﻿Here we're going to work through a kinetics problem that's very similar to one you'd see
on the free response section of the AP chemistry exam. This question is part of an AP chem mini
test you can download on my website right here to help you study for the AP chem exam.
Okay, let's get started. Part A shows three graphs and asks you to explain how the graphs suggest the
reaction above: that this one is second order with respect to NO2. Okay, many kinetics questions on
the AP exam will show you different graphs like this and ask you to correlate them with a reaction
order or a specific rate law. So if you know the pattern here, this is actually pretty easy.
These type of kinetics graphs always have time on the x-axis, right here it's in seconds, but the
y-axis is going to be a little different for each graph. Basically there are three possibilities,
okay? The first is the most straightforward—that’s what we have here—time versus the concentration
of the reactant. Generically speaking, we can think of this as A, the molar concentration of A.
Okay, the second graph plots time versus the natural log of the reactant or ln[A].
And the third graph plots time versus the inverse of the reactant, or 1/A. Now here's the thing,
only one of these graphs is going to be linear and the reaction order is determined by
which graph has a linear fit of the points that have been plotted. Okay, if the concentration
of the A graph is linear the reaction is zero order. If the natural log of a graph is linear,
we've got a first order reaction. And if the inverse of the A graph, the 1/A graph is linear,
we got a second order reaction. So that's the pattern. That's what we're looking for here. So,
there's mathematical reasons for the pattern but on most questions, AP doesn't want you to solve
equations; it just wants you to interpret the graphs. So, looking over at our three graphs here
we can see that only one has a linear fit and that is the graph of time versus the inverse of the
nitrogen dioxide concentration, NO2 here, that's the reactant that we're paying attention to. So,
this tells us that the reaction is second order. Now to correctly answer the question, right,
because we're explaining how the graphs suggest that the reaction is second order to correctly
answer the question, we want to write something like this. Okay, because the plot of time versus
1/NO2 is a linear fit, the reaction is second order with respect to NO2.
Okay, let's move on to part B. This one's pretty straightforward.
Write a rate law for the full reaction assuming it's zero order with respect to CO, or carbon
monoxide. Now here's the standard rate law equation: rate equals k, that's the rate constant,
times the concentration of each reactant raised to the exponent of its order.
For this reaction we can write the rate law like this where m and n are the order of NO2
and CO respectively. Now from part A we know that the reaction is second order with respect to NO2,
so the exponent on NO2 is 2. 2 for second order. And part B tells us right here in the question
that carbon monoxide is zero order, so its exponent is going to be zero.
Now since anything raised to the zero power is one, that means that the CO drops out
and our rate law is going to be rate equals [NO2] raised to the second power. So that's B.
Okay, part C. The initial rate of reaction is 1.4 times 10 to the negative fourth moles per
liter seconds. Using the graphs above, determine the rate constant k. To solve this, we need the
rate law that we determined in part B. And you can see that sometimes the parts of a free response
question are interrelated, okay? For this problem, part A was about reaction order, which led to part
B, where we wrote a rate law equation, and now in part C we're solving for k in that equation. So
be sure to work carefully as you work through the various steps. Okay, now we're solving for k here
so we can rearrange this equation, which gives us k equals rate divided by NO2 squared. Alright,
now let's start thinking about the values that we're going to plug in here. Okay, the problem
gives us the initial rate of the reaction that's this value here. We can plug it in right here.
Now we need the concentration of NO2. Where are we going to get that? Well this is the initial rate,
so we're going to need the initial or starting concentration of nitrogen dioxide. We can get that
from the graphs. Each one of these graphs starts at time zero, so we're talking about starting,
initial, but which one gives us the initial concentration of NO2 that we need? Okay,
well truthfully, we could get that answer from all three of them, but we'd have to do some extra
math. So why not use this one right here? For this one, the y-axis is simply the concentration
of NO2 so it's going to be the easiest to read and to work with. With no extra math required,
we can zoom in here and we'll see that when time equals zero, initial starting concentration to
match the initial rate t equals zero NO2 equals 0.020 molar. Perfect. We have this, plug it in
right there so now we have the two values we need to calculate k. Okay, when you're doing this math
to get the answer, don't forget to square this value. I don't know why it's a very common mistake
that students make, so don't make that. And the answer here is 0.35 liters per mole seconds.
Okay, before we move on let's talk about this kind of messy unit. The fact is k is a proportionality
constant, and the units on k are going to change depending on the reactants and their order.
This is the correct unit to use for this problem and you should have it, but what's
more important is the value, and that's what an AP grader is going to be looking for.
Okay, part D is about reaction mechanisms. This question proposes a two-step
mechanism and asks if it's consistent with the rate law that you proposed, or you determined,
back in part B. So, first things first, let's bring back that rate law. There it is.
Now there are few things you need to know about reaction mechanisms on the AP exam. The first is
the slow step is going to be the rate determining step. That means just what it says. It’s the step
in the sequence that determines the overall rate of the reaction because it's the one that takes
the longest to complete, right? If we think about an overall reaction as a series of smaller steps,
the overall reaction can't go any faster than the slowest step. You can have a variety of steps
that are really really fast, but it doesn't matter how fast those individual steps are.
The overall reaction can't go any faster than the slowest step, which is why the slow step
is the rate determining step. Okay, now this rate determining step, this slow step, must feature all
the species in the rate law. So all the chemicals, all the reactants for example for rate equals k
[NO2] squared here, the rate determining step has to include nitrogen dioxide. And if we know
that the reaction order is zero with respect to CO, it's unlikely that carbon monoxide would be
a reactant in the critical step, right? Because zero order means the concentration has no effect
on the rate at all. So, if it's zero order, it's not going to be involved with a critical
rate determining step. And lastly, we want to look at the molecularity
of the step. This is a fancy word that basically means: how does the number of molecules
impact the rate? For instance, our slow step here is bimolecular with respect to NO2. That means it
requires two NO2 molecules to react. And looking at our rate law we see that NO2 is second order,
which is consistent with a bimolecular step. So, putting all this together, this reaction mechanism
is an excellent candidate. Now you can explain this in many ways, but your answer should look
something like this: the mechanism is consistent with a rate law of this, the slow step is the rate
determining step and it has only NO2. Also, this step is bimolecular with respect to NO2,
which explains why it's second order. Okay, let's finish up. Part E reads the
reaction proceeds spontaneously. Sketch a relative potential energy diagram for the overall reaction.
A potential energy diagram is just a simple graph that charts the reaction progress on the x-axis,
and the energy of the reaction on the y-axis. Now here the problem specifically tells you that the
reaction is spontaneous, so that means that we're working with Gibbs free energy, or delta G. So
let's be even more specific, and we can label the y axis delta G. Now these kind of graphs
are pretty qualitative, okay? The AP exam just wants to make sure you understand the basic
concepts. There are two things to keep in mind when you're drawing energy diagrams. The first
is you want to be sure to depict whether the reaction has a positive or negative
change in energy, and this is usually represented by enthalpy, delta H, or Gibbs free energy,
delta G. So for our problem, we know that the reaction is spontaneous so it's going to have
a negative delta G. That means that it loses energy, so the reactants that we start with
are going to be higher on the y-axis than the products. Now, sometimes the AP exam asks about
enthalpy instead—whether the reaction is exothermic or endothermic. It’s the same basic
idea, we're just talking about enthalpy, or heat energy, as opposed to delta G.
Now, the second thing to keep in mind is activation energy. That's that
peak of energy required to kick start the reaction. The trick here is to remember that
each step in the reaction gets its own activation energy bump. The slow step
gets the higher bump, that's why it's slow because it requires a higher activation energy, so
we have the two humps for the slow and the fast. You have to be sure to put these in the correct
order. So for our reaction, this is the slow step. That should be the higher bump and it
should come first followed by a lower bump for the fast step. Okay, what does this actually look like
here? This is a pretty good example of what you could draw. You know this is pretty conceptual
and the AP graders will kind of accept anything that looks reasonable. Here are the basic
things that they're going to be looking for. Okay,
first of all, our reactants start at a higher energy than the products end up with, okay?
So we have a negative delta G. Second we got our two bumps here, okay? The slow step is the
higher bump that comes first, and our fast bump for our fast step is after, and it's smaller.
So as long as you have something that's reasonably similar to this, you should get full credit. So,
I hope this was helpful. This is how to tackle some of the most common kinetics questions on the
AP chemistry exam. Definitely check out my website for a bunch of resources that I really hope will
help you ace the test. And if the test is coming up in your future, I wish you the best of luck.