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⚗️ Chemistry Course
⏱️ Chemical Kinetics · Lesson 1 of 5

Reaction Rate & Rate Laws

How fast a reaction runs, what speeds it up, and why the recipe alone never tells you the answer.

High schoolIntro Gen ChemUni Year 1
⏱️ About 20 min
Reaction Rate & Rate Laws — illustration
Illustrative image (AI-generated) — decorative only.

Iron rusts over years; a firework reacts in milliseconds. Same kind of process — chemistry rearranging atoms — but wildly different speeds. Kinetics is the study of that speed: what sets it, and how a single number, the rate constant, lets you predict it.

💡
The big idea: The rate of a reaction depends on the concentrations of the reactants, captured by a rate law of the form rate = k[A]^m[B]^n. The exponents (the orders) are measured in the lab — you cannot read them off the balanced equation.
🎯 By the end, you'll be able to
  • Express reaction rate in terms of how fast a concentration changes over time
  • Write a rate law and identify the rate constant k and the reaction orders
  • Explain why orders are found by experiment, not from the balanced coefficients
  • Use the method of initial rates to find orders and the value of k
📎 Helpful to know first

What 'rate' actually measures

The rate of a reaction is how quickly a concentration changes with time. As reactants are consumed their concentration falls; as products form theirs rises. We usually track one species and divide by its coefficient so everyone agrees on a single number.

For a reaction aA → cC, the rate is the fall in [A] per second (or the rise in [C]), each scaled by its coefficient. The units are almost always molarity per second (mol L⁻¹ s⁻¹, or M/s).

\[ \text{rate} = -\frac{1}{a}\frac{\Delta[\text{A}]}{\Delta t} = +\frac{1}{c}\frac{\Delta[\text{C}]}{\Delta t} \]
The minus sign makes the reactant's rate positive (its concentration is dropping). Dividing by the coefficient gives one shared rate for the whole reaction.

The rate law: what the rate depends on

Experiments show that rate depends on how concentrated the reactants are. The relationship is the rate law:

\[ \text{rate} = k\,[\text{A}]^{m}\,[\text{B}]^{n} \]
k is the rate constant; m and n are the reaction orders in A and B. The overall order is m + n.

The exponents are the reaction orders

The exponent m is the order with respect to A, n is the order with respect to B, and their sum is the overall order. If a reaction is first order in A, doubling [A] doubles the rate. If it is second order in A, doubling [A] quadruples the rate (2² = 4). Zero order in A means [A] does not affect the rate at all.

⚠️ Orders are NOT the coefficients
This is the single most common kinetics mistake. In the rate law, the exponents come from experiment, not from the balanced equation. For 2N₂O₅ → 4NO₂ + O₂ the reaction is first order overall, even though N₂O₅ has a coefficient of 2. Only measured data can tell you the orders — they reflect the reaction mechanism, which the overall equation hides.
🔑 The rate constant k
k is a proportionality constant for a given reaction at a given temperature. A big k means a fast reaction. k does not depend on concentration, but it does rise with temperature (that's the Arrhenius lesson). Its units change with the overall order, so that rate always comes out in M/s.

Finding the orders: the method of initial rates

To measure an order, change one reactant's starting concentration, hold the others fixed, and see how the initial rate responds. Double [A] and watch: if the rate doubles, order 1; if it quadruples, order 2; if nothing changes, order 0. Repeat for each reactant, then back-solve for k.

📝 Worked example: For A + B → products, three experiments give: (1) [A]=0.10, [B]=0.10, rate=2.0×10⁻³ M/s; (2) [A]=0.20, [B]=0.10, rate=4.0×10⁻³ M/s; (3) [A]=0.10, [B]=0.20, rate=8.0×10⁻³ M/s. Find the rate law and k.
  1. Compare (1)→(2): [A] doubles, [B] fixed, and the rate doubles (×2). So rate ∝ [A]¹ — first order in A.
  2. Compare (1)→(3): [B] doubles, [A] fixed, and the rate quadruples (×4 = 2²). So rate ∝ [B]² — second order in B.
  3. Rate law: rate = k[A][B]². Overall order = 1 + 2 = 3.
  4. Solve for k with experiment (1): 2.0×10⁻³ = k(0.10)(0.10)² = k(0.10)(0.010) = k(1.0×10⁻³).
  5. k = (2.0×10⁻³) / (1.0×10⁻³) = 2.0, with units M⁻²s⁻¹ (so that k·M³ = M/s).
✓ rate = k[A][B]² with k = 2.0 M⁻²s⁻¹ (overall third order).
✏️ Practice: A reaction has the rate law rate = k[A][B]². What is its OVERALL order? (Add the exponents.)
Solution
  1. Overall order = sum of the exponents in the rate law.
  2. Order in A = 1, order in B = 2.
  3. 1 + 2 = 3 (overall third order).
✏️ Practice: For rate = k[A]²[B] with k = 0.20 M⁻²s⁻¹, [A] = 0.50 M and [B] = 0.10 M, what is the rate in M/s?
M/s
Solution
  1. Substitute into rate = k[A]²[B].
  2. = 0.20 × (0.50)² × (0.10) = 0.20 × 0.25 × 0.10.
  3. = 0.0050 M/s (that is 5.0×10⁻³ M/s).

Check your understanding

1. For the reaction 2NO₂ → 2NO + O₂, how do you know the reaction order in NO₂?
Reaction orders come from experimental rate data, never from the balanced coefficients. The coefficient of 2 tells you nothing about the order.
2. A reaction is second order in X. If you triple [X], the rate multiplies by:
Second order means rate ∝ [X]². Tripling [X] multiplies the rate by 3² = 9.
3. A reaction is fast (large k) but at equilibrium yields very little product. This shows that…
Kinetics tells you how FAST a reaction goes; equilibrium tells you how FAR. A reaction can be quick yet reach equilibrium with little product — rate ≠ yield.
🔎 Related lessons
Next: Integrated Rate Laws & Half-Life Collision Theory & Activation Energy Dynamic Equilibrium & K
✅ Key takeaways
  • Rate = how fast a concentration changes with time, in M/s, scaled by coefficients.
  • The rate law is rate = k[A]^m[B]^n; overall order = m + n.
  • Orders are found by experiment — they are NOT the balanced coefficients.
  • The method of initial rates: change one concentration, watch the rate, deduce each order, then solve for k.
  • A large k means a fast reaction; k rises with temperature but not with concentration.
➡️ The rate law tells you the rate at one instant. But concentrations fall as a reaction proceeds, so the rate keeps changing. Next we integrate the rate law to predict concentration at any time — and meet the elegant idea of half-life.
Want to test yourself on this? Try the Chemistry practice test →
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