10.2 Arrhenius Equation and Temperature Effects

Key Takeaways

  • The Arrhenius equation k = A e^(−E_a/RT) links the rate constant to absolute temperature and activation energy E_a.
  • Larger E_a means stronger temperature sensitivity: the same ΔT changes k more for high-E_a reactions.
  • A catalyst provides an alternative path with lower effective activation energy (or higher pre-exponential factor), increasing rate without changing equilibrium K.
  • The two-point form ln(k₂/k₁) = (E_a/R)(1/T₁ − 1/T₂) estimates how k scales between temperatures.
  • Raising T usually raises rate for elementary irreversible steps, but selectivity among parallel/series paths can improve or worsen depending on relative E_a values.
Last updated: August 2026

10.2 Arrhenius Equation and Temperature Effects

Quick Answer: (k = A e^{-E_a/RT}). Higher T or lower E_a raises k. Catalysts lower effective E_a (new path). Use (\ln(k_2/k_1) = (E_a/R)(1/T_1 - 1/T_2)) for two-temperature scaling. Catalysts do not change equilibrium K.

Section 10.1 fixed the concentration dependence of rate. Almost every industrial reaction rate also depends sharply on temperature through the rate constant. Fired heaters, steam jackets, quench exchangers, and interstage coolers on Qatar process units exist largely to put the reactor on a temperature that balances rate, selectivity, catalyst life, and materials limits.

The Arrhenius Equation

The Arrhenius form for the rate constant is:

[ k = A , e^{-E_a/(RT)} ]

where:

  • k — rate constant (units depend on order; Section 10.1)
  • A — pre-exponential (frequency) factor; same units as k
  • E_a — activation energy (J/mol or kJ/mol)
  • R — gas constant (8.314 J/(mol·K) if E_a in J/mol)
  • Tabsolute temperature (K)—never use °C inside the exponential

Physical picture: molecules must clear an energy barrier to react. The fraction of collisions (or transition-state attempts) with enough energy scales roughly as e^{−E_a/RT}. Larger E_a ⇒ fewer successful events at a given T ⇒ smaller k.

ParameterIf increasedEffect on k (typical irreversible step)
Tk ↑ (exponentially)
E_ak ↓ strongly
Ak ↑ proportionally

Linear form used to fit data:

[ \ln k = \ln A - \frac{E_a}{R},\frac{1}{T} ]

Plot ln k vs 1/T: slope = −E_a/R, intercept = ln A. This is the kinetic companion to the van ’t Hoff plot for equilibrium constants—do not confuse the two on the exam.

Activation Energy Meaning

Activation energy E_a is the empirical barrier height in the Arrhenius model. Rules of thumb for MCQ intuition (not universal laws):

E_a magnitude (order)Temperature sensitivity
Low (tens of kJ/mol)Mild; rate rises modestly with T
High (100+ kJ/mol)Strong; small ΔT can double/triple k
Very highMay need catalysis or extreme T to be practical

Rule-of-thumb doubling: for many liquid-phase organic reactions near ambient to moderate T, rate roughly doubles for a ~10 °C rise—but the exact factor depends on E_a and the temperature level. Always prefer the Arrhenius formula over a blind “10 °C rule” when numbers are given.

Endothermic vs exothermic confusion: E_a is a kinetic barrier. The heat of reaction ΔH_rxn is thermodynamic. An exothermic reaction can still have a large E_a (slow without heat/catalyst). Equilibrium shifts with T via van ’t Hoff; rate always cares about k(T) and concentrations.

Effect of Temperature on Rate

For irreversible power-law kinetics with k(T) Arrhenius:

[ (-r_A) = A e^{-E_a/(RT)} C_A^n \cdots ]

At fixed composition, raising T raises (−r_A). In a real reactor, heating also changes density, equilibrium limits (reversible reactions), side reactions, and catalyst deactivation—so “hotter is always better” is false for plant design. On UPDA conceptual items, for a single irreversible desired reaction with no selectivity issue stated, higher T ⇒ higher rate ⇒ smaller required V or t for a target conversion (within material limits).

SituationTypical T lever
Irreversible, desired reaction onlyHigher T speeds conversion
Reversible exothermic (e.g. many equilibrium-limited syntheses)High T helps rate but hurts equilibrium X_e—tradeoff
Reversible endothermicHigher T helps both rate and equilibrium
Parallel reactions with different E_aT shifts selectivity toward the higher-E_a path as T rises

Selectivity intro: if desired path has higher E_a than a waste path, higher T favors the desired rate relatively more—but absolute waste rate also rises. Exam stems that mention “activation energy of the desired reaction is higher” are cueing this relative Arrhenius effect.

Catalysts: Lower Effective E_a / New Path

A catalyst participates in the mechanism and is regenerated; it provides an alternative reaction path with a lower activation energy (and/or more favorable A), so k is larger at the same T.

FactExam-safe statement
RateCatalyst increases rate of approach to equilibrium
EquilibriumCatalyst does not change thermodynamic K or X_e
Both directionsCatalyst accelerates forward and reverse; equilibrium composition unchanged
AmountRate often scales with catalyst mass or active sites (heterogeneous: per mass or bed volume)
SelectivityDifferent catalysts can favor different products (different paths)

Not a catalyst: raising T, increasing concentration, or adding more reactant feed—these change rate without offering a new catalytic cycle. Inhibitors or poisons reduce active sites and lower observed rate.

Industrial context for Qatar-relevant plants: hydrotreating, reforming, Claus, ammonia, methanol, and FCC-type chemistry all live or die by catalyst activity, E_a of key steps, and temperature profiles—even when the exam stays at Arrhenius vocabulary level.

Two-Point Arrhenius Form and Worked Example

Between temperatures T₁ and T₂ with rate constants k₁ and k₂:

[ \ln\left(\frac{k_2}{k_1}\right) = \frac{E_a}{R}\left(\frac{1}{T_1}-\frac{1}{T_2}\right) ]

Equivalently:

[ \frac{k_2}{k_1} = \exp\left[\frac{E_a}{R}\left(\frac{1}{T_1}-\frac{1}{T_2}\right)\right] ]

If T₂ > T₁, then (1/T₁ − 1/T₂) > 0 and k₂ > k₁ for E_a > 0.

Worked numeric example

A first-order irreversible reaction has E_a = 60 kJ/mol = 60,000 J/mol. At T₁ = 300 K, k₁ = 0.020 min⁻¹. Estimate k₂ at T₂ = 320 K. Use R = 8.314 J/(mol·K).

[ \frac{1}{T_1}-\frac{1}{T_2} = \frac{1}{300}-\frac{1}{320} = 0.003333 - 0.003125 = 0.000208,\mathrm{K^{-1}} ]

[ \frac{E_a}{R}\left(\frac{1}{T_1}-\frac{1}{T_2}\right) = \frac{60000}{8.314}\times 0.000208 \approx 7217 \times 0.000208 \approx 1.50 ]

[ \frac{k_2}{k_1} = e^{1.50} \approx 4.48 \quad \Rightarrow \quad k_2 \approx 0.020 \times 4.48 \approx 0.090,\mathrm{min^{-1}} ]

A 20 K rise near 300 K roughly quadrupled k for this E_a—illustrating strong exponential sensitivity. If E_a were only 30 kJ/mol, the exponent would be about half (~0.75) and k₂/k₁ ≈ 2.1—much milder.

Conceptual check with reverse ratio

If someone reports k at 320 K and you need 300 K, invert: k₁/k₂ ≈ 1/4.48 ≈ 0.22. Cooling strongly slows high-E_a chemistry—useful for quench to freeze conversion or stop degradation.

Linking Arrhenius to Reactor Size

Because design time or volume scales with 1/(−r_A) in integral design equations, increasing k by a factor of four (example above) can cut required batch time or PFR volume by about four for first-order kinetics at the same conversion (constant density, isothermal idealization). CSTR volume also falls, but the algebraic factor still tracks 1/(−r_A exit).

LeverEffect on kEffect on required V or t (fixed X, isothermal ideal)
Raise T
Better catalyst (lower E_a or higher A)
Poisoned catalyst

UPDA Exam Checklist for Section 10.2

  1. Always use kelvin in Arrhenius exponentials.
  2. ln k vs 1/T slope = −E_a/R for data fitting language.
  3. Two-point formula for k₂/k₁ when E_a and two temperatures are given.
  4. Catalyst: faster rates, lower effective barrier/path change; K unchanged.
  5. Reversible exothermic systems: rate–equilibrium temperature tradeoff.
  6. Parallel paths: relative E_a values control how selectivity moves with T.

With k(T) and −r_A(C,T) in hand, Section 10.3 inserts them into batch and CSTR design equations.

Test Your Knowledge

In the Arrhenius equation k = A e^(−E_a/RT), increasing the activation energy E_a at fixed temperature and fixed A will:

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B
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Test Your Knowledge

A solid catalyst is introduced for a reversible reaction at fixed T and composition. The most accurate statement is:

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D
Test Your Knowledge

Using ln(k₂/k₁) = (E_a/R)(1/T₁ − 1/T₂), if T₂ > T₁ and E_a > 0, then:

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D