122ch14d - 42 14.61) 43 14.62) 44 14.63) 45 14.65) 46...

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Unformatted text preview: 42 14.61) 43 14.62) 44 14.63) 45 14.65) 46 14.68) The decomposition of hydrogen peroxide is catalyzed by iodide ion and is thought to proceed by a two-step mechanism: H2O2 (aq) + I& (aq) v H2O (R) + IO& (aq) (1) IO& (aq) + H2O2 (aq) v H2O (R) + O2 (g) + I& (aq) (2) -------------------------------------------------------------------2 H2O2 (aq) + I& (aq) + IO& (aq) v 2 H2O (R) + IO& (aq) + O2 (g) + I& (aq) Note: IO& and I& appear on both sides and will cancel (see b and c below). a) Write the rate law for each elementary step in the mechanism in terms of the reactants in that step. Step 1: Step 2: b) r = k1 [H2O2] [I&] r = k2 [H2O2] [IO&] Write the overall balanced equation for the reaction. Cancel things that appear on both sides of the equation above. 2 H2O2 (aq) v 2 H2O (R) + O2 (g) c) The intermediate is IO&(aq). Intermediates generally appear for the first time as a product in one step and used as a reactant in subsequent step. (It is produced and then consumed.) I& is a catalyst. Catalysts generally appear for the first time as a reactant and reproduced in a subsequent step as a product, so it can be used again and again. (It is consumed and reproduced.) d) If the first step is the rate-determining step (slow step) then the rate law will be the same as that for step 1 (as given above). r = k [H2O2] [I&] 47 14.69) The reaction 2 NO (g) + Cl2 (g) v 2 NOCl (g) has the following experimental rate law: r = k [NO]2 [Cl2] (Exp) The following mechanism has been proposed: NO (g) + Cl2 (g) v NOCl2 (g) (1) NOCl2 (g) + NO (g) v 2 NOCl (g) (2) (Slow) -----------------------------------------------2 NO (g) + Cl2 (g) + NOCl2 (g) + NO (g) v NOCl2 (g) + 2 NOCl (g) Note: NOCl2 appears on both sides and will cancel. It as an intermediate since it appears for the first time as a product in step 1 and then used as a reactant in step 2. a) If step (1) is the rate-determining step (slow step) the rate law would be: r = k1 [NO] [Cl2] b) However, this does not give the experimentally observed rate law, which is second-order in NO. This means step (1) can NOT be the slow step and step (2) must be the slow step. Write the rate law based on the reactants in step (2). r = k2 [NOCl2] [NO] There is a problem with this rate law. First of all it does not give the exp rate law. Also, there is an intermediate in the rate law, NOCl2, which is generally not allowed (since concentrations of intermediates usually can’t be measured). * continued on next page * 48 14.69) (cont.) You must solve for the concentration of NOCl2, [NOCl2], by using the previous fast step(s). We use the steady-state approximation. Since the intermediate NOCl2 can not be used up in step (2) as fast as it is produced in step (1) an equilibrium is established in step (1). NO (g) + Cl2 (g) W NOCl2 (g) (1) (fast equilibrium step) This means the rate of the forward reaction equals the rate of the reverse reaction. rfor = rrev k1 [NO][Cl2] = k&1 [NOCl2] (k1 = rate constant of forward rxn k&1 = rate constant of reverse rxn) Solve for [NOCl2]: k1 [NOCl2] = (------) [NO][Cl2] k&1 Substitute into the rate law from step 2: r = k2 [NOCl2] [NO] k1 r = k2 { (------) [NO] [Cl2] } [NO] k&1 k1 r = {k2 (------)} [NO]2 [Cl2] k&1 r = k [NO]2 [Cl2] where (exp rate law) k2 k1 k = --------k&1 49 14.70) 50 14.70) b) (cont.) 51 14.71) 52 14.71) a) (cont.) 14.72) 53 14.74) 54 14.76) 14.78) See Figure 14.21 55 14.81) ...
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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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