Q3: Week 7 – 3/21 – 3/25
Jump to: Tuesday, Wednesday, Thursday, Friday
3/21 – Monday – A Day – 2/3a Lab, 4
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
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2. Selected
3. Review Last nights Form:
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TODAY’s NOTES:
Thermodynamics: The end of it all!!!
Fire Syringe Demo –
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3/21 – Monday Homework: –
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1: Calorimetry Basics – (slide 16 and beyond)
3: Hess Law Calculations – (slide 16 and beyond)
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3/22 – Tuesday – B Day – 2, 3b/4 Lab
a) To perform constant volume calorimetry to determine the ∆Hrxn of a reaction.
b) To derive the state variable of enthalpy, ∆H.
Period 2 :
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Period 3/4:
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Lab 23 – Hess Law and Calorimetry
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.
Thermodynamics: The end of it all!!!
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3/22 – Tuesday Homework: –
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2: Hess Law Introduction and Enthalpy derivation –I derive the ∆H (ENTHALPY!) in the first 8 minutes.
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3/23 – Wednesday – A Day – 2/3a Lab, 4
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:
4. Apply Hess’s law to the Lab and complete the 2nd Objective.
5. Heat of formation – (intro and with demo)
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Period 4:
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!!!
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Lab 23 – Hess Law and Calorimetry
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:
Can you use the first 2 reactions ∆H values to obtain the third by manipulating the first 2 reactions? Is Hess Law supported?
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TODAY’s NOTES:
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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.
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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.
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END OF NOTES! 🙂
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3/23 – Wednesday Homework: –
4: Complete questions 1 – 4 of the Thermo 3 heat of formation. pdf worksheet and review with the key.
thermo 3 heat of formation.pdf View Download |
thermo 3 heat of formation key.pdf
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3/24 – Thursday – B Day – 2, 3b/4 Lab
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: –
thermo 4 – Hess Law ditto key.pdf
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4. Bond Energy – ∆H rxn = “bonds broken” – “bonds formed“.
Same as above except we will not Run the lab till tomorrow
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TODAY’s NOTES:


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.

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.
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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).
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![]() What is the bond enthalpy of H2? WHERE IS THE FAZ?
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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 absorbed Bonds being Broken always = +∆H Potential Energy increasing Bond being FORMED always = -∆H Potential Energy decreasing ∆H rxn = “bonds broken” – “bonds formed“. |
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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.
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END OF NOTES! 🙂
Decomposition of Hydrogen Peroxide –
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3/24 – Thursday Homework: –
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Bond Energy Lecture-
Heat of formation lecture-
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3/25 – Friday – A Day – 2/3a Lab, 4
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.
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3. Heat of dissolution/heat of formation review with CaCl2 and NaCl
4. Started the Phase change enthalpy.pdf

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TODAY’s NOTES:


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. |
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.
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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.
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