1 Chemical Kinetics: Rates of Reactions Chapter 13 Svante A. Arrhenius * Developed concept of activation energy; asserted solutions of salts contained ions.

2 Chemical Kinetics Kinetics is the study of how fast chemical reactions occur. There are 3 important factors which affect rates of reactions: –reactant concentration [molarity or pressure] –temperature –action of catalysts Goal: to understand chemical reactions at the molecular level.

3 Chemical Kinetics Reaction Rates Speed of a reaction is measured by the change in concentration with time [brackets means concentration, units of molarity, M] For a reaction A  B At the beginning of the reaction (t = 0), let us assume that [A] = 1.0 M and [B] = 0. Average rate = change in conc. of B change in time = Δ[B] Δ t

4 Reaction Rates A B Conc (M) [ ] Concentration of reactant and product as function of time

5 Reaction rates The rate for this reaction depends on the time. At beginning of reaction, rate is fast, and the rate slows down as the reaction proceeds. At any time, the rate is given by either: (a) change in [A] with change in time, or… (b) change in [B] with change in time. There are two ways of calculating rates: (a) average rate (rough and imprecise) (b) instantaneous rate (more exact ) Average rate = change in conc. of B change in time = Δ[B] Δ t - Δ[A] Δ t =

6 Average rate is rate between two time intervals At t=10 min, [B]=0.26 M; at t=30 min, [B]=0.60 M Average rate =  [B]/  t = ( )/(30-10) = (.34 M / 20 min) = M/min conc(M) [ ] What is rate between 10 min and 30 min?

7 The instantaneous rate is more precise than the average rate. It is the rate at some specific instance in time; the concept is derived from the calculus. It is the slope of a curve at a specific instance in time. This is obtained by constructing a tangent to the curve at the indicated point. “rise” = Δy “run” = Δx Δy Δx M time Slope =

8 0.6 M 38 min What is instantaneous rate at t = 20 min? Conc (M) = 0.96 M/s Rate = Δ[B] Δ t = 0.6 M 38 min M/min =

9 Reaction Rates A B Conc (M) [ ] When do we use (+) and (-) signs for rate? Slopes for reactants (A) are always negative (-) Slopes for products (B) are always positive (+) But, all rates are positive (+). Therefore…. If slope is (-), then Rate = - slope = (-)(-) = + If slope is (+), then Rate = +slope = (+)(+) = +

10 Reaction Rates and Stoichiometry For the reaction: N H 2 2 NH 3 we can expression reaction rate as: (a) (b) (N 2 is a reactant) (H 2 is a reactant) (c) (NH 3 is a product) But these rates are not equal! NH 3 is produced twice as fast as N 2 is reacted. So,        Δt ]Δ[N 2 Δt ]Δ[NH 23

11 The Dependence of Rate on Concentration In general rates increase as concentrations increase. NH 4 + (aq) + NO 2 - (aq)  N 2 (g) + 2H 2 O(l) x2 samex2 samex2 Let’s isolate these four experiments (1,2,5,6)

12 Initial [NH 4 + ] Initial [NO 2 - ] Rate (M/s) x x x x X2 same X2 same as [NH 4 + ] doubles (with [NO 2 - ] constant), the rate doubles, so: Rate  [NH 4 + ] Rate = k[NH 4 + ][ NO 2 - ] where k is the rate constant This reaction is 1 st order with respect to NH 4 + and 1 st order with respect to NO 2 - and is 1+1=2 or 2 nd order overall as [NO 2 - ] doubles (with [NH 4 + ] constant), the rate doubles, so: Rate  [NO 2 - ] We conclude: Rate  [NH 4 + ][NO 2 - ]

13 The Dependence of Rate on Concentration Determine rate expression and rate constant, k, for following: ___[A]___ ___[B]____ __Rate(M/s)____ ___[A]___ ___[B]____ __Rate____ ___[A]___ ___[B]____ __Rate____

14 The Change of Concentration with Time First-Order Reactions A plot of ln[A] t versus t is a straight line with slope -k and intercept ln[A] 0. In the above we use the natural logarithm, ln, which is log to the base e. Aproducts Rate = When integrated, it gives the 1 st order rate equation: kt [A] ln t o  OR: ot ln[A]ktln[A] 

15 The Change of Concentration with Time First-Order Reactions

16 The Change of Concentration with Time Half-Life Half-life is the time taken for the concentration of a reactant to drop to half its original value. That is, half life, t 1/2 is the time taken for [A] 0 to reach ½[A] 0. Mathematically, 2 1 kt [A] ln o 2 1 o  OR: k ln2 k 1 t 2 1 

17 The Change of Concentration with Time Half-Life

18 Data are for first order reaction of CH 3 NC. Calculate (a) rate constant and (b) half-life ln(A 0 /A t ) = kt ln(502/95.5) = k(8000 s) ln(5.26) = k(8000 s) 1.66 = k (8000 s) k = 2.07 x s -1 t 1/2 = 0.693/k = 0.693/2.07 x s -1 = 3340 sec Sample problem

19 The Change of Concentration with Time Second-Order Reactions For a second order reaction with just one reactant: A plot of 1/[A] t versus t is a straight line with slope k and intercept 1/[A] 0 For a second order reaction, a plot of ln[A] t vs. t is not linear.

20 The Change of Concentration with Time Second-Order Reactions

21 The Change of Concentration with Time Zero-Order Reactions For a zero order reaction with just one reactant: A 0 – A t = kt A plot of [A] t versus t is a straight line with slope -k and intercept [A] 0 Zero-order reactions are used in heterogeneous catalysis reactions, such as gasoline production in refineries and catalytic hydrogenation. For zero-order reactions, the rate does not change.

22 Temperature and Rate

23 Temperature and Rate As temperature increases, the rate increases. Since the rate law has no temperature term in it, the rate constant must depend on temperature. Activation Energy Arrhenius: molecules must possess a minimum amount of energy to react. Why? –In order to form products, bonds must be broken in the reactants. –Bond breakage requires energy. Activation energy, E a, is the minimum energy required to initiate a chemical reaction.

24 Temperature and Rate Activation Energy EaEa

25 Temperature and Rate Activation Energy

26 Temperature and Rate Activation Energy The change in energy for the reaction is the difference in energy between reactants and products. The activation energy, E a, is the difference in energy between reactants and transition state. The rate depends on E a. Notice that if a forward reaction is exothermic, then the reverse reaction is endothermic.

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28 Temperature and Rate The Arrhenius Equation Arrhenius discovered most reaction-rate data obeyed the Arrhenius equation: If we have a lot of data, we can determine E a and A graphically by rearranging the Arrhenius equation: If we do not have a lot of data, then we can use RT a E Aek   –Both A and E a are specific to a given reaction. –k is the rate constant –E a is the activation energy –R is the gas constant (8.314 J/K-mol) –T is the temperature in K. –A is the frequency factor.

29 Temperature and Rate The Arrhenius Equation If we have a lot of data, we can determine E a and A graphically by rearranging the Arrhenius equation: If we do not have a lot of data, then we can use

30 Data for a reaction. Use two points to calculate E a        11 ln 318 R E k a 288

31 Reaction Mechanisms First-order reactions are generally unimolecular, i.e., they depend only on the spontaneous reaction of one molecule. Second-order reactions are generally bimolecular, i.e., they depend upon the collision of one molecule with another to exchange atoms. Zero-order reactions depend upon a catalytic bottle- neck, such as catalytic hydrogenation or cracking of petroleum to form gasoline. Third, or fractional, order reactions are possible and involve very complex mechanisms.

32 Catalysis A catalyst changes the rate of a chemical reaction, by lowering the activation energy E a.

33 Catalysis – catalytic converter A platinum surface in the exhaust system of an automobile: Promotes the complete oxidation of fuel. Converts carbon monoxide to carbon dioxide. Converts nitrogen oxides to N 2 and O 2. Catalysis – catalytic hydrogenation Adds H 2 to C=C bonds of oils to convert unsaturated compounds to saturated compounds (fats are the typical examples). The most common catalyst is nickel.

34 Catalysis Hydrogenation of ethylene, C 2 H 4 C 2 H 4 (g) + H 2 (g)  C 2 H 6 (g),  H = -136 kJ/mol.

35 Catalysis - Enzymes An enzyme is a protein molecule, with a specific shape which catalyzes a specific reaction. The substance undergoing a reaction is called a substrate, which locks into an enzyme, resulting in a fast reaction (“lock and key” model).