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A rate equation such as rate = k[A]ᵐ[B]ⁿ is determined experimentally. Orders cannot generally be read from the balanced overall equation. Compare experiments in which one concentration changes while the others remain constant. If doubling A quadruples the rate, the order in A is two. Zero order means changing that reactant’s concentration has no effect under those conditions. Overall order is the sum of the individual orders, and k’s units depend on that total.
Video by Eliot Rintoul. The video’s own board label is retained; follow the specification reference and scope notes on this lesson for assessed content.
Viewing focus: Determine reaction orders from data
After watching, attempt these questions before opening the answers. You can also use the written lesson below.
After watching 1: What happens to rate if a second-order reactant concentration triples?
It increases ninefold. The concentration factor is squared.
After watching 2: What is constant in an ideal first-order decay?
Half-life. Successive halvings take equal times.
What you'll be able to do
A rate equation such as rate = k[A]ᵐ[B]ⁿ is determined experimentally. Orders cannot generally be read from the balanced overall equation. Compare experiments in which one concentration changes while the others remain constant. If doubling A quadruples the rate, the order in A is two. Zero order means changing that reactant’s concentration has no effect under those conditions. Overall order is the sum of the individual orders, and k’s units depend on that total.
A first-order concentration–time curve has a constant half-life. A zero-order concentration–time graph is linear while the rate law remains applicable. A rate–concentration graph can distinguish proportional, constant and quadratic behaviour. A proposed multistep mechanism must sum to the overall equation and agree with the observed rate law; species formed then consumed are intermediates. A catalyst is regenerated overall. Agreement supports a mechanism but does not prove it is the only possible microscopic route.
The Arrhenius relationship is . A plot of ln k against 1/T has gradient −Ea/R; temperature is in kelvin and energy is in joules if R is 8.31 J mol⁻¹ K⁻¹. Use the scale shown on the horizontal axis: a plot against 1000/T has a different numerical gradient. Keep other conditions controlled and estimate uncertainty from repeated rates. An inverse clock time is a relative-rate proxy, not automatically a rate constant.
Attempt these problems on paper before opening the worked reasoning. Explain your steps as well as giving a final answer.
Common mistakes to avoid
Key takeaways
Question 1: An ln k against 1/T plot has gradient −6000 K. Calculate activation energy.
49.9 kJ mol⁻¹.
Check the assessed scope and practical or mathematical requirements.
An external explanation or interactive activity to use alongside the lesson. It may include extension material.
Close the notes and answer from memory. Revisit any explanation you cannot yet reconstruct.
Test yourself
Answer a few questions, then review anything you missed.
Start with the quiz above to check the key ideas. You can also choose another topic from the practice library.