1. THERMOCHEMISTRY

2. Chapter 6
Thermochemistry

3. Definitions #1 Energy: The capacity to do work or produce heat
Potential Energy: Energy due to position or composition
Kinetic Energy: Energy due to the motion of the object

4. Radiant energy comes from the sun and is earth’s primary energy source
Some types Energy
Radiant energy comes from the sun and is earth’s primary energy source
Thermal energy is the energy associated with the random motion of atoms and molecules
Chemical energy is the energy stored within the bonds of chemical substances
Nuclear energy is the energy stored within the collection of neutrons and protons in the atom
Potential energy is the energy available by virtue of an object’s position
Which are Potential? Kinetic?
6.1

5. Definitions #2
Law of Conservation of Energy: Energy can neither be created nor destroyed, but can be converted between forms
The First Law of Thermodynamics: The total energy content of the universe is constant

6. First law of thermodynamics
DEsystem + DEsurroundings = 0 or
DEsystem = -DEsurroundings
C3H8 + 5O CO2 + 4H2O
Exothermic chemical reaction!
Chemical energy lost by combustion = Energy gained by the surroundings
system
surroundings
6.3

7. State Functions depend ONLY on the present state of the system
ENERGY IS A STATE FUNCTION
A person standing at the top of Mt. Everest has the same potential energy whether they got there by hiking up, or by falling down from a plane!
WORK IS NOT A STATE FUNCTION
WHY NOT???

8. , pressure, volume, temperature
Thermodynamics is the scientific study of the interconversion of heat and other kinds of energy.
State functions are properties that are determined by the state of the system, regardless of how that condition was achieved.
energy
, pressure, volume, temperature
DE = Efinal - Einitial
DP = Pfinal - Pinitial
DV = Vfinal - Vinitial
DT = Tfinal - Tinitial
Potential energy of hiker 1 and hiker 2 is the same even though they took different paths.
6.3

9. E = q + w E = change in internal energy of a system
q = heat flowing into or out of the system
-q if energy is leaving to the surroundings
+q if energy is entering from the surroundings
w = work done by, or on, the system
-w if work is done by the system on the
surroundings
+w if work is done on the system by the
surroundings

10. Work, Pressure, and Volume
Expansion
Compression
+V (increase)
-V (decrease)
-w results
+w results
Esystem decreases
Esystem increases
Work has been done by the system on the surroundings
Work has been done on the system by the surroundings

11. Work is not a state function!
Work Done On the System
w = F x d
w = -P DV
DV > 0
-PDV < 0
wsys < 0
P x V = x d3 = F x d = w F d2
Work is not a state function!
Dw = wfinal - winitial
initial
final
6.3

12. W = -0 atm x 3.8 L = 0 L•atm = 0 joules
A sample of nitrogen gas expands in volume from 1.6 L to 5.4 L at constant temperature. What is the work done in joules if the gas expands from a vacuum (0.0 atm) to a constant pressure of 3.7 atm?
w = -P DV
(a)
DV = 5.4 L – 1.6 L = 3.8 L
P = 0 atm
W = -0 atm x 3.8 L = 0 L•atm = 0 joules
(b)
DV = 5.4 L – 1.6 L = 3.8 L
P = 3.7 atm
w = -3.7 atm x 3.8 L = L•atm
w = L•atm x
101.3 J
1L•atm
= J
6.3

13. Enthalpy and the First Law of Thermodynamics
DE = q + w
q = DH and w = -PDV
At constant pressure:
DE = DH - PDV
DH = DE + PDV
6.4

14. DH = H (products) – H (reactants)
Enthalpy (H) is used to quantify the heat flow into or out of a system in a process that occurs at constant pressure.
DH = H (products) – H (reactants)
DH = heat given off or absorbed during a reaction at constant pressure
Hproducts < Hreactants
Hproducts > Hreactants
DH < 0
DH > 0
6.4

15. Energy Change in Chemical Processes
Endothermic:
Reactions in which energy flows into the system as the reaction proceeds.
+ qsystem
- qsurroundings
Exothermic:
Reactions in which energy flows out of the system as the reaction proceeds.
- qsystem
+ qsurroundings

16. 2H2 (g) + O2 (g) 2H2O (l) + energy
Exothermic process is any process that gives off heat – transfers thermal energy from the system to the surroundings.
2H2 (g) + O2 (g) H2O (l) + energy
H2O (g) H2O (l) + energy
Endothermic process is any process in which heat has to be supplied to the system from the surroundings.
energy + H2O (s) H2O (l)
energy + 2HgO (s) Hg (l) + O2 (g)
6.2

17. Endothermic Reactions

18. Exothermic Reactions

19. Review so far… Energy is…. 1st law of thermodynamics
Potential verses Kinetic Energy
State Function
Enthalpy (q, heat, …)
Work is…
E = q + w and w = -P DV
Endothermic
Exothermic

20. 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
1 calorie is the heat required to raise the temperature of 1 gram of water by 1 C
1 BTU is the heat required to raise the temperature of 1 pound of water by 1 F

21. Heat (q) absorbed or released:
The specific heat (s) of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius.
The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius.
C = m x s
Heat (q) absorbed or released:
q = m x s x Dt
q = C x Dt
Dt = tfinal - tinitial
6.5

22. 1 joule = 4.184 calories The Joule
The unit of heat used in modern thermochemistry is the Joule
1 joule = calories

23. A Bomb Calorimeter (commonly used by Oak Ridge HS)

24. A Cheaper Calorimeter (commonly used by WSHS)

25. Specific Heat
The amount of heat required to raise the temperature of one gram of substance by one degree Celsius.
Substance
Specific Heat (J/g·K)
Water (liquid)
4.184
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)

26. Calculations Involving Specific HeatOR
or  q = Cp * m * T
s = Specific Heat Capacity
q = Heat lost or gained
T = Temperature change
m = mass in grams

27. Chemistry in Action: Fuel Values of Foods and Other Substances
C6H12O6 (s) + 6O2 (g) CO2 (g) + 6H2O (l) DH = kJ/mol
1 cal = J
1 Cal = 1000 cal = 4184 J
Substance
DHcombustion (kJ/g)
Apple -2
Beef -8
Beer
-1.5
Gasoline
-34

28. Because there is no way to measure the absolute value of the enthalpy of a substance, must I measure the enthalpy change for every reaction of interest?
Establish an arbitrary scale with the standard enthalpy of formation (DH0) as a reference point for all enthalpy expressions. f
Standard enthalpy of formation (DH0) is the heat change that results when one mole of a compound is formed from its elements at a pressure of 1 atm. f
The standard enthalpy of formation of any element in its most stable form is zero.
DH0 (O2) = 0 f
DH0 (C, graphite) = 0 f
DH0 (O3) = 142 kJ/mol f
DH0 (C, diamond) = 1.90 kJ/mol f
6.6

29. Hess’s Law
“In going from a particular set of reactants to a particular set of products, the change in enthalpy is the same whether the reaction takes place in one step or a series of steps.”

30. Hess’s Law

31. Hess’s Law Example Problem
Calculate H for the combustion of methane, CH4:
CH4 + 2O2  CO2 + 2H2O
    Reaction
Ho  
C + 2H2  CH4 kJ
C + O2  CO2 kJ
H2 + ½ O2  H2O kJ
CH4  C + 2H kJ
Step #1: CH4 must appear on the reactant side, so we reverse reaction #1 and change the sign on H.

32. Hess’s Law Example Problem
Calculate H for the combustion of methane, CH4:
CH4 + 2O2  CO2 + 2H2O
    Reaction
Ho  
C + 2H2  CH4 kJ
C + O2  CO2 kJ
H2 + ½ O2  H2O kJ
CH4  C + 2H kJ
C + O2  CO kJ
Step #2: Keep reaction #2 unchanged, because CO2 belongs on the product side

33. Hess’s Law Example Problem
Calculate H for the combustion of methane, CH4:
CH4 + 2O2  CO2 + 2H2O
    Reaction
Ho  
C + 2H2  CH4 kJ
C + O2  CO2 kJ
H2 + ½ O2  H2O kJ
CH4  C + 2H kJ
C + O2  CO kJ
2H2 + O2  2 H2O kJ
Step #3: Multiply reaction #3 by 2

34. Hess’s Law Example Problem
Calculate H for the combustion of methane, CH4:
CH4 + 2O2  CO2 + 2H2O
    Reaction
Ho  
C + 2H2  CH4 kJ
C + O2  CO2 kJ
H2 + ½ O2  H2O kJ
CH4  C + 2H kJ
C + O2  CO kJ
2H2 + O2  2 H2O kJ
CH4 + 2O2  CO2 + 2H2O kJ
Step #4: Sum up reaction and H

35. Calculation of Heat of Reaction
Calculate H for the combustion of methane, CH4:
CH4 + 2O2  CO2 + 2H2O
Hrxn =  Hf(products) -   Hf(reactants)
    Substance
Hf  
CH4 kJ O2
0 kJ
CO2 kJ
H2O kJ
Hrxn = [ kJ + 2( kJ)] – [-74.80kJ]
Hrxn = kJ

36. The standard enthalpy of reaction (DH0 ) is the enthalpy of a reaction carried out at 1 atm.
rxn
aA + bB cC + dD
DH0
rxn
dDH0 (D) f
cDH0 (C) = [ + ] -
bDH0 (B)
aDH0 (A)
DH0
rxn
nDH0 (products) f = S
mDH0 (reactants) -
Hess’s Law: When reactants are converted to products, the change in enthalpy is the same whether the reaction takes place in one step or in a series of steps.
(Enthalpy is a state function. It doesn’t matter how you get there, only where you start and end.)
6.6

37. C (graphite) + 1/2O2 (g) CO (g)
CO (g) + 1/2O2 (g) CO2 (g)
C (graphite) + O2 (g) CO2 (g)
6.6

38. Calculate the standard enthalpy of formation of CS2 (l) given that:
C(graphite) + O2 (g) CO2 (g) DH0 = kJ
rxn
S(rhombic) + O2 (g) SO2 (g) DH0 = kJ
rxn
CS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = kJ
rxn
1. Write the enthalpy of formation reaction for CS2
C(graphite) + 2S(rhombic) CS2 (l)
2. Add the given rxns so that the result is the desired rxn.
rxn
C(graphite) + O2 (g) CO2 (g) DH0 = kJ
2S(rhombic) + 2O2 (g) SO2 (g) DH0 = x2 kJ
rxn +
CO2(g) + 2SO2 (g) CS2 (l) + 3O2 (g) DH0 = kJ
rxn
C(graphite) + 2S(rhombic) CS2 (l)
DH0 = (2x-296.1) = 86.3 kJ
rxn
6.6

39. 2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
Benzene (C6H6) burns in air to produce carbon dioxide and liquid water. How much heat is released per mole of benzene combusted? The standard enthalpy of formation of benzene is kJ/mol.
2C6H6 (l) + 15O2 (g) CO2 (g) + 6H2O (l)
DH0
rxn
nDH0 (products) f = S
mDH0 (reactants) -
DH0
rxn
6DH0 (H2O) f
12DH0 (CO2) = [ + ] -
2DH0 (C6H6)
DH0
rxn =
[ 12x– x–187.6 ] – [ 2x49.04 ] = kJ
-5946 kJ
2 mol
= kJ/mol C6H6
6.6

40. Chemistry in Action: Bombardier Beetle Defense
C6H4(OH)2 (aq) + H2O2 (aq) C6H4O2 (aq) + 2H2O (l) DH0 = ?
C6H4(OH)2 (aq) C6H4O2 (aq) + H2 (g) DH0 = 177 kJ/mol
H2O2 (aq) H2O (l) + ½O2 (g) DH0 = kJ/mol
H2 (g) + ½ O2 (g) H2O (l) DH0 = -286 kJ/mol
DH0 = – 286 = -204 kJ/mol
Exothermic!

41. DHsoln = Hsoln - Hcomponents
The enthalpy of solution (DHsoln) is the heat generated or absorbed when a certain amount of solute dissolves in a certain amount of solvent.
DHsoln = Hsoln - Hcomponents
Which substance(s) could be used for melting ice?
Which substance(s) could be used for a cold pack?
6.7

42. The Solution Process for NaCl
DHsoln = Step 1 + Step 2 = 788 – 784 = 4 kJ/mol
6.7