Arrhenius Equation Calculator for Two Temperatures

An Arrhenius equation calculator for two temperatures estimates how a reaction rate changes when temperature changes, using an activation energy you provide. Enter the activation energy and the two absolute temperatures in kelvins; the calculator returns the rate constant at the second temperature divided by the rate constant at the first.

Formula

The two-temperature form of the Arrhenius equation is:

ln(k₂/k₁) = (Eₐ/R)(1/T₁ − 1/T₂)

Equivalently, k₂/k₁ = exp[(Eₐ/R)(1/T₁ − 1/T₂)]. The calculator reports this ratio first, followed by its percentage change, calculated as (k₂/k₁ − 1) × 100%.

• Eₐ is activation energy in joules per mole (J/mol).
• R is the molar gas constant, 8.314462618 joules per mole per kelvin (J/(mol·K)).
• T₁ is the first, reference temperature in kelvins (K), and k₁ is the rate constant at that temperature.
• T₂ is the second temperature in kelvins, and k₂ is the rate constant there.

The ratio has no units, provided both rate constants use the same units. A ratio above 1 means the rate constant is larger at the second temperature; below 1 means it is smaller. This equation determines a relative change, not an absolute rate constant.

Worked Example

Using the calculator’s default inputs, Eₐ = 50,000 J/mol, T₁ = 300 K, and T₂ = 310 K:

1. Calculate the reciprocal-temperature difference: 1/300 − 1/310 = 0.0001075269 K⁻¹.
2. Calculate Eₐ/R: 50,000/8.314462618 ≈ 6,013.6 K.
3. Multiply: 6,013.6 × 0.0001075269 ≈ 0.6469.
4. Exponentiate: k₂/k₁ = exp(0.6469) ≈ 1.9098.
5. Calculate percentage change: (1.9098 − 1) × 100% = 90.98%.

Under the model and assumptions, the rate constant at 310 K is about 1.91 times its value at 300 K, an increase of about 90.98%. This is a prediction for the rate constant, not necessarily a prediction that every measured reaction will proceed exactly 90.98% faster in practice.

How the Two-Temperature Calculation Works

The full Arrhenius equation is k = A exp(−Eₐ/(RT)), where A is the pre-exponential factor. Writing this equation once at each temperature and dividing the two expressions cancels A, if it is the same at both temperatures. Taking the natural logarithm gives the two-temperature form above. That cancellation is why you can calculate a ratio without knowing A or either rate constant beforehand.

For a positive activation energy, raising the temperature makes 1/T smaller. Thus, if T₂ is higher than T₁, the term 1/T₁ − 1/T₂ is positive and the predicted ratio exceeds 1. Reversing the temperatures reverses the comparison: the ratio becomes less than 1. If both temperatures are equal, the ratio is exactly 1, whatever the activation energy.

When to Use the Calculator

Use this calculation when you have an activation energy and want to estimate a reaction’s relative rate constant at two temperatures. It is often useful for comparing laboratory conditions, interpreting a measured activation energy, or checking the temperature dependence assumed in a kinetic model.

Before entering values, check the data and units:

• Enter activation energy in J/mol. If it is given in kJ/mol, multiply by 1,000; for example, 50 kJ/mol is 50,000 J/mol.
• Enter temperatures in kelvins. Convert Celsius using K = °C + 273.15, or Fahrenheit using K = (°F + 459.67) × 5/9.
• Use the same reaction, rate-constant definition, and conditions other than temperature for both measurements as far as the model requires.
• Use an activation energy appropriate to the reaction and temperature range being considered.

Temperature differences in Celsius and kelvins have the same numerical size, but the formula uses reciprocal absolute temperatures. Substituting Celsius values directly is therefore incorrect. The same issue applies to Fahrenheit: convert to kelvins rather than using the displayed Fahrenheit number.

Interpreting Results and Common Mistakes

The rate-constant ratio is not an amount of material, a percentage, or a rate with units. A result of 2 means k₂ is twice k₁; the corresponding percentage increase is 100%, not 200%. A result of 0.5 means k₂ is half k₁, which is a 50% decrease relative to k₁.

Common errors include entering kJ/mol as though it were J/mol, mixing Celsius with kelvins, reversing T₁ and T₂, or interpreting the ratio as an absolute rate. Keep extra precision during intermediate calculations and round only the displayed answer. The calculator displays the ratio to four decimal places and percentage change to two; small differences in the last digit can result from rounding.

Assumptions and Limits

The calculation assumes Arrhenius behavior over the temperature interval: activation energy is treated as constant, and the pre-exponential factor does not change. Real reactions may depart from this model, particularly across a phase change, a change in reaction mechanism, or a range where other physical or chemical effects become important. Experimental uncertainty in activation energy can also noticeably affect the estimate.

The equation predicts the ratio of rate constants. Whether that translates directly into a reaction completing sooner by the same factor depends on the kinetic rate law, reactant concentrations, mixing, and other conditions. To obtain an absolute k₂, you also need k₁, then calculate k₂ = (k₂/k₁) × k₁. The units of k depend on the reaction’s rate law; the Arrhenius ratio itself is dimensionless.

Frequently Asked Questions

Can I use Celsius in the Arrhenius equation?
No. Convert both temperatures to kelvins first because the equation uses absolute temperature and reciprocal temperature.

How do I enter activation energy in kJ/mol?
Convert it to J/mol by multiplying by 1,000. For example, enter 72 kJ/mol as 72,000 J/mol.

Does the calculator give the new rate constant?
It gives k₂/k₁, the relative change. To find k₂ itself, multiply that ratio by the known k₁, keeping consistent rate-constant units.

What happens if the second temperature is lower?
For a positive activation energy, the predicted ratio is below 1, meaning the rate constant is lower at T₂ than at T₁.

Why does the calculation not need the pre-exponential factor?
When the two Arrhenius equations are divided, the common pre-exponential factor cancels. This relies on it being effectively unchanged between the two temperatures.

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in ea | Activation energy | J/mol | 50000 | 0
in t1 | First temperature | K | 300 | 0
in t2 | Second temperature | K | 310 | 0
out ratio | Rate constant ratio k2/k1 | - | exp(ea / 8.314462618 * (1 / t1 - 1 / t2)) | 4
out change | Percent change in rate constant | % | (ratio - 1) * 100 | 2