Arrhenius Equation Calculator

The Arrhenius equation: how temperature sets a rate constant

The Arrhenius equation, k = A e^(−Ea/RT), gives a reaction’s rate constant k from its activation energy Ea, the absolute temperature T and a pre-exponential factor A, with R = 8.314 J/(mol·K). Measure k at two temperatures and the activation energy follows from the two-point form, ln(k₂/k₁) = (Ea/R)(1/T₁ − 1/T₂). For the values loaded above, a rate constant rising from 0.010 s⁻¹ at 300 K to 0.035 s⁻¹ at 320 K, Ea = 8.314 × ln 3.5 ÷ (1/300 − 1/320) = 50.0 kJ/mol.

Enter any four of the five values and choose Solve for this on the fifth: the activation energy from two rate constants, the rate constant at a new temperature, or the temperature at which a reaction reaches a given rate constant. Temperatures can be typed in kelvin, °C or °F, and each is converted to kelvin before the arithmetic. The activation energy is in kJ/mol. The rate constants can be in any unit, s⁻¹ for a first-order reaction or M⁻¹ s⁻¹ for a second-order one, as long as both use the same one, and a solved k comes out in that unit.

Alongside the answer the calculator shows the pre-exponential factor A, how many times faster the reaction runs at T₂ than at T₁, and the factor a 10 K rise from T₁ would give. The working underneath puts your numbers into the natural-log form, step by step.

Worked example: activation energy and A from two rate constants

A reaction’s rate constant is 0.02 s⁻¹ at 500 K and 0.07 s⁻¹ at 700 K. What are its activation energy and its pre-exponential factor?

• Ea = R ln(k₂/k₁) ÷ (1/T₁ − 1/T₂) = 8.314 × 1.2528 ÷ 0.00057143 = 18,228 J/mol, which is 18.23 kJ/mol.
• Ea/RT₁ = 18,228 ÷ (8.314 × 500) = 4.385, so A = k₁ e^(Ea/RT₁) = 0.02 × e^4.385 = 1.60 s⁻¹.

Worked with common logarithms instead, as NCERT and many other courses do, the same equation reads log(k₂/k₁) = Ea/(2.303R) × (T₂ − T₁)/(T₁T₂), so Ea = 2.303 × 8.314 × log 3.5 × (500 × 700) ÷ (700 − 500) = 19.147 × 0.54407 × 1750 = 18,230 J/mol. The two routes agree to four figures, 18.23 kJ/mol. They part in the last digit mainly because 2.303 is ln 10 rounded up from 2.3026.

Why a few degrees change the rate so much

Temperature sits in an exponent, so a change of a few percent in T multiplies k rather than adding to it. The Factor for a 10 K rise readout shows by how much: for the values loaded above, 50 kJ/mol at 300 K, a rise of ten degrees multiplies k by 1.91. Near room temperature, at 298 K, a barrier of 52.9 kJ/mol doubles the rate for a rise of ten degrees, so that is the barrier the rule of thumb fits. It is not a law. At the same temperature a 100 kJ/mol barrier gives 3.7 times for ten degrees and a 5 kJ/mol barrier only 1.07 times, and at 500 K the 52.9 kJ/mol barrier itself gives only 1.28 times.

The collision theory simulator shows why: only molecules in the far tail of the energy distribution carry enough energy to react, and warming a gas grows that tail far faster than it moves the average. The kinetic theory gas simulator shows where the spread comes from, building the distribution of molecular speeds out of atoms colliding in a box.

The Arrhenius plot: ln k against 1/T

Taking logarithms turns the equation into a straight line, ln k = ln A − (Ea/R)(1/T). Plot ln k against 1/T and the slope is −Ea/R and the intercept ln A, which is how an activation energy is found from more than two measurements. These rate constants for the decomposition of hydrogen iodide, 2HI → H₂ + I₂, are the ones OpenStax Chemistry 2e uses to show the method.

The two end points give a slope of −22,302 K, so Ea = 22,302 K × 8.314 J/(mol·K) = 185 kJ/mol, and put into the calculator above they give 185.4 kJ/mol. A least-squares line through all five points gives 185.3 kJ/mol, and OpenStax rounds its own estimate to 180 kJ/mol. Neighbouring pairs of points give anything from 165 to 198 kJ/mol, because an error in k counts for more when the two temperatures are close together. That is why activation energies are measured over as wide a range of temperature as the reaction allows, and fitted through every point.

The linear regression calculator fits these five points, or your own 1/T and ln k, and gives the standard error of the slope as well. Ea is the slope times −R, so for this table Ea is 185.3 kJ/mol with a standard error of 2.2 kJ/mol.

Reading A and Ea from a table of rate data

Evaluated gas-phase data, such as IUPAC’s for atmospheric chemistry, usually give the equation as k = A exp(−B/T), where B is Ea/R in kelvin. For the reaction of oxygen atoms with ozone, O + O₃ → 2O₂, IUPAC recommends k = 8.0 × 10⁻¹² exp(−2060/T) cm³ molecule⁻¹ s⁻¹ between 200 and 400 K. So A is 8.0 × 10⁻¹² cm³ molecule⁻¹ s⁻¹ and Ea = 2060 K × 8.314 J/(mol·K) = 17.1 kJ/mol. Cooled from 298 K to 220 K, both inside that range, the reaction slows by a factor of 11.6.

Common mistakes

• Leaving temperatures in Celsius. The equation needs kelvin. Put 25 and 35 in place of 298.15 K and 308.15 K and a rate that doubles appears to have an activation energy of 0.5 kJ/mol instead of 52.9.
• Mixing joules and kilojoules. R is 8.314 J/(mol·K), so Ea goes into the equation in J/mol. Put a 50 kJ/mol barrier in as 50 rather than 50,000 and k changes by about a tenth of a percent between 300 K and 320 K instead of rising 3.5 times. This calculator takes kJ/mol and converts.
• Mixing log and ln. log(k₂/k₁) goes with 2.303R and ln(k₂/k₁) with R. Pair log with R and the activation energy comes out 2.303 times too small.
• Swapping a pair. k₁ belongs to T₁. Swap only the rate constants and the activation energy comes out negative, which few real reactions have.
• Giving the two rate constants different units. Only their ratio enters, so k₁ in s⁻¹ against k₂ in min⁻¹ puts a factor of 60 into it: for the values loaded above that turns 50 kJ/mol into 213 kJ/mol.
• Extrapolating far beyond the data. A and Ea are fitted over the temperatures that were measured. Well outside them a prediction can be out by more than a factor of two, as the next section shows.

What this calculator does not cover

It treats A and Ea as constants, which holds over the range they were measured across but not indefinitely: the IUPAC Green Book defines the activation energy from the slope of ln k against 1/T and notes that it can itself depend on temperature. For hydroxyl radicals reacting with hydrogen, HO + H₂ → H₂O + H, IUPAC’s evaluation recommends k = 7.7 × 10⁻¹² exp(−2100/T) cm³ molecule⁻¹ s⁻¹ between 200 and 450 K, while measurements up to 992 K were fitted by 4.12 × 10⁻¹⁹ T^2.44 exp(−1281/T). Carried out to 992 K, the first expression gives a rate constant 2.5 times smaller than the second. Forms with a power of T, k = A Tⁿ e^(−Ea/RT), are beyond this calculator.

Nor does it know the mechanism behind a rate constant. A negative activation energy is allowed, with a warning, because a few reactions have one: IUPAC gives 2NO + O₂ → 2NO₂ as k = 3.3 × 10⁻³⁹ exp(530/T) cm⁶ molecule⁻² s⁻¹ from 270 to 600 K, and records measurements in which k falls to a minimum near 600 K and then rises again. Reactions catalysed by enzymes speed up with temperature only until the enzyme starts to lose its shape. And it does not take A as an input: to find k from A and Ea at one temperature, work k = A e^(−Ea/RT) directly.

For a first-order reaction the half-life is t½ = ln 2 / k, so whatever a change of temperature does to k it does in reverse to the half-life, and the half-life calculator takes it from there. Radioactive decay is the exception: a nuclide’s half-life is, for practical purposes, independent of the temperature of the sample.

Converting units first? Use the temperature conversion table.