Rate of Reaction — The Factors and the Rate Law
Iron rusts over years; a firework reacts in milliseconds. Both are chemical reactions, and chemical kinetics is the part of chemistry that explains the difference. This article covers what "rate" precisely means, the factors that change it, and — the part exams actually test — how a rate law is found from data rather than guessed from the equation.
Defining the rate properly
Rate of reaction is the change in concentration of a reactant or product per unit time. Because reactants disappear, their concentration change is negative, and a minus sign is put in front so the rate itself stays positive. For a general reaction aA + bB → cC + dD:
Dividing by the stoichiometric coefficient is what makes the rate a single number for the whole reaction instead of four different numbers. The usual unit is mol L⁻¹ s⁻¹ (often written M s⁻¹).
Average rate uses a measured interval, Δ[A]/Δt. Instantaneous rate is the value at one instant — the slope of the tangent to a concentration–time curve. Initial rate is the instantaneous rate at t = 0, and it is the one used to determine order because no product has yet built up to interfere.
Worked example 1 — relating the rates of different species
For N₂ + 3H₂ → 2NH₃, ammonia is forming at 2.4 × 10⁻³ mol L⁻¹ s⁻¹. Find the rate of the reaction and the rate at which hydrogen is used up.
Rate of reaction = (1/2) × d[NH₃]/dt = (1/2)(2.4 × 10⁻³) = 1.2 × 10⁻³ mol L⁻¹ s⁻¹
−d[H₂]/dt = 3 × rate = 3 × 1.2 × 10⁻³ = 3.6 × 10⁻³ mol L⁻¹ s⁻¹
−d[N₂]/dt = 1 × rate = 1.2 × 10⁻³ mol L⁻¹ s⁻¹. Hydrogen disappears three times as fast as nitrogen, exactly as the equation says.
The five factors that change a rate
- Concentration (or pressure for gases). More particles per unit volume means more collisions per second. How strongly the rate responds is the order, and that is measured, not assumed.
- Temperature. The biggest effect. It does not simply make molecules move a little faster — it sharply increases the fraction of collisions with energy above the activation energy.
- Catalyst. Provides an alternative path with a lower activation energy. It speeds up the forward and reverse reactions equally, so it changes how fast equilibrium is reached, never the position of equilibrium and never ΔH.
- Surface area of a solid. Powdered calcium carbonate reacts with acid much faster than a single lump, because reaction happens only at the exposed surface.
- Radiation, for photochemical reactions. Light supplies the energy that starts reactions such as the chlorination of methane.
Note what is not on the list: the amount of a pure solid or a pure liquid solvent. Their "concentration" is fixed by their density, so they do not appear in the rate law.
The rate law — and why you cannot read it off the equation
Here k is the rate constant, m is the order with respect to A and n the order with respect to B. m and n are found by experiment. They are equal to the coefficients only for a single-step (elementary) reaction. A multi-step reaction runs at the pace of its slowest step, so the rate law reflects that step, not the balanced overall equation. Orders can be zero, fractional, or even negative.
Worked example 2 — finding the order from initial-rate data
For A + B → products, three experiments at the same temperature give:
| Experiment | [A] / M | [B] / M | Initial rate / M s⁻¹ |
|---|---|---|---|
| 1 | 0.10 | 0.10 | 2.0 × 10⁻³ |
| 2 | 0.20 | 0.10 | 8.0 × 10⁻³ |
| 3 | 0.10 | 0.20 | 4.0 × 10⁻³ |
Order in A: compare 1 and 2, where [B] is held constant. [A] doubles (0.10 → 0.20) and the rate goes 2.0 × 10⁻³ → 8.0 × 10⁻³, a factor of 4. Since 2m = 4, m = 2.
Order in B: compare 1 and 3, where [A] is held constant. [B] doubles and the rate doubles: 2n = 2, so n = 1.
Rate law: Rate = k[A]²[B], overall order = 2 + 1 = 3.
Rate constant from experiment 1:
k = rate / ([A]²[B]) = (2.0 × 10⁻³) / ((0.10)² × 0.10)
(0.10)² = 0.010; 0.010 × 0.10 = 1.0 × 10⁻³
k = (2.0 × 10⁻³) / (1.0 × 10⁻³) = 2.0 L² mol⁻² s⁻¹
Cross-check with experiment 3: k = (4.0 × 10⁻³)/((0.10)² × 0.20) = (4.0 × 10⁻³)/(2.0 × 10⁻³) = 2.0 — the same value, so the rate law is consistent.
Units of k — a free clue to the order
The units of the rate constant are fixed by the overall order, because the rate must always come out in mol L⁻¹ s⁻¹. This makes the units a useful check: if a question gives k in s⁻¹, the reaction is first order, whatever the equation looks like.
| Overall order | Units of k | Rate law |
|---|---|---|
| 0 | mol L⁻¹ s⁻¹ | Rate = k |
| 1 | s⁻¹ | Rate = k[A] |
| 2 | L mol⁻¹ s⁻¹ | Rate = k[A]² or k[A][B] |
| 3 | L² mol⁻² s⁻¹ | Rate = k[A]²[B] |
The pattern is (mol L⁻¹)1−order s⁻¹. For a first-order reaction only, the half-life is independent of concentration: t½ = 0.693 / k.
Worked example 3 — why 10 degrees matters so much
The temperature dependence sits in k, through the Arrhenius equation k = A e−Ea/RT. In two-temperature form:
Take Ea = 50.0 kJ mol⁻¹ and warm the reaction from 300 K to 310 K. R = 8.314 J K⁻¹ mol⁻¹.
1/300 = 3.3333 × 10⁻³ · 1/310 = 3.2258 × 10⁻³
difference = 1.0753 × 10⁻⁴ K⁻¹
Ea/R = 50000 / 8.314 = 6014
ln(k₂/k₁) = 6014 × 1.0753 × 10⁻⁴ = 0.6467
k₂/k₁ = e0.6467 = 1.9
A rise of just 10 K nearly doubles the rate constant. This is where the old rule of thumb "rate roughly doubles for every 10 °C" comes from — but treat it strictly as a rough guide, not a law. The factor depends on Ea and on the temperature range; a reaction with a much smaller activation energy responds far less.
Common mistakes that cost marks
- Reading the order off the balanced equation. Only legitimate for an elementary step. Otherwise the order must come from data.
- Confusing rate with rate constant. Rate falls as reactants are used up; k does not change with concentration at all. k changes only with temperature (and with a catalyst).
- Dropping the minus sign or the coefficient in the rate definition, which makes the rate negative or the wrong size by a factor of 2 or 3.
- Quoting k without units. Units carry marks and prove you understand the order.
- Saying a catalyst increases the yield. It does not shift the equilibrium; it only gets you there sooner.
- Putting a pure solid or the solvent into the rate law. Their concentration is effectively constant and is absorbed into k.
- Using t½ = 0.693/k for every reaction. That expression is for first order only.
Where kinetics appears in exams
| Exam | Typical question |
|---|---|
| CBSE / ICSE Class 12 | Define rate and order, derive the first-order integrated equation, factors affecting rate |
| NEET / JEE | Order from initial-rate tables, units of k, half-life problems |
| IIT-JAM / CUET-PG | Integrated rate laws, Arrhenius plots, activation-energy calculations |
| GATE / CSIR-NET | Mechanism and rate-determining step, steady-state approximation, enzyme kinetics |
Try the temperature calculation yourself. The Arrhenius calculator takes Ea, the two temperatures and one rate constant and returns the other, so you can reproduce the 1.9× result above and then see how the answer changes when Ea is 20 kJ mol⁻¹ instead of 50.
Open the Arrhenius Equation Calculator →Chemical kinetics rewards steady numerical practice more than memorising. ABC Chemistry teaches Class 11–12 chemistry at the Gurugram centre and online across India — details at abcchemistry.in.