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Tutorial on Monday and Wednesday • No. 2 (Chapter 22: The rates of chemical reactions) by Bryan Lucier • March 7 & 9 (from 2:30 to 3:50pm, Essex Hall room 182) 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: ~ k (C D) with hc v (C H ) CH 1 / 2 e 1 ( ) k (C H ) 2kT CD • Secondary kinetic isotope effect: the variation in reaction rate even though the bond involving the isotope is not broken to form product: k (C D) ~ ~ e hc{v (C H ) v (C H )} CH 1 / 2 k (C H ) 1 ( ) with 2kT CD Kinetic isotope effect (primary) Kinetics isotope effect (secondary) 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 d[ A*] k1[ A]2 dt The energized molecule may lose its excess energy by collision with another molecule: A + A* → A + A d [ A*] k1 [ A][ A*] ' dt The excited molecule might shake itself apart to form products P A* → P d[ A*] k 2 [ A*] dt The net rate of the formation of A* is d [ A*] k1 [ A]2 k1 [ A][ A*] k 2 [ A*] ' dt • 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 d [ A*] k1[ A]2 k1[ A][ A*] k2 [ A*] 0 ' dt Then k1[ A]2 [ A*] k1[ A] k2 ' The rate law for the formation of P could be reformulated as d [ P] k1k2 [ A]2 k2 [ A*] ' dt k1[ A] k2 Further simplification could be obtained if the deactivation of A* is much faster than A* P, i.e., k1' [ A ][A*] k2[ A*] then k1[ A] k2 ' d [ P ] k1 k 2 ' [ A] dt k1 in case k1[ A] k2 ' d[ P ] k1 [ A]2 dt d [ P] k1k2 [ A]2 • The equation dt k2 [ A*] ' k1[ A] k2 can be reorganized into d [ P] k1k 2 [ A] ( ' )[ A] dt k1[ A] k 2 • Using the effective rate constant k to represent k1k 2 [ A] k k1' [ A] k 2 • Then one has ' 1 k1 1 k k1k2 k1[ A] 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 1 E * P 1 E s 1 * E k b ( E ) 1 kb E • 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 (E1) A + A* → A + A (E1’) A* → P (E2) has an Arrhenius-like temperature dependence, one gets k1 k 2 A1e E1/ RT A2 e E2 / RT k1' ' E1 / RT ' A1e A1 A2 e ( E1 E2 E1 ) / RT ' ' A1 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: Chain reactions • Chain reactions: a reaction intermediate produced in one step generates an intermediate in a subsequent step, then that intermediate generates another intermediate, and so on. • Chain carriers: the intermediates in a chain reaction. It could be radicals (species with unpaired electrons), ions, etc. • Initiation step: • Propagation steps: • Termination steps: 23.1 The rate laws of chain reactions • Consider the thermal decomposition of acetaldehyde CH3CHO(g) → CH4(g) + CO(g) v = k[CH3CHO]3/2 it indeed goes through the following steps: 1. Initiation: CH3CHO → . CH3 + .CHO ki v = ki [CH3CHO] 2. Propagation: CH3CHO + . CH3 → CH4 + CH3CO. kp Propagation: CH3CO. → .CH3 + CO k’p 3. Termination: .CH3 + .CH3 → CH CH kt 3 3 • The net rates of change of the intermediates are: d [ .CH 3 ] ki [CH 3CHO ] k p [ .CH 3 ][CH 3CHO ] k ,p [CH 3CO . ] 2kt [ .CH 3 ]2 dt d [CH 3CO. ] k p [ .CH 3 ][CH 3CHO ] k ,p [CH 3CO. ] dt • Applying the steady state approximation: 0 k i [CH 3CHO] k p [ .CH 3 ][CH 3CHO] k ,p [CH 3CO. ] 2k t [ .CH 3 ]2 0 k p [ .CH 3 ][CH 3CHO] k ,p [CH 3CO. ] • Sum of the above two equations equals: ki [CH 3CHO] 2kt [.CH 3 ]2 0 • thus the steady state concentration of [.CH3] is: 1/ 2 k [.CH 3 ] i 2k [CH 3CHO ]1 / 2 t • The rate of formation of CH4 can now be expressed as 1/ 2 d [CH 4 ] k k p [ .CH 3 ][CH 3CHO] k p i 2k [CH 3CHO]3 / 2 dt t the above result is in agreement with the three-halves order observed experimentally. • Example: The hydrogen-bromine reaction has a complicated rate law rather than the second order reaction as anticipated. H2(g) + Br2(g) → 2HBr(g) Yield k[ H 2 ][Br2 ]3 / 2 v [ Br2 ] k '[ HBr ] The following mechanism has been proposed to account for the above rate law. 1. Initiation: Br2 + M → Br. + Br. + M ki 2. Propagation: Br. + H2 → HBr + H. kp1 H. + Br2 → HBr + Br. kp2 3. Retardation: H. + HBr → H2 + Br. kr 4. Termination: Br. + Br. + M → Br2 + M* kt derive the rate law based on the above mechanism. • The net rates of formation of the two intermediates are d[ H . ] k p1 [ Br . ][ H 2] k p 2 [ H . ][ Br2 ] k r [ H . ][ HBr ] dt d [ Br . ] 2k i [ Br2 ][ M ] k p1[ Br . ][ H 2 ] k p 2 [ H . ][ Br2 ] k r [ H . ][ HBr ] 2k t [ Br . ]2 [ M ] dt • The steady-state concentrations of the above two intermediates can be obtained by solving the following two equations: k p1 [ Br . ][ H 2] k p 2 [ H . ][ Br2 ] k r [ H . ][ HBr ] 0 2k i [ Br2 ][ M ] k p1 [ Br . ][ H 2 ] k p 2 [ H . ][ Br2 ] k r [ H . ][ HBr ] 2k t [ Br . ]2 [ M ] 0 1/ 2 k [ Br . ] i k [ Br2 ]1 / 2 t k p1 (k i / k t )1 / 2 [ H 2 ][ Br2 ]1 / 2 [H ] . k p 2 [ Br2 ] k r [ HBr ] • substitute the above results to the rate law of [HBr] d[ HBr ] k p1[ Br . ][ H 2 ] k p 2 [ H . ][ Br2 ] k r [ H . ][ HBr ] dt 1/ 2 3/ 2 d [ HBr ] 2k p1 (k i / k t ) [ H 2 ][ Br2 ] dt [ Br2 ] (k r / k p 2 )[ HBr ] continued • The above results has the same form as the empirical rate law, and the two empirical rate constants can be identified as 1/ 2 k kr k 2k p i k, k k p2 t • Effects of HBr, H2, and Br2 on the reaction rate based on the equation 1/ 2 3/ 2 d [ HBr ] 2k p1 (k i / k t ) [ H 2 ][ Br2 ] dt [ Br2 ] (k r / k p 2 )[ HBr ] • Self-test 23.1 Deduce the rate law for the production of HBr when the initiation step is the photolysis, or light-induced decomposition, of Br2 into two bromine atoms, Br.. Let the photolysis rate be v = Iabs, where Iabs is the intensity of absorbed radiation. • Hint: the initiation rate of Br. ? Exercises 23.1b: On the basis of the following proposed mechanism, account for the experimental fact that the rate law for the decomposition 2N2O5(g) → 4NO2(g) + O2(g) is v = k[N2O5]. (1) N2O5 ↔ NO2 + NO3 k 1, k 1’ (2) NO2 + NO3 → NO2 + O2 + NO k2 (3) NO + N2O5 → NO2 + NO2 + NO2 k3