Activation energy: the energy hill
Reactants do not turn into products directly. They first form a short-lived, high-energy arrangement called the activated complex (transition state). The extra energy needed to reach it is the activation energy, Ea.
On an energy profile diagram Ea is the height from the reactants to the top of the hill. The difference between reactants and products is ΔH. Low Ea → fast reaction; high Ea → slow reaction.
Effect of temperature: why heat speeds things up
At any temperature, molecules have a spread of energies (the Maxwell–Boltzmann distribution). Only the few in the high-energy tail have energy ≥ Ea. When the temperature rises, the whole spread shifts to higher energy and the tail above Ea becomes much bigger.
For many reactions, a 10 °C rise nearly doubles the rate. The ratio k(T+10)/k(T) is called the temperature coefficient, and it is usually between 2 and 3.
The Arrhenius equation
k = A e^(−Ea/RT)
- A = frequency (pre-exponential) factor: linked to how often molecules collide.
- e^(−Ea/RT) = fraction of molecules with energy ≥ Ea.
- R = 8.314 J K⁻¹ mol⁻¹, T in kelvin.
Take ln: ln k = ln A − Ea/RT. A graph of ln k against 1/T is a straight line with slope −Ea/R.
For two temperatures (derived step by step by subtracting the two ln equations):
log(k₂/k₁) = (Ea/2.303R) × (T₂ − T₁)/(T₁T₂)
Catalysts
A catalyst speeds up a reaction and is left unchanged in mass and chemical nature at the end. It works by giving an alternative path with lower Ea (often by forming a short-lived intermediate).
- It does not change ΔH or ΔG.
- It speeds up forward and backward reactions equally, so it does not change the equilibrium constant; equilibrium is just reached sooner.
- A small amount is enough. It cannot start a reaction that is not feasible (ΔG positive).
Examples: iron in making ammonia; enzymes like amylase in saliva.
Collision theory
Molecules are pictured as hard spheres. A reaction happens only on an effective collision, which needs two things:
- Enough energy: at least the threshold energy (Ea above the average).
- Proper orientation: the right atoms must meet. In CH₃Br + OH⁻, the OH⁻ must hit the carbon from the side opposite to Br.
Rate = P × Z_AB × e^(−Ea/RT), where Z_AB is the collision frequency and P is the probability (steric) factor for correct orientation.
Limit: it treats molecules as simple spheres and ignores their structure, so it works best for simple gas molecules.
Try it: predict, then check
In the 3D, set T = 300 K and Ea = 60 kJ/mol. Predict whether k goes up 2× or 10× when you move to 310 K, then check the readout. Turn on the catalyst and see the ball cross. At home: put one glow stick (or a cup of milk with a spoon of curd) in cold water and one in warm water. The warm one glows brighter (or sets faster): heat helps more molecules cross the hill.
Key formulas and definitions
- k = A e^(−Ea/RT)
- ln k = ln A − Ea/RT (slope of ln k vs 1/T = −Ea/R)
- log k = log A − Ea/(2.303RT)
- log(k₂/k₁) = (Ea/2.303R) × (T₂ − T₁)/(T₁T₂)
- Rate (collision theory) = P Z_AB e^(−Ea/RT)
- Temperature coefficient = k(T+10)/k(T) ≈ 2 to 3
Worked examples
1. A reaction's rate becomes 2 times when the temperature rises from 300 K to 310 K. Find Ea.
log 2 = (Ea/2.303 × 8.314) × 10/(300 × 310). 0.301 = Ea × 10/(19.147 × 93000). Ea = 0.301 × 19.147 × 9300 ≈ 53 600 J/mol ≈ 53.6 kJ/mol.
2. Ea = 0 for a reaction. What is k?
k = A e^0 = A. The rate constant equals the frequency factor and does not depend on temperature.
3. Find the fraction of molecules with energy ≥ Ea when Ea = 57.4 kJ/mol and T = 300 K.
Fraction = e^(−Ea/RT). Ea/RT = 57400/(8.314 × 300) = 23.0. Using log: log x = −23.0/2.303 = −10.0. So x ≈ 1 × 10⁻¹⁰.
4. A plot of ln k against 1/T has slope −6000 K. Find Ea.
Slope = −Ea/R. Ea = 6000 × 8.314 = 49 884 J/mol ≈ 49.9 kJ/mol.
5. A catalyst lowers Ea from 100 kJ/mol to 80 kJ/mol at 300 K (A unchanged). By how many times does k increase?
k₂/k₁ = e^((100000 − 80000)/(8.314 × 300)) = e^(8.02). log(k₂/k₁) = 8.02/2.303 = 3.48. k₂/k₁ ≈ 3.0 × 10³, about 3000 times.
6. k = 2 × 10⁻² s⁻¹ at 300 K and Ea = 60 kJ/mol. Find k at 320 K.
log(k₂/k₁) = 60000/(2.303 × 8.314) × 20/(300 × 320) = 3133.6 × 2.083×10⁻⁴ = 0.653. k₂/k₁ = 4.5. k₂ ≈ 9.0 × 10⁻² s⁻¹.
7. For log k = 14.34 − 1.25 × 10⁴ K/T, find Ea.
Compare with log k = log A − Ea/(2.303RT): Ea/2.303R = 1.25 × 10⁴. Ea = 1.25 × 10⁴ × 2.303 × 8.314 ≈ 239 × 10³ J/mol ≈ 239 kJ/mol.
Common mistakes
- Using °C instead of kelvin in the Arrhenius equation. Always convert: T(K) = t(°C) + 273.
- Mixing kJ and J: R = 8.314 J K⁻¹ mol⁻¹, so Ea must be in joules.
- Saying a catalyst changes ΔH or the equilibrium constant. It changes only the path and Ea.
- Thinking every collision gives a reaction. Only collisions with enough energy and the right orientation work.