KRISTEN CACCIATORE: Hello, welcome to AP Daily practice
sessions.
Today's session two, we're talking
about short free response questions.
And you know there's going to be four of those on the AP exam,
so let's get you ready for them.
If you would like to work through these problems
on your own, you can click on the link
you see in the video now or in the YouTube description,
and download a PDF of the problems.
All right, so number one, the following equation
represents the decomposition of N2 O5, for which the rate
law is rate equals k times the concentration of N2 O5.
Now, we have the decomposition reaction there, and it says,
a sample of pure N2 O5 gas is placed
in an evacuated container, and allowed
to decompose at a constant temperature of 300 kelvin.
The concentration of N2 O5 gas in the container
is measured over a period of time,
and the measurements are recorded in the following data
table.
Then we have some time and concentration information.
All right, so let's see what they want us to do.
Part A, determine the value of the rate constant,
k for the reaction.
Include units in your answer.
All right, so as we approach this,
we want to remember that the rate law for a first order
reaction, is related to the half life.
And the half life is the amount of time required,
for the concentration or the amount of the reactant
to decrease by half.
So looking at this data table, I can identify that between time
0 and 1.67 hours, the concentration of N2 O5 went
from 0.160 to 0.0800.
And that is going down by a factor of 2.
And so I can also see that in the next increment of time,
between 1.67 hours and 3.3 hours,
which is a period of 1.67 hours, the N2 O5 concentration went
down by half again, from 0.0800 to 0.0400.
So clearly the half life, the amount
of time required to go down by half is 1.67 hours.
And now I can use the relationship which is
on your equation sheet, k equals 0.693 divided by 2 1/2,
the half life.
And I can just plug in that half life
that I found in the data table, 1.67 hours.
And then I get 0.415, and the units are inverse hours.
Hours to the negative power.
All right, part B, the following mechanism
is proposed for the decomposition of N2 O5.
I have three steps here.
And then it says, identify which step
of the proposed mechanism 1, 2 or 3
is the rate determining step.
Justify your answer in terms of the rate law given.
Well, remember, they gave us a rate law,
rate equals k times the concentration of N2 O5.
The rate determining step should have the same rate law
as the overall reaction, because the rate determining step
determines the rate.
So remember that rate law there, K[N2 O5].
And if we look at our three steps,
we can see that step number one has N2 O5, one
molecule as the sole reactant.
And so step one has the same rate law,
as this overall rate law.
And so step one has to be the rate determining step,
because of that parallelism.
So step one, and then our explanation is just that
the rate law of elementary step one is rate times K[N2 O5],
which is the same as the overall reaction.
All right, part C. If this experiment
was repeated at the same temperature,
but with twice the initial concentration of N2 O5,
with the value of k, the rate constant increase,
decrease, or remain the same, explain your reasoning.
So k is concentration independent.
It does not depend on the concentration.
It does depend on the temperature,
but in this problem, they say it's the same temperature.
So if the temperature doesn't change, then k doesn't change.
And so that's what we're going to say.
It will remain the same.
It's independent of concentration,
and will remain the same at constant temperature.
So that's just something you need to know.
Let's look at another short free response problem.
This one has an equation 6H plus aq plus 2MnO4, minus 5h2C204aq,
produces 10CO2 gas plus H20 liquid,
plus 2 manganese2 aq or aqueous.
A student dissolved a 0.139 gram sample
of oxalic acid H2C204 in water, in an Erienmeyer flask.
Then the student titrated the H2C204 solution in the flask,
with a solution of KMnO[aq], which has a dark purple color.
The balanced chemical equation for the reaction
that occurred during the titration is shown above.
We just read through that.
All right, first part, identify the species
that was reduced in the titration reaction.
Justify your answer in terms of oxidation numbers.
Well, this is telling us we need to figure out
oxidation numbers, so I'm going to do that for each element.
Here's a table, and it shows us that the hydrogen
had the same oxidation number in both reactant and product.
The manganese went from plus 7 to plus 2, that
means that it gained electrons.
It gained five electrons, so the charge was reduced.
And oxygen stayed the same, and carbon went from plus 3
to plus 4, which means it was oxidized.
It gave up an electron.
So this is asking us about reduction,
so if manganese became less positive,
so it is the element reduced, which
means it gained electrons.
But we're not done yet, because notice
that the problem says, identify the species not the element.
So what species in the reactants was manganese present in?
And that is MnO4 minus.
So MnO4 minus is the species that is reduced,
and we need to justify that as the problem says,
in terms of oxidation numbers.
So it gained five electrons.
All right, part B. Ooh, we have some buret diagrams here.
The student used a 50.0 milliliter buret to add
the KMnO4[aq] to the H2C2O4aq until a faint lavender color
was observed in the flask.
An indication that the end point of the titration
had been reached.
The initial and final volume readings
of the solution in the buret are shown below.
Write down the initial reading and the final reading,
and use them to determine the volume of KMnO4[aq] that was
added during the titration.
All right, so first of all, we need
to figure out, how many significant figures are
our answers going to have?
And to do that, I need to look at the scale.
So the scale of the buret is 0.1 milliliters,
each tiny line represents 0.1 milliliters.
I need to read to one digit, further more precise than that.
So I need to read to 0.01 milliliters or the hundreds
place in milliliters.
And then I'm going to start off by reading the final volume,
and I'm going to read down from the number here.
So I see 29, and then it goes down
to 29.1, 0.2, 0.3, 0.4, 0.5.
And then the bottom of the meniscus
is between the 0.5 and the 0.6 line,
so I'm going to read that as 29.55 millimeters.
And then I'm going to go over to the initial.
I'm going to do the same thing, and I'm
going to see that the volume is 3.35 milliliters.
So those are the two numbers I need to write down.
And then I subtract, and I get the final volume
or the titration volume of 26.20 milliliters.
So that's the answer.
Note that you did not get credit on this part,
if you did not write down the readings from the diagram,
and also if you did not have the right number
of significant figures.
And let's look at part C. Given the concentration of KMnO4[aq]
was 0.0235 molar, calculate the number of moles of manganese
ions that reacted completely with the H2C204.
All right, so I need to calculate moles.
I'm going to multiply molarity which is given in this part,
times the volume that I calculated in the previous
part, 26.20 milliliters.
I need that to be in liters though, to do molarity.
So I get 0.02620 liters.
That's the volume from part B, in units of liters.
Times the given molarity, 0.0235 molar, and that
gives me an answer of 0.000616 molar
or you can write that in scientific notation.
Notice that there's three sig figs here,
because the concentration has three significant figures.
And last part of this problem says,
the student proposes to perform another titration using a 0.139
gram sample of H2C204, but this time using 0.00143 molar,
molarity KMnO4[aq] in the buret.
Would this titration concentration
be a reasonable choice to use, if the student
followed the same procedure and used the same equipment
as before?
Justify your response.
All right, so look at this concentration.
It is more than 10 times more dilute, than the 0.235
concentration that we used in part C.
So that means that we would have a titration volume more than 10
times greater.
We had 26.20 milliliters.
It would be more than 10 times that size, so
like up around 300 milliliters.
That would not fit in a 50 milliliter buret.
That would not work with this experiment.
So our correct answer is, no, that would not work.
And the answer is, the reason is that the titron is so diluted,
that the titration volume would be much greater than the 50
milliliter capacity of the buret, the instrument
that we're using to measure here.
All right, well, that's going to be all for this video.
But before I go, I want to give you
some tips, about how you can continue
your review for the exam.
So if you noticed on this video, that there
were some things you need to review, well, rate laws
and half life, you can find an AP Daily, topics 5.2 and 5.3.
Reaction mechanism and rate law is topic 5.8.
Oxidation numbers and redox reactions, topic 4.9.
And determining concentration through titration, topic 4.6.
In addition, we have so many more resources
to help you prepare.
There are seven more of these 2023 AP Daily practice
sessions.
And we also have AP Daily practice from 2021 and 2022,
to help you prepare for the exam.
And you can find all those on our YouTube channel.
Thanks so much for watching, and I'll catch you next time.