Archive – Q3 week 5 – 18-19

week 5

Week of 3/4 – 3/8 – Always REFRESH this page!

3/4  – Monday – Period 2

Key from Weekend’s Form
Hess Law lab form 1 key complete p.pdf

1: Calorimetry and Hess Law Lab – Complete take the L worksheet.
2. Enter into the form.
3. Start data collection – start with Ccal – Complete tomorrow.
Period 3/4-

1: Calorimetry and Hess Law Lab – Complete take the L worksheet.
2. Enter into the form.
3: Hess Law – complete lab data collection for all of the 3 reactions and the Ccal.

Lab packet:
Calorimetry and Hess law p.pdf

1:  Anyone who did not get a 100% on the Pre-lab calculations will not start lab until you redo the assignment and get a 100%. Some of you made small errors.  I made all of you redo it with a new version (“take the L”).

Those of you who did not do the assignment, for whatever reason, will have a zero for the assignment.

2. You are complete the lab with your lab partners to complete the Lab. You are to use the lab station computers with temperature probes and Logger Pro.

a) Your  experiment should be set to take temperatures at 5 sec intervals for about 180 seconds. You will need to make a regression line of the best linear part of your line and determine the y-intercept, which will be = Tmix.

b) Once you complete the Calibration of the calorimeter, (determine Ccal), you will need to complete 3 trials using 3 the different reactions and calculate the H of each reaction (just like the HW).

VERSION 2 – For those who did not did not get 100 or did not complete the homework – This will be completed by all students before we start the lab.  I will take the best out of 2 grades for those that did the assignment over the weekend.

I use this worksheet to model the problem this weekend.
Hess Law Lab Pre lab for lecture USE.pdf

I use this worksheet for the new form (“take the L” ) below:
Hess Law Lab Pre lab Version 2 – take the L p.pdf

#### Hess Law Lab Calculations Form Version 2 – take the L

3/4  – Monday Homework:  HESS LAW with Reactions!

Period 2:
1: Watch lecture on Hess Law (from 8:00 on!)  and complete with me the Hess Law problems that I model from the thermo 4 – Hess Law ditto.pdf worksheet.
2:  We will apply this to our lab tomorrow.
thermo 4 – Hess Law ditto.pdf

thermo 4 – Hess Law ditto key.pdf

Period 2/3:
1: Watch lecture on Hess Law (from 8:00 on!)  and complete with me the Hess Law problems that I model from the thermo 4 – Hess Law ditto.pdf worksheet.
2. Complete your Hrxn for each of the three reactions from your Lab packet using today’s data.

3. We will apply this Hess Law to our third reaction Tomorrow!
Homework Lecture from 8: 00 min mark on!
I derive the H in the first 8 minutes.
We defined H (enthalpy) Friday!

Is it Entropy???? NOOOOOOO its Enthalpy!!!!

3/5  – Tuesday – Hess Law – The interconnectivity of ΔH

Todays notes: Remember that we only obtain qrxn values from calorimetry and it is equal and opposite in value from the qcal.   qrxn = – qcal   Since most reactions occur in constant pressure conditions (NASA – Grade calorimeter) then qrxn values at constant pressure are equivalent to Enthalpy. qrxn = ∆H    ∆H is a state function that depends only on final conditions – initial conditions AND NOT THE pathway! Because ∆H is a state function then we can use its interconnectivity to solve for other ∆H values. In the simple example to the left if we know the direct distance from North Riverhead to Westhampton and the direct distance from Westhampton to Northhampton we can use both values to determine the Overall change of distance from North Riverhead to Northhampton.   The interconnectivity of the state function ∆H is called Hess’s Law!   This law allows for chemists to calculate ∆H values without doing calorimetry that might be too impractical or expensive.
We learn today that this interconnectivity works because all chemicals are formed in reactions that start from the same baseline that I will call a free atom zone.
The ∆H for formation of chemicals from their atoms (free atom zone) is called Heat of Formation, Hf .  The sum of all the Hf of the reactants equals the starting position in an energy diagram or in the case of the example above the starting position of the North Riverhead.
We will see how changes in the Heat of Formation of compounds (Hf0) can be connected to the overall change of the Enthalpy (H) of an entire reaction.  The power in this connectivity called Hess’s Law allows us to calculate the  Hrxn  from just a table of Hf(heats of formations) without having to do any calorimetry.  We need to keep in our heads that the reason for this interconnectivity is due to the using state functions (H) and that all Hfbegin from the same baseline of (elemental states).  This baseline leads us to the following formula that is posted in your reference table: Since ∆Hrxn is a state function we are not tied to just one way or pathway to calculates the change in Enthapy (∆Hrxn) of a chemical reaction.

Reactions in our lab:

1.                           HCl (aq)  +  NaOH (aq)       —>    NaCl (aq)   +   H2O (l)

Net Ion:            H+ (aq)    +    OH  (aq)     —->     H2O (l)

2.        NH4Cl (aq)  + NaOH (aq)    —–>   NH3 (g)   +     NaCl (aq)  +  H2O (l)

Net ion:           NH4+ (aq)  +  OH–  (aq)   —->  NH3 (aq)  +  H2O (l)

3.                                       NH3 (aq)  + HCl (aq) —-> NH4Cl (aq)

Net Ion:                 NH3 (aq)   +  H+ (aq)  —->   NH4 (aq)

Theoretical Values
1.                              HCl (aq)  +  NaOH (aq)     —–>    NaCl (aq)   +   H2O (l)                            ∆H = -55.9 kJ/mol

2.                         NH4Cl (aq)  + NaOH (aq)     —–>   NH3 (g)   +     NaCl (aq)  +  H2O (l)     ∆H = -3.7 kJ/mol

3.                                       NH3 (aq)  + HCl (aq)  —-> NH4Cl (aq)                                                  ∆H = -52.2 kJ/mol

Can you use the first 2 reactions ∆H values to obtain the third by manipulating the first 2 reactions? Is Hess Law supported?

When you rearrange and cancel:

1.                              HCl (aq)  +  NaOH (aq)     —–>    NaCl (aq)   +   H2O (l)                            ∆H = -55.9 kJ/mol

2.    flipped               NH3 (g)   +     NaCl (aq)  +  H2O (l)  —-> NH4Cl (aq)  + NaOH (aq)            ∆H = +3.7 kJ/mol
_____________________________________________________________________________
3.                                       NH3 (aq)  + HCl (aq)  —-> NH4Cl (aq)                                                  ∆H = -52.2 kJ/mol

first reaction    +     second reaction        =   third overall reaction
∆H = -55.9 kJ/mol  +  ∆H = +3.7 kJ/mol  =   ∆H = -52.2.9 kJ/m

Hess Law is supported!

So ∆H of reactions that cancel out to give the overall reaction clearly support the interconnectivity of Hess’s Law but the question remains why?

∆H is a state function yes but why does each reaction have a unique starting position  What explains its starting value?  We know the difference between the reactions creates an overall ∆H for the overall reaction as in the 3rd reaction above but it does not explain the positions in the energy diagram that creates the interconnectivity.

For instance from today’s Lab we get the following energy diagram for the 1st reaction using heats of formation values from the Thermo tables:

Heats of formation, ∆H0 or the enthalpy change of heat that occurs when a compound is made from its elements.  Every compound is made from elements combining to form bonds between atoms thus every compound in EVERY chemical reaction is contains a certain potential energy associated with how much energy is released if forming that(those) bond(s).

For example the ∆H for HCl that is one of the reactants in reaction 1 from our lab is represented by the following synthesis (composition or formation) reaction:

H2   +      Cl2   —>     2HCl

∆H  are always posted in tables in kJ/mol ( kilojoule per mol ).  Since the energy posted to form HCl from its elements is always PER 1 MOLE we reduce the above reaction to produce 1 mole of HCl.  Thus the reaction now becomes:

1/2 H2   +     1/2 Cl2   —>     HCl

When we look up the ∆H for HCl in thermodynamic tables we obtain ∆H = -167.2 kJ/mol.  The thermodynamic tables do not provide the reaction but only the producthus you need to be aware what the table is implying the energy change from the formation of the listed chemical from its elements.

∆H = -167.2 kJ/mol  for HCl means that -167.2 kJ/mol of energy is released when elements combine to form 1 mole of HCl.  Its negative because the energy is released as bonding of the elements creates increased stability.
Notice the arrow moving down in the diagram below for Every chemical in the reaction for reaction 1. Some move farther down than others because some compounds release more energy when they are formed WHICH MEANS THEY ARE MORE STABLE and have LOWER potential energy than the chemicals that do not release as much energy when they are formed.

So the starting point for the reactants in a chemical reaction is the SUM of the individual reactants heats of formations Notice the Sum of all of the ∆Hf of products – the Sum of all of the ∆Hf of products = ∆H rxn

That is the new skill tonight: determining the ∆H rxn using ∆H f values.

END OF NOTES…

Period 2/3-

1: Complete data Collections/hand out Lab/ key to weekend homework.

c) Verify Hess LAW:  (We complete this complete this once we learn about Hess Law)

Using the Hrxn of  reaction 1 and reaction 2 determine the Hrxn for reaction 3 using Hess Law.  Compare the your value using Hess Law with the value you determined through calorimetry.

Take the first 2 reactions and manipulate /cancel to get the overall reaction to be equal to the third reaction.
Use your Enthalpies to calculate the 3rd reaction (just like the worksheet.)  Compare it to the value of the Enthalpy that you directly obtained through calorimetry.
Show all work/ calculations/ Hess Law / Percent error and Hand in.
2. Heat of formation – (intro and with demo)
a) Discussion of reaction of Demo to discuss what Heat of formation (Hf0of NaCl.
b) Thermo tables, elemental states, thermodynamic standard conditions.
c) Compared Hf0 of Al2O3 and NaCl, and discussed strength of Bonds or lattice energy.
d) Drew diagram of the Free atom layer, which is the reference point to show Hf0 of reactants and products of another reaction.
e) Used Hf0 to determine the Hrxn of NaCl ionizing into water and determining whether the water will increase or decrease in temperature.
F) Used Hfto solve for the theoretical value of the third reaction.
thermo AP Tables.pdf

thermo AP Tables additions.pdf
Period 4 –

1) Verify Hess LAW:  (We complete this complete this once we learn about Hess Law)

Using the Hrxn of  reaction 1 and reaction 2 determine the Hrxn for reaction 3 using Hess Law.  Compare the your value using Hess Law with the value you determined through calorimetry.

Take the first 2 reactions and manipulate /cancel to get the overall reaction to be equal to the third reaction.
Use your Enthalpies to calculate the 3rd reaction (just like the worksheet.)  Compare it to the value of the Enthalpy that you directly obtained through calorimetry.
Show all work/ calculations/ Hess Law / Percent error and Hand in.

2) Heat of formation – new concept

a) Discussion of reaction of Demo to discuss what Heat of formation (Hf0of NaCl.
b) Thermo tables, elemental states, thermodynamic standard conditions.
c) Compared Hfof Al2O3 and NaCl, and discussed strength of Bonds or lattice energy.
d) Drew diagram of the Free atom layer, which is the reference point to show Hfof reactants and products of another reaction.
3/5  – Tuesday Homework:

1. READ MY NOTES posted at the start of today to it says END OF NOTES…

2: Lab 21 – Hess Law Lab Activity is due tomorrow .
Non-formal lab requirements posted below.
3.  Complete questions 1 – 4 of the Thermo 3 heat of formation. pdf worksheet and review with the key.

Look at the highlighted reference table above!
3:  Please complete question 5 on the  Thermo 1 calorimetry.pdf worksheet ( you have it already) and question 5 and 7 from the thermo 4 – Hess Law ditto.pdf worksheet.
* in question 5 in thermo 4 – Hess Law – the overall reaction is the heat of reaction of Mg(OH)2
Thermo 1 calorimetry.pdf

thermo 1 calorimetry key 0809.pdf
thermo 4 – Hess Law ditto.pdf

thermo 4 – Hess Law ditto key.pdf

 thermo 3 heat of formation.pdf thermo 3 heat of formation key.pdf

LAB 21  requirements:

Objectives:

Part 1:
Show all calculations clearly and neatly.  If you have to rewrite them then do so. I will take off for a lack of neatness. Show all units and sig figs correctly. There are 4 separate parts to your calculations.

Part 2: (3 parts)
a) On a separate piece of paper write the three reactions and their H/mole values.
Please use reaction 1 and reaction 2 and manipulate them to cancel out to get reaction 3.
Just like you did with Monday’s homework in thermo 4 – Hess Law ditto.

Thermo 4 – Hess Law ditto key.pdf
b) Calculate what the H/mole of the third reaction should be using Hess’s Law (reaction manipulation)  Compare it with the H/mole of that you obtained through calorimetry in lab for reaction 3

c) Calculate the known by using Heat of Formation Tablature for the 3rd reaction. Complete a percent error based on this value as your known or theoretical and run a percent error on both of your experimental values (value of H/mole using reactions and then with H/mole from calorimetry.

You should notice that the reaction manipulation H/mole value of reaction 3 will be closer to H/mole value obtained through calculations with heat of formations.

Error Analysis:  There is one major error with all forms of calorimetry.

Staple calculations and reactions. Hand in tomorrow.

Hess Law with Heats of Formation:
End of Tuesday…

3/6  – Wednesday – Period 2-

I am focused on you being able to visualize the FAZ diagram that illustrates the ∆H values that determine stability in the potential energy (stored energy of chemical bonds)  by how much energy is released to form individual compounds in a chemical reaction.

Hess Law works because all compounds are tied to the same Baseline = F.A.Z.  (Free Atom Zone).  Every chemical in every chemical reaction has a unique potential energy because of how much energy is released in its formation from the same Baseline (FAZ)!   THAT is how ∆H’s are interconnected in reactions!!

To obtain a ∆Hrxn all you need to do sum all of the ∆Hf of the reactants you will obtain the total energy lost to form the reactants.  This will tell what potential Energy is REMAINING!!!!!!

Do the same for the products in the and compare remaining potential energy from the products and reactants AND YOU HAVE THE ∆H of the entire reaction.

THIS IS HESS LAW! Today we will learn that this idea allows us to apply Hess law to many all values of ∆H, which has many names
including:

Heat of dissolution, Heat of decomposition, Heat of Fusion, Heat of vaporization,
Heat of sublimation, Heat of combustion, Heat of …..
These are all ∆H values!!!!

*Today was about linking the concepts of heat of formation (Hf)  with heat of dissolution (Hsol)
Table I in the Regents Reference Tables lists ionization reactions and their Hrxn.  Salts that dissolve water are endothermic (lower temperature of solution) or exothermic (increase temperature of solution).
1:  Table I – Heat of dissolution of NaCl demo – temperature fell (H dissolution = +4.2 KJ/mol)
Complete discussion of heat of formation into stability of bonds and H dissolution.
2. Started the Phase change enthalpy.pdf

* I really linked the idea of potential energy changes in the heating curve at the phase changes so that we better understand the Heat of Fusion and Heat of Vaporization that is used in Regents Chemistry.

This is an important concept as ∆H’s can also be used to determine the energy changes if you have a certain amount of the reactant and you know it’s ∆H . For instance if we calculate the ∆H fusion of water or Heat of Fusion (remember ∆H has a lot of names) from ∆H values for the physical process of H2O (s) —> H2O (l) we get the amount of Energy needed to be consumed (or absorbed) per mole of H2O (s).   We can use this value to determine calculate amount needed to melt (fusion) of a certain amount of ice.  We do this for chemical reactions as well.

We can determine the ∆Hvap (Heat of Vaporization) the same way with ∆H0  values for the physical process
H2O (l) —> H2O (g) and use it value to measure the energy it takes to vaporize the water with given amount of water in the liquid phase.

Also it was good to review why kinetic energy gets converted to potential energy at the phase changes.
Phase change enthalpy.pdf

Today’s Lesson videos:

Period 3/4-

1:  Table I – Heat of dissolution of NaCl demo – temperature fell (H dissolution = +4.2 KJ/mol)
Complete discussion of heat of formation into stability of bonds and H dissolution.
3:  Phase change enthaply worksheet review:
Concepts: heating and cooling curve, heat of fusion,heat of vaporization
Phase change enthalpy.pdf
∆H = -55.9 kJ/mol
3/6 Wednesday Homework:  Mock Part 2 question of an old AP question – The question I am giving you represents a previous released part 2 question. Remember that the first hour an a half of the AP Chemistry Exam will be the multiple choice. After a 10 minute break (after the completion of the MC) you will begin the part 2 portion which is 1 hour and 45 minutes. In this section you will have 7 part 2 free response questions.  Questions 1- 3 are usually 10 point questions, while questions 4 – 7 are 4 point questions.  Tonights homework represents a question from part 2 that is a 10 point question.

1. Complete Free Response AP question. Please give yourself about 20 -25 minutes and try to entirely complete this question before looking at the key.

2.  Please review your work with the key posted below and grade your work based on the AP Central Rubric posted below.

Free Response Thermo, Hess, Calorimetry AP problem.pdf
This is the key to the above problem
2013 Part 2 – AP Key p.pdf

3. Period 3/4 students:  Please also complete the short form by the tall teacher that does not count but reviews the concept that we learned from the block calorimetry problem:

Short from for period 3/4 students:

#### Calorimetry Practice Form ‎(MC)‎

Optional
Old Class lecture on Heats of formation:
End of Wednesday..

3/7  – Thursday – *Connections – We can use heat of formations from our tables to calculate the Hrxn . The Hrxn has many names tha describe the reaction.  In this case we are looking at a ionization reaction that occurs in water (aq) and thus we call this change in enthalpy (H) the Hdissolution or Hsol (solution).

The Hdissolution for CaCl2 (s) is exothermic (negative H) and thus the solution temperature rises as heat flows from the chemical (CaCl2 (s) ) to the water.  This means that the combined ions that result from the dissolving of CaCl2 (s) ARE MORE STABLE in water than water H-bonding to itself.  This results in a release of energy!

Today we observed NaCl dissolving and it was endothermic (positive H) and thus the solution temperature decreases as heat flows water the chemical (CaCl2 (s) ) to the water. (this results in an absorbtion of energy or I like to say consumption of energy).  This means that the combined ions that result from the dissolving of NaCl (s) ARE LESS STABLE in water than water H-bonding to itself.  This results in a absorption or consumption of energy!

Period 2/3 –

1. Hess law truth – ∆H0  values are the GOAT’s in HESS LAW!
slide 30 – slide 32 in Thermo presentation-
(canceling reactants in multiple chemical reaction Hess law problem is
really cancelling ∆H values! )
2.  Phase Change enthalpy worksheet –

Reviewed the potential energy conversion in the heating curves through ∆H0  values (Hess Law).

∆H fusion and ∆Hvap (Heat of Vaporization) are determined through calorimetry or through Hess Law using ∆H0  values AND THESE ∆H values (STATE FUNCTIONS) are constants that are used to obtain the energy changes at the phase changes without a temperature change (q = m C ∆T).  We use ∆H this way as in our homework.

Calculated the ∆Hvap and ∆Hfusion posted in Table B of the Regents reference table.

3. Heat of dissolution/heat of formation review with CaCl2
Hot Packs/Cold packs
4.  HW review

5. thermite Demo steel spheres /video
6:  Bond Enthalpies – Demo – “bonds broken” – bonds formed“.

Bond Enthalpy – 2H2O2  -> O2 +  2H2O
determined Hrxn with heat of formation then with bond enthalpies
modeled question 1 on Bonding 6 – Bond energies.pdf worksheet.

Bonding 6 – Bond energies Table.pdf

Bonding 6 – Bond energies.pdf

Bonding 6 – Bond energies key.pdf
Period 4

1. Homework Form – specific heat concept metal vs. water

1. Hess law truth – ∆H0  values are the GOAT’s in HESS LAW!
slide 30 – slide 32 in Thermo presentation-
(canceling reactants in multiple chemical reaction Hess law problem is
really cancelling ∆H0  values! )

3. Heat of dissolution/heat of formation review with CaCl2 and NaCl
Hot Packs/Cold packs – Enthalpy change in the dissolution is due to the
ENERGY NEEDed to breaking of the ionic bonds in the salt and the attractive forces in water
and the LOSS of Energy needed in forming new bonds or attractive forces in the products.

∆H rxn = “bonds broken” – bonds formed“.
4. Bond Enthalpy with demo – – 2H2O2  -> O2 +  2H2O
determined Hrxn with heat of formation then with bond enthalpies
modeled question 1 on Bonding 6 – Bond energies.pdf worksheet.
Today’s Video’s:

NaCl formation Demo:

Today’s demo:  I dream of genie reaction

3/7  Thursday Homework:
1. Please complete the front side of the Bonding 6 – Bond energies.pdf worksheet and review with the key.  You will need a table of Average Bond Energies!
We started question 1 in class with the soda bottles demo (I dream of genie demo).

Bonding 6 – Bond energies key.pdf
Bonding 6 – Bond energies.pdf

Bonding 6 – Bond energies key.pdf
2. Please complete a (Ques 1-3 example) part question. Give yourself 20 – 25 minutes and complete it in its entirety before scoring with key below.
2016 free response question 1.pdf
New key that is correct!
AP Chem 2016 ques 1 AP Key.pdf

3/8 – Friday –

Today’s Notes:

Yesterday I introduced a new way to obtain H of a chemical reaction using Average Bond Energy values.  Through careful experimentation of many chemists we have a comprehensive list of the bond energy or bond enthalpy’s of all covalent (nonmetal to nonmetal bonds).

If we know the energy it takes to break a bond then we know the energy it takes to form a bond because:
qsys    =     – qsurr
Energy Needed to break a bond     =       Energy needed to form a bond
Energy consumed  (+)     =        Energy released (-)

So bond energy values are always positive unlike heats of Formation values (Hf In this reaction the Hcomb of propane (C3H8) the reaction is exothermic (-H ) because the energy needed to form the bonds of the STABLE products is greater than the energy needed to break the bonds of the reactants. Remember that these values are Averages because bond lengths are constantly changing as they absorb and release infrared radiation.

Notice that all bond enthalpies (bond energy) are positive as it ALWAYS requires energy to break bonds which are the “covalent sharing electron conditions” that provide STABLE electron configurations for all atoms in the molecule!

Notice the C=C bond energy is NOT double the bond energy of C-C because a double bond has sigma and a pi bond. If you remember a pi bond is weaker than the direct overlap of a sigma bond.

Notice the *C-C bond energy is somewhere in between a single C-C bond and a C=C bond because of RESONANCE in Benzene (which is sp2 hybridized).

 Its Bond order is 1.5!

Lets not forget that the internuclear graph is plotted with potential energy! The potential energy drops as bonds are made because the attraction for the each other atoms valence electrons is provides stability as available orbitals are filled. In the case of Hydrogen below they will attain the stability of He when they bond diatonically.  Notice the potential energy drops because the ∆H bond (bond enthalpy or bond Energy) is Negative as heat is released from the atoms when they bond.  Can you identify the Free atom zone? What is the bond enthalpy of H2? Bond Enthalpies will be given in small tables in AP questions when needed but remember that Bond Enthalpy is always positive to describe the Energy needed to be absorbedor consumed or BREAK an existing bond. Bonds being Broken always = +∆H Potential Energy increasing Bond being FORMED always = -∆H Potential Energy decreasing ∆H rxn = “bonds broken” – “bonds formed“.
SO if we have Bond Enthalpy:

∆H rxn =        ∑ bonds broken               –         ∑ bonds formed
[ ∑ bond enthalpy of reactants]  –  [ ∑ bond enthalpy of products ]

This formula is NOT provided in your reference tables.

And if we have Heats of Formation (∆H):
∆H rxn =   ∑  ∆H0 of the Products ]  –  [ ∑  ∆Hof the Products ]

This formula IS provided in your reference tables.
Both of these equations ARE HESS LAW!!!

Today we start our discussion towards the Second Law of Thermodynamics:

0th Law:  Hold unto your seats!

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!”
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.

3rd law:   S˚ (absolute entropy of chemical) = Zero at 0 Kelvin “We are limited by 0K!”

This creates the AZZ zone!!!!

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 and whats worse is the 3rd Law  which states that we are limited in these possibilities by 0K and yet we cannot achieve it!

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).

– Period 2

1.  Review Free response question
2.  Bond energy relations to bond length and bond order.
Quick review of Internuclear Distance Diagram

3.  H, ∆S, and hopefully ∆G!!!!!!!!!

– Use the follow along notes as move throught the google Thermodynamics presentation:

THermo read along – Thermo notes.pdf

#### Thermodynamics

– Period 3/4

1.  Review Free response question
2.  Bond energy relations to bond length and bond order.
Quick review of Internuclear Distance Diagram

3.  H, ∆S, and hopefully ∆G!!!!!!!!!

– Use the follow along notes as move throught the google Thermodynamics presentation:

THermo read along – Thermo notes.pdf

4.  Interconnectivity between H, ∆S, and  ∆G.
Tablature review of ∆S, and  ∆G

5.  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: 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!!!!!!
thermo 6a – Enthalpy, Entropy , Gibbs of Glucose methane.pdf

thermo 6a – Enthalpy, Entropy , Gibbs of Glucose methane Key.pdf

3/8 – Friday – (Weekend) Homework

Please Pay ATTENTION TO THE HOMEWORK REQUIREMENTS FOR EACH CLASS.
They are color coded!

Period 2  Class: If you were absent today please watch both class lectures posted below.

1. Please View Todays’s Class Lecture 1: Optional (not if you were absent today)
If today was confusing I would review this video.

2.  Please view Today’s Class lecture 2: NOT OPTIONAL –
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.
3.  If you need further explanation of the signs of the thermodynamic quantities
please view the signs lecture.
4. Please complete the front side of the thermo 6a – Enthalpy, Entropy , Gibbs of Glucose methane.pdf worksheet with me with the lecture below.

THermo 6a worksheet lecture:

5.  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
Thermo 7 – Gibbs Free energy Free Response 1984 key.pdf

6. Complete the Thermodynamic Signs Form posted below:

_____________

Period 3/4 class: – If you were absent please watch both (today’s) class lectures posted and then follow the rest below.

1. Please complete the front side of the thermo 6a – Enthalpy, Entropy , Gibbs of Glucose methane.pdf worksheet with me with the THermo 6a worksheet lecture posted above.
thermo 6a – Enthalpy, Entropy , Gibbs of Glucose methane.pdf

thermo 6a – Enthalpy, Entropy , Gibbs of Glucose methane Key.pdf
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
Thermo 7 – Gibbs Free energy Free Response 1984 key.pdf

3.  Feel free to watch today’s lectures if today was fuzzy. I suggest that you watch the Signs lecture     that is posted below posted to review the concepts in the 2nd period video that is posted above.
These concepts  include the Signs of the thermodynamic values of H, ∆S, and ∆G in the Gibbs Free Energy equation.

4. Complete the Signs Form posted below.

The Signs Lecture:
Thermodynamic Sign form –  due Monday Morning 4:30 am:

End of week 5!