Chapter 14: Chemical Kinetics Lecture 5: Reaction Mechanisms

Collision Model Key Idea: Molecules must collide to react. However, only a small fraction of collisions produces a reaction. Why?

Collision Model Collisions must have sufficient energy to produce the reaction (equal to or exceeding the activation energy). Colliding particles must be correctly oriented to one another in order to produce a reaction.

Factors Affecting Rate Increasing temperature always increases the rate of a reaction. Particles collide more frequently Particles collide more energetically Increasing surface area increases the rate of a reaction Increasing Concentration USUALLY increases the rate of a reaction Presence of Catalysts, which lower the activation energy by providing alternate pathways

Reaction Mechanism Reactions that occur in a single step. ELEMENTARY REACTIONS: Reactions that occur in a single step. The number of molecules that participate is the molecularity: Unimolecular: One molecule involved A  B Bimolecular: Two molecules involved A + B  C Termolecular: Three molecules involved A + B + C  D

Reaction Mechanisms Unimolecular Termolecular Bimolecular

Rate Law for Elementary Reactions The order for the rate law for an elementary reaction is the molecularity. Unimolecular: Rate = k[A]1 Bimolecular: Rate = k[A]2 or Rate = k[A][B] Termolecular: Rate = k[A]3 or Rate = k[A][B][C]

Reaction Mechanism The reaction mechanism is the series of elementary steps by which a chemical reaction occurs. The sum of the elementary steps must give the overall balanced equation

Reaction Mechanisms Elementary step 1: H2O2 (aq) +I  H2O (l) +IO Elementary step 2: H2O2 (aq) +IO  H2O (l) + O2 (g) +I Overall rxn: 2 H2O2 (aq)  2 H2O (l) + O2 (g) IO is the intermediate. It forms in the first elementary reaction and then reacts to produce the final products. I is the catalyst. It is an initial reactant and a final product.

Reaction Mechanisms Elementary step 1: H2O2 (aq) +I  H2O (l) +IO Elementary step 2: H2O2 (aq) +IO  H2O (l) + O2 (g) +I Overall rxn: 2 H2O2 (aq)  2 H2O (l) + O2 (g) The rate law for each elementary step is: Elementary step 1: Rate = k [H2O2] [I] Elementary step 2: Rate = k [H2O2][IO] Rate = [concentration of reactants] coefficient

Reaction Mechanisms Elementary step 1: NO +NO  N2O2 + Elementary step 2: N2O2 + O2  2NO2 Overall step: 2 NO + O2  2NO2 The rate law for each elementary step is: Elementary step 1: Rate = k [NO]2 Elementary step 2: Rate = k [N2O2][O2] Rate = [concentration of reactants] coefficient

Rate-Determining Step In a multi-step reaction, the slowest step is the rate-determining step. The slowest step: Will have the HIGHEST activation energy. Determines the rate law for the reaction: Rate = [reactants]Coefficients for slow step

Reaction Mechanisms Elementary Rxn Reactants→Intermediates→Products

Reaction Mechanisms Reactants→Intermediates→Products Second step is rate limiting step: It has the higher activation energy.

Rate Law with Slow Initial Step For the reaction: 2H2(g) + 2NO(g)  N2(g) + 2H2O(g) Step #1 is the slowest step. What is the rate law? Step #1 H2(g) + 2NO(g)  N2O(g) + H2O(g) Step #2 N2O(g) + H2(g)  N2(g) + H2O(g) N2O(g) is the intermediate. The experimental rate law is the rate law for the slowest step: rate = k[NO]2[H2]

Slow Initial Step Rate = k [NO2]2 Step 1: NO2 + NO2  NO3 + NO (slow) Step 2: NO3 + CO  NO2 + CO2 (fast) The rate law is the one for the slow elementary step: Rate = k [NO2]2 Overall: NO2 (g) + CO (g)  NO (g) + CO2 (g)

Step 2: N2O(g) + O (g)  N2(g)+ O2 (g) (slow) Fast Initial Step Step 1: N2O (g)  N2(g)+ O(g) (fast) Step 2: N2O(g) + O (g)  N2(g)+ O2 (g) (slow) Write the rate law for the overall reaction: The rate law for the overall reaction is the rate law for the slow step: Rate = k2 [N2O] [O]  

Fast Initial Step Rate = k [NOBr2] [NO] Step 1: NO (g) + Br2 (g)  NOBr2(g) (fast) Step 2: NOBr2 (g) + NO (g)  2 NOBr(g) (slow) Overall: 2 NO (g) + Br2 (g)  2 NOBr(g) The rate law is the one for the slow elementary step: Rate = k [NOBr2] [NO] If the rate law contains a catalyst or intermediate, need to use fast elementary step to put in terms of initial reactants. Write the rate law in terms initial reactants or final products, because we can measure these. Intermediates are usually very difficult to measure.

Fast Initial Step Rate = k [NOBr2] [NO] Step 1: NO (g) + Br2 (g)  NOBr2(g) (fast) Step 2: NOBr2 (g) + NO (g)  2 NOBr(g) (slow) Overall: 2 NO (g) + Br2 (g)  2 NOBr(g) The rate law is the one for the slow elementary step: Rate = k [NOBr2] [NO]

Fast Initial Step NOBr2 can react two ways: With NO to form NOBr By decomposition to reform NO and Br2 The reactants and products of the first step are in equilibrium with each other. Therefore, Ratef = Rater

Fast Initial Step Because Ratef = Rater , k1 [NO] [Br2] = k−1 [NOBr2] Solving for [NOBr2] gives us k1 k−1 [NO] [Br2] = [NOBr2]

Fast Initial Step Substituting this expression for [NOBr2] in the rate law for the rate-determining step gives k2k1 k−1 Rate = [NO] [Br2] [NO] = k [NO]2 [Br2]