Chemical Kinetics Derived Rate Laws from Reaction Mechanisms

Reaction Mechanism CHEM 3310 Determine the rate law by experiment Devise a reaction mechanism If the predicted and experimental rate laws do not agree If the predicted and experimental rate laws agree Predict the rate law for the mechanism Look for additional supporting evidence CHEM 3310

Reaction Mechanism A sequence of one or more elementary reaction steps together forms a reaction mechanism. In a mechanism, elementary steps proceed at various speeds (governed by different rate constants, k). Elementary reaction steps must be balanced (as do all chemical reactions). The slowest step is the rate-determining step. It is the “bottleneck” in the formation of products. CHEM 3310

Reaction Mechanism A rate law derived from a set of mechanisms should only consist of concentrations of reactants and/or products, no intermediates. In predicting the rate law for an elementary step, the exponents for the concentration terms are the same as the stoichiometric coefficients. To propose a mechanism requires the knowledge of chemistry to give plausible elementary processes. In this course, you will not be asked to propose mechanisms, but you will be asked to derive the rate laws from given mechanisms. CHEM 3310

Uni-, Bi-, Termolecular Reaction Mechanism - Terminologies Molecularity – is the number of reacting species (i.e. atoms, ions or molecules) in an elementary reaction, which must collide simultaneously in order to bring about a chemical reaction. Uni-, Bi-, Termolecular Involving 1 species Involving 2 species Involving 3 species CHEM 3310

Reaction Mechanism 1. Unimolecular Elementary Step or There is only one molecule reacting, namely species "A" is reacting. This unimolecular reaction step implies the rate law, Involving a single species. Decomposition of ammonium nitrite NH4NO2 (g)  N2 (g) + 2 H2O (g) Examples: Decomposition of hydrogen peroxide H2O2 (aq)  H2O (l) + ½ O2 (g) Recall, this is a 1st order reaction. CHEM 3310

This is the definition of Keq. Reaction Mechanism 2. Reversible Unimolecular Elementary Step A B At equilibrium, the rate of the forward reaction equals the rate of the reverse reaction. Rforward = Rreverse k1[A] = k-1[B] Rearrange, This is the definition of Keq. CHEM 3310

A B Reaction Mechanism [A ] decreasing in forward reaction 2. Reversible Unimolecular Elementary Step (cont’d) A B Species "A" is in equilibrium with species “B". The forward reaction is governed by k1. The reverse reaction is governed by k-1. The rate, R, is equal to the rate of the forward step minus the rate of the reverse step. This implies the rate law, [A ] decreasing in forward reaction [A ] increasing in reverse reaction CHEM 3310

Reaction Mechanism 3. Bimolecular Elementary Step requires two molecules coming together at the same time. Implies this rate law. Implies this rate law. or Implies this rate law. Implies this rate law. CHEM 3310

collisions of two species. Reaction Mechanism 3. Bimolecular Elementary Step or Implies this rate law. Implies this rate law. Examples of reactions involving a bimolecular elementary step. Involves collisions of two species. NO + O3  NO2 + O2 Rate = k[NO][O3] 2 HI  H2 + I2 Rate = k[HI]2 CHEM 3310

A + B C A + B C + D Reaction Mechanism 4. Reversible Bimolecular Elementary Step A + B C Implies this rate law. Implies this rate law. or A + B C + D Implies this rate law. Implies this rate law. CHEM 3310

2 A C + D Reaction Mechanism At equilibrium, Rforward = Rreverse 4. Reversible Bimolecular Elementary Step (cont’d) 2 A C + D or Implies this rate law. Implies this rate law. At equilibrium, Rforward = Rreverse where Rforward = k2[A]2 Rreverse = k-2[C][D] k2[A]2 = k-2[C][D] Equilibrium constant CHEM 3310

Reaction Mechanism 2 HI H2 (g) + I2 (g) At equilibrium, the rate of the forward reaction equals the rate of the reverse reaction. Rf = k1[HI]2 Rforward = Rreverse k1[HI]2 = k-1[H2] [I2] Rr = k-1[H2] [I2] At the start At equilibrium CHEM 3310

Reaction Mechanism 5. Termolecular Elementary Step Termolecular reaction steps require three molecules coming together at the same time. Implies this rate law. Implies this rate law. or Implies this rate law. Implies this rate law. CHEM 3310

Rate = k [NO]2 [O2] Reaction Mechanism Example: 5. Termolecular Elementary Step or Implies this rate law. Implies this rate law. Example: The reaction mechanism for 2 NO (g) + O2 (g)  2 NO2 (g) involves a 3-body collision one-step mechanism. Rate = k [NO]2 [O2] CHEM 3310

Reaction Mechanism Recall , the equation in an elementary step represents the reaction at the molecular level, not the overall reaction. What about higher orders such as 4th or 5th orders? Simultaneous collision of 3 molecules is rare. In nature, we observe lots of 2-body collisions, very few 3-body collisions and not much else. CHEM 3310

Derive the rate law of a reaction mechanism Most balanced equations do not literally describe how a reaction occurs in terms of the collisions made or the actual sequence of events. The combustion of hexane illustrates this point: 2 C6H14 (g) + 19 O2 (g) → 12 CO2 (g) + 14 H2O (g) If it were to take things literally as written, the reaction is saying 2 hexane molecules and 19 oxygen molecules somehow collide simultaneously and fight among themselves until 12 molecules of carbon dioxides and 14 molecules of waters form. In nature, we observe lots of 2-body collisions, very few 3-body collisions and not much else. CHEM 3310

Derive the rate law of a reaction mechanism A one-step mechanism involving the collision of NO and O3. NO + O3  NO2 + O2 1 1 Since this is the only step in the reaction mechanism, then the rate law can be written directly from the stoichiometry of the step. 1 1 CHEM 3310

Suggestion of a bimolecular single step mechanism. Derive the rate law of a reaction mechanism Example: The overall reaction of  N2O5(g) + NO(g)  3 NO2(g)   occurs in a one-step mechanism where the two reactants collide to form the product. The derived rate law can be determined to be Rate = k [N2O5][NO] Suggestion of a bimolecular single step mechanism. CHEM 3310

Each step is governed by its rate constant. Derive the rate law of a reaction mechanism Mechanism: A sequence of one or more elementary reaction steps that proceed at various speeds. Each step is governed by its rate constant. What would the energy profile diagram look like? If k1 >> k2, If k1 << k2, Ea1 Ea1 Ea2 Ea2 Step 2 is the rate determining step (RDS). This implies that isolation of B is good. Step1 is the rate determining step (RDS). This implies that isolation of B is not easy. CHEM 3310

Derive the rate law of a reaction mechanism Mechanism: A sequence of one or more elementary reaction steps that proceed at various speeds. Consider the reaction mechanism for an overall exothermic reaction: 1. Identify A, B, C, D, and E. A and B are reactants C is the intermediate D and E are the products 2. What is the overall reaction? A + B  D + E 3. What is the rate law? In this mechanism, the rate law can be written directly from the slowest step. Rate = k1 [A] [B] CHEM 3310

Derive the rate law of a reaction mechanism Mechanism: A sequence of one or more elementary reaction steps that proceed at various speeds. Consider the reaction mechanism for an overall exothermic reaction: 4. Sketch the reaction coordinate of the reaction. Since step 1 is the rate determining step, k1 << k2. Ea2 Ea1 Ea1 > Ea2 CHEM 3310

Above 600K, a one-step mechanism. Below 600K, a two-step mechanism. Derive the rate law of a reaction mechanism Example: CO (g) + NO2 (g)  NO (g) + CO2 (g) This reaction proceeds via two reaction mechanisms. Above 600K, a one-step mechanism. Below 600K, a two-step mechanism. CHEM 3310

Derive the rate law of a reaction mechanism Experimentally determined Rate = k[CO][NO2] Example: CO (g) + NO2 (g)  NO (g) + CO2 (g) (balanced reaction) Above 600K, the reaction mechanism involves the collision between CO and NO2. A one-step elementary step which describes the collision of CO and NO2. 1 CO + NO2  NO + CO2 1 Reasonable - a single collision of two molecules (with correct orientation and minimum energy) would lead to the exchange of an oxygen between CO and NO2. 1 1 Derived rate law is consistent with the experimental rate law. CHEM 3310

CO (g) + NO2 (g)  NO (g) + CO2 (g) Derive the rate law of a reaction mechanism Example: CO (g) + NO2 (g)  NO (g) + CO2 (g) (balanced reaction) Below 600K, the experimentally determined rate law is Rate = k [NO2] 2 Proposed mechanism is a 2-step mechanism: CHEM 3310

Derive the rate law of a reaction mechanism Below 600K, experimentally determined rate law is Rate = k[NO2]2 Example: CO (g) + NO2 (g)  NO (g) + CO2 (g) 1. Identify the intermediate, if any. NO3 is the intermediate. 2. What is the overall reaction? NO2 + NO2 + NO3 + CO  NO + + NO2 + NO3 +CO2 Yes, the elementary steps add to give the overall reaction. NO2 + CO  CO2 + NO 3. Which is the rate determining step? Step 1 CHEM 3310

Derive the rate law of a reaction mechanism Below 600K, experimentally determined rate law is Rate = k[NO2]2 Example: CO (g) + NO2 (g)  NO (g) + CO2 (g) Proposed mechanism: NO3 is very reactive; consistent with step 1 being the RDS. 4. What is the rate law of the proposed mechanism? Is it consistent with the experimentally determined rate law? Since step 1 is the RDS, use step 1, the bimolecular step involving the collision of two NO2, to determine the rate law. Rate = rate of the slow step = k1 [NO2]2 The mechanism’s rate law is consistent with the experimentally determined rate law. Confirmed by reacting two NO2 molecules and look for yielding NO3 as a product. NO3 is highly reactive and is capable of transferring an oxygen atom to CO to give CO2. CHEM 3310

Derive the rate law of a reaction mechanism Consider the following reaction mechanisms proposed for the thermal decomposition of NO2. 2 NO2  2 NO + O2 Experimental rate law is Rate = k [NO2] 2 Two possible mechanisms: Mechanism 1 Mechanism 2 Which mechanism is consistent with the observed rate law? CHEM 3310

Derive the rate law of a reaction mechanism Consider the following reaction mechanism proposed for the overall reaction 2 NO2  2 NO + O2 Experimental rate law is Rate = k [NO2] 2 Mechanism 1 Derived rate law from Mechanism 1 is Rate = k1 [NO2] CHEM 3310

Derive the rate law of a reaction mechanism Consider the following reaction mechanism proposed for the overall reaction 2 NO2  2 NO + O2 Experimental rate law is Rate = k [NO2] 2 Mechanism 2 Derived rate law from Mechanism 2 is Rate = k1 [NO2]2 CHEM 3310

Derive the rate law of a reaction mechanism Consider the following reaction mechanism proposed for the overall reaction 2 NO + O2  2 NO2 Experimentally determined rate law is Rate = k [NO]2[O2] Consider a single step mechanism: Derived rate law appears to be consistent with the experimental rate law. Derived rate law is Rate = k1 [NO]2[O2] The mechanism invokes a termolecular step, which is very unlikely since three way collisions are less likely to take place. A mechanism Involving two species collision would be more probable. CHEM 3310

Derive the rate law of a reaction mechanism Consider the following reaction mechanism proposed for the overall reaction 2 NO + O2  2 NO2 Consider a two-step mechanism: Need to remove N2O2 from the rate law. Derived rate law is Rate = rate of the slow step = k2[N2O2][O2] N2O2 is an intermediate. The rate law cannot contain intermediates. Use the fast equilibrium step to find an expression for N2O2. k1[NO]2 = k-1[N2O2] Substitute this back into the above rate equation to remove N2O2. CHEM 3310

Derive the rate law of a reaction mechanism Consider the following reaction mechanism proposed for the overall reaction 2 NO + O2  2 NO2 Experimentally determined rate law Rate = k [NO]2[O2] Consider a two-step mechanism: Derived rate law is Rate = rate of the slow step = k2[N2O2][O2] Need to remove dinitrogen dioxide, N2O2, from the rate law. Need to remove dinitrogen dioxide, N2O2, from the rate law. Since Derived rate law is consistent with the experimental rate law. This two step mechanism each involving a bimolecular step is more plausible. CHEM 3310

Derive the rate law of a reaction mechanism Ozone depletion Experimental rate law: Example: 2 O3 (g)  3 O2 (g) Proposed mechanism: Derived rate law: Need to remove O from the rate law as O is an intermediate. Need to remove O from the rate law as O is an intermediate. From step 1, k1[O3] = k-1[O2][O] Derived rate law is consistent with the experimental rate law. Note k-1[O2][O] >> k2[O3][O] CHEM 3310

Derive the rate law of a reaction mechanism Steady State Approximation The steady-state approximation is a method used to derive a rate law. The method is based on the assumption that one intermediate in the reaction mechanism is consumed as quickly as it is generated. Its concentration remains unchanged for most of the reaction. The system reaches a steady-state, hence the name of the technique is called steady state approximation. When steady-state is reached, there is no change observed in the concentration of the intermediate. CHEM 3310

Derive the rate law of a reaction mechanism Steady State Approximation Example: H2 (g) + 2 ICl (g)  I2 (g) + 2 HCl (g) Experimental rate law: Rate = k[H2][ICl] Proposed mechanism: Since step 1 is the slow step, the derived rate law is consistent with the experimental rate law. Rate = k[H2][ICl] CHEM 3310

Derive the rate law of a reaction mechanism H2 (g) + 2 ICl (g)  I2 (g) + 2 HCl (g) Steady State Approximation Experimental rate law: Rate = k[H2][ICl] Proposed mechanism: Steady State Approximation Derived rate law: Need to remove HI from the rate law as HI is an intermediate. Need to remove HI from the rate law as HI is an intermediate. HI is being produced in step 1 and quickly removed in step 2 [HI] is pretty much constant throughout the reaction (with the exception of the beginning and the end of the reaction). Production of HI (step 1) Removal of HI (step 2) Substitute [HI] into the rate law. Substitute [HI] Into the rate law. Derived rate law is consistent with the experimental rate law. CHEM 3310

Derive the rate law of a reaction mechanism Chain Reactions Chain reactions are complex reactions that involve chain carriers, reactive intermediates which react to produce more intermediates. The elementary steps in a chain reaction may be classified into: initiation, propagation, inhibition, and termination steps. Example: Chlorofluorocarbons (CFCs) destruction of the ozone layer Initiation: Thermally or photochemically produces Cl radicals Propagation: Regenerates more Cl radicals Termination: Cl radicals deactivates by reacting to form an inactive product. Inhibition: A step involving product molecules being destroyed. CHEM 3310

Experiment 4: Does this mean that 5+1+6=12 particles must come together and collide? Reaction Mechanism What about this reaction in Experiment 4? 5 Br- (aq) + BrO-3 (aq) + 6 H+ (aq)  3 Br2 (aq) + 3 H2O (l) Experimentally determined rate law is Rate = k [Br- ][ BrO-3 ][H+ ]2 First order with respect to Br- and BrO3- ions Second order with respect to H+ ions Overall reaction is of 1+1+2 = 4. We observe that the reaction is found to be quite fast. It means that even though the balanced equation involves a large number of molecules, the reaction does not proceed by simultaneous collision of all these reacting particles. The mechanism can involve two or maximum three collisions simultaneous. CHEM 3310

Reaction Mechanism What about this reaction in Experiment 4? 5 Br- (aq) + BrO-3 (aq) + 6 H+ (aq)  3 Br2 (aq) + 3 H2O (l) This is a rather complex mechanism. According to Field et al., 1972; Pelle et al., 2004, the reaction is thought to occur by the following collection of bimolecular elementary reactions. The derived rate law for this mechanism is outside the scope of this course. It is found to be consistent with the experimental rate law. First order with respect to Br- and BrO3- ions Second order with respect to H+ ions CHEM 3310

Reaction Mechanism Integrated rate law from reaction mechanism This can get mathematical complicated Consider the 2 step reaction mechanism Step 1 : A  B Step 2 : B  C where k1 is the rate constant for step 1 k2 is the rate constant for step 2 at t =0: only A is present [A] = [A]o. [B]o = [C]o= 0 CHEM 3310

Reaction Mechanism 𝑑[𝐴] 𝑑𝑡 =− 𝑘 1 [𝐴] [𝐴] 𝑡 = [𝐴] 𝑜 𝑒 − 𝑘 1 𝑡 Consider the 2 step reaction mechanism Step 1 : A  B Step 2 : B  C where k1 is the rate constant for step 1 k2 is the rate constant for step 2 at t =0: only A is present [A] = [A]o. [B]o = [C]o = 0 Consider [A]: 𝑑[𝐴] 𝑑𝑡 =− 𝑘 1 [𝐴] [𝐴] 𝑡 = [𝐴] 𝑜 𝑒 − 𝑘 1 𝑡 CHEM 3310

𝑑[𝐵] 𝑑𝑡 = 𝑘 1 𝐴 − 𝑘 2 𝐵 = 𝑘 1 [𝐴] 𝑜 𝑒 − 𝑘 1 𝑡 − 𝑘 2 𝐵 Reaction Mechanism Consider the 2 step reaction mechanism Step 1 : A  B Step 2 : B  C where k1 is the rate constant for step 1 k2 is the rate constant for step 2 at t =0: only A is present [A] = [A]o. [B]o = [C]o = 0 [𝐴] 𝑡 = [𝐴] 𝑜 𝑒 − 𝑘 1 𝑡 Consider [B]: 𝑑[𝐵] 𝑑𝑡 = 𝑘 1 𝐴 − 𝑘 2 𝐵 = 𝑘 1 [𝐴] 𝑜 𝑒 − 𝑘 1 𝑡 − 𝑘 2 𝐵 𝑘 2 𝐵 + 𝑑[𝐵] 𝑑𝑡 = 𝑘 1 [𝐴] 𝑜 𝑒 − 𝑘 1 𝑡 CHEM 3310

𝑘 2 𝐵 + 𝑑[𝐵] 𝑑𝑡 𝑒 𝑘 2 𝑡 = 𝑘 1 [𝐴] 𝑜 𝑒 − 𝑘 1 𝑡 𝑒 𝑘 2 𝑡 Reaction Mechanism Consider the 2 step reaction mechanism Step 1 : A  B Step 2 : B  C 𝑘 2 𝐵 + 𝑑[𝐵] 𝑑𝑡 = 𝑘 1 [𝐴] 𝑜 𝑒 − 𝑘 1 𝑡 𝑘 2 𝐵 + 𝑑[𝐵] 𝑑𝑡 𝑒 𝑘 2 𝑡 = 𝑘 1 [𝐴] 𝑜 𝑒 − 𝑘 1 𝑡 𝑒 𝑘 2 𝑡 𝑑 𝑑𝑡 [𝐵] 𝑒 𝑘 2 𝑡 = 𝑘 1 [𝐴] 𝑜 𝑒 𝑘 2 − 𝑘 1 𝑡 𝑑 [𝐵] 𝑒 𝑘 2 𝑡 = 𝑘 1 [𝐴] 𝑜 𝑒 𝑘 2 − 𝑘 1 𝑡 𝑑𝑡 𝐵 𝑒 𝑘 2 𝑡 = 𝑘 1 𝐴 𝑜 𝑘 2 − 𝑘 1 𝑒 𝑘 2 − 𝑘 1 𝑡 +𝐷 D is the constant of integration 𝐵 = 𝑘 1 𝐴 𝑜 𝑘 2 − 𝑘 1 𝑒 − 𝑘 1 𝑡 +𝐷 𝑒 − 𝑘 2 𝑡 CHEM 3310

[𝐵] 𝑡 = 𝑘 1 𝐴 𝑜 𝑘 2 − 𝑘 1 𝑒 − 𝑘 1 𝑡 − 𝑒 − 𝑘 2 𝑡 Reaction Mechanism Consider the 2 step reaction mechanism Step 1 : A  B Step 2 : B  C 𝐵 = 𝑘 1 𝐴 𝑜 𝑘 2 − 𝑘 1 𝑒 − 𝑘 1 𝑡 +𝐷 𝑒 − 𝑘 2 𝑡 Using at t = 0, [B] = 0 0 = 𝑘 1 𝐴 𝑜 𝑘 2 − 𝑘 1 +𝐷 𝐷=− 𝑘 1 𝐴 𝑜 𝑘 2 − 𝑘 1 [𝐵] 𝑡 = 𝑘 1 𝐴 𝑜 𝑘 2 − 𝑘 1 𝑒 − 𝑘 1 𝑡 − 𝑒 − 𝑘 2 𝑡 Note shape of curve is the same if k1 and k2 are switched CHEM 3310

[𝐶] 𝑡 = [𝐴] 𝑜 1 − 𝑘 2 𝑘 2 − 𝑘 1 𝑒 − 𝑘 1 𝑡 + 𝑘 1 𝑘 2 − 𝑘 1 𝑒 − 𝑘 2 𝑡 Reaction Mechanism Consider the 2 step reaction mechanism Step 1 : A  B Step 2 : B  C where k1 is the rate constant for step 1 k2 is the rate constant for step 2 at t =0: only A is present; [A] = [A]o, [B]o = [C]o = 0 Consider the reaction’s stoichiometry: [A]o= [A]t + [B]t + [C]t or [C]t = [A]o – [A]t – [B]t [𝐶] 𝑡 = [𝐴] 𝑜 − 𝐴 𝑜 𝑒 − 𝑘 1 𝑡 − 𝑘 1 𝐴 𝑜 𝑘 2 − 𝑘 1 𝑒 − 𝑘 1 𝑡 − 𝑒 − 𝑘 2 𝑡 [𝐶] 𝑡 = [𝐴] 𝑜 1 − 𝑘 2 𝑘 2 − 𝑘 1 𝑒 − 𝑘 1 𝑡 + 𝑘 1 𝑘 2 − 𝑘 1 𝑒 − 𝑘 2 𝑡 CHEM 3310

[𝐵] 𝑡 = 𝑘 1 [𝐴] 𝑜 𝑘 2 − 𝑘 1 𝑒 − 𝑘 1 𝑡 − 𝑒 − 𝑘 2 𝑡 Reaction Mechanism Consider the 2 step reaction mechanism Step 1 : A  B Step 2 : B  C where k1 is the rate constant for step 1 k2 is the rate constant for step 2 at t =0: only A is present; [A] = [A]o, [B]o = [C]o = 0 [𝐴] 𝑡 = [𝐴] 𝑜 𝑒 − 𝑘 1 𝑡 [𝐵] 𝑡 = 𝑘 1 [𝐴] 𝑜 𝑘 2 − 𝑘 1 𝑒 − 𝑘 1 𝑡 − 𝑒 − 𝑘 2 𝑡 [𝐶] 𝑡 = [𝐴] 𝑜 1 − 𝑘 2 𝑘 2 − 𝑘 1 𝑒 − 𝑘 1 𝑡 + 𝑘 1 𝑘 2 − 𝑘 1 𝑒 − 𝑘 2 𝑡 CHEM 3310

Reaction Mechanism Consider the 2 step reaction mechanism Step 1 : A  B Step 2 : B  C CHEM 3310

Reaction Mechanism Determine the rate law by experiment Devise a reaction mechanism If the predicted and experimental rate laws do not agree If the predicted and experimental rate laws agree Predict the rate law for the mechanism Look for additional supporting evidence Rate laws can prove a mechanism is wrong, but can’t prove one right! CHEM 3310