Chemical Kinetics - General Chemistry II - Notes | CHEM 1220, Study notes of Chemistry

Download Chemical Kinetics - General Chemistry II - Notes | CHEM 1220 and more Study notes Chemistry in PDF only on Docsity! Chapter 13 - 1 CHEMISTRY 1220 NOTES CHAPTER 13 – CHEMICAL KINETICS Suppose we are interested in a chemical reaction, e.g., A + B C + D There are 2 chemical aspects that need to be considered: (1) Is the reaction favored to go? (Are the products “better” than the reactants?) Thermodynamics (if unfavored, thermodynamics might tell you how to “bend the rules”). (2) Even if it should go, will it go? Kinetics – often provides insight into how a reaction proceeds (i.e., mechanism of reaction). We will focus first on the 2nd question. So, how does kinetics give us such information? One measures rates of chemical reactions as a function of concentrations, time, and temperature (less often, pressure). Example, for the reaction: 2 N2O5 4 NO2 + O2 Chapter 13 - 2 We can measure the rate as: Δt ]OΔ[N 52      dt ]Od[N 52 (rate always +) Observe: [N2O5], M Rate of reaction slows with time. Two ways to estimate rate - average vs. instantaneous rates. time We’ll see a better way later. Note that the rate could also have been expressed as: Δt ]Δ[NO2 or Δt ]Δ[O2 (Not necessarily equal.) For the above, suppose that at t = 40 min., [N2O5] = 1.22 M; and then at t = 55 min., [N2O5] = 1.10 M: Average rate = Δt ]OΔ[N 52 = 40min.)(55 ]O[N]O[N if 5252   = min.15 M1.22)(1.10 = 0.008 min.L ONmoles 52  (at ~ 47.5 min.)  Chapter 13 - 5 Observe that when [N2O5] was doubled, so was the rate.  Rate = k[N2O5] (termed “first order”) In general, for a reaction: aA + dD  catalyst xX Rate = k[A]m[D]n[Cat]p (rate law) We will not consider.Rate constant (For reversible and/or stepwise reactions, [X] can also be involved – can get fractional or negative exponents.) m + n (+p) = overall order (mth order in A, etc.) 2nd Example: 2 NO(g) + Cl2(g) 2 NOCl(g) N Cl O 0.25 M 0.50 0.25 0.50 x 2 0.25 M 0.25 0.50 0.50 x 2 Rate, 1.4·10-6 5.7·10-6 2.9·10-6 11.4·10-6 x 4 x 2 [NOCl] t [Cl2][NO] (M/sec) Rate = k[NO]2[Cl2] (third order) (Might be interpreted as indicating that the reaction takes place by way of a collision of 2 NO and 1 Cl2 in a single step.) Chapter 13 - 6 Final Example, CH3C O OCH3 + OH - O CH3C O - + CH3OH 0.05 M 0.05 0.10 x 2 0.05 M 0.10 0.10 Rate, 0.00034 0.00069 0.00137 x 2 x 2 moles L - sec. [OH-]CH3C O OCH3 x 2 Rate = CH3C O OCH3 [OH -]k (likely single step; k = ?) Reflects dependence of reaction rate on collisions. OH- CH3C O OCH3 OH- CH3C O OCH3 1 collision 4 collisions Both concentrations doubled, rate quadrupled. Now that we have a general idea of how rate laws can be derived crudely from comparisons of rates at different concentrations, more sophisticated and detailed information can be obtained graphically. Chapter 13 - 7 INTEGRATED RATE LAWS: GRAPHICAL TREATMENT OF DATA Example: 1st Order Reactions Rate law, rate = dt d[R]- = k[R] (R for reactant) [R] d[R] = -kdt Integrate, ℓn[R]t - ℓn[R]o = -kt; ℓn[R]t = -kt + ℓn[R]o (k has time -1 units) t n[R] slope = -k n[Ro] 0 Straight line makes it easier to derive k. First order reactions (including radioactive decay) characterized by a half-life. 2nd Example: What is the half-life for , if k = 5.40·10-2hr-1 ? (cyclopropane) (propene; both C3H6) Chapter 13 - 10 Now for the second interval: t[R] 1 = o[R] 1 = kt          .500 1 .250 1 mole L = 2.00 mole L =       hrmole L1.90 t t = 1.90 2.00 hr = 1.052 hr (Note: There is no constant half-life.) Corresponding reaction rates: Initial, [CO] = 1.00 M rate - k[CO]2 =       hrmole L1.90 (1.00 M)2 = 1.90 hr-L mole When [CO] = 0.500 M, rate = k[CO]2 =       hrmole L1.90 (0.500 M)2 = 0.475 hr-L mole When [CO] = 0.25M, rate = k[CO]2 =       hrmole L1.90 (0.250 M)2 = 0.119 hr-L mole Chapter 13 - 11 2nd Example: The dimerization of butadiene, C4H6, to vinylcyclohexene has been followed by pressure measurements at 326ºC (reactants and products are gasses). Looking at the apparent half-lives of this reaction, decide whether the reaction is more likely first or second order. Then try to verify the conclusion via an appropriate plot, and also determine the rate constant. time, min 0 68.0 176.7 373 P(butadiene), mm Hg 316 158.6 89.3 41.1 * 1/P mm-1 0.00316 0.00630 0.01120 0.02433 * See it took ~ 68 min. for first “half-life.” Second “half-life” would be reached when P ~ 79 mm. However, 108.7 min. after 1st “half-life,” P = 89.3 mm, meaning the second “half- life” is much longer than the first.  this is not first order. Let’s see if we get a linear 1/P vs. t plot (expected for 2nd order). slope = k = (1/P) t  (0.02433 - 0.00316) mm-1 (373 - 0) min Chapter 13 - 12 = 5.67·10-5 mm-1 min-1 (better value via least-squares) Summary of Graphical Approaches n[R] d[R] = -kdt As we have just seen, graphical plots provide a much more effective method for determining k’s. Let’s summarize the results for 1st and 2nd order reactions (0 order relatively unimportant). First Order Second Order [R]t [R]o = -ktn n[R]t - n[R]o = -kt n[R]t = -kt + n[R]o plot n[R] vs. t, slope = -k intercept = n[R]o 1 [R]o = kt 1 [R]t - 1 [R]o = kt + 1 [R]t plot vs. t, slope = k intercept = 1 [R]o 1 [R] Since we have seen a 2nd order example above, let’s try a reaction expected to be 1st order: C6H5N2 +Cl- C6H5Cl + N2 Let’s try to confirm that it is first order, then determine k: Time, min mL(N2) [C6H5N2 +Cl-] ℓn[C6H5N2 +Cl-] 0 0 0.0700 -2.659 116 9.7 0.0587 -2.836 355 26.3 0.0393 -3.237 481 33.7 0.0307 -3.483 Chapter 13 - 15 NO + O3 NO2 + O2 Possible approaches (more than one could work): good bad worse worst ? However, even for correct approaches, there can be a substantial barrier – why? We have already seen that there can be attractive intermolecular interactions between molecules (dipole-dipole, dispersion forces, etc.) However, these occur at rather long distances compared to bond lengths. Example: C C 3.4 Å (1 Å = 10-8 cm) 3.4 Å separation (van der Waals distance) ideal for C - - C attractive intermolecular interaction. But, for a chemical bond: C C 1.54 Å So, in this case to get a normal bond to form, one must push the nonbonded atoms ~ 2 Å closer than optimal for a nonbonded interaction. In the process one could generate significant Chapter 13 - 16 repulsions between the electron clouds of the two atoms (i.e., an energy barrier is encountered). We can quantify such phenomena by use of Energy Profiles, e.g., (Reverse reaction unfavorable.) 132 kJ/mole 226 kJ/mole NO + CO2 NO2 + CO Typical diagram: Less common: 4 kJ/mole O3 + NO 200 kJ/mole (Reverse reaction again unfavorable.) O2 + NO2 Note: Very small barrier for second reaction  one would have to be careful to carry out such a reaction. Finally, C H H3C CH3 C H C H3C H CH3 C H (cis) (trans) A “reversible” reaction – barriers similar in either direction. Barriers are large, however – reaction generally very slow. 262 kJ/mole 4 kJ/mole Chapter 13 - 17 Of course, more complex behavior is possible, e.g., Intermediate species - may be observable or even isolable. Now we are in a position to understand the origin of the dependence of reaction rates on temperature: probability kinetic energy high T low T energy of activation* "Boltzmann distribution" * Suppose this is the amount of energy required to pass over the barrier (“energy of activation”) – clearly at higher T more molecules can overcome the barrier. Arrhenius (Svante) developed an effective model for the relationship between reaction rate and temperature (and proposed global Chapter 13 - 20 = 82.2 kJ/mole Alternatively, if data for only 2 temperatures are available, ℓnk2 = ℓnA - Ea RT2 ; ℓnk1 = ℓnA - Ea RT1 Subtracting, ℓnk2 - ℓnk1 = - Ea R 1 T2 - 1 T1 ℓn k2 k1 = - Ea R 1 T2 - 1 T1 (Try with above numbers.)  Now let’s ask if there could be a way of increasing reaction rates other than by increasing temperature (or [ ]). In some cases – yes. Through catalysis: Ea uncatalyzed unchangedcatalyzed Reduction of barrier with a catalyst increases rate of (forward and backward) reaction(s). Relative favorability of reactants and products unchanged. Chapter 13 - 21 Extremely important industrially. Heterogeneous: N2 + 3 H2 2 NH3 ~ 17·10 9 kg/yr in USA Haber Process (Fe2O3 + . . . . ) NH3  2O NO ( HNO3) ~ 8·10 9 kg/yr in USA (Pt) Fischer-Tropsch: CO + H2 hydrocarbons CO + H2O CO2 + H2 WGS (Water-Gas-Shift) Homogeneous – Monsanto: CH3OH + CO O CH3C OH ~ 2.1·10 9 kg/yr in USA via O O I Rh I C C (now, Ir) Ziegler-Natta catalysis – olefin (alkene) polymerization: Chapter 13 - 22 C H H H C H C C HH H H n (~ 15·109 kg/yr) ~107 C2H4 Ti Reaction Mechanisms We have seen how some simple reactions like: H2 + I2 2 HI seem to behave as expected kinetically, so that in this case, rate = k[H2][I2] suggesting a one-step process. In this case, the overall reaction might appear to be a single elementary step. In some other reactions, things are not so simple. For example: H2 + Br2 2 HBr rate = k[H2][Br2] ½ 1 + k'[HBr]/[Br2] Explainable via: Br2 2 Br elementary Br + H2 HBr + H steps H + Br2 HBr + Br Sometimes simple appearing rate laws can be deceptive, No absolute HBr or Br2 order. Chapter 13 - 25 (Seems similar to O3 + NO NO2 + O2) rate = k[NO2]2! Must involve 2NO2 ? as a rate-determining (slow) step. Perhaps: 2 NO2 NO + NO3 slow (disproportionation) NO3 + CO NO2 + CO2 fast  NO2 + CO NO + CO2 Now a more complicated example: 2 NO + O2 2 NO2 rate = k[NO]2[O2] Again, deceptive. At this stage, we could not be expected to predict actual mechanism. Evidence suggests the following: NO + O2 k1 k-1 OONO fast equilibrium NO + OONO k2 2 NO2 slow, rate determining step  2 NO + O2 2 NO2 From the second, slow step we can derive the rate equation, rate = k2[NO][OONO] (Could multiply by 2 for stoichiometry.) Chapter 13 - 26 How do we deal with this? It is a reactive intermediate, not necessarily detectable. Its presence is often inferred, not proven. Since OONO is present in equilibrium with NO and O2, we can use that relationship to substitute for [OONO] (steady- state approximation). Note: Rate of production of OONO = k1[NO][O2] Rate of reversion to NO and O2 = k-1[OONO] Because the second step is slow, these are virtually equal, i.e., k1[NO][O2] = k-1[OONO] k1 k-1 = [OONO] [NO][O2] = K or, [OONO] = K[NO][O2] So, rate = k2[NO][OONO] = k2[NO]K[NO][O2] = k2K[NO] 2[O2]  Get correct rate dependence even though there is no termolecular step. Chapter 13 - 27 Final example – recall: H2 + I2 2 HI H I I H ? rate = k[H2][I2] More recent proposed mechanism: I2 k1 k-1 2 I r.e.* K1 = k1 k-1 = [I]2 [I2] I + H2 k2 k-2 IH2 r.e.* K2 = k2 k-2 = [IH2] [I][H2] IH2 + I k3 2 HI * r.e. = rapid equilibrium  H2 + I2 2 HI Rate of formation of: HI = k3[I][IH2] (neglecting factor of 2) Need to substitute for each, using K1 and K2. Note: I readily related to I2, but IH2 is related to H2 and I, so after IH2 is removed, I will still need to be removed. Take out IH2 first: