Rate Laws & Reaction Order

Connecting measured rates to concentration dependence

kineticsrate lawsintro
⏱️ About 18 min

Two reactions can have the same balanced equation but completely different speed. The difference is hidden in the rate law.

💡
The big idea: A rate law is an experimentally determined relationship between reaction rate and species concentrations; its exponents define reaction orders and set the units of the rate constant.
🎯 By the end, you'll be able to
  • Write a power-law rate expression and identify k, reactant orders, and overall order
  • Compute a reaction rate from a given rate law and concentrations
  • Determine how the units of k depend on overall order
  • Distinguish stoichiometric coefficients from rate-law exponents
📎 Helpful to know first
  • Mass Transfer Case Study

What a rate law looks like

A common empirical form for many homogeneous reactions is the power-law rate law:

r = k[A]a[B]b

Here r is the reaction rate (often in mol/(L·s) or mol/(m3·s)), k is the rate constant, and a and b are the reaction orders with respect to A and B.

\[ r = k[A]^a[B]^b \]
Power-law rate law (empirical form).
🔑 Overall reaction order

The overall order for a power-law rate expression is the sum of exponents: n = a + b. It is a property of the rate law, not the balanced equation.

Rate law vs stoichiometry

The balanced equation tells you how moles relate when the reaction proceeds, but it does not automatically tell you how fast it proceeds.

For example, even if a reaction is written as A + B → products, the measured rate law might be r = k[A]2[B]. Those exponents (2 and 1) come from experiments and mechanism—not from stoichiometric coefficients.

✨ Units of k depend on the overall order

If rate r has units of concentration per time, then k must supply whatever extra units are needed so that k[A]a[B]b matches r.

For r in mol/(L·s) and concentration in mol/L, k has units (mol/L)1−(a+b)/s.

\[ \left[k\right] = \frac{(\mathrm{mol/L})^{\,1-(a+b)}}{\mathrm{s}} \]
Example unit relationship when r is in mol/(L·s) and concentrations are in mol/L.
📝 Worked example: A reaction follows r = k[A]^2[B]. At a given condition, k = 0.50 L^2/(mol^2·s), [A] = 0.20 mol/L, and [B] = 0.10 mol/L. Compute r.
  1. Write the rate law: r = k[A]^2[B].
  2. Compute [A]^2 = (0.20 mol/L)^2 = 0.040 (mol/L)^2.
  3. Multiply: r = (0.50 L^2/(mol^2·s))*(0.040 (mol/L)^2)*(0.10 mol/L).
  4. Combine numbers: 0.50*0.040*0.10 = 0.0020.
  5. Check units: (L^2/mol^2·s)*(mol^3/L^3) = mol/(L·s).
✓ 0.0020 mol/(L·s)
✏️ Practice: A reaction follows r = k[A]^1[B]^2. At a condition, k = 2.0 L^2/(mol^2·s), [A] = 0.30 mol/L, and [B] = 0.20 mol/L. Compute r.
mol/(L·s)
Solution
  1. Use r = k[A][B]^2.
  2. Compute [B]^2 = (0.20)^2 = 0.040 (mol/L)^2.
  3. Multiply: r = (2.0)*(0.30)*(0.040) = 0.024.
  4. Units: (L^2/(mol^2·s))*(mol/L)*(mol^2/L^2) = mol/(L·s).

Check your understanding

1. For r = k[A]^0.5[B]^1.5, what is the overall reaction order?
Overall order is the sum of exponents: 0.5 + 1.5 = 2.0.
2. Which statement is most accurate?
Only for truly elementary steps do stoichiometry and rate-law exponents necessarily match.
3. If r has units mol/(L·s) and the overall order is 2, what are the units of k (using mol/L for concentration)?
For overall order n=2: [k] = (mol/L)^(1−2)/s = (mol/L)^(-1)/s = L/(mol·s).
🔎 Related lessons
Next: The Arrhenius Equation
✅ Key takeaways
  • Power-law rate laws often take r = k[A]^a[B]^b.
  • Reaction orders are the exponents; overall order is a + b.
  • k’s units depend on the overall order to make units consistent.
  • Stoichiometry describes material balances; the rate law is determined experimentally.
➡️ Next we connect the rate constant k to temperature using the Arrhenius equation—one of the most useful relationships in reaction engineering.
Want to test yourself on this? Try the Chemical Aptitude test →