AP Chemistry AP Practice Exam: Free Response Part III

Hello, and welcome back to Educator.com; welcome back to AP Chemistry.0000

Today, we are going to close out our AP Chemistry course with the final instalment of the AP practice exam.0004

This is going to be the third part of the free response section.0010

In the last lesson, we actually did some reactions, and we did a question that involved experimental procedures.0014

This is the part of the free response section where you are actually not allowed to use calculators.0023

All you have to do is basically give good, solid explanations for why things are the way they are.0028

We are going to continue on with these types of questions.0035

The first one is going to discuss thermodynamics; so this is going to be question #6: let's just jump right on in.0037

OK, so answer the following questions in terms of thermodynamic principles and the concepts of kinetic-molecular theory.0044

Consider the reaction: CO₂ gas + 2 NH₃ gas goes to CO(NH₂)₂ solid + H₂O liquid; and we have a ΔH equal to -134 kilojoules.0051

We have this reaction: we have carbon dioxide gas and ammonia gas going to this compound (which is a solid, notice), and H₂O, which shows up as a liquid.0083

So now, let's go ahead and start answering some of the questions here.0094

This is A, and this is part 1 of A: so, for the reaction, indicate whether the standard entropy change is positive, negative, or 0; and they want you to justify your answer.0098

OK, so ΔS--that is the standard entropy change; well, take a look at what is happening in this reaction.0113

We have a gas, and we have another gas; we have 3 moles of gas in the reactants (which is a highly disordered system), going to a solid and a liquid (which are very highly ordered systems).0119

So, the entropy of the products is less than the entropy of the reactants; remember, entropy is disorder--it's a measure of the chaos in a system.0131

Gases are going to be a lot more disordered than a solid and a liquid; therefore, the entropy of the products is less than the entropy of the reactants.0139

Therefore, ΔS, which is products minus reactants, is actually going to end up being less than 0; it is going to end up being negative.0156

That is why: and your justification is "because reactants have a higher entropy than the products."0165

That is it; OK.0187

Part 2 says: Which factor (the change in enthalpy, ΔH, or the change in entropy, ΔS) provides the principal driving force for the reaction at 298 Kelvin?0190

Well, remember: they said that this reaction is spontaneous (right?--right up at the top, where you can read it, it said this is a spontaneous reaction).0200

Well, "spontaneous" implies that the ΔG, the free energy, is less than 0; in other words, it is negative.0210

Well, we know that ΔG is equal to ΔH minus TΔS.0218

Well, what do we know about ΔH?--it's negative.0226

Well, what do we know about ΔS?--ΔS is negative.0228

Well, if ΔS is negative, that means the -TΔS term is positive; well, if the -TΔS--if this term--is positive, and yet this is a spontaneous reaction (ΔG is negative), that means the driving force for the spontaneity of the reaction is the ΔH term.0231

That is negative; so, ΔG is negative; a negative number equals a negative number, plus something positive.0256

Because it's a spontaneous reaction, it says that that means that ΔG is negative; therefore, this negative term dominates under these conditions, so ΔH is the answer.0264

The reason is precisely this: write out the equation; you show that the -TΔS term is positive; it is positive because you know from part 1 that ΔS is negative; therefore, because it is spontaneous (negative ΔG), that means that ΔH is actually the driving force.0280

The enthalpy is the driving force for the spontaneity of this reaction at this temperature.0297

OK, part 3 says: For the reaction, how is the value of the standard free energy, ΔG, affected by an increase in the temperature?0304

Well, we use the equation again: ΔG equals ΔH minus TΔS.0315

Well, the ΔH term is negative; this term right here--the -TΔS--is positive; so, if we increase the temperature, as temperature increases, this -TΔS term increases.0324

In other words, it becomes more positive.0342

ΔH is fixed--it is fixed at -134; if I keep increasing the temperature (ΔS is also fixed, whatever it happens to be, but), this temperature, as it goes up--this whole term, -TΔS, goes up.0346

But -TΔS is positive, so it becomes more positive; or, ΔG becomes less negative and, if you want to say so, eventually positive.0361

Again, you can raise the temperature, eventually, so high that it is going to overrun the ΔH term, and it is going to make the reaction all of a sudden become non-spontaneous.0384

That is it--basic thermodynamic principles.0394

ΔG=ΔH-TΔS; ΔG is negative--that means spontaneous; ΔH negative means exothermic; ΔH positive means endothermic; if ΔS is positive, that means that entropy has increased from reactants to products; if ΔS is negative, that means entropy has decreased from reactants to products.0398

ΔG positive means non-spontaneous (or spontaneous in the reverse direction); that is what all of these mean.0423

All of these are accessible via this one equation.0428

OK, so now, let's talk about part B.0432

Let's see, what does part B say?--Some reactions that are predicted by their ΔG to be spontaneous at room temperature do not proceed at a measurable rate at room temperature.0438

#1: Account for this apparent contradiction.0454

Well, as it turns out, there is no contradiction here.0458

ΔG is a thermodynamic property; rate is a kinetic property--that is all you have to write.0461

ΔG is thermodynamics; rate is kinetics; so thermodynamics don't tell me how fast a reaction is going to go--they are telling me that, if I can get the reaction to go somehow, that it will go.0469

In other words, once I get it started, I don't have to do anything to it; rate is kinetics.0491

If you want to draw a little energy diagram, you have this; you have this; you have something like this; OK.0495

Now, ΔG--this part right here--that is thermodynamics; this part right here--the activation energy (remember, the energy of activation--it is the energy in order to get over this initial hump, in order for the reaction to proceed forward)--this is what controls the rate.0506

This is thermodynamics; this part here is what controls the rate.0526

The lower this is, the faster the reaction; in other words, more of the reactants have enough energy to get over this hill.0529

But, if this is really, really high, there aren't going to be a lot of particles that have enough energy to get over that hill, so the reaction rate is going to be slow.0536

ΔG is thermodynamics; rate is kinetics; a slow reaction (this is one thing that you definitely want to say) means a high activation energy; you definitely want to mention activation energy here.0545

You don't just want to leave it as "ΔG is thermodynamics, and rate is kinetics"; you want to say which energy actually controls the rate: it's the activation energy that controls the rate.0564

ΔG tells you about thermodynamics.0575

OK, so that is question #1; now, #2 says: A suitable catalyst increases the rate of such a reaction; what effect does the catalyst have on ΔG for the reaction?0578

OK, well, catalysts affect rates, not ΔG (and if you want to mention it) or K_eq.0592

A catalyst does not change the equilibrium constant; a catalyst does not change the thermodynamic property of a given reaction under a given set of conditions.0611

The only thing that a catalyst does is reduce the activation energy and/or provide an alternative pathway, which is tantamount to reducing the activation energy.0621

A catalyst makes reactions faster; it does not actually change the nature of the reaction.0632

That is thermodynamics; and again, a catalyst affects kinetics; ΔG is thermodynamics; the two are completely different domains.0639

Both are important, but they are actually handled separately; that is why we had different chapters--we had a kinetics chapter and a thermodynamics chapter.0648

Both are important, but a catalyst has no effect on the rate.0655

And again, if you want to use this same thing right here, you can do that; this energy difference is thermodynamics, ΔG; this energy difference is kinetics.0660

That is that; OK, now, let's go to question #7.0672

OK, question #7: Answer the following questions, which refer to 100-milliliter samples of aqueous solutions at 25 degrees Celsius in the stoppered flasks shown above.0682

In front of you, as you are looking at these questions, you should see some flasks (4 of them); they are all .1 Molar; one of them is sodium fluoride; one of them is magnesium chloride; one of them is C₂H₅OH (which is ethanol); and one of them is CH₃COOH, which is acetic acid, or vinegar.0697

So, we have sodium fluoride, magnesium chloride, ethanol, and acetic acid.0717

OK, so now, we are going to answer some questions about it: Which solution has the lowest electrical conductivity?0722

Well, the lowest electrical conductivity means the least dissociated into free ions.0728

Therefore, C₂H₅OH has the lowest conductivity.0756

The lowest conductivity means the least dissociated into free ions; of these things (the sodium fluoride, the magnesium chloride, the ethanol, or the acetic acid), the one that is dissociated the least is going to be the ethanol.0779

In fact, it is probably going to be not dissociated at all; it is a molecular compound--it isn't ionic at all.0793

Acetic acid will dissociate a little bit into H⁺ and acetate; magnesium chloride and sodium fluoride will dissociate completely, because they are ionic compounds.0798

This is a molecular compound: there is no dissociation (or very, very little).0809

It is true that this H is slightly acidic, but it is acidic on the order of water; so you are talking about 1x10^-14; it is not going to dissociate--there is no conductivity.0814

OK, now B says: Which has the lowest freezing point?--explain.0825

OK, colligative properties: when you add a solute to a solvent, you create a solution.0836

You lower the vapor pressure; you lower the freezing point; you elevate the boiling point; and you activate the osmolarity of that solution.0843

OK, lowest freezing point means highest number of solute particles.0854

The highest number of solute particles between sodium fluoride, magnesium chloride, ethanol, and acetate--well, sodium fluoride will dissolve into 1 mole of sodium ions and 1 mole of fluoride ion.0877

Magnesium chloride will dissociate and dissolve into 1 mole of magnesium ions and 2 moles of chloride ions, for a total of 3 moles of particles.0889

C₂H₅OH will not dissolve at all, or dissociate at all.0900

And the CH₃COOH will be partial dissociation into H⁺ and acetate ion.0905

Of those, magnesium chloride actually produces the most solute particles--free particles floating around in the solvent, which is water.0910

Again, the identity of the solute particles does not matter; all that matters is the number of particles floating around, which is why they are called colligative properties.0922

Identity is irrelevant; it just matters how many particles are there.0931

In other words, how many particles are interfering with the solvent molecules doing what they normally do?0935

That is it.0940

OK, C: Above which solution is the pressure of the water vapor greatest?--explain.0943

OK, here we go with our implications: more solute particles means a lower vapor pressure; therefore, a lower vapor pressure implies that C₂H₅OH is our particular answer to C, because C₂H₅OH has the fewest solute particles.0957

It has the fewest solute particles because it doesn't dissociate at all; it is just the amount of ethanol that you actually drop into water is that much ethanol; it doesn't break up into more particles like sodium fluoride, magnesium chloride, and acetic acid (to some extent).1000

OK, let's see what we have for part D--D says: Which solution has the highest pH?--explain.1016

OK, well, highest pH means the most basic; it doesn't necessarily mean that it is going to be basic--it just means you might have a solution of 6 (which is acidic), and another solution which is 4 (which is also acidic), but the 6 has a higher pH.1025

Let's take a look at acetic acid: CH₃COOH--that is going to be reasonably acidic; C₂H₅OH--well, it is not really going to dissociate much, in terms of hydrogen ion, so probably the pH is going to be reasonably neutral.1046

So yes, compared to those two, it is going to have a higher pH than the acetic acid.1061

The magnesium chloride--neither magnesium nor chloride is such that it is going to cause any sort of hydrolysis of the water; it is not going to do anything to the water molecules to release a hydrogen ion or hydroxide ion.1065

So, you are looking at a neutral pH there.1079

However, sodium fluoride--when sodium fluoride dissociates, what you are going to end up with is this: fluoride is the anion of a weak acid, HF (hydrofluoric acid).1081

So, when F^- is released into solution, what is going to happen: it is going to react with the water (HOH) to produce HF (hydrofluoric acid or hydrogen fluoride, really), plus OH^-.1105

This free OH^- in solution--that is going to create a basic solution; you are going to have a pH which is higher than 7.1119

So, your answer is sodium fluoride; and this is the reason why: because F^- is the anion of a weak acid; therefore, it will cause hydroxide ion to form.1125

If you write this, and just write this equation, that is all you need to write--very simple; very straightforward.1136

This is just recognition of chemistry; this part of the AP exam really tests the extent to which you understand the chemistry (regardless of the math).1141

OK, so we are down to our last problem in the AP chemistry exam; so this is kind of exciting--it has been a long road, but it has absolutely been worth it, I hope.1153

Question #8: Consider the carbon dioxide molecule and the carbonate ion (so we have CO₂ and CO₃^2-).1166

The first question says: Draw the complete Lewis electron dot structure for each species.1181

OK, so we have done this many times--these are the last lessons that we did, in fact, before we started the AP practice exam.1186

The Lewis dot structure for CO₂ is this (and remember, you have to include the lone pairs of electrons); that is CO₂.1194

Now, CO₃^- looks something like this.1202

OK, now, this is actually part of a resonance structure, so I am going to go ahead and actually draw all three resonance structures; it's not necessary to draw all three, but you know what, it is not a bad idea.1217

That way, they won't penalize you, because they might actually ask you a question regarding resonance.1228

This is going to be CO; I'm going to put the double bond on the left; I'm going to do 2-, 2, 4, 6, 2, 4, 6; and I'll go ahead and draw this other resonance structure over here--it doesn't matter--vertical or horizontal.1233

2-, CO; I'm going to draw the double bond on the left this time; there is 6; there is 2; there you go--that is CO₂, and this resonance structure; so this structure is actually an average of all three.1253

This is the Lewis structure; however, you would be OK if you just drew one of them--that is not a problem.1270

That is part 1; now, part 2 says: Account for the fact that the carbon-oxygen bond length in CO₃^2- is greater than the carbon-oxygen bond length in CO₂.1276

Here is the reason why: notice the Lewis structure here--these are fixed double bonds; CO₂ has fixed double bonds, which have a specific length.1293

Now, CO₃^2- has resonance structures, and each bond has both single-bond character and double-bond character.1318

Remember, when we have a resonance structure, it isn't that it is this or this or this; it is all three simultaneously.1338

It has both single-bond character and double-bond character: that is the reason why.1345

CO₂ has fixed double bonds, which have a specific length; CO₃^2- has resonance structures, and each bond has both single-bond character and double-bond character.1366

Therefore, because (let me see, how shall I write this) the bond is between single and double, it is longer than a straight double bond; there you go--something like that.1375

As long as you mention resonance structure, and you say that it has single-bond and double-bond character, the rest of it is actually pretty implicit; there you go.1418

OK, now part B--it says: Consider the molecules CF₄ and SF₄; so, we have CF₄, and we have SF₄: carbon tetrafluoride; sulfur tetrafluoride.1433

Part 1 says: Draw the complete Lewis structure for each molecule.1450

OK, so when we do a Lewis structure for CF₄, we are going to end up with something like this; and here we go with the lone pairs; that is CF₄.1455

Now, SF₄ is going to look almost the same, except we are going to have a lone pair on sulfur, also.1473

There you go: those are the two Lewis structures, CF₄ and SF₄.1490

The only difference: S has those extra two electrons; and those two electrons are actually on the sulfur itself.1494

Sulfur can accommodate more than an octet, because it is in the third row; it has d orbitals available to it.1501

OK, now, part 2 says: In terms of molecular geometry, account for the fact that the CF₄ molecule is nonpolar, whereas the SF₄ molecule is polar.1508

OK, molecular geometry: we have 4 objects around a central atom, and the central atom does not have any lone pairs, so remember that table that we were looking at.1519

CF₄ has a tetrahedral geometry; that geometry looks like this: C, this is F (I'm not going to draw the lone pairs); there is an F (no, I'm not going to have that; I can't have random lines on a Lewis structure--that is not going to work); F; F; F.1532

Here, as it turns out, because fluorine is more electronegative than carbon, you are going to have a dipole that way; you are going to have a dipole to the back and to the left; you are going to have a dipole to the back and to the right; and you are going to have a dipole pulling forward.1563

All of these dipoles are going to cancel (the bond dipoles--the individual bond dipoles); so all right, bond dipoles cancel, which makes it nonpolar.1578

Now, CF₄ looks something like...I'm sorry, SF₄ looks like this: SF₄...well, you have 5 objects around this central atom: you have 1, 2, 3, 4; your fifth object--remember, when we talk about an object, it includes lone pairs.1598

Now, because you have 5 objects around a central atom, it is actually going to be triagonal bipyramidal.1617

But because it has one lone pair, what we are going to have is a seesaw arrangement of the atoms.1624

SF₄ looks like this: there is an F here; there is an F there; there is an F going backward; there is an F coming forward; and there is a lone pair there.1631

The dipole bond, dipole bond...those two end up canceling.1645

However, now we have a dipole pulling to the left and back, and it is pulling to the left and front; the net dipole is going to be in that direction (to the left).1650

The vertical components of this dipole and this dipole cancel, but the horizontal ones--they are both to the left, so this is going to act like a bar magnet that has a negative side and a positive side.1663

Therefore, this is polar.1673

So here, the bond dipoles do not cancel; therefore, you have a polar molecule.1676

OK, and that, my friends, takes care of the AP practice exam.1699

It has been my pleasure presenting AP Chemistry and going through this AP Chemistry exam with you.1703

I wish you the best of luck in all that you do.1709

Take good care, and we will see you next time; thank you for joining us here at Educator.com; goodbye.1711