1. Chapter 17: Thermochemistry
Courtesy of lab-initio.com

2. Thermochemistry
The study of energy changes that occur during chemical reactions and changes in state.

3. Energy
Energy is the capacity to do work or supply heat, and can take many forms
Potential energy is stored energy or the energy of position
Chemical Potential energy is stored energy in bonds
Kinetic energy is the energy of motion
Thermal energy (heat) is an outward manifestation of movement at the atomic level
E can occur as either heat transfer or work, or both.

4. Heat (q)
Definition: The energy transfer due to a difference in temperature.
Heat always flows from a warm object to a cold object.

5. Important to identify the system and the surroundings.
System: the components of the chemical reaction.
Surroundings: everything else in the universe.

6. First Law of Thermodynamics: The energy of the universe is constant
First Law of Thermodynamics: The energy of the universe is constant. E = q + w E = change in system’s internal energy q = heat w = work
Flow of heat between the system and surroundings
Always defined from the point of view of the system

7. Exothermic Processes
Processes in which energy is released as it proceeds, and surroundings become warmer
Heat flow out of the system
q (negative)
Reactants  Products + energy
CaO(s) + H2O(aq)  Ca(OH) kJ energy
H = -65.2kJ
**thermochemical equation**

8. Endothermic Processes
Processes in which energy is absorbed as it proceeds, and surroundings become colder
Heat flows into the system
q (positive)
Reactants + energy  Products
129kJ + 2NaHCO3(s)  Na2CO3(s) + H2O(g) + CO2(g)
H = 129kJ

9. Extensive property-depends on the amount of the substance.
Heat of the Reaction
Extensive property-depends on the amount of the substance.
129kJ + 2NaHCO3(s)  Na2CO3(s) + H2O(g) + CO2(g)
H = 129kJ
**Double the moles of NaHCO3 the heat of the reaction doubles.**

10. Exothermic versus endothermic.
Endothermic * products have Higher PE
Exothermic* reactants have Higher PE
(Higher PE – weaker bonds)

11. Units for Measuring Heat
The calorie is the heat required to raise the temperature of 1 gram of water by 1 Celsius.
1Calorie (food energy) = 1kcal = 1000 calories
The Joule is the SI system unit for measuring heat:

12. Heat Capacity
Amount of heat needed to increase the temperature of an object exactly 1C.
Depends on mass and chemical composition
Example:
drop of water and cup of water.
Beach sand heats up quickly, but the ocean water needs all summer to raise its temp just a few degrees.

13. Specific Heat (C)
The amount of heat required to raise the temperature of one gram of substance by one degree Celsius.
Substance
Specific Heat (J/g·C)
Water (liquid)
4.18
Ethanol (liquid)
2.44
Water (solid)
2.06
Water (vapor)
1.87
Aluminum (solid)
0.897
Carbon (graphite,solid)
0.709
Iron (solid)
0.449
Copper (solid)
0.385
Mercury (liquid)
0.140
Lead (solid)
0.129
Gold (solid)

14. Now that we know what heat is, its good to know how much heat(q) can be given off by a system and potentially be used by another system.
This is know as…
We will look at both chemical reactions and change in state.
Enthalpy:H (change in enthalpy of a system when heat is given off or absorbed)

15. H = qP (system) at constant pressure
Is H = q ?
Enthalpy = H = E + PV
work=(-P∆V)
E = q + w
E = q - P∆V
q = E + P∆V
therefore…
H = qP (system) at constant pressure

16. Calculations Involving Specific HeatOR
C = Specific Heat
q = Heat lost or gained or
T = Temperature change
m = Mass
Units: J/(gC) or cal/(gC)

17. Sample Problem
The temperature of a 95.4-g piece of copper increases from 25.0C to 48.0 C when the oopper absorbs 849 J of heat. What is the specific heat of copper? Given: Unknown: m= 95.4 g Ccu = ? J/(gC) T = 48.0 C – 25.0 C q = 849 J

18. Calorimetry
The amount of heat absorbed or released during a physical or chemical change can be measured, usually by the change in temperature of a known quantity of water in a calorimeter.
Constant-pressure calorimetry
Used to determine heat
of a reaction in solution at
constant pressure.
Enthalpy (ΔH)

19. Calorimetry at constant volume
No work is done (V must change for PV-work to be done)
ΔE = q + w = q = qv ΔE = ΔT x heat capacity of calorimeter

20. Sample Problem
When 25.0 mL of water containing mol HCl at 25.0C is added to 25.0 mL of water containing mol NaOH at 25.0C in a foam cup calorimeter, a reaction occurs. Calculate the enthalpy change in kJ during this reaction if the highest temperature observed is 32.0 C. Assume the densities of the solutions are 1.00 g/mL. Given: Unknown: Cwater = 4.18J/(gC) H = ?? Vfinal = 50.0mL T = 7.0 C
H = -1500J

21. Enthalpy (Heat) of Combustion
ΔH for the complete burning of one mole of a substance

22. Heat of fusion and solidification
What happens when you place an ice cube on a table in a warm room?
absorb or release heat?
absorb

23. Latent Heat of Phase Change
Molar Heat of Fusion
The energy that must be absorbed in order to convert one mole of solid to liquid at its melting point.
Molar Heat of Solidification
The energy that must be removed in order to convert one mole of liquid to solid at its freezing point.
Temperature remains constant all heat is used to break bonds
△Hfus =-△Hsolid

24. Latent Heat of Phase Change #2
Molar Heat of Vaporization
The energy that must be absorbed in order to convert one mole of liquid to gas at its boiling point.
Molar Heat of Condensation
The energy that must be removed in order to convert one mole of gas to liquid at its condensation point.
△Hvap =-△Hcond

25. Water phase changes constant
Temperature remains __________ during a phase change.
Heat of vaporization
40.67 kJ/mol
Heat of fusion
6.01 kJ/mol

26. Latent Heat – Sample Problem
Problem: The molar heat of fusion of water is
6.009 kJ/mol. How much energy is needed to convert 60 grams of ice at 0C to liquid water at 0C?
Mass
of ice
Molar
Mass of
water
Heat of
fusion

27. Molar Heat of Solution
The Heat of Solution is the amount of heat energy absorbed (endothermic) or released (exothermic) when a mole of solute dissolves in a solvent.
Substance
Heat of Solution
(kJ/mol)
NaOH
-44.51
NH4NO3
+25.69
KNO3
+34.89
HCl
-74.84

28. Example: How much energy is needed to heat 55 grams of water ice from a temperature of -150 C to steam at a temperature of 1500 C?        

29. Solution: This problem requires several steps:   Step 1: Heat the ice from -150 C to 00 C   ∆H = mCp∆T = (55 grams)(2.03 J/g0C)(150) = 1,675 J      
Step 2: Undergo the phase change from a solid to liquid.   ∆H = n∆Hfus = (3.06 mol)(6.01 kJ/mol) = kJ

30. Step 3: Heat the liquid from 00 C to 1000 C ∆H = mCp∆T = (55 grams)(4
Step 3: Heat the liquid from 00 C to 1000 C   ∆H = mCp∆T = (55 grams)(4.184 J/g0C)(1000) = 23,012 J    
Step 4: Undergo the phase change from a liquid to gas.   ∆H = n∆Hvap = (3.06 mol)(40.7 kJ/mol) = kJ

31. Step 5: Heat the steam from 1000 C to 1500 C ∆H = mCp∆T = (55 grams)(2
Step 5: Heat the steam from 1000 C to 1500 C   ∆H = mCp∆T = (55 grams)(2.01 J/g0C)(500) = 5,528 J  
And when you’re done with all of this, just add up the different values you found above:   ∆Htotal = 1.68 kJ kJ kJ kJ kJ   ∆Htotal = kJ

32. Hess’s Law Enthalpy Reactants  Products
The change in enthalpy is the same whether the reaction takes place in one step or a series of steps.
Allows you to determine H of a reaction indirectly.
Reactions yield products in addition to the one of interest.
Reactions that are far too slow.

33. Oxidation of N2 to produce NO2
N2(g) + 2O2(g)  2NO2(g) ΔH1=68kJ
Or..
N2(g) + 2O2(g)  2NO (g) ΔH2=180kJ
2NO(g) + O2(g)  2NO2(g) ΔH3=-112kJ
N2(g) + 2O2(g)  2NO2(g) ΔH=68kJ

34. The principle of Hess's law
The principle of Hess's law. The same change in enthalpy occurs when nitrogen and oxygen react to form nitrogen dioxide, regardless of whether the reaction occurs in one (red) or two (blue) steps.

35. Calculations via Hess’s Law
1. If a reaction is reversed, H is also reversed.
N2(g) + O2(g)  2NO(g) H = 180 kJ
2NO(g)  N2(g) + O2(g) H = 180 kJ
2. If the coefficients of a reaction are multiplied by an integer, H is multiplied by that same integer.
6NO(g)  3N2(g) + 3O2(g) H = 3(180 kJ)
H = 540 kJ

36. Two forms of carbon are graphite, the soft, black, slippery material used in “lead” pencils and as a lubricant for locks, and diamond, the brilliant, hard gemstone. Using the enthalpies of combustion for graphite (-394 kJ/mol) and diamond (-396kJ/mol), calculate ∆H for the conversion of diamond to graphite: Cdiamond(s)  Cgraphite(s)

37. Cdiamond(s)  Cgraphite(s)
Cgraphite(s) +O2 (g)  CO2 (g) ∆H= kJ Cdiamond(s) + O2 (g)  CO2 (g) ∆H=-395.4kJ CO2  Cgraphite (s) + O2 (g) ∆H=-( kJ) Cdiamond (s) +O2 (g)  CO2 (g) ∆H= kJ Cdiamond(s)  Cgraphite(s)

38. Try this one!! (B2H6) is a highly reactive boron hydride, which was once considered as a possible rocket fuel for the US space program. Calculate ∆H for the synthesis of diborane from its elements, according to the equation 2B(s) + 3H2 (g)  B2H6(g)

39. Standard Enthalpies of Formation
ΔHf° change in enthalpy that accompanies the formation of 1 mol of compound from its elements w/all substances in their standard states
Provides an alternative to Hess’s Law
ΔHf° = ΔHf° (products) -ΔHf°(reactants)

40. Standard States Compound For a gas, pressure is exactly 1 atmosphere.
For a solution, concentration is exactly 1 molar.
Pure substance (liquid or solid), it is the pure liquid or solid.
Element
The form [N2(g), K(s)] in which it exists at 1 atm and 25°C.