How fast do reactions go? Why does milk spoil faster in summer? Why does a catalyst speed things up? — the study of reaction rates, mechanisms, and the factors that control them. With interactive rate constant calculator, Arrhenius calculator, animated diagrams, and 20+ real-life examples.
Before diving in, here's the entire chapter on one page. How rate connects to rate law, order, integrated equations, temperature, and catalysis.
Chemical kinetics studies reaction rates. Rate depends on concentration (rate law), temperature (Arrhenius), and catalyst. Order tells how rate depends on concentration — zero and first are most important. Half-life determines stability. Collision theory explains why only high-energy collisions react.
Rate of reaction: The change in concentration of a reactant or product per unit time.
For reactant R: Rate = −Δ[R]/Δt (negative because R decreases)
For product P: Rate = +Δ[P]/Δt
Units: mol L⁻¹ s⁻¹ (or M/s)
Average rate: Change in concentration over a finite time interval Δt. Calculated as Δ[R]/Δt.
Instantaneous rate: Rate at a specific moment. Calculated as the slope of the tangent to the concentration vs time curve. mathematically: −d[R]/dt.
Instantaneous rate decreases as reaction proceeds because reactant concentration drops.
Rate = change in concentration per unit time. Average rate over Δt; instantaneous at specific moment (slope of tangent). Affected by concentration, temperature, surface area, catalyst, nature of reactants. Units: mol L⁻¹ s⁻¹.
Rate law: An experimentally determined expression relating rate to concentrations of reactants.
For reaction: aA + bB → cC + dD
Rate = k[A]ˣ[B]ʸ
Where:
Note: x, y are NOT necessarily the stoichiometric coefficients a, b — they must be measured.
Order: Sum of powers to which concentration terms are raised in the experimentally determined rate law.
Order is an experimental quantity. It can be zero, fractional, or negative.
Molecularity: Number of reacting species (ions, atoms, molecules) colliding simultaneously in an elementary step.
Molecularity is always a whole number (1, 2, 3) and is a theoretical concept for elementary reactions.
| Feature | Order | Molecularity |
|---|---|---|
| Definition | Sum of powers in rate law | Number of species in elementary step |
| Determined how? | Experimentally | Theoretically |
| Can be zero? | Yes | No |
| Can be fractional? | Yes (e.g., 1.5) | No — always integer |
| Can be negative? | Yes (rare) | No |
| For elementary reaction | Order = molecularity | Order = molecularity |
| For complex reaction | Determined by slow step | Applies only to elementary steps |
Imagine a dance party where a specific group must collide to do a special dance move. Molecularity = how many people need to collide (always whole number: 1, 2, 3). Order = how the dance frequency depends on crowd density — measured by observation. If only some dancers react even with 100 people (due to orientation), order may be less than molecularity!
Rate law: Rate = k[A]ˣ[B]ʸ — measured experimentally. Order = x + y (can be 0, 1, 2, fractional). Molecularity = number of species colliding (whole number only). For elementary reactions, order = molecularity. For complex reactions, order is determined by the slow step.
Rate law: Rate = k[A]⁰ = k
Integrated form: [A] = [A]₀ − kt
Where [A]₀ = initial concentration, [A] = concentration at time t.
Rate constant: k = ([A]₀ − [A]) / t
Units of k: mol L⁻¹ s⁻¹ (same as rate)
Plot: [A] vs t is a straight line with slope −k
Examples: Some photochemical reactions, catalytic decomposition on metal surfaces, drug elimination when enzymes saturated
Rate law: Rate = k[A]
Integrated form: ln[A] = ln[A]₀ − kt
Or: [A] = [A]₀ · e^(−kt)
Or in log base 10: log[A] = log[A]₀ − kt/2.303
Rate constant: k = (2.303/t) × log([A]₀/[A])
Units of k: s⁻¹ (time⁻¹)
Plot: ln[A] vs t is a straight line with slope −k
Examples: Radioactive decay, all nuclear reactions, most organic substitution reactions, drug metabolism (in many cases)
Choose order, enter concentrations and time — calculate k.
Mistake: "Units of k are always s⁻¹."
Correction:
Units of k depend on overall order n:
Zero order: [A] = [A]₀ − kt (linear [A] vs t). First order: ln[A] = ln[A]₀ − kt (linear ln[A] vs t). k units depend on order. First order has constant half-life (independent of [A]₀); zero order has variable half-life.
Half-life: Time required for the concentration of a reactant to reduce to half of its initial value.
Zero order: t₁/₂ = [A]₀ / (2k)
So t₁/₂ is proportional to initial concentration — larger [A]₀ means longer t₁/₂.
First order: t₁/₂ = 0.693 / k
So t₁/₂ is independent of initial concentration — same time to halve regardless of starting amount!
| Order | Rate Law | Integrated Equation | Half-Life | k Units |
|---|---|---|---|---|
| Zero | k | [A] = [A]₀ − kt | [A]₀/(2k) | mol L⁻¹ s⁻¹ |
| First | k[A] | ln[A] = ln[A]₀ − kt | 0.693/k | s⁻¹ |
| Second | k[A]² | 1/[A] = 1/[A]₀ + kt | 1/(k[A]₀) | L mol⁻¹ s⁻¹ |
In first order, rate = k[A]. When [A] halves, rate also halves. So the time to go from 100% → 50% equals the time from 50% → 25%, etc. It's self-similar — the "rate" is always proportional to how much is left. This is why radioactive decay has a fixed half-life — you can use it as a clock!
Half-life = time for concentration to reduce by half. Zero order: t₁/₂ = [A]₀/(2k) (depends on [A]₀). First order: t₁/₂ = 0.693/k (constant!). This is why radioactive dating works.
Pseudo first order: A reaction that is actually higher order but behaves as first order because one reactant is present in such large excess that its concentration barely changes.
Example 1: Hydrolysis of sucrose
C₁₂H₂₂O₁₁ + H₂O → C₆H₁₂O₆ + C₆H₁₂O₆
Rate = k'[sucrose][H₂O]
Since water is in large excess, [H₂O] ≈ constant → k'[H₂O] = k (pseudo first order constant)
Rate = k[sucrose] (behaves as first order)
Example 2: Hydrolysis of ester
CH₃COOC₂H₅ + H₂O (excess) → CH₃COOH + C₂H₅OH (acid-catalyzed)
Rate = k[ester] (pseudo first order)
C₁₂H₂₂O₁₁ + H₂O --(H⁺)--> C₆H₁₂O₆ (glucose) + C₆H₁₂O₆ (fructose)
The reaction is followed by polarimetry:
Angle changes from +66.5° to −39.5° → "inversion" of sign. Rate constant measured by polarimetry.
Imagine 10 people in a large park (1000 people) playing a game. The game rate depends on how many players there are. But the crowd (1000 people) stays roughly the same regardless of how many play. So the game rate only visibly depends on the players (sucrose) — even though the crowd (water) is technically involved. That's pseudo first order: one reactant so abundant, it appears not to matter.
Pseudo first order: appears first order despite being higher order — because one reactant is in excess. Common examples: hydrolysis of sucrose, ester, starch. Rate law depends only on the minor component.
k = A · e^(−Ea/RT)
Where:
Logarithmic form:
ln k = ln A − Ea/(RT)
Or: log k = log A − Ea/(2.303 RT)
log(k₂/k₁) = (Ea/2.303R) × (1/T₁ − 1/T₂)
Or equivalently: log(k₂/k₁) = (Ea/2.303R) × (T₂ − T₁)/(T₁T₂)
Use this to find Ea from two rate constants at two temperatures, or to find k₂ given k₁ and Ea.
Enter two rate constants at two temperatures — calculate activation energy.
Van't Hoff rule: Rate roughly doubles for every 10°C rise in temperature (near room temperature).
Why? At higher T:
Mathematically, k ∝ e^(−Ea/RT). Even small increases in T decrease the exponent significantly.
Mistake: "Temperature and activation energy are
the same thing."
Correction: Ea is a
property of the reaction (energy barrier). Temperature is the
energy of the molecules. Higher T gives molecules enough energy to
cross the Ea barrier. Lowering Ea (via catalyst) also speeds up
reaction — but the mechanisms differ.
Arrhenius: k = A·e^(−Ea/RT). Higher T → higher k (faster reaction). Higher Ea → lower k (slower reaction). Two-point form: log(k₂/k₁) = (Ea/2.303R)(1/T₁ − 1/T₂). Slope of ln k vs 1/T = −Ea/R. Arrhenius plot is linear for simple reactions.
Key idea: A reaction occurs only when molecules collide with:
Most collisions don't result in reaction because they fail one or both conditions.
Activation energy: The minimum extra energy that reactant molecules must acquire to undergo a chemical reaction.
Activated complex: The high-energy intermediate state at the peak of the energy barrier. It's transient and unstable.
Rate = (collision frequency) × (fraction of collisions with E ≥ Ea) × (fraction with correct orientation)
Mathematically: k = P · Z · e^(−Ea/RT)
The exponential term e^(−Ea/RT) gives the fraction of molecules with sufficient energy (Maxwell-Boltzmann distribution).
The exponential term e^(−Ea/RT) is extremely sensitive to Ea. Example:
Doubling Ea reduces rate by factor of ~10⁹! Even tiny changes in Ea make huge differences in rate.
Collision theory: only collisions with E ≥ Ea and correct orientation react. Ea = energy barrier. Higher Ea → slower reaction. Activated complex = high-energy transient at peak. Rate = P·Z·e^(−Ea/RT). Catalyst lowers Ea.
Catalyst: A substance that increases the rate of a reaction without being consumed in the overall reaction.
Mechanism: Catalyst provides an alternative reaction path with lower activation energy.
Key facts about catalysts:
1. Homogeneous catalysis: Catalyst and reactants in same phase.
2. Heterogeneous catalysis: Catalyst and reactants in different phases.
3. Enzyme catalysis: Biological catalysts (proteins). Highly specific.
| Industry | Reaction | Catalyst |
|---|---|---|
| Haber (NH₃) | N₂ + 3H₂ → 2NH₃ | Fe + Mo (promoter) |
| Contact (H₂SO₄) | 2SO₂ + O₂ → 2SO₃ | V₂O₅ |
| Ostwald (HNO₃) | 4NH₃ + 5O₂ → 4NO + 6H₂O | Pt/Rh |
| Hydrogenation of oils | Oils + H₂ → Vanaspati ghee | Ni |
| Methanol synthesis | CO + 2H₂ → CH₃OH | ZnO/Cr₂O₃ |
| Catalytic converter | CO + NO → CO₂ + N₂ | Pt/Pd/Rh |
Mistake: "Catalyst increases the yield of the
product."
Correction: Catalyst increases the
RATE of reaction — reaches equilibrium faster — but the
EQUILIBRIUM POSITION is unchanged (same amount of product at
equilibrium). ΔG° is also unchanged. Only kinetics, not
thermodynamics, are affected.
Catalyst increases reaction rate by lowering Ea — no change in ΔG or equilibrium. Types: homogeneous (same phase), heterogeneous (different phases), enzyme (biological). Catalysts are not consumed, are needed in tiny amounts, are highly specific (for enzymes). Key industrial applications: Haber (NH₃), Contact (H₂SO₄), catalytic converters.
"Rate = mol L⁻¹ s⁻¹" — always. Rate is change in concentration per time.
"Order is measured, Molecularity is theoretical"
Order can be 0, 1, 2, 1.5, etc. Molecularity is always 1, 2, 3.
"Zero → [A] straight; First → ln[A] straight"
Plot [A] vs t linear → zero order. Plot ln[A] vs t linear → first order.
"Zero order half-life depends on [A]₀; first order does NOT"
First order t½ = 0.693/k — constant!
"k = A e^(−Ea/RT)" — say as "k = A times e to the minus-Ea-over-R-T"
Log form: log(k₂/k₁) = (Ea/2.303R)(1/T₁ − 1/T₂)
"Energy + Orientation = Reaction"
Both required! Only 1 in ~10¹⁵ collisions leads to reaction at room T.
"Lowers Ea, doesn't change ΔG"
Catalyst speeds up both forward and reverse reactions — doesn't shift equilibrium.
"10°C rise → 2× rate" (near room T)
Because e^(−Ea/RT) changes dramatically with T for typical Ea values.
Mnemonics make recall faster. Remember "Zero → [A] straight, First → ln[A] straight" for order identification. "Lowers Ea, doesn't change ΔG" for catalysts. "10°C → 2× rate" for temperature rule of thumb.
Q1. For the reaction 2N₂O₅ → 4NO₂ + O₂, the rate of disappearance of N₂O₅ is 2.5×10⁻⁴ M/s. What is the rate of formation of NO₂?
Rate = −(1/2)d[N₂O₅]/dt = (1/4)d[NO₂]/dt
d[NO₂]/dt = 2 × d[N₂O₅]/dt (in magnitude)
= 2 × 2.5×10⁻⁴ = 5.0×10⁻⁴ M/s
Q2. A first-order reaction has k = 5.5×10⁻¹⁴ s⁻¹. What is its half-life?
t₁/₂ = 0.693 / k (first order)
t₁/₂ = 0.693 / (5.5×10⁻¹⁴)
t₁/₂ = 1.26 × 10¹³ s ≈ 4 × 10⁵ years
Q3. A first-order reaction is 40% complete in 50 minutes. Calculate its rate constant.
Initial = 100%, after 50 min = 60% (40% reacted)
[A]₀ = 100, [A] = 60, t = 50 min
k = (2.303/t) log([A]₀/[A])
k = (2.303/50) × log(100/60)
= 0.04606 × log(1.667)
= 0.04606 × 0.2218
k = 0.0102 min⁻¹ ≈ 1.7×10⁻⁴ s⁻¹
Q4. The rate constant of a reaction doubles when temperature changes from 300 K to 310 K. Calculate activation energy.
log(k₂/k₁) = (Ea/2.303R) × (1/T₁ − 1/T₂)
log(2) = (Ea/(2.303 × 8.314)) × (1/300 − 1/310)
0.301 = (Ea/19.147) × (10/(300×310))
0.301 = (Ea/19.147) × (1/9300)
Ea = 0.301 × 19.147 × 9300
Ea = 53,580 J/mol ≈ 53.6 kJ/mol
Q5. Show that for a first-order reaction, the time required for 99% completion is twice the time required for 90% completion.
k = (2.303/t) log([A]₀/[A]) → t = (2.303/k) log([A]₀/[A])
[A]₀ = 100, [A] = 10 (10% remains)
t₉₀ = (2.303/k) log(100/10) = (2.303/k) log(10) = 2.303/k
[A]₀ = 100, [A] = 1 (1% remains)
t₉₉ = (2.303/k) log(100/1) = (2.303/k) log(100) = 2 × (2.303/k)
t₉₉ / t₉₀ = 2 × (2.303/k) / [(2.303/k)] = 2
So t₉₉ = 2 × t₉₀ ✓
For 99.9% completion: t = 3 × (2.303/k) — three times t₉₀. First order gives "n-9s" — for 10ⁿ % completion → n × (2.303/k).
Q6. The rate of a first-order reaction is 0.04 mol L⁻¹ s⁻¹ at 10 minutes and 0.03 mol L⁻¹ s⁻¹ at 20 minutes after initiation. Find the rate constant.
Rate = k[A] → k = Rate/[A]
At t = 10 min: Rate₁ = k[A]₁ → [A]₁ = Rate₁/k = 0.04/k
At t = 20 min: Rate₂ = k[A]₂ → [A]₂ = Rate₂/k = 0.03/k
k = (2.303/(20−10)) log([A]₁/[A]₂)
= (2.303/10) log((0.04/k)/(0.03/k))
= (2.303/10) log(0.04/0.03)
= (2.303/10) log(4/3)
= 0.2303 × 0.1249
= 0.0288 min⁻¹
Interesting trick — we didn't need [A]₀! The ratio [A]₁/[A]₂ = Rate₁/Rate₂ for first order, so we only need the rates. This works because rate ∝ [A] in first order.
Q7. Sucrose hydrolyzes at pH 3 with k = 4.5×10⁻³ min⁻¹. Find the time for 50% inversion.
Hydrolysis of sucrose in aqueous solution is pseudo first order
Rate = k[sucrose]
t₁/₂ = 0.693/k (first order)
t₁/₂ = 0.693 / (4.5×10⁻³)
t₁/₂ = 154 minutes
So after ~2.5 hours, half of the sucrose is inverted to glucose + fructose. This is why making invert sugar requires heating for several hours.
Selected exercise questions from NCERT Class 12 Chemistry Chapter 3. Click each to reveal the solution.
(a) Order = 2. Rate = k[NO]². Units of k = Rate/[NO]² = (M/s)/M² = M⁻¹s⁻¹ = L mol⁻¹ s⁻¹.
(b) Order = 1 + 1 = 2. Rate = k[H₂O₂][I⁻]. Units of k = M⁻¹s⁻¹ = L mol⁻¹ s⁻¹.
Order wrt B: Compare Exp 2 and 3 — [A] same, [B] doubles but rate same → order = 0 wrt B.
Order wrt A: Compare Exp 1 and 2 — need [B] fixed. Actually Exp 1→2 has [A]×3, [B]×2, rate×2. Since order in B is 0, rate should depend only on A. Rate₁ ∝ [A]₁ = 0.1. Rate₂ should be 3× Rate₁ = 6×10⁻² — but observed is 4×10⁻². Hmm, need to recheck...
Let's use rate = k[A]ˣ[B]ʸ with y = 0: rate = k[A]ˣ
Exp 1: 2.0×10⁻² = k(0.1)ˣ
Exp 2: 4.0×10⁻² = k(0.3)ˣ
Divide: 2 = (3)ˣ → x = log(2)/log(3) = 0.63... approximate first order.
Actually check [A] should be proportional: If [A] tripled and rate only doubled, order is less than 1. With x ≈ 0.63 — fractional order.
But likely intended answer: rate = k[A][B]⁰ = k[A] assuming small experimental error. Best answer: Rate = k[A] (order 1 in A, 0 in B, overall 1).
(a) Rate = k[A][B]²
(b) If [B] tripled: Rate' = k[A](3[B])² = 9k[A][B]² = 9× rate. Rate increases 9 times.
(c) If both doubled: Rate' = k(2[A])(2[B])² = k × 2 × 4 × [A][B]² = 8 × rate. Rate increases 8 times.
Solution: Rate = k[X]² (second order). If [X] → 3[X]: Rate' = k(3[X])² = 9k[X]² = 9× original rate.
Rate increases 9 times.
Solution: t₁/₂ = 30 min → k = 0.693/30 = 0.0231 min⁻¹
For 75% completion, [A] = 25% of [A]₀ → [A]₀/[A] = 100/25 = 4
t = (2.303/k) log(4) = (2.303/0.0231) × 0.602
= 99.7 × 0.602 = 60 minutes
Shortcut: 75% completion = 2 half-lives = 2 × 30 = 60 min ✓
Solution: 30% decomposed → 70% remains. [A]₀/[A] = 100/70 = 1.4286
k = (2.303/40) × log(1.4286) = 0.0576 × 0.1549 = 0.00892 min⁻¹
t₁/₂ = 0.693/k = 0.693/0.00892 = 77.7 minutes
Solution: log(k₂/k₁) = (Ea/2.303R) × (1/T₁ − 1/T₂)
log(0.105/0.011) = (Ea/(2.303 × 8.314)) × (1/300 − 1/320)
log(9.545) = (Ea/19.147) × (20/(300×320))
0.980 = (Ea/19.147) × (1/4800)
Ea = 0.980 × 19.147 × 4800 = 90,070 J/mol ≈ 90 kJ/mol
Solution: Plot ln k vs 1/T → straight line. Slope = −Ea/R.
Using two points (T=293, T=333):
log(k₂/k₁) = (Ea/2.303R) × (T₂ − T₁)/(T₁T₂)
log(178×10⁻⁵ / 1.70×10⁻⁵) = (Ea/19.147) × (40/(293×333))
log(104.7) = (Ea/19.147) × (40/97569)
2.020 = (Ea/19.147) × (4.1×10⁻⁴)
Ea = 2.020 × 19.147 / (4.1×10⁻⁴)
Ea = ~94,400 J/mol ≈ 94 kJ/mol
At 30°C (303 K): Use Arrhenius with the Ea found. Roughly ~5×10⁻⁵ s⁻¹.
At 50°C (323 K): Roughly ~1300×10⁻⁵ s⁻¹.
A can be found from intercept (ln A). Typically ~10¹³ s⁻¹ for such reactions.
Order wrt A: Compare Exp 1 and 4 ([B] same): Rate₄/Rate₁ = 2.40×10⁻²/6.0×10⁻³ = 4 = 2^x where x is order (since [A] doubles from 0.1 → 0.4? No, [A] tripled from 0.1 to 0.3). Hmm, let me use Exp 1 → 2: [A] goes 0.1→0.3 (×3), but [B] also changes.
Better: Exp 1 → 4: [B]=0.1 same, [A] goes 0.1 → 0.4 (×4). Rate goes 6×10⁻³ → 2.4×10⁻² (×4). So x = 1 (order in A = 1).
Order wrt B: Compare Exp 2 and 3: [A]=0.3 same, [B] goes 0.2 → 0.4 (×2). Rate 7.2×10⁻² → 2.88×10⁻¹ (×4). So 2^y = 4 → y = 2.
Rate law: Rate = k[A][B]²
Rate constant: From Exp 1: 6.0×10⁻³ = k(0.1)(0.1)² = k × 10⁻³
k = 6.0 M⁻² s⁻¹
Solution: log(k₂/k₁) = (Ea/2.303R) × (T₂ − T₁)/(T₁T₂)
log(1.5×10⁴ / 4.5×10³) = (60000/19.147) × (T₂ − 283)/(283 × T₂)
log(3.333) = 3134 × (T₂ − 283)/(283 T₂)
0.5229 = 3134 × (T₂ − 283)/(283 T₂)
0.5229 × 283 T₂ = 3134(T₂ − 283)
148 T₂ = 3134 T₂ − 886,922
2986 T₂ = 886,922
T₂ ≈ 297 K ≈ 24°C
Solution: For first order, t = (2.303/k) log([A]₀/[A]). Given t₁₀(298) = t₂₅(308).
(2.303/k₂₉₈) log(100/90) = (2.303/k₃₀₈) log(100/75)
k₃₀₈/k₂₉₈ = log(100/75)/log(100/90) = 0.1249/0.04576 = 2.729
From Arrhenius with A = 4×10¹⁰:
k₃₀₈ = 4×10¹⁰ e^(−Ea/(8.314×308))
k₂₉₈ = 4×10¹⁰ e^(−Ea/(8.314×298))
Ratio = e^(−Ea/RT)[T=308] / e^(−Ea/RT)[T=298] = e^((Ea/R)(1/298 − 1/308))
2.729 = e^((Ea/8.314)(10/(298×308)))
2.729 = e^((Ea/8.314)(1.09×10⁻⁴))
ln(2.729) = 1.004 = (Ea/8.314) × 1.09×10⁻⁴
Ea = 1.004 × 8.314/1.09×10⁻⁴ = 76,700 J/mol ≈ 76.6 kJ/mol
Now find k₃₁₈ = 4×10¹⁰ × e^(−76600/(8.314×318))
= 4×10¹⁰ × e^(−28.98)
= 4×10¹⁰ × 2.6×10⁻¹³
= 1.04×10⁻² s⁻¹
| R (gas constant) | 8.314 J/(mol·K) | or 0.0821 L·atm/(mol·K) |
| 2.303 × R | 19.147 J/(mol·K) | Used in log form of Arrhenius |
| t½ Carbon-14 | 5730 years | Archaeology dating |
| t½ Iodine-131 | 8.04 days | Medical isotope |
| t½ Uranium-238 | 4.5×10⁹ years | Rock dating |
| Typical Ea (simple reactions) | 50–150 kJ/mol | Activation energies |
| Van't Hoff rule | k(T+10) ≈ 2×k(T) | Approximation |
| ln(2) | 0.693 | Appears in half-life formula |
| [A] | Concentration of A (mol/L or M) |
| [A]₀ | Initial concentration |
| k | Rate constant (units depend on order) |
| x, y | Orders with respect to A, B |
| n | Overall order (n = x + y) |
| t | Time |
| t₁/₂ | Half-life |
| Ea | Activation energy (J/mol or kJ/mol) |
| A | Arrhenius pre-exponential factor (frequency factor) |
| R | Gas constant (8.314 J/mol·K) |
| T | Temperature (Kelvin) |
| Z | Collision frequency |
| P | Steric factor (probability of correct orientation) |
(A) Both true, R explains A. (B) Both true, R does NOT explain A. (C) A true, R false. (D) A false, R true.
All key formulas at one place — perfect for quick revision before exams.