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Concept 06 / 06Chemical Kinetics

Collision Theory of Reaction Rates

Collision theory: collision frequency, effective collisions requiring threshold energy and proper orientation, the steric (probability) factor, rate = P Z e^(-Ea/RT), energy distribution of molecules and why a 10 K rise roughly doubles the rate, and limitations of the theory.

Figure 1
Collision theory (Arrhenius, Max Trautz and William Lewis) explains rates at the molecular level, treating reacting molecules as hard spheres. Main ideas. 1. Molecules must collide to react. The number of collisions per second per unit volume is the collision frequency ZZ (for example ZABZ_{\mathrm{AB}} for A and B). 2. Only effective collisions lead to products. A collision is effective if (a) the colliding molecules have kinetic energy at least equal to the threshold energy (EaE_{a} above the average), and (b) they collide with the proper orientation, so that the atoms that must bond come into contact. 3. Rate =PZABe−Ea/RT= P Z_{\mathrm{AB}} e^{-E_{a}/RT}, where PP is the probability or steric factor (the fraction of collisions with suitable orientation), and e−Ea/RTe^{-E_{a}/RT} is the fraction with enough energy. Comparing with Arrhenius, A=PZABA = P Z_{\mathrm{AB}}. Orientation. In CH3Br+OH−→CH3OH+Br−\mathrm{CH_{3}Br + OH^{-} \rightarrow CH_{3}OH + Br^{-}}, the hydroxide must approach the carbon from the side opposite bromine; collisions at the bromine end do nothing, however energetic. Energy distribution and temperature (see figure). The kinetic energies of molecules follow the Maxwell-Boltzmann distribution. The fraction with energy at least EaE_{a} is the area under the curve beyond EaE_{a}. On heating, the curve flattens and shifts to higher energy; the area beyond EaE_{a} roughly doubles for a 10 K10\,\mathrm{K} rise near room temperature (for typical Ea≈50 kJ mol−1E_{a} \approx 50\,\mathrm{kJ\,mol^{-1}}), which is why rate roughly doubles. The total number of collisions increases only slightly with temperature (about 2%2\% per 10 K10\,\mathrm{K}), so it is the energy factor that matters. A catalyst lowers EaE_{a}, which moves the threshold to the left and greatly enlarges the fraction of effective collisions at the same temperature. Limitations. The theory treats molecules as hard spheres and ignores their structure, so it cannot predict PP; calculated rates agree with experiment only for simple reactions. Transition state theory refines it. Key Takeaways: - Effective collisions need energy ≥Ea\ge E_{a} and proper orientation. - Rate =PZe−Ea/RT= PZe^{-E_{a}/RT}; A=PZA = PZ. - Temperature mainly raises the fraction of energetic molecules, not the collision frequency. - A catalyst increases the fraction of effective collisions by lowering EaE_{a}. - Pitfall: not every collision is a reaction; typically only a tiny fraction succeeds. Brain-Teaser 1: For Ea=50 kJ mol−1E_{a} = 50\,\mathrm{kJ\,mol^{-1}}, what fraction of collisions has enough energy at 300 K300\,\mathrm{K}, and by what factor does it increase at 310 K310\,\mathrm{K}? Solution: EaRT=500008.314×300=20.05\frac{E_{a}}{RT} = \frac{50000}{8.314 \times 300} = 20.05, so the fraction is e−20.05=1.96×10−9e^{-20.05} = 1.96 \times 10^{-9}. At 310 K310\,\mathrm{K}: 500008.314×310=19.40\frac{50000}{8.314 \times 310} = 19.40, fraction 3.76×10−93.76 \times 10^{-9}. The factor is about 1.91.9: the rate roughly doubles. Brain-Teaser 2: Why does increasing temperature from 300300 to 310 K310\,\mathrm{K} increase collision frequency by only about 1.7%1.7\% yet nearly double the rate? Solution: Collision frequency is proportional to average speed, which scales as T\sqrt{T}: 310300=1.017\sqrt{\frac{310}{300}} = 1.017. The rate depends mostly on the exponential factor e−Ea/RTe^{-E_{a}/RT}, which changes by a factor of about 22 over the same range (see Brain-Teaser 1). Brain-Teaser 3: A catalyst lowers EaE_{a} from 100100 to 75 kJ mol−175\,\mathrm{kJ\,mol^{-1}} at 300 K300\,\mathrm{K}. By what factor does the rate increase (assuming AA unchanged)? Solution: kcatk=e(100000−75000)/(8.314×300)=e10.02≈2.3×104\frac{k_{\text{cat}}}{k} = e^{(100000 - 75000)/(8.314 \times 300)} = e^{10.02} \approx 2.3 \times 10^{4}. Lowering the barrier by 25 kJ mol−125\,\mathrm{kJ\,mol^{-1}} makes the reaction over twenty thousand times faster.

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