ReactionmechanismsActivity

ReactionmechanismsActivity - 351 ChemActivity 60 Reaction...

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Unformatted text preview: 351 ChemActivity 60 Reaction Mechanisms (I!) Chemical reactions occur on the molecular level by a sequence of one or more steps known as a mechanism. Every step is either a unimolecular or a bimolecular event. Typically, a single-headed arrow (w—I») is used to indicate a unimolecular or bimolecular event. - A Unirnolecular Step. A molecule undergoes a decomposition whereby a bond breaks and the molecule becomes two molecular fragments. The rate of a unimolecular step depends on the number of molecules present or, more simply, the concentration of the species. Two examples of unimolecular events are shown below: Br2(g) —> Br(g) + Br(g) rate = k’ (Brg) (CH3)3CBr(aq) —> (CH3)3C+(aq) + Br—(aq) rate=k" ((CH3)3CB1') - A Bimolecular Step. Two molecules collide, one or more bonds are broken, new bonds may or may not form. The rate of the step depends on the number of collisions between the two molecules (remember that not all collisions lead to a reaction). The number of collisions depends on the product of the concentrations of the two species. Three examples of bimolecular events are shown below: rate =15" (CH3Cl) (1e) rate = 15'" ((31131) (Cl‘) rate = k‘"" (Brg) (H) CH3Cl(aq) '+ I—(aq) —> CH3I(aq) + Cl—(aq) CH3I(aq) + c1—(aq) —a-» CH3Cl(aq) + 1-(aq) Br2(s) 4“ Me) —> HBr(s) + Bug) The "1:" for a bimolecular or unimolecular step is called a specific rate constant; the value of a specific rate constant depends on the molecular event specified and the temperature. Note that the first two bimolecular events above are simply the reverse of each other. In this case, 16'" I 5.2 x 10—7/Ms and k"" = 1.5 x 10‘11/Ms. sane—7 Ms 1.5 x10—11 ' Ms forward rate = reverse rate = (CH3I) (CIT) These steps can be combined using a double arrow, and the specific rate constants can be used to calculate an equilibrium constant for the steps: CH3Cl(aq) + I—(aq) 2 CH31(aq) + Clt(aq) K: 3.5 x 104 352 ChemActtvity 60 Reaction Mechanisms (ll) Every step in a mechanism is reversible. Often, however, the reverse reaction is negligible and not explicitly included in the mechanism. The rate law for a proposed mechanism is determined hy the sequence of steps that comprise the mechanism. HoWever, the rate law is often dominated by the slowest step, called the rate-limiting step. Model 1: A Proposed Three-Step Mechanism for a Chemical Reaction. Overall reaction: (CH3)3CBr(aq) + OHfaq) 2 (CH3)3COH(aq) + Br‘(aq) It is found experimentally that when the initial concentration of (CH3)3CBr is doubled (keeping the initial hydroxide concentration constant) the rate of the reaction doubles. Furthermore, it is found that when the initial concentration of hydroxide is doubled (keeping the initiai concentration of (CH3)3CBr constant) the rate of the reaction remains the same. Experimental rate law: rate = kexp ((CH3)3CBr) (OH‘)0 = kexp ((CH3)3CBr) Proposed Mechanism: Step Molecular Event Rate of Forward Step Relative Rate 1 (CH3)3CBr(aq) 2 (CH3)3C+(aq) + snag) k1((CH3)3CBr) fofgxd fast 2 (Climates + H20 4:” (CH3)3COH2+(a€3J k2((CH3)3C+)(H20) equilibrium fast 3 (CH3)3congt(aq)+0Hr(aq) «at—m” (CH3)3COH(aq)+H20 k3((CH3)3COH2+)(0H*) 6 Mining] Rate law for proposed mechanism :5 rate for step 1 = k} ((CH3)3CBr) Critical Thinking Questions 1. How do we know that the rate law for the overall reaction does not depend on (OH)? 2. For this proposed mechanism, which forward steps are unimolecular and which forward steps are bimolecular? ChemActivity 60 Reaction Meehanismsfll) 353 examples given, how is the rate of a fowvard step determined from the molecular 1 g 3. Based on the information in Model 1 and the unimoleeular and bimolecular event? l 4. Given the stoichiometry of the reaction, why isn't the rate law as follows? rate = kexperimental ((CH3)3CBT) (Off) 5. Show that the sum of the three steps in the mechanism gives the stoichiometry of it the overall reaction. 7 . Is the rate law for the proposed mechanism consistent with the experimentai rate law? If not, why not? 354 ChemActivity 60 Reaction Mechanisms (Ii) Model 2: A Proposed Two-“Step Mechanism for a Chemical Reaction. Overall reaction: 2 N0(g) + 02(g) Z 2 N02(g) Experimental rate law: rate = irexp (NO)2 (02) Proposed Mechanism: Step Molecular Event Rate of Forward Step Relative l N0(g) + N0(g) 4:” N202(g) 1:1 (NO)2 1:51;: equilibrium 2 N202(g) + 02(g) Z 2 N02(g) kg (N202) (02) slow forward Rate law for the proposed mechanism 3 rate for step 2 = [Q (N202) (02) Note that the species N202 does not appear in the overall reaction. Normally, it is quite difficult to measure the concentration of a reactive species that is not one of the reactants or productsécalled an intermediate species. For this reason, intermediate species are not normally included in rate laws. In this mechanism, note that step 1 is fast, contains N202 and NO (3 reactant), and is at equilibrium. Step 1 should remain in an equilibrium state as step 2 slowly consumes 02. Thus, x. Step 1 : NO + NO 47—" N202 fast equilibrium rate of forward step = rate of reverse step 1C1 (N0? = 1L1 (N202) k ' - Thus, (N202) = tic—1(NO)2 = K (NO)2 , where K is the equilibrium constant for step 1. —1 Rate law for the proposed mechanism 2 rate for step 2 2 k2 (N202) (02) z 7‘72 K (NO)2 (02) “1 kexp (NO)2 (02) Note that it is not necessary to know the values of kg or K to show that this mechanism leads to a rate law that is consistent with experimental data. However, to prove that this mechanism is the correct mechanism, it might be necessary to experimentally determine [(2 and K and to show that k2 x K = kexp . ChemActivity 60 Reaction Mechanismsfll) 355 Critical Thinking Questions 8. For this mechanism, which forward steps are unimolecular and which forward steps are bimolecular? 9. Is the "rate of forward step" given for each step consistent with your answer to CTQ 3? 10. Why is the rate law for the proposed mechanism approximately equal to the rate of forward step 2? 11. Is the rate law for the proposed mechanism consistent with the experimental rate \ law for the overall reaction? 12. Show that the sum of the two steps in the mechanism gives the stoichiometry of the overall reaction. 356 ChemActivity 60 Reaction Mechanisms (i!) Exercises 1. Indicate the molecularity (unimclecular or bimolecular) of each of the following steps. Give the rate for each step (the first process is shown as an example): 03(g) —» 02(g) + 0(g) unimolecular rate:k' (03) 0NBr(g) + 0NBr(g) —* N0(g) + N0(g) + Br2(g) N202(g) —’* N0(g) + N0(g) N069) + N0(g) —> N2(s) + 02(g) Kg) + H2(g) —> We) + He) 2. The following reaction is first order with respect to both NO and F2:1 2 N02(g) + F2(g) Z 2 N02F(g) This rate law is consistent with which of the following mechanisms? a) N02 + F2 :5 N02F + F fast N02 + F 2 N02F slow b) N02 + F2 :1“ NOgF i- F slow N02 + F 4:” NOZF fast \ 0) F2 2 F + F slow 2N0; + 2F 2 2N02F fast Add the molecular species for the two steps in each of the mechanisms. How is this sum related to the stoichiometry of the overall reaction? 3. J. N. Spencer, G. M. Bodner, and L. H. Rickard, Chemistry: Structure dis Dynamics, ’ Fourth Edition, John Wiley & Sons, 2009. Chapter 14: Problems: 37-39. 1.? . N. Spencer, G. M. Bodner, and H. Rickard, Chemistry: Structure & Dynamics, Fourth Edition, John Wiley & Sons,.2009, Chapter 14, Problem 35. ...
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