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Reaction Mechanism - stoichiometry

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A reaction occurs with the following stoichiometry

A + P -> AP

The concentration of A was measured versus time after mixing, the data are the following:

[A](nM) Time (s)

50 0
40 100
30 229
20 411
10 721

(a) There is no dependence of rate on P. What is the order of the reaction? Calculate the rate constant and give the units.

This part, I understand. Once the graph is plotted [A](M) vs. time(s). I can calculate the rate by taking the the slope at an instantaneous point. This reaction based on the graph is first order.

(b) Propose a two step mechanism that is consistent with the kinetics and stoichiometry of the reaction. Be sure to lable your steps descriptively, using words such as : slow, fast, and/or equilibrium. Show how your mechanism is consistent with the measured rate law. Relate the measured k to the k's in your mechanism.

(c) The data given above were measured at 10C. When the reaction was studied at 10C, that rate constant doubled. Calculate the activation energy in KJ/mol.

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Solution Preview

(b)Reaction mechanism:

A -> A* (Slow)
A* + P -> AP (Fast)

Since the first reaction is the rate ...

Solution Summary

The solution is short answers with equations.

See Also This Related BrainMass Solution

Rate Law mechanism observed

The rate law for the Br—- catalyzed reaction
C6H5NH2 + HNO2 + H+ C6H5N2+ + 2H2O

is observed to be rate = = k [H+][HNO2][Br -].

A proposed mechanism is
(1) H+ + HNO2 H2NO2+ (rapid equilibrium)

(2) H2NO2+ + Br- NOBr + H2O (slow)

(3) NOBr + C6H5NH2 C6H5N2+ + H2O + Br- (fast)

Deduce the rate law for this mechanism and relate the observed rate constant k to the rate constants appearing in the assumed mechanism.

1. Step (1) is a rapid equilibrium, meaning k1 and k-1 are much faster than k2.
2. The forward and reverse rates are equal for step (1) since it is an equilibrium; this fact can
be used to get an expression for the concentration of the intermediate H2NO2+.
3. Recall, the reaction orders for elementary reactions are determined by the reaction
4. Step (2) is sufficiently slow that it can be considered rate-determining.

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