molecalculator

Reactions & Stoichiometry

Arrhenius Equation Derivation and Calculator

Calculator No. 05-0134 · Updated · Formula and assumptions shown below

The calculator on this page is switched off while it is checked. The formula is below, and the worked example shows how to do the sum by hand.

For study and lab planning. The result follows from the numbers you enter and the formula shown. Check it against your method before you weigh, mix or dilute anything, follow the safety data sheet and your supervisor, and do not use it for clinical, dosing or medication work.

The Arrhenius equation predicts how a reaction’s rate constant changes with absolute temperature, using the activation energy and a pre-exponential factor. Enter those two substance-specific parameters and a temperature to calculate the rate constant; the result is meaningful only when the parameters and rate-constant units match the reaction being studied.

Formula

The Arrhenius equation is k = A exp(−Ea/(RT)). In the calculator it is written as k = A × exp(−Ea × 1000/(8.31446261815324 × T)), because activation energy is entered in kJ/mol and is converted to J/mol by multiplying by 1000.

  • k is the reaction rate constant, here reported in s−1. Its proper units depend on the reaction’s rate law; this calculator’s stated units therefore assume a first-order rate constant.
  • A is the pre-exponential factor, entered in the same units as k (s−1 here).
  • Ea is activation energy in kJ/mol.
  • R is the ideal gas constant, exactly 8.31446261815324 J/(mol·K).
  • T is absolute temperature in kelvin (K).

The exponential’s argument must be dimensionless. That is why energy and temperature units must agree with R. Do not enter Celsius as though it were kelvin: convert using T(K) = T(°C) + 273.15 before using this equation.

Worked Example

Using the calculator’s exact defaults, A = 1,000,000,000,000 s−1, Ea = 50 kJ/mol, and T = 298.15 K:

  1. Convert the activation energy: 50 kJ/mol × 1000 = 50,000 J/mol.
  2. Calculate the exponent: −50,000 J/mol ÷ (8.31446261815324 J/(mol·K) × 298.15 K) ≈ −20.159.
  3. Evaluate the exponential: exp(−20.159) ≈ 1.758 × 10−9.
  4. Multiply by A: 1.000 × 1012 s−1 × 1.758 × 10−9 ≈ 1.758 × 103 s−1.

Rounded to three decimal places, the calculated result is 1758.000 s−1. Intermediate values are approximate, so small differences in the final digits can result from rounding them; the calculator evaluates the full expression before rounding its display.

Assumptions

  • The entered A and Ea describe the same reaction pathway and temperature range.
  • A is expressed in s−1, consistent with a first-order rate constant; other reaction orders need different rate-constant units.
  • Temperature is an absolute temperature in kelvin, and activation energy is a molar quantity.
  • The simple Arrhenius relationship is an adequate model over the temperatures being considered.

How the Arrhenius Equation Is Derived

The equation is an empirical relationship: it summarizes measured rate constants, rather than following as an exact consequence of the basic chemical rate laws. Its exponential form has a useful physical interpretation rooted in molecular energy distributions. At a given temperature, molecules have a range of energies. A reaction can proceed through a particular pathway only when reacting molecules can reach an activated configuration associated with an energy barrier.

In a simplified picture, the fraction of molecules with enough energy to cross a barrier Ea is proportional to the Boltzmann factor exp(−Ea/(RT)). If the number of suitably energetic encounters and the probability of successful reaction are collected into a temperature-dependent prefactor, the rate constant is modeled as that prefactor multiplied by the energetic fraction:

k ∝ exp(−Ea/(RT))

Replacing the proportionality with a constant A gives k = A exp(−Ea/(RT)). A is called the pre-exponential or frequency factor. It can incorporate factors such as encounter frequency, molecular orientation, and details of the reaction pathway; it is not generally just a count of collisions.

Taking natural logarithms makes the relationship linear:

ln k = ln A − Ea/(RT)

This is the common route for estimating activation energy from experimental data. Plot ln k vertically against 1/T horizontally. If the data follow a straight line, its slope is −Ea/R and its vertical intercept is ln A. Thus Ea = −(slope)R. The reciprocal-temperature axis is in K−1; using Celsius on that axis gives an invalid slope.

What the Parameters Mean

Activation energy describes how strongly the rate constant responds to temperature in this model. A larger Ea means a stronger temperature dependence, all else equal. A is the extrapolated rate constant as 1/T approaches zero in the linearized form; that limit corresponds to extremely high temperature and is not usually a practical measurement. Neither A nor Ea is a universal constant: both are determined for a particular reaction and set of conditions.

The equation predicts a rate constant, not directly the amount of reactant consumed or product formed. To turn k into a concentration-versus-time prediction, the reaction order and integrated rate law are also needed. For a first-order reaction, for example, concentration changes according to [A]t = [A]0exp(−kt); that extra relation is not part of this calculator.

Using the Equation and Interpreting Results

Use the calculator when A and Ea have been reported or obtained from suitable kinetic measurements, and you want a model-based k at a specified temperature. Check the original source for the reaction conditions, rate-law order, units, and temperature interval. The defaults are illustrative values for calculation, not claims about a particular chemical reaction.

As a useful comparison, for the same A and Ea, raising T makes −Ea/(RT) less negative, increasing the exponential and therefore k. The increase is nonlinear. A common two-temperature form, obtained by subtracting the logarithmic equation at two temperatures, is:

ln(k2/k1) = (Ea/R)(1/T1 − 1/T2)

This form estimates a rate-constant ratio without knowing A, provided the same Arrhenius parameters apply at both temperatures. Do not interpret a faster rate constant as proof that a reaction is safer or more desirable; this is a kinetic calculation, not a hazard assessment. For bench work, follow your lab method, the safety data sheet (SDS) and your supervisor.

Limits and Common Mistakes

  • Mixing units: R is in joules here, so convert kJ/mol to J/mol. Using 50 directly in place of 50,000 makes the exponent wrong by a factor of 1000.
  • Using Celsius: the denominator requires kelvin. A Celsius temperature must be shifted by 273.15.
  • Assuming A is universal: it depends on the reaction and on the units used for k. A value reported for one reaction cannot automatically be transferred to another.
  • Extrapolating too far: a straight line in an Arrhenius plot is an approximation over a tested range. Mechanism changes, phase changes, or other effects can produce curvature.
  • Confusing activation energy with reaction enthalpy: Ea is a kinetic barrier parameter; it is not simply the net heat released or absorbed by the overall reaction.

Real reactions can involve multiple pathways, diffusion limits, catalysts, or changing mechanisms. In such cases, a single A and Ea may only describe an apparent rate over a limited range. The calculation should therefore be read as a model prediction, not a substitute for measured kinetics.

Frequently Asked Questions

Why does the Arrhenius equation use kelvin?
Its exponential contains energy divided by RT, so temperature must be on an absolute scale. Kelvin makes the Boltzmann factor physically meaningful and keeps the exponent dimensionless.

How do you find activation energy from an Arrhenius plot?
Plot ln k against 1/T in kelvin. For a straight-line fit, activation energy is −R times the slope, using a consistent energy unit for R.

Does a catalyst lower activation energy?
A catalyst can provide a reaction pathway with a lower effective activation barrier, changing the rate. It does not change the equilibrium constant simply by speeding the forward reaction; it also affects the reverse pathway.

Can activation energy be negative?
An apparent negative activation energy can be observed in some complex, multistep systems over a limited temperature range. It does not mean molecules cross a literal negative energy barrier; it signals that the simple one-step interpretation may not apply.

What are the units of the pre-exponential factor?
A has the same units as k. Those units depend on the rate law: a first-order k has units of inverse time, while other orders have concentration-dependent units.

Related

Formula used
in a | Pre-exponential factor A | s^-1 | 1000000000000 | 0
in ea | Activation energy Ea | kJ/mol | 50 | 0
in t | Temperature T | K | 298.15 | 0
out k | Rate constant k | s^-1 | a * exp(-ea * 1000 / (8.31446261815324 * t)) | 3

Each line is one input (in, sel, list) or one result (out: label, unit, formula, decimals). How the calculators work

Method: the standard formula for this calculation with the constants and conversion factors written on the page, rounded as each result states. See method and constants, the disclaimer or report an error.

More in this section

The whole Reactions & Stoichiometry section →