Reaction rates
A + B → C by collision: concentration and temperature set how fast product appears.
Product vs time
Readouts
What's happening
Molecules only react when they collide with enough energy. Raising concentration packs more molecules into the same space, so collisions happen more often.
Raising temperature makes each collision more likely to succeed (more molecules clear the activation energy). Both show up as a steeper product curve: fast at the start, flattening as reactants run out.
- Rate equationrate = k [A] [B]
- Arrhenius equationk ∝ e−Ea / RT
On the syllabus
- GCSE Chemistry AQA 4.6.1, Edexcel, OCRrates of reaction and collision theory
- A-Level Chemistrythe Arrhenius equation and activation energy
Challenge
Predict first: which speeds the reaction up more here: doubling the amount of A, or raising the temperature by 100 K? Run both and compare the product curves.
Raising the temperature by 100 K wins: the rate constant grows exponentially with temperature, while doubling A only doubles the collision frequency.
Learn this properly
Lessons from GCSE Combined and Separate Science on this topic: worked steps, then exam-style questions with new numbers every time.
- Rate graphs and tangentsRates of reaction and equilibriumFree
- Collision theory and the factors that change a rateRates of reaction and equilibriumPlus
- Required practical: concentration and rateRates of reaction and equilibriumPlus
See the whole courseThe full course, topic tests and mock papers come with Plus. See Plus
FAQ
What is collision theory?
Particles react only when they collide with at least the activation energy and the right orientation. Anything that makes successful collisions more frequent speeds the reaction up.
Why does temperature speed up a reaction so much?
Particles move faster and collide more often, but the bigger effect is that far more collisions have enough energy to react. The rate constant grows exponentially, k ∝ e−Ea/RT.
How does a catalyst work?
It gives the reaction a different route with a lower activation energy, so more collisions succeed. It is not used up.
How do I find the activation energy from two rate constants?
Use ln(k₂/k₁) = (Eₐ/R)(1/T₁ − 1/T₂). If k doubles between 300 K and 310 K, Eₐ = 8.314 × ln 2 / (1/300 − 1/310) = 53.6 kJ/mol. That is the old rule of thumb that 10 K doubles the rate. Pick Activation energy under Solve for to see each step.