The Arrhenius equation

The Arrhenius equation shows how the rate constant of a reaction depends on temperature and activation energy: k = Ae^−Ea/RT. Here k is the rate constant, A is the pre-exponential factor (the Arrhenius constant), Ea is the activation energy in J mol⁻¹, R is the gas constant, 8.31 J K⁻¹ mol⁻¹, and T is the temperature in kelvin. A higher temperature or a lower activation energy gives a bigger rate constant, and so a faster reaction.

The Arrhenius equation

The equation is named after the Swedish chemist Svante Arrhenius. AQA's A-level specification describes A as a constant known as the Arrhenius constant, and the equation sits in the rate equations topic, which is A-level only.

What the terms mean

The equation splits the rate constant into two parts.

• e^−Ea/RT is linked to the fraction of collisions with at least the activation energy. It is always between 0 and 1, and usually tiny. Raising T or lowering Ea makes it bigger.
• A takes account of how often the particles collide and whether they collide the right way round. OpenStax calls it the frequency factor, which reflects how well the conditions favour properly oriented collisions.

So k is big when collisions are frequent, well aimed and energetic enough. That is collision theory in one equation.

Arrhenius equation units

The exponent, Ea ÷ RT, must have no unit. With Ea in J mol⁻¹, R in J K⁻¹ mol⁻¹ and T in K, all the units cancel. That only works if Ea is in joules, so 75 kJ mol⁻¹ must go in as 75 000 J mol⁻¹. Because e^−Ea/RT has no unit, A always has the same unit as k: s⁻¹ for a first-order reaction, mol⁻¹ dm³ s⁻¹ for a second-order one.

Why temperature has such a big effect

Raising the temperature does two things. The particles move faster, so they collide a little more often. Much more important, a bigger fraction of them have enough energy to react, at least Ea, because the Maxwell-Boltzmann distribution spreads out and shifts to the right. The Arrhenius equation captures this: T sits inside an exponential, so a small rise in temperature can make a big difference to k.

For a typical activation energy of 50 kJ mol⁻¹, warming from 298 K to 308 K nearly doubles the rate constant:

That is where the rule of thumb "a 10 °C rise roughly doubles the rate" comes from. It is only a rough guide: the bigger the activation energy, the more the rate changes for each degree.

Test the idea. Choose the Maxwell-Boltzmann curve that matches each change, then press Mark to check it in the course.

The Arrhenius equation graph

Taking natural logs of both sides gives the linear form of the Arrhenius equation:

This has the shape y = mx + c. Plot ln k on the y-axis against 1/T on the x-axis, and you get a straight line with:

• gradient = −Ea ÷ R, so Ea = −gradient × R;
• y-intercept = ln A, so A = e^intercept.

Here x is 1/T in units of 10⁻³ K⁻¹ (so x = 3.333 means 1/300 K⁻¹) and y is ln k. The line slopes down because the gradient, −Ea ÷ R, is negative: the hotter the reaction, the further left it sits and the bigger ln k is. A steeper line means a bigger activation energy, so a reaction more sensitive to temperature. This line is for Ea = 50 kJ mol⁻¹ and A = 1.0 × 10¹⁰ s⁻¹, so its gradient is −50 000 ÷ 8.31 = −6017 K and its intercept is ln(1.0 × 10¹⁰) = 23.03.

The Arrhenius equation rearranged

To rearrange for Ea, start from the log form, ln k = ln A − Ea/RT. Move the Ea term to one side: Ea/RT = ln A − ln k = ln(A ÷ k). Then multiply by RT.

Two temperatures: the two-point form

When you know k at two temperatures but not A, subtract the log forms for the two temperatures. The ln A cancels:

This is the quickest way to find an activation energy from two experiments. The Rates and Equilibrium Calculator has an Arrhenius mode that finds k from A, Ea and T, or Ea from two pairs of k and T, with every step shown.

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Rates and Equilibrium Calculator

Work with rate equations, the Arrhenius equation and Kc

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Arrhenius equation worked examples

Example 1: finding k

A first-order reaction has A = 2.0 × 10¹¹ s⁻¹ and Ea = 85 kJ mol⁻¹. Find k at 310 K.

Convert first: Ea = 85 000 J mol⁻¹. Work out the exponent on its own:

Example 2: how much faster when hotter?

The same reaction is warmed to 330 K. By what factor does k increase?

At 330 K, k = 2.0 × 10¹¹ × e^{−85 000/(8.31 × 330)} = 6.91 × 10⁻³ s⁻¹. Dividing, 6.91 × 10⁻³ ÷ 9.36 × 10⁻⁴ = 7.4. A 20 K rise makes this reaction about 7 times faster.

Example 3: activation energy from two rate constants

A reaction has k = 1.2 × 10⁻³ s⁻¹ at 300 K and 4.8 × 10⁻³ s⁻¹ at 320 K. Find its activation energy.

Use the two-point form. The ratio k₂ ÷ k₁ is 4, so:

The activation energy is 55.3 kJ mol⁻¹.

Example 4: from an Arrhenius plot

A student plots ln k against 1/T and draws a line of best fit with gradient −7200 K and y-intercept 22.5. Find Ea and A.

Ea is 59.8 kJ mol⁻¹, and A is 5.9 × 10⁹ in the same unit as k. Measure the gradient from two points far apart on the line, not from two data points.

Example 5: the effect of a catalyst

A catalyst lowers a reaction's activation energy from 90 kJ mol⁻¹ to 70 kJ mol⁻¹. Assuming A is unchanged, by what factor does k increase at 298 K?

Divide the two Arrhenius equations. The A cancels, leaving e to the power of the difference:

Lowering Ea by 20 kJ mol⁻¹ makes the reaction about 3200 times faster at room temperature. This is why catalysts, and enzymes in living things, are so powerful.

Example 6: finding the temperature

A reaction has A = 5.0 × 10⁹ s⁻¹ and Ea = 60 kJ mol⁻¹. At what temperature is k = 0.10 s⁻¹?

The temperature is 293 K, which is 20 °C.

Arrhenius equation derivation

At A-level you do not need to derive the equation, but you can see where it comes from. Experiments show that a graph of ln k against 1/T is a straight line. Any straight line on those axes can be written as ln k = ln A − (Ea/R)(1/T), and raising e to the power of both sides gives k = Ae^−Ea/RT. Collision theory and the Maxwell-Boltzmann distribution then explain what A and the exponential term mean.

Common mistakes

• Temperature in °C. T must be in kelvin: add 273. Using 25 instead of 298 gives nonsense.
• Ea in kJ mol⁻¹. Convert to J mol⁻¹, or the exponent is 1000 times too small.
• Using log instead of ln. The equation uses natural logs, ln, the inverse of e^x. Using log₁₀ gives the wrong gradient.
• Losing the minus sign. The gradient of ln k against 1/T is negative, but Ea is positive: Ea = −gradient × R.
• Plotting k against T. That gives a curve. Only ln k against 1/T gives a straight line.
• Rounding the exponent too early. e^−33.0 and e^−32.5 differ by a factor of about 1.65. Keep at least 4 significant figures in the exponent.

The Arrhenius equation in A-level Chemistry

In AQA A-level Chemistry, students should be able to explain the qualitative effect of temperature on the rate constant, do calculations with k = Ae^−Ea/RT, and use ln k = −Ea/RT + ln A with experimental data to plot a straight line with gradient −Ea/R. AQA's specification says these equations and the gas constant R will be given when required, so you do not have to memorise them, but you do need to use them fluently. This was checked against AQA's specification in November 2026; other boards publish their own specifications, so check yours.

The Reaction Kinetics course builds up to this from collision theory, energy profiles and the Maxwell-Boltzmann distribution, then covers rate equations and orders.

Practise kinetics in a short timed paper

A-level Chemistry: Year 2 gives you a short timed paper every day, with 8 questions worth 25 to 35 marks across the course, so rate equations turn up regularly but not in every paper.

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Chemistry · Rates and equilibriumRates and equilibrium calculator: rate equations and KcRate equations, orders from initial rates, activation energy with the Arrhenius equation, and Kc with an ICE table, all with step-by-step working.