Download presentation

Presentation is loading. Please wait.

Published byJaron Grewell Modified about 1 year ago

1
Chapter 15

2
Chemical kinetics is the study of the rates of chemical reactions. Macroscopic level: rates of reactions, reaction rate, factors affecting rates Particulate level: reaction mechanism (the detailed pathway taken by atoms and molecules as a reaction proceeds) Kinetics

3
reaction rate is a measure of the speed of a reaction rate of a chemical reaction refers to the change in concentration of substance per unit of time amounts of reactants decrease and amounts of products increase often in units of M/s 15.1 Rates of Chemical Reactions

4
2N 2 O 5 4NO 2 + O 2 rate of rxn = - Δ [N 2 O 5 ]/ Δ t rate is expressed as a positive value; the sign on the rate indicates a decrease (reactant concentration) or increase (product concentration) Rates of Chemical Reactions

5
2N 2 O 5 4NO 2 + O 2 The graph of concentration vs. time is not straight because the rate of reaction changes during the course of the reaction. (largest at time = 0) Over a time period, the average rate can be calculated. For a single point in time, an instantaneous rate can also be calculated from the slope of the tangent line. Rates of Chemical Reactions

6
Compare the rates of appearance or disappearance of each product and reactant, in the decomposition of nitrosyl chloride, NOCl. 2NOCl(g) 2NO(g) + Cl 2 (g) Practice Problem

7
Sucrose decomposes to fructose and glucose in acid solution. A plot of the concentration of sucrose as a function of time is given on page 675. What is the rate of change of the sucrose concentration over the first 2 h? What is the rate of change over the last 2 h? Estimate the instantaneous rate at 4 h. Practice Problem

8
Reaction rate increases with more collisions and more energy. Factors affecting the speed of reaction: concentration of reactants – increasing [reactant] causes more collisions therefore increasing reaction rate surface area – more surface area results in more collisions (powdered medicine works more quickly than pill forms) physical state of reactants – homogeneous mixtures of liquids and gases tend to react faster (tied to surface area) temperature – higher temp increases KE therefore increases collisions catalysts – accelerate reaction but are not transformed themselves; changes reaction mechanism bio catalysts are called enzymes (BEANO!) 15.2 Reaction Conditions and Rate

9
After reading sections , you should be able to do the following… p. 712 (1-5) Homework

10
the rate of reaction is proportional to the reactant concentration may be proportional to the concentrations of more than one reactant catalyst can affect the rate 15.3 Effect of Concentration on Rate

11
A rate equation or rate law expresses the relationship between reactant concentration and rate rate of reaction = k [N 2 O 5 ] where k is the rate constant. Rate Equations

12
For a reaction such as aA + bB xX the rate equation is rate = k [A] m [B] n where m and n are experimentally determined exponents NOT necessarily the stoichiometric coefficients (usually positive whole numbers) Rate Equations

13
The order with respect to a reactant is the exponent of its concentration term in the rate expression. Total reaction order is the sum of the exponents in the concentration terms. Order of a Reaction

14
A reactant is first order if doubling it (while holding the other reactants constant) doubles the rate. A reactant is second order if doubling it quadruples the rate. A reactant is third order if doubling it increases the rate by 8. A reactant is zero order if doubling it has no effect on the rate. Order of a Reaction

15
2NO(g) + Cl 2 (g) 2NOCl(g) Rate = k [NO] 2 [Cl 2 ] This reaction is second order in NO, first order in Cl 2, and third order overall. Compare experimental evidence to figure out the order respective to reactants. Exp.[NO] mol/L [Cl 2 ] mol/L Rate mol/L∙s e e e e -6

16
The rate constant, k, is a proportionality constant relating rate and concentrations at a given temp. Units of the rate constant have to match; depends on order of reaction. 0 order = mol/L∙time 1 st order = time -1 2 nd order = L/mol∙time 3 rd order = L 2 /mol 2 ∙time Rate Constant, k

17
A rate equation must be determined experimentally. “method of initial rates” initial rate is the instantaneous reaction rate at the start of the reaction (at t = 0) Determining a Rate Equation

18
The initial rate of the reaction of nitrogen monoxide and oxygen 2NO(g) + O 2 (g) 2NO 2 (g) was measured at 25 o C for various initial concentrations of NO and O 2. Data are collected in the table on p Determine the rate equation from these data. What is the value of the rate constant and what are the appropriate units of k ? Practice Problem

19
The rate constant k is h -1 for the reaction Pt(NH 3 ) 2 Cl 2 (aq) + H 2 O(l) [Pt(NH 3 ) 2 (H 2 O)Cl] + (aq) + Cl - (aq) and the rate equation is rate = k [Pt(NH 3 ) 2 Cl 2 ] Calculate the rate of reaction when the concentration of Pt(NH 3 ) 2 Cl 2 is 0.020M. What is the rate of change in the concentration of Cl - under those conditions? Practice Problem

20
After reading section 15.3, you should be able to do the following… P. 713 (9-13) Homework

21
Integrated rate equation: ln( [R] t /[R]o ) = - kt [R] o and [R] t are concentrations of reactant at time t = 0 and at a later time, t. The ratio of concentrations is the fraction of reactant that remains after a given time has elapsed Concentration-Time Relationships

22
ln [R] t /[R]o = - kt For a first order reaction, k has units of time -1, which means that you can choose any unit for [R] such as M, moles, grams, atoms, molecules, or pressure. First Order Reactions

23
Sucrose, a sugar, decomposes in acid solution to glucose and fructose. The reaction is first order in sucrose, and the rate constant at 25 o C is k =0.21h -1. If the initial concentration of sucrose is M, what is its concentration after 5.0 h. Practice Problem

24
Gaseous NO 2 decomposes when heated 2NO 2 (g) 2NO(g) + O 2 (g) The disappearance of NO 2 is a first order reaction with k = 3.6x10 -3 s -1 at 300 o C. a.A sample of gaseous NO 2 is placed in a flask and heated at 300 o C for 150 s. What fraction of the initial sample remains after this time? b.How long must a sample be heated so that 99% of the sample has decomposed? Practice Problem

25
For second-order reactions, the rate equation becomes 1/[R] t – 1/[R] o = kt same symbolism applies and k has units of L/mol∙time. Second-Order Reaction

26
Using the rate constant for HI as k = 30. L/mol∙min, calculate the concentration of HI after 12 min if [HI] 0 = mol/L. Practice Problem

27
For a zero-order reaction, the integrated rate equation is [R] 0 – [R] t = kt where the units of k are mol/(L∙time). Zero-Order Reactions

28
Each of the integrated rate laws can be rearranged to take the form of a straight line y = mx + b where m is the slope and b is the y-intercept. Graphical Methods

29
Zero Order [R] t = - k t + [R] o First Order ln[R] t = - k t + ln[R] o Second Order 1/[R] t = k t + 1/[R] o Graphical Methods

30
The half-life, t 1/2, of a reaction is the time required for the concentration of a reactant to decrease to one-half its initial value. Longer half-life equals slower reaction. For first order reactions: t 1/2 = 0.693/ k half life is independent of concentration! Half-Life and First-Order Reactions

31
Zero Order t 1/2 = [R] o /2 k Second Order t 1/2 = 1/( k [R] o ) Half-Life and Other Reactions

32
Americium is used in smoke detectors and in medicine for the treatment of certain malignancies. One isotope of americium, 241 Am, has a rate constant, k, for radioactive decay of year -1. In contrast, radioactive iodine-125, which is used for studies of thyroid functioning, has a rate constant for decay of day -1. a. What are the half-lives of these isotopes? b. Which element decays faster? c. If you begin a treatment with iodine-125, and have 1.6e 15 atoms, how many remain after 2.0 days? Practice Problem

33
After reading section 15.4, you should be able to do the following… P. 714 (19-23) Homework

34
For a reaction to occur: Collision Theory Reacting molecules must collide with each other. Reacting molecules must collide with sufficient energy. Molecules must collide in an orientation that can lead to rearrangement of the atoms Particulate View of Reaction Rates

35
Increasing the concentration of the reactant particles will increase the number of collisions. The number of collisions between the reactant molecules is directly proportional to the concentrations of each reactant. Collision Theory

36
An “energy barrier” must be overcome for a reaction to occur. The energy required to overcome this barrier is called activation energy, E a. Fast reactions usually have low activation energies, and slow reactions have high activation energies. Activation Energy

37
As a reaction passes over the activation energy barrier, the transition state (also called an activated complex) is the arrangement of reactant molecules and atoms at the maximum point in the reaction coordinate diagram. (see page 695) Must be inferred! Activation Energy

38
Raising temperature increases the reaction rate by increasing the fraction of molecules with enough energy to surmount the activation energy barrier. Effect of Temperature Increase

39
“steric factor” The lower the probability of achieving the proper alignment, the lower the value of k and the slower the reaction. Effect of Molecular Orientation

40
The observation that reaction rates depend on the energy and frequency of collisions between reacting molecules, on the temperature, and on whether the collisions have the correct geometry is summarized by the Arrhenius equation: k = Ae -Ea/RT Arrhenius Equation

41
k = Ae -Ea/RT where R is the gas constant ( e -3 kJ/Kmol) and T is the temperature in kelvins. The parameter A is called the frequency factor and is related to the number of collisions that have correct geometry. The other factor refers to the fraction of molecules having at least the minimum energy required for reaction. Arrhenius Equation

42
Can be used to calculate the value of activation energy from the temperature dependence of the rate constant AND calculate the rate constant for a given temperature if the activation energy and A are known. Conceptual aspects of the Arrhenius equation and interpreting graphs are tested on the National Exam, but calculations will not be. Arrhenius Equation

43
Can be rearranged to form an equation for a straight line relating ln k to 1/T. k = Ae -Ea/RT ln k = ln A – (E a /R)(1/T) y = mx + b Arrenius Equation

44
We can also solve for E a by comparing k at two different temperatures… ln k 2 /k 1 = - E a /R[1/T 2 – 1/T 1 ] Arrhenius Equation

45
The colorless gas N 2 O 4 decomposes to the brown gas NO 2 in a first-order reaction: N 2 O 4 (g) 2NO 2 (g) The rate constant k = 4.5 x 10 3 s -1 at 274K and 1.00 x 10 4 s -1 at 283K. What is the energy of activation? Practice Problem

46
Catalysts speed up the rate of a chemical reaction but are not consumed in the reaction. Reaction intermediates are species formed in one step of a reaction that are then consumed in a later step. Effect of Catalysts

47
Acid-base catalysis: a reactant either gains or loses a proton; changing the rate of the reaction Surface catalysis: either a new reaction intermediate is formed, or the probability of successful collisions is modified Some enzymes accelerate reactions by binding to the reactants in a way that lowers activation energy; others react with reactant species to form a new intermediate Catalysis

48
After reading section 15.5, you should be able to do the following… P. 715 (32-36) Homework

49
Reaction mechanisms refer to the sequence of bond- making and bond-breaking steps that occurs during the conversion of reactants to products Reaction Mechanisms

50
Step 1: Br 2 + NO Br 2 NO Step 2: Br 2 NO + NO 2BrNO Overall: Br 2 + 2NO 2BrNO Each step is an elementary step. Each step has its own E a and k. Intermediates are substances formed in one step and then consumed in another (don’t appear in overall reaction). Reaction Mechanisms

51
Although the rate constant for a reaction MUST be determined experimentally, for an elementary step it is determined stoichiometrically. The molecularity (unimolecular, bimolecular, termolecular) is the same as its order. Classified by the number of reactant molecules that come together. NOT true for the overall reaction! Elementary Steps

52
Nitric oxide is reduced by hydrogen to give nitrogen and water, 2NO(g) + 2H 2 (g) N 2 (g) + 2H 2 O(g) and one possible mechanism is 2NO(g) N 2 O 2 (g) N 2 O 2 (g) + H 2 (g) N 2 O(g) + H 2 O(g) N 2 O(g) + H 2 (g) N 2 (g) + H 2 O(g) What is the molecularity of each of the three steps? What is the rate equation for the third step? Show that the sum of these elementary steps gives the net equation. Practice Problem

53
Products can never be produced faster than the rate of the slowest elementary step, which is therefore called the rate-determining step. Rate-Determining Steps

54
The Raschig reaction produces the industrially important reducing agent hydrazine, N 2 H 4, from NH 3 and OCl - in basic, aqueous solution. A proposed mechanism is given on page 707. What is the overall stoichiometric equation? Which step is rate-determining? Write the rate equation for the rate- determining step. What reaction intermediates are involved? Practice Problem

55
After reading section 15.6, you should be able to do the following… P. 716 (39-43) Homework

Similar presentations

© 2017 SlidePlayer.com Inc.

All rights reserved.

Ads by Google