In short
Chemical kinetics is the study of how fast a chemical reaction proceeds, what the rate depends on (concentration, temperature, catalyst, pressure) and what mechanism lies behind the observed rate expression. It supplies the rate law -r_A = f(C_A, T) that every reactor sizing calculation then integrates, so no reactor volume, no catalyst loading and no runaway-safety margin can be computed without it. In Indian universities it is the opening block of Chemical Reaction Engineering I and carries 8-12 marks in GATE (CH) every year, usually as a rate-law/order problem, an Arrhenius problem or a mechanism-reduction problem. Employers test it in placement rounds at Reliance Industries, IOCL, BPCL, HPCL, ONGC, GAIL, Tata Chemicals, UPL, Aarti Industries, Deepak Nitrite, Sun Pharma, Dr. Reddy's and process-design firms such as Technip Energies, L&T Technology Services and Toyo Engineering India. It also underlies PSM/HAZOP work governed by the Manufacture, Storage and Import of Hazardous Chemical Rules, 1989 and the Static and Mobile Pressure Vessels Rules.
Explained
Rate of reaction: definitions and conventions
- Intensive rate for species A:
r_A = (1/V) dN_A/dt(mol per m³ per s) - rate per unit volume of reacting fluid. This is the definition used for homogeneous reactions. - Sign convention:
r_Ais negative for a reactant, so textbooks write-r_Aas a positive number, "the rate of disappearance of A". - Other bases are used when volume is not the natural measure: per mass of catalyst
-r'_A(mol/kg·s), per unit interfacial area-r''_A(mol/m²·s), per unit volume of reactor-r'''_A. Mixing bases is a classic exam trap. - Stoichiometric coupling: for
aA + bB → cC + dD,(-r_A)/a = (-r_B)/b = r_C/c = r_D/d. Rates of different species are never numerically equal unless the coefficients are. - Conversion
X_A = (N_A0 - N_A)/N_A0; for constant-density systemsX_A = (C_A0 - C_A)/C_A0, soC_A = C_A0(1 - X_A). - Variable-volume (gas-phase) systems:
V = V_0(1 + ε_A X_A)at constant T and P, whereε_A = y_A0 δandδ = (d + c - b - a)/ais the change in moles per mole of A reacted. ThenC_A = C_A0(1 - X_A)/(1 + ε_A X_A). With T and P changing,C_A = C_A0 (1 - X_A)/(1 + ε_A X_A) · (T_0/T)(P/P_0). - Homogeneous vs heterogeneous: homogeneous reactions occur in one phase (liquid-phase esterification); heterogeneous ones occur at an interface (gas over a solid catalyst), where observed rates may be masked by transport. Intrinsic kinetics must be measured free of diffusion limitation.
Rate laws, order and molecularity
- Power-law rate law:
-r_A = k C_A^α C_B^β;αis the order with respect to A,βwith respect to B, and the overall order n = α + β. - Order is experimental, obtained by fitting data; molecularity is the number of molecules taking part in an elementary step and is a mechanistic integer. They coincide only for elementary reactions.
- Order may be zero, fractional or negative. Zero order: rate independent of concentration (surface-saturated catalysis, photochemical reactions). Negative order: a species inhibits the reaction (product inhibition).
- Elementary reaction: the rate law follows the stoichiometry directly -
A + B → productswith-r_A = k C_A C_B. Non-elementary: the rate law does not match stoichiometry, e.g.H₂ + Br₂ → 2HBrwithr_HBr = k C_H2 C_Br2^0.5 /(1 + k' C_HBr/C_Br2). - Units of k: from
-r_A = k C_A^n,khas units(concentration)^(1-n) (time)^(-1). For n = 0: mol/m³·s; n = 1: s⁻¹; n = 2: m³/mol·s. Writing k without units loses marks. - Pseudo-order: when one reactant is in large excess its concentration barely changes, so it is absorbed into the constant -
k' = k C_B0. Hydrolysis of an ester in excess water is second order but behaves first order (pseudo-first-order). - Reversible reactions:
A ⇌ Bgives-r_A = k_f C_A - k_b C_B; at equilibrium the net rate is zero, soK_C = k_f/k_b. Thermodynamics fixes the ratio, kinetics fixes the magnitudes.
Temperature dependence
- Arrhenius equation:
k = A exp(-E/RT).Ais the pre-exponential (frequency) factor with the same units as k,Ethe activation energy (J/mol or kJ/mol),R = 8.314 J/mol·K,Tabsolute temperature in K. - Linear form
ln k = ln A - E/(RT): a plot ofln kagainst1/Tis a straight line of slope-E/R- the standard graphical question. - Interpretation: E is the minimum energy barrier colliding molecules must cross; a high E means a strongly temperature-sensitive reaction, a low E a weakly sensitive one. Most industrial reactions have E in the range 40-250 kJ/mol.
- The thumb rule "rate doubles for every 10 K" holds only near E ≈ 50-60 kJ/mol around room temperature; it is not a general law.
- Collision theory (gas phase):
k = p Z exp(-E/RT)- collision frequency Z times a steric/probability factor p times the fraction of sufficiently energetic collisions; predictsA ∝ T^0.5. - Transition state (activated complex) theory: reactants form an activated complex in quasi-equilibrium;
k = (k_B T/h) exp(ΔS‡/R) exp(-ΔH‡/RT), givingA ∝ T. Over a modest temperature range the exponential dominates, so the Arrhenius form fits both. - Why temperature is the main handle: doubling concentration doubles a first-order rate; a 20 K rise with E = 100 kJ/mol (350 K to 370 K) multiplies it about 6.4 times. That is also why exothermic reactions can run away - an important safety point in HAZOP studies.
- Catalysts lower E by offering an alternative path; they change k, never
K_C, so they cannot shift equilibrium conversion.
Integrated rate laws for constant-volume batch systems
Starting from -dC_A/dt = k C_A^n for a batch reactor of constant volume:
- Zero order:
C_A = C_A0 - k t; linear inC_Avs t; reaction stops abruptly when C_A reaches zero;t_1/2 = C_A0/(2k)- proportional to initial concentration. - First order:
ln(C_A0/C_A) = k t, i.e.C_A = C_A0 e^(-kt); in terms of conversion-ln(1 - X_A) = k t;t_1/2 = ln2/k = 0.693/k- independent of initial concentration. - Second order, single reactant (2A → P or A + B with C_A0 = C_B0):
1/C_A - 1/C_A0 = k t;X_A/(1 - X_A) = k C_A0 t;t_1/2 = 1/(k C_A0)- inversely proportional to initial concentration. - Second order, unequal initial concentrations
A + B → P, withM = C_B0/C_A0 ≠ 1:ln[(M - X_A)/(M(1 - X_A))] = C_A0 (M - 1) k t. - nth order (n ≠ 1):
C_A^(1-n) - C_A0^(1-n) = (n - 1) k t;t_1/2 = [2^(n-1) - 1] / [k (n-1) C_A0^(n-1)]. - General half-life diagnostic:
t_1/2 ∝ C_A0^(1-n), so a log-log plot of half-life against initial concentration gives slope(1 - n)- a fast way to find order from two experiments. - First-order reversible
A ⇌ B:-ln[1 - X_A/X_Ae] = (k_f + k_b) t, whereX_Aeis the equilibrium conversion; the slope gives the sum of constants andK_Cgives the ratio, so both are recovered. - Variable-volume batch (constant-pressure, gas phase): use
V = V_0(1 + ε_A X_A); for first order,-ln(1 - X_A) = k tstill holds because the first-order form cancels the volume change, while higher orders do not.
Analysis of kinetic data: how order and k are found
Three standard methods, all examined:
- Integral method - assume an order, plot the corresponding linearised group (
C_A,ln C_A,1/C_A) against t, and accept the order that gives the best straight line;kcomes from the slope. Simple and robust to scattered data, but trial-and-error and poor for fractional orders. - Differential method - obtain
-dC_A/dtfrom theC_Avs t curve (graphical slopes or finite differences), then plotln(-dC_A/dt)againstln C_A: slope = order n, intercept =ln k. Handles any order including fractional, but differentiating noisy data amplifies error. - Method of initial rates - measure the initial slope at several different
C_A0;ln(-r_A0)vsln C_A0gives n. Avoids product-inhibition effects entirely, but needs many runs.
Supporting techniques:
- Half-life method: run at two or more
C_A0and uset_1/2 ∝ C_A0^(1-n). - Method of excess (isolation): flood all but one reactant to collapse the rate law to pseudo-order in that one species, then repeat for each species.
- Experimental reactors: constant-volume batch (liquid, follow concentration or conductivity), constant-pressure batch (follow volume), differential (packed-bed) reactor with small conversion so
-r_A ≈ F_A0 X / Wis read directly, and a recycle/CSTR (mixed-flow) reactor where the algebraic mass balance gives the rate without any differentiation - the most accurate for intrinsic kinetics. - Pressure-based gas data: for a constant-volume gas reaction, total pressure is linear in the extent of reaction, so
P_totaldata converts toC_Athrough stoichiometry. - Fit quality is judged by linearity/residuals, not by a single point; and a rate law valid over 10 K must never be extrapolated 100 K.
Mechanisms, rate-determining step and approximations
- A mechanism is a set of elementary steps whose sum is the overall stoichiometry; it must also reproduce the observed rate law, which is the real test.
- Intermediates (free radicals, ions, complexes, adsorbed species) appear in steps but not in the overall equation and cannot stay in the final rate law.
- Rate-determining step (RDS) approximation: if one step is far slower, the overall rate equals that step's rate with the faster steps assumed at equilibrium.
- Pseudo-steady-state hypothesis (PSSH): for a reactive intermediate present at tiny concentration, set
r_intermediate ≈ 0, solve algebraically for it and substitute. This is the standard GATE method. - Lindemann mechanism for unimolecular reactions:
A + M ⇌ A* + M,A* → Pgives-r_A = k1 k2 C_A C_M/(k_-1 C_M + k2). At high pressure (largeC_M) it is first order; at low pressure, second order. It explains why "unimolecular" decompositions change order with pressure. - Chain reactions: initiation → propagation → (branching) → termination; thermal cracking of naphtha and ethane in petrochemical plants, and polymerisation, follow this pattern. Branching chains underlie explosions.
- Autocatalysis: a product accelerates the reaction,
-r_A = k C_A C_P; the rate-vs-conversion curve passes through a maximum, which later governs reactor choice. - Note the limit of all of this: a mechanism consistent with the data is not proved by it - several mechanisms can give the same rate law.
Key points
- Rate of reaction - moles reacted per unit volume per unit time; always state the basis and the species.
- Rate law - the experimentally determined dependence
-r_A = k(T) f(C); it is an input to reactor design, never an output. - Order - the exponent on concentration in the fitted rate law; experimental, possibly fractional, zero or negative.
- Molecularity - number of species in an elementary step; a positive integer, equal to order only for elementary reactions.
- Units of k -
(conc)^(1-n)(time)^(-1); use them to check that a quoted order is consistent. - Elementary reaction - rate law mirrors stoichiometry; a bimolecular step is second order.
- Arrhenius equation -
k = A exp(-E/RT); slope ofln kvs1/Tis-E/R. - Activation energy - energy barrier; large E means high temperature sensitivity, not necessarily a slow reaction.
- Catalyst - lowers E and raises k for both directions;
K_Cand equilibrium conversion are untouched. - Half-life - time for 50% consumption; constant for first order,
∝ C_A0for zero order,∝ 1/C_A0for second order. - First-order plot -
ln C_Avs t linear; second-order -1/C_Avs t linear; zero-order -C_Avs t linear. - Conversion relations -
C_A = C_A0(1 - X_A)only at constant volume; gas phase needs(1 + ε_A X_A). - Expansion factor -
ε_A = y_A0 δ; ignoring it is the commonest gas-phase error. - Pseudo-first order - achieved by large excess of one reactant;
k_obs = k C_B0. - Reversible kinetics -
K_C = k_f/k_b; only(k_f + k_b)is obtained from the approach-to-equilibrium slope. - PSSH - set the net rate of a trace intermediate to zero to eliminate it from the rate expression.
- Lindemann - unimolecular reactions are first order at high pressure, second order at low pressure.
- Differential vs integral method - differential finds any order but magnifies data noise; integral is stable but trial-and-error.
Formulas and examples
-r_A = -(1/V) dN_A/dt - rate of disappearance of A, mol/(m³·s); V in m³, N_A in mol, t in s(-r_A)/a = (-r_B)/b = r_C/c - stoichiometric rate ratios for aA + bB → cC-r_A = k C_A^α C_B^β, overall order n = α + β; [k] = (mol/m³)^(1-n) s^(-1)X_A = (N_A0 - N_A)/N_A0; C_A = C_A0(1 - X_A) (constant volume)V = V_0(1 + ε_A X_A), ε_A = y_A0 δ, δ = (c - a - b)/a; C_A = C_A0(1 - X_A)/(1 + ε_A X_A)Zero order:
C_A = C_A0 - k t; t_1/2 = C_A0/(2k)First order:
ln(C_A0/C_A) = k t; t_1/2 = 0.693/kSecond order (equal C_0):
1/C_A - 1/C_A0 = k t; t_1/2 = 1/(k C_A0)nth order (n ≠ 1):
C_A^(1-n) - C_A0^(1-n) = (n-1) k t; t_1/2 ∝ C_A0^(1-n)Second order,
M = C_B0/C_A0: ln[(M - X_A)/(M(1 - X_A))] = C_A0(M - 1) k tReversible first order:
-ln[1 - X_A/X_Ae] = (k_f + k_b) t; K_C = k_f/k_b = X_Ae/(1 - X_Ae)Arrhenius:
k = A exp(-E/RT); ln(k_2/k_1) = -(E/R)(1/T_2 - 1/T_1); R = 8.314 J/(mol·K)Transition state:
k = (k_B T/h) exp(ΔS‡/R) exp(-ΔH‡/RT); k_B = 1.381×10⁻²³ J/K, h = 6.626×10⁻³⁴ J·sLindemann:
-r_A = k_1 k_2 C_A C_M /(k_-1 C_M + k_2)Example 1 - activation energy from two rate constants. A liquid-phase reaction has k = 0.0025 s⁻¹ at 300 K and k = 0.0400 s⁻¹ at 340 K. Find E and A.
k_2/k_1 = 0.0400/0.0025 = 16, soln 16 = 2.7726.1/T_1 - 1/T_2 = 1/300 - 1/340 = 0.00333333 - 0.00294118 = 3.9216×10⁻⁴ K⁻¹.E = R ln(k_2/k_1)/(1/T_1 - 1/T_2) = 8.314 × 2.7726 / 3.9216×10⁻⁴ = 23.050/3.9216×10⁻⁴ ≈ 58,776 J/mol ≈ **58.8 kJ/mol**.A = k_1 exp(E/RT_1) = 0.0025 × exp(58,776/(8.314×300)) = 0.0025 × exp(23.564) = 0.0025 × 1.72×10¹⁰ ≈ **4.3×10⁷ s⁻¹**.- Check: at 340 K,
E/RT = 58,776/2,826.8 = 20.792,exp(-20.792) = 9.3×10⁻¹⁰;k = 4.3×10⁷ × 9.3×10⁻¹⁰ ≈ 0.040 s⁻¹. Consistent.
Example 2 - order and k by the integral method. Batch data for A → products, constant volume, C_A0 = 2.0 mol/L:
| t (min) | C_A (mol/L) | ln C_A | 1/C_A (L/mol) |
|---|---|---|---|
| 0 | 2.000 | 0.693 | 0.500 |
| 10 | 1.000 | 0.000 | 1.000 |
| 20 | 0.667 | -0.405 | 1.500 |
| 30 | 0.500 | -0.693 | 2.000 |
| 40 | 0.400 | -0.916 | 2.500 |
ln C_Avs t is not linear (drops 0.693, 0.405, 0.288, 0.223 per interval) - not first order.1/C_Avs t rises by exactly 0.500 L/mol every 10 min - a straight line, so the reaction is second order.k = slope = 0.500/10 = **0.05 L/(mol·min)**.- Cross-check with half-life:
t_1/2 = 1/(k C_A0) = 1/(0.05 × 2.0) = 10 min, and C_A is indeed 1.0 mol/L at t = 10 min.
Example 3 - order from half-lives. For a reaction, t_1/2 = 50 min at C_A0 = 0.10 mol/L and t_1/2 = 12.5 min at C_A0 = 0.40 mol/L. Find the order and k.
t_1/2 ∝ C_A0^(1-n):ln(12.5/50) = (1 - n) ln(0.40/0.10)→ln 0.25 = -1.3863,ln 4 = 1.3863.1 - n = -1.3863/1.3863 = -1, son = **2**(second order).k = 1/(t_1/2 C_A0) = 1/(50 × 0.10) = **0.20 L/(mol·min)**; check at the second run:1/(0.20 × 0.40) = 12.5 min. Correct.
Example 4 - gas-phase variable volume. A → 3B in a constant-pressure, constant-temperature batch reactor charged with 60 mol% A and 40 mol% inert; first order, k = 0.02 min⁻¹, C_A0 = 0.030 mol/L. Find the volume ratio and C_A at 80% conversion, and the time taken.
δ = (3 - 1)/1 = 2;y_A0 = 0.60, soε_A = 0.60 × 2 = 1.2.- At X_A = 0.80:
V/V_0 = 1 + 1.2 × 0.80 = **1.96**(the gas nearly doubles in volume). C_A = 0.030 × (1 - 0.80)/1.96 = 0.030 × 0.10204 = **3.06×10⁻³ mol/L**.- First order is unaffected by expansion:
t = -ln(1 - 0.80)/k = 1.6094/0.02 = **80.5 min**. - Had the expansion been ignored, C_A would have been quoted as 6.0×10⁻³ mol/L - a 96% error in concentration, and a large error in any higher-order rate.
Example 5 - reversible first-order reaction. A ⇌ B, liquid phase, starts with pure A. Equilibrium conversion is 0.75, and 50% conversion is reached in 8 min. Find k_f, k_b and K_C.
K_C = X_Ae/(1 - X_Ae) = 0.75/0.25 = **3.0**.-ln(1 - X_A/X_Ae) = (k_f + k_b) t→-ln(1 - 0.50/0.75) = -ln(0.33333) = 1.0986.k_f + k_b = 1.0986/8 = 0.13733 min⁻¹.- With
k_f = 3 k_b:4 k_b = 0.13733→k_b = **0.0343 min⁻¹**,k_f = **0.1030 min⁻¹**. - Note: an irreversible first-order reaction with the same k_f would need
-ln(0.5)/0.1030 = 6.7 minfor 50% - the reverse step slows the approach.
Comparison of order diagnostics
| Order n | Linear plot | k units | t_1/2 | Dependence on C_A0 |
|---|---|---|---|---|
| 0 | C_A vs t | mol/(L·min) | C_A0/2k | ∝ C_A0 |
| 1 | ln C_A vs t | min⁻¹ | 0.693/k | none |
| 2 | 1/C_A vs t | L/(mol·min) | 1/(k C_A0) | ∝ 1/C_A0 |
| 3 | 1/C_A² vs t | L²/(mol²·min) | 1.5/(k C_A0²) | ∝ 1/C_A0² |
Where it is used
- Refining and petrochemicals: naphtha and ethane cracking, catalytic reforming, hydrotreating and alkylation kinetics at Reliance Jamnagar, IOCL Panipat/Paradip, BPCL Kochi and HMEL Bathinda - cracking severity is set from first-order coil kinetics.
- Fertilisers and bulk chemicals: ammonia synthesis, urea, sulphuric acid (contact process) and chlor-alkali plants at IFFCO, RCF, Coromandel and Tata Chemicals, where temperature-conversion trade-offs follow directly from k(T) against equilibrium.
- Pharmaceuticals and agrochemicals: API batch kinetics, impurity-formation rates and shelf-life (accelerated stability testing by Arrhenius extrapolation) at Sun Pharma, Cipla, Dr. Reddy's, UPL and Aarti Industries - documented for CDSCO and ICH-aligned filings.
- Process safety: reaction calorimetry (DSC, ARC, RC1) measures heat-release kinetics and time to maximum rate, feeding HAZOP and MAH-site risk assessment under the Manufacture, Storage and Import of Hazardous Chemical Rules, 1989; many Indian runaway incidents trace to an unrecognised exotherm's activation energy.
- Environment and utilities: BOD decay, chlorination and ozonation kinetics in effluent treatment plants designed to CPCB discharge norms; NOx reduction kinetics in SCR units on thermal power plants.
- Job roles: process engineer, R&D/kinetics engineer, catalyst development, process-simulation engineer (Aspen Plus/HYSYS reaction models), technical services, and the engineer-trainee kinetics interview at PSU refineries.
Interview questions
- Q: What is the difference between order and molecularity? A: Order is the experimentally fitted exponent on concentration and can be zero, fractional or negative; molecularity is the integer number of molecules in an elementary step. They are equal only for elementary reactions.
- Q: How do you find the units of a rate constant? A: From
-r_A = k C_A^n, k has units(concentration)^(1-n)(time)^(-1)- s⁻¹ for first order, m³/(mol·s) for second order. - Q: Why is a first-order half-life independent of initial concentration? A: Integration gives
ln(C_A0/C_A) = kt, so the time to halve depends only on the ratio, not the starting value - which is also why radioactive decay has a fixed half-life. - Q: A catalyst speeds a reaction up - does it increase equilibrium conversion? A: No. It lowers the activation energy of both forward and reverse steps equally, so
K_C = k_f/k_band the equilibrium conversion are unchanged; only the time to reach equilibrium falls. - Q: What does a high activation energy tell you about operating the reactor? A: The rate is strongly temperature-sensitive, so small temperature excursions cause large rate changes - good for selectivity control by temperature, but risky for thermal runaway and demanding on temperature control.
- Q: What is the pseudo-steady-state hypothesis and when is it used? A: For a reactive intermediate present in very small concentration, its net rate of formation is set to zero and it is eliminated algebraically; it is used to derive non-elementary rate laws from proposed mechanisms.
- Q: Why does a "unimolecular" gas decomposition change order with pressure? A: By the Lindemann mechanism, activation needs a collision; at high pressure activation is fast and the decomposition step controls (first order), at low pressure activation controls (second order).
- Q: How is order determined when one reactant's effect is masked by others? A: By the method of excess - flood the others so only one concentration varies, giving a pseudo-order; repeat for each species and assemble the full rate law.
- Q: Which experimental reactor gives the cleanest intrinsic kinetics and why? A: A well-mixed CSTR (or recycle reactor), because the steady-state mass balance gives the rate algebraically from measured concentrations with no differentiation of data and at uniform temperature and composition.
- Q: For a gas-phase reaction
A → 2Bin a constant-pressure batch reactor, what changes in the analysis? A: The volume expands asV = V_0(1 + ε_A X_A), so concentration must be computed asC_A0(1 - X_A)/(1 + ε_A X_A); only a first-order rate law keeps its usual integrated form. - Q: Can two different mechanisms give the same rate law? A: Yes - agreement with the observed rate law is necessary but not sufficient, so mechanisms need independent evidence such as intermediate detection or isotope labelling.
- Q: Why can't a rate law measured over 10 K be extrapolated 100 K? A: The dominant mechanism or rate-determining step can change (and mass-transfer or equilibrium limitations can set in), so A and E are no longer valid; the Arrhenius fit is only an interpolation within the tested range.
Common mistakes
- Assuming the rate law from the balanced equation - stoichiometry gives order only for elementary reactions, and most industrial reactions are non-elementary.
- Using
C_A = C_A0(1 - X_A)for a mole-changing gas-phase reaction, i.e. forgettingε_Aand the volume change. - Quoting k without units, or carrying mol/L and mol/m³ (and min with s) through the same calculation.
- Mixing up rates of different species: for
A + 2B → C,-r_B = 2(-r_A), not-r_A. - Substituting a temperature in °C into the Arrhenius equation, or dropping the minus sign so E comes out negative.
- Treating
-dC_A/dtas the rate in a variable-volume system - the correct definition is(1/V) dN_A/dt. - Claiming a catalyst shifts equilibrium, or that a higher-E reaction is necessarily slower at the operating temperature.
- Over-fitting: declaring an order from two or three scattered points, or differentiating noisy batch data without smoothing.
Related topics
Reactor design: batch, CSTR and PFR · Multiple reactions and selectivity · Catalysis · Non-ideal reactors and residence time distribution · Chemical reaction thermodynamics and equilibrium · Biochemical engineering basics