Chemical Kinetics · Lesson 7 of 10
Temperature, Activation Energy and Energy Distributions
“Explain temperature-dependent reaction rates through energy distributions and distinguish an activation barrier from the net energy change.”
• Explain why temperature often changes a rate constant strongly. • Interpret an energy profile and distinguish barriers, intermediates and reaction enthalpy. • Describe the effect of warming on a normalised molecular-energy distribution. • Interpret the Arrhenius equation with consistent energy and temperature units.
Why a warmer reaction can proceed much faster
A dilute permanganate solution can lose its colour more rapidly in reaction with oxalic acid when conditions are warmer. Many food changes also accelerate on warming. A temperature change can alter the rate constant itself, rather than merely change the reactant concentrations. For many common thermal reactions, an increase of about 10 K produces a noticeable rate increase, but the multiplier is not a universal constant.
A measured rate is not the same as the rate constant. Comparing rates at different temperatures also requires comparable concentrations, solvent conditions, catalyst and pathway. A simple statement such as “the rate doubles for every 10 K” is an empirical approximation over a particular range, not a law applicable to every reaction or every starting temperature.
The energy barrier differs from the endpoint change
In an Arrhenius description, the parameter controlling the exponential temperature dependence of k. A simple reaction-profile picture represents a barrier between the reactant level and an activated arrangement, while more detailed theories distinguish several related activation quantities.
On a reaction-coordinate diagram, the horizontal coordinate represents progress in atomic arrangement, not elapsed time or distance travelled by one molecule. The vertical coordinate must be identified as a kind of energy. A peak illustrates an activated arrangement or transition state. Reactants can reach that high-energy region and either continue toward products or return; forming products is not guaranteed simply by reaching a certain geometry.
The net enthalpy change compares products with reactants. In an exothermic schematic, products lie below reactants even though the forward route still has a positive barrier. This explains why a reaction can release considerable energy yet start slowly. For an endothermic schematic, products lie higher. Neither endpoint difference alone determines the rate.
Problem
A schematic one-step enthalpy profile has reactants at 0 kJ mol⁻¹, a barrier top at 60 kJ mol⁻¹ and products at −30 kJ mol⁻¹. Interpret the energy differences.
- 1.Forward barrier height is 60−0=60 kJ mol⁻¹. Reverse barrier height is 60−(−30)=90 kJ mol⁻¹.
- 2.Reaction enthalpy is −30−0=−30 kJ mol⁻¹; the conversion is exothermic in this diagram.
- 3.The difference between the forward and reverse schematic barrier heights is −30 kJ mol⁻¹, matching the endpoint change.
- 4.These are a consistent simple-profile illustration, not an assertion that measured Arrhenius E_a, activation enthalpy and every potential-energy barrier are exactly identical quantities.
The balanced conversion H₂(g)+I₂(g)→2HI(g) can be represented schematically with H–H and I–I bonds being reorganised while H–I bonds form. A barrier-top drawing is an activated arrangement, not a stable intermediate automatically implied by the overall equation. The actual mechanism requires evidence beyond this simplified picture.
At one temperature molecules have a range of energies
A gas sample does not consist of molecules all moving with the same kinetic energy. A normalised energy distribution assigns an area fraction to each energy interval. Its total area is one. The most probable energy is at the peak, but many molecules have smaller or larger energies. Probability density at a single energy and probability in an interval are different quantities: an area under the curve represents the latter.
When temperature rises, a normalised Maxwell–Boltzmann translational-energy curve broadens, its peak moves to higher energy and generally becomes lower. The total area remains one because probabilities still sum to one. The area above a fixed high-energy threshold becomes larger. The curve does not simply move upward everywhere, and the barrier itself does not have to fall merely because the sample is warmed.
The Arrhenius equation captures the temperature dependence
The exponent must be dimensionless: E_a and RT both have energy-per-mole units. For positive E_a at fixed A, increasing T makes −E_a/(RT) less negative, so the exponential and k increase. At fixed T and A, a lower E_a also increases k. A may reflect encounter, orientation and dynamical factors; it is not automatically equal to the number of collisions per second per unit volume.
The factor exp(−E_a/RT) is the simplified activation or Boltzmann factor used in this model. It gives the strong exponential sensitivity to the barrier. It should not be presented as the exact integral of every three-dimensional molecular translational-energy distribution: that tail area also includes the distribution’s shape. For learning the rate-constant law, distinguish this activation factor from the qualitative tail-area picture that motivates it.
Problem
Evaluate exp(−E_a/RT) for E_a=209.5 kJ mol⁻¹ and T=581 K.
- 1.Convert 209.5 kJ mol⁻¹ to 209500 J mol⁻¹ so it matches R=8.314 J mol⁻¹ K⁻¹.
- 2.E_a/(RT)=209500/(8.314×581)≈43.3708, a dimensionless positive number.
- 3.The activation factor is exp[−209500/(8.314×581)]=1.45976e-19. It is tiny because the barrier is large compared with RT.
- 4.This is the simplified Arrhenius energy-selection factor, rather than an exact measured fraction of all molecules or a complete prediction of the observed rate without A.
Problem
Using an Arrhenius model with E_a=50.0 kJ mol⁻¹ and nearly constant A, compare the effect of warming 300→310 K with warming 600→610 K.
- 1.At the lower temperatures, the rate-constant ratio is exp[(50000/8.314)(1/300−1/310)]≈1.91.
- 2.At the higher temperatures, the same expression gives exp[(50000/8.314)(1/600−1/610)]≈1.18.
- 3.The same energy barrier and same 10 K increase give different multipliers because reciprocal temperature, not temperature difference alone, controls the exponent.
- 4.The next lesson derives the ratio equation systematically and uses it for quantitative prediction.
The simple Arrhenius form is useful over many ranges but can fail when the dominant pathway, solvent state, catalyst or mechanism changes. Apparent negative activation energies and non-Arrhenius behaviour can occur in more complex systems; they do not justify ignoring the conditions of a positive-barrier example.
Quick Quiz
Quiz
For positive E_a and fixed A, warming usually changes k how?
Explanation: The negative exponent becomes less negative as absolute temperature rises, increasing the exponential factor.
What remains one for a normalised molecular-energy distribution at any temperature?
Explanation: The area represents total probability, which remains one even as the distribution broadens.
An exothermic reaction can still be slow because:
Explanation: Endpoint energy release does not eliminate the barrier that must be negotiated along a pathway.
Which temperature and energy units are compatible with R=8.314 J mol⁻¹ K⁻¹?
Explanation: The ratio E_a/(RT) must be dimensionless, requiring matching energy units and absolute temperature.
Does warming automatically lower the energy barrier of the same schematic pathway?
Explanation: The standard temperature explanation changes energy populations and the Arrhenius factor rather than requiring a lowered barrier.
Practice Problems
- Explain why a molecular-energy distribution peak can fall while a reaction becomes faster on warming.
- A profile has reactants at 20, a barrier top at 75 and products at 5 kJ mol⁻¹. Find forward barrier height, reverse barrier height and reaction enthalpy.
- Why must 60.0 kJ mol⁻¹ be converted before insertion with R=8.314 J mol⁻¹ K⁻¹, and what is its activation factor at 300 K?
- Does a large negative reaction enthalpy imply a large k? Explain with the energy-profile picture.
- Interpret exp(−E_a/RT) and distinguish it from an exact Maxwell–Boltzmann tail probability.
A warmer normalised distribution broadens and shifts probability toward higher energies. Its peak can therefore fall while the high-energy area above a fixed threshold grows. Reaction speed depends on the relevant effective encounters and barrier crossing, not on the maximum height of the plotted distribution.
Key Takeaways
• Temperature can change k itself, independently of concentration changes. • A 10 K temperature increase does not have a universal fixed multiplier. • A reaction barrier is distinct from the enthalpy difference between products and reactants. • An activated arrangement at a peak is distinct from an intermediate between steps. • Warming broadens the normalised energy distribution and increases a relevant high-energy population. • Arrhenius behaviour is k=Aexp(−E_a/RT) over a suitable range. • Use kelvin and matching molar-energy units so the exponent is dimensionless. • The Arrhenius activation factor is not the exact integral of every molecular-energy distribution.