Skip to content

Chemistry · Ch 7 — Chemical Kinetics

Collision Theory

7.7

Collision Theory

The core idea. Proposed independently by Max Trautz (1916) and William Lewis (1918), and grounded in the kinetic theory of gases: a chemical reaction happens because reacting molecules physically COLLIDE. For A2(g)+B2(g)→2AB(g)A_2(g)+B_2(g)\rightarrow2AB(g), if reaction proceeds purely via collision, the rate should track the number of collisions occurring per second, and since collision frequency rises with how crowded each reactant is, Collision rate∝[A2][B2]\text{Collision rate}\propto[A_2][B_2], written as Collision rate=Z[A2][B2]\text{Collision rate}=Z[A_2][B_2] where Z (obtainable from the kinetic theory of gases) is the theoretical total-collision-frequency constant.

Why most collisions do NOT lead to reaction. At room temperature (298 K) and 1 atm, kinetic theory predicts roughly 10910^9 collisions per molecule per second -- if every single one of these led to product, essentially every gas-phase reaction would finish in about a billionth of a second. Real reactions plainly take far longer, so most collisions must be INEFFECTIVE. The missing ingredient is energy: to actually react, colliding molecules must together carry at least a minimum energy called the activation energy, EaE_a (visualised as the energy barrier separating reactants from products in Fig 7.5) -- collisions with less kinetic energy than EaE_a simply bounce the molecules apart unchanged.

Quantifying the energetic fraction. The FRACTION of collisions that are energetic enough is given by the Boltzmann-type expression

f=e−Ea/RTf=e^{-E_a/RT}

Working this out for a typical activation energy of 100 kJ mol−1100\ \text{kJ mol}^{-1} at 300 K: f=e−(100×1038.314×300)=e−40.09≈4×10−18f=e^{-\left(\frac{100\times10^3}{8.314\times300}\right)}=e^{-40.09}\approx4\times10^{-18} -- meaning out of every 101810^{18} collisions, only about FOUR are energetic enough to have any chance of reacting. This single number explains why real reactions, despite billions of collisions per second, still take a measurable amount of time.

Even energy is not enough -- orientation matters too. Even a sufficiently energetic collision fails to react if the colliding molecules are not aligned correctly to form the transition state (Fig 7.6 shows this directly: 'proper alignment' between A2_2 and B2_2 gives an effective collision that splits cleanly into two AB product molecules, while 'improper alignment' gives an ineffective collision where the same reactant molecules simply bounce apart unreacted). The fraction of energetic collisions that ALSO have the right orientation is captured by the steric (orientation) factor p, so …

Figure 7.5Progress of the reaction (potential energy diagram)

What this figure shows. A potential-energy-vs-reaction-progress curve: potential energy on the y-axis, reaction progress on the x-axis. The curve starts flat and low, labelled 'Reactants', rises smoothly to a single rounded peak, then falls back down to a flat, lower plateau labelled 'Products' (an exothermic profile, products lower than reactants). A double-headed vertical arrow labelled EaE_a spans from the reactants' energy level up to the peak, marking the activation energy as the minimum energy barrier that must be crossed -- regardless of colliding with enough energy, mol …

Figure 7.6Orientation of reactants -- schematic representation

What this figure shows. Two side-by-side collision sequences for a hypothetical A2+B2→2ABA_2+B_2\rightarrow2AB. TOP row ('Proper alignment'): two A atoms and two B atoms approach with A facing A and B facing B; they collide to form a tight four-atom cluster ('Effective collision'), which then splits cleanly into two separate AB product molecules. BOTTOM row ('improper alignment'): the same four atoms approach but misaligned (A-A pointed at B-B side-on rather than end-on); they collide ('ineffective collision') but simply bounce apart again as the SAME unreacted A2+B2A_2+B_2 reactants, with no product formed -- illustrating that energy alone is not suff …

Misc collision-factor-worked-calcWorked calculation -- magnitude of the effective-collision fraction f

Worked out. For a reaction with activation energy 100 kJ mol−1100\ \text{kJ mol}^{-1} at 300 K: f=e−(100×103 J mol−18.314 J K−1mol−1×300 K)=e−40.09≈4×10−18f=e^{-\left(\frac{100\times10^3\ \text{J mol}^{-1}}{8.314\ \text{J K}^{-1}\text{mol}^{-1}\times300\ K}\right)}=e^{-40.09}\approx4\times10^{-18}. Interpreted literally: of every 101810^{18} molecular collisions at this temperature, only about 4 carry enough energy to convert reactants into products -- the other ∼1018\sim10^{18} collisions are simply too gentle, which is exactly why bulk chemical reactions, despite billions of collisions per second, still take a measurable, finite time to go to completion …