The Circuitous Route to an Overall Reaction

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Presentation transcript:

The Circuitous Route to an Overall Reaction Reaction Mechanisms The Circuitous Route to an Overall Reaction

Elementary Steps A series of simple reactions that represent the progress of the overall reaction at the molecular level. The reaction order can be determined by inspection. Can not be proven, only supported. However, some mechanisms can be disproved.

Example 2NO(g) + Cl2(g)  2NOCl(g) * I have know idea if NOCl2 is observed as the reaction progresses. I am merely stating this so as to illustrate how a mechanism might be proposed. Any similarity between the above statement and reality is purely coincidental. It is observed that NOCl2(g) is present during the course of the reaction.*

Possible mechanism for 2NO(g) + Cl2(g)  2NOCl(g) Step 1: NO + Cl2  NOCl2(g) slow Step 2: NOCl2 + NO  2NOCl fast

The elementary steps must add up to the overall reaction. Two requirements must be satisfied for the proposed mechanism to be possible: The elementary steps must add up to the overall reaction. The rate law of the overall reaction must agree with the rate law of the “rate determining step” (the slowest step) of the proposed mechanism.

Do the elementary steps add up? Step 1: NO + Cl2  NOCl2(g) Step 2: NOCl2 + NO  2NOCl 2NO(g) + Cl2(g)  2NOCl(g) YES!!

Does the rate law of the overall reaction agree with the slow elementary step? Experiment [NO] [Cl2] Rate M/s 1 1.0 3.0 4.6 x 102 2 1.5 2.3 x 102 3 2.0 9.2 x 102 For overall reaction, rate = k[NO][Cl2]

Determine the rate law of the slow elementary step Step 1: NO + Cl2  NOCl2(g) slow The reaction step is bimolecular since there are two molecules reacting. Remember that the reaction order for an elementary step can be determined by inspection. For this elementary step, rate = k[NO][Cl2]