Nyb F09 Unit 2 Slides 26 57


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  • Write general rate law equation for: H2(g) + I2(g)  2 HI(g) 2 H2(g) + 2 NO(g)  N2(g) + 2 H2O(g)
  • The rate constant k has different units depending on the order of the reaction. You either have to memorize them, or it is possible to work out what the k would be in any situation in case you forget. If you remember than the rate of a reaction is always M/s (concentration per second). Then the left side units must always be the same as the right side units of an equation. For zero order reactions, if rate = k , then clearly k has units of M/s as well (or mol/L . s) Examples of zero order reactions are catalysed reactions where the catalyst is completely saturated.
  • Notice how the units of the k constant for a first order reaction is different. Again, you will either have to memorize it from practice, or deduce it yourself according to the rate law equation. For first order reactions: rate = k [A] M/s ? x M Therefore, it is easy to see that k for first order reactions must be s-1 in order for both sides of the equation to equate.
  • The integrated rate law is derived from equating two different rate terms that can be used to express the rate of reaction. The reaction of A  B can be expressed as –Δ[A]/Δt. And according to rate law, the rate can also be expressed as k [A]m. Therefore, –Δ[A]/Δt = k[A]m You do not need to know the derivation of the integrated rate law equations is not necessary for this course, nor is it necessary to actually memorize them. The formulas will be given to you on tests and on the final exam. The integrated rate law equation simply relates how the concentration of reactant(s) changes over time (according to the rate constant) in a LINEAR fashion.
  • Reverse Haber is zero order (wiki): 2 NH3  N2 + 3 H2 If rate constant is 0.00124 M/s, how long will it take 2.00 M NH3 to completely react?
  • ln([A]o/[A]t) when separated is ln[A]o – ln[A]t. Rearranging the equation gives a form resembling the straight line equation.
  • Figure: 13-08-01UN Title: EXAMPLE 13.3 The First-Order Integrated Rate Law: Using Graphical Analysis of Reaction Data Caption: Consider the following equation for the decomposition of SO 2 Cl 2 . The concentration of SO 2 Cl 2 was monitored at a fixed temperature as a function of time during the decomposition reaction and the following data were tabulated: SO 2 Cl 2 ( g ) -> SO 2 ( g ) + Cl 2 ( g ) { Time (s) } { [SO 2 Cl 2 ] (M) } { 0 } { 0.100 } { 100 } { 0.0971 } { 200 } { 0.0944 } { 300 } { 0.0917 } { 400 } { 0.0890 } { 500 } { 0.0865 } { 600 } { 0.0840 } { 700 } { 0.0816 } { 800 } { 0.0793 } { 900 } { 0.0770 } { 1000 } { 0.0748 } { 1100 } { 0.0727 } { 1200 } { 0.0706 } { 1300 } { 0.0686 } { 1400 } { 0.0666 } { 1500 } { 0.0647 } Show that the reaction is first order and determine the rate constant for the reaction.
  • Using the integrated rate laws with actual experimental data: From the graph of [N2O5] vs time, we can see that it is curved. Therefore, we can immediately come to the conclusion that it is not a zero order reaction. However, it is impossible to currently know if it is first or second order.
  • Some trial and error is required. The original curved data has a linear relationship according to the first order integrated rate law, and therefore means that the reaction is obeying first order kinetics.
  • Half life is a convenient term that quickly gives an indication of how fast a particular reaction occurs. The half life equations are simply converted from the integrated rate law equations. These equations AREN’T supplied on tests and the final, so you will either have to memorize them, or solve for the half life equations through substitution. Take the zero order half life equation for example: The integrated rate law for zero order reactions is [A]t – [A]o = - k t If we substitute t for t½ and [A]t for ½[A]o We get ½[A]o - [A]o = - k t½  -½[A]o = -k t½  t½ = [A]o/2 k Problem 16.5 (modified) For the reaction of cyclopropane  propene at 1000oC. The rate constant is 9.2 s-1. What is the half life for this reaction? How long does it take for the original concentration to become ¼ the original? a) From the information given you need to realize that this is a first order reaction. The units of the rate constant give this away. Therefore, to solve for the half-life, you simply plug into the equation. t½ = 0.075 s, an extremely fast reaction if half of the reactants is gone in a split second! b) This is more of a logic problem. Actual numbers help to see this. Let’s pretend the original concentration is 1.0 M of cyclopropane. We want to know how long it takes to become 0.25 M, a quarter of the original. Some of you realized that a quarter remaining is essentially 2 half-lives of time. M  0.50 M  0.25 M 0.075 s + 0.075 s = 0.15 s The other way to do this is to simply put the two values into the integrated rate law equation. ln(1.0M/0.25M) = k t solving for t gives 0.15 s as well.
  • Just for interest sake, a method for determining the age of ancient artifacts is carbon 14 decay, which is a first order reaction. Since the kinetics of the reaction is known, it is possible to see how long ago the item stopped living by seeing how much carbon 14 is left.
  • Keep in mind that the initial rate of a reaction can be taken ONLY after graphing kinetics data of concentration versus time. Initial rate is the instantaneous rate at time = 0 of a reaction.
  • The most useful change between different experimental trials is to double the concentration of one of the reactants. If we compare the rate of experiment #2 with experiment #1, we’ll see how doubling the concentration of O2 affects the rate of reaction. Experiment #2: Rate2 = k [O2]2m[NO]2n Experiment #1: Rate1 = k [O2]1m[NO]1n Since the concentration of NO was not altered between the two trials, and since k remains the same due to the fact that it’s still the same reaction, these terms will cancel each other out. Therefore: From the comparison of the data from Experiment #2 and Experiment #1, it can be concluded that the order with respect to [O2] is 1. If this same comparison is conducted between Experiment #3 and Experiment #1, you will see that doubling the NO concentration quadruples the rate. This tells you that the order with respect to [NO] is 2. Quadrupled rate: 4 = 2n Therefore, n = 2
  • We learned early on that temperature can affect the rate of a reaction. According to studies made by Swedish chemist Svante Arrhenius, the reason is because temperature changes the rate constant of a reaction.
  • The relationship of rate constant, k and temperature can be made linear by manipulating the Arrhenius equation so that plotting ln k versus 1/T yields a straight line. (Would have to run multiple experiments at different temperatures to generate this plot) The minimum amount of Problem 16.6. Because k and T are related according to the Arrhenius equation, kinetics data can be used to determine the activation energy of a reaction. For the reaction 2 HI(g)  H2(g) + I2(g), the rate constant is 9.51 x 10-9 L/mol . s at 500 K, and 1.10 x 10-5 L/mol . s at 600 K. R = 8.314 J/mol.K If we subtract equation 1 from equation 2, the ln A term will drop out and the equation will simplify to Solving for Ea gives 1.76 x 105 J/mol
  • As we’ve learned, the order of a reaction must be determined experimentally from kinetics data. Why are some reactions zero order, 1st order, or second order, etc? Why doesn’t the stoichiometric coefficients affect the order? A simple answer to these questions is that it is simply the nature of the reaction. However, chemists hypothesize that each reaction has a “mechanism” that determines its rate order. A reaction mechanism is essentially multiple “little” steps a reaction takes to produce the final product(s). Breaking down an overall reaction to multiple manageable steps provides possible insight into why any given reaction has its given kinetic profile. Notice on the slide that termolecular reactions are unlikely. According to collision theory, 3 different molecules must collide into each other simultaneously with the right orientation and amount of energy for the reaction to occur. It’s highly improbably, but not impossible. Example: Example: H2(g) + 2 ICl(g)  2 HCl(g) + I2(g) According to the stoichiometry, one H2 molecule must collide with 2 ICI molecules simultaneously, in the correct orientation, and with enough energy for the reaction to occur. Instead of one step, it seems more reasonable that the reaction may be happening in stages, called elementary steps or elementary reactions. All the elementary reactions make up the overall reactions “mechanism”. Elementary step (1): H2 + ICl  HI + HCl (bimolecular reaction): Rate(1) = Elementary step (2): HI + ICl  HCl + I2 (bimolecular reaction): Rate(2) = The steps of the mechanism add up to the overall balanced reaction. HI is an intermediate: produced during the reaction but consumed such that it does not appear in the overall equation or the rate law
  • Nyb F09 Unit 2 Slides 26 57

    1. 1. The Rate Law: How Concentration Affects Rate • The rate law expresses the rate of any reaction at a fixed temperature on the reactant concentration , AND the order of reaction. Rate law is determined from experimental data . - For a general reaction, aA + bB + …  cC + dD … etc. - The rate law has the form: Rate = k [A] m [B] n … etc. Where k is the rate constant, and m and n are the reaction orders (could be the number 0, 1, or 2) - The overall order of the reaction is m + n +…etc
    2. 2. • Explaining Reaction Order - Using the simple general reaction of A  B The rate law would be: Rate = k [A] n - Depending on the nature of the reaction, the order with respect to [A] can be 0, 1, or 2. The order (n) as well as the k value must be experimentally determined. - In the rate law equation, the order represents the power of [A] (the reactant concentration). Mathematically, the order tells you to what extent the concentration affects the rate.
    3. 3. - From the equation, you can see that any concentration of [A] 0 equals 1. - The rate of a zero order reaction is solely determined by the k constant, which has units of mol . L -1. time -1 The reaction would always proceed at a constant rate. A plot of [A] vs. time would be linear. • If m = 0, or a zero order reaction , the concentration has no affect on rate of the reaction The rate law equation would be: Rate = k [A] 0
    4. 4. • If m = 1, it is a first reaction order . - The rate of reaction will be directly proportional to the concentration of the reactant [A]. The higher the concentration, the faster the rate - The k constant for a 1 st order reaction has units of time -1 The rate of a 1 st order reaction slows down linearly as reactant is used up. A plot of [A] vs. time would be curved. The rate law equation would be: Rate = k [A] 1
    5. 5. • If m = 2, it is a second reaction order . - The rate of reaction will be proportional to the concentration of the reactant [A] squared . The higher the concentration, the faster the rate - The k constant for a 2 nd order reaction has units of L . mol -1. time -1 The rate of a 2 nd order reaction would slow down exponentially as reactant is used up. A plot of [A] vs. time would be curved The rate law equation would be: Rate = k [A] 2
    6. 6. For the reaction A  B
    7. 7. Integrated Rate Law: How Concentration Changes Over Time - The rate expression defined rate as: For the general reaction of A  B or - The rate law is defined rate as: k [A] n • Equating the two terms produces what is called the differential rate law . - Integrating this equation produces integrated rate law equations depending on the order of reaction (n value) - The integrated rate law equation simply relates how the concentration of reactant(s) changes over time (according to the rate constant) in a LINEAR fashion Δ [A] Δ t d [A] d t Δ [A] Δ t k [A] n =
    8. 8. - Through methods of calculus, the integrated rate law equation for zero order reactions is: [A] t – [A] o = - k t Concentration of A at time t Concentration of A originally Rearranging: [A] t = - k t + [A] o y = m x + b Therefore, in a plot of [A] vs t, slope = - k Zero Order: • Zero Order Integrated Rate Law
    9. 9. If we observed the change in concentration of A over time for a zero order reaction [A] (mol/L) Time (unit) Slope = - k - Zero order reactions are relatively easy to identify straight away since a plot of reactant concentration over time is linear .
    10. 10. - 1st order reactions are more difficult to interpret since the plot of [reactant] vs time would be curved . - Through methods of calculus, the integrated rate law equation for a first order reaction is: Rearranging: ln [A] t = - k t + ln[A] o y = m x + b Therefore, in a plot of ln[A] vs t, slope = - k [A] o [A] t ln = k t First Order: • First Order Integrated Rate Law
    11. 11. [A] (mol/L) Time (units) ln [A] Time (units) Slope = - k If the reaction is first order , transforming the plot of [A] vs t, to ln[A] vs t will yield a linear plot where the slope = - k
    12. 13. - 2nd order reactions are also hard to interpret since the plot of [reactant] vs. time is curved, and looks very similar to first order plots - Through methods of calculus, the integrated rate law equation for a second order reaction is: Rearranging: • Second Order Integrated Rate Law [A] o [A] t = k t 1 - 1 y = m x + b [A] t 1 = k t + [A] o 1 Therefore, in a plot of 1/[A] vs. t, slope = k
    13. 14. [A] (mol/L) 1/[ A] Time (units) If the reaction is second order , transforming the plot of [A] vs t, to 1/[A] vs t will yield a linear plot where the slope = k Slope = k Time (units)
    14. 15. N 2 O 5 (g)  N 2 O 4 (g) + ½ O 2 (g) • Example: Decomposition of dinitrogen pentoxide Rate = k[ N 2 O 5 ] a What is the order of the reaction?
    15. 16. Since it fits the ln plot, it is a 1 st order Reaction with respect to [N 2 O 5 ] Original data transformed mathematically Data will fit a straight line equation for only ONE of the plots.
    16. 17. Reaction Half-Life • The half-life (t 1/2 ) is the time required for the reactant concentration to reach half its initial value. At t 1/2 , [A] t = ½[A] o - By substituting ½[A] o into the integrated rate law equations, we get: t 1/2 Zero Order: t 1/2 t 1/2 = [A] o 2 k = ln 2 k = 1 k [A] o First Order: Second Order:
    17. 18. Carbon 14’s half-life is 5, 700 years. Taking the ratio of C-12 to remaining C-14 gives an estimate of an objects age, good for about 60,000 years. Method can date only living/organic artifacts.
    18. 19. • Technique of Swamping - Kinetics analysis of reactions becomes more complicated when there are more than 1 reactant Eg. A + B  Products Rate = k[A] m [B] n - As the reaction goes, both [A] and [B] will decrease over time and BOTH are affecting the rate (as long as the orders with respect to [A] and [B] are not 0) - In “swamping”, we purposely make the starting concentration of one of the reactants so high compared to the other, we assume the reactant of high concentration remains the same concentration throughout the experiment.
    19. 20. • Initial Rates Method For Determining Order Reaction: O 2 (g) + 2 NO(g)  2 NO 2 (g) Rate = k [O 2 ] m [NO] n - Multiple experimental runs are required for this method. [NO 2 ] time Trial #1 [NO 2 ] time Trial #2 (Concentration of reactant changed) Initial rate changes
    20. 21. Rate = k [O 2 ] m [NO] n
    21. 22. Effect of Temperature on Reaction Rate • A plot of rate constant versus temperature (in K) shows an exponential increase. The relationship between k and T is defined by the Arrhenius equation : k = Ae -Ea/RT A is related to the colliding molecules, E a is the activation energy, R is the universal gas constant (J/mol . K), and T is the temperature in Kelvin.
    22. 23. • Taking the natural log of the Arrhenius equation gives a linear relationship of k and T. y = b + m x Activation energy can be calculated from a minimum of two experiments done at different temperatures. - Two-Point form of Arrhenius equation ln k = ln A - E a R 1 T ln k 2 k 1 = E a R 1 T 1 1 T 2 -
    23. 24. Reaction Mechanisms: Multiple Steps in the Overall Reaction • Many reactions do not occur in one single step. Rather, multiple reaction steps occur to eventually convert the reactant(s) to the product(s) • Each reaction step in an overall reaction is called an elementary step Rare • Elementary reaction steps have predictable rate law equations, where the order with respect to a reactant is the coefficient of the reactant
    24. 25. • A proposed mechanism must meet 3 criteria: 1. The elementary steps must add up to the overall balanced equation. 2. The elementary steps must by physically reasonable. This means that they should either be unimolecular or bimolecular. Termolecular is very unlikely. 3. The mechanism must correlate with the actual rate law for the overall reaction. Rate-determining step: The slowest elementary step. Since it limits how fast the overall reaction can proceed, the rate law for the rate-determining step defines the rate law for the overall reaction.
    25. 26. • Reaction mechanisms are hypothetical , but are based on chemical evidence of intermediates and data from rate experiments 2 NO 2 (g) + F 2 (g)  2NO 2 F(g ) Example: - Experimentally, it has been determined that the rate law is first order in both reactants. (1) NO 2 (g) + F 2 (g)  NO 2 F(g) + F(g) (slow; rate determining) (2) NO 2 (g) + F(g)  NO 2 F(g) (fast) • The mechanism provides insight into how and why any given reaction has its characteristic kinetics Rate = k [NO 2 ][F 2 ] The proposed mechanism: Is it reasonable?
    26. 27. Reaction profile for: 2 NO 2 (g) + F 2 (g)  2NO 2 F(g ) The rate determining step is the transition state with the highest energy, not the step with the greatest E a At the transition state , the molecular entity Is believed to look like a combination of both reactant and product.
    27. 28. Reaction Coordinate I II III IV 1. How many transition states? 2. How many intermediates 3. Which is the rate determining step?
    28. 29. • If the first elementary step is not the slowest (rate determining) step, determination of the rate law equation is more complicated Observed rate law: Rate = k[H 2 ][NO] 2 - Proposed mechanism is 3 elementary steps, where the 2 nd step is the slowest. - Does the mechanism fit the observed rate law? 2 H 2 (g) + 2 NO(g)  2 H 2 O(g) + N 2 (g) Eg.
    29. 30. Riddle: 4 men went camping in the woods and became lost for days. They have no more food or water and some are seriously injured. They will all die within an hour, unless they can cross a bridge that leads them to salvation. A sign on the bridge says that as soon as it is stepped on, it will collapse in 17 minutes. It’s dark and since they have only 1 flashlight, only 2 men at most can cross at any given time. The time it takes for 2 men to cross is limited by the slowest man. Once 2 have crossed, one person must come back to bring back the flashlight before more can cross. Man 1 is fine and can cross in 1 minutes. Man 2 has blisters on his feet and can cross in 2 minutes. Man 3 has a sprained ankle and can cross in 5 minutes. Man 4 has a broken leg and can crawl across in 10 minutes. How can they all make it across in 17 minutes?
    30. 31. • Example of a 1-step overall reaction: CH 3 Br + OH -  CH 3 OH + Br - The topmost point of energy is called the transition state , where the molecular entity looks like a combination of both reactant and product. Rate = k [ CH 3 Br ][ OH - ] Observed rate law:
    31. 32. Where R-X represents CH 3 Br • Example of a multi-step overall reaction: CH 3 Br + H 2 O  CH 3 OH + HBr The reaction step of highest energy is the rate- determining step Observed rate law: Rate = k[ CH 3 Br ]