FAD1018 Chemical Kinetics — Intuition Note
1. Thermochemistry vs Kinetics — Two Different Questions
| Thermochemistry | Kinetics | |
|---|---|---|
| Question | Can it happen? | How fast does it happen? |
| Focus | Energy (enthalpy, entropy) | Movement and collisions |
| Controls | $\Delta G$, $\Delta H$, $\Delta S$ | $k$, $E_a$, temperature, concentration |
Diamond turning into graphite is thermodynamically spontaneous ($\Delta G < 0$). But you'd never see it happen — the activation energy is enormous. Kinetics is the gatekeeper: thermodynamics says whether you can get through the door, kinetics says how fast you actually do.
2. Reaction Rate — What Does "Fast" Even Mean?
Rate = change in concentration per unit time. Units: M s⁻¹.
The steeper the concentration-time curve, the faster the reaction. Three ways to measure:
| Type | What it is | How to get it |
|---|---|---|
| Average | Rate over an interval | Slope of the chord between two points |
| Instantaneous | Rate at one instant | Slope of the tangent at that point |
| Initial | Rate at $t = 0$ | Slope of the tangent at the very start |
Initial rate is the cleanest measurement — no products have accumulated yet to complicate things (no reverse reaction, no intermediate buildup).
3. Differential Rate Equation — Relating Everyone's Rates
For $aA + bB \rightarrow cC + dD$:
$$\text{Rate} = -\frac{1}{a}\frac{d[A]}{dt} = -\frac{1}{b}\frac{d[B]}{dt} = +\frac{1}{c}\frac{d[C]}{dt} = +\frac{1}{d}\frac{d[D]}{dt}$$
The intuition:
- Reactants disappear → negative sign ($-\frac{d[A]}{dt}$)
- Products appear → positive sign ($+\frac{d[C]}{dt}$)
- The $1/a$, $1/b$, etc. are just scaling factors — if 2 moles of A are consumed for every 1 mole of B, then A disappears twice as fast as B, so we divide by 2 to make them equal
[!tip] Think of it like a factory assembly line If 3 bolts (A) and 2 nuts (B) make 1 widget (C), then bolts are used up 1.5× faster than nuts. The $1/a$ and $1/b$ normalise these different speeds to a single "reaction rate" that means the same thing for every species.
4. Rate Law — The Central Concept
$$\text{Rate} = k[A]^m[B]^n$$
The single most important thing the lecturer drilled:
"Takde kaitan dengan coefficient" — The exponents $m$ and $n$ have nothing to do with the stoichiometric coefficients $a$ and $b$.
The rate law is experimental, not theoretical. You cannot look at a balanced equation and write the rate law. You must measure it.
The Three Questions the Rate Law Answers
- $k$ (the rate constant): How inherently fast is this reaction? Big $k$ = fast reaction (at this temperature).
- $m$, $n$ (the orders): How sensitive is the rate to each reactant's concentration?
- $m + n$ (overall order): What's the combined sensitivity?
What Does the Order Actually Mean Physically?
| Order | Meaning | Doubling [A] makes rate... |
|---|---|---|
| 0 | Changing [A] does nothing to the rate | ×1 (no change) |
| 1 | Rate is directly proportional to [A] | ×2 |
| 2 | Rate is proportional to [A]² | ×4 |
| 3 | Rate is proportional to [A]³ | ×8 |
Zero order means the reactant isn't involved in the rate-determining step (or the step is saturated — like an enzyme active site that's full). Even if $H^+$ has a coefficient of 2 in $H_2O_2 + 3I^- + 2H^+ \rightarrow I_3^- + 2H_2O$, its order is zero — it doesn't appear in the rate law at all.
5. Method of Initial Rates — How We Find Orders Experimentally
The logic: Hold everything constant except one reactant, see how the rate changes.
| If you... | And rate... | Then order = |
|---|---|---|
| Double [A] | Stays the same | 0 |
| Double [A] | Doubles | 1 |
| Double [A] | Quadruples | 2 |
| Double [A] | 8× | 3 |
General rule: If [A] is multiplied by $f$ and rate changes by $f^n$, then $n$ is the order.
The trick is picking the right pair of experiments. Find two where only one concentration changes. If no perfect pair exists, set up a ratio: $\frac{\text{Rate}_2}{\text{Rate}_1} = \left(\frac{[A]_2}{[A]_1}\right)^m$ and solve for $m$.
6. The Rate Constant $k$ — What Controls It
$k$ is the proportionality constant between rate and concentration. Its unit changes with overall order:
| Overall order | Unit of $k$ |
|---|---|
| 0 | M s⁻¹ |
| 1 | s⁻¹ |
| 2 | M⁻¹ s⁻¹ |
A first-order $k$ has units of "per time" because Rate = $k[A]$ means $k = \text{Rate}/[A] = (\text{M s}^{-1})/\text{M} = \text{s}^{-1}$.
$k$ depends on:
- Temperature — higher $T$ → larger $k$ (Arrhenius)
- Activation energy — higher $E_a$ → smaller $k$
- Catalyst — lowers $E_a$ → larger $k$
7. Integrated Rate Laws — Why We Integrate
The rate law tells you the instantaneous rate. But in the lab, you measure concentration over time, not instantaneous rate. Integrated rate laws connect the two.
The Three Linear Plots — Your Order-Detection Toolkit
| Order | Equation | Linear plot | Slope | $y$-intercept |
|---|---|---|---|---|
| 0 | $[A]_t = -kt + [A]_0$ | $[A]$ vs $t$ | $-k$ | $[A]_0$ |
| 1 | $\ln[A]_t = -kt + \ln[A]_0$ | $\ln[A]$ vs $t$ | $-k$ | $\ln[A]_0$ |
| 2 | $\frac{1}{[A]_t} = kt + \frac{1}{[A]_0}$ | $1/[A]$ vs $t$ | $+k$ | $1/[A]_0$ |
[!tip] The only one with a positive slope is second order. If your $1/[A]$ vs $t$ plot goes up, it's second order.
The Core Intuition — What Each Order "Looks Like"
Imagine watching concentration drop over time:
- Zero order: Linear decline. Constant rate. Like a leaky bucket with a fixed hole — water drains at the same speed regardless of how full the bucket is.
- First order: Exponential decay. The rate drops as the reactant is used up. Like radioactive decay — a fixed fraction decays per unit time, not a fixed amount.
- Second order: Faster-than-exponential decay at the start, then a long tail. The rate depends on two molecules finding each other — when there's lots, they find each other easily; when there's little, it gets very slow.
8. Half-Life — What It Reveals About Order
Half-life ($t_{1/2}$) is the time for [A] to fall to half its value.
| Order | $t_{1/2}$ formula | What happens to $t_{1/2}$ as reaction proceeds? |
|---|---|---|
| 0 | $\frac{[A]_0}{2k}$ | Decreases (each successive half-life is shorter) |
| 1 | $\frac{0.693}{k}$ | Constant (independent of concentration) |
| 2 | $\frac{1}{k[A]_0}$ | Increases (each successive half-life is longer) |
[!tip] Memory aid from the lecturer
- 0th: Masa ↓ (half-life gets shorter)
- 1st: Masa sama (half-life stays the same)
- 2nd: Masa ↑ (half-life gets longer)
Why First-Order Half-Life Is Constant
Radioactive decay is first order. No matter how much radioactive material you start with, the time for half of it to decay is always the same. That's why carbon dating works — the half-life is a fixed property of the isotope.
For zero order, the rate is constant (doesn't depend on [A]), so less concentrated samples hit half in less time. For second order, the rate drops as [A] drops, so each half-life takes longer.
9. Collision Theory — What Needs to Happen for a Reaction
Molecules must collide to react, but not all collisions work. For an effective collision:
- Sufficient energy — the collision must have kinetic energy $\geq$ activation energy ($E_a$)
- Proper orientation — the molecules must hit each other the right way
Think of it like a lock and key. You can smash the key into the lock all day with the wrong orientation — it won't open. But with the right orientation and enough force, it clicks.
10. Activation Energy — The Energy Hill
$E_a$ is the minimum energy that colliding molecules must have to react. It's the height of the hill between reactants and products.
- Low $E_a$: Many molecules have enough energy → fast reaction
- High $E_a$: Few molecules have enough energy → slow reaction
Energy Profile Diagrams
| Exothermic | Endothermic | |
|---|---|---|
| $\Delta H$ | Negative (reactants higher) | Positive (products higher) |
| $E_{a,\text{reverse}}$ | $E_{a,\text{forward}} + |\Delta H|$ | $E_{a,\text{forward}} - \Delta H$ |
| Appearance | Reactants higher than products | Reactants lower than products |
The activated complex (transition state) sits at the peak — it's not a stable molecule, it's a fleeting configuration where old bonds are partially broken and new ones partially formed.
[!warning] Always appears in exams ("Selalu keluar exam")
11. Factors That Affect Rate — The Four Dials
| Factor | Effect on rate | Why? |
|---|---|---|
| ↑ Concentration | ↑ | More particles per volume → more collisions |
| ↑ Pressure (gases) | ↑ | Same as concentration — squeezes gas into smaller space |
| ↓ Particle size (solids) | ↑ | More surface area → more collision sites |
| ↑ Temperature | ↑ | More kinetic energy → more collisions above $E_a$ |
| Catalyst | ↑ | Lowers $E_a$ → more collisions are effective |
Surface Area Intuition
Solids are immobile ("tak boleh bergerak"). Only the surface is available for reaction. If you break a rock into dust, the same mass now has vastly more surface area exposed. That's why powdered zinc reacts faster with acid than a zinc pellet.
12. Maxwell-Boltzmann — Why Temperature Matters So Much
The Maxwell-Boltzmann distribution shows the spread of molecular kinetic energies at a given temperature.
Key insight: At higher temperature:
- The curve shifts right (higher average energy)
- The curve flattens (broader spread)
- The fraction of molecules with energy $\geq E_a$ increases dramatically
This is why a 10°C increase can double or triple a reaction rate. The increase in the "above $E_a$" area is not linear — it's exponential (that's what the $e^{-E_a/RT}$ in Arrhenius is capturing).
13. Arrhenius Equation — Putting Numbers to It
$$k = Ae^{-E_a/RT}$$
| Symbol | Meaning | Intuition |
|---|---|---|
| $k$ | Rate constant | How fast the reaction actually goes |
| $A$ | Pre-exponential factor / frequency factor | The rate if every collision had enough energy ($E_a = 0$ or $T \to \infty$) |
| $e^{-E_a/RT}$ | Boltzmann factor — fraction of molecules with $E \geq E_a$ | The bottleneck — most collisions fail |
| $E_a$ | Activation energy | Height of the energy hill |
| $R$ | Gas constant (8.314 J mol⁻¹ K⁻¹) | Unit conversion |
| $T$ | Temperature (K) | How much thermal energy is available |
13.1 What Is $A$? Deep Intuition
$A$ is the pre-exponential factor (also called the frequency factor or Arrhenius constant). It packages together everything about a reaction except the energy barrier.
From collision theory:
$$A = p \times Z$$
| Component | Symbol | What it captures |
|---|---|---|
| Collision frequency | $Z$ | How many collisions happen per second per unit volume. Depends on molecular size, speed ($\propto \sqrt{T}$), and concentration. |
| Steric factor | $p$ | The fraction of collisions with the right orientation for bonds to form. A number between 0 and 1. |
So $A$ answers: "If energy were no object, how fast could this reaction possibly go?"
- Big $A$ → molecules collide frequently and/or orient favourably
- Small $A$ → molecules rarely meet, or need a very precise collision angle
Real-world anchor: Iodine atom recombination $I + I \to I_2$ in the gas phase has $A \approx 10^{10}\ \text{M}^{-1}\text{s}^{-1}$. Every collision is effective — two atoms just need to touch, no orientation needed. That's the upper limit for bimolecular reactions.
Where does $A$ come from experimentally?
From the linearised Arrhenius plot ($\ln k$ vs $1/T$):
$$\ln k = -\frac{E_a}{R}\left(\frac{1}{T}\right) + \ln A$$
- Slope → $-E_a/R$ (gives $E_a$)
- $y$-intercept → $\ln A$ (gives $A$)
$A$ is the value of $k$ when $1/T = 0$, i.e., $T \to \infty$ — where $e^{-E_a/RT} \to 1$ and every collision overcomes the barrier.
Does $A$ depend on temperature?
Strictly, yes — mildly. Collision frequency $Z$ scales with $\sqrt{T}$ (molecules move faster at higher $T$). But $e^{-E_a/RT}$ changes exponentially with $T$, so the $\sqrt{T}$ dependence of $A$ is usually negligible. That's why we treat $A$ as constant when plotting $\ln k$ vs $1/T$.
Units of $A$
$A$ has the same units as $k$ (since $e^{-E_a/RT}$ is dimensionless):
| Overall order | Unit of $k$ (and $A$) |
|---|---|
| 0 | M s⁻¹ |
| 1 | s⁻¹ |
| 2 | M⁻¹ s⁻¹ |
13.2 The Exponential Term — The Real Bottleneck
$e^{-E_a/RT}$ is the fraction of collisions that have enough energy to react.
| Condition | Value of $e^{-E_a/RT}$ | Meaning |
|---|---|---|
| $E_a = 0$ | $1$ | Every collision works (no barrier) |
| $E_a \gg RT$ | $\approx 0$ | Almost no collisions have enough energy |
| $T \to \infty$ | $\to 1$ | Temperature high enough that the barrier is irrelevant |
| $T \to 0$ | $\to 0$ | Frozen — no reactive collisions |
Key insight: For a typical reaction at room temperature, $E_a \approx 50\ \text{kJ/mol}$ and $RT \approx 2.5\ \text{kJ/mol}$, so $e^{-E_a/RT} = e^{-20} \approx 2 \times 10^{-9}$. That means only 2 in a billion collisions have enough energy. The other 999,999,998 accomplish nothing. This is why reactions are nowhere near as fast as the collision frequency would suggest.
Why Logs?
The exponential form is hard to work with, so we linearise it:
$$\ln k = -\frac{E_a}{R}\left(\frac{1}{T}\right) + \ln A$$
Plot $\ln k$ vs $1/T$ → straight line with slope $= -E_a/R$. This is how $E_a$ is determined experimentally.
Two-Point Form
$$\ln\frac{k_2}{k_1} = \frac{E_a}{R}\left(\frac{1}{T_1} - \frac{1}{T_2}\right)$$
Use this when you have rate constants at two temperatures and need $E_a$ (or vice versa).
Units trap: $E_a$ is often given in kJ/mol but $R = 8.314$ J mol⁻¹ K⁻¹. Convert: multiply kJ by 1000.
14. Reaction Mechanisms — The Steps Behind the Curtain
Most reactions don't happen in one step. They go through a sequence of elementary steps (the mechanism).
| Overall reaction | Elementary step | |
|---|---|---|
| What it is | The net equation | One actual molecular event |
| Rate law | Must be determined experimentally | Can be written from molecularity |
| Example | $2NO + O_2 \rightarrow 2NO_2$ | $2NO \rightarrow N_2O_2$ |
Intermediates are species formed in one step and consumed in the next — they don't appear in the overall equation (e.g., $N_2O_2$).
15. Rate-Determining Step — The Bottleneck
The slowest step controls the overall rate. Everything else is waiting on it.
Two cases for deriving the rate law from a mechanism:
Case 1: RDS is the first step
The rate law comes directly from the RDS. Easy.
Case 2: RDS involves an intermediate
You need to replace the intermediate using a fast equilibrium step before it.
The pattern:
- Write the rate law for the slow step (using its molecularity)
- If it contains an intermediate, use the fast equilibrium to express [intermediate] in terms of reactants
- Substitute and collect constants
Example from the lecture — Mechanism B for $2NO + 2H_2 \rightarrow N_2 + 2H_2O$:
Slow step: $N_2O_2 + H_2 \rightarrow \dots$ → Rate $= k_2[N_2O_2][H_2]$
$N_2O_2$ is an intermediate. Fast equilibrium gives $K = [N_2O_2]/[NO]^2$, so $[N_2O_2] = K[NO]^2$.
Substitute: Rate $= k_2 K [NO]^2 [H_2] = k_{obs}[NO]^2[H_2]$ — matches the experimental rate law.
[!tip] The steady-state approximation (Tutorial 2) is a more general version — it assumes [intermediate] stays constant rather than at equilibrium. For FAD1018, the equilibrium approach is usually sufficient.
16. Molecularity — How Many Molecules Meet?
| Molecularity | What happens | Rate law | How common? |
|---|---|---|---|
| Unimolecular | One molecule rearranges or breaks apart | $k[A]$ | Common (isomerisation, decomposition) |
| Bimolecular | Two molecules collide | $k[A][B]$ or $k[A]^2$ | Most common |
| Termolecular | Three molecules collide simultaneously | $k[A][B][C]$ | Very rare (three-way collision is unlikely) |
Key distinction: Molecularity is theoretical (from the mechanism, always a small integer). Reaction order is experimental (from data, can be fractional). Don't confuse them.
17. Catalysis — The Shortcut
A catalyst provides an alternative pathway with lower $E_a$. It changes the mechanism, not the thermodynamics.
| Uncatalysed | Catalysed | |
|---|---|---|
| $E_a$ | Higher | Lower |
| $\Delta H$ | Same | Same |
| Products | Same | Same |
| Mechanism | Direct path | Path through catalyst-bound intermediates |
A catalyst does not change $\Delta H$, $\Delta G$, or the equilibrium constant. It only affects how fast you get there.
Homogeneous vs Heterogeneous
| Homogeneous | Heterogeneous | |
|---|---|---|
| Phase | Same as reactants | Different (usually solid) |
| Example | $I^-$ catalysing $H_2O_2$ decomposition | Fe catalysing the Haber process |
| Separation | Hard | Easy (filter it out) |
18. Enzyme Kinetics — Nature's Catalysts
Enzymes are protein catalysts. They follow the Michaelis-Menten model:
$$E + S \rightleftharpoons ES \rightarrow E + P$$
$$v = \frac{V_{max}[S]}{K_m + [S]}$$
The Two Regimes
| Condition | Rate simplifies to | What's happening |
|---|---|---|
| $[S] \ll K_m$ | $v \approx \frac{V_{max}}{K_m}[S]$ | First order — plenty of free enzyme, rate limited by substrate availability |
| $[S] \gg K_m$ | $v \approx V_{max}$ | Zero order — enzyme is saturated, rate maxed out |
What $K_m$ and $V_{max}$ Actually Mean
- $K_m$: The $[S]$ at which $v = V_{max}/2$. Low $K_m$ = enzyme loves its substrate (high affinity, reaches half-speed at low $[S]$).
- $V_{max}$: Maximum rate when every active site is full. $V_{max} = k_{cat}[E]{total}$, where $k{cat}$ (turnover number) is how many substrates one enzyme molecule can process per second.
The Lineweaver-Burk plot ($1/v$ vs $1/[S]$) linearises the curve so you can read $K_m$ and $V_{max}$ from the intercepts. It's just the Michaelis-Menten equation flipped upside down.
Problem-type Mapping
| Type | What you're given | What to do |
|---|---|---|
| Differential rate equation | Balanced eq + one species' rate | Scale by stoichiometric coefficients |
| Rate law from data | Initial rates table | Compare experiments pairwise to find orders, then solve for $k$ |
| Integrated rate law | $k$ + $[A]_0$ + time or concentration | Pick the right order's equation, plug and solve |
| Order from graph | $[A]$ vs $t$ data | Test all three plots — which one is linear? |
| Half-life | $k$ or concentration data | Use the correct $t_{1/2}$ formula for that order |
| Arrhenius (two-point) | $k$ at two $T$ | Use $\ln(k_2/k_1) = (E_a/R)(1/T_1 - 1/T_2)$ |
| Arrhenius (graphical) | $k$ at multiple $T$ | Plot $\ln k$ vs $1/T$, slope $= -E_a/R$ |
| Mechanism → rate law | Proposed elementary steps | RDS determines rate law; substitute intermediates via fast equilibrium |
| Michaelis-Menten | $[S]$, $K_m$, $V_{max}$ | Plug into $v = V_{max}[S]/(K_m + [S])$ |
Idea Hierarchy — What to Memorise vs What to Derive
Everything in kinetics flows from a handful of axioms. Here's the dependency tree.
Must Memorize (the axioms — cannot be derived)
These are the seeds. Everything else grows from them.
- Rate definition: Rate $= -\frac{1}{a}\frac{d[A]}{dt} = ...$ — this is what "rate" means
- Rate law form: Rate $= k[A]^m[B]^n$ — empirical, orders are experimental
- Half-life definition: $t_{1/2}$ is when $[A] = [A]_0/2$
- Arrhenius equation: $k = Ae^{-E_a/RT}$ — nature's rule for how $k$ depends on $T$ and $E_a$
- Collision theory: effective collision needs (i) energy $\geq E_a$, (ii) proper orientation
- Molecularity → rate law for an elementary step: unimolecular → $k[A]$, bimolecular → $k[A][B]$, termolecular → $k[A][B][C]$
- Catalyst definition: provides alternative pathway with lower $E_a$, not consumed
- Michaelis-Menten model: $v = \frac{V_{max}[S]}{K_m + [S]}$ — the enzyme kinetics model itself
Derive (can work out, but memorise for speed)
| Derived from | What | How |
|---|---|---|
| Rate law form | Effect of doubling $[A]$ on rate: rate changes by $2^n$ | Substitute $[A] \rightarrow 2[A]$ into Rate $= k[A]^n$ |
| Rate law form | Unit of $k$: $M^{1-n}s^{-1}$ | Solve $k = \text{Rate}/[A]^n$, plug in units |
| Rate definition + calculus | Zero-order integrated: $[A]_t = -kt + [A]_0$ | Integrate $-d[A]/dt = k$ |
| Rate definition + calculus | First-order integrated: $\ln[A]_t = -kt + \ln[A]_0$ | Integrate $-d[A]/dt = k[A]$ |
| Rate definition + calculus | Second-order integrated: $1/[A]_t = kt + 1/[A]_0$ | Integrate $-d[A]/dt = k[A]^2$ |
| Integrated rate law | Linear plot for each order | Rearrange to $y = mx + c$ form |
| Integrated rate law + half-life defn | $t_{1/2}$ formulas: zero: $[A]_0/2k$, first: $\ln2/k$, second: $1/k[A]_0$ | Substitute $[A]_t = [A]_0/2$ into integrated form |
| Arrhenius equation | Two-point form: $\ln\frac{k_2}{k_1} = \frac{E_a}{R}(\frac{1}{T_1} - \frac{1}{T_2})$ | Write $\ln k = \ln A - E_a/RT$ for two temps, subtract |
| Arrhenius equation | Graphical method: plot $\ln k$ vs $1/T$, slope $= -E_a/R$ | Linearise $k = Ae^{-E_a/RT}$ |
| Energy profile diagram | $\Delta H = E_{a,f} - E_{a,r}$ | Definition: $\Delta H = E_{products} - E_{reactants}$ |
Derivable (know it can be done, but just memorise the result)
These are one-liners from the derivations above — not worth re-deriving in an exam.
| Result | Source derivation |
|---|---|
| $f = 2^n$ when $[A]$ is multiplied by $f$ | Rate law substitution |
| $k$ units: M s⁻¹ (0th), s⁻¹ (1st), M⁻¹ s⁻¹ (2nd) | Rate law unit algebra |
| First-order $t_{1/2}$ is constant | $t_{1/2} = \ln2/k$ has no $[A]_0$ |
| Zero-order $t_{1/2}$ decreases over time | $t_{1/2} = [A]_0/2k$ depends on current $[A]$ |
| Second-order $t_{1/2}$ increases over time | $t_{1/2} = 1/k[A]_0$ depends on $[A]_0$ |
| $[S] \ll K_m$ → first order in $[S]$ | Factor $K_m$ out of $V_{max}[S]/(K_m + [S])$ |
| $[S] \gg K_m$ → zero order in $[S]$ | Cancel $[S]$ from $V_{max}[S]/(K_m + [S])$ |
| Lineweaver-Burk: $1/v = (K_m/V_{max})(1/[S]) + 1/V_{max}$ | Reciprocal of Michaelis-Menten |
| Mechanism rate law with intermediate substitution | RDS rate law + equilibrium expression |
Derivation Drills (optional — practice if derivations feel shaky)
No separate drill pack exists for this. To internalise the derivations:
- Cover the "Derive" column and derive each from the "Must Memorize" row above it
- For integrated rate laws: start with $-d[A]/dt = k[A]^n$, separate variables, integrate
- For half-life: take the integrated law, set $[A]_t = [A]_0/2$, solve for $t$
- For two-point Arrhenius: write $\ln k_1 = \ln A - E_a/RT_1$ and $\ln k_2 = \ln A - E_a/RT_2$, subtract
- For mechanism rate law: start from the RDS's molecularity, then hunt for intermediates and eliminate them
[!tip] Exam reality You can derive most things from the axioms if you panic-forget a formula. But in a 2-hour exam, you don't want to be integrating $\int d[A]/[A]^2$ from scratch. Memorise the Derive column. The Derivable column is automatic once you have.