Linear Arrhenius Equation: Formula and Calculator

The linear form of the Arrhenius equation is ln(k) = ln(A) − E_a/(RT); it relates a reaction’s rate constant to absolute temperature. This calculator uses a rate constant at each of two temperatures to find the activation energy, then returns the fitted linear equation’s slope and intercept.

Formula

The two-point Arrhenius equation is ln(k) = ln(A) − E_a/(RT). For measurements at two temperatures, subtracting the equations gives:

ln(k₂/k₁) = (E_a/R)(1/T₁ − 1/T₂)

Therefore, E_a = R ln(k₂/k₁) / (1/T₁ − 1/T₂). The calculator reports E_a in kJ/mol, dividing the result in J/mol by 1000. Here k is the rate constant, with the same units for both measurements; T is absolute temperature in kelvin; R is the gas constant, 8.31446261815324 J/(mol·K); and A is the pre-exponential factor, with the same units as k.

In a plot of ln(k) against 1/T, the slope is −E_a/R and the vertical intercept is ln(A). The slope’s unit is kelvin because ln(k) is dimensionless as a plotted quantity only after a consistent reference for its units; in practical work, the same rate-constant unit must be used throughout. The intercept is the logarithm of a numerical rate constant expressed in that chosen unit.

Worked Example

Using the defaults, k₁ = 0.01 s⁻¹ at T₁ = 300 K, and k₂ = 0.1 s⁻¹ at T₂ = 310 K:

• Rate-constant ratio: k₂/k₁ = 0.1/0.01 = 10.
• Logarithm of the ratio: ln(10) = 2.3025851.
• Reciprocal-temperature difference: 1/300 − 1/310 = 0.00010752688 K⁻¹.
• E_a = 8.31446261815324 × 2.3025851 / 0.00010752688 = 178000 J/mol, or 178.000 kJ/mol to the calculator’s displayed precision.
• Slope = −178000/8.31446261815324 = −21409.3 K (rounded).
• Intercept = ln(0.01) − (−21409.3/300) = 66.141 (rounded).

The resulting fitted line is approximately ln(k) = 66.141 − 21409.3/T, with T in kelvin. The intercept corresponds to ln(A) when the rate constant is expressed in s⁻¹.

Assumptions

• The rate constants describe the same reaction and the same rate law at both temperatures.
• Both temperatures are in kelvin, and both rate constants use the same units.
• The Arrhenius model is an adequate description over the temperature interval, so activation energy is treated as constant.
• The inputs are positive, and the two temperatures are distinct. A zero rate constant or equal temperatures makes the logarithm or calculation undefined.

What the Linear Form Means

The original Arrhenius equation is k = A exp(−E_a/(RT)). Taking the natural logarithm converts its exponential dependence into a straight-line relationship. Temperature itself is not the horizontal coordinate in the standard linear plot: the x-axis is reciprocal temperature, 1/T, usually in K⁻¹. As temperature rises, 1/T falls; for a positive activation energy, ln(k) generally rises, producing a negative slope.

The linear form is useful because it makes parameter estimation and model checks straightforward. Experimental measurements across several temperatures can be plotted as ln(k) versus 1/T. A straight-line fit estimates the slope and intercept; the slope gives activation energy, while the intercept estimates the pre-exponential factor. This calculator uses two points, so it gives the exact line through those two inputs rather than a regression that averages measurement scatter across a dataset.

Using the Result and Choosing Data

Use rate constants determined under comparable conditions and with a consistent kinetic model. A rate constant is not simply the observed reaction rate: its units depend on the overall order of the rate law. For example, a first-order rate constant commonly has units of s⁻¹, while a second-order constant may have units such as L mol⁻¹ s⁻¹. The calculator’s example uses s⁻¹, but any consistent rate-constant unit can be used for both inputs.

For better parameter estimates, collect reliable measurements at multiple temperatures and fit a straight line rather than relying on only two points. Consider these checks:

• Use kelvin, not Celsius. Convert with T(K) = T(°C) + 273.15 before entering values.
• Keep the rate-constant units and their definition consistent across temperatures.
• Use a temperature range where the reaction mechanism and rate-limiting behavior are not changing.
• Inspect the plot and residuals. Curvature or systematic deviations suggest that a single Arrhenius line may not describe the data well.

The calculator’s displayed activation energy is in kJ/mol. A positive value is common for thermally activated reactions, but the equation itself does not require the value to be positive. If one measured rate constant is smaller at the higher temperature, the calculation can produce a negative apparent activation energy; that may signal complex kinetics, a mechanism shift, or inconsistent measurements rather than a simple elementary barrier.

Common Errors and Limits

A frequent mistake is plotting ln(k) against temperature instead of 1/T. That plot is generally not linear under the Arrhenius model. Another is using log base 10 but interpreting its slope as though it came from a natural-log plot. For log₁₀(k), the slope has an additional conversion factor: it equals −E_a/(2.303R).

Temperature errors are especially consequential because the equation uses reciprocal absolute temperature. Entering a Celsius number as though it were kelvin can substantially distort the result. Also avoid rounding temperatures or rate constants too early; logarithms and reciprocal temperatures can magnify small input changes. The output precision does not imply that the underlying measurements are equally precise.

Two-point estimates are sensitive to experimental error and provide no independent test of linearity. A fitted activation energy is an apparent, interval-dependent parameter if the mechanism changes or if multiple processes contribute. The pre-exponential factor can also depend on the chosen units for k, so state those units when reporting A.

Frequently Asked Questions

What is the linear form of the Arrhenius equation?
It is ln(k) = ln(A) − E_a/(RT). Written as y = mx + b for a plot of ln(k) against 1/T, the slope is −E_a/R and the intercept is ln(A).

Why does the Arrhenius plot use 1/T?
The temperature term in the exponent is E_a/(RT). Taking the natural logarithm leaves a term proportional to 1/T, which produces a straight line when plotted on that horizontal axis.

How do I calculate activation energy from two rate constants?
Use E_a = R ln(k₂/k₁)/(1/T₁ − 1/T₂). Use kelvin and the same rate-constant units at both temperatures.

Should I use ln or log in an Arrhenius plot?
The standard linear form uses the natural logarithm, ln. A base-10 plot can also be linear, but its slope is −E_a/(2.303R), so the conversion must be included.

Can I use Celsius in the Arrhenius equation?
No. Use absolute temperature in kelvin. Convert Celsius to kelvin by adding 273.15.

Why might an Arrhenius plot not be a straight line?
Possible causes include measurement error, a change in reaction mechanism, a changing rate-limiting step, or a temperature range over which one constant activation energy is not an adequate model. Follow your lab method, the safety data sheet (SDS) and your supervisor.

Related

• Percent Yield Calculator: Actual Yield ÷ Theoretical Yield × 100
• Mole Calculator From a Chemical Equation: Moles of Product From the Mole Ratio

in k1 | Rate constant at T1 | s^-1 | 0.01 | 0
in t1 | Temperature T1 | K | 300 | 0.000001
in k2 | Rate constant at T2 | s^-1 | 0.1 | 0
in t2 | Temperature T2 | K | 310 | 0.000001
out ea | Activation energy | kJ/mol | 8.31446261815324 * ln(k2 / k1) / (1 / t1 - 1 / t2) / 1000 | 3
out slope | Slope of ln(k) vs 1/T | K | -ea * 1000 / 8.31446261815324 | 3
out intercept | Intercept ln(A) | - | ln(k1) - slope / t1 | 3