Chemical Kinetics - General Chemistry and Qualitative Analysis - Lecture Slides, Slides of Chemistry

Download Chemical Kinetics - General Chemistry and Qualitative Analysis - Lecture Slides and more Slides Chemistry in PDF only on Docsity! Chapter 12, Chemical Kinetics This chapter is about: 1. numerical descriptions of how fast rxns. occur 2. the intermediates that form during a rxn (re. mechanism) 3. applying thermodynamics & the kinetic molecular theory to go from the descriptive learning to understanding. Our focus will be on gas and liquid phase (soln.) reactions. Why care? Your health! 1 I. Reaction Rates A. Defining rate: change per unit time rate = concentration change time change B. Example: 2 N2O5 (g) ÷ 4 NO2 (g) + O2 (g) How could you measure the rate? 1. Measure [N2O5], [NO2], or [O2] at various times. 2. Method: Pressure change, color change, etc. 2 So, if the O2 formation rate was 0.80 M/s, the NO2 formation rate should be , and the N2O5 decomposition rate should be . (Remember, these rates are always positive numbers.) F. Does the rxn. rate change with time? Would you have predicted this? Logic? Try problem 12.2, p. 434. 5 II. Rate Laws & Rxn. Order (Demolition derby?) A. We are going to relate our numerical analysis of the rxn. rate to a mental picture (model) of what sort of collisions are occurring during the rxn. B. Consider the following general rxn.: a A + b B ÷ Products 1. I can write a rate law for this reaction in the form: Rate = k [A]m [B]n k: rate constant specific for this rxn., T, etc. 6 2. Why does this equation work? (Collision-theory) a) Why does a k value have to be included? b) Why [A] times [B]? (Boltzmann?) c) Why is [A] raised to the m power? (Aside: Is a sometimes equal to m [and b to n]?) C. Reaction order 1. The overall reaction order = the sum of all the exponents in the rate law. 2. The reaction order “with respect to a specific reactant” = the exponent for that reactant. Do prob. 12.3, p. 436 to illustrate. 7 A. To get m, compare exp. #1 & #2, where [O2] is constant, but [NO] changes by a factor of 2 (0.015 M vs. 0.030 M). (Do math on board!) If m = 1, the rate should double, if m = 2, the rate should quadruple, if m = 3, the rate should go up by factor of 8, etc. Logic 21 = 2, 22 = 4, 23 = 8, etc. From the data, we see that the rate quadrupled: 0.192 / 0.048 = 4, therefore, m = 2. 10 B. Following same logic, we can determine value of n. What exps. should we compare in Table 12.3 to accomplish this? Exp. # and exp. # . Note: In this case we want [NO] to be constant. (Do math on board!) C. Now that we know m = 2 and n = , do we have enough information to determine k? Solve for k in the space below:(Do math on board!) 11 Should we do Worked Key Concept Ex. 12.4 (p. 439) in class? Comment on Key Concept problems!!! IV. Integrated Rate Law: 1st-order Reaction A. For rxn, A ÿ products, that is 1st order overall: rate = !ΔA / Δt = k [A] (m = 1) Recall from your calculus: !ΔA / Δt = k [A] is ~same as:!dA / dt = k [A] which we can rearrange to get: dA / [A] = !k dt which we can integrate to get: ln [A]t /[A]0 = !k t 12 1/k C. Rxn. half-life: time required for amount of reactant to decrease to ½ of its original value. (Fig. 12.7, p. 445) D. Mathematically: log [A]t /[A]0 = !k t /2.303 log (½) = !k t½ /2.303 !0.3010 = !k t½ /2.303 So, t½ = (!0.3010 × !2.303)/ k = 0.693/ k 15 E. This means that: t½ % 1'k Surprising? (Do Prob.12.9, p 446, Key Concept Prob 12.10) VI. Radioactive Decay Rates A. Applications: 1. Medicine, particularly internal exposure. 2. Archeology, etc. 14C dating 3. Geology: How old are rocks, age of the earth? B. Try an application with 14C dating: prob. 12.13, p. 449. 16 VII. Second-order Reactions Many reactions do not fit 1st order kinetics. Of these, many do fit 2nd order kinetics. A. For rxn that is 2nd order in A: ΔA / Δt = k [A]2 Example: 2 NO2 (g) ÿ 2 NO (g) + O2 (g) B. The integrated rate law is: 1/[A]t = k t + 1/ [A]0 C. Does this look like y = m x + b? Look at Figure 12.8 & the figure on p. 452 in Ex. 12.12. 17 3. Enzymes can sometimes be viewed as surface catalysts. Assays to determine the amount of a given enzyme (ex.: the enzyme PAH for PKU diagnosis) are often set up to be zeroeth order with respect to the reactants. Would the slope change if we changed [A]0? 20 IX. Reaction Mechanisms (major shift in approach) Time to see if there is more to the analysis that we have been doing than meets the eye. A rxn mechanism describes the specific changes that occur during a chemical rxn. This stuff is massively useful in drug development! A. Terms 1. A single step in mechanism is called an elementary reaction or an elementary step. 2. The molecularity of a rxn refers to the # of separate molecules/atoms on reactant side of elementary rxn. a) unimolecular = one b) bimolecular = two c) termolecular = three (These are uncommon.) 21 B. Let’s look at a mechanism for: NO2 (g) + CO (g) ÿ NO (g) + CO2 (g) 1. Experiments have suggested a two step process: 1st step: NO2(g) + NO2(g) ÿ NO(g) + NO3(g) 2nd step: NO3(g) + CO(g) ÿ NO2(g) + CO2(g) See Fig. 12.11, p. 454 (Think about molecules colliding and/or breaking apart!) 22 X. Rate Laws for Elementary Reactions (simple) A. 1st order example, decay of O3: O3 ÿ O2 + O If this is 1st order in O3, what is the rate law? B. Second order example, rxn. of CH3Br w/ OH!: See SN2 Molecular Movies animation. Also on 4th floor computers. CH3Br + OH! ÿ CH3OH + Br! If this is 1st order in both CH3Br & OH! what is the rate law? 25 Important: When the overall rxn. occurs in a single step, then the rate law is determined by the molecular equation. Because this rxn. has a single step mechanism: CH3Br(aq) + OH!(aq) ÿ CH3OH(aq) + Br!(aq) rate = !Δ [CH3Br] / Δt = k [CH3Br] [OH!] Do prob. 12.17, p. 459. 26 XI. Rate Laws for Overall Reactions Back to the crossword puzzle analogy. The balanced molecular equation for a rxn. only gives information about stoichiometry, not necessarily about the mechanism. However, the rate law for the rate limiting step in a mechanism does define the rate law for the overall rxn. A. Introductory comments: 1. Find an every day analogy of a multistep process with a clear rate limiting step. The rate limiting step is the slowest step. (Shopping?) Ever worked on or seen an assembly line? When a rxn. occurs with more than one elementary step, usually one of the elementary steps is much slower than the others. This is the rate limiting step. 27 What is the order of the NO2-CO rxn.: Overall? With respect to NO2? With respect to CO? Does this seem logical? In other words: does it bother you that the rate law for the rxn.: NO2(g) + CO(g) ÿ NO(g) + CO2(g) has no [CO] term in it? Try Prob. 12.18, p. 463 30 C. Multi step Reactions with Initial Fast Step New! What if step 1 in the mechanism isn’t the rate limiting step? 1. Example: 2 NO(g) + 2 H2(g) ÿ N2(g) + 2 H2O(g) 2. Proposed mechanism: k1 Step 1: 2 NO W 2 N2O2 fast, reversible k!1 k2Step 2: N2O2 + H2 ÿ N2O + H2O slow, rate limiting k3Step 3: N2O + H2 ÿ N2 + H2O fast Do these steps sum to give the overall balanced equation? 31 3. Based on our previous logic, the rate law should be: rate = k2 [N2O2] [H2] This isn’t practical or useful. a) Not practical because we can’t measure [N2O2] easily. b) Not useful because we can’t control [N2O2] easily. 4. Is there a way to cope? a) Because the 1st rxn is fast, it is essentially at equilibrium. (By definition: at equilibrium, rateforward = ratereverse) b) Therefore ratef = k1[NO]2 and rater = k!1[N2O2] c) Since ratef = rater then k1[NO]2 = k!1[N2O2] d) Rearrange k1[NO]2 = k!1[N2O2] to solve for [N2O2]: 32 2. Do electrons normally repel each other? (Y or N?) 3. The intermediates in a rxn. are usually very unstable & high in potential energy. (See Fig 12.15) Important terms: a) activation energy (Ea) b) transition state (synonym: activated complex) Try to relate the process of going from reactants to the transition state to some human activity. (Mountain climbing?) 35 B. Some of the thoughts behind this part of kinetics started with analysis of collision rates in the gas phase. 1. Air, at normal T & P: each gas molecule has about 109 collisions/sec. Imagine a reaction occurring with each collision. While this does seem consistent with the rates of chemical rxns. that cause explosions, it clearly is not consistent with the rates of the majority of chemical rxns. Therefore, for most rxns., only a small percentage of the collisions lead to a reaction occurring. These productive collisions occur with appropriate energy and orientation. 36 2. The fraction of collisions with enough energy (see Fig. 12.15) to reach Ea is: f = e!Ea/RT (Units for T??) At T= 298K, a rxn. with Ea = 75 kJ/mol has f = 7 x 10!14. With a collision rate of 109 /s, how long (ave)would you have to wait to get a collision with sufficient energy? What would f be for this reaction at 373 K? Can you see why T has such a large effect on rxn rate? This takes care of part of the energy consideration mentioned before. What about orientation? 37 º » a) Recall the O (red) has to hit the H (white) b) Is the surface area of the H greater than the surface area of the Cl (green)? 40 D. We can now try to set up an equation predicting rxn. rates w/ the collision theory described here: 1. collision rate = Z [A] [B] (This was 109 above.) 2. rxn rate = p × f × collision rate = p × f × Z [A][B] Remind you of our general rate law? rate = k [A][B] Clearly, p × f × Z = k. Therefore: k = p Z e!Ea'RT We now can relate structure, velocity, energy, etc. to understanding & predicting rxn rates. Thanks Arrhenius! p Z is sometimes called A, the frequency factor. 41 XIII. Using the Arrhenius equation A. Rearrange k = A e!Ea/RT to: ln k = ! Ea/RT + lnA Can you see y = mx + b ? By measuring k a range of different T values we can get a good estimate of Ea. (Do prob. 12.21 on your own.) B. Why would anyone want to do this? 42 3-D view of inhibitor-enzyme interactions: (From PDBSum code 1hpx) The thicker tube structure represents the inhibitor (KNI-272). The thinner tubes represent those parts of the enzyme that interact directly with the inhibitor. Standard CPK color-coding for elements. grey: carbon red: oxygen blue: nitrogen yellow: sulfur What element appears to be strangely absent? 45 The active site of the enzyme is a different environment than bulk solvent. Hence, different energy pathway. 3. Catalytic antibodies = made-to-order catalysts. XIV. Catalysis (PKU & PAH? HIV protease?) A. Catalysts work by decreasing activation energy. (See Fig. 12.18) B. Enzymes (and catalysts in general) do this by creating a different environment (different from what?), which allows for a different (lower energy) reaction pathway. 46 For some enzymes, the environment and pathway is very well understood. See HIV protease. C. Do catalysts speed up the rate of the reverse rxn. too? Do Key Concept Prob. 12.22, p. 473-4 rxn: 2 A + C2 ÿ 2 AC A = red B = blue C = yellow Relatives rates: 1 1 2 3 a) What is the order of the rxn in A, B, C? b) Write the rate law. c) Write a mechanism that agrees with the rate law. d) Identify all catalysts & intermediates in mechanism. 47