DENA LEGGETT: Welcome to AP daily practice session,
where we're going to be focusing on some short free response.
My kids call me Doc Dena, and I am so
happy to be here to help you.
The first one we're going to look at
is focusing on Lewis dot structures and equilibria.
Now, we have a question here about two isomers
with big names that I'm not even going to think about right now.
The key is that they want us to do a Lewis dot
structure of this big molecule in the box.
It's one of the two isomers, meaning same molecular formula.
And the key here is it says to show all valence electrons.
Very important.
Now, we have an organic molecule-- carbon, hydrogen,
and oxygen. And I want to show you
an approach that can get you to that Lewis dot structure
quickly.
It's called HONC, 1234.
Hydrogen has one bond.
Oxygen typically two bonds, two pairs.
Nitrogen typically three bonds, one pair.
And the carbon has four bonds.
And so what we're going to do is we're
going to start by making bonds everywhere.
So those are my bonds with carbon and hydrogen,
and hydrogen is done.
We don't have to worry about those hydrogens.
Now I'm going to make my bonds between the carbons
and the oxygens.
And you notice that I have, with the middle oxygen there,
that makes up its two bonds on either side to the carbons,
but the one on top had to be a double bond.
I had to double bond that to get the four bonds to that carbon,
as well as the two bonds to oxygen.
Now, you can't quit there.
The question says show all valence electrons.
And that means you must draw in those pairs as well,
or you will not get credit.
OK.
The next one.
There's so much information here.
Circle, circle, circle.
Here, I've got the reaction that was given to me,
and I've got two ethanoic acid forming a dimer.
And it says that when equilibrium is reached,
50% of my molecules have reacted.
And it asks me to calculate the total pressure.
So let's look at the information I have.
I have mols.
I have a liter vessel, 450 Kelvin,
and it gives me an initial pressure of 0.74.
It also tells me that my measured pressure
is lower than 0.74.
And it doesn't ask me about that yet.
It just asks me to calculate the total pressure at equilibrium.
So I've got mols.
It helps to label these with the symbols you'd
find in the math equation.
Because if you label those with an n, a V, and a P, and a T,
hopefully, your mind goes to [INAUDIBLE]..
And you notice I have constant volume, constant temperature,
and my pressure decreases.
That indicates I have to have a decrease in mols.
So, I've got-- let's take a look at what we have.
This just repeats what we had before.
Let's calculate that total pressure.
You notice that in my RICE box, I
put two ethanoic in the reaction instead of one and one.
That's going to help us with our mol ratio a little bit.
Now, it told me in the question that 50% reacts.
That's why it took away 50% of my ethanoic acid.
I'm going to gain that same amount in my product,
but I have to multiply by the mol ratio, that 1 over 2
mol ratio.
At the end, that gives me 0.01 of my ethanoic acid
and 0.005 of that dimer.
So now, I can add these up.
This was with mols.
You can use a RICE table with mols.
So now I add those up.
And I have my total mols in the vessel
after the reaction reaches equilibrium.
So the pressure is nRT over V.
We've listed our givens.
You want to check your units.
And so we want mols.
We want liters.
We've got atmospheres.
And we need Kelvin in temperature,
which is fortunately given.
And so you've got to grab that right R value, the correct R
value of 0.082, or 0.08206.
And that would calculate my pressure at 0.55 atmospheres.
So that was Bi.
OK.
Now I'm going to use that to calculate the value
of my equilibrium constant, Kp.
So I've already done my RICE.
Kp is equal to the partial pressure of my products,
over the partial pressure of my reactant,
squared, because there were two of those ethanoic acids.
Now here's a tricky part.
I have to find--
I know my total pressure.
I got that in the previous slide.
But this time I need my partial pressures.
And to do that, we have to calculate the mol fractions.
And a mol fraction is part of our whole.
So my part mols that were the dimer is out of the RICE table.
Look at the E in the RICE table.
And that gives me my partial pressure.
The part over the whole.
That was 0.0150, times the total pressure,
and that gives me my partial pressure.
Now I need to do that for my reactant, as well.
It's part, from that RICE table, was 0.01 mols, over the total,
and that Is times the total pressure.
And so that gives me my partial pressure of my ethanoic acid.
Plug that into Kp, and I get 1.4.
That was fast.
Remember, you can pause.
This next one is another short one,
involving photoelectron spectroscopy
and electromagnetic radiation.
So right away we see a photoelectron spectrum.
Those are related to your electron configurations.
They give me my peaks and my peak energy.
They also give me frequency ranges.
Now, it says that the frequency ranges
to generate the spectrum has an energy of 7 times 10
to the fourth kilojoules per mol.
It says, which region of the electromagnetic spectrum
is this?
Justify your answer with a calculation.
This is a little tricky because you've
got to be well familiar with your units here.
We are given energies in kilojoules per mol,
and frequencies in seconds to the minus 1.
The issue is that the energy of the photon
is not kilojoules per mol.
So we have energy.
We need to convert it to frequency.
To do that, we need joules per photon.
So we're going to have to do a unit conversion here.
So I've got kilojoules per mol.
I want to get rid of kilojoules and go to joules.
That should be a no-brainer, but I need
to go kilojoules per photon.
That's never going to be explicit in the question.
It'll just say joules.
You have to know that it's joules per photon.
So I'm going to use Avogadro's number.
Mols and mols and mols, no matter what I'm counting.
My units cancel.
Now I have joules per photon.
Now I can plug that in to calculate frequency.
Energy is Planck's constant, h, times frequency.
And so I plug those values in and my frequency
is 2 times 10 to the 17th.
You can't stop there.
You have to stake your claim.
The frequency alone doesn't answer
the question that was asked.
The question says, what's the region.
So you have to answer that question.
And you notice that our frequency
that we calculated lands squarely
in the X-ray region of the electromagnetic spectrum.
It's really heart wrenching to see great justifications
and no claim made.
So always make sure you answer the question.
Now, this next one shows a photoelectron spectrum.
But this time we're going to be analyzing the peaks
in that spectrum, and the peaks are
going to match up with your electron configurations.
So, we have a proposal, and we need to refute the proposal.
Now, don't try to support the proposal.
We've seen that, as well.
It's telling us that the proposal is incorrect.
So we want to refute that proposal.
This one says the subshell from which
an electron is removed in the second ionization energy.
And so we need to identify that electron,
so we know which peak it's from, so we
know which energy it matches.
OK?
And identify means, don't explain.
It means, simply say the answer.
Write the answer.
And it's the, remember, not the first ionization,
which would be the first electron,
it's going to be the second electron removed.
So let's assign those peaks.
1s2.
2s2.
2p5.
So the first electron would be removed from the 2p.
The second electron is also removed from the 2p.
So that means we're dealing with peak three at 1.68 times 10
to the third.
That's why the person was wrong.
Don't try to tell us the person was right
when we told you the person was wrong.
So now we have a similar with part
B. This time it asks us to identify the subshell
for the second peak.
Well, if you've written your electron configuration
above the peaks, which I strongly recommend you do,
you've already answered that question.
We've already identified that peak as a 2s electron.
Don't explain why.
Just tell them what it is.
All right.
Thank you so much.
I know that was quick, and there's a lot of information,
but we have even more valuable information
if you want to head on over to the College Board YouTube site.
There's a QR code for AP Chemistry
reviews in 2021 and 2022.
We have so much help for you there.
In the meantime, I am so grateful that you had me
along for your journey as you learn some more chemistry.