Reaction Kinetics Calculator

Set the reaction order, Arrhenius parameters, and temperature to see how concentration changes over time.

Starting values — Reaction Kinetics Calculator

Activation energy (kJ/mol)	50000
Temperature (K)	298
Pre-exponential factor A (s⁻¹)	10000000000000
Initial concentration (M)	1
Reaction order	First order

Change any control above and the results recompute from these.

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Sixteen degrees makes food spoil four times faster 🖖

The exponential term in the Arrhenius equation is a Boltzmann factor: it estimates the fraction of molecular collisions energetic enough to cross the activation barrier. Because the high-energy tail of the Maxwell-Boltzmann distribution falls away exponentially, a modest temperature change can produce a large change in rate. Take this tool's food-spoilage preset, with an activation energy of 60 kJ/mol: move it from a 4 °C fridge to a 20 °C room and the temperature rises by only 16 K, but the rate constant it reports goes up by a factor of 4.1. That same sensitivity is why refrigeration slows spoilage, why fevers shift enzyme rates, and why high-barrier industrial reactions need heat or a catalyst.

Reaction order sets the pace 🖖

The rate law rate = k·[A]^m tells you how a reaction's speed depends on how much reactant is left. Zero, first, and second order give very different concentration curves: a zero-order reaction consumes reactant at a steady rate until it runs out, while a first-order reaction slows continuously yet keeps a half-life that never changes. Switch the order in this tool and watch the [A]-versus-time curve change shape.

Sobering up is zero-order 🖖

Most reactions slow down as the reactant runs low, but your body clears alcohol at a nearly constant rate — about 0.015 %BAC per hour — regardless of how much you drank. The enzyme alcohol dehydrogenase is fully saturated at ordinary blood-alcohol levels, so it works flat-out, exactly like the zero-order curve in this tool. That is why waiting is the only real cure: doubling the dose doubles the time, not the speed.

Problem solved in full

A first order reaction at 298 K with E a = 50 kJ/mol 5 steps
Test the rule that a reaction runs twice as fast for every 10 degrees. The state is first order, with an activation energy E_a = 50 kJ/mol, a pre-exponential factor A = 10¹³ s⁻¹, and T = 298 K.
1. A barrier height means nothing on its own; it matters only against the thermal energy available, so the two enter as a ratio. Here the barrier is about twenty times RT, and the exponential of minus that ratio is the fraction of collisions arriving with enough energy to cross.
  
  $k = A e^{- E_{a} / R T}, \frac{E _{a}}{R T} = \frac{50 000}{8.314 \times 298} = 20.181$
2. One collision in 5.8 × 10⁸ clears the barrier. A counts how often the attempt is made, so the rate constant is a frequency multiplied by a probability — which is why a sluggish collision rate and a low barrier can produce the same k as the opposite pair.
  
  $e^{- 20.181} = 1.720 \times 1 0^{- 9} ⟹ k = 1 0^{13} \times 1.720 \times 1 0^{- 9} = 1.720 \times 1 0^{4} s^{- 1}$
3. First order means the fraction consumed per second does not depend on how much is left, so the half-life is fixed by k alone and by nothing about the starting concentration.
  
  $t_{1/2} = \frac{ln 2}{k} = \frac{0.693147}{1.720 \times 1 0 ^{4}} = 4.030 \times 1 0^{- 5} s$
4. Q₁₀ is the same expression evaluated at two temperatures and divided, so A cancels — collision frequency plays no part in temperature sensitivity at all. Combining the two reciprocal temperatures leaves one exponential with T(T + 10) sitting in the denominator.
  
  $Q_{10} = \frac{k ( T + 10 )}{k ( T )} = exp [\frac{10 E _{a}}{R T ( T + 10 )}] = exp [\frac{500 000}{8.314 \times 298 \times 308}] = e^{0.6552} = 1.93$
5. That denominator is the finding: Q₁₀ falls roughly as 1/T², so one unchanging molecule has a different Q₁₀ at every temperature. Running the formula backwards gives the barrier height that would make the doubling exact.
  
  $Q_{10} (277) = 2.13, Q_{10} (298) = 1.93, Q_{10} (373) = 1.52; Q_{10} = 2 ⟺ E_{a} = \frac{R T ( T + 10 ) ln 2}{10} = 52.9 kJ/mol$
Answer

The tool prints k = 1.720 × 10⁴ s⁻¹, t½ = 4.030 × 10⁻⁵ s and Q₁₀ = 1.93×. The rule of thumb survives here by arithmetic accident: 50 kJ/mol sits just below the 52.9 kJ/mol that makes Q₁₀ exactly 2 at room temperature, and nothing in the chemistry pins activation energies to that value. The rule also carries a temperature that is almost never quoted with it — this same reaction gives 2.13 in a 4 °C fridge and 1.52 in boiling water. So a Q₁₀ measured in cold storage cannot be carried into a fermenter, and the error has a known sign: a cold Q₁₀ always overstates the sensitivity at high temperature. Q₁₀ is a property of a reaction at a temperature, not a property of the reaction.

$Q_{10} = exp [\frac{10 E _{a}}{R T ( T + 10 )}] \propto e^{c / T^{2}}$

References (2)

The temperature dependence the tool computes: S. Arrhenius, "Über die Reaktionsgeschwindigkeit bei der Inversion von Rohrzucker durch Säuren." Zeitschrift für Physikalische Chemie 4U(1), 226–248, 1889.
The Boltzmann factor and the high-energy tail behind it: P. Atkins and J. de Paula, Atkins’ Physical Chemistry, 10th ed., ch. 20–21. Oxford University Press, 2014. ISBN 978-0-19-969740-3.

By the 42Math team · Updated 2026-08-17

Reviewed against published sources. Corrections welcome. Editorial policy · Corrections

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Example problems

Body temperature - At body temperature a 10 K rise multiplies the rate by 1.83× — the Q10 rule of thumb, computed rather than assumed.
High energy barrier - A tall barrier at 500 K still gives a half-life of 0.2386 s, with a Q10 of 1.76×.
Food spoilage - Refrigeration is this number: a Q10 of 2.48× means every 10 K costs you more than half the shelf life.
Fast reaction - A low barrier makes the temperature sensitivity almost flat, Q10 1.30× — the reaction is already fast enough not to care.