Self-test 22.8 Derive the rate law for the decomposition of ozone in the reaction 2O3(g) → 3O2(g) on the basis of the following mechanism O3 → O2 + O k1 O2 + O → O3 k1’ O + O3 → O2 + O2 k2 Solution: First write the rate law for the reactant O3 and the intermediate product O Applying the steady state approximation to [O] Plug the above relationship back to the rate law of [O3]

Rate determining step

Simplifications with the rate determining step Suppose that k2 >> k1, then k2 – k1 ≈ k2 therefore concentration C can be reorganized as [C] ≈ (1 - )[A]0 The above result is the same as obtained with steady state approximation

Kinetic and thermodynamic control of reactions Consider the following two reactions A + B → P1 rate of formation of P1 = k1[A][B] A + B → P2 rate of formation of P2 = k2[A][B] [P1]/[P2] = k1/k2 represents the kinetic control over the proportions of products.

Problems 22.6 The gas phase decomposition of acetic acid at 1189 K proceeds by way of two parallel reactions: CH3COOH → CH4 + CO2, k1 = 3.74 s-1 CH3COOH → H2C=C=O + H2O, k2 = 4.65 s-1 What is the maximum percentage yield of the ketene CH2CO obtainable at this temperature.

Pre-equilibrium Consider the reaction: A + B ↔ I → P when k1’ >> k2, the intermediate product, I, could reach an equilibrium with the reactants A and B. Knowing that A, B, and I are in equilibrium, one gets: and When expressing the rate of formation of the product P in terms of the reactants, we get

Self-test 22.9: Show that the pre-equilibrium mechanism in which 2A ↔ I followed by I + B → P results in an overall third-order reaction. Solution: write the rate law for the product P Because I, and A are in pre-equilibrium so [I] = K [A]2 Therefore, the overall reaction order is 3.

Kinetic isotope effect Kinetic isotope effect: the decrease in the rate of a chemical reaction upon replacement of one atom in a reactant by a heavier isotope. Primary kinetic isotope effect: the kinetic isotope effect observed when the rate-determining step requires the scission of a bond involving the isotope: with Secondary kinetic isotope effect: the variation in reaction rate even though the bond involving the isotope is not broken to form product:

Kinetic isotope effect

22.8 Unimolecular reactions The Lindemann-Hinshelwood mechanism A reactant molecule A becomes energetically excited by collision with another A molecule: A + A → A* + A The energized molecule may lose its excess energy by collision with another molecule: A + A* → A + A The excited molecule might shake itself apart to form products P A* → P The net rate of the formation of A* is

If the reaction step, A + A → A If the reaction step, A + A → A* + A, is slow enough to be the rate determining step, one can apply the steady-state approximation to A*, so [A*] can be calculated as Then The rate law for the formation of P could be reformulated as Further simplification could be obtained if the deactivation of A* is much faster than A*  P, i.e., then in case

The equation can be reorganized into Using the effective rate constant k to represent Then one has

The Rice-Ramsperger-Kassel (RRK) model Reactions will occur only when enough of required energy has migrated into a particular location in the molecule. s is the number of modes of motion over which the energy may be dissipated, kb corresponds to k2

The activation energy of combined reactions Consider that each of the rate constants of the following reactions A + A → A* + A A + A* → A + A A* → P has an Arrhenius-like temperature dependence, one gets Thus the composite rate constant also has an Arrhenius-like form with activation energy, E = E1 + E2 – E1’ Whether or not the composite rate constant will increase with temperature depends on the value of E, if E > 0, k will increase with the increase of temperature

Combined activation energy

Theoretical problem 22.20 The reaction mechanism A2 ↔ A + A (fast) A + B → P (slow) Involves an intermediate A. Deduce the rate law for the reaction. Solution: