Reading Notes
for Chapter 12
These are Dr. Bodwin's reading notes for Chapter 12 of "Chemistry
2e" from OpenStax.
I am using a local .pdf copy that was downloaded in May 2020.
Chapter
Summary:
Kinetics is the study of rates and mechanisms. Rates are how fast a reaction happens, mechanisms
are how a reaction happens.
Start
with Theory:
Different books and sources have different approaches when it comes to
presenting theory vs application. I lean toward a "theory first"
approach, especially when looking at kinetics, because starting with
"why" can often give context to a bunch of the "what" topics. If we
want to study kinetics, the first question that we really have to ask
is, "Why do chemical reactions occur?"
Collision
Theory (Section 12.5):
The name kind of gives it away, but chemical reactions occur because
collisions occur. For the generic reaction:
A + B --> C
It is not possible to form C if A does not interact with B, and we call
that interaction a collision. To make this easier to visualize, let's
start with a gas-phase reaction where there is just one A particle and
one B particle held inside of a container. A and B will bounce around
the container (look back at the Gas chapter to review some of this
info) until they hit one another. Does that mean that C will form?
Maybe. There are a couple specific condition we have to kepe in mind:
- Collisions must occur
- The collisions must be sufficiently energetic to allow reaction
- The colliding particles must be oriented in such a manner that
the desired reaction can occur
Keeping those points in mind will help us understand a LOT of kinetics!
Collision Theory (and kinetics) is all about probability. Let's take a
closer look at each point:
1. Collisions must occur
This is the basic part. A and B have to interact. How can we affect how
often they interact? Or stated differently, how do we increase the
probability of a collision occuring?
- What if the container is smaller? A and B will have less space to
move, so they will collide more often. On the macroscale, "make the
container smaller" is the same as "increase the pressure".
- What if we increase the temperature? A and B will move faster,
which means that for a given amount of time, A and B will collide more
often.
- What if we put another A and another B in the container? We
should get more collisions if there are more A and B bouncing around.
On the macroscale, this means increasing the concentration. {Side Note:
"make the container smaller" also "increases the concentration".}
2. Collisions must be energetic
Just because 2 people happen to bump shoulders in a crowd doesn't mean
they will react to one another. Similarly, A & B have to collide
with sufficient energy to (momentarily) mash their electron clouds
togetherto allow an electron rearrangement to occur. {This "electron
rearrangement" is a chemical reaction!} Again, this is a probability
step... there are a LOT of collision, but only a certain percentage of
them are sufficiently energetic to cause reaction. If we increase the
temperature, the reactant particles will move faster, and the faster
they move, the more energetic their collisions will be. Similarly, if
we increase the number of A
& B particles, there is a higher probability that a collision will
be sufficiently energetic to react.
3. Collisions must be oriented
Have you ever tried to connect a trailer to a vehicle? There's only one
way to orient the trailer and the vehicle that can lead to a successful
coupling. Pulling your truck up to the side of the boat just doesn't work.
Similarly with chemical reactions, the colliding particles have to be
oriented in such a way that reaction is possible. If a sodium atom is
reacting with a chlorine atom, there's probably not much of an
orientation effect because both are (mostly) spherically symmetric, but
most reactions have a strongly preferred orientation. Again, it's a
probability thing... if the "front" of A has to collide with the "side"
of B, then there are a bunch of collisions that will not be productive.
Increase the number of collisions, increase the number of properly
oriented collisions.
For the sake of argument, let' say that A & B collide 10% of the
time. Of those collisions, let's say that 30% are sufficiently
energetic to cause reaction. That's (0.1*0.3)=0.03=3% of the time that
collisions occur and are sufficiently energetic. Now let's further say
that 40% of the collisions are correctly oriented to cause reaction.
That's (0.1*0.3*0.4)=0.012=1.2% of the time that collisions occur and
are sufficiently energetic and oriented for reaction to occur.
Reaction
Rates:
What is a "rate"? The most general description is that a rate is a
change in some measurable property over a change in time. There are a
bunch of "measurable properties" that we could measure, but for
chemical kinetics we usually measure the change in concentration of
reactants and products.
One of the biggest benefits of a balanced chemical equation is that
gives us a relationship between reactants and products. If it takes
2.83minutes to react 1 mole of A, then it also takes 2.83minutes to react 1
mole of B, and 2.83minutes to form
1 mole of C! This gives us a HUGE advantage because it is often
difficut to measure the concentration of a chemical substance during a
reaction, but as long as we can observe and measure any one reactant or
product, we can know the rate of consumption or formation of all the reactants and products.
Average
vs Instantaneous Rates:
This may sound a little like calculus, but don't worry, it's not!
{Well, it is, but don't let
that get you all twisted up in knots...}
An "average rate" is the rate over a "long" period of time. If you
change the period of time, the rate is different. In some cases,
average rates work great, but they all have the challenge that changing
the time period changes the rate.
An "instantaneous rate" is the rate at a single moment in time. For the
calculus folks, it's the tangent line to the concentration curve...
also known as the derivative. But you don't really have to know
derivatives to understand instantaneoud rates. Instantaneous rates also change at every moment in
time, but they give a nice snapshot of what is happening as the
reaction occurs.
SIDE NOTE: There's a magic negative sign that appears and disappears in
reaction rates. Why is that? Is it truly magic? Nope. By its nature, a
"rate" of a chemical reaction is always thought of a positive. Or,
perhaps more specifically, a chemical reaction rate is a magnitude of change over time, not
a direction of change over
time. When we look at a "delta function", the formal definition of that
is "final condition minus initial condition". For a reactant, the final
concentration is smaller than
the initial condition, so the "final minus initial" is negative. But
rates can't be negative, so we add a negative sign. If we're looking at
a product, the final
concentration is larger than
the initial concentration, so "final minus initial" is positive, so we
don't need the negative sign.
Rate
Laws:
Look at Figure 12.5 in your textbook. This is a ncie example of a
"concentration vs. time" graph for a chemical reaction. At any point on
those curves, we can determine an instantaneous rate. That's a lot of
rates. But there is one unique point (or rate) on those curves... every
point in those curves has another point to the left of it and the right
of it except... the initial point. As long as we know where to start,
we can calculate the curvature of the concentration vs. time graph, but
that initial point is crucial.
To start our discussion of rate laws, let's go to an even simpler
reaction:
D --> E
That's it. One reactant forming one product. It doesn't get much
simpler. If I have a single D particle in a bottle, it will bounce
around until it has the right collision with the wall of the container
for a reaction to occur. The rate of this reaction is directly related
to the size of the container; or another way to say that is it's
directly proportional to the concentration of D.
Raterxn ∝ [D]i
Proportionalities are nice in concept, but they're not so useful in
math... we can change that to an equality with a proportionality
constant, let's call it "k"
Rateinitial = k[D]initial
That's a rate law
expression. A rate law expression describes the initial rate of
a reaction in terms of its reactants raised to a power that is
determined by the collisions that occur in the reaction. For any
reaction, we can write a rate law expression if we know the reactants
involved, although to start with there will be a LOT of unknown
variables. For our favorite test reaction, A + B --> C, the rate law
expression is:
Rateinitial = k[A]ix[B]iy
For our purposes in this class, "x" and "y" will only have the values
of 0, 1, or 2, and these values must be experimentally determined. Let
me repeat that: MUST
BE EXPERIMENTALLY DETERMINED. These exponents are called orders
of the reaction, and indicate the number of each particle that is
involved in the collision that determines the rate of the reaction.
Hold on! What?!?!
OK, let's go back to collisions for a moment and talk about probability
again. If a particle has to collide with a wall, that seems like it
might not be too hard, but what if A has to collide with B? That seems
like it would have a smaller probability than if A just had to collide
with a wall. And don't forget, the collision has to occur, AND it has
to be energetic, AND it has to be oriented. What it we propose that three
particles have to collide to make a reaction occur? That seems less
likely. And f those 3 particles have to collide, AND the collision has
to be energetic, AND the collision has to be oriented, then it seems
VERY unlikely that it will occur... the probability is likely so low
that we can say it "never" happens. SO the orders of the reaction tell us how
many of each particle are involved in the collision that determines the
rate of the reaction.
Reaction
Mechanisms:
What if we have a reaction such as:
2 F + 3 G + 2 H ---> products
{Because rate law expressions only deal with reactants, it's not really
that important to include specific products here.}
From our collision discussion, we know
that these particles can't all react by a single collision... that
would require SEVEN particles to collide, AND be energetic, AND be
properly oriented... that's not going to happen. So there must be a
series of collisions involving fewer particles that add up to the
overall reaction. In fact, that is the definition of a reaction
mechanism... A reaction mechanism is a series of elementary reactions (unimolecular
and/or bimolecular collisional steps) that:
- When added together yield the overall chemical reaction
- Is consistent with the observed rate law expression for the
reaction
So for the above reaction, the rate law expression is:
Rateinitial = k[F]ix[G]iy[H]iz
The values of x, y, and z must be determined experimentally. For every
reaction, there is a rate
limiting step or a rate
determining step that ultimately defines the rate law
expression. I'll use both terms when I'm talking about this, but I will
use "RDS" as shorthand when I'm identifying a Rate Determining Step. Think of the RDS like
a construction zone on a highway... Traffic might be moving at 60mph before the construction zone and
70mph after the construction
zone, but the total time of your trip is going to be determined by the
construction zone where traffic is moving at 5mph. Or if you start a
cooking-making business and your oven can only hold 1 pan of cookies at
a time... you might be able to mix up the dough pretty quickly, and
package the cookies quickly, but the baking and cooling step will slow
down your overall process - it's the RDS. That's why the order of a
reaction can be zero with respect to some reactants... if those
reactants are not part of the RDS, they may not have any impact on the
rate!
Determining
Rate Laws:
The easiest way to experimentally determine the orders in a rate law is
by the method of multiple trials under different concentration
conditions. Process takes practice, so make sure you take advantage of
any opportunities. In your textbook, Example 12.5 walks through a
relatively simply example. A couple notes about solving these problems:
- I already said that for our purposes the orders will either be 0,
1, or 2, so if you're doing a "textbook" problem you might be able to
determine the rate without using logarithms. For example, if you get to
something like 3x=9, I don't think you need a logarithm to
see that x must equal 2.
- Use logarithms if you're assessing "real" data, or data that's
not textbook-perfect. Even though we know that the orders are supposed
to be 0, 1, or 2, if we compare a couple experiments and find that one
has a logarithm-calculated order of 1.98 and the other experiment has a
logarithm-calculated order of 1.83, we can be pretty confident that the
first experiment's data is a little more reliable.
- Don't bother memorizing rules for "correct" units of k. The units
of k need to be whatever units make sense in the problem.
- Speaking of k, you always
need to include a value (with correct units) for k. If these are
missing, your answer isn't complete.
- Remember, any number raised to the zero power is equal to 1. It
doesn't matter what the number is, x0=1.
Don't take shortcuts until you're confident in the "long way" to do the
problem. After you set up a few dozen of these problems, you'll start
to get comfortable enough to estimate the orders, afterall, they are
limited to only 0, 1, or 2. Qualitatively, if you see that the rate
triples when the concentration of A triples (and no other
concentrations or conditions change), then you know that the reaction
must be 1st order with respect to A. But don't try to use the shortcut
without setting up a bunch of these problems mathematically first.
NOTE: There's nothing magical about the number 2. Many textbooks will
systematically change concentrations by multiplying or dividing by 2.
They do this because they are lazy and it's easy. The exact same rules
and process apply for factors of 3 or 4 or 1.25 or 3.374.
Integrated
Rate Laws (IRLs):
Your brain should be tingling here... yout textbook shows a chemist
doing a bunch of math to a perfectly lovely expression... why? Why do
chemists do "extra" math, especially extra calulus math? Because they're
trying to force something to be a line because lines are easy to
understand.
"Regular" rate law expressions are packed full of information, but one thing they are not
good at is predicting concentrations as time passes because there is no
time component in the rate law expression. {Yes, time is part of the
rate, but that's difficult to use in a regular rate law expression...}
By doing a little calculus, we can get an explicit time factor in the
equation. That's what an integrated rate law expression is good for.
You need to know the order of the reaction with respect to the the
reactant of interest so you know which integrated rate law expression
to use.
Using IRLs, you can also determine the orders with respect to a given reactant. Check out Example 12.7 for a good example.
Recognizing
Different Problem Types:
This chapter has a lot of different problem types that look just
similar enough that it can be hard to know what type of problem you're
doing. Here are some things to look at...
- If you're given an initial concentration, a final concentration,
and a "long" time period, you're probably looking at an average rate
type of problem.
- If you're given the rate with respect to one reactant or product
and asked about a rate with respect to another reactant or product,
it's a relative rate problem("Relative Rates of Reaction", textbook page 661
- If you're given a few different experiments where the reactant
concentrations are systematically modified (often in a short data
table), you're probably doing a rate law problem.
- If you're given a bunch
of time and concentration data, you're probably expected to make an
integrated rate law plot to determine the order(s) and value of k.
- If you're given time, be suspicious that you might be looking at some version of an integrated rate law problem.
These are just a couple things to look for. the best way to recognize these is to practice.
Activation Energy:
Activation energy is the hill that we must get over when going from
reactants to products. It determines the rate of the reaction. If
there's a big hill, the reaction is slow; a small hill and the reaction
is fast. Figure 12.14 in your textbook shows this hill as part of a reaction coordinate diagram.
Arrhenius Equation - The "short" form of the Arrhenius equation, as
shown on page 686 of your textbook, is a lovely little equation, but it
is minimally useful in that form. It's much more useful in its linear
(p688 of your textbook) or comparative form.
Catalysts:
Catalysts increase reaction rates but are not consumed by the reaction.
They do this by changing the mechanism in a such a way that the
activation energy is lower.
Homogeneous catalysts are in the same phase as the reaction. They tend
to work well because they can mix very effectively with the reactants.
Big disadvantage - they can be hard to remove from the reaction mixture.
Heterogeneous catalysts are a different phase or are immobilized on a
different phase. Steps must be taken to help them mix effectively with
the reactants, but they are (relaively) easy to remove when the
reaction is done.
In biological systems, enzymes are catalysts.
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