Archive – Q3 week 7 – 19-20
Week of 3/16 – 3/20 – REFRESH EVERY TIME!
*Okay we are out for 2 weeks but we need to finish this course and get ready for the AP test. I intend to do that but I will need your help on your end. For those that are not very good at putting in the time out of school you need to step it up! I have a plan and we will execute it together.
My 2 AP Chemistry classes have been outstanding this year and we are so close to the finish line. I thank you for your hard work and your trust in me. Please continue your efforts as I will! The school is nothing without its students!
Please be patient and refresh this page often as I working to get everything ready for each day. Let’s go!
I plan on doing live teaching later this week and plan on doing so during out allotted class time, 8:15 – 10:30. Please make the necessary accommodations to keep that time free for class time. We are out of school but school is open digitally! This is the time we will use for live timed assessments and tests.
Test on Thursday this week – I will give out more details but we will take in together at the same time. I am planning on giving it Thursday or Wednesday at 9:00am – 10:00 am. It will be a shared doc and it will open at 9:00am and close at 10:00 am. Please plan accordingly. More information to follow.
3/16 – Monday – period 2/3/4
Classwork:
1. Please review your Form from over the weekend. I will have them graded this morning. Make one more
submission today and that submission will NOT be on auto-grade as I need to see if I need to give you another submission.
– Some hints that you may have missed in lectures:
A)
Exothermic reactions can be reported 2 ways:
A + C —> AC + 40. kJ
A + C —> AC ∆H = –40. kJ
*Notice the heat or energy is written in the product side of the reaction to show
that there is a NET release of energy.
Endothermic reactions can be reported 2 ways:
40. kJ + AC —> A + C
AC —> A + C ∆H = +40. kJ
*Notice the heat or energy is written in the reactant side of the reaction to show
there is a NET absorption of energy.
B) Careful with what you are discussing in terms of bonds or IMF in dissolution/ionization reactions.
C) Solving for the q of a reaction and calculation the heat released is not the same. It just how it is worded.
q = – HEAT for exothermic reactions
the negative means released (direction)
HEAT is positive value so if I ask for the heat released I am
looking for the absolute value of the q as I am already implying
the negative sign by saying released.
Its the same if I ask for the decrease in temperature. You still give me a positive temperature.
2. Today we start our discussion towards the Second Law of Thermodynamics:
Please read through my notes regarding the 3* Laws of Thermodynamics below.
Class notes: Some people have expressed that they like the written notes even if there are a some grammatical errors!
Today’s Notes:
0th Law: Hold unto your seats, this will be revealed in the lectures below!!
1st Law: ∆Esys = –∆Esurr – Law of Conservation Mass – Uplifting Law!!!
Because of this law it seems that we should be able to create 100% efficient engines and possibly perpetual energy systems that recycle energy forever!!! Uplifting Law!
2nd Law: ∆Suniv > 0 – Dispersion of Heat – “What a Bummer Law!”
disorder of energy
∆S = entropy! Is it enthalpy (∆H)??? Noooo! Its ENTROPY (∆S) !!!!!!!!!
The second Law tell us that entropy ∆S of the universe must increase for a spontaneous process.
Entropy is the movement of concentrated energy (free energy) to less concentrated energy and thus we cannot create any pathway we desire for energy. Because of this the amount of concentrated (free energy) must move (disperse) to less concentrated areas of Energy. A lot teachers will discuss entropy as the change of the disorder of a system, surroundings or total universe (system + surroundings = universe) but technically that is not completely true as it is really about the change of the disorder of the energy (dispersion of energy). Changes in overall disorder do parallel the energy dispersion (spreading out of the energy) so overall disorder explanations do work but to truly understand entropy you need to think about ∆S as the change in the dispersion of energy.
3rd law: S˚ (absolute entropy of chemical) = Zero at 0 Kelvin “We are limited by 0 K !”
This creates the AZ zone!!!! (Absolute Zero) zone. This is the lowest amount of energy zone. At absolute zero there is no more movement as Temperature is proportional to kinetic energy. We say absolute entropy S˚ as we are evaluating the dispersion of energy by how molecules move or vibrate. *Remember that bond enthalpies are averages because the stretch and recoil around the average bond length ( and stretch an recoil around their average bond enthalpy!)
So at 0 K there is no more kinetic energy and every chemical that is in a crystal lattice will have the lowest amount of ability to move along each bond (stretch and recoil) and thus have the lowest ability to “disperse energy”! This is the starting point at which we measure the absolute entropy of each chemical as we increase the temperature to the standard thermodynamic conditions denoted by the S˚ (degree sign after the S). Remember that the standard thermodynamic condition in tables are 25˚ C and 1 atm of pressure. The absolute entropy given in our tables IS LIKE THE REVERSE OF HEAT OF FORMATION! Its the the total value of the chemicals ability to move as a molecules or move along its bonds.
Making a graphic here…
So the first law is so uplifting because every pathway is possible but the 2nd law is a downer because it tells us that every pathways IS not possible as there are pathways that are favored (allowable) over other pathways by the universe. What is worse is the 3rd Law which states that we are limited in these possibilities by 0K and yet we cannot achieve it!
*We cannot get to O K because the energy needed to be removed by the system to get to O K will wind up reheating that system (2nd Law = Hot —-> Cold!)
This Energy that must disperse for the second law includes :
kinetic energy (∆H) and Potential Energy (bonds or degree of freedom of particles)
A Spontaneous process is an allowable pathway that the Universe provides because the process is moving in the direction that takes Concentrated Energy and is dispersing it OVERALL in the Universe. A better way to say this is that a spontaneous process or reaction ALWAYS increases the entropy ∆S or the universe. A spontanenous has a pathway that requires no continual input of energy. Example: Riding a pony uphill. Or burning of paper!
Spontaneous reactions or processes provide FREE Energy to do work.
If the reaction or process does not increase the ∆S of the universe then the reaction has no pathway to occur and it is Non-spontaneous. It only occurs if there is constant supply of energy to force it forward. Example: Carrying the pony uphill! Also photosynthesis!
Non-Spontaneous reactions REQUIRE Free Energy to Make IT OCCUR!
Because ∆S MUST increases we cannot create engines that are 100 efficient because we must lose concentrated Energy (Free Energy) to the universe with every energy conversion. This is reason we cannot have a perpetual energy device that recycles energy! There can never really be renewable energy!!! “What a Bummer!”
Spontaneous processes are Pathways that the Universe allows, however we can manipulate chemical reactions by Le Chateliers Principle OR WE CAN FORCE an nonspontaneous to occur IF WE USE FREE ENERGY to make it happen. Last year we learned about energy coupling where we use spontaneous process (cellular respiration to make ATP) to Drive non-spontaneous processes (anabolically making large macro molecules like enzymes from amino acids).
3. Use the follow along by taking notes. If you want to use the Thermo pages I gave you last week or your
own way to learn it is up to you. You know how you best learn!
THermo read along – Thermo notes.pdf
a) Interconnectivity between ∆H, ∆S, and ∆G.
Tablature review of ∆S, and ∆G
Presentation that will be shown in the 2 lectures below:
Today’s Class Lectures:
Lecture 1: Connecting Heats of formation, Bond Enthalpies and Hess Law with explanation of the Laws of Thermodynamics. Here is where I start discussing ENTROPY, ∆S (for reactions) and S˚ (from tables).
Everything we do from Thermodynamic is based on the 2nd law which is the measure of whether a process is spontaneous! Remember we started this in SEPTEMBER with our voltaic cells (batteries)!!!
Lecture 2: I derive Gibbs Free Energy Formula which measures FREE ENERGY of the Universe using
the systems ∆H and ∆S. I also discuss the signs of the thermo quantities that you need.
The key here is that we ARE ABLE TO MEASURE THE ∆S of the universe by measuring the systems ∆H, T (K), and ∆S (of the system = degree of motion of molecules or ions)! We have a quantity that can measure whether a reaction has a pathway to succeed (Spontaneity!!!!!) and that quantity is ∆G!!!
Since everything we do is based on the 2nd Law, and ∆G is the measure of the second law, everything we do is based on ∆G! I derive the equation for ∆G below!
We will now be using the following formulas:
 |
If you notice in your Thermodynamic tables there are 2 other thermodynamic quantities, S(absolute) and ∆G.
We can calculate the ∆S and ∆G through tablature by using the following formulas.
Also ∆G can be determined through a formula that measures the Universe’s Entropy Change by just looking at the system!
|
∆G = ∆H – T∆S!!!!!!
4. Please complete the thermo 6a – Enthalpy, Entropy , Gibbs of Glucose methane.pdf worksheet and
review with the key. We will now be using the following formulas:
THermo 6a worksheet lecture:
Take a break and please come back later to complete the homework!! Yes you have homework too!
I feel bad… now I am over it!!
Oh and the Forms count as tests now 🙂
3/16 – Monday – Homework –
1. Please complete Thermo 7 – Gibbs Free energy Free Response 1984.pdf worksheet.
It very much the same skills as the thermo 6a worksheet above. Review with key below.
I gave this out last Friday:
2. View the Signs Lecture below and Then complete the Form posted below.
If you need more help with the thermodynamic signs of the ∆G formula please view
the screencast lecture after the form below.
The Signs Lecture:
Thermodynamic Sign form – due Tuesday Morning 4:30 am:
Additional resources:
Gibbs free Energy Lecture Screencast:
I suggest that you watch the Signs lecture that is posted below posted to review the concepts in the 2nd class lecture posted today that is posted above. These concepts include the Signs of the thermodynamic values of ∆H, ∆S, and ∆G in the Gibbs Free Energy equation.
End of Monday…..I hope you did not use up all of your Free Energy today!
3/17 – Tuesday – period 2/3/4 – Happy St. Pat’s day!
THERE ARE NO SHORTCUTS TO A GREAT JOB. Those of you trying to do the least amount will pay a price in your understanding and eventually in your grade. Listen to every minute of every lecture, complete every worksheet (and review with keys), and read everything that I post here. Follow my plan. Everything has a purpose. I know it harder out of class but that is just how it is.
Think. Do you think other AP Chemistry classes are doing what you are doing? Ultimately your AP grade is based on (statistics) how you do compared to others that take it.
1. Please review your Heat of Dissolution Bond Enthalpy Form with the key that I
emailed to you or with link below:
https://drive.google.com/file/d/16fXfwWxTKwlIAfYYJUIes-QxBrMIKEfB/view?usp=sharing
2. Please review Thermo Quantities and Sign Form. I have graded it and emailed it back you. A much better job by many in this Form and a better job by me with a correct key! However, there are many that did not score at the class average of 84% and I will give those one more submission that will count as the average of your 2 submissions.
– I will post a a written tomorrow on this form.
3. Please view today’s lecture that will review our Thermodynamic quantities, ∆G, ∆H, and ∆S, especially the Second Law of Thermodynamics!!
Today’s Lecture:
The most beautiful Thermodynamic equation that measures the entropy of the universe by measuring the systems ∆H and ∆S (degrees of freedom of particles). You need to understand its derivation!
TOTAL Entropy of the Universe change is measured by the systems:
∆H – ENTHAPLY:
– the heat releases from an exothermic reaction actually speeds up the surrounding matter and makes them disperse the heat outward so that a systems release of heat (-∆H) will actually increase the Entropy of the surroundings. –∆H or exothermic values are favorable to the universe as in increases entropy!
+∆H or endothermic values are unfavorable because the consume or absorb thermal energy away from the surroundings. Endothermic reactions create products with higher potential energy and thus create concentrated sources or energy, which decreases entropy.
∆S – ENTHAPLY:
– The change of entropy of the system (∆S) is a measure of the absolute entropy changes in chemical reaction that will either increase or decrease the degrees of freedom of molecules or ions ability to move and spread out or disperse the heat. If in the course of a chemical or physical process the products have greater change in their degrees of freedom or Entropy (∆S) then the entropy increases in the system, thus +∆S is a favorable condition, while a –∆S is an unfavorable condition that lead to less ability of molecules or ions to dissipate heat with less degrees of freedom.
∆G = ∆H – T∆S
If the entropy of the universe increases then ∆G negative and the chemical reaction is SPONTANEOUS! THERE IS A PATHWAY AND FREE ENERGY (∆G) decreases as it is being released so that you can do work without a constant supply of energy!! (riding a pony or a kayak moving with the current).
Hand Warmer Demo: The crystallization or precipitate must give off heat to be spontaneous!!
Entropy Driven!!! The favorable entropy (-∆H) can overcome the unfavorable (–∆S) if enough heat is released.
–∆H = the hand warmer reaction releases Heat to the surroundings!
–∆S = a solid is being formed and solids have less ability to move than other
phases of matter.
∆G = –∆H – T –∆S
∆G can be negative = if ∆H is – T to overcome
large enough the unfavorable
negative value negative value
Sometimes Rule ∆G = (-) – T (-)
favorable unfavorable
∆G = (-) + T (+)
So as long the ∆H is negative enough to overcome the positive value at the end the ∆G will be negative (which means the reaction increases the universes entropy and the reaction is spontaneous!). If that is the case like in the hand warmer demo below than the reaction is spontaneous and is ENTHALPY driven!
And so any solidification or crystallization MUST be exothermic. Ask Gavin A if his arm got warm when they put a cast on his arm last year? When it snows the air gets a bit warmer! Freezing is exothermic. Cement solidifying in exothermic!
4. Thermodynamics test practice – We will have a Thermodynamics Test on Thursday 3/19
It will look very much like the Review Test you will take today.
You will receive an email with the test at 9:30 – and it will close at 9:55!
You will do your work on a separate piece of paper and I will give you time to take photos of your work and paste into the doc.
Please plan accordingly!
I have posted a blank example of todays classwork (Review Thermo Test) – I know many have trouble printing at home so use your computer to view and write your answers on a separate piece of paper. Please take out you AP reference table and calculator.
Give yourself 25 minutes only!!!!!!
(Blank)
(Key)
Test Topics: (Thermo Review Test is a Great Review!)
A. Calorimetry – thermo 1 calorimetry key 0809.pdf
B. Hess law with – Heat of Formation – Thermo 3 heat of formation.pdf
C. Hess Law with Reactions – Thermo 4 – Hess Law ditto key.pdf
D. Hess Law with Bond Energies – Bonding 6 – Bond energies.pdf
E.
∆S from absolute entropies – T
hermo 7 – Gibbs Free energy Free Response 1984 key.pdf
F.
∆G calculations –
thermo 6a – Enthalpy, Entropy , Gibbs of Glucose methane.pdf
thermo 7 – Gibbs Free energy Free Response 1984 key.pdf
3/17 – Tuesday Homework –
1. Please view the lecture below on the derivation. I am asking that you follow along and try to pick up the major understandings of the derivation.
2. Complete the form that reviews the lecture.
3. Review the practice Thermo test –Test on Thursday-
Nonstandard G derivation formula:
Derivation lecture FORM:
End of Tuesday..
3/18 – Wednesday – We are having a live meeting today! At about 9:30 am!
*Please if you can watch today’s lecture with your graded form from last night before we meet.
I know its early but I am tall.
1. I will take any questions that you are have. I will email you the following information to invite you. As of right mow you will have to use your home computer, phone, or tablet device to connect with me as you do not have admin access to your computers.
LET’s Try with school computers first as I have been told that there is a option to use a browser instead of the app!
I am using Zoom to web conference with you and you will be prompted by your device to download a free app to your device to connect. I am working on getting you access through your school computers. The reason I am using Zoom is that I think it is superior in its quality of audio and its video that it broadcasts. We will see how it works today.
We are having a live web meeting today-
Mr. Grodski is inviting you to a scheduled Zoom meeting.
Topic: AP Chemistry – Thermo 1
Time: Mar 18, 2020 09:00 AM Eastern Time (US and Canada)
Join Zoom Meeting
https://zoom.us/j/805247577?pwd=Q0x3bFo2b056cGNWeEhSVFIrTG5rQT09
Meeting ID: 805 247 577
Password: 017905
One tap mobile
+16465588656,,805247577# US (New York)
Dial by your location
+1 646 558 8656 US (New York)
Meeting ID: 805 247 577
Find your local number: https://zoom.us/u/ackiT22Vle
Today’s Notes:
Okay we have seen that that changes in concentrations (molarity) of solutes in a solution or partial pressures in a mixture of gases can gauge or measure the change in a systems entropy (∆S) from nonstandard values we get from tablature. This means the ∆S0 (from standard conditions) can be converted into ∆S (from nonstandard conditions) by how much the concentration of molarities change from 1M (for c) or 1 atm (for p) (standard conditions).
S = S0 – R ln c or S = S0 – R ln p
We track how much the c or p changes from 1 by using the reaction quotient (Q).
S = S0 – R ln Q
Q = [Products]x
[Reactants]y
Since the systems entropy is connected to overall entropy of the universe in the Gibbs free energy equation:
∆G0 = ∆H0 – T∆S0
When we substitute the nonstandard entropy (S) into the above equation and simplify we get an equation that measures nonstandard ∆G!!! We are not bound by standard conditions no more!!
∆G = ∆G0 + RT ln Q
This means that we can evaluate chemicals reactions spontaneity based on a evaluation of the what the current concentrations or partial pressures are (Q). Changing the concentrations of the products and reactants changes the ∆G or spontaneity!
This is the equation of Le Chateliers Principle!
This equation is your reference tables:
1. Listen to Today’s Class Lecture:
Based on Last nights Form review – Non standard ∆G
I am using the files below in the video to review:
Tomorrows Test will only have a bonus question from this material.
2. Review the Thermo signs Form with the key that will be emailed to you.
I will try to accomplish this Live:
3. Nonstandard ∆G with equilibrium derivation!
Keq vs Q
4. Le Chateliers Principle returns!!!
– What are really the Shifts?
5. Complete/ review the THermo Test 1 – 7a – THermo Review Test.pdf
(Blank)
(Key)
3/18 Wednesday’s Homework –
Here is a link to today’s meeting:https://zoom.us/rec/play/tJArcOihrz03HNWcsASDCv4qW9W-Jqys0HIXqfMKyUvnWyYDZ1GnZeEXYLFsRFUMsNbwrQMygF11szNQ?continueMode=true
1. Study for a test on Thermodynamics. Topics Posted under Tuesday.
There is no topics on the test that we covered today/or last night. The homework below is based mostly on today’s material but you will get some overlap that will help, especially calculating ∆Gfº from tablature.
2. Complete thermo 6 – Gibbs Free Energy 1617.pdf with me with another lecture that is posted below.
Wednesday’s Homework Lecture:
End of Wednesday….
3/19 – Thursday – Thermo Test – 10:00 am – 10:40 am
The Practice test key had some issues and questions:
Test – 10:00 am – 10:40 am
1. Thermo test – 1 period test (timed) – (only concepts from the practice test)
*Please have you calculator and AP Chemistry Reference Tables handy! Do not look for them AFTER I start the timed test!
You will have to join the live web meeting during your test. That way I can monitor you during your test so that you are not doing it as a group.
You will get a shared doc with test questions on it. You are to write your work on a separate piece of paper. When the test time is over, I will give you 10 minutes to cut and paste your work into the shared doc. Use your phones to take CLEAR pictures of your work and paste them into the correct place of the shared doc. Once that is complete and I have word through Zoom that you are finished taking photos of your work and pasting them into the google doc I will lock the test. You will not have a lot of time to copy and paste for obvious reasons,
The purpose of the test is to give you experience of a part 2 – free response question AND evaluate your New Thermo skills. This is really an exercise that will help you prepare for the AP in May.
1 Thermo test – 1 period test (timed) – (only concepts from the practice test)
2. Derivation of ∆G formula that ties spontaneity with equilibrium (Keq).
Short lecture below!
Today’s small lecture from a tall teacher:
3/19 – Thursday Homework –
1. Read the posted Notes that are in the separate page in 3rd quarter:
2. View the in class lecture (up to question 27:00 min mark only) about how ∆G, Keq, and Q are connected. (Le Chateliers Principle!!)
3: View the tank demonstration
4. Complete the form below.
5.Complete the both sides ofEquilibrium 1 – Gibbs.pdf worksheet and review with key. Careful with the math. If you are not accustomed to working with natural log please use the key to help. The calculator does all the work really and I will explain tomorrow if the math gets you down.
Lecture for the form(up to question 27:00 min mark only):
Now Read:
Ok so we should understand that the larger the Keq the more more negative the ΔG and the smaller the Keq the more positive ΔG is.
This is intuitive but it relates to the following formula:
ΔGo=−RTlnK
SO a very large K value (really a Q at equilibrium position) will make the ln K part of the equation VERY LARGE and thus that would make the – RT ln K become a LARGER negative value.
ΔGo=−RTlnK
ΔGo=− value = VERY Spontaneous!
Why? Because the equilibrium position favors the formation or production of products MUCH more than the reactants. If the products are favored so much more than the reactant than the reaction HAS A PREFERRED PATHWAY TO MOVE FORWARD! Isn’t that what spontaneity is!!!!!!!
If the K value is small the reactants are favored and the REVERSE reaction in more favored and a small K value (which would be a number less than 1 would make the ln (number less than 1) = a negative number.
ΔGo=−RTlnK
ΔGo=− ( negative value due to the log of a number less than 1 = – #)
ΔGo=− ( negative)
ΔGo= positive and non spontaneous in the forward reaction
So if your not very good in the forward direct you become good in the reverse and that is why the reactants are favored in small K values!
Hey that formula is in our reference tables!!! In Blue?
Form:
Tank Demonstration:
End of Thursday….
3/20 – Friday – We will meet at 10:00 am today to review the test that will be emailed back to you
this morning.
1. Please review last nights form, and make another submission if needed. The average of the 2 grades will be
entered into Powerschool.
2. THermo test review: live
I graded it based on a AP grade and a SCALED percentage score:
3. Please watch both lectures:
Today’s lecture (#1):
Todays lecture (#2) ( 27:00 – to the end):
3. ∆G and equilibrium (Keq), Completion to equilibrium reactions
– thermo google slides – (slides 57 – 59)
5. Le Chatelier’s Principle (Q vs. Keq) with solutions, Cobalt complex Lab Demo –
6. Le Chateliers’s Principle with Heat, N2O4 ——> 2 NO2 demo
a) Heat changes the Keq position.
b) Derivation
We know that increases and decreases of Temperature CHANGE Keq position or value. An Endothermic Reaction is favored in higher temperatures and an Exothermic Reaction is favored in lowered temperatures. We understand this by setting our standard ΔG formulas together and looking at the ln K equals (specifically at the ΔH part of the equation).
Cobalt Complex Le Chateliers (Regents Chemistry) Lab:
Thermo google slides: (slides 57 – 59)
3/20 – Friday – Homework
1. Please complete the Le Chateliers Principle Side 1 and review with the key. Think while you are doing the worksheet, “What are we doing with Q in relation to K? “When does Keq move in relation to Q?
THis worksheet should look familiar!!!!
2. Please use the worksheet below to follow along with me in the lecture below. ONLY THE FIRST SIDE!
THis will make perfect sense if you are on it and have been fighting through all my forms!
3. We will do only the first side!!! The second side is the Quadratic catastrophe that we will discuss
tomorrow.
4. Complete form below:
To be posted…
Equilibrium Ice table / intro to KSP / Q vs K in action:
I made mistake in question 2 (although the answer of the Ksp is correct):
I should have (1.3 x 10-4 /2) in the value for the equilibrium concentration for the CrO4-2 ion in both the ice table and inside the formula for the solving of the Ksp value.
I made a mistake (imagine that!) on number three. In the ICE table in for the second reactant I should of calculated 1.06 x 10-3 not .065 x 10-3. THanks Caroline!!!!
The final Keq is calculated correct but I had the wrong value there.
Weekend form:
Please Read before completing:
So what does all of this mean?
When the temperature increases the Endothermic reaction (pathway) is favored.
So when the temperature increases the endothermic pathway or direction will become more spontaneous.
This means that a reaction will “shift” towards the endothermic reaction when the temperature increases BECAUSE the Keq (K) CHANGES!!! The equilibrium constant is sensitive to temperature. The “shift” occurs BEcAUSE the Keq changes. This is unlike all other stresses.
In the demo with the NO2/N2O4 tubes we figured out that the chemical reaction looked like this:
2 NO2(g) —-> N2O4 (g) + Heat
Brown clear
The reaction shifts to left when we heat the tubes AND we saw this in the lecture. The tube became DARKER when heated because When The Temperature Increases the Reaction Shifts to the Endothermic Pathway which is in the Left direction. In this case, the Keq became smaller with increasing temperature.
**Temperature increases ——> Shift in the Endothermic Direction (in the left direction above)
And when considering an exothermic reaction with a ΔH= negative value
**Temperature decreases ——> Shift in the Exothermic Direction (in the right direction)
which is why the tubes get clearer when placed in ice.
Weekend Form:
End of week 7!
This not your homework: