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   Q3 : Week 7 – 3/20 – 3/24

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Main focus –                                                                                                                                                         
                                                 
a) Review Heat of dissolution and the IMF’s formed.     
 
b) Review heat of Fusion and Heat of Vaporization as well as cooling curves
 
c) To define entropy (∆S) and the second law of thermodynamics.

d) To derive Gibb’s Free energy (∆G) and how it measures the ∆S universe by using                   system’s  values

 

Period 2/3a, 4:  

 
1. Lets review the Free Response Question from the homework:
 
AP Chem 2016 ques 1 AP Key.pdf
 
2.  Homework Form Review:
 
Heat of Dissolution, Bond Enthalpy Form – Key- P.pdf
View Download
        

3. Review of the 2nd FRQ

2022 FRQ Question question 1 AP Central Key.pdf
View Download 
 
2022 FRQ Question question 1 Grodski Key.pdf
View Download 

a) Micro lesson on Specific heat and heating and cooling curves problems.

4.  Today we start our discussion towards the Second Law of Thermodynamics: No one got to this in class.

      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                                                                                                                    
Thermo AP Tables.pdf
View Download                        

   

_____________________

TODAY’s NOTES: 

 
1. Review of some enthalpy of reaction basics that was demonstrated in your homework: 
 
                                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.
 
                                                                   It’s the same if I ask for the decrease in temperature. You still give                                                                                                     me a positive temperature.

 

1. Review of Laws of Thermodynamics: 

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 process has a pathway that requires no continual input of energy.  Example: Riding a pony downhill.  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 the 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 = coupling.  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).
 

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

 

                                                                       ∆G = ∆H – T∆S!!!!!!

 

Thermodynamics: The end of it all!!!

Ion Dipole or Molecule ion IMF of a Salt Solution- p can waters compete with the Ionic Bonds of the ions in the crystal lattice.  Is the Lattice energy too large?  

Heating Curve Demo- The potential Energy changes at the phase diagrams can be measured by enthalpy!

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3/20 – Monday’s Homework: –

 

STEP #1 (all classes)

1.  a) Please View Lecture 1 above From 17:27 – to the end – LAWS OF THERMODYNAMICS!
 
      b) Please view the complete Lecture 2 above.
 
      c)  Please complete the thermo 6a – Enthalpy, Entropy , Gibbs of Glucose methane.pdf worksheet (side 1) with me using the posted video below:
                                         
   We will now be using the following formulas:
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
thermo 6a – Enthalpy, Entropy , Gibbs of Glucose methane.pdf
View Download
 
thermo 6a – Enthalpy, Entropy , Gibbs of Glucose methane Key.pdf
View Download 
 
Thermo 6a worksheet lecture: 
 
2. 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.
  
Thermo 7 – Gibbs Free energy Free Response 1984.pdf
View Download
 
Thermo 7 – Gibbs Free energy Free Response 1984 key.pdf
View Download
 
3.  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.

 

 3: The signs lecture: 

 
3 : THermo Quantities and Sign Form 
      Thermodynamic Sign form –  due Tuesday Morning 4:30 am:

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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.

 ________________________________________________________________________                                 Jump to: Tuesday Homework / TOP

3/21 – Tuesday – B Day – 2, 3b/4 Lab

Main focus –                                                                                                                                                         
                                                  
a) To define entropy (∆S) and the second law of thermodynamics.

b) To derive Gibb’s Free energy (∆G) and how it measures the ∆S universe by using                   system’s  values

c) To identify the values of  ∆G in terms of a spontaneous and nonspontaneous                          process.

d) To identify an entropy or enthalpy driven reaction.

e) To calculate the ∆G and ∆S through tablature.

 

Period 2,3b/4:  

1. We will review Thermo Quantities and Sign Form at the end of the lesson.
 
THermo Quantities and Sign Form Key p.pdf
View Download
 
2.  We will review our Thermodynamic quantities, ∆G, ∆H, and ∆S, especially the Second Law of Thermodynamics!!
 
    a) 2nd law of Thermodynamics: ∆S universe must increase!
   
    b) Derivation of ∆G = which is really a measure of the ∆S universe.
         – The cool thing is that we can relate the ∆S universe by relating the 2 system’s quantities:
           ∆H system and ∆S system .
    
    c) ∆S system  = the difference of the absolute entropy of molecules in the chemical reactions from reactants                                    to products.
                                   
     d) calculation of ∆G from equation or tablature.
 
     e) Always Rule, Never Rule, Sometime Rule (Enthalpy driven or Entropy driven)

     f) Values of ∆G, demos 

Thermodynamics: The end of it all!!!

_____________________

TODAY’s NOTES: 

 
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 at constant pressure and temperature.
 
The 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 – ENTROPY:
– 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.
 
                                                                   
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). 
                 And according the 2nd Law of Thermodynamics the ΔSuniverse must increase if there is a             
                                         pathway for the process to proceed = spontaneous reactions.
 
                                                       The ΔSuniverse is measured by ΔG (Free Energy).
     
The Entropy  ΔSuniverse  is affected by two factors:   
There are two ways that energy can be “dispersed” or released.
            1.  Heat or thermal energy is released into the surroundings. ΔH = heat of reaction
                    
                    This thermal energy will increase the temperature of the surroundings and increase the motion of
                    surrounding molecules. This increased motion will make these surrounding molecules “move away”
                    and disperse their motion energy “away” from the system.
                    ΔH = negative if HEAT energy is released – (increases  ΔSuniverse )
                   ΔH = positive if HEAT energy is absorbed – (decreases  ΔSuniverse ) 
 
           2.  The degrees of freedom of the particles or molecules increases. ΔS = entropy of particles
 
                   The chemical reaction or physical process increases the number of moving “parts” so that motion
                    energy is dispersed as the chemical or physical process proceeds.
                    ΔS  = positive if degree of freedom increases for molecules as process proceeds.
                   ΔS  =  negative if degree of freedom decreases for molecules as process proceeds.
 
To measure a if the Entropy ΔSuniverse is decreasing or increasing we use Gibbs Free Energy Formula:
                                                                                             
                                                                               ΔG =  ΔH – TΔS
                         
                         IF ΔG = negative then the Entropy ΔSuniverse  is increasing = Spontaneous
                      IF ΔG = positive  then the Entropy ΔSuniverse  is decreasing = Non – Spontaneous

 

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!

  

Sometime Rule Demo: Hand warmers

Whoosh bottle Demo Notes:
 
Once the I have the alcohol in the gas phase, which has now mixed with the oxygen in the air inside the container I ignite the flammable mixture by providing a spark and the combustion reaction begins. This is an example of an spontaneous reaction that releases energy exothermically, (exergonic reaction).   Here is the chemical change that takes place:
 

                                     2CH3OH(l)   +  3O2(g)  →. 2CO2(g) +   4H2O(g)  + Energy                                                                                                                                           Alcohol                Oxygen          carbon dioxide       water                  

                                                                        

                                                                                        (l) = liquid, (g) = gas

 

Notice the Coefficients that balance the reaction due to the Law of Conservation of Mass.  We have the same number of Carbon (C), Hydrogen (H), and Oxygen (O) atoms on both sides. This is chemical change due to the fact that there are new substances (compounds) produced which means bond have been broken and reformed.

 

Notice that initially 3 gas molecules (3O2(g)) are used with 2 liquid molecules .

These are called Reactants as they on the left side of the chemical reaction and this is what we start with before the chemical reaction begins.
 

The products (2CO2(g) +   4H2O(g) ) are the chemicals that are on the right side and they include 6 gases. So 3 gas molecules and 2 liquid molecules become 6 gas molecules .  

The entropy is increasing and the chemical reaction is spontaneous. 

 

Uplifting COOL DEMO NOTES:

 

 

     

Whoosh Bottle Demo: (always rule!)

A COOL Demo!: (Entropy Driven!)

3/21 – Tuesday’s Homework: – 

 

1.  Please complete the FRQ as a test to hand in tomorrow: 

THermo Test 1 – 7a – THermo Test 1.pdf
View Download

2.  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.

 
I used the following notes to derive this:
 
Thermo 9 Notes for nonstandard.pdf
View Download
 
3.  Complete the form that reviews the lecture.   
  

 

 

2: Nonstandard ∆G Derivation (this gets us to equilibrium!!!!)

 
3 : Nonstandard delta G derivation Form 
      Only one submission! –  due Wednesday Morning 4:30 am:

Please use Private email accounts as there seems to be some new restrictions that are not allowing whbschools accounts to send graded forms back. 

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Additional Resource: Classroom lecture reviewing Last nights Form:

____________________________________________________________________________                     Jump to: Wednesday Homework / top

3/22 – Wednesday – A Day – 2/3a Lab, 4

Main focus –                                                                                                                                                         
                                                  

a) To derive the non-standard ∆G equation

b) To write the Law of Mass Action (@ equilibrium, Keq and when not at equlibrium Q)

c) To calculate Gibbs free energy with tablature.

    

Period 2/3a,4:  

1. Review Yesterdays Demo, especially the “uplifting demo” in terms of the signs in the form and the ∆G                 changes and that we could also calculate the Standard ∆G using tablature also.

 

2. Collect the take-home -review AND REVIEW with the key. 

3. Derivation Review of the NON-Standard  ∆G.

Thermo 9 Notes for nonstandard.pdf
View Download

NONstandard Form most wrong 2019 form p.pdf
View Download

 
4.  Nonstandard  ∆G with equilibrium derivation!
 
      Keq vs Q 
 
5. Le Chateliers Principle returns!!!
 
– What are really the Shifts?
 
LeChateliers principle worksheet.pdf
 
LeChateliers principle worksheet KEY 2012 p.pdf
View Download

                                                              

 

Thermodynamics: The end of it all!!!

   _____________________

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:
     

 

Nonstandard Gibbs Free Energy Lecture with form review:

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3/22 – Wednesday Homework:

1.  Please make one more submission to last nights form.
 
2. Complete thermo 6 – Gibbs Free Energy 1617.pdf with me with another lecture that is posted below. 
 
thermo 6 – Gibbs Free Energy 1617.pdf
View Download
 
thermo 6 – Gibbs Free Energy key.pdf
View Download

 

3. Complete the Thermo 2 test.This is due tomorrow. 

THermo Test 2 .pdf
View Download

 

 

2. Non-standard Gibbs Free energy formula review:

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3/23 – Thursday – B Day – 2, 3b/4 Lab 

Main focus –                                                                                                                                                         
                                                 
a) To determine the difference between Q and Keq 
b) To compare and contrast Q vs K in shifts in equilibrium.

c) To identify changes in equilibrium at different temperatures that cause shifts.

 

Period 2,3b/4:  

Key to last nights Thermo 1 Test –
THermo 7a – THermo Test complete KEY.pdf
View Download

 

1.  Collect Thermo Test 2 and Review with key.
2.  Return Thermo Test 1

3 . Review of the last 2 questions of Tuesday form – 

NONstandard Form most wrong 2019 form p.pdf
View Download

 
4. Complete – Derive the Standard ∆G formula with Keq (equilibrium constant)

     ∆G formula that ties spontaneity with equilibrium (Keq).

5.  Using our new formula of ∆G that relates to Keq (equilibrium constants):

The value of ∆G depends on the value of the Keq. 

  LeChateliers Principle!

 Demo of the NO2 tubes:
 
LeChateliers principle worksheet.pdf
 
LeChateliers principle worksheet KEY 2012 p.pdf
View Download

 

_____________________

TODAY’s NOTES:

 
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 direction 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?
 

 

Today’s Lecture Non standard ∆G to Keq

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3/22 – Thursday Homework: –

                                                                                                                                                                                                                               

*Period 3/4 does not have to do 1 and 2!

1. Read the posted Notes that are in the separate page in 3rd quarter:

https://mrgrodskichemistry.com/non-std-gibbs-equilibrium-notes/

 
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 of Equilibrium 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. I did review this in class.
 
*Also these are old FRQ’s that use Kilocalories instead of Kilojoules. That is an older heat unit. It does not change anything. You can pretend Kilojoules is Kilocalories. 
 
Equilibrium 1 – Gibbs- 1819.pdf
View Download
 
Equilibrium 1 – Gibbs- 819 Key p.pdf
View Download
       
6. Complete the FRQ that was handed out. It will be collected.   
 
AP Chemistry – FRQ question 1 practice                                                                                                            View Download   

 
 

  .

2: Lecture for the form(up to question 27:00 min mark only):

3: Equilibrium Tank Demo- 

  
4 : Equilibrium , Q Gibbs….. Form –

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__________________________________________________________________                                   Jump to: Friday Homework / top      3/23 – Friday – A Day – 2/3a Lab, 4

Main focus –                                                                                                                                                         

   

a) To identify the shifts in Le Chateliers Principle are actually the spontaneity changes that occur when you compare Q vs. Keq.

b) To recognize that Temperature stress on a reversible reaction actually moves the equilibrium position.  Equilibrium temperature sensitive.

c) To identify the basic concepts of equilibrium.

d) Demonstrate the common ion effect,  and temperature changes that affect pH.

Period 2,3a/4:  

1.  Slides 61 – 62 – to review the shifts due to Q vs, K and the size of the Keq with size of the  ΔGo

 

2.  LeChateliers Worksheet – classwork 
 -Focus here is to discuss the “shift” with temperature or the repositioning of the Keq position. 
 
LeChateliers principle worksheet.pdf
 
LeChateliers principle worksheet KEY 2012 p.pdf
   
      a) What are the shifts?
      b) Treat heat as a reactant or products
      c) pressure factor 
      b) common ion factor – backside 
 
            Fizz keeper SCAM!!!!!!!!
 
3.  Q vs. K demo with NO2 Tubes
       a) Shifting of Keq position
       b) Increase in Temperature favors Endothermic position
       c) Decrease in temperature favors Exothermic position 
 
4.  Cobalt complex Lab Demo – 
 
5.   pH, equilibrium, and the common ion effect.
 

 

 Cobalt complex Lab Demo – 
Cobalt Complex Le Chateliers (Regents Chemistry) Lab:
 

 

_____________________

TODAY’s NOTES: 

 
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).
 
Please Read before completing Homework:

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.

Review of the Last nights form – questions 1 – 5- 

Class lecture (From 27:00 min and to the end):

∆G Derivation to determine how Temperatures affects an Exothermic or Endothermic  

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3/24 – Weekend 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 and was given out earlier this week.

You have completed this worksheet in September. Remember that with pressure you need to count the moles of the gas to see if your shift needs to increase the pressure or decrease the pressure.
 
LeChateliers principle worksheet.pdf
 
LeChateliers principle worksheet KEY 2012 p.pdf
View Download 
 
2.Please use the worksheet below to follow along with me in the homework lecture below.
    We are ONLY working on THE FIRST SIDE!
    This will make perfect sense if you are on it and have been fighting through all my forms!
 
Equilibrium 3 – Equilibrium constants 1.pdf
 
Equilibrium 3 – Equilibrium constants 1 key.pdf
View Download
 
3. Complete form below:  It will be posted later..
 
Equilibrium Ice table / intro to KSP / Q vs K in action:
 
 Homework Video Disclaimer:
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.  Thank you Caroline!!!!
The final Keq is calculated correct but I had the wrong value there.
IN SPITE OF ME!!!!!!!! 

Homework Lecture: Equilibrium Basics

  
3 : LeChats.. Form –  Fix #3 and Fix #7 key

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