Reaction rate and rate expressions

IB Chemistry HL· 20 min read

1. Defining Reaction Rate★★☆☆☆HL only⏱ 5 min

📘 Definition

Reaction rate

The change in concentration of a reactant or product per unit of time, always reported as a positive value.

Example:

For the generic reaction $A \to B$ , rate is defined as: $r a t e = - \frac{Δ [ A ]}{Δ t} = \frac{Δ [ B ]}{Δ t}$

The negative sign for reactants accounts for the fact that reactant concentration decreases over time, so the final rate value remains positive. Reaction rate can be measured as an average rate over a time interval, or an instantaneous rate at a specific time.

📐 Worked Example

In the first 15 seconds of the decomposition reaction $2 N_{2} O_{5} \to 4 N O_{2} + O_{2}$ , the concentration of $N_{2} O_{5}$ decreases from 0.25 mol dm⁻³ to 0.16 mol dm⁻³. Calculate the average reaction rate with respect to $N_{2} O_{5}$ .

1
Calculate the change in concentration of $N_{2} O_{5}$ :
2
$\Delta[N_2O_5] = 0.16 - 0.25 = -0.09 \ mol \ dm^{-3}$
3
Apply the rate definition, substituting values:
4
$rate = -\frac{\Delta[N_2O_5]}{\Delta t} = -\frac{(-0.09 \ mol \ dm^{-3})}{15 \ s} = 0.006 \ mol \ dm^{-3} \ s^{-1}$

Exam tip:

Always include units for reaction rate. Examiners consistently penalize missing units in kinetics questions.

2. Rate Expressions and Reaction Order★★★☆☆HL only⏱ 6 min

📘 Definition

Rate expression (rate law)

For $a A + b B \to products$ , $rate = k [A]^{m} [B]^{n}$

An experimentally derived equation that relates reaction rate to the concentration of reactants, where $k$ is the rate constant, $m$ is the order with respect to $A$ , and $n$ is the order with respect to $B$ .

The overall order of a reaction is the sum of the individual orders with respect to each reactant ( $overall order = m + n + ...$ ).

📐 Worked Example

Experimental data for the reaction $2 N O + C l_{2} \to 2 N O C l$ shows rate is proportional to $[N O]^{2}$ and $[C l_{2}]^{1}$ . Write the rate expression and state the overall order.

1
Write the general form of the rate expression:
2
$rate = k[NO]^m[Cl_2]^n$
3
Substitute the experimentally determined orders:
4
$rate = k[NO]^2[Cl_2]^1$
5
Calculate the overall order by summing the individual orders:
6
$\text{overall order} = 2 + 1 = 3$

3. Deducing Rate Expression from Initial Rate Data★★★☆☆HL only⏱ 7 min

The most common IB exam question provides a table of initial concentrations and initial rates for multiple experimental runs. To find the order for each reactant, compare two runs where all other reactant concentrations are constant, then use the rate ratio method.

📐 Worked Example

Use the data below to deduce the rate expression for the reaction $A + B \to C$ : Run 1: $[A] = 0.10$ , $[B] = 0.10$ , rate = $2.0 \times 1 0^{- 5}$ Run 2: $[A] = 0.20$ , $[B] = 0.10$ , rate = $4.0 \times 1 0^{- 5}$ Run 3: $[A] = 0.10$ , $[B] = 0.20$ , rate = $1.6 \times 1 0^{- 4}$ (Concentrations in mol dm⁻³, rates in mol dm⁻³ s⁻¹)

1
Find order with respect to $A$ , comparing runs 1 and 2 where $[B]$ is constant:
2
$\frac{rate_2}{rate_1} = \left(\frac{[A]_2}{[A]_1}\right)^m \rightarrow \frac{4.0 \times 10^{-5}}{2.0 \times 10^{-5}} = \left(\frac{0.20}{0.10}\right)^m \rightarrow 2 = 2^m \rightarrow m=1$
3
Find order with respect to $B$ , comparing runs 1 and 3 where $[A]$ is constant:
4
$\frac{rate_3}{rate_1} = \left(\frac{[B]_3}{[B]_1}\right)^n \rightarrow \frac{1.6 \times 10^{-4}}{2.0 \times 10^{-5}} = \left(\frac{0.20}{0.10}\right)^n \rightarrow 8 = 2^n \rightarrow n=3$
5
Write the final rate expression:
6
$rate = k[A]^1[B]^3$

✓ Quick check

Test your understanding of the rate ratio method:

Doubling the concentration of reactant X does not change the initial rate of reaction. What is the order with respect to X?
- 0
- 1
- 2
- 3
Reveal answer
0 —
If doubling concentration leaves rate unchanged: $2^{m} = 1 = 2^{0}$ , so $m = 0$ . A zero order reactant does not affect the reaction rate.

4. Units of the Rate Constant $k$★★☆☆☆HL only⏱ 4 min

The units of the rate constant $k$ depend on the overall order of the reaction. You can always derive units by rearranging the rate expression to solve for $k$ , then substituting the standard units of rate (mol dm⁻³ s⁻¹) and concentration (mol dm⁻³).

📐 Worked Example

Find the units of $k$ for a reaction with rate expression $r a t e = k [A]^{2} [B]$ , overall order 3.

1
Rearrange the rate expression to isolate $k$ :
2
$k = \frac{rate}{[A]^2[B]}$
3
Substitute standard units for each quantity:
4
$\text{Units of } k = \frac{mol \ dm^{-3} \ s^{-1}}{(mol \ dm^{-3})^2 (mol \ dm^{-3})} = \frac{mol \ dm^{-3} \ s^{-1}}{mol^3 \ dm^{-9}} = mol^{-2} \ dm^6 \ s^{-1}$

Overall order of reaction	Common units of $k$
0	mol dm⁻³ s⁻¹
1	s⁻¹
2	dm³ mol⁻¹ s⁻¹
3	dm⁶ mol⁻² s⁻¹
4	dm⁹ mol⁻³ s⁻¹

Exam tip:

Checking your units for $k$ is a quick way to verify you calculated the overall order correctly. If your units don't match the expected pattern, you made a mistake in finding the order.

5. Common Pitfalls

Wrong move:

Assuming reaction orders match the stoichiometric coefficients of the balanced overall equation.

Why:

Only elementary reaction steps have orders matching stoichiometry; overall reactions are almost never elementary.

Correct move:

Always determine reaction orders from the experimental data provided in the question, never from the balanced equation alone.

Wrong move:

Getting a negative reaction rate when calculating from reactant concentration change.

Why:

Reactant concentration decreases over time, so $Δ [reactant]$ is inherently negative.

Correct move:

Always add a negative sign to the $Δ [reactant]$ term to get a positive reaction rate.

Wrong move:

Flipping the ratio of rates or concentrations when calculating reaction order.

Why:

Mixing up the order of runs leads to incorrect values for the reaction order.

Correct move:

Keep the ratio order consistent: $\frac{r a t e _{2}}{r a t e _{1}} = (\frac{[ X ] _{2}}{[ X ] _{1}})^{m}$ , with run 2 on top for both sides.

Wrong move:

Memorizing units of $k$ instead of deriving them, leading to incorrect units for higher overall orders.

Why:

Different overall orders have different units, and memorization often leads to mistakes.

Correct move:

Always derive units by rearranging the rate expression and substituting units for rate and concentration.

Wrong move:

Using average rate from the end of the reaction instead of initial rate to calculate $k$ .

Why:

Initial rate data uses the rate at $t = 0$ , when product concentration is zero and reactant concentrations are the starting values given.

Correct move:

Always use the initial rates provided in the experimental table to calculate orders and the rate constant.

6. Quick Reference Cheatsheet

Concept	Key Formula/Rule	Common Units
Average reaction rate	$r a t e = - \frac{Δ [ reactant ]}{Δ t} = \frac{Δ [ product ]}{Δ t}$	$m o l d m^{- 3} s^{- 1}$
General rate expression	$r a t e = k [A]^{m} [B]^{n}$	$k$ units depend on order
Overall reaction order	$n = m + n + ...$	Unitless
Rate constant units (n = overall order)	$(m o l d m^{- 3})^{(1 - n)} s^{- 1}$	Varies by order
Order from rate ratio	$\frac{r a t e _{2}}{r a t e _{1}} = (\frac{[ X ] _{2}}{[ X ] _{1}})^{m}$	Solve for $m$

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2025 · Paper 1
Deduce rate expression from experimental data
2024 · Paper 2
Calculate rate constant and its units
2023 · Paper 1
Find overall order of reaction

Going deeper

syllabus documentIB Chemistry HL Syllabus: KineticsTopic 5 & 16 content reference

What's Next

Reaction rate and rate expressions are the foundation for all further topics in IB HL kinetics. Mastery of these concepts is required to identify the rate-determining step in reaction mechanisms, derive rate laws from reaction mechanisms, use the Arrhenius equation to calculate activation energy, and analyze integrated rate law graphs for zero, first, and second order reactions. Rate expressions are also a common theme in extended response questions across both papers 1 and 2, so solid understanding of this sub-topic will directly contribute to your overall exam score.