kine - Homework Solutions\Kinetics 1. This problem...

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Homework Solutions\Kinetics 1. This problem illustrates how the differential rate law (DRL) can be simplified when the concentration of one or more of the participating species remains effectively constant during the course of the reaction. (i) Since the initial concentration of CH 3 COCl is 0.1 M, for convenience we will assume 1 liter of solution. Thus, we have 0.1 mols of CH 3 COCl in 1000 mL of H 2 O. Since the density of H 2 O (assume that the reaction is run at room temperature) is ~1.0 g/ml, the weight of H 2 O present initially is 1000 g. No. of mols H 2 O present at time t (=0) = (1000 g)/(18 g mol –1 ) = 55.6 mols (ii) From the reaction stoichiometry, when the reaction is complete 0.1 mols CH 3 COCl will have reacted with 0.1 mols H 2 O. Final [H 2 O] = 55.5 M Thus, since H 2 O is present in great excess, [H 2 O] remains effectively constant during the course of the reaction. We will take the effective constant concentration of H 2 O to be the average of the initial and final values; i.e. [H 2 O] = 55.55 M. Therefore, under the conditions of the experiment, the DRL – d[CH 3 COCl]/dt = k[CH 3 COCl][H 2 O] (1) may be simplified to: – d[CH 3 COCl] / dt = k' [CH 3 COCl] (2) where k' = k [H 2 O] = k (55.55 M) (3) Thus, under these conditions, the overall order of the reaction has been effectively reduced from second to first -- we say that the reaction is PSEUDO FIRST-ORDER . (iii) Rearranging equation (2) we may write that: – d[CH 3 COCl]/[CH 3 COCl] = k'dt (4) Integrating equation (4) we obtain: – ln{[CH 3 COCl] t /[CH 3 COCl] o } = k't
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2 When [CH 3 COCl] t = 0.05 M, t = t 1/2 i.e. the reaction half life Then t 1/2 = (1/k') { – ln (0.05 M /0.1 M)} = {ln 2}/k' = 0.693/k' From equation (3), k' = k (55.55 M) = 1.16 x 10 –3 M –1 s –1 ) (55.55 M) = 6.44 x 10 –2 s –1 t 1/2 = {0.693 / 6.44 x 10 –2 s –1 } = 10.7 s 2 .T he experimental rate law for the reaction 2 NO + Cl 2 2 NOCl is: (1/2) d[NOCl]/dt = k [NO] 2 [Cl 2 ] (the factor of (1/2) arises from the stoichiometric coefficient of NOCl) Since step (ii) is the rate determining step (rds), we may write: Reaction Rate = (1/2) d[NOCl]/dt = k 2 [N 2 O 2 ] [Cl 2 ]( 1 ) (Recall that the rate law for an ELEMENTARY REACTION may be written down by inspection) Then, since Step(i) is a fast equilibrium step, we may use it to determine [N 2 O 2 ]. Note -- the final expression for the rate law must only contain concentrations of species appearing in the stoichiometric equation for the overall reaction. From Step (i): K = {[N 2 O 2 ] / [NO] 2 }( 2 ) Thus, [N 2 O 2 ] = K [NO] 2 (3) Then, using the result of equation (3) in equation (1) we find: (1/2) d [NOCl] / dt = K k 2 [NO] 2 [Cl 2 4 ) Equation (4) agrees with the experimental rate law if k = K k 2 . Thus the proposed mechanism is consistent with the experimental rate law.
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3 3. An examination of the mechanism reveals that Step (ii) is the rds. Reaction Rate = (1/2) d [NOCl] / dt = k' 2 [NO .
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This document was uploaded on 03/03/2012.

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kine - Homework Solutions\Kinetics 1. This problem...

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