Rate laws, order, molecularity and Arrhenius equation
Rate expressions, order versus molecularity, units of k and the Arrhenius temperature dependence, with worked calculations of k and activation energy.
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Why it matters
Every reactor calculation starts from a rate expression: without −rA as a function of concentration and temperature you cannot size a batch vessel, a CSTR or a PFR. The order of reaction tells you how strongly conversion responds to dilution or pressure, and the Arrhenius equation tells you how much a 10 K change in temperature will speed up (or run away with) the reaction.
Key ideas
Rate of reaction. For a homogeneous reaction the rate of disappearance of A is the moles of A consumed per unit volume per unit time, written −rA (mol/m³·s). For aA + bB → cC + dD the rates are linked by stoichiometry: −rA/a = −rB/b = rC/c = rD/d. So if 2A + B → C and −rA = 4 mol/m³·s, then −rB = 2 mol/m³·s and rC = 2 mol/m³·s. Always state which species a rate (and its k) refers to.
Rate law. An empirical expression −rA = kA·CA^α·CB^β. The exponents α and β are the orders with respect to A and B; n = α + β is the overall order. Orders are found by experiment and need not match stoichiometric coefficients; they may be zero, fractional or negative (e.g. an inhibiting product). Some rate laws have no single order at all, for example −rA = k1·CA/(1 + k2·CA), which is first order at low CA and zero order at high CA.
Elementary and non-elementary reactions. A reaction is elementary when the rate law follows directly from the stoichiometry of a single molecular event (A + B → products with −rA = k·CA·CB). Most industrial reactions are non-elementary: the observed rate law is the net result of a sequence of elementary steps (a mechanism), often involving short-lived intermediates handled with the pseudo-steady-state hypothesis.
Molecularity. The number of molecules that take part in one elementary step: uni-, bi- or (rarely) termolecular. It is always a small whole number and is defined only for elementary steps. Order is an experimental quantity defined for the overall rate law; molecularity is a theoretical count for one step. For an elementary step they coincide.
Units of k. From −rA = k·CA^n, k has units (mol/m³)^(1−n)·s⁻¹. Zero order: mol/m³·s; first order: s⁻¹; second order: m³/mol·s. Reading the units of k in a problem is the fastest way to identify the order.
Temperature dependence — Arrhenius. k = k0·exp(−E/RT). E is the activation energy (J/mol), the minimum energy barrier between reactants and products; k0 (also written A) is the frequency or pre-exponential factor with the same units as k. A plot of ln k against 1/T is a straight line of slope −E/R. A large E means high temperature sensitivity: the rate rises steeply with T. Collision theory predicts k ∝ T^0.5·exp(−E/RT) and transition-state theory k ∝ T·exp(−E/RT), but over normal ranges the exponential term dominates, so the simple Arrhenius form is used in design.
Catalysts lower the effective activation energy by providing a different mechanism. They change k (and may change the rate law itself because the mechanism changes), but they never change the equilibrium conversion.
Reversible reactions. For A ⇌ R, −rA = k1·CA − k2·CR, and at equilibrium Kc = k1/k2. The van 't Hoff equation d ln K/dT = ΔHr/RT² links equilibrium to temperature in the same way Arrhenius links k to temperature.
Formulas
−rA/a = −rB/b = rC/c = rD/d
- For aA + bB → cC + dD; each r in mol/m³·s.
−rA = k·CA^α·CB^β, n = α + β
- CA, CB: concentrations (mol/m³); k: rate constant, units (mol/m³)^(1−n)·s⁻¹; α, β: orders (dimensionless).
k = k0·exp(−E/(R·T))
- k0: frequency factor (same units as k); E: activation energy (J/mol); R = 8.314 J/mol·K; T: absolute temperature (K).
ln(k2/k1) = (E/R)·(1/T1 − 1/T2)
- Two-temperature form; use to find k at a new temperature or to find E from two measured rate constants. Valid when E is constant over T1–T2.
ln k = ln k0 − (E/R)·(1/T)
- Linear (Arrhenius plot) form: slope = −E/R.
Kc = k1/k2
- Equilibrium constant of an elementary reversible reaction A ⇌ R.
Worked examples
Example 1 (standard). A first-order reaction has k = 2.5 × 10⁻³ s⁻¹ at 300 K and E = 50 kJ/mol. Find k at 350 K.
- Formula: ln(k2/k1) = (E/R)·(1/T1 − 1/T2).
- E/R = 50 000 / 8.314 = 6014 K.
- 1/T1 − 1/T2 = 1/300 − 1/350 = 4.762 × 10⁻⁴ K⁻¹.
- ln(k2/k1) = 6014 × 4.762 × 10⁻⁴ = 2.864, so k2/k1 = e^2.864 = 17.53.
- k2 = 2.5 × 10⁻³ × 17.53 = 4.38 × 10⁻² s⁻¹.
k at 350 K = 4.38 × 10⁻² s⁻¹ (a 50 K rise speeds the reaction about 17.5 times).
Example 2 (GATE level). The rate of a reaction doubles when the temperature is raised from 300 K to 310 K at the same composition. Find the activation energy.
- At fixed composition the rate ratio equals the rate-constant ratio: k2/k1 = 2.
- Formula rearranged: E = R·ln(k2/k1) / (1/T1 − 1/T2).
- 1/300 − 1/310 = 1.0753 × 10⁻⁴ K⁻¹.
- E = 8.314 × 0.6931 / 1.0753 × 10⁻⁴ = 53 594 J/mol.
E ≈ 53.6 kJ/mol. This is the origin of the rule of thumb that "the rate doubles for every 10 °C rise near room temperature" — it holds only for E around 50 kJ/mol.
Example 3 (stoichiometry). For 2A + B → 3C, −rA = 0.6 mol/m³·s. Then −rB = 0.6/2 = 0.3 mol/m³·s and rC = (0.6/2)·3 = 0.9 mol/m³·s.
Common mistakes
- Using °C instead of K in the Arrhenius equation, or E in kJ/mol with R in J/mol·K.
- Writing the two-temperature form with the sign reversed: if T2 > T1, k2 must come out larger than k1 for positive E — use that as a check.
- Taking orders from stoichiometric coefficients for a non-elementary reaction.
- Quoting molecularity for an overall reaction, or calling a fractional order a "molecularity".
- Forgetting that k depends on which species the rate is written for (kA = 2kB for 2A + B → C).
- Claiming a catalyst shifts equilibrium; it only changes how fast equilibrium is reached.
For GATE CH
Expect NAT questions on finding E from rate constants at two temperatures, finding k at a new temperature, identifying order from the units of k or from half-life data, and converting rates between species via stoichiometry. Conceptual MCQs test order versus molecularity, elementary versus non-elementary reactions and the effect of a catalyst. Practise the logarithm arithmetic quickly and carry R = 8.314 J/mol·K consistently.
Quick check
- What are the units of k for a second-order reaction with concentration in mol/m³?
- Can the order of a reaction be 1.5? Can the molecularity?
- For 2A → R, −rA = 0.8 mol/m³·s. What is rR?
- On an Arrhenius plot, what does a steeper line mean?
Answers: 1. m³/mol·s; 2. order yes, molecularity no; 3. 0.4 mol/m³·s; 4. a larger activation energy (greater temperature sensitivity).
Interview questions
All Chemical Reaction Engineering interview questionsTry answering each one aloud before you open it.
1.What is a rate law in chemical reaction engineering?Concept
A rate law is an equation that relates the rate of a chemical reaction to the concentration of the reactants. It typically takes the form: rate = k[A]^m[B]^n, where k is the rate constant, [A] and [B] are the concentrations of the reactants, and m and n are the reaction orders with respect to each reactant.
2.Explain the difference between reaction order and molecularity.Concept
Reaction order is an empirical parameter that indicates how the rate of reaction depends on the concentration of reactants. It can be a whole number, fraction, or zero. Molecularity, on the other hand, is a theoretical concept that refers to the number of molecules colliding in an elementary reaction step. It is always a whole number.
3.What is the Arrhenius equation and what does it describe?Concept
The Arrhenius equation is a formula that describes how the rate constant (k) of a reaction changes with temperature. It is given by k = A·e^(-Ea/RT), where A is the pre-exponential factor, Ea is the activation energy, R is the universal gas constant, and T is the temperature in Kelvin. This equation shows that the rate constant increases with temperature.
4.Why is the Arrhenius equation important in chemical reaction engineering?Application
The Arrhenius equation is important because it helps predict how changes in temperature affect the rate of a chemical reaction. This is crucial for designing reactors and optimizing reaction conditions to achieve desired reaction rates and yields.
5.What happens to the rate of a reaction if the activation energy is decreased?Application
If the activation energy of a reaction is decreased, the rate of the reaction increases. This is because a lower activation energy means that more molecules have sufficient energy to overcome the energy barrier, leading to more frequent successful collisions.
6.How does a catalyst affect the rate law of a reaction?Application
A catalyst opens a different reaction pathway (mechanism) with a lower activation energy, so the rate constant at a given temperature is much larger. Because the mechanism changes, the form of the rate law and the apparent orders can also change; catalytic rate laws often take a Langmuir-Hinshelwood form instead of a simple power law. What a catalyst cannot change is the equilibrium constant or equilibrium conversion, since it accelerates the forward and reverse reactions equally.
7.What is the significance of the pre-exponential factor (A) in the Arrhenius equation?Concept
The pre-exponential factor (A) in the Arrhenius equation represents the frequency of collisions and the orientation of reactant molecules. It is a measure of the number of times reactants approach the activation energy barrier per unit time. A higher value of A indicates more frequent and effective collisions.
8.Calculate the rate constant at 350 K for a first-order reaction with an activation energy of 50 kJ/mol and a pre-exponential factor of 1.5 × 10^12 s^-1.Numerical
Use k = k0·exp(−E/RT). E/RT = 50 000 / (8.314 × 350) = 17.18, so exp(−17.18) = 3.45 × 10^-8. Then k = 1.5 × 10^12 × 3.45 × 10^-8 ≈ 51.7 s^-1. An interviewer will also want you to note that E must be in J/mol with R in J/mol·K and T in kelvin.
9.If the concentration of a reactant is doubled in a first-order reaction, what happens to the rate of the reaction?Application
In a first-order reaction, the rate of the reaction is directly proportional to the concentration of the reactant. Therefore, if the concentration of the reactant is doubled, the rate of the reaction will also double.
10.Determine the overall order of a reaction with the rate law: rate = k[A]^2[B]^1.Numerical
The overall order of a reaction is the sum of the exponents of the concentration terms in the rate law. For the given rate law, rate = k[A]^2[B]^1, the overall order is 2 + 1 = 3.
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