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Chapter 16 Rate of reactions / Kinetics: Rates and Mechanisms of Chemical Reactions Lecture 1 Lecture 2 Lecture 3 16.1 Meaning of Reaction Rate 16.2 Reaction.

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Presentation on theme: "Chapter 16 Rate of reactions / Kinetics: Rates and Mechanisms of Chemical Reactions Lecture 1 Lecture 2 Lecture 3 16.1 Meaning of Reaction Rate 16.2 Reaction."— Presentation transcript:

1 Chapter 16 Rate of reactions / Kinetics: Rates and Mechanisms of Chemical Reactions Lecture 1 Lecture 2 Lecture Meaning of Reaction Rate 16.2 Reaction Rate and Concentration 16.3 Reactant Concentration and Time 16.4 Models for Reaction Rate 16.5 Reaction Rate and Temperature 16.6 Catalyst 16.7 Reaction Mechanisms Lecture 4

2 Highlights: Rates and Mechanisms of Chemical Reactions Factors That Influence Reaction Rates: concentration, physical state, temperature, use of catalyst Expression of the Reaction Rate: average, instantaneous and initial reactions rates The Rate Law and Its Components: initial rate, reaction order terminology, reaction orders Integrated Rate Laws: Concentration Changes over Time First, second, and zero-order reactions, reaction order determination Reaction Mechanisms: Steps in the Overall Reaction: elementary reactions & molecularity, rate-determining step, mechanism correlation with the rate law Catalyst: Speeding Up a Chemical Reaction: reaction energy, homogeneous, and heterogeneous catalyst The Effect of Temperature on Reaction Rate: Arrhenius equation, activation energy Explanation of the Effects of Concentration and Temperature: collision theory - temperature effect, molecular structure effect

3 Meaning of the reaction rate Reaction: aA + bB cC + dDThis is a balanced equation. If the reactants mix, the products will form. Definition of reaction rate or chemical kinetics: 1.Is the study of a reaction and the reaction rate is a positive quantity. 2.Is the change in concentration for reactants as a function of time. 3.Is the change in concentration for products as a function of time. 4.Therefore, the reaction rate is the central focus of the chemical kinetics.

4 aA + bB cC + dD balanced equation 1.The reactant concentration decreases while the product concentration increases. 2.The change in reactant concentration is always negative,  0. 3.The change in product concentration is always positive,  0. 4.[ ] square bracket expresses the concentration in moles per liter (mol/L = M). Rate = Meaning of the reaction rate

5 1. Average reaction rate The change in concentration of a reactant/product over a change in time  [A] /  t. 2. Instantaneous reaction rate at a certain time, the slope of a tangent to [A] vs time 3.Initial reaction rate t = 0, the reactants just mix, no products accumulate. Expressing the reaction rate Point 1 aA + bB cC + dD a balanced equation Point 2 Point 3

6 Average reaction rate O 3 (g) + C 2 H 4 (g)  C 2 H 4 O (g) + O 2 (g) Time sec [O 3 ] M Within 10 sec Rate 1 = -( )/(10-0) = 4.1  (M/sec) Between 40 and 50 secs Rate 2 = -( )/(50-40) = 1.4  (M/sec) Between 50 and 60 secs Rate 3 = -( )/60-50) = 1.0  (M/sec) During the reaction course, the average rate decreases since the reactants were used up and there were fewer of them present. Measurement of reaction rate

7 Instantaneous reaction rate O 3 (g) + C 2 H 4 (g)  C 2 H 4 O (g) + O 2 (g) If we choose the time interval infinitely small, the average rate becomes the instantaneous rate. The slope of line A is the instantaneous rate for 20 sec. Slope = - ([O 3 ] 2 -[O 3 ] 1 )/(t 2 -t 1 ) = - ( )/(33-6) = 2.72  (M/s) M M 6 sec 33 sec [O 3 ] 1 [O 3 ] 2

8 Initial reaction rate O 3 (g) + C 2 H 4 (g)  C 2 H 4 O (g) + O 2 (g) time = 0 sec, instantaneous reaction rate is called the initial reaction rate. Slope = -([O 3 ] 2 -[O 3 ] 1 )/(t 2 -t 1 ) =-( )/(12-0) =5.41  (M/s) M M 12 sec

9 -- Each reaction has its own characteristic rate. -- The reaction rate is determined by the chemical nature of the reactants. -- Four factors will influence the reaction rate. 1.Concentration of reactants 2.Physical state of reactants 3.Temperature at which the reaction occurs 4.The Use of catalyst Four factors which influence on the reaction rate

10 NO x O3O3 1.Reactants crash into each other. 2.Reactants collide the vessel walls. 3.Only the reaction happens when the O 3 and NO x molecules collide. 4.The more molecules present, the more frequently the collide, the more often reaction occurs. A. Concentration influence The reaction rate is proportional to the reactants concentration. The energetic collision leads to the reaction. Rate  collision frequency  concentration of the reactants

11 1.Reactant molecules must mix to collide. 2.Collision frequency between molecules depends on the state of reactants. 3.In aqueous solution, the thermal motion brings the reactants contact. 4.Heterogeneous phase– the contact mainly occurs at the interface. 5.Agitation and grinding will favor the reaction. 6.Enough surface area is needed for solid chemicals to react B. Physical state influence It is safe to say: the finer the solid/liquid reactant is, the greater the surface is, the more contact reactants make, the faster the reaction occurs.

12 1.Energetic collision (effective collision) results in the reaction. 2.Enough energy is required for the effective collision. 3.Temperature is the major effect on the energy. C. Temperature influence It is safe to say: At a higher temperature, more collision occurs. Rising temperature will increase the number of collisions and the energy of collisions. So rising temperature enhance the reaction rate. Rate  collision energy  temperature Temperature affects the kinetic energy of a molecule. Thus, temperature influences the collision.

13 Reaction rate and concentration Rate expression and rate constantOrder of reaction For a balanced reaction: aA + bB  cC + dD The rate law (rate expression) indicates: the rate law 1. In the rate law, we assume the product does not appear. 2. The proportionality k is the rate constant. 3. k is specific for a given reaction at a certain temperature. 4. k is NOT influenced by proceed but temperature. 5. m, n is the individual reaction order with respect to the A and B reactants. 6. m, n is NOT necessarily related to the a, and b. 7. The exponent m and n determines how the concentration influences on the rate. Rate = k [A] m [B] n

14 Reaction rate, reaction orders, and rate constant. The order of a reaction involving a single reactant: Zero order : m = 0 First order : m = 1 Second order : m = 2 The order of a reaction involving a multi-reactants: Zero order : m + n = 0 First order : m + n = 1 Second order : m + n = 2 We use the experimental approach to determine the component of the rate law: 1. Using concentration measurement to find initial rate. 2. Using initial rate to find the reaction order. 3. Using initial rate and reaction orders to calculate the rate constant. Rate = k [A] m [B] n Three components for the rate law: Rate = k [A] m Rate = k [A] m [B] n Overall reaction order: m + n

15 Example:NO (g) + O 3 (g)  NO 2 (g) + O 2 (g) Rate = k [NO] m [O 3 ] n The individual reaction order in NO : m The individual reaction order in O 3 : n The overall reaction order : m + n From the experiment, the rate law : Rate = k [NO][O 3 ]. Q: Determine the individual order in NO and O 3, as well the over all order. The individual order in NO : 1 The individual order in O 3 : 1 The overall reaction order : 1+1 = 2 Determine reaction order from the rate law

16 SOLUTION: Determining Reaction Order from Rate Laws PROBLEM:For each of the following reactions, determine the reaction order with respect to each reactant and the overall order from the given rate law. (a) 2NO(g) + O 2 (g) 2NO 2 (g); rate = k[NO] 2 [O 2 ] (b) CH 3 CHO(g) CH 4 (g) + CO(g); rate = k[CH 3 CHO] 3/2 (c) H 2 O 2 (aq) + 3I - (aq) + 2H + (aq) I 3 - (aq) + 2H 2 O(l); rate = k[H 2 O 2 ][I - ] PLAN:Look at the rate law and not the coefficients of the chemical reaction. (a) The reaction is 2nd order in NO, 1st order in O 2, and 3rd order overall. (b) The reaction is 3/2 order in CH 3 CHO and 3/2 order overall. (c) The reaction is 1st order in H 2 O 2, 1st order in I - and zero order in H +, while being 2nd order overall.

17 Write the rate law for each experiment: rate = k [O 2 ] n [NO] m Take the ratio of rate law for expt 1 and 2 to determine m. Take the ratio of rate law for expt 1 and 3 to determine n. Sum of n and m is the overall reaction order m + n. Use the rate law for one of the expts to determine k. Experiment Initial Reactant Concentrations (mol/L) Initial Rate (mol/L*s) O2O2 NO 1.10          Determining Reaction Orders from the experimental data

18 Determining Reaction Orders using initial rates O 2 (g) + 2NO(g) 2NO 2 (g)rate = k [O 2 ] m [NO] n Compare 2 experiments in which the concentration of one reactant varies and the concentration of the other reactant(s) remains constant. k [O 2 ] 2 m [NO] 2 n k [O 2 ] 1 m [NO] 1 n = rate 2 rate 1 = [O 2 ] 2 m [O 2 ] 1 m = 6.40x10 -3 mol/L*s 3.21x10 -3 mol/L*s [O 2 ] 2 [O 2 ] 1 m = 1.10x10 -2 mol/L 2.20x10 -2 mol/L m ; 2 = 2 m m = 1 Do a similar calculation for the reactant, NO. n = 2. Overall reaction order: m+n =1+2 = 3 Experiment Initial Reactant Conc. (M)Initial Rate (mol/L*s) O2O2 NO 1.10          Use any expt result to determin k : k = initial rate/([O 2 ][NO] 2 ) = 1.73  10 3 L 2 /(mol 2  s) Unit for k depends on the reaction order.

19 Integrated rate law: -- Expresses the concentration change for reactants with time. Rate = -  [A]/  t From the rate law: Rate = k [A] m The expression for rate equals.  [A]  t t - = k [A] m [1] By using calculus, the equation [1] can be integrated. Reactant Concentration and Time

20 Integrated Rate Laws: express the concentration change with time rate = -  [A] tt = k [A]rate = -  [A] tt = k [A] 0 rate = -  [A] tt = k [A] 2 first order rate equation second order rate equation zero order rate equation- rate will not change with the concentration ln [A] 0 [A] t = - k t ln [A] t = ln [A] 0 -k t 1 [A] t 1 [A] 0 - = k t 1 [A] t 1 [A] 0 += k t [ A] t - [A] 0 = - k t [A] t is the concentration at any time [A] 0 is the initial concentration

21 Units of the Rate Constant k for Several Overall Reaction Orders Overall Reaction OrderUnits of k (t in seconds) 0 mol/L*s (or mol L -1 s -1 ) 11/s (or s -1 ) 2L/mol*s (or L mol -1 s -1 ) 3L 2 / mol 2 *s (or L 2 mol -2 s -1 ) General formula: Unit of k = (L/mol) order-1 Unit of time Important information

22 Integrated rate laws and reaction order ln[A] t = -kt + ln[A] 0 [A] t = -kt + [A] 0 1/[A] t = kt + 1/[A] 0

23 An Overview of Zero-Order, First-Order, and Simple Second-Order Reactions Zero OrderFirst OrderSecond Order Plot for straight line Slope, y-intercept Half-life Rate law rate = k rate = k [A]rate = k [A] 2 Units for k mol/L*s 1/sL/mol*s Integrated rate law in straight-line form [A] t = k t + [A] 0 ln[A] t = -k t + ln[A] 0 1/[A] t = k t + 1/[A] 0 [A] t vs. t ln[A] t vs. t1/[A] t = t k, [A] 0 -k, ln[A] 0 k, 1/[A] 0 [A] 0 /2kln 2/k1/k [A] 0 Important information

24 PLAN: SOLUTION: Determining Reaction Concentration at a Given Time PROBLEM:At C, cyclobutane (C 4 H 8 ) decomposes in a first-order reaction, with the very high rate constant of 87 s -1, to two molecules of ethylene (C 2 H 4 ). (a) If the initial C 4 H 8 concentration is 2.00M, what is the concentration after s? (b) What fraction of C 4 H 8 has decomposed in this time? Find the [C 4 H 8 ] at time, t, using the integrated rate law for a 1st order reaction. Once that value is found, divide the amount decomposed by the initial concentration. ; ln 2.00 [C 4 H 8 ] = (87 s -1 )(0.010s) [C 4 H 8 ] = 0.83mol/L ln [C 4 H 8 ] 0 [C 4 H 8 ] t = k t (a) (b)[C 4 H 8 ] 0 - [C 4 H 8 ] t [C 4 H 8 ] 0 = 2.00M M 2.00M = 0.58

25 Reaction Half-life Reaction Half-life (t 1/2 ) : 1.is defined as the time required for the reactant to reach half its initial concentration. 2.Is expressed in time units for a give reaction. 3.Is characteristic of that reaction at a certain temperature. The equation to calculate half-life for reactions: Zero order: First order: Second order From the equation, it can be seen the half-life for first order reaction is not dependent on the initial concentration.

26 PLAN: SOLUTION: Determining Reaction Concentration at a Given Time PROBLEM:At C, cyclobutane (C 4 H 8 ) decomposes in a first-order reaction, with the very high rate constant of 87 s -1, to two molecules of ethylene (C 2 H 4 ). (a) If the initial C 4 H 8 concentration is 2.00M, what is the concentration after s? (b) What fraction of C 4 H 8 has decomposed in this time? Find the [C 4 H 8 ] at time, t, using the integrated rate law for a 1st order reaction. Once that value is found, divide the amount decomposed by the initial concentration. ; ln 2.00 [C 4 H 8 ] = (87 s -1 )(0.010s) [C 4 H 8 ] = 0.83mol/L ln [C 4 H 8 ] 0 [C 4 H 8 ] t = k t (a) (b)[C 4 H 8 ] 0 - [C 4 H 8 ] t [C 4 H 8 ] 0 = 2.00M M 2.00M = 0.58

27 Reaction Half-life Reaction Half-life (t 1/2 ) : 1.is defined as the time required for the reactant to reach half its initial concentration. 2.Is expressed in time units for a give reaction. 3.Is characteristic of that reaction at a certain temperature. The equation to calculate half-life for reactions: Zero order: First order: Second order From the equation, it can be seen the half-life for first order reaction is not dependent on the initial concentration.

28 A plot of [N 2 O 5 ] vs. time for three half-lives. t 1/2 = for a first-order process ln 2 k k =

29 PLAN: SOLUTION: Determining the Half-Life of a First-Order Reaction PROBLEM:Cyclopropane is the smallest cyclic hydrocarbon. Because its 60 0 bond angles allow poor orbital overlap, its bonds are weak. As a result, it is thermally unstable and rearranges to propene at C via the following first-order reaction: The rate constant is 9.2 s -1, (a) What is the half-life of the reaction? one-quarter (b) How long does it take for the concentration of cyclopropane to reach one-quarter of the initial value? Use the half-life equation, t 1/2 = k, to find the half-life. One-quarter of the initial value means two half-lives have passed. t 1/2 = 0.693/9.2s -1 = s(a)2 t 1/2 = 2(0.075s) = s second half life(b)

30 Determining the rate constant for a First-Order Reaction from half-life. Iodine-123 is used to study thyroid gland function. This radioactive isotope breaks down in a first order process with a half-life of 13.1 hr. What is the rate constant? PLAN: use the equation: t = ½ PROBLEM: k = = = 5.29  hr -1 = ? S -1 (please convert unit from hr -1 to s -1 ) SOLUTION: ½

31 Temperature does affect on the reaction rate. Higher temperature gives higher reaction rate. Temperature influences the rate by affecting the rate constant. A plot of k vs temperature gives exponential equation. The relationship between temperature and rate constant is expressed by Arrhenius equation. k = A e -Ea/RT Lnk = LnA - ( ) k - rate constant, A - frequency factor, R - ideal gas constant, E a - activation energy Higher temperature,  larger k,  increased rate The effect of temperature on reaction rate

32 1.E a is the minimum amount of energy for a molecule to have for reaction. 2.A chemical reaction between two substances occurs only when an atom, ion, or molecule collides with others. 3.Only a fraction of the total collisions result in a reaction, because usually only a small percentage of the substances interacting have the minimum amount of kinetic energy a molecule must possess for it to react. Activation energy: k = A e -Ea/RT ln k = ln A - ( )

33 Dependence of the rate constant on temperature k = A e -Ea/RT

34 Graphical determination of the activation energy ln k = - + ln A

35 The Arrhenius Equation The Effect of Temperature on Reaction Rate ln k = ln A - E a /RT ln k2k2 k1k1 = EaEa R - 1 T2T2 1 T1T1 - where k is the kinetic rate constant at T E a is the activation energy R is the energy gas constant T is the Kelvin temperature A is the collision frequency factor Re-arrange

36 PLAN: SOLUTION: Determining the Energy of Activation PROBLEM:The decomposition of hydrogen iodide, 2HI(g) H 2 (g) + I 2 (g) has rate constants of 9.51x10 -9 L/mol*s at 500 K and 1.10x10 -5 L/mol*s at 600. K. Find E a. Use the modification of the Arrhenius equation to find E a. ln k2k2 k1k1 = EaEa - R 1 T2T2 1 T1T1 - E a = - Rln k2k2 k1k1 1 T2T2 1 T1T K 1 500K - ln 1.10x10 -5 L/mol*s 9..51x10 -9 L/mol*s E a = - (8.314J/mol*K) E a = 1.76x10 5 J/mol = 176 kJ/mol

37 Determining the rate constant from the known Energy of Activation The reaction 2NOCl (g)  2NO (g) + Cl 2 (g) has an activation energy E a of 1.00  10 2 kJ/mol and rate constant L/mol  s at 500 K. What is the rate constant for 600K? ln k2k2 k1k1 = EaEa R - 1 T2T2 1 T1T1 - Lnk 2 /k 1 = k 2 /k 1 = e k 2 = L/mol  s (second order) use the Arrhenius equation: PROBLEM: PLAN: SOLUTION:

38 11.4 Models for Reaction Rate The dependence of possible collisions on the product of reactant concentrations. AA AA BB BB AA AA BB BB AA 4 collisions Add another molecule of A 6 collisions Add another molecule of B AA AA BB BB AABB Collision theory: reactants must collide to react. The number of collision depends on the reactant concentration. At higher T, more collision have enough energy to exceed the E a. The relative size for the E a depends on whether the overall reaction is exothermic or endothermic. An effective orientation for reaction is important. The structure complexity decreases the reaction rate.

39 Energy-level diagram for a reaction REACTANTS PRODUCTS ACTIVATED STATE Collision Energy E a (forward) E a (reverse) The forward reaction is exothermic because the reactants have more energy than the products.

40 The importance of molecular orientation to an effective collision. NO + NO 3 2 NO 2 A is the frequency factor A = pZ where Z is the collision frequency p is the orientation probability factor

41 Transition state theory If the potential energy is less than activation energy, the molecules recoil. The repulsion decreases, speed increases, and molecules move apart without reacting. Kinetic energy of molecules push them together with enough force to overcome the repulsion and then molecules react. Molecules are oriented effectively and moving at high speed. Nuclei in one atom attracts the electrons from another, so the atomic orbital will overlap, electron density shifts. Some bonds lengthen and weaken, some bonds start to form. The reactant molecules gradually change their bonds and shapes, while turning into the products. There exists a smooth transformation, neither a reactant nor a product, but a transitional species with partial bonds. Transition state theory depicts the kinetic energy of particles changing to the potential energy during a collision. Transition state reaches when a sufficiently energetic collision and effective molecular orientation is given. The reaction energy diagram indicates the changing energy of chemical system as it progresses from reactants through transition state to products.

42 Reaction energy diagrams and possible transition states.

43 SOLUTION: Drawing Reaction Energy Diagrams and Transition States PROBLEM:A key reaction in the upper atmosphere is O 3 (g) + O (g) 2O 2 (g) The E a(fwd) is 19 kJ, and the  H rxn for the reaction is -392 kJ. Draw a reaction energy diagram for this reaction, postulate a transition state, and calculate E a(rev). PLAN:Consider the relationships among the reactants, products and transition state. The reactants are at a higher energy level than the products and the transition state is slightly higher than the reactants. E a = 19kJ  H rxn = -392kJ E a(rev) = ( ) kJ = 411 kJ transition state Reaction progress Potential Energy

44 Reaction energy diagram of a catalyzed and an uncatalyzed process. Catalysts provide an (alternative) mechanism involving a different transition state and lower activation energy. More molecular collisions have the energy needed to reach the transition state. Catalysts can perform reactions much faster, more specific, or at lower temperatures. Catalysts reduce the amount of energy needed to start a chemical reaction. Catalysts cannot make energetically unfavorable reactions possible The net free energy change of a reaction is the same whether a catalyst is used or not. Generic graph indicates the effect of a catalyst in an hypothetical exothermic chemical reaction.

45 CATALYSTS Each catalyst has its own specific way of functioning. In general a catalyst lowers the energy of activation. Lowering the E a increases the rate constant, k, and thereby increases the rate of the reaction A catalyst increases the rate of the forward AND the reverse reactions. A catalyzed reaction yields the products more quickly, but does not yield more product than the uncatalyzed reaction. A catalyst lowers E a by providing a different mechanism, for the reaction through a new, lower energy pathway.

46 Rate Laws for General Elementary Steps Elementary StepMolecularityRate Law A product 2A product A + B product 2A + B product Unimolecular Bimolecular Termolecular Rate = k [A] Rate = k [A] 2 Rate = k [A][B] Rate = k [A] 2 [B] REACTION MECHANISMS A sequence of single reaction steps that sum to the overall reaction. Elementary Step is defined as an individual step, which together makes up the proposed reaction mechanism. Elementary step is NOT made of single step and must be physically reasonalbe. Elementary step is characterized by its molecularity involving the reactant particles in the steps. The overall rate of a reaction is related to the rate of the slowest, or rate-determining step (RDS).

47 PLAN: SOLUTION : Determining Molecularity and Rate Laws for Elementary Steps PROBLEM:The following two reactions are proposed as elementary steps in the mechanism of an overall reaction: (1)NO 2 Cl(g)NO 2 (g) + Cl (g) (2)NO 2 Cl(g) + Cl (g)NO 2 (g) + Cl 2 (g) (a) Write the overall balanced equation. (b) Determine the molecularity of each step. (a) The overall equation is the sum of the steps. (b) The molecularity is the sum of the reactant particles in the step. 2NO 2 Cl(g)2NO 2 (g) + Cl 2 (g) (c) Write the rate law for each step. rate 2 = k 2 [NO 2 Cl][Cl] (1)NO 2 Cl(g)NO 2 (g) + Cl (g) (2)NO 2 Cl(g) + Cl (g)NO 2 (g) + Cl 2 (g) (a)Step(1) is unimolecular. Step(2) is bimolecular. (b) rate 1 = k 1 [NO 2 Cl](c)

48 Models for Reaction Rate (Optional) Two primary models: Collision model – Activation energy Transition-state model – Activation energy diagrams Collision theory: reactants must collide to react. The number of collision depends on the reactant concentration. At higher T, more collision have enough energy to exceed the E a. The relative size for the E a depends on whether the overall reaction is exothermic or endothermic. An effective orientation for reaction is important. The structure complexity decreases the reaction rate.


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