IB DP Chemistry Reactivity 2.2: How fast? Rate equations, mechanisms and the Arrhenius equation – Revision Notes

These notes boil down Reactivity 2.2.6 to 2.2.13, the additional higher level end of Reactivity 2.2 (How fast?). They are matched to the IB’s Diploma Programme Chemistry guide issued in February 2023, used for exams from 2025. Everything here is HL only. The SL and HL material that comes first (Reactivity 2.2.1 to 2.2.5) is in the rates and collision theory revision notes.

Full explanations and longer worked examples are in the study guide, while the practice set holds original exam-style questions. Our IB DP chemistry hub links each unit, the checklist prints the course on one page, and our free diagnostics help you spot gaps.

Key definitions

Term	Meaning
Elementary step	One single molecular event in a mechanism: a collision, or one particle breaking apart
Mechanism	The sequence of elementary steps; the steps must add up to the overall equation
Rate-determining step (RDS)	The slowest step; it limits the overall rate
Intermediate	Made by one step and consumed by a later step; it cancels out of the overall equation
Transition state	The highest-energy arrangement within one step; bonds part-made and part-broken
Molecularity	Number of reacting particles in one elementary step: uni- (1), bi- (2), termolecular (3)
Rate equation	rate = k[A]ᵐ[B]ⁿ, found only by experiment
Order with respect to A	The exponent m on [A] in the rate equation
Overall order	m + n: add the orders for all the reactants
Rate constant, k	The constant linking rate to the concentration terms; it changes with temperature, not with concentration
Arrhenius factor, A	Term that accounts for the frequency of collisions with proper orientation

Mechanisms (2.2.6)

Method: testing a proposed mechanism

1. Add the steps. Intermediates must cancel, leaving the overall equation.
2. Find the slow step. Write a rate equation from the species that react in it.
3. If the slow step contains an intermediate made in an earlier fast step, replace the intermediate with the reactants that made it.
4. Compare with the experimental rate equation. Match = consistent (not proven). No match = reject.

Small reminder. Mechanism: L + M → LM (slow); LM + M → LM₂ (fast). Overall: L + 2M → LM₂. Intermediate: LM. Rate equation: rate = k[L][M]. Only one M is in the slow step, so the reaction is first order in M even though the equation has 2M.

RDS not the first step. If a fast step 1 makes an intermediate that is then used in a slow step 2, all the particles that went into making the intermediate show up in the rate equation.

Energy profiles for multistep reactions (2.2.7)

• One hump per elementary step.
• Peaks = transition states. Troughs between humps = intermediates.
• The RDS is the step whose transition state is highest above the reactants.
• Build a profile from data level by level: transition state = previous level + Ea of that step; next level = previous level + ΔH of that step.
• Overall ΔH = products level − reactants level.

Molecularity (2.2.8)

Molecularity	Particles in the step	Note
Unimolecular	1	One particle breaks apart or rearranges
Bimolecular	2	Most common; two particles collide
Termolecular	3	Rare: three-body collisions with enough energy and right geometry are unlikely

Molecularity applies to an elementary step. Order applies to the rate equation.

Rate equations and orders (2.2.9 and 2.2.10)

• Orders are found by experiment, never from equation coefficients.
• Only integer orders are assessed; the graphs cover orders 0, 1 and 2.
• The order in a reactant can show how many of its particles are in the RDS (counting fast steps before it).

Method: initial rates

1. Pick a pair of experiments in which just one reactant’s concentration is different.
2. Note how many times bigger that concentration is, and how many times bigger the rate is.
3. Rate factor = (concentration factor)^order. Solve for the order.
4. Do the same for every other reactant and combine the orders into a rate equation.
5. Substitute one experiment’s values to find k, with units.

Order	[A] × 2	[A] × 3	[A] × 0.5
0	rate × 1	rate × 1	rate × 1
1	rate × 2	rate × 3	rate × 0.5
2	rate × 4	rate × 9	rate × 0.25

Graph shapes

Order	Concentration–time	Rate–concentration
0	Falling straight line	Flat line, parallel to the [A] axis
1	Curve; concentration halves in equal time intervals	Rising straight line from (0, 0)
2	Curve, steep at first then flattening more	Rising curve from (0, 0), getting steeper

To go from a concentration–time graph to rate data, draw tangents at several concentrations; each gradient is the rate there.

Units of k (2.2.11)

units of k = mol dm⁻³ s⁻¹ ÷ (mol dm⁻³)^(overall order)

Units of k	Overall order	Example
mol dm⁻³ s⁻¹	zero	rate = k
s⁻¹	first	rate = k[X]
mol⁻¹ dm³ s⁻¹	second	rate = k[X]²
mol⁻² dm⁶ s⁻¹	third	rate = k[X][Y]²

• k does not change with concentration.
• k does change with temperature (and is larger with a catalyst, which lowers Ea).

Halogenoalkanes link (Reactivity 3.4). Primary + OH⁻: rate = k[RX][OH⁻], k in mol⁻¹ dm³ s⁻¹ (SN2, one step). Tertiary + OH⁻: rate = k[RX], k in s⁻¹ (SN1, carbocation intermediate).

Arrhenius equation (2.2.12 and 2.2.13)

Form	Equation
Exponential	k = A e^(−Ea/RT)
Linear	ln k = −(Ea/R)(1/T) + ln A
Two temperatures (from the linear form)	ln(k₂/k₁) = (Ea/R)(1/T₁ − 1/T₂)

According to the guide, the exponential and straight-line forms both appear in the data booklet; the two-temperature form is simply derived from the straight-line one. Use R = 8.31 J K⁻¹ mol⁻¹, T in kelvin and Ea in J mol⁻¹ inside the equation.

Graph: plot 1/T horizontally and ln k vertically.

• Gradient = −Ea/R, so Ea = −gradient × R.
• y-intercept = ln A, so A = e^(intercept).
• A has the same units as k.

Method: Ea and A from a table of k and T

1. Convert each T to 1/T (K⁻¹) and each k to ln k.
2. Put 1/T on the x-axis and ln k on the y-axis; draw the straight line that fits the points best.
3. Measure its gradient with two points far apart on your line (not table values, if the points scatter).
4. Multiply the gradient by −8.31 to get Ea in J mol⁻¹, then divide by 1000 for kJ mol⁻¹.
5. Read any point off the line and use ln A = ln k − gradient × (1/T); then A = e^(ln A).

Qualitative points

• Higher T gives larger k, and the rise is roughly exponential.
• Larger Ea gives a steeper line, so k is more sensitive to temperature.
• A reflects collision frequency and orientation; e^(−Ea/RT) is the fraction of collisions with at least Ea.

Must-know distinctions

• Intermediate vs transition state: trough vs peak; can be detected vs cannot be isolated.
• Order vs molecularity: experimental, for the rate equation vs a count of particles in one step.
• Rate vs rate constant: rate changes with concentration; k does not.
• Order vs stoichiometry: coefficients in the overall equation say nothing about orders.
• Gradient vs Ea: the gradient is −Ea/R, not Ea.

Quick self-test

1. State the units of k for a reaction that is third order overall.
2. rate = k[A][B]². By what factor does the rate change if [A] is tripled and [B] is doubled?
3. A first-order reaction has rate 6.0 × 10⁻⁴ mol dm⁻³ s⁻¹ when [A] = 0.20 mol dm⁻³. Calculate k.
4. The best-fit line of ln k versus 1/T has a gradient of −6.00 × 10³ K. Calculate Ea in kJ mol⁻¹.
5. The same plot has y-intercept 25.0. Calculate A.
6. k = 2.0 × 10⁻³ s⁻¹ at 300 K and 8.0 × 10⁻³ s⁻¹ at 320 K. Calculate Ea.
7. State the molecularity of the step Cl + CH₄ → HCl + CH₃.
8. Mechanism: L + M → LM (slow); LM + M → LM₂ (fast). Identify the intermediate and write the rate equation.
9. rate = k[A]², k = 0.25 mol⁻¹ dm³ s⁻¹. Calculate the rate when [A] = 0.040 mol dm⁻³.
10. Describe the concentration–time graph for a zero-order reactant.

Answers

1. mol⁻² dm⁶ s⁻¹ (third order overall).
2. 3 × 2² = × 12.
3. k = 6.0 × 10⁻⁴ ÷ 0.20 = 3.0 × 10⁻³ s⁻¹.
4. Ea = 6.00 × 10³ × 8.31 = 4.99 × 10⁴ J mol⁻¹ = 49.9 kJ mol⁻¹.
5. A = e^25.0 = 7.2 × 10¹⁰ (same units as k).
6. ln(8.0 × 10⁻³ ÷ 2.0 × 10⁻³) = ln 4 = 1.386; 1/300 − 1/320 = 2.083 × 10⁻⁴ K⁻¹; Ea = 8.31 × 1.386 ÷ 2.083 × 10⁻⁴ = 5.53 × 10⁴ J mol⁻¹ = 55.3 kJ mol⁻¹.
7. Bimolecular (two particles).
8. Intermediate LM; rate = k[L][M].
9. 0.25 × 0.040² = 4.0 × 10⁻⁴ mol dm⁻³ s⁻¹.
10. A straight line sloping down with constant gradient, until the reactant runs out.

Where marks are usually lost

• Writing the rate equation from the coefficients of the balanced equation.
• Forgetting that a fast equilibrium before the slow step brings extra particles into the rate equation.
• Labelling an intermediate as a transition state, or drawing it at a peak.
• Rejecting a mechanism on kinetics alone without checking that its steps add up, or vice versa.
• Giving k with no units, or with units that do not fit the overall order.
• Using T in °C in the Arrhenius equation.
• Mixing kJ and J: Ea in kJ mol⁻¹ with R in J K⁻¹ mol⁻¹.
• Taking the gradient of the Arrhenius plot as Ea, or losing its minus sign.
• Using data points from the table rather than from the best-fit line when the points scatter.
• Saying a termolecular step is impossible: it is possible but rare.

Official syllabus

Source document: the IB’s Chemistry guide for the Diploma Programme (International Baccalaureate, February 2023; first assessment 2025). These notes cover Reactivity 2.2 understandings 2.2.6 to 2.2.13, all of them HL only.