- Define and explain Rates of Reaction in your own words
- Use key terms such as dynamic equilibrium accurately
- Apply what you have learned to new examples and questions
- Avoid the common mistakes learners make with this topic
This lesson focuses on Rates of Reaction: use collision theory to explain how reactions speed up.
Use collision theory to explain how reactions speed up.
Key ideas
Rates depend on successful collisions
For particles to react they must collide with enough energy (the activation energy) and in the correct orientation. Raising the temperature increases both the collision frequency and, more importantly, the fraction of collisions with energy at or above the activation energy — which is why rates roughly double for each 10 °C rise. Catalysts speed reactions by providing an alternative route with lower activation energy, without being used up.
Equilibria respond to change
At dynamic equilibrium the forward and backward rates are equal, so concentrations stay constant even though both reactions continue. Le Chatelier's principle predicts the response to change: raising the temperature shifts equilibrium in the endothermic direction, and raising pressure favours the side with fewer moles of gas. Kc quantifies the position: a large Kc means the equilibrium lies far to the right.
Key term — dynamic equilibrium: The state where forward and backward reactions occur at equal rates in a closed system, so concentrations stay constant.
Explain why increasing temperature increases the rate of reaction.
Particles move faster, colliding more often, but the key effect is that many more collisions have energy at or above the activation energy, so far more collisions are successful.
Answer: Particles move faster, colliding more often, but the key effect is that many more collisions have energy at or above the activation energy, so far more collisions are successful.
- Saying a catalyst changes the position of equilibrium A catalyst speeds up both forward and backward reactions equally, so equilibrium is reached faster but Kc and the final concentrations do not change.
- Forgetting the minus sign on exothermic ΔH Exothermic enthalpy changes are always negative — heat leaving the system. Writing +209 kJ/mol for a combustion is simply wrong, however good the arithmetic.
Practice
High pressure shifts equilibrium to the right (fewer moles of gas: 4 → 2), raising yield. Low temperature would favour the exothermic forward reaction but make it too slow, so a moderate temperature with an iron catalyst is the compromise.
ΔHc = −25.0 ÷ 0.0200 = −1250 kJ/mol.
ΔHf = (−394) + 2(−286) − (−890) = −394 − 572 + 890 = −76 kJ/mol.
Kc = 0.80² ÷ (0.50 × 1.20³) = 0.64 ÷ 0.864 = 0.741, units mol⁻² dm⁶.
Quick check
Which of these best defines "dynamic equilibrium"?
A Maxwell–Boltzmann distribution is drawn at a higher temperature. Describe two changes to the curve.
- Rates of Reaction: use collision theory to explain how reactions speed up.
- Rates depend on successful collisions: For particles to react they must collide with enough energy (the activation energy) and in the correct orientation.
- activation energy: The minimum energy particles need for a collision to cause reaction.
- Watch out for: saying a catalyst changes the position of equilibrium