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Finding reaction velocity Physical Chemistry Lecture 7 Multi-step reaction mechanisms One knows the form of the reaction rate for every elementary reaction Express disappearance of reactant or appearance of product in terms of the rates of these elementary steps Include terms for every step that affects reactants (or products), with stoichiometry Example from Bodenstein’s work: d[H2 ] d[HBr] v2 v4 v2 v3 v4 dt dt d[Br2 ] v1 v3 v5 dt Elementary reactions Reactions are often more complex than indicated by the beginning and ending states alone Some reactions do occur in a single step -elementary reactions Generally involve simple mono- or bimolecular interactions Order in elementary reactions is the stoichiometry number Examples: H2 + O H2O H + Br HBr Reaction mechanism A description of the course of a reaction in terms of elementary reactions Example from Bodenstein’s work: H 2 (g) Br2 Br H H Br Br2 ( g ) 2 HBr ( g ) H2 Br2 HBr Br 2 Br HBr H HBr Br Br H2 Br2 v1 v2 v3 v4 v5 Reaction mechanisms are not necessarily unique Reducing rate equations Eliminate dependences on reactive species (reactive intermediates such as radicals) Use approximations Steady state for reactive species Fast-equilibrium approximation when two fast steps precede a slow one Must be able to identify which are reactive intermediates Steady-state approximation Reactive species have very low concentrations After initiation, this concentration is presumed to be independent of time d [reactive species] 0 dt Gives relation among velocities of elementary steps Example from Bodenstein’s work: d [ Br ] 2 v1 dt v2 v3 v4 2 v5 0 1 Fast-equilibrium approximation Summary One step containing an intermediate is ratelimiting Prior step is reversible (1) A B C fast ( 2) C A B fast ( 3) C Product slow One presumes a quasi-equilibrium in the first two steps to relate the concentrations K eq k1 [C ] k 2 [ A][ B] d [ Product ] k3 [C ] k3 K eq [ A][ B ] dt Every reaction is described by an equation H2 ( gas) Br2 ( gas) 2 HBr ( gas) “Simple” reaction sequences solved exactly Generally, equation like above does NOT describe reaction course Often cannot get exact time dependence of concentrations for reactions Use a mechanism Overall reaction expressed in terms of elementary steps Not unique “Solve” mechanism, using approximations if necessary Rate-limiting steps Steady-state approximation Fast-equilibrium approximation Solving mechanism of formation of HBr From the reaction mechanism, identify time changes of reactive species and use the steady-state approximation d [ Br ] 2 v1 v2 v3 v4 dt d[ H ] v2 v3 v4 0 dt 2 v5 0 These two equations give relations 1/ 2 v1 k v5 [ Br ] 1 [ Br2 ]1/ 2 k5 k k2 [H] 1 2 k5 1/ 2 [ H2 ][ Br2 ]1/ 2 k 3[ Br2 ] k 4 [ HBr ] Example: H2 + Br2 continued Identify the reaction velocity in terms of change of a reactant or product v d [ Br2 ] v1 dt v3 v5 v3 Substitution gives the final prediction of the reaction mechanism k k 2k 2 v 1 2 3 k5 1/ 2 [ H2 ][ Br2 ]3/ 2 k 3[ Br2 ] k 4 [ HBr ] 2