CHEMICAL KINETICS. Chapter 13. The Rates of Chemical Reactions. Rates of Reaction

Transcription

1 Chapter 13 CHEMICAL KINETICS The Rates of Chemical Reactions Hill, Petrucci, McCreary & Perry 4 th Ed. Rates of Reaction Definitions of Reaction Rate: The Increase in Molar Concentration of a Product per unit of time. The Decrease in Molar Concentration of a Reactant per unit of time. Rates are always Positive! Where: rate General Reaction: A B t [B] t t final - t initial [A] t [B] [A] [B] final - [B] initial more - less => positive [A] final - [A] initial less - more => negative.

2 Rates are Related by the Reaction Stoichiometry N O 5(g) 4NO (g) + O (g) rate =? [O ] t = 1 4 [NO ] t = - 1 [N O 5 ] t 4 times larger than O rate neg & times larger than O rate To Determine a Reaction Rate the experimentalist must determine the concentration of a reactant or a product over the course of a reaction Average Rate for: A B rate [B] t [B] final - [B] initial t final - t initial Experimentalist needs two [B] at two different times to get an average change in [B] over time. Experimantally Measuring the Rate of a Chemical Reaction H O (aq) H O(l) + 1 O (g) What can we measure? Easiest: Evolution of Oxygen! Others, the concentration changes of Hydrogen peroxide or water can be obtained by stoichiometry!

3 Measuring Reaction Rates In general the greater the reactant concentration the faster the reaction. The average rate is the rate of reaction over the duration of the whole experiment. An instantaneous rate is equal to the tangent to the curve at a given point. Think of this like a speedometer reading at any point in a trip from A to B, while the average rate was for the whole journey. Decomposition of H O H O (aq) H O(l) + 1 O (g) Time, s Mass O, g [H O ], M Mathematical Determination of [H O ] H O (aq) H O 1 (l) + O (g) 1.00 L, 0.88 M -- To calculate [H O ] at 60 sec from. Number of moles of O produced at 60 s: 1 mol O.960 g O x = mol O 3.00 g O Number of moles of H O consumed at 60 s: mol H mol O x O = mol H O 1 mol O Number of moles of H O present at 0 s: 1.00 L x 0.88 M H O = 0.88 mol H O Number of moles of H O remaining at 60 s: 0.88 mol H O mol H O consumed = mol H O mol H O / 1.00 L = M H O at 60 sec.

4 The General Rate Law a A + b B C d D + e E Rate = k [A]m [B]n [C]p Where: k = rate constant m = order in [A] n = order in [B] p = order in [C] Note: m, n, p do not appear in the equation. They must be determined by experiment! Usually they are small whole numbers and may coincidentally be the same as a or b. m + n + p = overall reaction order Determination of Reaction Order The Initial Rate Method! Conduct multiple experiments in which one reactant concentration is doubled and all the other reactant concentrations and the catalyst concentration is held constant. The rate is measured before and after the concentration doubling and the effect on the rate of reaction is measured. This process is repeated for each reactant being doubled. The others remaining constant. Effect on Reaction Rate for Different Orders! Expt. 1: a A + b B C d D + e E Rate = k [A]m [B]n [C]p k, [B], [C] are constant If [A] is doubled and: m is: rate is: order is: -1 halved -1 0 unchanged zero 1 doubled first quadrupled second

5 Reaction Order from Initial Rates Example Problem: H O (aq) + 3I - (aq) + H+ (aq) I 3 - (aq) + H O (l) Initial Concentrations, M Expt: H O I - H + Initial Rate mol/(l x s) x 10-6 doubled constant doubled x 10-6 When the rate doubles on doubling concentration: n = 1! x 10-6 Now compare Expt 1 & 3 for [ I - ]! m =? x 10-6 Now compare Expt 1 & 4 for [ H + ]! p =? Eqn. Calculation of Order by Initial Rates H O (aq) + 3I - (aq) + H+ (aq) Compare Expt 1 with Expt Rate = k [H O ] m [I-] n [H+] p Rate 1 = k [H O ] m [I-] n [H+] p Rate = k [0.00] m [0.010] n [ ] p Rate 1 = k [0.010] m [0.010] n [ ] p Rate = Rate 1 [0.00] m [0.010] m =.30 x x 10-6 products []m = m = 1 Then compare Expt 1 with Expt 3 & then Expt 4 Example Reactions and Orders CH CH H C CH 3 CH CH Rate = k [cc 3 H 6 ]1 NO(g) + H (g) N (g) + H O(g) Rate = k [NO][H ]1 CH 3 COCH 3(aq) + I (aq) H + CH 3 COCH I(aq) + HI(aq) Rate = k [CH 3 COCH 3 ]1[I ]0[H + ]1 Rate = k [CH 3 COCH 3 ][H + ]

6 Change of Concentration with Time First Order of type: a A products δ[a] = Rate = k [A]1 δ t Can t be used to calculate an elapsed time or a concentration change. The Integrated Rate Equation: ln [A] t [A] 0 = -kt or log [A] t [A] 0 = -kt.303 ln [A]t = -k t + ln [A]o equation y = m X + b straight line Half-life & the Integrated Rate Equation Half-life = Elapsed Time necessary to reduce [A] t to one-half of [A] o. ln [A] t 1 = -kt but [A]t = [A] 0 [A] 0 1 ln [A] 0 1 = -kt = ln [A] 0 or ln = kt t 1/ = k For a first order reaction: A products The half-life is a constant! Relating Half-life to the Rate Constant EXAMPLE: SO Cl (g) SO (g) + Cl (g) Experimental Data: At 30 o k =.0 x 10-5 s-1 The reaction is determined to be first order. Calculating half life: t =.0 x 10-5 s-1 t 1 = 3.15 x 104 s or 8.75 h Half-Lives: Fraction Remaining: Percent Remaining: [Percent Reacted:] [ ]

7 First Order Reactions: If the half-life, t 1/, is known then the rate constant, k, is known. The half-life, t 1/, does NOT depend on the initial concentration of the reactant! All Radioactive Decays are First Order! The half-life, t 1/, is a constant which is commonly tabulated and used to characterize radioactive nuclides. Effect of Temperature on Reaction Rate Arrhenius Equation Rate constant, k = A e -Ea/RT Rates of reaction double or triple with every 10K increase in temperature. The precise amount depends on E a, the activation energy for the reaction. Reaction Mechanism A series of elementary reactions leading from reactants to products. The sum of these elementary reactions = net chemical equation. Example: NO (g) + CO(g) NO(g) + CO (g) Rate Law: rate = k[no ] Mechanism: slow NO (g) + NO (g) NO(g) + NO 3(g) a reaction intermediate NO 3(g) + CO(g) NO (g) + CO (g) NO (g) + CO(g) NO(g) + CO (g) Note: A reaction intermediate may be detected but cannot be isolated.

8 Molecularity & Reaction Order Unimolecular Cl Cl + Cl Bimolecular Bimolecular Cl + CHCl 3 HCl + CCl 3 Bimolecular Cl + CCl 3 CCl 4 Termolecular Br + Br + Ar Br + Ar * The Asterisk means that the Argon atom carries away energy, otherwise the two bromine atoms would simply separate again. Rate Equations for Elementary Reactions Unimolecular: A B + C rate = k [A] Bimolecular: A + B C + D rate = k [A][B] Termolecular: A + B + C D + E rate = k [A][B][C] The rate law can be written directly from the elementary reaction. Kinetics and the Reaction Mechanism The Experimental Rate Law is an important clue to the Chemical Reaction Mechanism! This is why kinetics is so important to chemistry! The Experimental Rate Law is directly related to the Rate Determining Step, the RDS, of the reaction. The Rate Determining Step is the Slowest! Step in a chemical reaction mechanism.

9 Kinetics and the Reaction Mechanism Net Chemical Eqn: NO (g) + F (g) NO F(g) Experimental Rate Law: Rate = k [NO ][F ] Reaction Mechanism: RDS k > k 1 slow NO (g) + F (g) NO F(g) + F(g) k 1 fast NO (g) + F(g) NO F(g) k NO (g) + F (g) NO F(g) RDS Rate = k 1 [NO ][F ] Mechanisms Involving a Fast Equilibrium Step Chemical Reaction: N O 5(g) 4 NO (g) + O (g) Expt. Rate Law: Rate = k [N O 5 ] Mechanism: k 1 fast N O 5 NO + NO 3 fast k-1 RDS k NO + NO 3 NO + NO + O slow NO + NO 3 k 3 fast NO + NO Experimental versus Theoretical Rate Law Experimental and Theoretical Rate Laws must be equivalent! From the previous reaction: Expt. Rate Law: Rate = k [N O 5 ] Theor. Rate Law: Rate = k [NO ][NO 3 ] RDS The Meaning of Chemical Equilibrium: Forward Rate = Reverse Rate Rate 1 = Rate -1 Rate 1 = k 1 [N O 5 ] = k -1 [NO ][NO 3 ] = Rate -1.

10 Reconciliation of the Theor. & Expt. Rate Laws From the Equilibrium: [NO 3 ] Theoretical Rate Law: = k 1 [N O 5 ] k -1 [NO ] Rate = k [NO ][NO 3 ] k Rate = k [NO ] 1 [N O 5 ] k-1 [NO ] Rate = k k 1 k -1 [N O 5 ] Rate = k expt [N O 5 ] k expt = k k 1 k -1 Mechanisms Involving a Fast Equilibrium Step: Derive the theoretical rate equation and show it is equivalent to the Experimental Rate Law: Chemical Reaction: NO(g) + O (g) NO (g) Mechanism: Rate = k [NO] [O ] k 1 fast NO + O NO 3 fast k-1 k slow RDS NO + NO 3 NO + NO Catalysis A Catalyst increases the Rate of a chemical reaction but is not consumed in the reaction. A Catalyst changes the Mechanism of a chemical reaction. This change results in a lower Energy of Activation. A Catalyst can be present in trace or in stoichiometric amounts depending on its activity and mode of action.

11 Catalysis: Points to Note Catalysis can be homogeneous (same phase) of heterogeneous (not same phase). Inhibitors are molecules which block an available mechanism forcing a reaction to use a mechanism with a higher energy of activation. Inhibitors slow or prevent reactions from occurring and are useful as food preservatives, stabilizers, etc. Sometimes inhibitors deactivate natural catalysts formed in a system. Ozone Decomposition Chemical Reaction: O 3(g) 3 O (g) Mechanism: O 3 k 1 k-1 O + O k RDS O + O 3 O slow * concentrations are very low in air. concentration in the air is high. equilibrium shifted by LeChatelier's P. Catalysed Mechanism: O 3 + Cl fast ClO + O ClO + O 3 fast Cl + O Heterogeneous Adsorption Catalysis Reactant molecules are adsorbed on catalyst surface. Reactant molecules diffuse along the surface. Reactant molecules react to form product molecules. Product molecules are desorbed from the surface.