## Q3: Week 7  – 3/21 – 3/25

3/21 – Monday A Day – 2/3a Lab, 4

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 2/3, 4:

1. 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

2.  Selected

3.  Review Last nights Form:

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TODAY’s 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 up:

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:

*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.

Thermodynamics: The end of it all!!!

Fire Syringe Demo –

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3/21 – Monday Homework: –

1:  Please watch the calorimetry basics lecture and follow along with me with the THermo read along – Thermo notes.pdf worksheet that we started in class.

2:  Complete questions  1 – 3 in the thermo 1 calorimetry.pdf  worksheet and review with the key.
I completely review these questions with you in the lecture below.

Thermo 1 calorimetry.pdf

thermo 1 calorimetry key 0809.pdf

3.   PRE-LAB – Calorimetry / Hess Law Lab

Watch Lecture on Lab calculations, and complete worksheet given and answer in form below.

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

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

Hess Law Calculations Lab:

****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…

1: Calorimetry Basics – (slide 16 and beyond)

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

3 : Hess law lab – calorimetry
>

_________________________________________________________________________________________________________________________

3/22 – Tuesday – B Day – 2, 3b/4 Lab

Main focus –

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

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

Period 2 :

1. Review of Hess Lab Form –

Hess Law lab form 1 key complete p.pdf
2. Derivation of Enthalpy (∆H)
3. Introduction to Hess’s Law

Period 3/4:

1. Review of Hess Lab Form –
2. Derivation of Enthalpy (∆H)
3. Introduction to Hess’s Law
4. Perform Lab 23

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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?

Thermodynamics: The end of it all!!!

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3/22 – Tuesday Homework: –

1.  Complete all Calculations in the Lab toattain the  qrxn = (we will convert this to enthalpy tomorrow!)
IF you were not able to complete your data collection (almost all of period 2) then you can skip this          first step.)

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):

Hess Law lab form 1 key complete p.pdf

2.  HESS LAW with Reactions!

*Remember that ΔH is a really important thermodynamic state function that will help us interconnect other Δquantities 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!!!!

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3/23 – Wednesday A Day – 2/3a Lab, 4

Main focus –

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

b) To define enthalpy.

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

Period 2/3:

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.

5. 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

Period 4:

1. Define  ∆H and review the concept of Hess’s Law.

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

3. Heat of formation – (intro and with demo

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
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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!

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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! 🙂

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3/23 – 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 3/25 .
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 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)  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 the ∆H/mole using reactions and then with ∆H/mole from calorimetry.

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

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

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.

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3/24 – Thursday – B Day – 2, 3b/4 Lab

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 3/4:  –

1.  Review a selected homework problems:

Questions 5 & 7:

thermo 4 – Hess Law ditto key.pdf

Questions 5:
thermo 1 calorimetry key 0809.pdf

2.  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!!

3. 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 3/4:  –

Same as above except we will not Run the lab till tomorrow

_____________________

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!
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?  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 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! 🙂

Decomposition of Hydrogen Peroxide –

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3/24 – Thursday 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.

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.

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

Bond Energy Lecture-

Heat of formation lecture-

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3/25 – Friday A Day – 2/3a Lab, 4

Main focus –

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

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

Period 2/3, 4:
1.  Review of homework Mock AP Question:

2013 Part 2 – AP Key p.pdf

2.  Bond Enthalpy – revisited

3. Heat of dissolution/heat of formation review with CaCland NaCl

Table I –

Hot Packs/Cold packs

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

4. Started the Phase change enthalpy.pdf

– phase diagram of water and carbon dioxide: _____________________

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

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

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

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?

Heat of Dissolution Solution Classroom lecture-

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3/25- Weekend Homework: –

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

AP Chemistry – 2016 Free Response question 1.pdf

AP Chem 2016 ques 1 AP Key.pdf