HW #8 - Chemistry 132, Winter 2008 Solutions to Homework...

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Chemistry 132, Winter 2008 Solutions to Homework No. 8 Problem 1. See Levine problem 17.44. (a) Assume a rate law of the form r = k [ A ] α [ B ] β [ E ] ± (Levine eq. 17.5), where A , B , and E represent ClO - , Cl - , and OH - , respectively. Let r i denote the initial rate reported in column i = 1 ,..., 4 of the given table. Similarly, let A i , B i , and E i denote the the initial concentrations of species A , B , and E , given in column i of the table. Then, according to the initial-rate method (Levine p. 542), the exponents in the rate law can be determined by taking ratios between initial rates from appropriate paris of columns in the table r i r j = ± A i A j ² α ± B i B j ² β ± E i E j ² ± Specifically, columns 1 and 3 give r 1 r 3 = 0 . 48 0 . 24 = 2 r 1 r 3 = ± 4 . 00 2 . 00 ² α ± 2 . 00 2 . 00 ² β ± 1000 1000 ² ± = 2 α α = 1 . Similarly, columns 2 and 3 give r 2 r 3 = 0 . 50 0 . 24 2 r 2 r 3 = ± 2 . 00 2 . 00 ² α ± 4 . 00 2 . 00 ² β ± 1000 1000 ² ± = 2 β β = 1 . Lastly, columns 3 and 4 give r 3 r 4 = 0 . 24 0 . 94 1 4 r 3 r 4 = ± 2 . 00 2 . 00 ² α ± 2 . 00 2 . 00 ² β ± 1000 250 ² ± = 4 ± ± = - 1 . Therefore, the rate law is r = [ A ][ B ] / [ E ], or r = k [ClO - ][Cl - ] [OH - ] . (b) One of the concentrations in the rate law has a negative exponent and the reaction takes place in aqueous solution, so the solvent is S = H 2 O. Then, rule 3 in Levine p. 550 applies in this case and can be used to predict the overall atomic composition of the reactants present in the rate-limiting step (RLS) αA + βB + ±E + xS = ClO - + Cl - - OH - + x H 2 O = IClH 2 x - 1 O - x , where x is an integer that must be greater than zero to make the number of H atoms non-negative. Also, according to rule 3, the species OH - does not enter the RLS, but is produced in equilibria preceding the RLS. In the simplest case there is only one equilibrium step preceding the RLS and OH - could be produced in this step by deprotonation of water. Some possible realizations of this step using the available species are 2H 2 O * ² H 3 O + + OH - (1) ClO - + H 2 O * ² HOCl + OH - (2) I - + H 2 O * ² HI + OH - . (3) Also, in the simplest case x = 1 and the atomic composition of the reactants in the RLS would be IClHO - . The following elementary reactions are consistent with this atomic compo- sition HOCl + I - HOI + Cl - (4) HI + ClO - HCl + OI - (5) I - + ClO - + H 3 O + HOI + Cl - + H 2 O (6) I - + ClO - + H 3 O + HCl + OI - + H 2 O (7) The reaction chosen for the first step determines the possible reactions for the RLS. For example, if reaction (2) is assumed to take place in the first step, then reaction (4) should fol- low in the RLS. However, none of the above reactions for the RLS yields both of the overall products Cl - and OI - at once. Thus, at least another step is required to complete the mechanism. In the simplest case, there is only one addi- tional step after the RLS. The choice of the reaction for this additional step depends on the choice of the reaction for the RLS. For example, if reaction (4) was chosen for the RLS,
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This note was uploaded on 05/22/2008 for the course CHEM 132 taught by Professor Lindenberg during the Winter '08 term at UCSD.

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HW #8 - Chemistry 132, Winter 2008 Solutions to Homework...

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