Potential Energy Diagrams
We have studied EP diagrams when looking at enthalpy. They still apply, but extra information is added to the graph.
Following the numerical labels on the diagram, in the forward direction for the reaction,
H2 + I2 ⇄ 2 HI:
- H2 & I2 molecules are present as reactants
- the H2 & I2 molecules approach each other with high EK; as they get closer together, they slow (EK↓, EP↑)
- the transition state (the place on the diagram of highest potential energy - at this point the reactants are making the transition into products) occurs here when H2 & I2 form the activated complex (a species in which old reactant bonds are breaking and new product bonds are forming; literally, the reactants have been activated to become products)
- the 2 HI molecules move away from each other (EK↑, EP↓)
- the final EP < initial EP, thus reaction is exothermic
Reaction Mechanism
Chemical reactions usually take place as a series of 2 particle collisions.
The sequence of these collisions or elementary processes is called the reaction mechanism.
These individual steps in the mechanism add up to the
overall reaction. The mechanism is determined experimentally.
ex. HBr(g) + O2 →
HOOBr(g) slow elementary step (RDS)
HBr(g) + HOOBr(g) →
HOBr(g) + HOBr(g) fast elementary step
HBr(g) + HOBr(g) →
H2O(g) + Br2(g) fast elementary step
HBr(g) + HOBr(g) →
H2O(g) + Br2(g) fast elementary step
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4HBr(g) + O2 →
2H2O(g) + 2Br2(g) overall reaction
Rate Determining Step
Recall that we can only determine the rate law by experiment.
It is the elementary steps or more importantly, their
speeds, which determines the rate law.
The slowest elementary process determines the rate law (thus, for the above reaction,
rate = k[HBr][O2] since step 1 is the slowest elementary step and this step includes 1 HBr and 1 O2).
This is because the overall reaction rate cannot exceed the rate of the slowest elementary step or rate determining step***Looking for more info? This video is good (don't go beyond the 7:00 mark). This video (don't go beyond the 3:35 mark), too. ***
Homework #19-21
Answers to Student Questions:
1. Can you go over question #19?
(a)
To find the exponent for [A2], use trials 1 & 2 (since [A2] changes, but [B] does not):
[A2] x 2 and r x 2, so since 21= 2, the exponent is 1
To find the exponent for [B], use trials 1 & 3 (since [B] changes, but [A2] does not):
[B] x 2 and r x 1, so since 20=1, the exponent is 0
Thus, the rate law is r = k[A2]1[B]0 or more simply written, r = k[A2].
(b) Since the rate law (r = k[A2]1[B]0) and the rate determining step (RDS) are
linked, we can determine that the RDS must have 1 A2 as the
reactant, since the rate law shows [A2]1 (there will be
no B because the rate law shows [B]0). Thus, the RDS must be
A2 →
products
Since there is only
one way a diatomic molecule (A2) can react (by decomposition), the
RDS must be:
A2 → A + A
(c) Recall that all the elementary steps must add up to give the
overall reaction. The reaction mechanism
must include the RDS and at least one other step (since the RDS is the same as
the overall reaction, there must be other elementary steps as well).
A2 → A + A
?
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A2 + 2B → 2AB
Since A2 must show up in the overall reaction,
the unknown elementary step cannot have A2 as a product or it would
cancel out. Also, the overall reaction
does not have 2A, but it does have 2B and 2AB.
Creating steps that will include what we want and cancel (I have highlighted the species that get cancelled) what we don’t want
gives:
A2 → A + A
A + B → AB
A + B → AB
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A2 + 2B → 2AB
