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Transcript of Thermodynamics A.Energy – Capacity to do work or transfer heat B.Types of Energy 1. Kinetic –...
![Page 1: Thermodynamics A.Energy – Capacity to do work or transfer heat B.Types of Energy 1. Kinetic – Energy of motion a. KE = ½ mv 2 b. Unit – Joules (kg –m.](https://reader035.fdocuments.in/reader035/viewer/2022070412/56649e975503460f94b9a6e0/html5/thumbnails/1.jpg)
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Thermodynamics
A. Energy – Capacity to do work or transfer heat
B. Types of Energy1. Kinetic – Energy of motion
a. KE = ½ mv2
b. Unit – Joules (kg –m2/s2)1 calorie = 4.184 J1000 cal = 1 Cal (nutr)
Types of Energy
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Thermodynamics
F = ma
W = Fd
K = ½ mv2
U = mgh
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Thermodynamics
c. Macroscale KE – energy
of movement (car, baseball)
d. Microscale KE – Temp.
measure of average KE of molecules
Types of Energy
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Thermodynamics
2. Potential Energy – stored energy
a. Macroscale PE – energy of position (rollercoaster ex.)
b. Microscale PE – Energy in chemical bonds.
(Food example)
Types of Energy
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Thermodynamics
System and Surroundings
1. System – What we are studying
2. Surroundings – rest of the universe
a. Open system – Car, earth
b. Closed system – Vacuum
Types of Energy
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Thermodynamics
The First Law of Thermodynamics
Energy is neither created nor destroyed. It only changes form. Energy is conserved.
E = q + w
E = change in energy
q = heat
w = work
The First Law
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Thermodynamics
B. Examples
1. Car
2. Vacuum Cleaner
The First Law
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Thermodynamics
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
The First Law
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Thermodynamics
Exothermic Endothermic• Gives off heat• Surroundings get hotter
(transfer of heat from molecules)
• H –• Hot pack, burning gas,
exercise
• Absorbs heat from surroundings
• System gets hotter, surroundings get colder
• H +
• Cold Pack, cooking
Exo and Endothermic
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Thermodynamics
A. Heat – Energy transferred because of a difference in temperature (Cooking a turkey example)
B. Extensive Property – depends on amount of materialGlass of water vs. iceberg“Which has more heat”
Enthalpy – Heat Energy
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Thermodynamics
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 Kelvin1. Unit – J/goC2. Higher the sp. heat, more energy needed to raise temp
Calorimetry
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Thermodynamics
3. Example – Wooden Spoon (~2 J/goC) vs. metal spoon (~0.50 J/goC)
4. Water
a. Very high specific heat - 4.18 J/goC
b. Oceans
c. Our bodies
Calorimetry
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Thermodynamics
C. Heat for one substance
1. Formula
q = mCpT
q = heat
m = mass (grams)
Cp = specific heat (J/goC)
T = Tfinal – Tinitial
Calorimetry
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Thermodynamics
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)
Calorimetry
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Thermodynamics
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?
Calorimetry
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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.
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Thermodynamics
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
Calorimetry
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Thermodynamics
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)
Calorimetry
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Thermodynamics
1. Calorimeter – Coffee Cup2. Used to measure heats of a chemical
reaction3 H = -q
H = -mCpTm = mass of BOTH reactants
Cp = often 4.18 J/goC for aqueous solns
Heat of Reaction
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Thermodynamics
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)
Heat of Reaction
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Thermodynamics
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)
Heat of Reaction
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Thermodynamics
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”
Standard State
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Thermodynamics
3. Examples Hfo
O2(g) 0H2(g) 0N2(g) 0F2(g), Cl2(g), Br2(l), I2(s) 0
C(graphite) 0Fe(s) 0
Standard State
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Thermodynamics
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
Standard State
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Thermodynamics
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Thermodynamics
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Thermodynamics
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Thermodynamics
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Thermodynamics
5. Writing Standard State Equations
Examples:
CO(g)
H2O(g)
KClO3(s)
Standard State
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Thermodynamics
NaNO3(s)
NH4CN(s)
Al2(CO3)3(s)
Standard State
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Thermodynamics
Fe(NO3)3(s)
CH3COOH(l)
NaOH (s)
Standard State
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Thermodynamics
6. Using Standard Enthalpies of Formation
Hor = nHf
oprod – mHf
oreactants
n,m = coefficients
Hfo = values in text
H of Reaction
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Thermodynamics
Example 1.
Calculate H for the combustion of liquid C6H6 to CO2 and water
(Ans:-3267 kJ/mol C6H6)
H of Reaction
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Thermodynamics
Example 2.
Calculate H for the combustion of liquid C2H5OH to CO2 and water
(Ans:-1367 kJ/mol C2H5OH)
H of Reaction
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Thermodynamics
Example 3.
Calculate Hfo for CaCO3(s) if you are given:
CaCO3(s) CaO(s) + CO2(g)
Hrxno = +178.1 kJ
(Ans:-1207.1 kJ/mol CaCO3)
H of Reaction
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Thermodynamics
Example 4.
Calculate Hfo for CuO(s) if you are
given:
CuO(s) + H2(g) Cu(s) + H2O(l)
Hrxno = -129.7 kJ
(Ans:-156.1 kJ/mol CuO)
H of Reaction
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Thermodynamics20.0 mL of 0.100 M NaOH(aq) is neutralized with 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 Hrxn/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?
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Thermodynamics
A. Example
2H2(g) + O2(g) 2H2O(g) H = -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
Enthalpy – Heat Energy
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Thermodynamics
B. Example 1.
How much heat is released from the combustion of 10.0 g of H2?
2H2(g) + O2(g) 2H2O(g) H = -483.6 kJ
(Ans: -1210 kJ)
Enthalpy – Heat Energy
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Thermodynamics
Example 2.
How much heat is released when 5.00 g of H2O2 decomposes?
2H2O2 2H2O + O2 H= -196 kJ
Answer: -14.4 kJ
Enthalpy – Heat Energy
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Thermodynamics
Example 3
How much heat is required to produce 25.0 g of H2O2?
Reverse above reaction
Answer: +72.1 kJ
Enthalpy – Heat Energy
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Thermodynamics
Example 4
How much H2O2 is produced if 300.0 kJ of heat are absorbed?
2H2O + O2 2H2O2 H= +196 kJ
Answer: 104 grams
Enthalpy – Heat Energy
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Thermodynamics
Example 5
How many grams of water are produced if 8437 kJ of heat are released?
C6H6 + 15/2O2 6CO2 +3H2O H= -3267 kJ
Answer: 139.5grams
Enthalpy – Heat Energy
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Thermodynamics
A. Hess’s Law - If a reaction is carried out in a series of steps, you add the H’s for the individual steps
B. Can calculate H without having to do the experiment.
C. Rules
1. If you flip the reaction, flip the sign of H
2. If you multiply the reaction, multiply theH.
Hess’s Law
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Thermodynamics
D.Example 1.
C(gr) C(diamond)
C(gr) + O2(g)CO2(g) H = -393.5 kJ
C(dia) + O2(g)CO2(g) H = -395.4 kJ
(Ans: +1.9 kJ)
Hess’s Law
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Thermodynamics
Example 2.
NO (g) + O(g) NO2(g)
NO(g) + O3(g) NO2(g) + O2 (g) H=-198.9 kJ
O3(g) 3/2 O2 (g) H=-142.3kJ
O2(g) 2 O(g) H=+495.0kJ
(Ans: -304.1 kJ)
Hess’s Law
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Thermodynamics
Example 3.
2C(s) + H2(g) C2H2(g)
C2H2(g) + 5/2O2(g) 2CO2(g) + H2O (l)
H= -1299.3 kJ
C(s) + O2(g) CO2(g) H= -393.5 kJ
H2(g) + ½O2(g) H2O(l) H= -285.8 kJ
(Ans: +226.5 kJ)
Hess’s Law
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Calculate the enthalpy for the reaction:
1/2N2(g) + 2H20(l)NO2(g) + 2H2(g)
2NH3(g) N2(g) + 3H2(g) H=161KJ
NO2(g) + 7/2H2(g) 2H2O(l) + NH3(g)
H=-378KJ
(ANS: 297.5 kJ)
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Thermodynamics
A. Food
1. Carbohydrates – break down into glucose, then CO2 and H2O
C6H12O6 + O2 CO2 + H2O
H = -2803 kJ/mol
Provide 17 kJ/g
About 4 Cal/g (4 per raisin)
Food
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Thermodynamics
2. Fats – produce CO2 and H2O as waste
38 kJ/g
9 Cal/g (9 Calories per 2 M&M’s)
Food
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Thermodynamics
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
Food
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Thermodynamics
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
Fuels
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Thermodynamics
Acid rain affected this statue at the Field Museum in Chicago (botttom is after restoration)
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Thermodynamics
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
Fuels
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Thermodynamics42.a) endo b) 127 kJ c) 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
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Thermodynamics68.
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 c) -822.2 kJ
d) -487.0 kJ
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Thermodynamics72. a) -426.74 kJ b) -382.5 kJ c) -433.7 kJ
d) -150 kJ
74. -60.6 kJ
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Mr. Fredericks’ old ID card was getting a bit dated.