1 / 27

Rates of Reactions

2. Factors affecting rates. TemperatureIncreasing temperature increases rate.More molecules have sufficient energy to react.Reactant concentrationsDependence on concentration must be determined experimentally.Can be used to deduce mechanism.Catalystsspeed up reactionsheterogeneous (e.g. sol

bobby
Download Presentation

Rates of Reactions

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


    1. 1 Rates of Reactions Why study rates? We have learned in first term how to determine the direction of a spontaneous reaction under any temperature and concentration conditions, but we noted that this told us nothing about how quickly a reaction may occur. The rate is often an important factor. For example, industry is unlikely to adopt a process that takes weeks, or years, to form a few moles of product. If more than one reaction is possible using some of the same reactants, the one with the fastest rate often uses up all the reactant before a more desirable process has a chance. In other cases the most stable product wins; this depends on the rates of the reverse reactions also. We would like to know how the rate depends on the factors we can control, so that we can optimized those conditions to give the best mixture of products in the most efficient manner. We will look at both differential and integrated rate laws.Why study rates? We have learned in first term how to determine the direction of a spontaneous reaction under any temperature and concentration conditions, but we noted that this told us nothing about how quickly a reaction may occur. The rate is often an important factor. For example, industry is unlikely to adopt a process that takes weeks, or years, to form a few moles of product. If more than one reaction is possible using some of the same reactants, the one with the fastest rate often uses up all the reactant before a more desirable process has a chance. In other cases the most stable product wins; this depends on the rates of the reverse reactions also. We would like to know how the rate depends on the factors we can control, so that we can optimized those conditions to give the best mixture of products in the most efficient manner. We will look at both differential and integrated rate laws.

    2. 2 Factors affecting rates Temperature Increasing temperature increases rate. More molecules have sufficient energy to react. Reactant concentrations Dependence on concentration must be determined experimentally. Can be used to deduce mechanism. Catalysts speed up reactions heterogeneous (e.g. solids) or homogeneous (same phase) 15.215.2

    3. 3 Dependence of concentration on time, in solution For the simple reaction A ? C, starting with [A] = 1.0 and [C] = 0 Concentration of A decreases. Concentration of C increases at the same rate. Reaction slows, but continues until A runs out, or until equilibrium is established. At completion, or equilibrium, concentrations of A and C are constant. 15.1 The plot is for a hypothetical irreversible reaction that occurs in one unimolecular step (we will define these terms later).15.1 The plot is for a hypothetical irreversible reaction that occurs in one unimolecular step (we will define these terms later).

    4. 4 Dependence of concentration on time, in solution Rate of reaction can be expressed as the rate of disappearance of A or as the rate of appearance of C. For the reaction A ? C, Rate = -D[A]/D t = +D[C]/Dt Once we see how the rate of reaction depends on the concentrations, we will write mathematical expressions for the concentrations as a function of time, these are the integrated rate laws. 15.1 I have written the rate as a derivative of the concentration with respect to time, this is an instantaneous rate. If you are unfamiliar with differential calculus, you can simply think of this as a difference in concentration over the difference in time, or an average rate. This plot is the same as the previous, but the initial concentration of the product has been increased, with no effect on the rate of reaction. Note that rate for the forward reaction does not usually depend on concentration of products, even when expressed as the change in these concentrations. When the reaction is reversible, the net rate of change for a given species concentration will depend on both products and reactants, since these are arbitrary distinctions in a reversible reaction. 15.1 I have written the rate as a derivative of the concentration with respect to time, this is an instantaneous rate. If you are unfamiliar with differential calculus, you can simply think of this as a difference in concentration over the difference in time, or an average rate. This plot is the same as the previous, but the initial concentration of the product has been increased, with no effect on the rate of reaction. Note that rate for the forward reaction does not usually depend on concentration of products, even when expressed as the change in these concentrations. When the reaction is reversible, the net rate of change for a given species concentration will depend on both products and reactants, since these are arbitrary distinctions in a reversible reaction.

    5. 5 Dependence of concentration on time, in solution For the reaction A ? 2C, The rate of appearance of C is twice the rate of disappearance of A + D[C]/ D t = 2(- D[A]/ D t ) In general, for any reaction a A + b B ? c C - D[A] = - D[B] = + D[C] a D t b D t c D t 15.1 The plot shows that the rate of appearance of C is twice the rate of disappearance of A, in this case. This is true purely because of the stoichiometry of the reaction and has nothing to do with how the rate depends on the concentrations. a, b and c are the stoichiometric coefficients of A, B and C. Note the negative sign in front of each of the reactants, since as the reaction proceeds, their concentrations decrease. Note that each rate is divided by the associated stoichiometric coefficient to make the expressions equal. This may be a little counterintuitive, but can be better understood by considering the case of A ? 2C described above. 15.1 The plot shows that the rate of appearance of C is twice the rate of disappearance of A, in this case. This is true purely because of the stoichiometry of the reaction and has nothing to do with how the rate depends on the concentrations. a, b and c are the stoichiometric coefficients of A, B and C. Note the negative sign in front of each of the reactants, since as the reaction proceeds, their concentrations decrease. Note that each rate is divided by the associated stoichiometric coefficient to make the expressions equal. This may be a little counterintuitive, but can be better understood by considering the case of A ? 2C described above.

    6. 6 Measuring the rate of a reaction The rate is often measured as D[X]/Dt, where X may be a reactant or product. Depending on the nature of X, the change in concentration may be monitored by a change in colour (intensity of some wavelength) pressure (for gases) pH (for OH- or H3O+) conductivity (ions) radioactivity, etc. 15.1 The rate may be measured as a finite difference (average rate), or as a differential (instantaneous rate), depending on how many points on the concentration vs. time plot are known. We mentioned in class last week that the colour of a substance is due to the light it absorbs to excite electrons. The higher the concentration of absorbing molecules, the more intense the colour. The pressure can be used to measure the progress of a reaction as long as the number of gas molecules changes from products to reactants. The pressure may increase or decrease as the reaction proceeds. If there is more than one gaseous species involved we will be measuring the extent of reaction, rather than the concentration of one species directly. If we detect alpha particles, beta particles, or other products of radioactive decay, we are typically measuring individual occurrences of radioactive decay processes, the number of detections per unit time will tell us about the rate of decay. This may be the rate we are trying to measure, or it may tell us about the concentration of some radioactive species, which is a product or reactant in some other reaction of interest.15.1 The rate may be measured as a finite difference (average rate), or as a differential (instantaneous rate), depending on how many points on the concentration vs. time plot are known. We mentioned in class last week that the colour of a substance is due to the light it absorbs to excite electrons. The higher the concentration of absorbing molecules, the more intense the colour. The pressure can be used to measure the progress of a reaction as long as the number of gas molecules changes from products to reactants. The pressure may increase or decrease as the reaction proceeds. If there is more than one gaseous species involved we will be measuring the extent of reaction, rather than the concentration of one species directly. If we detect alpha particles, beta particles, or other products of radioactive decay, we are typically measuring individual occurrences of radioactive decay processes, the number of detections per unit time will tell us about the rate of decay. This may be the rate we are trying to measure, or it may tell us about the concentration of some radioactive species, which is a product or reactant in some other reaction of interest.

    7. 7 Measuring the initial rate The rate of the forward reaction depends on the concentration of reactants, not products. The dependence may be linear, quadratic, etc., this must be determined experimentally. The rate is measured at the beginning of the reaction (the initial rate), as a function of the initial reactant concentrations This determines the reaction order. 15.3 The rate may also be determined at some other point, or continuously. The initial rate is common, because the initial concentrations (the independent variable) can be most easily controlled by the experimenter. If the concentration is measured continuously, then a derivative can be used to find the instantaneous rate. If many points are known, a curve may be fit to the data. Otherwise, the concentration is measured at just two, or a few, points and the rate is measured as a finite difference.15.3 The rate may also be determined at some other point, or continuously. The initial rate is common, because the initial concentrations (the independent variable) can be most easily controlled by the experimenter. If the concentration is measured continuously, then a derivative can be used to find the instantaneous rate. If many points are known, a curve may be fit to the data. Otherwise, the concentration is measured at just two, or a few, points and the rate is measured as a finite difference.

    8. 8 Initial reaction rates 2 NO (g) + 2 H2 (g) ? N2 (g) + 2 H2O (g) 15.3 Note that the reaction order does NOT relate directly to the stoichiometry of the reaction. Therefore the rate law must be determined experimentally. Later we will see some special cases, where the reaction order is determined by the stoichiometry, but this is not generally the case.15.3 Note that the reaction order does NOT relate directly to the stoichiometry of the reaction. Therefore the rate law must be determined experimentally. Later we will see some special cases, where the reaction order is determined by the stoichiometry, but this is not generally the case.

    9. 9 Initial reaction rates reaction order 15.3 Here is a simplified, generalized view of the experimental data we just saw. The general procedure is to change one concentration by some factor and measure the change in the rate. We equate the ratio of the rate change to the ratio of the concentration change, raised to some power, and solve for the exponent. (You will use a similar method to determine the order of a reaction with respect to KIO3 concentration, in Lab 8.) This procedure is repeated for each species which might affect the rate of the reaction. It is possible to have fractional reaction orders.15.3 Here is a simplified, generalized view of the experimental data we just saw. The general procedure is to change one concentration by some factor and measure the change in the rate. We equate the ratio of the rate change to the ratio of the concentration change, raised to some power, and solve for the exponent. (You will use a similar method to determine the order of a reaction with respect to KIO3 concentration, in Lab 8.) This procedure is repeated for each species which might affect the rate of the reaction. It is possible to have fractional reaction orders.

    10. 10 CONSIDER THE RATE DATA FOR THE REACTION: 2NO + O2 2NO2

    11. 11 Reaction order & the rate law The rate law is: Rate = k[A]x[B]y The order of a reaction is equal to the value of the exponent in the rate law. The reaction has an order with respect to each reactant (and each catalyst). The overall reaction order is the sum of the individual orders. k is the rate constant, which depends on T. 15.3 A and B represent any species which may affect the rate, these are generally limited to reactants and catalysts. Note that the rate constant, k, is a constant in the same sense that the equilibrium constant, K, is a constant they each depend on the reaction of interest, and on the temperature, but not on concentrations. Note also the upper vs. lower case k used to represent these two types of constants.15.3 A and B represent any species which may affect the rate, these are generally limited to reactants and catalysts. Note that the rate constant, k, is a constant in the same sense that the equilibrium constant, K, is a constant they each depend on the reaction of interest, and on the temperature, but not on concentrations. Note also the upper vs. lower case k used to represent these two types of constants.

    12. 12 The rate constant Once the form of the rate law is known, we can fill in the data from any one run of our determination to find the rate constant. e.g. 2 NO (g) + 2 H2 (g) ? N2 (g) + 2 H2O (g) Rate = 0.0339 Ms-1 = k (.210 M)2(.122 M) k = 6.30 M-2 s-1 The units of k depend on the order of the reaction 15.3 Alternatively, all the data may be plotted to get a better estimate, which averages out experimental error. 15.3 Alternatively, all the data may be plotted to get a better estimate, which averages out experimental error.

    13. 13 The rate constant The value of k depends on the nature of the reactants and on the temperature. Arrhenius found that the temperature dependence could be expressed as k = Ae-Ea / RT The preexponential factor A, and the activation energy, Ea, are relatively independent of temperature. What are these parameters? Why does k have this dependence? Preview of 15.5 Ea is almost always a positive number, and R is the gas constant. Therefore increasing the temperature always increases the rate of a reaction. Of course k is always a positive number, it refers to the reaction proceeding in the forward direction only, so A is always a positive number.Preview of 15.5 Ea is almost always a positive number, and R is the gas constant. Therefore increasing the temperature always increases the rate of a reaction. Of course k is always a positive number, it refers to the reaction proceeding in the forward direction only, so A is always a positive number.

    14. 14 Consider the combination reaction of NO and O2 to produce NO2 : 2 NO(g) + O2(g) ? 2 NO2 (g) Determination of the Rate Law (via Methods of Initial Rates) Initial Concentrations (mol/ L) Initial Rate Experiment [NO] [O2] (mol/L s) 1 0.020 0.010 0.028 2 0.020 0.020 0.057 3 0.020 0.040 0.114 4 0.040 0.020 0.227 5 0.010 0.020 0.014 Based on these data, what is the rate equation? What is the value of the rate constant k? Rate Law Determination

    15. 15 Rate Law; Solving for rate Constant

    16. 16 Microscopic view In order to understand our macroscopic observations about temperature and concentration dependence, we should look at the reaction microscopically - on the size scale of atoms and molecules. The rates of chemical reactions are explained by collision theory, which is based on kinetic theory. Collision theory views a reaction as the result of a successful collision between two or more reactants and/or catalysts. A few reactions occur without any collision. 15.515.5

    17. 17 Collision Theory The number of collisions between two or more species is proportional to the product of their concentrations. When the reaction is the result of a single collision an elementary step then the concentration dependence is directly related to the stoichiometry of that collision. The probability that A will collide with B is proportional to [A][B]. The probability that A will collide with A is proportional to [A]2 For more complicated processes, the rate law is some combination of these elementary steps. In order to react, the molecules must collide in a favourable orientation and with sufficient energy. These factors are accounted for in the rate constant. 15.5 We will look at relating complex reaction processes back to elementary steps a little later on. The orientation problem is related to entropy, and is accounted for by the preexponential factor. The energy barrier is the activation energy (enthalpy) in the exponential factor of the Arrhenius expression.15.5 We will look at relating complex reaction processes back to elementary steps a little later on. The orientation problem is related to entropy, and is accounted for by the preexponential factor. The energy barrier is the activation energy (enthalpy) in the exponential factor of the Arrhenius expression.

    18. 18 Molecularity of elementary steps For an elementary step (arising from one collision), the rate law depends on the stoichiometry of the collision. A step involving only one molecule is called unimolecular. Rate = k[A] A step involving two molecules is called bimolecular. Rate = k[A][B], or Rate = k[A]2 A step involving three molecules is called termolecular. Rate = k[A][B][C], etc. Very few elementary steps involve more than 3 molecules. 15.6 (a look ahead)15.6 (a look ahead)

    19. 19 Reaction progress & Ea For an elementary process we can plot the potential (chemical) energy of the molecules as they approach each other, collide, react and move apart. For an elementary process which involves only one molecule, we can plot the potential energy as some internal coordinate, such as bond length or angle, changes. This plot is sometimes called a reaction coordinate diagram, or an energy plot. There is typically a maximum near the collision. Molecules move along this reaction coordinate with some initial kinetic energy. K.E. is converted to P. E. to overcome the energy barrier, the activation energy. Those molecular collisions starting with enough kinetic energy can overcome the barrier and react. 15.5 See Figure 15.11 Reaction coordinate for the conversion of 2-butene from the cis form to the trans form. Collisions or vibrations with insufficient initial energy will not be able to reach the maximum in the potential energy barrier. These will collide without reacting, or return to their original configurations. Molecules which collide with sufficient energy, but in the wrong orientation are unlikely to react.15.5 See Figure 15.11 Reaction coordinate for the conversion of 2-butene from the cis form to the trans form. Collisions or vibrations with insufficient initial energy will not be able to reach the maximum in the potential energy barrier. These will collide without reacting, or return to their original configurations. Molecules which collide with sufficient energy, but in the wrong orientation are unlikely to react.

    20. 20 Arrhenius and Boltzmann We saw in chapter 13 that only a certain proportion of molecules had enough energy to remain in the gas phase. The same type of energy distribution is at play here. The Boltzmann distribution tells us that at any particular temperature a certain percentage of the molecules are above some energy cut-off. The cut-off of interest in this case is the activation energy. The percentage of molecules with energy above Ea is related to the factor exp(-Ea/RT) in the Arrhenius expression in the rate. As the temperature increases, so does the percentage of molecules above the cut-off. 15.5 See figure 13.17 and/or 15.1215.5 See figure 13.17 and/or 15.12

    21. 21 Calculations with Ea & T k = Ae-Ea / RT ln k = ln A (Ea/RT) Increasing the temperature from 300 K to 310 K increases the rate by a factor of 2. What is the activation energy? Given a set of T and k data, a plot of ln k vs. 1/T has a slope of -Ea/R 15.5 To be done in class. Since these are typical reaction temperatures and a typical activation energy, it is a good rule of thumb that a 10 degree increase in temperature leads to a doubling of the rate. You will use the graphical method of determining Ea in Lab 8.15.5 To be done in class. Since these are typical reaction temperatures and a typical activation energy, it is a good rule of thumb that a 10 degree increase in temperature leads to a doubling of the rate. You will use the graphical method of determining Ea in Lab 8.

    22. 22 Rate determining step When the reaction is a series of elementary steps, rather than a single step, the rate of reaction is determined by the slowest step, which is typically the step with the highest activation barrier. This step is called the rate determining step, and the rate law for a known mechanism can be written in terms of the rate for this step. If the rate determining step is not the first step, the rate may depend on some species which do not appear as reactants in the overall reaction equation. 15.615.6

    23. 23 Reaction mechanism Chemists often study reaction rates in order to deduce or confirm a reaction mechanism the stepwise progress of the reaction. A proposed mechanism is written as a sum of elementary steps, which may be reversible. If the rate law derived from the proposed mechanism matched the observed rate law, then we are more confident in our proposal, but still unsure. If the rate laws do not match, we must come up with a different proposal. 15.615.6

    24. 24 2 NO (g) + Br2 (g) ? 2 BrNO (g) Step 1 Rate = k1[Br2][NO] Br2 (g) + NO (g) ? NOBr2 (g) Step 2 Rate 2 = k2[Br2][NOBr2] NOBr2 (g) + NO (g) ? 2 BrNO (g) NOBr2 is an intermediate it is formed and then used up. The overall rate will depend on which step is rate determining, and on whether either step is reversible. 15.6 Note the error in the original notes: reactant 2 in step 2 should be NO, not Br2 !! This is a good reminder that the proposed mechanism must add up to the correct overall reaction stoichiometry, and each step must be balanced.15.6 Note the error in the original notes: reactant 2 in step 2 should be NO, not Br2 !! This is a good reminder that the proposed mechanism must add up to the correct overall reaction stoichiometry, and each step must be balanced.

    25. 25 2 NO2 (g) + F2 (g) ? 2 FNO2 (g) Step 1 rate = k1[NO2][F2] NO2 + F2 ? FNO2 + F slow Step 2 rate = k2[NO2][F] NO2 + F ? FNO2 fast Overall rate = k1[NO2][F2] Rate of reaction = rate of the slowest step k2 >> k1 15.615.6

    26. 26 2 NO (g) + O2 (g) ? 2 NO2 (g) Step 1 is reversible K1 = [NO3] / [NO][O2] NO + O2 NO3 fast equilibrium Step 2 rate = k2[NO3][NO] NO3 + NO ? 2 NO2 slow Overall rate = k2 [NO3][NO] Rate of reaction = rate of the slowest step, but NO is an intermediate difficult to determine its concentration. Want to replace [NO] with known quantities: K1 = [NO3] / [NO][O2] [NO3] = K1 [NO] [O2] Rate = k2(K1 [NO] [O2]) [NO] = k [NO]2[O2] Extension of 15.6 to reaction mechanisms involving fast equilibria not covered in K&T k is the observed, or apparent, rate constant. In this case it equals k2 K1 Note the correction to original notes: I solved for the wrong species in the step 1 equilibrium expression, so replaced the wrong reactant in step 2 rate law !! Extension of 15.6 to reaction mechanisms involving fast equilibria not covered in K&T k is the observed, or apparent, rate constant. In this case it equals k2 K1 Note the correction to original notes: I solved for the wrong species in the step 1 equilibrium expression, so replaced the wrong reactant in step 2 rate law !!

    27. 27 Equilibria in reaction mechanisms Note that this topic is not covered in Kotz and Treichel In principle all reaction are reversible, but only some are reversible on a time scale relevant to the overall process. A reaction, or step, which is fast in both the forward and reverse direction will come to equilibrium rapidly. Dynamic equilibrium is reached when the rate of the forward reaction equals the rate of the reverse reaction. For an elementary step 2A B + C, at equilibrium rate forward = k1[A]2 = k-1[B][C] = rate reverse = equilibrium constant. More extension to 15.6 Note that the rate of forward and reverse can only be written in this way for elementary steps. Note the k-1 is the notation commonly used for the rate constant of the reverse of step 1. Note addition of missing equals signs in last equation!!More extension to 15.6 Note that the rate of forward and reverse can only be written in this way for elementary steps. Note the k-1 is the notation commonly used for the rate constant of the reverse of step 1. Note addition of missing equals signs in last equation!!

More Related