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## Q3 : Week 6 – 3/13 – 3/17

Week 7 ->                                                                                                                             Jump toTuesday,   Wednesday,  Thursday,   Friday

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Jump toMonday Homework     _____________________________________________________                                                                                         3/13 – Monday B Day – 2, 3b/4 Lab

Main focus –

a) To Review State Functions and Path functions

b) To identify the 1st Law of Thermodynamics and apply it to calorimetry

c) To calculate heat transferred in calorimetry problems.

Period 2:

1. TLC of Universal Indicator – gather data

Period 3b/4:

1. TLC of Universal Indicator – gather data

2. Review the Homework Form:  Slides 1 – 15 of the Thermodynamics presentation

– state functions, path functions, delta E  – 1st Law of. Thermodynamics

Thermodynamics Form 1 21-22- Form Key.pdf

THermo read along – Thermo notes.pdf

– making of dry – Ice – deposition – JOULE-THompson effect – IMF’s!!!!!
3: CO2 Demo – Universal solution and dry ice:
warm water and Universal (green) plus few drops of .1 M NaOH solution- add solid CO2
on overhead projector with light hole (from sunset demo)

– phase diagram of water and carbon dioxide:

______________________

Lab 22 – TLC of Universal Indicator

TLC of Universal Indicator

In this lab we will perform a physical separation technique based on the solubility of the individual components in a mixture.

This is Lab will be written up on a google doc.  I would like for you to write a background on how TLC works and relative strength of the different IMF’s in the molecular compounds (chemical indicators) that make up a industry standard mixture called Universal Indicator.

Universal Indicator Acid/Base Indicator that has a unique color for each pH value:

A lot of pool kits that measure pH use Universal Indicator!
Universal Indicator can accomplish this broad range of colors because it is a homogeneous solution of three Acid/base indicators (CHROMOPHORES!!!!!!!!!!!!) that change their amount of conjugation based on the concentration of available protons (H+) or hydroxides (OH).

Also do not forget the acid/base indicators ARE CONJUGATE ACID and CONJUGATE BASE solutions which are Buffer solutions.

HIn                             —->                            H+           +         In

Conjugate Acid                                                                                      Conjugate Base
one color                                                                                                 New color

Phenolphthalein is an example:
There are slightly different types of Universal Indicators that use slightly different mixtures of indicators. We will using the BOGEN Universal Indicator mixture.

Bogen Universal Indicator Mixture:                      1.  Methyl Red                       pKa = 4.95
2.  Bromothymol Blue         pKa = 7.1
3.  Phenolphthalein             pKa = 9.4

So the colors of each pH from above are due to the mixture colors of each indicator at a certain pH levels:
(negative theory of light)

Okay so the point of this lab it separate each of these 3 individual molecular substances by using TLC (Thin Layer Chromatography).  Thin Layer Chromatography is another physical way we can separate mixtures.

– We have separated homogenous mixtures before using evaporation in the carbonate lab where we added HCl to an unknown white powder (that was either NaHCO3 or Na2CO3) and created NaCl (aq) and CO2 that left as bubbles.  We boiled off the water (separated the H-Bonds of water) and obtained dry NaCl (s) which we used by stoichiometry to determine our unknown. Connections!!!!

– we also studied distillation (of Coca-Cola) as another physical separation technique of mixtures!

– we also used filtration of a precipitate to separate a mixture in our gravimetric titrations!

Remember that solutions are mixtures that are soluble that are held together by IMF’s!  When we separate a mixture it is a physical change because we are not breaking bonds we are ONLY BREAKING IMF’s between MOLECULES!!!

TLC just like all chromatography separates a mixtures based on the idea that in a mixture the individual molecules retain their individual properties since their chemical formulas are not changing in the process.  This means that each chemical indicator will retain their individual solubilities (their percentage of non-polarity due to the amount of predominant LDF forces or their percentage of polar nature due to the amount of dipolar and H-boding) in the mixture.

Thus in TLC we will separate the 3 acid/base indicators by their solubility to either a mostly non- polar solvent (mobile phase – hexanes) or polar solvent (stationary phase – Silica Gel).

We will spot a TLC plate with Universal Indicator in a starting position close to the bottom of a Silica TLC plate (stationary polar phase) and add a liquid solvent – hexanes (non-polar mobil phase) that move up the Silica TLC plate by capillary action.  When the solvent reaches the “spot” or mixture EACH component of the mixture,

Methyl RedBromothymol Blue, and Phenolphthalein will have a choice to either stay with the polar Silica Gel on the TLC plate or move with the non-polar mobil hexanes.  BASED on their individual molecular structures (that define the IMF’s that they have), each chemical indicator will have a unique amount of polar or non-polar characteristics AND they will each move a unique distance up the TLC plate.

– If one of the chemicals is more polar (have stronger dipolar forces or have H-bonding) than the other three than they will prefer the Silica TLC plate and MOVE LESS than the other chemical indicators of the mixture that must have more non-polar characteristics.

– The chemical indicator that is the most non-polar (have mostly LDF’s with less dipolar forces), would prefer to be attracted to the mobile non-polar hexane solvent and thus move up the plate more.

*Remember this a separation technique based on solubility. Solubility is based on how similar the IMF’s have with the mobile non-polar solvent (hexane) or the polar stationary solvent (TLC silica Plate). Look at the two solvent molecular structure above to identify what IMF are most similar with each solvent!

The distance that the individual chemical indicators move compared to the original spot will be measured.

Yellow layer – Carotenoids –

Green layers – Chlorophyll A and Chlorophyll B – is one slightly more polar?

– Notice these pigments are chromophores!!!

– There were three layers separated in the Time Lapse Chromatography below.  Looking at the pigments separate in the video below, and the molecular structure above can you identify the chemicals in the chromatograph? Why was the yellow layer almost at the top while the 2 green layers barely moved at all?

– Also notice that we did not spot this TLC plate, we rolled a coin on a spinach leaf unto the TLC plate to make line. The green line was our starting point.

So the assignment is to rank the polarity and non-polarity of the three chemicals in the mixture of Universal Indicator based on what you know of IMF’s AND Predict which chemicals will travel the greatest distance up the TLC plate and which chemicals with travel the shortest distance.  This discussion is the subject of your Background of the lab. Use these structural formulas in your Background.
THEN Write a hypothesis based on your background.

Hypothesis Example: Chemical A will have the largest Rate of Flow (move the farthest on the TLC plate) while the Chemical B will have the smallest Rate of flow.

Bogen Universal Indicator Mixture:                      1.   Methyl Red                      pKa = 4.95
2.  Bromothymol Blue         pKa = 7.1
3.  Phenolphthalein             pKa = 9.4

Methyl red:
Bromothymol Blue:

Phenolphthalein:

Thermodynamics: The end of it all!!!

Fire Syringe Demo –

3/13 – Monday’s Homework: –

1. Please write the procedure/materials section of the TLC of the Universal Indicator Lab in shared google doc that was linked to you last week.

2.  PRE-LAB – Calorimetry / Hess Law Lab

Watch Lecture on Lab calculations, and complete worksheet given.

I use this worksheet to model the problem –  the Blue one or Lavender one!
Hess Law Lab Pre lab for lecture USE.pdf

3.  Use the yellow copy and complete the Form based on the this worksheet. Also use the homework notes to guide you to complete the form.

This is the yellow copy -: (which the form is based on)
Hess Law Lab Pre lab p.pdf

Hess Law Calculations Lab tomorrow:

*In the video I make a small mistake. The density of all the solutions used in the calorimetry are 1.02 g/ml !
**I also made a mistake where I did not carry over a negative sign from my q —-> ΔH

In spite of me you can learn…

3: Hess Law Calculations – (slide 16 and beyond)

_____________________________

HOMEWORK NOTES:

1. Review the basics of the Calorimetry Lab.

a) Calibration of the Ccal:
After Hot water exchanged energy with Cold water the temperature of the water should be the average between temperature of the room temperature water and the heated water of equal volumes.

That temperature of the water is Tavg

NOW water at Tavg is higher in temperature of the styrofoam cup that is at room temperature. So there must be some heat flowing into the cup.  That is why YOUR Tmix is lower than the Tavg due to the loss of heat from the water into the cup.

To calculate the heat lost to the cup we set the following thermal equal:

Inside the system = Water and the Cup

Assume that the system is isolated*  then     cal (CUP)  +  qwater  =  0

cal (CUP)  = – qwater

heat absorbed  =  heat released

* we use the change of energy of the water to relate it to the change of energy to the cup
That is the basis of calorimetry due to the 1st law of thermodynamics (conservation of energy)

cal (CUP)  = – qwater

m C ΔT   =   m C ΔT
^
|*we combine mass with specific heat
|
Ccal ΔT  =  – qwater

Ccal  =   – qwater  /  ΔT  —> (temp change of Hot and cold water )
^
|

(100g)(4.18 J/g C ) (Tf – Ti)
^
|
Tmix – Tavg

bCalculation of the q rxn:

heat released  =  heat absorbed

q rxn = -(q solution + q cal(cup) )
^                    ^
|                    |
|                   Ccal x (Tmix – T initial = room temperature)
|
|
(102 g)(4.18 J/g C) (Tmix – T initial = room temperature)

c) Calculation of ΔH: (We have not defined this yet)

*from last nights derivation = qp =  q rxn =  ΔE  +  PΔV

A brand new Thermodynamic quantity H (enthalpy!) = E  +   PV

NOW we do not need to know specifics of the pathway to attain energy changes!

We can use the ΔH to CONNECT other reactions together = Hess Law!

Now to calculate the ΔH for each reaction we recognize that qp =  q rxn = ΔHrxn

We just need to make some adjustments to the q reaction. We need to relate it to amount of some mole quantity of a reactant or product.

q rxn is just the amount of heat released or absorbed but ΔH needs to know the heat per amount of reactant that got consumed or was produced.  We need to express it it terms of:

(KJ) energy / mole = kJ per mole (of some chemical)

In this way it becomes a ratio of energy per amount of chemical which will be constant.

2 : Hess law lab – calorimetry

3/14 – Tuesday – A Day – 2/3a Lab, 4

Main focus –

a) To perform constant volume calorimetry to determine the ∆Hrxn of a reaction.

b) To derive the state variable of enthalpy, ∆H.

c) To collect data from the TLC lab.

Period 2/3a:

1. Universal Indicator demo’s –
– rainbow tube –
making of dry – Ice – deposition – JOULE-THompson effect – IMF’s!!!!!
– CO2 Demo – Universal solution and dry ice:
warm water and Universal (green) plus few drops of .1 M NaOH solution- add solid CO2
on overhead projector with light hole (from sunset demo)

2. Complete TLC Lab Data Collection
a) Calculate Rate of flow for each spot.

3. Review of Hess Lab Form –

Hess Law lab form 1 key complete p.pdf

4. Derivation of Enthalpy (∆H)
5. Introduction to Hess’s Law
6. Perform Lab 23 – Calorimetry Lab

Period 4:

1. Perform Lab 23

TLC Lab Data:

*Notice that the phenolphthalein trails the methyl red.  It is probably faint in your chromatograph but it is there.
They do not use a lot of it in the Bogen Universal indicator mixture and thus it is not a concentrated spot.

Our next step is to calculate the RATE of Flow for each SPOT. This is a quantitative value that we can actual use to identify the chemical that is being separated from the mixture.

RATE of FlowR =   distance each spot traveled from the starting position (bottom line)
distance to solvent font

__________________________________________________________

Lab 23 – Hess Law and Calorimetry

Lab packet:
Calorimetry and Hess law p.pdf

Objectives:

1.  To use constant pressure calorimetry to calculate the kJ of heat per mole reactant for three chemical reactions.

2. To calculate the ∆Hrxn for the third reaction by using Hess’s Law using the enthalpies determined through calorimetry for the first 2 reactions.

3. To dertermine the accuracy of your results by comparing your Hess law value of the the third reaction and the calorimetry value.

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?

Time Lapse of the TLC of Universal Indicator –

Thermodynamics: The end of it all!!!

3/14 – Tuesday’s Homework: –

TLC lab should have the background, hypothesis, materials, procedure and data added to the shared google doc.

1. Complete all Calculations in the Calorimetry Lab to attain the Ccal and the qrxn 1 (1st Page)

We will convert this to enthalpy tomorrow!

Write on the lab packet (it is an informal lab).
Have this ready for Wednesday..We will complete the lab in class tomorrow if you have all the calculation
completed.

If you need a review of the calculations you can use the key from last nights homework (form or the video from last night.

Hess Law lab form 1 key complete p.pdf

also the notes yesterday could also be helpful.

2. HESS LAW with Reactions!

*Remember that ΔH is a really important thermodynamic state function that will help us interconnect other ΔH quantities from other reaction if these reactions are connected.

Please watch lecture on Hess Law and complete with me the Hess Law problems that I model from the thermo 4 – Hess Law ditto.pdf worksheet. We will apply this to our lab tomorrow.

thermo 4 – Hess Law ditto.pdf

thermo 4 – Hess Law ditto key.pdf

We will apply this Hess Law to our third reaction Tomorrow!

2: Hess Law Introduction and Enthalpy derivation –I derive the H  (ENTHALPY!) in the first 8 minutes.

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

____________________________________________________________________________                     Jump to: Wednesday Homework / top

3/15 – Wednesday – B Day – 2, 3b/4 Lab

Main focus –

a) To complete the calculations of the Calorimetry Lab including the enthalpy of the                reaction, ∆Hrxn.

b) To define enthalpy (∆H).

c) To determine the ∆Hrxn using Hess’s law.

Period 2:

1. Complete collecting Data for Lab 23 – calorimetry and Hess’s Law
2. Review of the calculations for the lab.
3. Define  ∆H and review the concept of Hess’s Law.

4. Apply Hess’s law to the Lab and complete the 2nd Objective.

a) 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.

Period 3b/4:

Same 1 – 4 above.

5. Universal Indicator demo’s –
– rainbow tube –
making of dry – Ice – deposition – JOULE-THompson effect – IMF’s!!!!!
– CO2 Demo – Universal solution and dry ice:
warm water and Universal (green) plus few drops of .1 M NaOH solution- add solid CO2
on overhead projector with light hole (from sunset demo)

6. 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 Hfof Al2O3 and NaCl, and discussed strength of Bonds or lattice energy.
d) Drew diagram of the Free atom Zone (FAZ), which is the reference point to show Hfof reactants and products of another reaction.
e) Used Hfto 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

Thermodynamics: The end of it all!!!

_____________________

Lab 23 – Hess Law and Calorimetry

Lab packet:
Calorimetry and Hess law p.pdf

Objectives:

1.  To use constant pressure calorimetry to calculate the kJ of heat per mole reactant for three chemical reactions.

2. To calculate the ∆Hrxn for the third reaction by using Hess’s Law using the enthalpies determined through calorimetry for the first 2 reactions.

3. To dertermine the accuracy of your results by comparing your Hess law value of the the third reaction and the calorimetry value.

Calculations in this Lab:

a) Calculation of Ccal – posted on Monday above
b) Calculation of qrxn – posted on Monday above
c) Calculation of ΔH:

*from last nights derivation = qp =  q rxn =  ΔE  +  PΔV

A brand new Thermodynamic quantity H (enthalpy!) = E  +   PV

NOW we do not need to know specifics of the pathway to attain energy changes!

We can use the ΔH to CONNECT other reactions together = Hess Law!

Now to calculate the ΔH for each reaction we recognize that qp =  q rxn = ΔHrxn

We just need to make some adjustments to the q reaction. We need to relate it to amount of some mole quantity of a reactant or product.

q rxn is just the amount of heat released or absorbed but ΔH needs to know the heat per amount of reactant that got consumed or was produced.  We need to express it it terms of:

(KJ) energy / mole = kJ per mole (of some chemical)

In this way it becomes a ratio of energy per amount of chemical which will be constant.

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!

_____________________

TODAY’s NOTES:

1. Hess Law – The interconnectivity of ΔH

 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 than 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 Northampton we can use both values to determine the Overall change of distance from North Riverhead to Northampton.   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.
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.

 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.

END OF NOTES! 🙂

3/15 – Wednesday Homework:

Please pay particular attention to the questions I am asking you to complete.  I have a reason for every problem that I assign.  Please complete them in the correct order.

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

2. Watch the Hess law lecture with Heats of FORMATION

Hess Law with Heats of Formation:

3.  Lab 23 – Hess Law Lab Activity is due Friday 3/17 .
Non-formal lab requirements posted below.

4.  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!

5.  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 formation of Mg(OH)2

Thermo 1 calorimetry.pdf

thermo 1 calorimetry key 0809.pdf                                                                                                                              View Download

thermo 4 – Hess Law ditto.pdf

thermo 4 – Hess Law ditto key.pdf

LAB 23  requirements:

Part 1
Show all calculations clearly and neatly.  If you have to rewrite them then do so.  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)  You will compare it with the ∆H/mole of that you obtained through calorimetry in lab for reaction 3 in step c) below.

c) Calculate the known or the theoretical value of the ∆H/mole of the third reaction by using Heat of Formation Tablature for the 3rd reaction. This is the new skill learned today in class or through the lecture above.

*Use the net ion equation of the final reaction to get the theoretical value from heats of formation:

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

d) Complete 2 percent error calculations (using Heats of Formation value as Theoretical for both):
1. based on the ∆H/mole using reactions manipulations = experimental
2. then with ∆H/mole from calorimetry from reaction 3 = experimental

*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. Was this an isolated system?

Staple calculations and reactions. Hand in tomorrow.

Carbon Dioxide and Universal Indicator Demo –

________________________________________________________________________                              Jump to: Thursday Homework / top

3/16 – Thursday – A Day – 2/3a Lab, 4

Main focus –

a) To calculate the ∆H from ∆Hf0.

b) To build a potential energy diagram through ∆Hf0 values and identify the path                   functions and the state functions.

Period 2/3a:

1.  Review notes from yesterday  and Slides 26 – 32 to review the Calorimetry lab

The starting point of a reaction IS THE SUM OF THE Heats of formation of all reactants.
The ending point of a reaction IS THE SUM OF THE Heats of formation of all products.

2 Review a selected homework problems:

Questions 5 & 7:

thermo 4 – Hess Law ditto key.pdf

Questions 5:

thermo 1 calorimetry key 0809.pdf

3.  Used Hfto determine the Hrxn of NaCl ionizing into water and determining whether the water will increase or decrease in temperature. Heat of dissolution.

a) Define Exothermic and Endothermic
b) Related the energy changes in its dissolution to IMF. Why are some salts exothermic or endothermic?
Table I – ionization Energies

c) Compared the Hf of NaCl and Al2O3 in terms of bond strength and Lattice Energy.
d) Lattice Energy is measured through Hess Law!!

4. Complete comparison of ∆Hof Al2O3 and NaCl with thermite reaction.

– Modeled the calculation of the ∆Hrxn for the thermite reaction using heats of formation
– Wrote a a potential energy diagram for the Thermite reaction and used the idea of potential
energy to describe the formula of :

∆Hrxn   =  Sum of the ∆H Products – Sum of the ∆H Reactants

∆Hrxn =    Potential Energy of the Products – Potential Energy of the Reactants
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.

4. Bond Energy –   ∆H rxn = “bonds broken” – “bonds formed“.

Period 4:

1.  Quick repeat of carbon dioxide demo.

2.  Calorimetry Lab review with slides from presentation.

Same as above skipping thermite and getting to Bond energy!

_____________________

TODAY’s NOTES:

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!

So let’s revisit the reaction that we used to determine the enthalpy of the reactions below:      Hf is the GOAT !!!

If we consider that the energy lost to form each reaction from the FAZ (Free Atom Zone) we can determine the Potential energy of the reactants and products and develop the following potential energy diagram.  Remember that the level of potential energy is due to what is left after the energy is released to form the compound from its elements.

Now we can also use this potential energy diagram to identify a NEW way to obtain the ∆H of a chemical reaction which through average bond energies or average bond enthalpies.

Today I have introduced a new way to obtain ∆H of a chemical reaction using Average Bond Enthalpy 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 (-)

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!
Let’s 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?  WHERE IS THE FAZ? 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 above for BOND ENTHAPy’s  is NOT provided in your reference tables.

And if we have Heats of Formation (∆H):

∆H rxn =   ∑  ∆Hof the Products ]  –  [ ∑  ∆Hof the Products ]

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

END OF NOTES! 🙂

Thermite Micro Reaction –  Tremendous energy released due to the stability of aluminum oxide!

Thermite Reaction –  Tremendous energy released due to the stability of aluminum oxide!

Decomposition of Hydrogen Peroxide –

3/16 – Thursday Homework: –

1. Read my Notes above.

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.

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

3. 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

4. Complete the 1st side of the Bond Energy Worksheet and review with the key.

Bonding 6 – Bond energies Table.pdf

Bonding 6 – Bond energies.pdf

Bonding 6 – Bond energies key.pdf

.

Heat of formation lecture-

Bond Energy Lecture-

__________________________________________________________________                                    Jump to: Friday Homework / top      3/17 – Friday – B Day – 2, 3b/4 Lab

Main focus –

a) To find the pot of Au.

b) To identify ionization reactions and predict the relative strengths of reactant IMF’s           and Products IMF’s through heat of dissolution.

c) To Use Heat of Formation as a measure of Lattice Energy.

Period 2:

1. Review of Bond Enthalpy’s the – Demo of hydrogen peroxide – draw potential energy diagram of reaction
a) Revisted the internuclear distance diagram – PE changes and Bond Enthalpy’s.

2 Review a selected homework problems from previous nights homework:

Questions 5 & 7:

thermo 4 – Hess Law ditto key.pdf

Questions 5:

thermo 1 calorimetry key 0809.pdf

3 Review of homework Mock AP Question (calorimetry):
2013 Part 2 – AP Key p.pdf

4.  Used Hfto determine the Hrxn of NaCl ionizing into water and determining whether the water will increase or decrease in temperature. Heat of dissolution.

a) Define Exothermic and Endothermic
b) Related the energy changes in its dissolution to IMF. Why are some salts exothermic or endothermic?
Table I – ionization Energies

Period 3b/4:

Same as 1 – 3.

4. Complete comparison of ∆Hof Al2O3 and NaCl with thermite reaction.
a) Compared the Hf of NaCl and Al2O3 in terms of bond strength and Lattice Energy.
b) Lattice Energy is measured through Hess Law!!

– Modeled the calculation of the ∆Hrxn for the thermite reaction using heats of formation
– Wrote a a potential energy diagram for the Thermite reaction and used the idea of potential
energy to describe the formula of :

∆Hrxn   =  Sum of the ∆H Products – Sum of the ∆H Reactants

∆Hrxn =    Potential Energy of the Products – Potential Energy of the Reactants
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.

4.  Used Hfto determine the Hrxn of NaCl ionizing into water and determining whether the water will increase or decrease in temperature. Heat of dissolution.

a) Define Exothermic and Endothermic
b) Related the energy changes in its dissolution to IMF. Why are some salts exothermic or endothermic?
Table I – ionization Energies

_____________________

TODAY’s NOTES:

Bond Enthalpy – revisted

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

*So in every reaction there is absorption of energy to break bonds   and a release of energy to form new bonds.

In the example above we would write reactions as:

C3H8   +   5O2      3CO2   +   4H2O  +  2220 kJ      ∆H = -2220 kJ/mole

This represents an exothermic reaction Because there is a NET release of energy.
This means there was more energy released in forming new bonds than energy absorbed to break old bonds!
The heat is written on the product side to show the NET release of Heat.
The surroundings increase in energy and the Temperature increases.

183 kJ   +   N2    +    O2       2NO        ∆H = +183 kJ/mole

This represents an endothermic reaction Because there is a NET absorption of energy.
This means there was more energy absorbed to break old bonds that energy released in forming new bonds!
The heat is written on the reactant side to show the NET absorption of Heat.
The surroundings decrease in energy and the Temperature decreases.

3. Heat of dissolution/heat of formation review with CaCl2 and NaCl
Table I –

First lets Review the concept of Lattice Energy WITH HESS LAW!

#### Lattice Energy – Born Haber Cycle:

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)
CaCl2 demo – temperature rose (∆H dissolution = -82.4 KJ/mol)

Complete discussion of heat of formation into stability of bonds and ∆H dissolution.
*Connections We can use heat of formations from our tables to calculate the ∆Hrxn . The ∆Hrxn has many names that 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!

 Ion – dipole IMF’s The IMF’s that soluble ions have with water (solvent) are called ion – dipole.  IF these ion- dipole IMF’s are more significant in strength than the combined                H – Bonding that water has for itself and the lattice energy of the salt then the ∆Hdissolution  = negative like the case of CaCl2 (aq). It has an exothermic heat of dissolution.    CaCl2 could be used in HOT packs. In the case of NaCl, the ion dipole IMF’s are less significant in strength than the combined H-Bonding that water has for itself and the lattice energy of the salt thus the ∆Hdissolution  = positive. NaCl(aq) has an endothermic heat of dissolution.NaCl could be used in Cold packs.

We can describe whether a salt when dissolved is exothermic (raising the temperature of the solution) or endothermic (lowering the temperature of the solution )  by using our concepts of Bond Enthalpy!

If it takes MORE energy to break the existing IMF’s that the water has for itself and the ionic bonds of the salt THAN the energy released to form new IMF’s (ion – dipole) between the water (solvent) and the ions (solute) then the process is ENDOTHERMIC.

If it takes LESS energy to break the existing IMF’s that the water has for itself and the ionic bonds of the salt THAN the energy released to form new IMF’s (ion – dipole) between the water (solvent) and the ions (solute) then the process is EXOTHERMIC.

Also CaCl2 is called ice melt because of its exothermic Heat of Dissolution and its the preferred salt that is put on the roads because of its chemistry.

4. 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 ∆H0  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.

Phase change enthalpy.pdf

Heat of Dissolution Solution Classroom lecture-

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!

3/17 – Weekend Homework: –

1.  Please complete the TLC of Universal Indicator.  You need the following components:
a) Title Page
b) Background
c) Hypothesis
d) Procedure/materials
e) Data
f)  Results
g) Conclusion – Why did your results differ your expected outcomes? Who had more similar IMF’s with the solvent and the stationary phase? What were the differences in the structures that lead to your outcomes. Did any of the compounds of the mixtures actually dissolve?
f)  Sources

3. Please complete  Free Response part 2 questions. Give yourself 20 – 25 minutes and complete it in its entirety before scoring with key below.

1st FRQ
AP Chemistry – 2016 Free Response question 1.pdf

AP Chem 2016 ques 1 AP Key.pdf

2nd FRQ:
2022 FRQ Question 1.pdf

2022 FRQ Question question 1 AP Central Key.pdf

2022 FRQ Question question 1 Grodski Key.pdf