Chemistry 1011 Slot 51 Chemistry 1011 TOPIC Rate of Reaction TEXT REFERENCE Masterton and Hurley Chapter 11

Chemistry 1011 Slot 52 Reviewing Rates of Reaction Reaction rates are affected by: 1.reactant concentrations, 2.temperature, 3.catalysts, 4.physical state of reactants Rate of reaction must be determined by experiment Relation between the rate and the concentrations of reactants is the –rate law expression

Chemistry 1011 Slot 53 Reviewing Rates of Reaction If aA + bB  xX Then rate = k[A] m [B] n 1.k is the rate constant 2.exponents m and n are the order of reaction with respect to A and B 3.overall order of reaction is m + n Collision theory postulates an energy barrier, the activation energy Transition state theory postulates the formation of an activated complex

Chemistry 1011 Slot 54 Reviewing Rates of Reaction Catalysts alter the rate of reaction without being consumed The Arrhenius equation is derived from collision theory and links the rate constant to the activation energy k = Ae -E a /RT Reaction mechanisms are developed to explain experimental results. 1.They consist of a number of elementary steps. 2.The slowest step is the rate determining step 3.This step determines the overall rate of reaction

Chemistry 1011 Slot 55 Review - Reaction Mechanisms YOU ARE EXPECTED TO BE ABLE TO: Define reaction mechanism and show how the reaction order is dependent upon the mechanism by which a reaction takes place. For a reaction taking place in more than one step, identify the rate determining step and identify reaction intermediates. Determine if a proposed reaction mechanism is consistent with experimental rate data.

Chemistry 1011 Slot 56 Review – Reaction Mechanism A reaction mechanism is a suggested path or sequence of steps by which a reaction occurs The suggested mechanism must explain the experimental determined rate law expression and order of reaction The individual steps that make up a reaction pathway are called elementary steps For elementary reactions, the rate law can be determined from the equation

Chemistry 1011 Slot 57 Reaction of CO (g) with NO 2(g) CO (g) + NO 2(g)  NO (g) + CO 2(g) At low temperatures, the experimentally determined rate law expression is: Rate = k [NO 2 ] 2 A two step reaction mechanism is suggested Step 1 SLOW NO 2(g) + NO 2(g)  NO 3(g) + NO (g) Step 2 FAST CO (g) + NO 3(g)  CO 2(g) + NO 2(g) CO (g) + NO 2(g)  NO (g) + CO 2(g)

Chemistry 1011 Slot 58 Reaction of CO (g) with NO 2(g) For the elementary step #1: NO 2(g) + NO 2(g)  NO 3(g) + NO (g) The rate law is: Rate = k 1 [ NO 2 ] 2 For the elementary step #2: CO (g) + NO 3(g)  CO 2(g) + NO 2(g) The rate law is: Rate = k 2 [ CO ][ NO 3 ]

Chemistry 1011 Slot 59 Reaction of CO (g) with NO 2(g) In this mechanism, SLOW step #1 is the rate determining step Step 1 SLOW: NO 2(g) + NO 2(g)  NO 3(g) + NO (g) The rate law expression for this step is Rate = k 1 [ NO 2 ] 2 This is the rate law predicted for the overall reaction by the proposed mechanism This is consistent with experiment

Chemistry 1011 Slot 510 Review - Mechanism with a Fast Initial Step Sometimes the first step in a reaction mechanism, which results in the creation of a reaction intermediate, will be FAST, and the second step, where the reaction intermediate is a reactant, may be SLOW The rate determining step will be the second step The rate law expression should then include the concentration of the reaction intermediate, but this cannot be measured The final rate law expression can only include species occurring in the balanced equation

Chemistry 1011 Slot 511 Reaction of NO with Cl 2 Step 1: NO (g) + Cl 2(g) NOCl 2(g) FAST Step 2: NOCl 2(g) + NO (g)  2NOCl (g) SLOW Overall: 2 NO (g) + Cl 2(g)  2NOCl (g) Rate of overall reaction = rate of step 2 Rate = k 2 [NOCl 2 ][NO] The first (fast) elementary step in the reaction is reversible; The reactants and products are in equilibrium rate forward = rate reverse

Chemistry 1011 Slot 512 Reaction of NO with Cl 2 For the first (fast) elementary step: rate forward = rate reverse k 1 [NO][Cl 2 ] =k  [NOCl 2 ] [NOCl 2 ] = k 1 [NO][Cl 2 ] k  Substitute in overall rate law expression Rate of reaction = rate of step 2 = k 2 [NOCl 2 ][NO] Rate = k 2 k 1 [NO] 2 [Cl 2 ] = k exp [NO] 2 [Cl 2 ] k 

Chemistry 1011 Slot 513 Limitations of Mechanism Studies Mechanisms are suggested in order to explain observed rate laws and orders of reaction Often more than one mechanism can explain experimental results

Chemistry 1011 Slot 514 Review Problem #1 Hydrogen peroxide decomposes: 2H 2 O 2(aq)  2H 2 O (l) + O 2(g) Initial rate data: [H 2 O 2 ] Initial Rate (mol/L.min) x x x Determine the order of reaction 2.Write the rate law expression 3.Calculate the rate constant, k

Chemistry 1011 Slot 515 Review Problem #2 Two mechanisms are proposed for the reaction: 2NO (g) + O 2(g)  2NO 2(g) Mechanism #1: NO + O 2 NO 3 (fast) NO 3 + NO  2NO 2 (slow) Mechanism #2: NO + NO N 2 O 2 (fast) N 2 O 2 + O 2  2NO 2 (slow) Show that each is consistent with the rate law: Rate = k[NO] 2 [O 2 ]

Chemistry 1011 Slot 516 Review Problem #2 Mechanism #1 NO + O 2 NO 3 (fast) NO 3 + NO  2NO 2 (slow) Step 2 is rate determining Rate 2 = k 2 [NO 3 ][NO] –But NO 3 is an intermediate The reactants and products in Step 1 are in equilibrium –rate forward = rate reverse k 1 [NO][O 2 ] = k -1 [NO 3 ] Substitute for [NO 3 ] in rate law expression Rate 2 = Rate overall = k 2.k 1 /k -1 [NO][O 2 ][NO] Rate = k[NO] 2 [O 2 ]

Chemistry 1011 Slot 517 Review Problem #2 Mechanism #2 NO + NO N 2 O 2 (fast) N 2 O 2 + O 2  2NO 2 (slow) Step 2 is rate determining Rate 2 = k 2 [N 2 O 2 ][O 2 ] –But N 2 O 2 is an intermediate The reactants and products in Step 1 are in equilibrium –rate forward = rate reverse k 1 [NO][NO] = k -1 [N 2 O 2 ] Substitute for [N 2 O 2 ] in rate law expression Rate 2 = Rate overall = k 2.k 1 /k -1 [NO][NO][O 2 ] Rate = k[NO] 2 [O 2 ]

Chemistry 1011 Slot 518 Review Problem #3 Hydrogen peroxide in basic solution oxidizes iodide ions to iodine The proposed mechanism for the reaction is: Step 1 (SLOW): H 2 O 2(aq) + I  (aq)  HOI (aq) + OH  (aq) Step 2 (FAST): HOI (aq) + I  (aq)  I 2(aq) + OH  (aq) Write the overall equation H 2 O 2(aq) + 2I  (aq)  I 2(aq) + OH  (aq) Write an expression for the overall rate law Rate = k [ H 2 O 2 ][ I  ]