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Finding reaction order from rate data

Orders come from experiments, not from the balanced equation.

Appears onAP Chem

The common mistake

Reading the rate law straight off the balanced equation's coefficients — assuming 2A + B → C means rate = k[A]²[B].

Instead: Compare two trials where only one concentration changes. If doubling it doubles the rate, that reactant is first order; ×4 is second order; no change is zero order.

Worked example with Guide me

Doubling [A] with [B] fixed makes the rate ×4. Doubling [B] with [A] fixed leaves the rate unchanged. Write the rate law.

  1. Sugar · Doubling A gives ×4. What power of 2 is 4?
    Student · 2², so A is second order.
  2. Sugar · Doubling B changes nothing. What order is B?
    Student · Zero order.
  3. Sugar · So what's the rate law?
    Student · rate = k[A]².

Answer: rate = k[A]², overall second order.

Try 3 practice questions

  1. 1. Tripling [X] triples the rate. What order is X?

  2. 2. Doubling [Y] makes the rate ×8. What order is Y?

  3. 3. rate = k[A][B]². What is the overall order?

Quick answers

What is the most common mistake with finding reaction order from rate data?
Reading the rate law straight off the balanced equation's coefficients — assuming 2A + B → C means rate = k[A]²[B].
How do I get finding reaction order from rate data right?
Compare two trials where only one concentration changes. If doubling it doubles the rate, that reactant is first order; ×4 is second order; no change is zero order.
Can you show a worked example?
Doubling [A] with [B] fixed makes the rate ×4. Doubling [B] with [A] fixed leaves the rate unchanged. Write the rate law. rate = k[A]², overall second order.

Related concepts

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