Topic 5.2 Notes – Introduction to Rate Law

1. What a Rate Law Is

A rate law connects how fast a reaction happens to the concentrations of its reactants.

General form:

$$\text{Rate} = k[A]^m[B]^n$$

• Rate = change in concentration over time (M/s)
• k = rate constant (depends on temperature)
• [A], [B] = molar concentrations
• m, n = reaction orders (found experimentally)

The key word is proportional. If a reactant concentration changes, the rate changes by some predictable factor based on its exponent.

One thing students mix up all the time: you cannot take the exponents from the coefficients in the balanced equation (unless it’s explicitly an elementary step, which AP does not assume here). Rate laws come from data, not from stoichiometry.

2. Reaction Orders and Overall Order

The exponents tell you how sensitive the rate is to each reactant.

Order with Respect to One Reactant

If the rate law contains $[A]^m$, then:

• Zero order (m = 0)
  Rate = k
  Changing [A] does nothing to the rate.
• First order (m = 1)
  Double [A] → rate doubles.
• Second order (m = 2)
  Double [A] → rate increases by 4.
• Third order (m = 3)
  Double [A] → rate increases by 8.

General idea:
If concentration changes by a factor of $x$, rate changes by $x^m$.

Here’s the pattern clearly:

Overall Reaction Order

Add the exponents.

If:
$$\text{Rate} = k[A]^{2}[B]$$

Overall order = 2 + 1 = 3

Overall order matters because:

• It determines units of k
• It affects how dramatically rate responds to concentration

Students often forget to add the exponents when asked for overall order. Easy point to lose.

3. Determining a Rate Law from Experimental Data

Rate laws are determined using initial rate experiments.

Chemists run trials with different starting concentrations and measure the initial rate before concentrations change significantly.

Method of Initial Rates

Imagine the general form:

$$\text{Rate} = k[A]^m[B]^n$$

Then:

1. Compare two trials where only one reactant changes.
2. See how much that concentration changed.
3. See how much the rate changed.
4. Match the pattern.

Example logic:

• If [A] doubles and rate stays the same → zero order in A.
• If [A] triples and rate increases by 9 → second order in A (since $3^{2} = 9$).

After finding all exponents, plug numbers from any one trial into the full rate law to solve for k.

A common trap on tests: students compare two trials where both reactants change. That gives you nothing useful. Always isolate one variable at a time.

4. The Rate Constant k

What k Represents

k tells you how fast the reaction is at a specific temperature.

• Larger k → faster reaction.
• If temperature increases → k increases.
• For the same reaction at different temperatures, k changes.

k does not depend on concentration. Only temperature changes it.

Units of k Depend on Overall Order

Rate always has units:

M/s

Since
$$\text{Rate} = k(\text{concentration terms})$$

k must balance the units.

Pattern:

• Zero order → k has units M/s
• First order → k has units s⁻¹
• Second order → k has units M⁻¹·s⁻¹
• Third order → k has units M⁻²·s⁻¹

General rule:

If overall order = $n$,
k has units $\text{M}^{1-n}\text{s}^{-1}$

If you’re ever unsure, write out the units and solve algebraically. That always works.

5. Connecting Rate Laws to Experimental Measurement

Experimentally, rate is measured as:

$$\text{Rate} = \frac{\Delta[\text{reactant or product}]}{\Delta t}$$

• Reactant concentrations decrease
• Product concentrations increase

The rate law captures how those measured rates depend on concentration.

Connecting this back to collision theory:

• Higher concentration → more collisions
• Reaction order tells you how strongly those collisions influence rate

On the AP exam, they love giving you a data table and asking for:

• The order with respect to each reactant
• The overall order
• The value and units of k

Each step builds from understanding that the exponents come from comparing how rate responds to controlled concentration changes.