Presentation is loading. Please wait.

Presentation is loading. Please wait.

Chemical Kinetics Chapter 13 13.1-13.6. Chemical Kinetics  In learning chemical kinetics, you will learn how to: –Predict whether or not a reaction will.

Similar presentations


Presentation on theme: "Chemical Kinetics Chapter 13 13.1-13.6. Chemical Kinetics  In learning chemical kinetics, you will learn how to: –Predict whether or not a reaction will."— Presentation transcript:

1 Chemical Kinetics Chapter

2 Chemical Kinetics  In learning chemical kinetics, you will learn how to: –Predict whether or not a reaction will take place. –Once started, determine how fast a reaction will proceed. –Learn how far a reaction will go before it stops.

3 Rate of a Reaction  Thermodynamics- Does a reaction take place?  Kinetics- How fast does a reaction proceed?  Chemical Kinetics- the area of chemistry concerned with the speeds or rates at which a chemical reaction occurs.  Reaction Rate- the change in the concentration of a reactant or product with time. (M/s)

4 Rate of a Reaction  Why do we need to know the rate of a reaction? –Practical knowledge is always useful –Preparation of drugs –Food processing –Home repair

5 Rate of a Reaction  General equation for a reaction: –A → B –Reactant → Product  In order to monitor a reaction’s speed or rate, we can look at one of two things: –Decrease in [ reactant ] –Increase in [ product ] –Can be represented as: rate = - Δ [A] / Δ tor rate = Δ [B] / Δ t

6 Rate of a Reaction

7

8  How do we measure this experimentally? –For reactions in solution:  Changes in concentration can be measured spectroscopically –For reactions involving gases:  Changes in pressure can be measured –For reactions in solution with ions present:  Change in concentrations can be measured through electrical conductance

9 Rate of a Reaction  So if we have an aqueous solution of molecular bromine and formic acid, how do we determine the reaction rate? Br 2 (aq) +HCOOH (aq) → 2Br – (aq) +2H + (aq) +CO 2 (g) time

10 Rate of a Reaction  Look for color changes  Molecular bromine is usually reddish- brown in color. Formic acid is colorless.  As the reaction progresses, the color of the solution changes.  It fades until it becomes colorless.  What does this mean?

11 Rate of a Reaction

12 Rate Calculations  How do we calculate the rate of a reaction? –We first need this information:  Time (s)  [reactant]

13 Rate Calculations Br 2 (aq) + HCOOH (aq) → 2Br – (aq) + 2H + (aq) + CO 2 (g)

14 Rate Calculations  Instantaneous Rate– rate of a reaction for a specific point in time.  Average Rate vs. Instantaneous rate –Examples????

15 Rate Calculations  Average Rate = -Δ [Br 2 ] / Δt = - [Br 2 ] final – [Br 2 ] initial / [t] final – [t] initial  Instantaneous Rate = rate for specific instance in time [Br 2 ] / t

16 Rate Calculations  Using this information, calculate the average rate of the bromine reaction over the first 50s of the reaction.

17 Rate Calculations Average Rate = - [Br 2 ] final – [Br 2 ] initial / [t] final – [t] initial Average Rate = - ( )M / (50s – 0s) Average Rate = M / 50s Average Rate = 3.80 x M/s

18 Average Rate

19 Reaction Rates and Stoichiometry  For reactions more complex than A → B we cannot use the rate expression initially described.  Example: –2A → B –Disappearance of A is twice as fast formation of B –Rate = - ½ Δ[A] /Δt

20 Reaction Rates and Stoichiometry  In general, for the reaction –aA + bB → cC + dD –Rate = - 1/a Δ[A] /Δt = - 1/b Δ[B] /Δt = 1/c Δ[C] /Δt = 1/d Δ[D] /Δt

21 Reaction Stoichiometry Write the rate expression for the following reaction: CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O (g) rate = -  [CH 4 ] tt = -  [O 2 ] tt 1 2 =  [CO 2 ] tt =  [H 2 O] tt 1 2

22

23

24

25 Rate Constant  Look back to molecular bromine chart.  What is k? –K- the rate constant. A constant of proportionality between the reaction rate and the concentration of the reactant. –K may change slightly over time. –K is represented as:  K = rate/ [reactant]  K is not affected by the [reactant] or rate alone, since it is a ratio of these two. At any given point on a graph, k should be similar in value to it’s value at other points in the same graph.

26 Rate Constant

27 The Rate Law  Rate Law- expresses the relationship of the rate of a reaction to the rate constant and the concentrations of the reactants raised to some power.  Using the general reaction: aA + bB → cC + dD aA + bB → cC + dD Rate Law is: Rate Law is: rate = k [A] x [B] y rate = k [A] x [B] y

28 The Rate Law aA + bB cC + dD Rate = k [A] x [B] y reaction is xth order in A reaction is yth order in B reaction is (x + y)th order overall

29 Reaction Order  Reaction Order- the sum of the powers to which all reactant concentrations appearing in the rate law are raised.  Reaction order is always defined in terms of reactant concentration.  Overall reaction order- x + y  Example:  Rate = k [F 2 ] [ClO 2 ]  Reaction order = first  Overall reaction order = second

30 Reaction Order  What is the rate expression for aA + bB → cC + dD where x=1 and y=2? –Rate = k[A][B] 2  What is the reaction order? –First in A, second in B  Overall reaction order? –2 +1 = 3

31 Reaction Order F 2 (g) + 2ClO 2 (g) 2FClO 2 (g) rate = k [F 2 ] x [ClO 2 ] y

32 Reaction Order If initially [F 2 ] = 1.0M and [ClO 2 ]=1.0M, what will happen to the reaction rate if F 2 is doubled? If initially [F 2 ] = 1.0M and [ClO 2 ]=1.0M, what will happen to the reaction rate if F 2 is doubled? Rate 1 = k(1.0M)(1.0M) 2 Rate 1 = k(1.0M 3 )[F 2 ] = 1.0M Rate 2 = k(2.0M)(1.0M) 2 Rate 2 = k(2.0M 3 )[F 2 ] = 2.0M Rate 2 = 2 x Rate 1

33 Reaction Order What will happen in the same reaction if the [ClO 2 ] is doubled? What will happen in the same reaction if the [ClO 2 ] is doubled? Rate 1 = k(1.0M)(1.0M) 2 Rate 1 = k(1.0M)(1.0M) 2 Rate 1 = k(1.0M 3 )[ClO 2 ] = 1.0M Rate 2 = k(1.0M)(2.0M) 2 Rate 2 = k(4.0M 3 )[ClO 2 ] = 2.0M Rate 2 = 4 x Rate 1 Rate 2 = 4 x Rate 1

34 Determination of Rate Law Experiment [F 2 ] [ClO 2 ] Rate (M/s) x x x x10 -2 F 2 (g) + 2ClO 2 (g) 2FClO 2 (g)

35 Determination of Rate Law  Experiments 1 & 4 As [F 2 ] doubles, so does the rate  Experiments 2 & 3 As [ClO 2 ] doubles, so does the rate  2:2 ratio…..1:1 ratio x = 1 and y = 1  Rate = k [F 2 ] [ClO 2 ]

36 Rate law/Expression Calculations Experimen t [S 2 O 8 2- ] [I - ] Initial Rate (M/s) x x x rate = k [S 2 O 8 2- ] x [I - ] y Double [I - ], rate doubles (experiment 1 & 2) y = 1 Double [S 2 O 8 2- ], rate doubles (experiment 2 & 3) x = 1 k = rate [S 2 O 8 2- ][I - ] = 2.2 x M/s (0.08 M)(0.034 M) = 0.08/Ms rate = k [S 2 O 8 2- ][I - ] Determine the rate law and calculate the rate constant for the following reaction from the following data: S 2 O 8 2- (aq) + 3I - (aq) 2SO 4 2- (aq) + I 3 - (aq)

37 Rate Law/Reaction Order Rate laws are always determined experimentally Reaction order is always defined in terms of reactant Reactant order is not related to the stoichiomteric coefficient in the overall reaction. F 2 (g) + 2ClO 2 (g) 2FClO 2 (g) rate = k [F 2 ][ClO 2 ]

38 Relation between Reactant Concentration and Time  First Order Reaction- a reaction whose rate depends on the reactant concentration raised to the first power. Reaction Type: A → B Rate of: -Δ [A]/Δt or k[A] Combining and simplifying these equations brings us to the following rate equation: ln[A] t = -kt + ln[A 0 ]

39 Relation between Reactant Concentration and Time

40 Reaction Time The reaction 2A B is first order in A with a rate constant of 2.8 x s -1 at 80 0 C. How long will it take for A to decrease from 0.88 M to 0.14 M ? ln[A] = ln[A] 0 - kt kt = ln[A] 0 – ln[A] t = ln[A] 0 – ln[A] k = 66 s [A] 0 = 0.88 M [A] = 0.14 M ln [A] 0 [A] k = ln 0.88 M 0.14 M 2.8 x s -1 =

41

42

43 Decomposition of Nitrogen Pentoxide  Data on page 560  Plot of ln[N2O5] (M) vs. t (s) will allow us to see and calculate more information about the reaction taking place

44 Decomposition of Nitrogen Pentoxide

45 Gas Phase Reactions  First order gas phase reactions have a linear relationship between partial pressure of gas and time. lnP t = -kt + lnP 0 lnP t = -kt + lnP 0

46

47

48 Gas Phase Reactions

49 Reaction Half-life  As a reaction proceeds, the concentrations of the reactants decreases.  Another way to measure [reactant] over time is to use the half-life.  Half-life, t 1/2 – the time required for the concentration of a reactant to decrease to half of its initial concentration.

50 Reaction Half-life  Expression for half-life of a first order reaction is: t 1/2 = ln2/k t 1/2 = ln2/k or or t 1/2 = 0.693/k t 1/2 = 0.693/k

51 Reaction Half-life

52 What is the half-life of N 2 O 5 if it decomposes with a rate constant of 5.7 x s -1 ? t½t½ ln2 k = x s -1 = = 1200 s = 20 minutes t½t½ t½t½ t½t½

53

54 Second-Order Reactions  Second-order reaction- a reaction whose rate depends on the concentration of one reactant raised to the second power OR on the concentrations of two different reactants, each raised to the first power.  Simple Type: A → B – rate = k[A] 2  Complex Type: A + B → C –rate = k[A][B]

55 Second-order Reactions  For A → B, the following expression is used: 1 [A] = 1 [A] 0 + kt

56 Half-life of a Second-order Reaction  Equation for half-life  What is the difference between this equation and the equation for half- life of first-order reactions? t ½ = 1 k[A] 0

57

58

59 Zero-order Reactions  Very rare reactions  Usually occur on metallic surfaces  Half-life Equation:  Reaction rate is described by: –Rate = k –Why? t ½ = [A] 0 2k2k

60 Summary

61 Activation Energy and Temperature Dependence of Rate Constants

62 The Collision Theory of Chemical Kinetics  Gas molecules frequently collide with one another  Expect that the rate of a reaction is equivalent to the number of collisions  Reaction rate is dependent on concentration

63 The Collision Theory of Chemical Kinetics  Activation Energy (E a )- the minimum amount of energy required to initiate a chemical reaction.  Activated Complex (Transition State)- a temporary species formed by the reactant molecules as a result of the collision before they form the product.

64 The Collision Theory of Chemical Kinetics  What does this have to do with temperature? –High energy molecules –High temperatures –Increased product formation

65 The Collision Theory of Chemical Kinetics  Factors that affect rate –1. –2. –3.

66 The Arrhenius Equation  Relation between activation energy and temperature. lnk = (E a /R) x (1/T) + lnA lnk = (E a /R) x (1/T) + lnA

67

68

69

70 Rate Constants and Temperature lnK 1 = E a x (T 1 – T 2 ) lnK 1 = E a x (T 1 – T 2 ) lnK 2 R (T 1 T 2 )

71

72 Activation Energy, Reaction Rates and Temperature  As stated earlier, for a reaction to take place, molecules must posses enough kinetic energy.  Kinetic energy must be higher than E a.  Each reaction takes place at a specific temperature……but what happens if we adjust this temp.?

73 Activation Energy, Reaction Rates and Temperature  Increasing Temperature leads to: –Molecules reach high ke faster –Number of molecules with high enough ke increases –Reaction rate increases

74 Catalysts  A catalyst is defined by the ability of a substance to do each of the following: –Catalysts increase the rate of reaction. –Catalysts are not consumed by the reaction. –A small quantity of catalyst should be able to affect the rate of reaction for a large amount of reactant. –Catalysts do not change the equilibrium constant for the reaction.

75 Catalysts  Heterogeneous catalyst- the reactants and the catalyst are in different phases. catalyst = solid reactants = liquid/gas  Homogeneous catalyst- catalyst and reactants are in the same phase, usually liquid.

76 Catalysts

77 Enzyme Catalysts

78

79 The End!!!!!!!


Download ppt "Chemical Kinetics Chapter 13 13.1-13.6. Chemical Kinetics  In learning chemical kinetics, you will learn how to: –Predict whether or not a reaction will."

Similar presentations


Ads by Google