
|---|---|---|
| k | Rate constant | mol⁻¹·L·s⁻¹ or similar |
| A | Pre-exponential factor (frequency factor) | same unit as k |
| E_A | Activation energy | J/mol |
| R | Gas constant (8.314) | J/(mol·K) |
| T | Temperature | K |
Derivation & background – Arrhenius Equation
In 1889, Svante Arrhenius formulated the equation on the basis of empirical observations. The temperature rule of thumb states approximately that a rise of +10 K doubles the reaction rate. The activation energy is the energy barrier that must be overcome.
Exam blueprint
Validity range
Applies to many reactions dominated by an activation barrier; A and E_A are often treated as constant over limited temperature ranges.
Derivation steps
Only a temperature-dependent fraction of particles has enough energy to overcome the activation barrier.
- 1The Boltzmann factor gives the fraction e^{-E_A/(RT)}.
- 2The pre-exponential factor A scales collision frequency and orientation.
Rearrangements
Linearized form
A plot of ln(k) versus 1/T has slope -E_A/R.
Task variant
Why does temperature strongly increase reaction rate?
The exponential term is sensitive to T; many more particles exceed E_A.
Common mistakes
Using E_A in kJ/mol while R is in J/(mol·K).
Match units: usually use E_A in J/mol.
Exam context
- Often with Arrhenius plots, temperature comparison and catalyst effects.
These mistakes cost points in real exams. The set drills them until they stick.
Formula cluster
Kinetics and thermodynamics
Clearly separates reaction rate from equilibrium position.
Worked example
A reaction with E_A = 50 kJ/mol at T₁ = 300 K and T₂ = 310 K: k₂/k₁ = e^(50000/8.314 × (1/300 − 1/310)) ≈ 1.9 → roughly a doubling.
Applications
Chemical process engineering, food preservation (refrigeration), pharmacy (shelf life), geology (dating)
Quanta exam set
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Which formula describes Arrhenius Equation?
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How do you rearrange k = A·e^(−EA/RT) for Linearized form?
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Which common mistake happens with Arrhenius Equation?
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Scientific sources
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Frequently asked questions about Arrhenius Equation
How do you calculate the rate constant with the Arrhenius equation?+
Multiply the pre-exponential factor A by the Boltzmann factor e^(−E_A/(R·T)): k = A·e^(−E_A/(R·T)). Insert the activation energy E_A and the gas constant R = 8.314 J/(mol·K) in matching units and the temperature T in kelvin. The exponent is dimensionless, so E_A and R·T must have the same energy unit. A common mistake is to leave E_A in kJ/mol while R is in J/(mol·K); then convert E_A to J/mol, that is with a factor of 1000. The pre-exponential factor A has the same unit as k and describes the collision frequency with suitable orientation.
Why does a small temperature increase raise the reaction rate so strongly?+
Because the temperature sits in the exponent of the Arrhenius equation and the exponential term reacts very sensitively. The Boltzmann factor e^(−E_A/(R·T)) gives the fraction of particles that have enough energy to overcome the activation barrier. Even a small temperature increase shifts the energy distribution so that markedly more particles lie above the threshold. As a rule of thumb the reaction rate often doubles to triples per 10 kelvin. Example: from 300 K to 310 K, k rises by about a factor of 1.9 at E_A = 50 kJ/mol. This strong temperature dependence explains why many reactions proceed dramatically faster when heated.
How do you determine the activation energy from an Arrhenius plot?+
Take the logarithm of the Arrhenius equation, then you get the linear form ln k = ln A − (E_A/R)·(1/T). Plotting ln k against 1/T gives a straight line with slope −E_A/R. From the measured slope m it follows that E_A = −m·R. The intercept yields ln A and thus the pre-exponential factor. You therefore need at least two measured values of k at different temperatures; more points make the evaluation more accurate. Make sure to insert the temperature as the reciprocal 1/T in kelvin. This linearization is the standard method to determine activation energies experimentally.
What is the difference between activation energy and reaction enthalpy?+
The activation energy E_A is the energy barrier the reactants must overcome to reach the transition state and then the products. It determines how fast a reaction proceeds, that is the kinetics. The reaction enthalpy ΔH, by contrast, is the energy difference between products and reactants and determines whether the reaction releases or absorbs energy, that is the thermodynamics. E_A affects the rate, not the position of equilibrium; ΔH affects the energy balance, not the rate. A catalyst lowers E_A and thus speeds up the reaction, but leaves ΔH unchanged. Both quantities are independent and describe different aspects of a reaction.
How does a catalyst act in the Arrhenius equation?+
A catalyst lowers the activation energy E_A by providing an alternative reaction path with a lower energy barrier. In the Arrhenius equation k = A·e^(−E_A/(R·T)) the E_A sits negatively in the exponent, so a smaller barrier raises the Boltzmann factor and thus the rate constant k noticeably. Because E_A acts in the exponent, even a moderate lowering leads to a large acceleration. The catalyst changes neither the position of equilibrium nor the reaction enthalpy ΔH; it speeds up the forward and reverse reactions equally. After the reaction it is left unconsumed. This is exactly why catalysts are so important in industry and biochemistry.
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How do you calculate with Arrhenius Equation?
Here is how to work through a typical Arrhenius Equation (k = A·e^(−EA/RT)) task step by step:
- 1
Task
Why does temperature strongly increase reaction rate?
Solution path
The exponential term is sensitive to T; many more particles exceed E_A.