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Chapter 16

Chapter 16. Rate of reactions / Kinetics: Rates and Mechanisms of Chemical Reactions. Lecture 1. Lecture 2. Lecture 3. Lecture 4. Highlights: Rates and Mechanisms of Chemical Reactions. Expression of the Reaction Rate: average, instantaneous and initial reactions rates.

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Chapter 16

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  1. Chapter 16 Rate of reactions / Kinetics: Rates and Mechanisms of Chemical Reactions Lecture 1 Lecture 2 Lecture 3 Lecture 4

  2. Highlights: Rates and Mechanisms of Chemical Reactions Expression of the Reaction Rate: average, instantaneous and initial reactions rates Factors That Influence Reaction Rates: concentration, physical state, temperature, use of catalyst 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 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 Catalyst: Speeding Up a Chemical Reaction: reaction energy, homogeneous, and heterogeneous catalyst Reaction Mechanisms: Steps in the Overall Reaction: elementary reactions & molecularity, rate-determining step, mechanism correlation with the rate law

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

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

  5. Expressing the reaction rate aA + bBcC + dD a balanced equation • 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 • Initial reaction rate • t = 0, the reactants just mix, no products accumulate. Point 1 Point 3 Point 2

  6. Measurement of reaction rate Average reaction rate O3 (g) + C2H4 (g)  C2H4O (g) + O2 (g) Within 10 sec Rate1 = -(0.124-0.165)/(10-0) = 4.110-3 (M/sec) Between 40 and 50 secs Rate2 = -(0.039-0.053)/(50-40) = 1.4 10-3 (M/sec) Between 50 and 60 secs Rate3 = -(0.029-0.039)/60-50) = 1.010-3 (M/sec) During the reaction course, the average rate decreases since the reactants were used up and there were fewer of them present.

  7. Instantaneous reaction rate O3 (g) + C2H4 (g)  C2H4O (g) + O2 (g) [O3]1 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 = - ([O3]2-[O3]1)/(t2-t1) = - (0.0600-0.128)/(33-6) = 2.7210-3 (M/s) 0.128 M [O3]2 0.0600 M 33 sec 6 sec

  8. Initial reaction rate O3 (g) + C2H4 (g)  C2H4O (g) + O2 (g) 0.165 M time = 0 sec, instantaneous reaction rate is called the initial reaction rate. Slope = -([O3]2-[O3]1)/(t2-t1) =-(0.100-0.165)/(12-0) =5.4110-3 (M/s) 0.100 M 12 sec

  9. Four factors which influence on the reaction rate • -- 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. • Concentration of reactants • Physical state of reactants • Temperature at which the reaction occurs • The Use of catalyst

  10. NOx O3 A. Concentration influence • Reactants crash into each other. • Reactants collide the vessel walls. • Only the reaction happens when the O3 and NOx molecules collide. • The more molecules present, the more frequently the collide, the more often reaction occurs. • The reaction rate is proportional to the reactants concentration. • The energetic collision leads to the reaction. • Rate  collision frequency  concentration of the reactants

  11. B. Physical state influence • Reactant molecules must mix to collide. • Collision frequency between molecules depends on the state of reactants. • In aqueous solution, the thermal motion brings the reactants contact. • Heterogeneous phase– the contact mainly occurs at the interface. • Agitation and grinding will favor the reaction. • Enough surface area is needed for solid chemicals to react 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. C. Temperature influence • Energetic collision (effective collision) results in the reaction. • Enough energy is required for the effective collision. • Temperature is the major effect on the energy. Temperature affects the kinetic energy of a molecule. Thus, temperature influences the collision. 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

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

  14. Three components for the rate law: Rate = k [A]m [B]n • 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: • Using concentration measurement to find initial rate. • Using initial rate to find the reaction order. • Using initial rate and reaction orders to calculate the rate constant. Rate = k [A]m Rate = k [A]m [B]n Overall reaction order: m + n

  15. Determine reaction order from the rate law • Example: NO (g) + O3 (g)  NO2 (g) + O2 (g) • Rate = k [NO]m[O3]n • The individual reaction order in NO : m • The individual reaction order in O3 : n • The overall reaction order : m + n • From the experiment, the rate law : • Rate = k [NO][O3]. • Q: Determine the individual order in NO and O3, as well the over all order. • The individual order in NO : 1 • The individual order in O3 : 1 • The overall reaction order : 1+1 = 2

  16. 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) + O2(g) 2NO2(g); rate = k[NO]2[O2] (b) CH3CHO(g) CH4(g) + CO(g); rate = k[CH3CHO]3/2 (c) H2O2(aq) + 3I-(aq) + 2H+(aq) I3-(aq) + 2H2O(l); rate = k[H2O2][I-] PLAN: Look at the rate law and not the coefficients of the chemical reaction. Determining Reaction Order from Rate Laws SOLUTION: (a) The reaction is 2nd order in NO, 1st order in O2, and 3rd order overall. (b) The reaction is 3/2 order in CH3CHO and 3/2 order overall. (c) The reaction is 1st order in H2O2, 1st order in I- and zero order in H+, while being 2nd order overall.

  17. Determining Reaction Orders from the experimental data • Write the rate law for each experiment: rate = k [O2]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. Initial Reactant Concentrations (mol/L) Initial Rate (mol/L*s) Experiment O2 NO 1 1.10  10-2 1.30 10-2 3.21 10-3 2 2.20 10-2 1.30 10-2 6.40 10-3 3 1.10 10-2 2.60 10-2 12.8 10-3

  18. Determining Reaction Orders using initial rates rate = k [O2]m[NO]n O2(g) + 2NO(g) 2NO2(g) rate2 m [O2]2 [O2]1 rate1 m 2.20x10-2mol/L 1.10x10-2mol/L Initial Reactant Conc. (M) Initial Rate (mol/L*s) Experiment O2 NO 1 1.10 10-2 1.30 10-2 3.21 10-3 2 2.20 10-2 1.30 10-2 6.40 10-3 3 1.10 10-2 2.60 10-2 12.8 10-3 Compare 2 experiments in which the concentration of one reactant varies and the concentration of the other reactant(s) remains constant. Do a similar calculation for the reactant, NO. n = 2. Overall reaction order: m+n =1+2 = 3 k [O2]2m[NO]2n [O2]2m = = = [O2]1m k [O2]1m[NO]1n 6.40x10-3mol/L*s 2 = 2m m = 1 = ; 3.21x10-3mol/L*s Use any expt result to determin k : k = initial rate/([O2][NO]2 )= 1.73 103 L2/(mol2 s) Unit for k depends on the reaction order.

  19. Reactant Concentration and Time 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] - = k [A]m [1]  t By using calculus, the equation [1] can be integrated.

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

  21. Important information 1 1/s (or s-1) 2 L/mol*s (or L mol -1 s-1) General formula: Unit of k = 3 L2 / mol2 *s (or L2 mol-2 s-1) (L/mol)order-1 Unit of time Units of the Rate Constant k for Several Overall Reaction Orders Overall Reaction Order Units of k (t in seconds) 0 mol/L*s (or mol L-1 s-1)

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

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

  24. PROBLEM: At 10000C, cyclobutane (C4H8) decomposes in a first-order reaction, with the very high rate constant of 87 s-1, to two molecules of ethylene (C2H4). [C4H8]0 ln 2.00 ; ln = (87 s-1)(0.010s) = k t [C4H8]t (a) [C4H8] (b) [C4H8]0 - [C4H8]t 2.00M - 0.87M = 2.00M [C4H8]0 Determining Reaction Concentration at a Given Time (a) If the initial C4H8 concentration is 2.00M, what is the concentration after 0.010 s? (b) What fraction of C4H8 has decomposed in this time? PLAN: Find the [C4H8] 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. SOLUTION: [C4H8] = 0.83mol/L = 0.58

  25. Reaction Half-life • Reaction Half-life (t1/2) : • is defined as the time required for the reactant to reach half its initial concentration. • Is expressed in time units for a give reaction. • 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. PROBLEM: At 10000C, cyclobutane (C4H8) decomposes in a first-order reaction, with the very high rate constant of 87 s-1, to two molecules of ethylene (C2H4). [C4H8]0 ln 2.00 ; ln = (87 s-1)(0.010s) = k t [C4H8]t (a) [C4H8] (b) [C4H8]0 - [C4H8]t 2.00M - 0.87M = 2.00M [C4H8]0 Determining Reaction Concentration at a Given Time (a) If the initial C4H8 concentration is 2.00M, what is the concentration after 0.010 s? (b) What fraction of C4H8 has decomposed in this time? PLAN: Find the [C4H8] 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. SOLUTION: [C4H8] = 0.83mol/L = 0.58

  27. Reaction Half-life • Reaction Half-life (t1/2) : • is defined as the time required for the reactant to reach half its initial concentration. • Is expressed in time units for a give reaction. • 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. for a first-order process ln 2 0.693 t1/2 = = k k A plot of [N2O5] vs. time for three half-lives.

  29. 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 1000 0C via the following first-order reaction: 0.693 Use the half-life equation, t1/2 = , to find the half-life. k (a) t1/2 = 0.693/9.2s-1 = 0.075 s Determining the Half-Life of a First-Order Reaction The rate constant is 9.2 s-1, (a) What is the half-life of the reaction? (b) How long does it take for the concentration of cyclopropane to reach one-quarter of the initial value? PLAN: One-quarter of the initial value means two half-lives have passed. SOLUTION: (b) 2 t1/2 = 2(0.075s) = 0.150 s second half life

  30. Determining the rate constant for a First-Order Reaction from half-life. PROBLEM: 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 SOLUTION: k = = = 5.2910-2 hr-1 = ? S-1 (please convert unit from hr-1 to s-1) ½

  31. The effect of temperature on reaction rate • 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/RTLnk = LnA - ( ) • k - rate constant, A - frequency factor, • R - ideal gas constant, Ea - activation energy • Higher temperature, larger k, increased rate

  32. Activation energy: • Ea is the minimum amount of energy for a molecule to have for reaction. • A chemical reaction between two substances occurs only when an atom, ion, or molecule collides with others. • 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. 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. k2 Ea 1 1 ln = - - k1 R T2 T1 The Effect of Temperature on Reaction Rate The Arrhenius Equation Re-arrange ln k = ln A - Ea/RT where k is the kinetic rate constant at T Recall : Ea is the activation energy R is the energy gas constant T is the Kelvin temperature A is the collision frequency factor

  36. PROBLEM: The decomposition of hydrogen iodide, 2HI(g) H2(g) + I2(g) -1 k2 Ea= - R ln k2 Ea 1 1 1 1 k1 - - ln = - k1 R T2 T2 T1 T1 1.10x10-5L/mol*s 1 1 Ea = - (8.314J/mol*K) - ln 500K 9..51x10-9L/mol*s 600K Determining the Energy of Activation has rate constants of 9.51x10-9 L/mol*s at 500 K and 1.10x10-5 L/mol*s at 600. K. Find Ea. PLAN: Use the modification of the Arrhenius equation to find Ea. SOLUTION: Ea = 1.76x105 J/mol = 176 kJ/mol

  37. k2 Ea 1 1 ln = - - k1 R T2 T1 Determining the rate constant from the known Energy of Activation PROBLEM: The reaction 2NOCl (g)  2NO (g) + Cl2 (g) has an activation energy Ea of 1.00  102 kJ/mol and rate constant 0.286 L/mols at 500 K. What is the rate constant for 600K? PLAN: use the Arrhenius equation: SOLUTION: Lnk2/k1 = 0.2406 k2/k1 = e0.2406 k2 = 0.364 L/mol  s (second order)

  38. 11.4 Models for Reaction Rate A A A A A A A A B B B B B B B The dependence of possible collisions on the product of reactant concentrations. • 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 Ea. • The relative size for the Ea depends on whether the overall reaction is exothermic or endothermic. • An effective orientation for reaction is important. • The structure complexity decreases the reaction rate. 4 collisions Add another molecule of A 6 collisions Add another molecule of B

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

  40. NO + NO3 2 NO2 The importance of molecular orientation to an effective collision. Z is the collision frequency p is the orientation probability factor A = pZ where A is the frequency 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. PROBLEM: A key reaction in the upper atmosphere is O3 (g) + O (g) 2O2 (g) 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. Ea= 19kJ transition state Hrxn = -392kJ Drawing Reaction Energy Diagrams and Transition States The Ea(fwd) is 19 kJ, and the  Hrxn for the reaction is -392 kJ. Draw a reaction energy diagram for this reaction, postulate a transition state, and calculate Ea(rev). SOLUTION: Ea(rev)= (392 + 19) kJ = 411 kJ Potential Energy Reaction progress

  44. Reaction energy diagram of a catalyzed and an uncatalyzed process. Generic graph indicates the effect of a catalyst in an hypothetical exothermic chemical reaction. 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.

  45. CATALYSTS • Each catalyst has its own specific way of functioning. • In general a catalyst lowers the energy of activation. • Lowering the Ea 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 Ea by providing a different mechanism, for the reaction through a new, lower energy pathway.

  46. REACTION MECHANISMS A sequence of single reaction steps that sum to the overall reaction. A product 2A product A + B product 2A + B product The overall rate of a reaction is related to the rate of the slowest, or rate-determining step (RDS). Rate Laws for General Elementary Steps 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. Elementary Step Molecularity Rate Law Unimolecular Rate = k [A] Bimolecular Rate = k [A]2 Rate = k [A][B] Bimolecular Termolecular Rate = k [A]2[B]

  47. PROBLEM: The following two reactions are proposed as elementary steps in the mechanism of an overall reaction: (1) NO2Cl(g) NO2(g) + Cl(g) (2) NO2Cl(g) + Cl(g) NO2(g) + Cl2(g) (b) Step(1) is unimolecular. Step(2) is bimolecular. (a) (1) NO2Cl(g) NO2(g) + Cl(g) (2) NO2Cl(g) + Cl(g) NO2(g) + Cl2(g) (c) rate1 = k1 [NO2Cl] 2NO2Cl(g) 2NO2(g) + Cl2(g) Determining Molecularity and Rate Laws for Elementary Steps (a) Write the overall balanced equation. (b) Determine the molecularity of each step. (c) Write the rate law for each step. PLAN: (a) The overall equation is the sum of the steps. (b) The molecularity is the sum of the reactant particles in the step. SOLUTION: rate2 = k2 [NO2Cl][Cl]

  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 Ea. • The relative size for the Ea 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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