Thermodynamics
Types of Energy
A. Energy – Capacity to do work or transfer
heat
B. Types of Energy
1. Kinetic – Energy of motion
a. KE = ½ mv2
b. Unit – Joules (kg –m2/s2)
1 calorie = 4.184 J
1000 cal = 1 Cal (nutr)
Thermodynamics
F = ma
W = Fd
K = ½ mv2
U = mgh
Thermodynamics
Types of Energy
c. Macroscale KE – energy
of movement (car, baseball)
d. Microscale KE – Temp.
measure of average KE of
molecules
Thermodynamics
Types of Energy
2. Potential Energy – stored energy
a. Macroscale PE – energy of
position (rollercoaster ex.)
b. Microscale PE – Energy in
chemical bonds.
(Food example)
Thermodynamics
Types of Energy
System and Surroundings
1. System – What we are studying
2. Surroundings – rest of the universe
a. Open system – Car, earth
b. Closed system – Vacuum
Thermodynamics
The First Law
The First Law of Thermodynamics
Energy is neither created nor destroyed. It only
changes form. Energy is conserved.
A. DE = q + w
DE = change in energy
q = heat
w = work
Thermodynamics
The First Law
B. Examples
1. Car
2. Vacuum Cleaner
Thermodynamics
The First Law
C. Transfer of Energy
1. Car – Chemical to heat to mechanical
D. Consequences of the First Law
1. No machine is 100% efficient
2. Always lose some energy to heat
Thermodynamics
Exo and Endothermic
Exothermic
•
•
•
•
Gives off heat
Surroundings get hotter
(transfer of heat from
molecules)
DH –
Hot pack, burning gas,
exercise
Endothermic
• Absorbs heat from
surroundings
• System gets hotter,
surroundings get colder
• DH +
• Cold Pack, cooking
Thermodynamics
Enthalpy – Heat Energy
A. Heat – Energy transferred because of a
difference in temperature (Cooking a turkey
example)
B. Extensive Property – depends on amount of
material
Glass of water vs. iceberg
“Which has more heat”
Thermodynamics
Calorimetry
A. Calorimetry – Measuring heat by
measuring a temperature change
B. Specific Heat – Amount of heat to raise
temperature of one gram of a substance
one degree Celsius or Kelvin
1. Unit – J/goC
2. Higher the sp. heat, more energy
needed to raise temp
Thermodynamics
Calorimetry
3. Example – Wooden Spoon (~2 J/goC) vs. metal
spoon (~0.50 J/goC)
4. Water
a. Very high specific heat b. Oceans
c. Our bodies
4.18 J/goC
Thermodynamics
Calorimetry
C. Heat for one substance
1. Formula
q = mCpDT
q = heat
m = mass (grams)
Cp = specific heat (J/goC)
DT = Tfinal – Tinitial
Thermodynamics
Calorimetry
1. How much heat is needed to warm 250.0 grams of
water from 22.0oC to 98.0oC to make tea? (Ans:
79.5 kJ)
2. Calculate the specific heat of copper if 12.0 grams of
copper cools from 98.0oC to 96.2 oC and tranfers
8.32 J of heat. (0.385 J/goC)
Thermodynamics
Calorimetry
a. Calculate how much heat is absorbed by 50.0 kg of
rocks if their temperature increases by 12.0 oC?
Assume the specific heat is 0.82 J/g oC. (492 kJ)
b. Calculate how many calories of heat they absorbed.
(117 kcal)
c. Calculate the final temperature of the rocks if they
absorbed 450 kJ of heat and began at a temperature
of 26.9 oC. (37.9 oC)
d. 50 kg of Copper (0.385 J/goC)absorbs the same
amount of heat as the rocks in part c. Would the
temperature change be greater or less for copper?
Thermodynamics
A student heats 60.0 grams of iron to 98.0oC in a
hot water bath. They then submerge the metal in
150.0 g of water (Cp = 4.18 J/goC) at 20.0oC.
After a few minutes, the cup and metal reach a
final temperature of 23.2oC.
a. Calculate the heat gained by the water.
b. Calculate the specific heat (Cp) of the metal.
Thermodynamics
Calorimetry
3. Heat Capacity – heat to raise
temperature of a particular object by 1K;
unit of J/K
a. Used only for one object
b. Car part
c. heat capacity= mCp
Thermodynamics
Calorimetry
a. How many Joules of heat are required to raise
1 gram of water by 1 degree Celsius?
b. Calculate the molar heat capacity of water?
(Ans: 75.2 J/ oC)
c. Calculate the heat capacity of 250.0 grams of
water? (Ans: 1050 J/K)
d. Calculate the heat capacity of 6.20 mole of
iron? (Ans: 156 J/oC)
Thermodynamics
Heat of Reaction
1. Calorimeter – Coffee Cup
2. Used to measure heats of a chemical
reaction
3. DH = -q
DH = -mCpDT
m = mass of BOTH reactants
Cp = often 4.18 J/goC for
aqueous solns
Thermodynamics
Heat of Reaction
Example 1.
When a student mixes 50.0 mL of 1.00 M HCl
and 50.0 mL of 1.00 M NaOH in a calorimeter,
the temperature rises from 21.0oC to 27.5oC.
Calculate the change in enthalpy for the
reaction.
(Ans: -2.7 kJ and –54 kJ/mol NaOH)
Thermodynamics
Heat of Reaction
Example 2.
When 50.0 mL of 0.100 M AgNO3 and 50.0
mL of 0.100 M HCl are mixed, the temperature
rises from 22.30oC to 23.11oC. Calculate the
change in enthalpy for the reaction per mole of
AgNO3.
(Ans: –68 kJ/mol AgNO3)
Thermodynamics
Standard State
A. Standard State – State of an element
@25oC and 1 atm.
1. Standard Enthalpy of formation of an
element in its st. state is zero
2. Takes no energy to make an element
assume st. state.
“Exists in this form naturally”
Thermodynamics
Standard State
3. Examples
O2(g)
H2(g)
N2(g)
F2(g), Cl2(g), Br2(l), I2(s)
C(graphite)
Fe(s)
DHfo
0
0
0
0
0
0
Thermodynamics
Standard State
4. Carbon Allotropes – different physical forms of the
same element
a. Graphite
b. Diamond
c. Buckyball (Buckminster Fuller)
C(gr)
0
C(dia)
1.88 kJ/mol
C60(s)
2269 kJ/mol
Thermodynamics
Thermodynamics
Thermodynamics
Thermodynamics
Thermodynamics
Standard State
5. Writing Standard State Equations
Examples:
 CO(g)
 H2O(g)
 KClO3(s)
Thermodynamics
Standard State
 NaNO3(s)
 NH4CN(s)
 Al2(CO3)3(s)
Thermodynamics
Standard State
 Fe(NO3)3(s)
 CH3COOH(l)
 NaOH (s)
Thermodynamics
DH of Reaction
6. Using Standard Enthalpies of Formation
DHor = SnDHfoprod – S mDHforeactants
n,m = coefficients
DHfo = values in text
Thermodynamics
DH of Reaction
Example 1.
Calculate DH for the combustion of liquid
C6H6 to CO2 and water
(Ans:-3267 kJ/mol C6H6)
Thermodynamics
DH of Reaction
Example 2.
Calculate DH for the combustion of liquid
C2H5OH to CO2 and water
(Ans:-1367 kJ/mol C2H5OH)
Thermodynamics
DH of Reaction
Example 3.
Calculate DHfo for CaCO3(s) if you are
given:
CaCO3(s)  CaO(s) + CO2(g)
DHrxno = +178.1 kJ
(Ans:-1207.1 kJ/mol CaCO3)
Thermodynamics
DH of Reaction
Example 4.
Calculate DHfo for CuO(s) if you are
given:
CuO(s) + H2(g)  Cu(s) + H2O(l)
DHrxno = -129.7 kJ
(Ans:-156.1 kJ/mol CuO)
20.0
mL of 0.100 M NaOH(aq) is neutralized with
Thermodynamics
0.0950 M HNO3(aq).
a. Write the balanced chemical reaction for this
process.
b. Calculate the volume of HNO3 required.(21.1 mL)
c. When these two volumes are mixed, the
temperature rises from 22.3oC to 23.0 oC.
Calculate the DHrxn/mole of NaOH. (-60.1 kJ/mol)
d. Calculate heat of reaction the using the heats of
formation values found in the appendix (not the
pancreas). (-55.83 kJ/mol)
e. How do your two answers compare?
Thermodynamics
Enthalpy – Heat Energy
A. Example
2H2(g) + O2(g)  2H2O(g) DH = -483.6 kJ
1. Hindenberg
2. Heat is released to the surroundings.
3. -483.6 kJ released per 2 mole of H2, 1 mole
O2, or 2 mole of H2O
Thermodynamics
Enthalpy – Heat Energy
B. Example 1.
How much heat is released from the
combustion of 10.0 g of H2?
2H2(g) + O2(g)  2H2O(g) DH = -483.6 kJ
(Ans: -1210 kJ)
Thermodynamics
Enthalpy – Heat Energy
Example 2.
How much heat is released when 5.00 g of
H2O2 decomposes?
2H2O2  2H2O + O2 DH= -196 kJ
Answer:
-14.4 kJ
Thermodynamics
Enthalpy – Heat Energy
Example 3
How much heat is required to produce
25.0 g of H2O2?
Reverse above reaction
Answer: +72.1 kJ
Thermodynamics
Enthalpy – Heat Energy
Example 4
How much H2O2 is produced if 300.0 kJ
of heat are absorbed?
2H2O + O2  2H2O2 DH= +196 kJ
Answer: 104 grams
Thermodynamics
Enthalpy – Heat Energy
Example 5
How many grams of water are produced
if 8437 kJ of heat are released?
C6H6 + 15/2O2  6CO2 +3H2O
DH= -3267 kJ
Answer: 139.5grams
Thermodynamics
Hess’s Law
A. Hess’s Law - If a reaction is carried out in a
series of steps, you add the DH’s for the
individual steps
B. Can calculate DH without having to do the
experiment.
C. Rules
1. If you flip the reaction, flip the sign of DH
2. If you multiply the reaction, multiply the DH.
Thermodynamics
Hess’s Law
D.Example 1.
C(gr)  C(diamond)
C(gr) + O2(g)CO2(g) DH = -393.5 kJ
C(dia) + O2(g)CO2(g) DH = -395.4 kJ
(Ans: +1.9 kJ)
Thermodynamics
Hess’s Law
Example 2.
NO (g) + O(g)  NO2(g)
NO(g) + O3(g) NO2(g) + O2 (g) DH=-198.9 kJ
O3(g) 3/2 O2 (g)
DH=-142.3kJ
O2(g) 2 O(g)
DH=+495.0kJ
(Ans: -304.1 kJ)
Thermodynamics
Hess’s Law
Example 3.
2C(s) + H2(g)  C2H2(g)
C2H2(g) + 5/2O2(g) 2CO2(g) + H2O (l)
DH= -1299.3 kJ
C(s) + O2(g) CO2(g)
DH= -393.5 kJ
H2(g) + ½O2(g) H2O(l)
DH= -285.8 kJ
(Ans: +226.5 kJ)
Calculate the enthalpy for the reaction:
1/2N2(g) + 2H20(l)NO2(g) + 2H2(g)
2NH3(g)  N2(g) + 3H2(g)
DH=161KJ
NO2(g) + 7/2H2(g) 2H2O(l) + NH3(g)
DH=-378KJ
(ANS: 297.5 kJ)
Thermodynamics
Food
A. Food
1. Carbohydrates – break down into glucose,
then CO2 and H2O
C6H12O6 + O2  CO2 + H2O
DH = -2803 kJ/mol
Provide 17 kJ/g
About 4 Cal/g (4 per raisin)
Thermodynamics
Food
2. Fats – produce CO2 and H2O as waste
38 kJ/g
9 Cal/g (9 Calories per 2 M&M’s)
Thermodynamics
Food
3. Proteins – Produce CO2, H2O and (NH2)2CO as
waste
a. Amino Acid structure responsible for the
urea.
b. 17 kJ/g
4 Cal/g
Thermodynamics
Fuels
1. Fossil Fuels (coal, oil, natural gas, propane,
diesel, gasoline)
2. Coal
a. All anthracite in PA
b. Fossilized plant matter
c. Sulfur often in coal (SO3)
SO3 + H2O  H2SO4
Thermodynamics
Acid rain affected this statue at
the Field Museum in Chicago
(botttom is after restoration)
Thermodynamics
Fuels
3. Hydrogen
a. Can be produced by breaking down water
or fossil fuels (Maybe solar powered)
b. Burns to produce water
c. 20 to 50 % more efficient than gasoline
d. Explosive
42.a) endo
b) 127 kJ c)
Thermodynamics
1.15 g d) -165 kJ
44.a) -18.8 kJ b) -5.14 kJ c) not spontaneous
46.a) -630 kJ b) + 630 kJ c) reverse favored
50.a) Hg(l)
b) 70 J
52. 4110 J or 4.11 kJ
54.a) 25 kJ/mol b) endothermic
62. -867.7kJ
64.155.7 kJ
68.
Thermodynamics
½ H2(g) + ½ Br2(l)  HBr(g)
Ag(s) + ½ N2(g) + 3/2 O2(g)  AgNO3(s)
2Fe(s) + 3/2 O2(g)  Fe2O3(s)
2C(gr) + 2H2(g) + O2(g)  CH3COOH(l)
a) -36.23 kJ
b) -124.4 kJ
d) -487.0 kJ
c) -822.2 kJ
72. a) -426.74 kJ
b) -382.5 kJ
Thermodynamics
d) -150 kJ
74. -60.6 kJ
c) -433.7 kJ
Mr. Fredericks’ old ID card was getting a bit dated.