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Chemical Reaction Kinetics

Collision theory, activation energy, and rate laws

Every second, the molecules in this flask slam into each other billions of times — and almost nothing happens. Most collisions are too gentle: they bounce off, unchanged. A reaction is a lottery where only the most violent crashes count, and temperature is how you rig the odds. Warm the flask by a barely noticeable ten degrees and you don't speed things up a little — you nearly double the winners. Chemistry answers to the exponential, not the thermometer.

What you'll be able to do

  • Use collision theory and the Maxwell-Boltzmann distribution to explain why only a fraction of collisions react
  • Apply the Arrhenius equation to quantify how temperature and activation energy set the rate constant
  • Explain catalysis as barrier lowering: faster path, same endpoints, catalyst unconsumed

Formulas

k=Ae−Ea/RT(Arrhenius equation)k = A e^{-E_a/RT} \quad (\text{Arrhenius equation})
Rate constant k increases exponentially with temperature
rate=k[A]m[B]n(rate law)\text{rate} = k[A]^m[B]^n \quad (\text{rate law})
Rate law: rate depends on concentration raised to reaction order
t1/2=ln⁡2k(first-order half-life)t_{1/2} = \frac{\ln 2}{k} \quad (\text{first-order half-life})
Half-life of a first-order reaction (independent of concentration)

Make a prediction

A reaction with Ea = 50 kJ/mol creeps along at 300 K. You warm it to 310 K — barely 3% hotter. Roughly what happens to the rate?

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Answer: It nearly doubles — reaction rate grows exponentially with temperature

The Arrhenius equation: ln(k₂/k₁) = (Ea/R)(1/T₁ − 1/T₂) = (50000/8.314)(1/300 − 1/310) ≈ 0.65, so k₂/k₁ ≈ e^0.65 ≈ 1.9. Warmer molecules do collide more often — but that effect scales with √T, a mere 1.6%. What actually explodes is the fraction of collisions carrying enough energy to clear Ea: the Boltzmann tail past the barrier grows exponentially with T. How often molecules collide matters far less than how hard.

Quiz (0/4)

A reaction has Ea = 50 kJ/mol. Raising the temperature from 300 K to 310 K increases the rate by approximately what factor? (R = 8.314 J/mol·K)

A first-order reaction has k = 0.1 s⁻¹. What is the half-life, and what fraction remains after 30 seconds?

The rate doubles when [A] is doubled, and doubles again when [B] is doubled. What is the rate law?

A catalyst reduces Ea from 80 kJ/mol to 50 kJ/mol at 300 K. By what factor does it increase the rate?

You can now

  • Use collision theory and the Maxwell-Boltzmann distribution to explain why only a fraction of collisions react
  • Apply the Arrhenius equation to quantify how temperature and activation energy set the rate constant
  • Explain catalysis as barrier lowering: faster path, same endpoints, catalyst unconsumed

Step-by-step

  1. Adjust temperature and watch molecules move faster, collide more energetically.
  2. The reaction rate counter increases.
  3. Switch to Energy Diagram view to see the activation energy barrier and how catalysts lower it.
  4. Use the Maxwell-Boltzmann graph to see what fraction of molecules can react.

Key formulas

  • k=Ae−Ea/RT(Arrhenius equation)k = A e^{-E_a/RT} \quad (\text{Arrhenius equation})Rate constant k increases exponentially with temperature
  • rate=k[A]m[B]n(rate law)\text{rate} = k[A]^m[B]^n \quad (\text{rate law})Rate law: rate depends on concentration raised to reaction order
  • t1/2=ln⁡2k(first-order half-life)t_{1/2} = \frac{\ln 2}{k} \quad (\text{first-order half-life})Half-life of a first-order reaction (independent of concentration)

Frequently asked questions

A reaction has Ea = 50 kJ/mol. Raising the temperature from 300 K to 310 K increases the rate by approximately what factor? (R = 8.314 J/mol·K).
You can work it out this way: use Arrhenius: ln(k2/k1) = (Ea/R)(1/T1 - 1/T2). = (50000/8.314)(1/300-1/310) ≈ 0.65, so k2/k1 ≈ e^0.65 ≈ 1.9 (about doubles).
A first-order reaction has k = 0.1 s⁻¹. What is the half-life, and what fraction remains after 30 seconds?
T₁/₂ = ln2/k = 0.693/0.1 = 6.93 s. After 30 s ≈ 4.3 half-lives: fraction = (1/2)^4.3 ≈ 0.05 (5%).
The rate doubles when [A] is doubled, and doubles again when [B] is doubled. What is the rate law?
Rate ∝ [A]¹[B]¹ → rate = k[A][B] (second-order overall, first-order in each reactant).
A catalyst reduces Ea from 80 kJ/mol to 50 kJ/mol at 300 K. By what factor does it increase the rate?
K_cat/k_uncat = e^((Ea_old - Ea_new)/RT) = e^(30000/(8.314×300)) = e^12.0 ≈ 1.6×10⁵. The catalyst speeds up the reaction 160,000-fold!