(i) for a multi-step reaction, prediction of,
(i) a rate equation that is consistent with the rate-determining step
(ii) possible steps in a reaction mechanism from the rate equation and the balanced equation for the overall reaction
In a two step reaction there is a good chance that one step proceeds very slowly and the other much more quickly.
If this is the case then only the very slow step really affects the rate - it is the rate-determining step (rds).
By looking at the rate equation it is possible to suggest a mechanism.
Given the choice it is always easier to answer a question where the first step is the rds.
eg for the overall reaction: A + 2B --> C + D with a rate equation Rate = k[A][B]
We can assume that if the first step is the rds then it involves a collision between one molecule of A and one molecule of B.
We don't know what is created but the question will only ask us to suggest a mechanism.
So as long as the first step involves A + B --> and the overall equation is correct we will still gain marks.
Possible solution 1
Step 1: A + B --> E ----- rds
Step 2: E + B --> C + D Add the two steps together and cancel to get the overall reaction:
Overall: A + B + E + B --> E + C + D
A + 2B --> C + D
Possible solution 2
Step 1: A + B --> C + E ---- rds
Step 2: E + B --> D Add the two steps together and cancel to get the overall reaction:
Overall: A + B + E + B --> E + C + D
A + 2B --> C + D
eg for the overall reaction: A + 2B --> C + D with a rate equation Rate = k[B]2
We can assume that if the first step is the rds then it involves a collision between two molecules of B .
Again e don't know what is created but the question will only ask us to suggest a mechanism.
So as long as the first step involves B + B --> and the overall equation is correct we will still gain marks.
Possible solution 1
Step 1: B + B --> E
Step 2: E + A --> C + D Add the two steps together and cancel to get the overall reaction:
Overall: B + B + E + A --> E + C + D
A + 2B --> C + D
Possible solution 2
Step 1: B + B --> C + E
Step 2: E + A --> D Add the two steps together and cancel to get the overall reaction:
Overall: B + B + E + A --> E + C + D
A + 2B --> C + D
This can be more complicated as shown in the below example.
eg for the overall reaction: A + 2B --> C + D with a rate equation Rate = k[A]2[B]
It's highly unlikely to be a one step reaction involving a collision of three molecules - although that's not entirely out of the question.
If we know that two steps are involved then we can assume that the second step is the rds purely because the rate-equation is rather complicated.
Possible solution 1
We might suggest that the first step involves a collision between two molecules of B to form an intermediate, BB, that is effectively two A molecules conjoined (eg NO2 + NO2 --> N2O4)
The intermediate, BB, then collides with a molecule of A in the rds.
Step 1: B + B --> BB
Step 2: BB + A --> C + D ----- rds - The rds involves a molecule of A a "two molecules of B"
Overall: B + B + BB + A --> BB + C + D
A + 2B --> C + D
Possible solution 2
We might suggest that the first step involves a collision between a molecule of B collides with a molecule of A to form an intermediate, AB, that is effectively an A molecule and B molecule conjoined.
The intermediate, BB, then collides with a molecule of A in the rds.
Step 1: A + B --> AB
Step 2: AB + B --> C + D ----- rds - The rds involves a molecule of B a "molecule containing both A & B"
Overall: A + B + AB + B --> AB + C + D
A + 2B --> C + D
It really doesn't matter how weird, ridiculous or ungainly your suggested intermediates look - they are intermediates and so only existed for moments anyway.
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