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Chemical kinetics

Chemical Engineering (B.E./B.Tech) · Reaction engineering · free study notes for exams, placements and interviews

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

Rate laws, order and molecularity

Temperature dependence

Integrated rate laws for constant-volume batch systems

Starting from -dC_A/dt = k C_A^n for a batch reactor of constant volume:

Analysis of kinetic data: how order and k are found

Three standard methods, all examined:

  1. 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; k comes from the slope. Simple and robust to scattered data, but trial-and-error and poor for fractional orders.
  2. Differential method - obtain -dC_A/dt from the C_A vs t curve (graphical slopes or finite differences), then plot ln(-dC_A/dt) against ln C_A: slope = order n, intercept = ln k. Handles any order including fractional, but differentiating noisy data amplifies error.
  3. Method of initial rates - measure the initial slope at several different C_A0; ln(-r_A0) vs ln C_A0 gives n. Avoids product-inhibition effects entirely, but needs many runs.

Supporting techniques:

Mechanisms, rate-determining step and approximations

Key points

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/k
Second 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 t
Reversible 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·s
Lindemann: -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.

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_A1/C_A (L/mol)
02.0000.6930.500
101.0000.0001.000
200.667-0.4051.500
300.500-0.6932.000
400.400-0.9162.500

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.

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.

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.

Comparison of order diagnostics

Order nLinear plotk unitst_1/2Dependence on C_A0
0C_A vs tmol/(L·min)C_A0/2k∝ C_A0
1ln C_A vs tmin⁻¹0.693/knone
21/C_A vs tL/(mol·min)1/(k C_A0)∝ 1/C_A0
31/C_A² vs tL²/(mol²·min)1.5/(k C_A0²)∝ 1/C_A0²

Where it is used

Interview questions

Common mistakes

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

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