122ch14e - 56 14.83) The balanced chemical equation shows...

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Unformatted text preview: 56 14.83) The balanced chemical equation shows the overall, net change during a chemical reaction and only tells us what the reactants and products are and how much of each is involved. It does NOT tell us HOW the reaction occurs. Most reactions occur as a series of steps (elementary, single step, reactions). The rate law is initially written in terms of the reactants from the rate-determining step. If any intermediates are present (in a multi-step mechanism) in this initial rate law we try to “get rid” of them by algebraic solving for their concentrations in terms of reactants, products and/or catalyst. For any elementary process (single-step process) the rate law can be written directly from the balanced equation for the step (using the coefficients). This is ONLY true for an elementary process. All steps in a multistep mechanism are elementary processes. For a reaction that occurs in a single step (an elementary process), the rate law can be written directly from the balanced equation. However, you can NOT tell that a reaction is an elementary process simply by looking at it. You must either be told or have experimental evidence (such as the experimental rate law). Our experimental rate law tells us something about the mechanism. If the exponents of the substances in the rate law agree with the coefficients in the balanced equation, then we can theorize that the reaction may be an elementary process (occurs as a single step). However, it does not mean that it does, it is just one possibility. If the exponents of the substances in the rate law do NOT agree with the coefficients in the balanced equation, then we can say the reaction does NOT occur as an elementary process and must proceed as a series of steps, MUST have a multistep mechanism. 57 14.85) 58 14.85) (cont.) 59 14.86) 60 14.86) a) (cont.) 61 14.86) a) (cont.) c) Use the rate law to find the rate [HgCl2] = 0.100 M [C2O42&] = 0.25 M r = k [HgCl2] [C2O42&]2 r = (8.672 x 10&3 M-2s-1) [0.100 M] [0.25 M]2 r = 5.420 x 10&5 M/s = 5.4 x 10&5 M/s 62 14.88) The following are the half-life relationships for the reactions we’ve studied: Zero-order ------------[A]o t1/2 = ------2k 1st-order ------------0.693 t1/2 = --------k DECREASES as rx. proceeds (successive ½-lifes get smaller - ½ previous e.g. t1/2,1 = 40 s t1/2,2 = 20 s t1/2,3 = 10 s ) CONSTANT as rx. proceeds (successive ½-lifes are same e.g. t1/2,1 = 40 s t1/2,2 = 40 s t1/2,3 = 40 s ) t1/2 2nd-order ----------1 = ---------k [A]o INCREASES as rx. proceeds (successive ½-lifes get larger - double e.g. t1/2,1 = 10 s t1/2,2 = 20 s t1/2,3 = 40 s ) For a first-order reaction the half-life is constant, whereas the half-life for zeroorder and 2nd-order are NOT constant. The half-life for zero-order gets shorter as the reaction proceeds (successive half-lifes are cut in half each time). The halflife for 2nd-order gets longer as the reaction proceeds (successive half-lifes double each time). For rxn (1) the half-life is constant as the rxn proceeds. This implies the reaction is first-order. For rxn (2) the half-life gets longer as the rxn proceeds. This implies the reaction is NOT first-order. It is also NOT zero-order (half-life gets shorter). Reaction (2) must be a higher order than first-order. We can not say for sure whether it is a 2nd-order but it could be. A reaction that is 3rd order in reactant A would have the rate law, r = k [A]3. This would then have an integrated rate equation and half-life of the following forms: 1 1 ------- = k t + -------[A]t2 [A]02 3 t 1/2 = ---------k [A]02 This half-life increases as the reaction proceeds (gets longer; each successive halflife is 4 times the previous). 63 14.90) We have two radioactive particles. The decay process for radioactivity is first order. The rate constants for the radioactive decay are given. We want the halflifes for each, which one decays at a faster rate and the amount present after 3 halflives. The rate law for both is r = rate = k [substance]. Americium-241: 241 Am k = 1.6 x 10!3 yr!1 Iodine-125: 125 I k = 1.1 x 10!2 day!1 The rate law for each would be: r = (1.6 x 10!3 yr!1) [Am] r = (1.1 x 10!2 day!1) [I] a) Half-lifes: 241 Am: t1/2 = 0.693/(1.6 x 10!3 yr!1) = 433.1 yr = 4.3 x 102 yr 125 I: t1/2 = 0.693/(1.1 x 10!2 day!1) = 63.0 days = 63 days b) Which decays at a faster rate? For a given sample size, half of the 241Am decays in 433 years, whereas half of the 125 I sample decays in 63 days. It takes less time for the 125I to decay to half of its initial conc. The 125I decays at a much faster rate. *** continued next page *** 64 14.90) (cont.) c) How much of a 1.00-mg sample of each remains after 3 half-lives? There are two ways to determine this answer. 1) Method 1: really easy We simply want what remains after 3 half-lives. Remember a half-life means the time it takes to get to ½ of the original concentration. That means 50%, 25%, 12.5%, 6.25%, etc. Since we are going 3 half-lives this is 12.5% of the original. Thus, each will be 12.5% of the original 1.00 mg sample, 0.125 mg. The real difference is in how long it will take the concentration of each radioactive substance to decrease to 12.5% (0.125 mg). 241 Am: t = 3 t1/2 = 3 (433.1 yr) = 1299 yr = 1.3 x 103 yr 125 I: t = 3 t1/2 = 3 (63.0 days) = 189 days = 1.9 x 102 days 2) Method 2: use the 1st order integrated rate equation ln[A]t = !kt + ln[A]0 Want conc. after 3 half-lives (see time above). 241 Am: ln[Am]t = ! (1.6 x 10!3 yr!1)(1299 yr) + ln(1.00) = ! 2.079 + 0.000 [Am]t = e!2.079 = 0.1250 = 0.13 mg 125 I: ln[I]t = ! (1.1 x 10!2 day!1)(189 days) + ln(1.00) = ! 2.079 + 0.000 [I]t = e!2.079 = 0.1250 = 0.13 mg 65 14.94) Use the Arrhenius equation that relates k to Ea and T. k = A e&Ea/RT a) where, A = frequency factor Ea = activation energy T = temperature of Kelvin Even if Earx1 = Earx2 the k’s do NOT have to be equal. The rate constant, k, also depends on A, the frequency factor, which is related to the frequency of collisions and their effectiveness (proper orientation). So if Arx1 … Arx2 then krx1 … krx2 b) Two similar reactions have the same rate constant at 25°C, but at 35°C one of the reactions has a higher rate constant than the other. Why? From the two-point form of the Arrhenius equation one sees that the temperature dependance of the rate constant, k, depends on the activation energy, Ea. Reactions with different behaviors for k with respect to temperature must have different activation energies, Ea. The fact the two reactions have the same rate constant at 25°C is a coincidence. The reaction with the higher rate constant at 35°C has the larger activation energy. Temperature has a bigger effect on reactions with higher activation energies. Temperature increases the kinetic energy of the particles and the reaction with the larger Ea can use this increase in KE more effectively. 66 14.96) The following mechanism has been proposed for the reaction of NO with H2 to form N2O and H2O. a) Show the steps in the mechanism will add to give the overall balanced equation. NO (g) + NO (g) v N2O2 (g) (1) (Slow) N2O2 (g) + H2 (g) v N2O (g) + H2O (g) (2) -----------------------------------------------------2 NO (g) + N2O2 (g) + H2 (g) v N2O (g) + H2O (g) + N2O2 (g) Note: The N2O2 appears on both sides of the equation and will cancel out. It is an intermediate. Overall balanced eqn: 2 NO (g) + H2 (g) v N2O (g) + H2O (g) * continued on next page * 67 14.96) (cont.) 68 14.99) 69 14.99) (cont.) 70 14.102) Enzymes are often described using the following two-step mechanism: E+S k1 » ES (fast) k!1 ES v E + P ---------------------------E+Sv E+P k2 E = enzyme (catalyst) (slow) S = substrate (reactant) P = product What is the rate law? Write the initial rate law in terms of the reactants from the slow (rate-det.) step. r = k2 [ES] However, ES is in intermediate. We need to try to get rid of it using step 1. rate of forward step 1: rfor = k1 [E][S] rate of reverse step 1: rrev = k!1 [ES] Since step 1 is faster than step 2, step 1 will come to equilibrium. rate forward rxn = rate everse rxn rfor = rrev k1 [E][S] = k!1 [ES] k1 [ES] = ------ [E][S] k!1 k2 k1 r = k2 [ES] = ------- [E][S] = k [E][S] k!1 so r = k [E][S] where k = (k2 k1/k!1) 71 14.105) 72 14.107) The gas phase reaction of NO and F2 to form NOF and F, NO + F2 -----> NOF + F is believed to be bimolecular and has an activation energy Ea = 6.3 kJ/mol and a frequency factor A = 6.0 x 108 M!1s!1. a) Calculate k at 100 °C. Use the Arrhenius equation (as below or straight-line form): k = A e!Ea/RT ln k = ! Ea/(RT) + ln A 3 k = (6.0 x 108 M!1s!1) e!(6.3 x 10 J/mol)/((8.314 J/molCK)(373.15 K)) k = (6.0 x 108 M!1s!1) e!2.0307 = 7.874 x 107 M!1s!1 = 7.9 x 107 M!1s!1 b) Draw the Lewis structures for NO and NOF (ONF, N in middle) 73 14.107) (cont.) c) NOF (ONF) has 3 things around it (3 regions of electron density) so its electron domain is trigonal planar. However, it has 2 atoms and 1 lpe! on the N so its molecular shape (geometry) is bent, with a bond angle of .120°. d) Possible transition state e) Give a possible reason for the low activation energy. The NO molecule is an odd-electron system with one unpaired e!. That makes it e! deficient. It is trying to pair up that e!. It is attracted to the e! rich F2. This attraction between the molecules means they are more likely to collide than by just random collisions. Also, you aren’t breaking a bond between the N and O in order to form the NOF (ONF). This unpaired e! on the N is attracted to the e! in the bond between the F atoms. You are breaking the bond between the F atoms, which requires energy, but then forming a bond between the F atom (with its now unpaired e!) and the unpaired e! on the N, which releases energy (nearly as much as required 74 14.110) ...
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This note was uploaded on 06/11/2011 for the course CHEM 122 taught by Professor Zellmer during the Spring '07 term at Ohio State.

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