Chapter 14 Chemical Kinetics

Transcription

1 Chapter 14 Chemical Kinetics Factors that Affect Reaction rates Reaction Rates Concentration and Rate The Change of Concentration with Time Temperature and Rate Reactions Mechanisms Catalysis

2 Chemical Kinetics Is the study of the rate at which reactions occur and also gives us information on how the reaction occurs (the Reaction Mechanism) 14.1 Factors that Affect Reaction Rates Reaction rates depend on several factors On a molecular level Reaction rates depend on the frequency with which molecules collide

3 14.2 Reaction Rates Require a quantitative definition of the Reaction rate of a chemical reaction This is defined in terms of product(s) forming and reactant(s) disappearing per unit time N 2 (g) + 3H 2 (g) 2NH 3 (g) This is the average rate, it doesn t gives us an actual rate at a given moment in time e.g. Drive the 30km to the Ferries in 30 minutes was I driving 60km/hr all the way?

4 Instantaneous Rate Gives information on the rate at a particular moment, for this we plot the concentration of (product) with time and determine the slope at our time of interest Sample Exercise 14.2 Reaction Rates and Stoichiometry For the reaction plotted 1 mole of C 4 H 9 Cl forms 1mole of C 4 H 9 OH. So rate of products formed is equal to the rate of disappearance of reactants. What happens when this is not the case? e.g 2HI(g) H 2 (g) + I 2 (g)

5 In this case the rate of appearance (formation) of H 2 and I 2 is equal, but the rate of HI disappearing is given by This leads to the generalisation that in a given reaction aa + bb cc + dd Sample Exercise 14.3

6 14.3 Concentration and Rate How does the rate of reaction change as the concentration of (initial reactants is changed)? For the reaction NH 4+ (aq) + NO 2 (aq) N 2 (g) + 2 H 2 O(l) The initial concentrations of both reactants are changed and the observed rate is measured

7 The rate law is expressed in the general form Rate = k[a] m [B] n Where k is the rate constant (depends on temp.) Big k characterizes fast reaction m and n are the reaction orders with respect to the concentration of (A and B), Not stoichiometric coefficients, must be determined experimentally The exponents in a rate law indicate how the rate is affected by the concentration of each reactant, usually 0, 1 or 2 If m and n are both = 1 then the rate is first order with respect to [A] and first order with respect to[b]. So the rate law has an overall reaction order of 1+1 = 2, and the reaction is second order Sample Exercise14.5 Determining orders and units

8 Determining Rate Laws For the Reaction: A + B C + D The experimentally data was tabulated as Expt [A] M [B] M Init.Rate M s

9 So now we could find the value of k by substituting in our values from, say, expt. 1, starting with initial concentrations. The rate of a reaction depends on concentration. The rate constant k does not, the rate constant is affected by temperature and a catalyst Given a rate law we can calculate the rate of reaction using the rate constant and initial reactant concentrations. Or given initial concentrations and initial rate we can calculate a rate law and the rate constant Sample Exercise 14.6 Determining a rate law from Initial rate data

10 14.4 The change in Concentration with Time A rate law is an equation that tells us how a reaction rate depends on reactant concentrations. But what if we are interested in how the reactant and product concentrations vary with time? From a rate law we can calculate the rate of reaction using the rate constant and reactant concentrations. We now need an equation that allows us to determine the concentration of reactants and products at any particular time First Order Reactions

11 Since ln[a] t /[A] o = ln[a] t - ln[a] o We can get this equation in the form y = mx +c by re-arranging We can test data to see if it fits this equation.yes? Then first order Sample exercise 14.7 Using the Integrated First order rate law

12 Second Order Reactions Sample Exercise 14.8 Determining Reaction Order from the Integrated Rate Law Given the concentration of [A] at various times (t), tabulate to include ln[a] and 1/[A] Plot both vs. t, which gives the linear slope?

13 Half-Life The half-life of a reaction t ½ is the time required for the concentration of a reactant to reach half its original value It is a convenient way of expressing how fast a reaction occurs First Order Reactions

14 Half-Life Second Order Reactions In contrast to first order reactions the half-life of second order reactions Does depend on the initial concentrations The half-life of a second order reaction increases as the reaction progresses Summary of kinetics from an experimental perspective Rate of reaction can be obtained from the rate law (if it is known) Without the rate law, determine the rate of a reaction by

15 Determine the order of reaction Use the initial rates if they are available (no products to affect the rate) OR, Find the graph of the rate data that yields a straight line OR,Test for constancy of the half-life Substitute rate data into integrated rate laws to find the law that yields constant values for k, the rate constant Time (s) [A] M 0 A o t 1 A 1 t 2 A 2 t 3 A 3

16 14.5 Temperature and Rate Why is a reaction first or second order and why does increasing the temperature speed up a reaction? The Collision Model, explains what is happening on a molecular level

17 In a gas phase reaction there may be collisions per second, only a small number of collisions lead to reaction The Molecules must also have sufficient energy to react (breaking bonds)

18 The arrangement of atoms at the top of the energy barrier is the transition state The Activation energy is the minimum amount of energy required to initiate a chemical reaction, (form the transition state) Arrhenius developed a relationship between the Activation Energy and the rate constant k

19 Determining the Activation Energy Determining k at several different temperatures and plotting ln k vs. 1/T the Ea can be found from the slope Remember to plot T in Kelvin as the gas constant R has units of J / mol-k! Sample exercises and The Lower the Activation Energy, the faster the reaction The higher the temperature the greater the proportion of molecules with KE Ea

20 14.6 Reaction Mechanisms A reaction mechanism tells us the sequence of events that describes the process of forming products from reactants The reaction may occur in a series of steps The number of molecules participating as reactants in an elementary process is defined by the molecularity The rate law can not generally be deduced from the overall balanced equation for a reaction as it depends on the rate law of the mechanism steps (Slowest)

21 Rate Laws for Elementary Reactions are determined by molecular proportions For this elementary reaction (step 1 in a 2 step mechanism) H 2 + ICl HI + HCl The rate Law is given as Sample exercise and 14.13

22 Multi-step Mechanisms A balanced chemical equation often occurs in a multi-step mechanism through a sequence of Elementary reactions The sequence of Elementary reactions produces intermediates One step is slower Rate determining step A valid mechanism must meet three criteria Sample exercise 14.14

23 Two types of Mechanisms Mechanisms with a slow initial step 2NO 2 (g) + F 2 (g) 2NO 2 F(g Sample exercise 14.14

24 Mechanisms with a fast initial step If the slow step is not the first step, an intermediate is a reactant in the rate determining step Example: 2NO(g) + Br 2 (g) 2NOBr(g) Has an experimentally determined Rate law of Rate = k[no] 2 [Br 2 ] The proposed mechanism is: NO(g) + Br 2 (g) NOBr 2 (g) NOBr 2 (g) + NO(g) 2NOBr(g) Because step 2 is the slow step the overall rate law is governed by that step Rate = k 2 [NOBr 2 ][NO] But NOBr 2 is an unknown and unstable intermediate Because the forward and reverse reactions occur faster than step 2 an equilibrium is established

25 In any dynamic equilibrium the rate of the forward reaction is equal to the rate of the reverse reaction, so When a fast step precedes a slow step, we can solve for the concentration of an intermediate by assuming that an equilibrium is established in the fast step Sample exercise 14.15

26 14.7 Catalysis A catalyst increases the rate of a reaction (by lowering the Activation Energy), without being consumed in the reaction Homogeneous Catalyst Heterogeneous Catalyst

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28 The reaction I 3- (aq) + 2N 3- (aq) 3I - (aq) + 3 N 2 (g) is catalyzed by CS 2 (aq) The mechanism is: CS 2 + N - 3 S 2 CN - 3 SLOW 2S 2 CN I - 3 2CS 2 + 3N 2 + 3I - FAST What is the rate law?

29 Heterogeneous Catalysts Exist in a different Phase to the reactants For example catalytic hydrogentation of alkenes with a Ni catalyst and Catalytic Converters.

30 Enzyme Catalysis Substrate (yellow) binds to enzyme (purple) in an active site Each enzyme catalyses a specific reaction in the same way a key fits a given lock The product is produced and the enzyme is unchanged. An enzyme can increase a reaction rate up to times! The product(s) leaving the active site is the rate determining step, once the products have left the site can be filled by another substrate molecule

31 At A, reduction Catalyst Pt and Rh, rips the N out of the molecule, N + N gives N 2 de-adsorbed 2NO => N 2 + O 2 or 2NO 2 => N 2 + 2O 2 At B, oxidation Catalyst Pt and Pd oxidize CO and hydrocarbons to CO 2 with remaining O 2 in exhaust g 2CO + O 2 => 2CO 2