Rate Constant Calculator

The barrier sits in an exponent, so a small change in it moves the rate by orders of magnitude.

Rate ratio between these temperatures4 ×

Activation energy52.9056 kJ/mol52,905.6 J/mol

Pre-exponential factor2.6715 × 10⁹in the same units as the rate constants

Fraction of molecules over the barrier at 20 °C3.7432 × 10⁻¹⁰

Effect of a further 10 degrees1.8754 ×from 40 to 50 °C

Effect of lowering the barrier by 10 kJ/mol60.51 ×what a catalyst does

This barrier is in the range where the old rule of thumb — roughly double the rate per 10 degrees — happens to hold.

The formula

k = A exp(-Ea / RT) ; ln(k2/k1) = -(Ea/R)(1/T2 - 1/T1)

The two-point form

Measuring a rate constant at two temperatures is enough to recover the activation energy, because the pre-exponential factor cancels between the two readings. That is why the two-point form is the one used in practice: A is awkward to measure directly and rarely needed.

Where the rule of thumb comes from

Chemists say a reaction roughly doubles in rate for every 10 degree rise near room temperature. Put a doubling between 298 and 308 K through the equation and it returns an activation energy of about 53 kJ/mol — which is simply a typical value for ordinary solution chemistry. The rule is a consequence of that coincidence, not a law, and it fails badly for reactions with very high or very low barriers.

The barrier dominates

The activation energy sits in an exponent, so modest changes in it produce enormous changes in rate. At room temperature, lowering the barrier by 6 kJ/mol multiplies the rate by more than ten. That is the whole business of catalysis: not making the reaction more favourable, but lowering the hill in front of it.

Balanced first, always

Every quantity on this page depends on a balanced equation. The coefficients set the molar ratio, the ratio sets the theoretical yield, and the yield is what the percentage is measured against. An unbalanced equation does not give a slightly wrong answer — it gives an answer to a different question.