Presentation is loading. Please wait.

Presentation is loading. Please wait.

Chapter 13 Chemical Kinetics

Similar presentations


Presentation on theme: "Chapter 13 Chemical Kinetics"— Presentation transcript:

1 Chapter 13 Chemical Kinetics
Roy Kennedy Massachusetts Bay Community College Wellesley Hills, MA

2 Chemical Kinetics The speed of a chemical reaction is called its reaction rate The rate of a reaction is a measure of how fast the reaction makes products or uses reactants The ability to control the speed of a chemical reaction is important Tro: Chemistry: A Molecular Approach, 2/e

3 Defining Rate Rate is how much a quantity changes in a given period of time The speed you drive your car is a rate – the distance your car travels (miles) in a given period of time (1 hour) so the rate of your car has units of mi/hr Tro: Chemistry: A Molecular Approach, 2/e

4 Defining Reaction Rate
The rate of a chemical reaction is generally measured in terms of how much the concentration of a reactant decreases in a given period of time or product concentration increases For reactants, a negative sign is placed in front of the definition Tro: Chemistry: A Molecular Approach, 2/e

5 Reaction Rate Changes Over Time
As time goes on, the rate of a reaction generally slows down because the concentration of the reactants decreases At some time the reaction stops, either because the reactants run out or because the system has reached equilibrium Tro: Chemistry: A Molecular Approach, 2/e

6 Reaction Rate and Stoichiometry
In most reactions, the coefficients of the balanced equation are not all the same H2 (g) + I2 (g)  2 HI(g) For these reactions, the change in the number of molecules of one substance is a multiple of the change in the number of molecules of another for the above reaction, for every 1 mole of H2 used, 1 mole of I2 will also be used and 2 moles of HI made therefore the rate of change will be different To be consistent, the change in the concentration of each substance is multiplied by 1/coefficient Tro: Chemistry: A Molecular Approach, 2/e

7 Average Rate The average rate is the change in measured concentrations in any particular time period linear approximation of a curve The larger the time interval, the more the average rate deviates from the instantaneous rate Tro: Chemistry: A Molecular Approach, 2/e

8 Instantaneous Rate The instantaneous rate is the change in concentration at any one particular time slope at one point of a curve Determined by taking the slope of a line tangent to the curve at that particular point first derivative of the function for you calculus fans Tro: Chemistry: A Molecular Approach, 2/e

9 H2 (g) + I2 (g)  2 HI (g) Using [H2], the instantaneous rate at 50 s is Using [HI], the instantaneous rate at 50 s is Tro: Chemistry: A Molecular Approach, 2/e

10 Example 13. 1: For the reaction given, the [I] changes from 1
Example 13.1: For the reaction given, the [I] changes from M to M in the first 10 s. Calculate the average rate in the first 10 s and the D[H+]. H2O2 (aq) + 3 I(aq) + 2 H+(aq)  I3(aq) + 2 H2O(l) solve the rate equation for the rate (in terms of the change in concentration of the given quantity) solve the rate equation (in terms of the change in the concentration for the quantity to find) for the unknown value Tro: Chemistry: A Molecular Approach, 2/e

11 Practice – If 2. 4 x 102 g of NOBr (MM 109. 91 g) decomposes in a 2
Practice – If 2.4 x 102 g of NOBr (MM g) decomposes in a 2.0 x 102 mL flask in 5.0 minutes, find the average rate of Br2 production in M/s 2 NOBr(g)  2 NO(g) + Br2(l) Tro: Chemistry: A Molecular Approach, 2/e

12 Practice – If 2. 4 x 102 g of NOBr decomposes in a 2
Practice – If 2.4 x 102 g of NOBr decomposes in a 2.0 x 102 mL flask in 5.0 minutes, find the average rate of Br2 production 2 NOBr(g)  2 NO(g) + Br2(l) Given: Find: 5.45 M Br2, 3.0 x 102 s D[Br2]/Dt, M/s 240.0 g NOBr, L, 5.0 min. D[Br2]/Dt, M/s 240.0 g NOBr, mL, 5.0 min. D[Br2]/Dt, M/s Conceptual Plan: Relationships: D M D mole Br2 Rate D min D s } 2 mol NOBr:1 mol Br2, 1 mol = g, 1 mL=0.001 L Solve: Tro: Chemistry: A Molecular Approach, 2/e

13 Measuring Reaction Rate
To measure the reaction rate you need to be able to measure the concentration of at least one component in the mixture at many points in time There are two ways of approaching this problem 1. for reactions that are complete in less than 1 hour, it is best to use continuous monitoring of the concentration 2. for reactions that happen over a very long time, sampling of the mixture at various times can be used when sampling is used, often the reaction in the sample is stopped by a quenching technique Tro: Chemistry: A Molecular Approach, 2/e

14 Factors Affecting Reaction Rate: Nature of the Reactants
Nature of the reactants means what kind of reactant molecules and what physical condition they are in small molecules tend to react faster than large molecules gases tend to react faster than liquids, which react faster than solids powdered solids are more reactive than “blocks” more surface area for contact with other reactants certain types of chemicals are more reactive than others e.g. potassium metal is more reactive than sodium ions react faster than molecules no bonds need to be broken Tro: Chemistry: A Molecular Approach, 2/e

15 Factors Affecting Reaction Rate: Temperature
Increasing temperature increases reaction rate chemist’s rule of thumb—for each 10 °C rise in temperature, the speed of the reaction doubles for many reactions There is a mathematical relationship between the absolute temperature and the speed of a reaction discovered by Svante Arrhenius, which will be examined later Tro: Chemistry: A Molecular Approach, 2/e

16 Factors Affecting Reaction Rate: Catalysts
Catalysts are substances that affect the speed of a reaction without being consumed Most catalysts are used to speed up a reaction; these are called positive catalysts catalysts used to slow a reaction are called negative catalysts. Homogeneous = present in same phase Heterogeneous = present in different phase How catalysts work will be examined later Tro: Chemistry: A Molecular Approach, 2/e

17 Factors Affecting Reaction Rate: Reactant Concentration
Generally, the larger the concentration of reactant molecules, the faster the reaction increases the frequency of reactant molecule contact concentration of gases depends on the partial pressure of the gas higher pressure = higher concentration Concentrations of solutions depend on the solute-to-solution ratio (molarity) Tro: Chemistry: A Molecular Approach, 2/e

18 The Rate Law The rate law of a reaction is the mathematical relationship between the rate of the reaction and the concentrations of the reactants and homogeneous catalysts as well The rate law must be determined experimentally!! The rate of a reaction is directly proportional to the concentration of each reactant raised to a power For the reaction aA + bB  products the rate law would have the form given below n and m are called the orders for each reactant k is called the rate constant Tro: Chemistry: A Molecular Approach, 2/e

19 Reaction Order The exponent on each reactant in the rate law is called the order with respect to that reactant The sum of the exponents on the reactants is called the order of the reaction The rate law for the reaction 2 NO(g) + O2(g) ® 2 NO2(g) is Rate = k[NO]2[O2] The reaction is second order with respect to [NO], first order with respect to [O2], and third order overall Tro: Chemistry: A Molecular Approach, 2/e

20 Example: The rate equation for the reaction of NO with ozone is Rate = k[NO][O3]. If the rate is 6.60 x 10−5 M/sec when [NO] = 1.00 x 10−6 M and [O3] = 3.00 x 10−6 M, calculate the rate constant Given: Find: [NO] = 1.00 x 10−6 M, [O3] = 3.00 x 10−6 M, Rate = 6.60 x 10−6 M/s k, M−1s−1 Conceptual Plan: Relationships: Rate, [NO], [O3] k Rate = k[NO][O3] Solve: Tro: Chemistry: A Molecular Approach, 2/e

21 Practice – The rate law for the decomposition of acetaldehyde is Rate = k[acetaldehyde]2. What is the rate of the reaction when the [acetaldehyde] = 1.75 x 10−3 M and the rate constant is 6.73 x 10−6 M−1•s−1? Tro: Chemistry: A Molecular Approach, 2/e

22 Practice – What is the rate of the reaction when the [acetaldehyde] = 1.75 x 10−3 M and the rate constant is 6.73 x 10−6 M−1•s−1? Given: Find: [acetaldehyde] = 1.75 x 10−3 M, k= 6.73 x 10−6 M−1•s−1 Rate, M/s Conceptual Plan: Relationships: k, [acetaldehyde] Rate Rate = k[acetaldehyde]2 Solve: Tro: Chemistry: A Molecular Approach, 2/e

23 Finding the Rate Law: the Initial Rate Method
The rate law must be determined experimentally The rate law shows how the rate of a reaction depends on the concentration of the reactants Changing the initial concentration of a reactant will therefore affect the initial rate of the reaction then doubling the initial concentration of A does not change the initial reaction rate then doubling the initial concentration of A quadruples the initial reaction rate then doubling the initial concentration of A doubles the initial reaction rate if for the reaction A → Products Tro: Chemistry: A Molecular Approach, 2/e

24 Rate = k[A]n If a reaction is Zero Order, the rate of the reaction is always the same doubling [A] will have no effect on the reaction rate If a reaction is First Order, the rate is directly proportional to the reactant concentration doubling [A] will double the rate of the reaction If a reaction is Second Order, the rate is directly proportional to the square of the reactant concentration doubling [A] will quadruple the rate of the reaction Tro: Chemistry: A Molecular Approach, 2/e

25 Determining the Rate Law When there Are Multiple Reactants
Changing each reactant will effect the overall rate of the reaction By changing the initial concentration of one reactant at a time, the effect of each reactant’s concentration on the rate can be determined In examining results, we compare differences in rate for reactions that only differ in the concentration of one reactant Tro: Chemistry: A Molecular Approach, 2/e

26 Comparing Expt #1 and Expt #2, the [NO2] changes but the [CO] does not
Example 13.2: Determine the rate law and rate constant for the reaction NO2(g) + CO(g)  NO(g) + CO2(g) given the data below Comparing Expt #1 and Expt #2, the [NO2] changes but the [CO] does not Tro: Chemistry: A Molecular Approach, 2/e

27 Example 13.2: Determine the rate law and rate constant for the reaction NO2(g) + CO(g)  NO(g) + CO2(g) given the data below Tro: Chemistry: A Molecular Approach, 2/e

28 Example 13.2: Determine the rate law and rate constant for the reaction NO2(g) + CO(g)  NO(g) + CO2(g) given the data below Tro: Chemistry: A Molecular Approach, 2/e

29 Example 13.2: Determine the rate law and rate constant for the reaction NO2(g) + CO(g)  NO(g) + CO2(g) given the data below Tro: Chemistry: A Molecular Approach, 2/e

30 Example 13.2: Determine the rate law and rate constant for the reaction NO2(g) + CO(g)  NO(g) + CO2(g) given the data below n = 2, m = 0 Tro: Chemistry: A Molecular Approach, 2/e

31 Example 13.2: Determine the rate law and rate constant for the reaction NO2(g) + CO(g)  NO(g) + CO2(g) given the data below Substitute the concentrations and rate for any experiment into the rate law and solve for k Expt. Number Initial [NO2], (M) Initial [CO], (M) Initial Rate (M/s) 1. 0.10 0.0021 2. 0.20 0.0082 3. 0.0083 4. 0.40 0.033 Tro: Chemistry: A Molecular Approach, 2/e

32 Practice – Determine the rate law and rate constant for the reaction NH4+ + NO2− ® N2 + 2 H2O given the data below Tro: Chemistry: A Molecular Approach, 2/e

33 Practice – Determine the rate law and rate constant for the reaction NH4+ + NO2− ® N2 + 2 H2O given the data below Rate = k[NH4+]n[NO2]m Tro: Chemistry: A Molecular Approach, 2/e

34 Finding the Rate Law: Graphical Methods
The rate law must be determined experimentally A graph of concentration of reactant vs. time can be used to determine the effect of concentration on the rate of a reaction This involves using calculus to determine the area under a curve – which is beyond the scope of this course Later we will examine the results of this analysis so we can use graphs to determine rate laws for several simple cases Tro: Chemistry: A Molecular Approach, 2/e

35 Reactant Concentration vs. Time A  Products
insert figure 13.6 Tro: Chemistry: A Molecular Approach, 2/e

36 Integrated Rate Laws For the reaction A  Products, the rate law depends on the concentration of A Applying calculus to integrate the rate law gives another equation showing the relationship between the concentration of A and the time of the reaction – this is called the Integrated Rate Law Tro: Chemistry: A Molecular Approach, 2/e

37 Half-Life The half-life, t1/2, of a reaction is the length of time it takes for the concentration of the reactant to fall to ½ its initial value The half-life of the reaction depends on the order of the reaction Tro: Chemistry: A Molecular Approach, 2/e

38 Zero Order Reactions [A] Rate = k[A]0 = k [A] = −kt + [A]initial
constant rate reactions [A] = −kt + [A]initial Graph of [A] vs. time is straight line with slope = −k and y–intercept = [A]initial t ½ = [Ainitial]/2k When Rate = M/sec, k = M/sec [A]initial [A] slope = −k time Tro: Chemistry: A Molecular Approach, 2/e

39 First Order Reactions Rate = k[A]1 = k[A] ln[A] = −kt + ln[A]initial
Graph ln[A] vs. time gives straight line with slope = −k and y–intercept = ln[A]initial used to determine the rate constant t½ = 0.693/k The half-life of a first order reaction is constant When Rate = M/sec, k = s–1 Tro: Chemistry: A Molecular Approach, 2/e

40 ln[A] slope = −k time ln[A]initial
Tro: Chemistry: A Molecular Approach, 2/e

41 The Half-Life of a First-Order Reaction Is Constant
Tro: Chemistry: A Molecular Approach, 2/e

42 Rate Data for C4H9Cl + H2O ® C4H9OH + HCl
Tro: Chemistry: A Molecular Approach, 2/e

43 C4H9Cl + H2O  C4H9OH + 2 HCl Tro: Chemistry: A Molecular Approach, 2/e

44 C4H9Cl + H2O  C4H9OH + 2 HCl Tro: Chemistry: A Molecular Approach, 2/e

45 C4H9Cl + H2O  C4H9OH + 2 HCl slope = −2.01 x 10−3 k = 2.01 x 10−3 s-1
Tro: Chemistry: A Molecular Approach, 2/e

46 Second Order Reactions
Rate = k[A]2 1/[A] = kt + 1/[A]initial Graph 1/[A] vs. time gives straight line with slope = k and y–intercept = 1/[A]initial used to determine the rate constant t½ = 1/(k[A0]) When Rate = M/sec, k = M−1s−1 Tro: Chemistry: A Molecular Approach, 2/e

47 1/[A] slope = k time 1/[A]initial
Tro: Chemistry: A Molecular Approach, 2/e

48 Rate Data For 2 NO2 ® 2 NO + O2 Tro: Chemistry: A Molecular Approach, 2/e

49 Rate Data For 2 NO2 ® 2 NO + O2 Tro: Chemistry: A Molecular Approach, 2/e

50 Rate Data Graphs for 2 NO2 ® 2 NO + O2
Tro: Chemistry: A Molecular Approach, 2/e

51 Rate Data Graphs for 2 NO2 ® 2 NO + O2
Tro: Chemistry: A Molecular Approach, 2/e

52 Rate Data Graphs for 2 NO2 ® 2 NO + O2
Tro: Chemistry: A Molecular Approach, 2/e

53 Tro: Chemistry: A Molecular Approach, 2/e

54 the new concentration is less than the original, as expected
Example 13.4: The reaction SO2Cl2(g)  SO2(g) + Cl2(g) is first order with a rate constant of 2.90 x 10−4 s−1 at a given set of conditions. Find the [SO2Cl2] at 865 s when [SO2Cl2]initial = M Given: Find: [SO2Cl2]init = M, t = 865, k = 2.90 x 10-4 s−1 [SO2Cl2] Conceptual Plan: Relationships: [SO2Cl2] [SO2Cl2]init, t, k Solution: Check: the new concentration is less than the original, as expected Tro: Chemistry: A Molecular Approach, 2/e

55 Practice – The reaction Q  2 R is second order in Q
Practice – The reaction Q  2 R is second order in Q. If the initial [Q] = M and after 5.0 x 102 seconds the [Q] = M, find the rate constant Tro: Chemistry: A Molecular Approach, 2/e

56 Practice – The reaction Q  2 R is second order in Q
Practice – The reaction Q  2 R is second order in Q. If the initial [Q] = M and after 5.0 x 102 seconds the [Q] = M, find the rate constant Given: Find: [Q]init = M, t = 5.0 x 102 s, [Q]t = M k Conceptual Plan: Relationships: k [Q]init, t, [Q]t Solution: Tro: Chemistry: A Molecular Approach, 2/e

57 Graphical Determination of the Rate Law for A  Product
Plots of [A] vs. time, ln[A] vs. time, and 1/[A] vs. time allow determination of whether a reaction is zero, first, or second order Whichever plot gives a straight line determines the order with respect to [A] if linear is [A] vs. time, Rate = k[A]0 if linear is ln[A] vs. time, Rate = k[A]1 if linear is 1/[A] vs. time, Rate = k[A]2 Tro: Chemistry: A Molecular Approach, 2/e

58 Practice – Complete the Table and Determine the Rate Equation for the Reaction A ® 2 Product
What will the rate be when the [A] = M? Tro: Chemistry: A Molecular Approach, 2/e

59 Practice – Complete the Table and Determine the Rate Equation for the Reaction A ® 2 Product
Tro: Chemistry: A Molecular Approach, 2/e

60 Tro: Chemistry: A Molecular Approach, 2/e

61 Tro: Chemistry: A Molecular Approach, 2/e

62 Tro: Chemistry: A Molecular Approach, 2/e

63 Practice – Complete the table and determine the rate equation for the reaction A ® 2 Product
Because the graph 1/[A] vs. time is linear, the reaction is second order, Rate  k[A]2 k  slope of the line  0.10 M–1 s –1 Tro: Chemistry: A Molecular Approach, 2/e

64 Relationship Between Order and Half-Life
For a zero order reaction, the lower the initial concentration of the reactants, the shorter the half-life t1/2 = [A]init/2k For a first order reaction, the half-life is independent of the concentration t1/2 = ln(2)/k For a second order reaction, the half-life is inversely proportional to the initial concentration – increasing the initial concentration shortens the half-life t1/2 = 1/(k[A]init) Tro: Chemistry: A Molecular Approach, 2/e

65 the new concentration is less than the original, as expected
Example 13.6: Molecular iodine dissociates at 625 K with a first-order rate constant of s−1. What is the half-life of this reaction? Given: Find: k = s−1 t1/2 Conceptual Plan: Relationships: t1/2 k Solution: Check: the new concentration is less than the original, as expected Tro: Chemistry: A Molecular Approach, 2/e

66 Practice – The reaction Q  2 R is second order in Q
Practice – The reaction Q  2 R is second order in Q. If the initial [Q] = M and the rate constant is 1.8 M−1•s−1 find the length of time for [Q] = ½[Q]init Tro: Chemistry: A Molecular Approach, 2/e

67 Practice – The reaction Q  2 R is second order in Q
Practice – The reaction Q  2 R is second order in Q. If the initial [Q] = M and the rate constant is 1.8 M−1•s−1 find the length of time for [Q] = ½[Q]init Given: Find: [Q]init = M, k = 1.8 M−1s−1 t1/2, s Conceptual Plan: Relationships: t1/2 [Q]init, k Solution: Tro: Chemistry: A Molecular Approach, 2/e

68 The Effect of Temperature on Rate
Changing the temperature changes the rate constant of the rate law Svante Arrhenius investigated this relationship and showed that: where T is the temperature in kelvins R is the gas constant in energy units, J/(mol•K) A is called the frequency factor, the rate the reactant energy approaches the activation energy Ea is the activation energy, the extra energy needed to start the molecules reacting Tro: Chemistry: A Molecular Approach, 2/e

69 Tro: Chemistry: A Molecular Approach, 2/e

70 Activation Energy and the Activated Complex
There is an energy barrier to almost all reactions The activation energy is the amount of energy needed to convert reactants into the activated complex aka transition state The activated complex is a chemical species with partially broken and partially formed bonds always very high in energy because of its partial bonds Tro: Chemistry: A Molecular Approach, 2/e

71 The Arrhenius Equation: The Exponential Factor
The exponential factor in the Arrhenius equation is a number between 0 and 1 Tt represents the fraction of reactant molecules with sufficient energy so they can make it over the energy barrier the higher the energy barrier (larger activation energy), the fewer molecules that have sufficient energy to overcome it That extra energy comes from converting the kinetic energy of motion to potential energy in the molecule when the molecules collide increasing the temperature increases the average kinetic energy of the molecules therefore, increasing the temperature will increase the number of molecules with sufficient energy to overcome the energy barrier therefore increasing the temperature will increase the reaction rate Tro: Chemistry: A Molecular Approach, 2/e

72 Tro: Chemistry: A Molecular Approach, 2/e

73 Arrhenius Plots The Arrhenius Equation can be algebraically solved to give the following form: this equation is in the form y = mx + b where y = ln(k) and x = (1/T) a graph of ln(k) vs. (1/T) is a straight line (−8.314 J/mol∙K)(slope of the line) = Ea, (in Joules) ey–intercept = A (unit is the same as k) Tro: Chemistry: A Molecular Approach, 2/e

74 Example 13.7: Determine the activation energy and frequency factor for the reaction O3(g)  O2(g) + O(g) given the following data: Tro: Chemistry: A Molecular Approach, 2/e

75 Example 13.7: Determine the activation energy and frequency factor for the reaction O3(g)  O2(g) + O(g) given the following data: Tro: Chemistry: A Molecular Approach, 2/e

76 slope, m = 1.12 x 104 K y–intercept, b = 26.8
Example 13.7: Determine the activation energy and frequency factor for the reaction O3(g)  O2(g) + O(g) given the following data: slope, m = 1.12 x 104 K y–intercept, b = 26.8 Tro: Chemistry: A Molecular Approach, 2/e

77 Arrhenius Equation: Two-Point Form
If you only have two (T,k) data points, the following forms of the Arrhenius Equation can be used: Tro: Chemistry: A Molecular Approach, 2/e

78 Example 13.8: The reaction NO2(g) + CO(g)  CO2(g) + NO(g) has a rate constant of 2.57 M−1∙s−1 at 701 K and 567 M−1∙s−1 at 895 K. Find the activation energy in kJ/mol Given: Find: T1 = 701 K, k1 = 2.57 M−1∙s−1, T2 = 895 K, k2 = 567 M−1∙s−1 Ea, kJ/mol Conceptual Plan: Relationships: Ea T1, k1, T2, k2 Solution: most activation energies are tens to hundreds of kJ/mol – so the answer is reasonable Check: Tro: Chemistry: A Molecular Approach, 2/e

79 Practice – It is often said that the rate of a reaction doubles for every 10 °C rise in temperature. Calculate the activation energy for such a reaction. Hint: make T1 = 300 K, T2 = 310 K and k2 = 2k1 Tro: Chemistry: A Molecular Approach, 2/e

80 Practice – Find the activation energy in kJ/mol for increasing the temperature 10 ºC doubling the rate Given: Find: T1 = 300 K, T1 = 310 K, k2 = 2 k1 Ea, kJ/mol Conceptual Plan: Relationships: Ea T1, k1, T2, k2 Solution: Check: most activation energies are tens to hundreds of kJ/mol – so the answer is reasonable Tro: Chemistry: A Molecular Approach, 2/e

81 Collision Theory of Kinetics
For most reactions, for a reaction to take place, the reacting molecules must collide with each other on average about 109 collisions per second Once molecules collide they may react together or they may not, depending on two factors – 1. whether the collision has enough energy to "break the bonds holding reactant molecules together"; and 2. whether the reacting molecules collide in the proper orientation for new bonds to form Tro: Chemistry: A Molecular Approach, 2/e

82 Effective Collisions Kinetic Energy Factor
For a collision to lead to overcoming the energy barrier, the reacting molecules must have sufficient kinetic energy so that when they collide the activated complex can form Tro: Chemistry: A Molecular Approach, 2/e

83 Effective Collisions Orientation Effect
Tro: Chemistry: A Molecular Approach, 2/e

84 Effective Collisions Collisions in which these two conditions are met (and therefore lead to reaction) are called effective collisions The higher the frequency of effective collisions, the faster the reaction rate When two molecules have an effective collision, a temporary, high energy (unstable) chemical species is formed – the activated complex Tro: Chemistry: A Molecular Approach, 2/e

85 Collision Theory and the Frequency Factor of the Arrhenius Equation
The Arrhenius Equation includes a term, A, called the Frequency Factor The Frequency Factor can be broken into two terms that relate to the two factors that determine whether a collision will be effective Tro: Chemistry: A Molecular Approach, 2/e

86 Collision Frequency The collision frequency is the number of collisions that happen per second The more collisions per second there are, the more collisions can be effective and lead to product formation Tro: Chemistry: A Molecular Approach, 2/e

87 Molecular Interpretation of Factors Affecting Rate – Temperature
Increasing the temperature raises the average kinetic energy of the reactant molecules There is a minimum amount of kinetic energy needed for the collision to be converted into enough potential energy to form the activated complex Increasing the temperature increases the number of molecules with sufficient kinetic energy to overcome the activation energy Tro: Chemistry: A Molecular Approach, 2/e

88 Molecular Interpretation of Factors Affecting Rate – Concentration
Reaction rate generally increases as the concentration or partial pressure of reactant molecules increases except for zero order reactions More molecules leads to more molecules with sufficient kinetic energy for effective collision distribution the same, just bigger curve Tro: Chemistry: A Molecular Approach, 2/e

89 Molecularity The number of reactant particles in an elementary step is called its molecularity A unimolecular step involves one particle A bimolecular step involves two particles though they may be the same kind of particle A termolecular step involves three particles though these are exceedingly rare in elementary steps Tro: Chemistry: A Molecular Approach, 2/e

90 Catalysts Catalysts are substances that affect the rate of a reaction without being consumed Catalysts work by providing an alternative mechanism for the reaction with a lower activation energy Catalysts are consumed in an early mechanism step, then made in a later step mechanism without catalyst O3(g) + O(g)  2 O2(g) V. Slow mechanism with catalyst Cl(g) + O3(g)  O2(g) + ClO(g) Fast ClO(g) + O(g)  O2(g) + Cl(g) Slow Tro: Chemistry: A Molecular Approach, 2/e

91 Molecular Interpretation of Factors Affecting Rate – Catalysts
Catalysts generally speed a reaction They give the reactant molecules a different path to follow with a lower activation energy heterogeneous catalysts hold one reactant molecule in proper orientation for reaction to occur when the collision takes place and sometimes also help to start breaking bonds homogeneous catalysts react with one of the reactant molecules to form a more stable activated complex with a lower activation energy Tro: Chemistry: A Molecular Approach, 2/e

92 Catalysts Homogeneous catalysts are in the same phase as the reactant particles Cl(g) in the destruction of O3(g) Heterogeneous catalysts are in a different phase than the reactant particles solid catalytic converter in a car’s exhaust system Tro: Chemistry: A Molecular Approach, 2/e

93 Types of Catalysts Tro: Chemistry: A Molecular Approach, 2/e

94 Enzymes Because many of the molecules are large and complex, most biological reactions require a catalyst to proceed at a reasonable rate Protein molecules that catalyze biological reactions are called enzymes Enzymes work by adsorbing the substrate reactant onto an active site that orients the substrate for reaction 1) Enzyme + Substrate  Enzyme─Substrate Fast 2) Enzyme─Substrate → Enzyme + Product Slow Tro: Chemistry: A Molecular Approach, 2/e


Download ppt "Chapter 13 Chemical Kinetics"

Similar presentations


Ads by Google