1. Thermochemistry

2. Chapter Outline Overview of Energy in Chemical Reaction
First Law of Thermodynamics
Heat, Specific heat, Calorimetry
Enthalpy of Reaction
Standard Enthalpy of Formation
Hess’s law
Tro: Chemistry: A Molecular Approach, 2/e

3. Chemical Hand Warmers and much more
Most hand warmers uses the heat from rxn:
4 Fe(s) + 3 O2(g) → 2 Fe2O3(s)
The temperature change in your hand depends on
the size of the hand warmer
the size of your glove, etc.
mainly, the amount of heat from the reaction
Most chemical rxn involves energy change
Chemical energy provides nearly 70% electricity in the US. Nuclear 11%, Hydroelectric 17%.
All batteries (car, portable electronic device)
Tro: Chemistry: A Molecular Approach, 2/e

4. About Energy
Even though chemistry is the study of matter, energy affects matter
Energy is anything that has the capacity to do work.
Example:
Kinetic energy in moving water;
Potential energy: meteorite near the Earth
Chemical energy (a form of potential energy) in gasoline or battery
Energy exchange through doing work or heat exchange.
Tro: Chemistry: A Molecular Approach, 2/e

5. Units of Energy
Joule (J): the amount of energy for 1 Newton force for a distance of 1 meter
1 J = 1 N∙m = 1 kg∙m2/s2
calorie (cal): the amount of energy needed to raise the temperature of one gram of water 1°C
kcal = energy needed to raise 1000 g of water 1°C
food Calories = kcals
Energy Conversion Factors
1 calorie (cal) =
4.184 joules (J) (exact)
1 Calorie (Cal)
1000 cal = 1 kcal = 4184 J
1 kilowatt-hour (kWh)
3.60 x 106 J
Tro: Chemistry: A Molecular Approach, 2/e

6. Classification of Energy
Kinetic energy is energy of motion or energy that is being transferred
Thermal energy is the energy associated with temperature
thermal energy is a form of kinetic energy
Tro: Chemistry: A Molecular Approach, 2/e

7. Some Forms of Energy Electrical Heat or thermal energy
kinetic energy associated with the flow of electrical charge
Heat or thermal energy
kinetic energy associated with molecular motion
Light or radiant energy
kinetic energy associated with energy transitions in an atom
Nuclear
potential energy in the nucleus of atoms
Chemical
potential energy due to the structure/bonding of the atoms/molecules
Tro: Chemistry: A Molecular Approach, 2/e

8. Exchange of Energy Work = force acting over a distance
Energy = Work = Force x Distance
Example: Compressor convert electric energy to take away heat from room.
Heat is the flow of energy caused by a difference in temperature.
Energy can be exchanged between objects through contact
How? Fast-moving particles pass the kinetic energy to slow moving particles
Atomic level: Heat exchange through molecular collisions
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9. Energy Exchange: System vs. Surroundings
System: the material or process within which we are studying the energy changes within. Example: Chemical Rxn.
Surroundings: everything else with which the system can exchange energy with. Example: Solvent where rxn occurs.
Left: System gives energy to the Surroundings.
Right: System takes energy from the Surroundings.
Surroundings
System
Surroundings
System
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10. The First Law of Thermodynamics: Law of Conservation of Energy
Thermodynamics is the study of energy and its interconversions
The First Law of Thermodynamics: the Law of Conservation of Energy
This means that the total amount of energy in the universe is constant
You can therefore never design a system that will continue to produce energy without some source of energy
Tro: Chemistry: A Molecular Approach, 2/e

11. Energy Flow and Conservation of Energy
Conservation of energy requires that the sum of the energy changes (DE) in the system and the surroundings must be zero
DEnergyuniverse = 0 = DEnergysystem + DEnergysurroundings
D Is the symbol that is used to mean change
final amount – initial amount
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12. Internal Energy
The internal energy is the sum of the kinetic and potential energies of all of the particles that compose the system
The change in the internal energy of a system only depends on the amount of energy in the system at the beginning and end
DE = Efinal – Einitial
Internal energy is a state function, a mathematical function whose result only depends on the initial and final conditions, not on the process used
DEreaction = Eproducts − Ereactants
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13. State Function
From the foothill to the peak, the change in potential energy is the same regardless of pathways.
Potential energy is a state function. It depends only on the difference in states (elevation between the base and the peak)
Traveling distance is NOT a state function
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14. Energy Diagrams
Energy diagrams: showing the direction of energy flow during a process
Internal Energy
initial
final
energy increase
If the final condition has a larger amount of internal energy than the initial condition, the change in the internal energy will be +
Internal Energy
initial
final
energy decrease
If the final condition has a
smaller amount of internal
energy than the initial
condition, the change in the
internal energy will be ─
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15. Energy Flow: System  Surrounding
When energy flows out of a system, it must all flow into the surroundings
When energy flows out of a system, DEsystem is ─
Surroundings gains energy, DEsurroundings is +
So ─ DEsystem= DEsurroundings
Example: Hot water (system) lose heat to cool air in the room (surroundings)
Surroundings
DE +
System
DE ─
Tro: Chemistry: A Molecular Approach, 2/e

16. Energy Flow: System  Surroundings
When energy flows into a system, it must all come from the surroundings
System gains energy, DEsystem is +
Energy flows out of the surroundings, DEsurroundings is ─
So DEsystem= ─ DEsurroundings
Example: Rechargeable battery (system) receives energy from charger (surroundings)
Surroundings
DE ─
System
DE +
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17. Work and Heat in Energy Exchange
Energy is exchanged between the system and surroundings through heat and work
q = heat (thermal) energy
w = work energy
q and w are NOT state functions, their value depends on the process
DE = q + w
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18. Energy Exchange in real life
Example:
Combustion reaction in engine: System (reaction) loses heat (q < 0) and does work (w < 0) to the surroundings.
Charging battery: System (battery) loses heat (gets warm, q < 0) but gains work (w > 0) from the surroundings
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19. Heat Exchange
Heat is the exchange of thermal energy between the system and surroundings
Heat exchange occurs when system and surroundings have a difference in temperature
Temperature is the measure of the amount of thermal energy within a sample of matter
Heat flows from matter with high temperature to matter with low temperature until both objects reach the same temperature
thermal equilibrium
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20. Temperature change and Heat Capacity
When a system absorbs heat q, its temperature increases
The increase in temperature DT is directly proportional to the amount of heat absorbed: More heat  higher temperature increase.
The proportionality constant is called the heat capacity, C. C = q / DT
units of C are J/°C or J/K
Given same amount of heat, object with larger heat capacity will experience smaller temperature rise. Example, same heat given, 1 kg water gets warmer than 2 kg water.
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21. Factors Affecting Heat Capacity
The heat capacity of an object depends on its amount of matter
usually measured by its mass
200 g of water requires twice as much heat to raise its temperature by 1°C as does 100 g of water
The heat capacity of an object depends on the type of material
1000 J of heat energy will raise the temperature of 100 g sand by 12°C, but only raise the temperature of 100 g water by 2.4 °C
Tro: Chemistry: A Molecular Approach, 2/e

22. Specific Heat Capacity
The specific heat (c) is the amount of heat energy required to raise the temperature of one gram of a substance by 1°C
Measure of a substance’s intrinsic ability to absorb heat
Unit = J/(g∙°C)
c = q/(mass x DT)
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23. How to measure Heat Energy?
The heat capacity of an object is proportional to its mass and the specific heat of the material
Heat calculation from specific heat: Heat absorbed or released can be calculated as if we know the mass, the specific heat, and the temperature change of the object
Heat = (mass) x (specific heat) x (temp. change)
q = m x c x DT
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24. Find Heat: How much heat is absorbed by a copper penny with mass 3
Find Heat: How much heat is absorbed by a copper penny with mass 3.10 g whose temperature rises from −8.0 °C to 37.0 °C? Specific heat of copper is J/g °C
q = 53.7 J
DT = 45.0°C

25. Heat Transfer and Temperature Change
When two objects are placed in contact, heat flows from the material at the higher temperature to the material at the lower temperature
Heat flows until both materials reach the same final temperature
Heat lost by the hot material equals the heat gained by the cold material
qhot = −qcold
mhotCs,hotThot = −(mcoldCs,coldTcold)
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26. Measure Heat from Chem. Rxn: Bomb Calorimeter
Calorimeter for Heat measurement
Used to measure DE because it is a constant volume system
The heat capacity of the calorimeter = heat absorbed by the calorimeter per degree rise in temperature. Also called the calorimeter constant
Ccal, kJ/ºC
Tro: Chemistry: A Molecular Approach, 2/e

27. DHreaction = qreaction at constant pressure
Enthalpy
The enthalpy, H, of a system is the sum of the internal energy E of the system and the product of pressure and volume
H is a state function
H = E + PV
The enthalpy change, DH, of a reaction is the heat evolved in a reaction at constant pressure
DHreaction = qreaction at constant pressure
Usually DH and DE are similar in value, the difference is largest for reactions that produce or use large quantities of gas
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28. Endothermic and Exothermic Reactions
DH < 0 (or Hfinal < Hinitial): heat is released by the system
Heat-releasing reactions: exothermic reactions
Examples:
Chemical heat packs
Combustion reaction, Explosion reaction
DH > 0 (or Hfinal > Hinitial): heat is absorbed by the system
Heat-absorbing reactions: endothermic reactions
Chemical cold packs contain NH4NO3 that dissolves in water in an endothermic process ─ your hands get cold because the pack is absorbing your heat
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29. C3H8(g) + 5 O2(g) → 3 CO2(g) + 4 H2O(g) DH = −2044 kJ
Enthalpy of Reaction
The enthalpy change in a chemical reaction is an extensive property (aka, depending on the amount)
the more reactants you use, the larger the enthalpy change. More C3H8? More enthalpy change! 
By convention, we calculate the enthalpy change for the number of moles of reactants in the reaction as written
C3H8(g) + 5 O2(g) → 3 CO2(g) + 4 H2O(g) DH = −2044 kJ
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30. Example: How much heat is evolved in the complete combustion of 13
Example: How much heat is evolved in the complete combustion of 13.2 kg of C3H8(g)?
C3H8(g) + 5O2(g)→3CO2(g) + 4H2O(g) DH = −2044 kJ
mol kJ kg g

31. Measuring DH Calorimetry at Constant Pressure
Reactions done in aqueous solution are at constant pressure
open to the atmosphere
Instrument: Foam cup calorimeter contains the solution (foams provides great insulation! )
qreaction = ─ qsolution
= ─(masssolution x Cs, solution x DT)
DHreaction = qconstant pressure = qreaction
to get DHreaction per mol, divide by the number of moles of limiting reagent
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32. More Example: What is DHrxn/mol Mg for the reaction Mg(s) + 2 HCl(aq) → MgCl2(aq) + H2(g) if g Mg reacts in mL of solution and the solution changes the temperature from 25.6°C to 32.8 °C? Assume dsoln = 1.00 g/mL, Csoln = 4.18 J/g∙°C
DHrxn/mol Mg = -4.6E+5 J/mol Mg
Qsoln = 3.0E+3 J
Mol Mg = 6.499E-3 mol

33. Relationships Involving DHrxn
When reaction is multiplied by a factor, DHrxn is multiplied by that factor
because DHrxn is extensive
C(s) + O2(g) → CO2(g) DH = −393.5 kJ
2 C(s) + 2 O2(g) → 2 CO2(g) DH = ____________
If a reaction is reversed, then the sign of DH is changed
CO2(g) → C(s) + O2(g) DH = ____________
−787.0 kJ, kJ
Tro: Chemistry: A Molecular Approach, 2/e

34. Finding DHrxn for new reaction: Hess’s Law
If a reaction can be expressed as a series of steps, then the DHrxn for the overall reaction is the sum of the heats of reaction for each step
A + 2B → C DH1
C + E → F DH2
F → G DH3
A + 2B + C + E + F → C + F + G
Total reaction:
A + 2B + E → G DH4 = DH1 + DH2 + DH3
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35. How to use Hess’s Law from given DH’s
Compare the target reaction and given reactions
Two basic manipulations: Reverse and/or Multiply
If one formula exists in more than one of the given rxns, do not attempt to manipulate eqn using this formula.
Example: Find enthalpy change for A + 2B → 3F
Given: A + 2B → 2C DH1
3G → 2C DH2
F → G DH3
2C → 3G DH2
3G → 3F x DH3
A + 2B → 3F DH = DH1 + (-DH2) + (-3DH3)
A + 2B → 2C DH1
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36. Example:
Calculating ΔHºrxn from given ΔHº Values
Find DHrxn for the rxn: 4NH3(g) + 5O2(g) → 4NO(g) + 6H2O(g)
Given: N2(g) + 3H2(g) → 2NH3(g) DH1 = kJ
N2(g) + O2(g) → 2NO(g) DH2 = kJ
2H2(g) + O2(g) → 2H2O(g) DH3 = kJ
DHrxn= kJ

37. Using Hess’ Law to Calculate an Unknown DH
The catalytic converter in automobiles ultilizes the following reaction to convert pollutants CO and NO into harmless gases:
CO (g) + NO (g) → CO2 (g) + ½N2 (g)
Given the following information, calculate the unknown DH:
Equation A: CO (g) + ½O2 (g) → CO2 (g) DHA = –283.0 kJ
Equation B: N2 (g) + O2 (g) → 2NO (g) DHB = kJ
DHrxn = –373.3 kJ

38. Standard Conditions
The standard state is the state of a material at a defined set of conditions
pure gas at exactly 1 atm pressure
pure solid or liquid in its most stable form at exactly 1 atm pressure and temperature of interest
usually 25 °C
substance in a solution with concentration 1 M
The standard enthalpy change, DH°, is the enthalpy change when all reactants and products are in their standard states
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39. Standard enthalpy of formation, DHf°,
Formation Reaction: Reactions of elements in their standard state to form 1 mole of a pure compound
The standard states of all metals are solid except Hg; Nonmetals: Only Br2 is liquid, low atomic mass nonmetals are gas whilst heavy nonmetals are solid
The standard enthalpy of formation, DHf°: the enthalpy change for the reaction forming 1 mole of a pure compound from its constituent elements
the elements must be in their standard states
the DHf° for a pure element in its standard state = 0 kJ/mol
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40. Standard Enthalpies of Formation
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41. Writing Formation Reactions Write the Formation Reaction for CO(g)
The formation reaction is the reaction between the elements in the compound
C + O → CO(g)
The elements must be in their standard state
C(s, graphite) + O2(g) → CO(g)
The equation must be balanced, but the coefficient of the product compound must be 1
C(s, graphite) + ½ O2(g) → CO(g)
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42. Write the formation reactions for the following:
CO2(g)
Hg2SO4(s)
C(s, graphite) + O2(g)  CO2(g)
2 Hg(l) + S(s) + 2 O2(g)  Hg2SO4(s)
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43. More example:
Writing Formation Equations
Write balanced equations for the formation of 1 mol of the following compounds from their elements in their standard states and include DHºf.
(a) Silver chloride, AgCl, a solid at standard conditions.
(b) Calcium carbonate, CaCO3, a solid at standard conditions.
(c) Hydrogen cyanide, HCN, a gas at standard conditions.
Ag (s) + ½Cl2 (g) → AgCl (s) DHºf = –127.0 kJ
Ca (s) + C(graphite) + O2 (g) → CaCO3 (s) DH°f = – kJ 3 2
½H2 (g) + C(graphite) + ½N2 (g) → HCN (g) DHf = 135 kJ

44. Calculating Standard Enthalpy Change for a Reaction
Any reaction can be written as the sum of formation reactions (or the reverse of formation reactions) for the reactants and products
The DH° for the reaction is then the sum of the DHf° for the component reactions
DH°reaction = S n DHf°(products) − S n DHf°(reactants)
S means sum
n is the coefficient of the reaction
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45. 2. DH°reaction = S n DHf°(products) − S n DHf°(reactants)
Calculate the enthalpy change in the reaction 2 C2H2(g) + 5 O2(g) ® 4 CO2(g) + 2 H2O(l)
1. Collect the DHf° for each compound in the equation
2 C(s, gr) + H2(g) ® C2H2(g) DHf° = kJ/mol
C(s, gr) + O2(g) ® CO2(g) DHf° = − kJ/mol
H2(g) + ½ O2(g) ® H2O(l) DHf° = − kJ/mol
2. DH°reaction = S n DHf°(products) − S n DHf°(reactants)
DHrxn = − kJ
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46. 4NH3(g) + 5O2 (g) → 4NO (g) + 6H2O (g)
Calculating ΔHºrxn from ΔHºf Values
For the following reaction:
4NH3(g) + 5O2 (g) → 4NO (g) + 6H2O (g)
Calculate DHrxn from DHf values.
DHf (kJ/mol): NH3(g) -45.9, NO(g) 90.3, H2O(g)
DHrxn = SmDHºf (products) – SnDHºf (reactants)
m and n are the coefficients for products and reactants, respectively.
DHrxn = –906 kJ

47. 1. Collect the DHf° for each compound in the equation
Example: Determine the standard enthalpy of formation for nitromethane from DH°rxn 2CH3NO2(l) + 3/2 O2(g) ® 2CO2(g) + N2(g) + 3H2O(g) DHrxn = kJ/mol
1. Collect the DHf° for each compound in the equation
C(s, gr) + O2(g) ® CO2(g) DHf° = − kJ/mol
H2(g) + ½ O2(g) ® H2O(g) DHf° = − kJ/mol
N2(g) and O2(g) are the element at the standard state,
DHf° = 0
2. Set up equation based on
DH°rxn = S n DHf°(products) − S n DHf°(reactants)
DHCH3NO2 = −401.6 kJ
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48. Practice: Determine the standard enthalpy of formation for TNT (C7H5N3O6). The heat of combustion for TNT is -3400kJ/mol. C7H5N3O6(s) + ?O2(g) ® 7CO2(g) + 3/2 N2(g) + 5/2 H2O(l) DHrxn = kJ/mol find DHf°(C7H5N3O6)
DHf°[CO2(g)] = − kJ/mol
DHf°(H2O(l)) = − kJ/mol
N2(g) and O2(g) as element at the standard state, DHf° = 0
DH°rxn = S n DHf°(products) − S n DHf°(reactants)
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49. Practice: How many kg of octane must be combusted to supply 1
Practice: How many kg of octane must be combusted to supply 1.0 x 1011 kJ of energy?
MMoctane = g/mol
DH°rxn = − kJ
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50. Table Selected Standard Enthalpies of Formation at 25°C (298K)
Formula
DH°f (kJ/mol)
Calcium
Ca(s)
CaO(s)
CaCO3(s)
Carbon
C(graphite)
C(diamond)
CO(g)
CO2(g)
CH4(g)
CH3OH(l)
HCN(g)
CSs(l)
Chlorine
Cl(g)
–635.1
–1206.9
1.9
–110.5
–393.5
–74.9
–238.6
135
87.9
121.0
Hydrogen
Nitrogen
Oxygen
Formula
DH°f (kJ/mol)
H(g)
H2(g)
N2(g)
NH3(g)
NO(g)
O2(g)
O3(g)
H2O(g)
H2O(l)
Cl2(g)
HCl(g)
–92.3
218.0
–45.9
90.3
143
–241.8
–285.8
107.8
Formula
DH°f (kJ/mol)
Silver
Ag(s)
AgCl(s)
Sodium
Na(s)
Na(g)
NaCl(s)
Sulfur
S8(rhombic)
S8(monoclinic)
SO2(g)
SO3(g)
–127.0
–411.1
0.3
–296.8
–396.0

51. The two-step process for determining DHºrxn from DHºf values.

52. Find Heat: How much heat is absorbed by a copper penny with mass 3
Find Heat: How much heat is absorbed by a copper penny with mass 3.10 g whose temperature rises from −8.0 °C to 37.0 °C?
Sort Information
Given:
Find:
T1 = −8.0 °C, T2= 37.0 °C, m = 3.10 g
q, J
q = m ∙ Cs ∙ DT
Cs = J/g•ºC (Table 6.4)
Strategize
Conceptual Plan:
Relationships:
Cs m, DT q
Follow the conceptual plan to solve the problem
Solution:
Check
Check:
the unit is correct, the sign is reasonable as the penny must absorb heat to make its temperature rise
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53. Find Specific heat: A 100. -g of unknown metal alloy is heated to 99
Find Specific heat: A 100.-g of unknown metal alloy is heated to 99.8°C and submerged into 200. g water at 25.9°C. The final temperature of the water becomes 32.9°C. What is the specific heat of the metal alloy?
qm = −qH2O
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54. Find Specific heat: A 100. -g of unknown metal alloy is heated to 99
Find Specific heat: A 100.-g of unknown metal alloy is heated to 99.8°C and submerged into 200. g water at 25.9°C. The final temperature of the water becomes 32.9°C. What is the specific heat of the metal alloy?
Given:
Find:
mm=100.g, Tmi = 99.8°C, mH20 =200.g, TH2Oi =25.9°C, Tfinal= 32.9°C
find Cm
qm = −qH2O
CH2O= 4.18 J/g•ºC
Plan:
Solution:
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55. Find Specific heat (contd. ): A 100
Find Specific heat (contd.): A 100.-g of unknown metal alloy is heated to 99.8°C and submerged into 200. g water at 25.9°C. The final temperature of the water becomes 32.9°C. What is the specific heat of the metal alloy?
Given:
Find:
mm = 100. g, Tmi = 99.8°C, mH20 = 200. g, TH2Oi = 25.9°C, Tfinal = 32.9°C find Cm
Solution:
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56. Predict DT: A 32. 5-g cube of aluminum initially at 45
Predict DT: A 32.5-g cube of aluminum initially at 45.8 °C is submerged into g of water at 15.4 °C. What is the final temperature of both substances at thermal equilibrium?
qAl = −qH2O
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57. Predict DT: A 32. 5-g cube of aluminum initially at 45
Predict DT: A 32.5-g cube of aluminum initially at 45.8 °C is submerged into g of water at 15.4 °C. What is the final temperature of both substances at thermal equilibrium?
Given:
Find:
mAl = 32.5 g, Tal = 45.8 °C, mH20 = g, TH2O = 15.4 °C
Tfinal, °C
q = m ∙ Cs ∙ DT
Cs, Al = J/g•ºC, Cs, H2O = 4.18 J/g•ºC(Table 6.4)
Conceptual Plan:
Relationships:
Cs, Al mAl, Cs, H2O mH2O
DTAl = kDTH2O
qAl = −qH2O
Tfinal
Solution:
Check:
the unit is correct, the number is reasonable as the final temperature should be between the two initial temperatures
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58. (Continued): A 32. 5-g cube of aluminum initially at 45
(Continued): A 32.5-g cube of aluminum initially at 45.8 °C is submerged into g of water at 15.4 °C. What is the final temperature of both substances at thermal equilibrium?
Given:
Find:
mAl = 32.5 g, Tal = 45.8 °C, mH20 = g, TH2O = 15.4 °C
Tfinal, °C
q = m ∙ Cs ∙ DT
Cs, Al = J/g•ºC, Cs, H2O = 4.18 J/g•ºC(Table 6.4)
Conceptual Plan:
Relationships:
Cs, Al mAl, Cs, H2O mH2O
DTAl = kDTH2O
qAl = −qH2O
Tfinal
Solution:
Check:
the unit is correct, the number is reasonable as the final temperature should be between the two initial temperatures
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59. the units are correct, the sign is as expected for combustion
Molar Energy Change: When g of sugar is burned in a bomb calorimeter, the temperature rises from °C to °C. If Ccal = 4.90 kJ/°C, find DE for burning 1 mole of sugar.
Given:
Find:
1.010 g C12H22O11, T1 = °C, T2 = °C, Ccal = 4.90 kJ/°C
DErxn, kJ/mol
Conceptual Plan:
Relationships:
Ccal, DT
qcal
qrxn
qrxn, mol DE
qcal = Ccal x DT = −qrxn
MM C12H22O11 = g/mol
Solution:
Check:
the units are correct, the sign is as expected for combustion
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60. When g of sugar (C12H22O11) is burned in a bomb calorimeter, the temperature rises from °C to °C. If heat capacity of calorimeter Ccal = 4.90 kJ/°C, find DE for burning 1 mole of sugar.
qcal = kJ
qrxn = kJ
Mol sugar = mol
E = x 103 kJ
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61. Example: How much heat is evolved in the complete combustion of 13
Example: How much heat is evolved in the complete combustion of 13.2 kg of C3H8(g)?
Given:
Find:
13.2kg C3H8, C3H8(g) + 5O2(g)→3CO2(g) + 4H2O(g) DH = −2044 kJ
q, kJ/mol
combustion of 1 mole C3H8 gives 2044 kJ heat.
1 kg = 1000 g, Molar Mass = g/mol
Conceptual Plan:
Relationships:
mol kJ kg g
Solution:
Tro: Chemistry: A Molecular Approach, 2/e

62. Example:
Determining the Enthalpy Change of an Aqueous Reaction
PROBLEM:
50.0 mL of M NaOH is placed in a coffee-cup calorimeter at 25.00ºC and 25.0 mL of M HCl is carefully added, also at 25.00ºC. After stirring, the final temperature is 27.21ºC. Calculate qsoln (in J) and the change in enthalpy, DH, (in kJ/mol of H2O formed).
Assume that the total volume is the sum of the individual volumes, that d = 1.00 g/mL, and that c = J/g∙K
PLAN:
Heat flows from the reaction (the system) to its surroundings (the solution). Since –qrxn = qsoln, we can find the heat of the reaction by calculating the heat absorbed by the solution.

63. SOLUTION:
(a) To find qsoln:
Total mass (g) of the solution = (25.0 mL mL) x 1.00 g/mL = 75.0 g
DTsoln = 27.21ºC – 25.00ºC = 2.21ºC = 2.21 K
qsoln = csoln x masssoln x DTsoln = (4.184 J/g∙K)(75.0 g)(2.21 K)
= 693 J
(b) To find DHrxn we first need a balanced equation:
HCl (aq) + NaOH (aq) → NaCl (aq) + H2O (l)

64. Example (contd):
For HCl:
25.0 mL HCl x
1 L
103 mL
0.500 mol x
1 mol H2O
1 mol HCl
= mol H2O
For NaOH:
50.0 mL NaOH x
1 L
103 mL x
1 mol H2O
1 mol NaOH
= mol H2O
0.500 mol
HCl is limiting, and the amount of H2O formed is mol.
DHrxn =
qrxn
mol H2O
mol
–693 J =
1 kJ
103J x
= –55.4 kJ/mol H2O

65. Cs, DT, m qsol’n qrxn qrxn, mol DH
Example 6.8: What is DHrxn/mol Mg for the reaction Mg(s) + 2 HCl(aq) → MgCl2(aq) + H2(g) if g Mg reacts in mL of solution and changes the temperature from 25.6 °C to 32.8 °C?
Given:
Find:
0.158 g Mg, mL, T1 = 25.6 °C, T2 = 32.8 °C,
assume d = 1.00 g/mL, Cs = 4.18 J/g∙°C
H, kJ/mol
Conceptual Plan:
Relationships:
Cs, DT, m
qsol’n
qrxn
qrxn, mol DH
qsol’n = m x Cs, sol’n x DT = −qrxn, MM Mg= g/mol
Solution:
Check:
the heat released from reaction (-) is transferred to the solution, so solution gains heat (+)
Tro: Chemistry: A Molecular Approach, 2/e

66. Cs, DT, m qsol’n qrxn qrxn, mol DH
Example 6.8: What is DHrxn/mol Mg for the reaction Mg(s) + 2 HCl(aq) → MgCl2(aq) + H2(g) if g Mg reacts in mL of solution and changes the temperature from 25.6 °C to 32.8 °C?
Given:
Find:
0.158 g Mg, mL, T1 = 25.6 °C, T2 = 32.8 °C,
assume d = 1.00 g/mL, Cs = 4.18 J/g∙°C
H, kJ/mol
Conceptual Plan:
Relationships:
Cs, DT, m
qsol’n
qrxn
qrxn, mol DH
qsol’n = m x Cs, sol’n x DT = −qrxn, MM Mg= g/mol
Solution:
Check:
the sign is correct and the value is reasonable because the reaction is exothermic
Tro: Chemistry: A Molecular Approach, 2/e 66 66

67. Determining the Enthalpy Change of an Aqueous Reaction
50.0 mL of M NaOH is placed in a coffee-cup calorimeter at 25.00ºC and 25.0 mL of M HCl is carefully added, also at 25.00ºC. After stirring, the final temperature is 27.21ºC. Calculate qsoln (in J) and the change in enthalpy, DH, (in kJ/mol of H2O formed). Assume that the total volume is the sum of the individual volumes, that d = 1.00 g/mL, and that c = J/g∙K
PLAN:
Heat flows from the reaction (the system) to its surroundings (the solution): –qrxn = qsoln. Start with heat absorbed by the solution.
Find Limiting reactant: HCl; the amount of product H2O = mol.
qsoln = 693 J
qrxn = -693 J
DHrxn = qrxn/mol L.R. = kJ/mol H2O

68. Calculating ΔHºrxn from given ΔHº Values
More example:
Calculating ΔHºrxn from given ΔHº Values
Find DHrxn for the rxn: 4NH3(g) + 5O2(g) → 4NO(g) + 6H2O(g)
Given: N2(g) + 3H2(g) → 2NH3(g) DH1 = kJ
N2(g) + O2(g) → 2NO(g) DH2 = kJ
2H2(g) + O2(g) → 2H2O(g) DH3 = kJ
NH3: Rxn# 1 reverse and x NH3(g) → 2N2(g) + 6H2(g) DH = kJ
O2: in more than one rxn, skip
NO: Rxn# 2 times N2(g) + 2O2(g) → 4NO(g) DH = kJ
H2O: Rxn# 3 times H2(g) + 3O2(g) → 6H2O(g) DH = kJ
Sum of all reactions: DHrxn= kJ kJ + ( kJ) = kJ
4NH3(g) + 2N2(g) + 2O2(g) + 6H2(g) + 3O2(g) → 2N2(g) + 6H2(g) + 4NO(g) + 6H2O(g)

69. Example 6. 12: How many kg of octane must be combusted to supply 1
Example 6.12: How many kg of octane must be combusted to supply 1.0 x 1011 kJ of energy?
Given:
Find:
1.0 x 1011 kJ
mass octane, kg
DH°rxn = − kJ
Write the balanced equation per mole of octane
Conceptual Plan:
Relationships:
MMoctane = g/mol, 1 kg = 1000 g
DHf°’s
DHrxn° kJ
mol C8H18
g C8H18
kg C8H18
from
above
Solution:
C8H18(l) + 25/2 O2(g) → 8 CO2(g) + 9 H2O(g)
Look up the DHf°
for each material
in Appendix IIB
Check:
the units and sign are correct,
the large value is expected
Tro: Chemistry: A Molecular Approach, 2/e

70. Example 6. 12: How many kg of octane must be combusted to supply 1
Example 6.12: How many kg of octane must be combusted to supply 1.0 x 1011 kJ of energy?
Given:
Find:
1.0 x 1011 kJ
mass octane, kg
DH°rxn = − kJ
Write the balanced equation per mole of octane
Conceptual Plan:
Relationships:
MMoctane = g/mol, 1 kg = 1000 g
DHf°’s
DHrxn° kJ
mol C8H18
g C8H18
kg C8H18
from
above
Solution:
C8H18(l) + 25/2 O2(g) → 8 CO2(g) + 9 H2O(g)
Check:
the units and sign are correct,
the large value is expected
Tro: Chemistry: A Molecular Approach, 2/e 70 70

71. Example 6. 12: How many kg of octane must be combusted to supply 1
Example 6.12: How many kg of octane must be combusted to supply 1.0 x 1011 kJ of energy?
Given:
Find:
1.0 x 1011 kJ
mass octane, kg
DH°rxn = − kJ
Write the balanced equation per mole of octane
Conceptual Plan:
Relationships:
MMoctane = g/mol, 1 kg = 1000 g
DHf°’s
DHrxn° kJ
mol C8H18
g C8H18
kg C8H18
from
above
Solution:
C8H18(l) + 25/2 O2(g) → 8 CO2(g) + 9 H2O(g)
Check:
the units and sign are correct,
the large value is expected
Tro: Chemistry: A Molecular Approach, 2/e 71 71

72. More examples:
Using Hess’s Law to Calculate an Unknown DH
PROBLEM:
Two gaseous pollutants that form in auto exhausts are CO and NO. An environmental chemist is studying ways to convert them to less harmful gases through the following reaction:
CO (g) + NO (g) → CO2 (g) + ½N2 (g) DH = ?
Given the following information, calculate the unknown DH:
Equation A: CO (g) + ½O2 (g) → CO2 (g) DHA = –283.0 kJ
Equation B: N2 (g) + O2 (g) → 2NO (g) DHB = kJ
PLAN:
Manipulate Equations A and/or B and their DH values to get to the target equation and its DH. All substances except those in the target equation must cancel.

73. More example:
Writing Formation Equations
PROBLEM:
Write balanced equations for the formation of 1 mol of the following compounds from their elements in their standard states and include DHºf.
(a) Silver chloride, AgCl, a solid at standard conditions.
(b) Calcium carbonate, CaCO3, a solid at standard conditions.
(c) Hydrogen cyanide, HCN, a gas at standard conditions.
PLAN:
Write the elements as reactants and 1 mol of the compound as the product formed. Make sure all substances are in their standard states. Balance the equations and find the value of DHºf in Table 6.3 or Appendix B.

74. (a) Silver chloride, AgCl, a solid at standard conditions.
SOLUTION:
(a) Silver chloride, AgCl, a solid at standard conditions.
Ag (s) + ½Cl2 (g) → AgCl (s) DHºf = –127.0 kJ
(b) Calcium carbonate, CaCO3, a solid at standard conditions.
Ca (s) + C(graphite) + O2 (g) → CaCO3 (s) DH°f = – kJ 3 2
(c) Hydrogen cyanide, HCN, a gas at standard conditions.
½H2 (g) + C(graphite) + ½N2 (g) → HCN (g) DH°f = 135 kJ 74