1. Principles of Chemistry: A Molecular Approach, 1st Ed. Nivaldo Tro
Chapter 6
Thermochemistry

2. Nature of Energy
Even though Chemistry is the study of matter, energy affects matter.
Energy is anything that has the capacity to do work.
Work is a force acting over a distance.
energy = work = force × distance
Energy can be exchanged between objects through contact.
collisions

3. Energy, Heat, and Work
You can think of energy as a quantity an object can possess.
or collection of objects
You can think of heat and work as the two different ways that an object can exchange energy with other objects.
either out of it, or into it

4. Classification of Energy
Kinetic energy is energy of motion or energy that is being transferred.
Thermal energy is kinetic.

5. Classification of Energy
Potential energy is energy that is stored in an object, or energy associated with the composition and position of the object.
Energy stored in the structure of a compound is potential.

6. Exchanging Energy Energy cannot be created or destroyed.
Energy can be exchanged between objects.
Energy can also be transformed from one form to another.
heat → light → sound

7. Some Forms of Energy electrical
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 of the atoms, the attachment between atoms, the atoms’ relative position to each other in the molecule, or the molecules’ relative positions in the structure

8. System and Surroundings
We define the system as the material or process that contains the energy changes we are studying.
We define the surroundings as everything else in the universe.
What we study is the exchange of energy between the system and the surroundings.
Surroundings
System
Surroundings
System

9. Units of Energy
The amount of kinetic energy an object has is directly proportional to its mass and velocity.
KE = ½mv2
When the mass is in kg and velocity in m/s, the unit for kinetic energy is
One joule of energy is the amount of energy needed to move a 1-kg mass at a speed of 1 m/s.
1 J = 1

10. Units of Energy
One joule (J) is the amount of energy needed to move a 1-kg mass a distance of 1 meter.
1 J = 1 N∙m = 1 kg∙m2/s2
One calorie (cal) is the amount of energy needed to raise one gram of water by 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 calories (cal)
1 kilowatt-hour (kWh)
3.60 × 106 joules (J)

11. Energy Use

12. The First Law of Thermodynamics— Law of Conservation of Energy
When energy is exchanged between objects or transformed into another form, all the energy is still there.
The total amount of energy in the universe before the change has to be equal to the total amount of energy in the universe after.
You can, therefore, never design a system that will continue to produce energy without some source of energy.

13. Energy Flow and Conservation of Energy
Conservation of energy requires that the total energy change 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

14. Internal Energy
The internal energy is the total amount of kinetic and potential energy a system possesses.
The change in the internal energy of a system depends only on the amount of energy in the system at the beginning and end.
A state function is a mathematical function whose result depends only on the initial and final conditions, not on the process used.
DE = Efinal – Einitial
DEreaction = Eproducts – Ereactants

15. State Function
You can travel either of two trails to reach the top of the mountain. One is long and winding; the other is shorter but steeper. Regardless of which trail you take, when you reach the top you will be 10,000 ft. above the base.
The height of the mountain is a state function. It depends only on the distance from the base to the peak, not on how you arrive there!

16. Energy Diagrams
Energy diagrams are a “graphical” way of showing the direction of energy flow during a process.
Internal Energy
initial
final
energy added
DE = +
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
If the final condition has a
smaller amount of internal
energy than the initial
condition, the change in the
internal energy will be –.
energy removed
DE = –

17. –DEsystem = DEsurroundings
Energy Flow
When energy flows out of a system, it must all flow into the surroundings.
When energy flows out of a system, DEsystem is –.
When energy flows into the surroundings, DEsurroundings is +.
therefore:
–DEsystem = DEsurroundings
Surroundings
DE +
System
DE –

18. DEsystem = –DEsurroundings
Energy Flow
When energy flows into a system, it must all come from the surroundings.
When energy flows into a system, DEsystem is +.
When energy flows out of the surroundings, DEsurroundings is –.
therefore:
DEsystem = –DEsurroundings
Surroundings
DE –
System
DE +

19. Energy Flow in a Chemical Reaction
The total amount of internal energy in 1 mol of C(s) and 1 mole of O2(g) is greater than the internal energy in 1 mole of CO2(g).
at the same temperature and pressure
In the reaction C(s) + O2(g) → CO2(g), there will be a net release of energy into the surroundings.
−DEreaction = DEsurroundings
In the reaction CO2(g) → C(s) + O2(g), there will be an absorption of energy from the surroundings into the reaction.
DEreaction = −DEsurroundings
Internal Energy
CO2(g)
C(s), O2(g)
energy
absorbed
DErxn = +
energy
released
DErxn = –
Surroundings
System
C + O2 → CO2
System
CO2 → C + O2

20. 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

21. Energy Exchange
Energy is exchanged between the system and surroundings through either heat exchange or work being done.

22. Heat and Work
On a smooth table, most of the kinetic energy is transferred from the first ball to the second—with a small amount lost through friction.

23. Heat and Work
On a rough table, most of the kinetic energy of the first ball is lost through friction—less than half is transferred to the second.

24. Heat, Work, and Internal Energy
In the previous billiard ball example, the DE of the white ball is the same for both cases, but q and w are not.
On the rougher table, the heat loss, q, is greater.
q is a more negative number.
But on the rougher table, less kinetic energy is transferred to the purple ball, so the work done by the white ball, w, is less.
w is a less negative number.
The DE is a state function and depends only on the velocity of the white ball before and after the collision.
In both cases, it started with 5.0 kJ of kinetic energy and ended with 0 kJ because it stopped.
q + w is the same for both tables, even though the values of q and w are different.

25. The unit and sign are correct.
Example 6.1 If burning fuel in a potato cannon gives 855 J of kinetic energy to the potato and releases 1422 J of heat from the cannon, what is the change in internal energy of the fuel?
Given:
Find:
qpotato = 855 J, wpotato = 1422 J
DEfuel, J
Concept Plan:
Relationships:
q, w
potato
fuel
q, w DE
qsystem = −qsurroundings, wsystem = −wsurroundings, DE = q + w
Solution:
Check:
The unit and sign are correct.

26. Practice—Reacting 50 mL of H2(g) with 50 mL of C2H4(g) produces 50 mL of C2H6(g) at 1.5 atm. If the reaction produces 3.1 × 102 J of heat and the decrease in volume requires the surroundings to do 7.6 J of work on the system, what is the change in internal energy of the system?

27. The unit and sign are correct.
If the reaction produces 3.1 × 102 J of heat and the decrease in volume requires the surroundings to do 7.6 J of work on the system, what is the change in internal energy of the system?
Given:
Find:
qreaction = −310 J, wsurroundings = −7.6 J
DEfuel, J
Concept Plan:
Relationships: w
surrounding
reaction
q, w DE
qsystem = −qsurroundings, wsystem = −wsurroundings, DE = q + w
Solution:
Check:
The unit and sign are correct.

28. Heat Exchange
Heat is the exchange of thermal energy between the system and surroundings.
It occurs when the system and surroundings have a difference in temperature.
Heat flows from matter with high temperature to matter with low temperature until both objects reach the same temperature.
thermal equilibrium

29. Quantity of Heat Energy Absorbed Heat Capacity
When a system absorbs heat, its temperature increases.
The increase in temperature is directly proportional to the amount of heat absorbed.
The proportionality constant is called the heat capacity, C.
Units of C are J/°C or J/K.
q = C × DT
The heat capacity of an object depends on 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 of sand 12 °C, but only raise the temperature of 100 g of water by 2.4 °C.

30. Specific Heat Capacity
measure of a substance’s intrinsic ability to absorb heat
The specific heat capacity is the amount of heat energy required to raise the temperature of one gram of a substance 1 °C. Cs
units are J/(g∙°C)
The molar heat capacity is the amount of heat energy required to raise the temperature of one mole of a substance 1 °C.

31. Specific Heat of Water
The rather high specific heat of water allows water to absorb a lot of heat energy without a large increase in its temperature.
The large amount of water absorbing heat from the air keeps beaches cool in the summer.
Without water, the Earth’s temperature would be about the same as the moon’s temperature on the side that is facing the sun (average 107 °C or 225 °F).
Water is commonly used as a coolant because it can absorb a lot of heat and remove it from the important mechanical parts to keep them from overheating.
or even melting
It can also be used to transfer the heat to something else because it is a fluid.

32. Quantifying Heat Energy
The heat capacity of an object is proportional to its mass and the specific heat of the material.
So we can calculate the quantity of heat absorbed by an object if we know the mass, the specific heat, and the temperature change of the object.
heat = (mass) × (specific heat capacity) × (temp. change)
q = (m) × (Cs) × (DT)

33. Example 6. 2 How much heat is absorbed by a copper penny with mass 3
Example 6.2 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 (Table 6.4)
Strategize
Conceptual Plan:
Relationships:
Cs m, DT q
Follow the conceptual plan to solve the problem.
Solution:
Check
Check:
The unit and sign are correct.

34. Practice—Calculate the amount of heat released when 7
Practice—Calculate the amount of heat released when 7.40 g of water cools from 49° to 29 °C.

35. The unit and sign are correct.
Practice—Calculate the amount of heat released when 7.40 g of water cools from 49° to 29 °C.
Sort information
Given:
Find:
T1 = 49 °C, T2 = 29 °C, m = 7.40 g
q, J
q = m ∙ Cs ∙ DT
Cs = 4.18 J/gC (Table 6.4)
Strategize
Conceptual Plan:
Relationships:
Cs m, DT q
Follow the conceptual plan to solve the problem
Solution:
Check
Check:
The unit and sign are correct.

36. –work = external pressure × change in volume
Pressure–Volume Work
PV work is work that is the result of a volume change against an external pressure.
When gases expand, DV is +, but the system is doing work on the surroundings, so wgas is –.
As long as the external pressure is kept constant,
–work = external pressure × change in volume
w = –PDV
To convert the units to joules, use J = 1 atm∙L.

37. The unit and sign are correct.
Example 6.3 If a balloon is inflated from L to 1.85 L against an external pressure of 1.00 atm, how much work is done?
Given:
Find:
V1 = L, V2 = 1.85 L, P = 1.00 atm
w, J
Conceptual Plan:
Relationships:
P, DV w
101.3 J = 1 atm∙L
Solution:
Check:
The unit and sign are correct.

38. Practice—A certain process results in a gas system releasing 68
Practice—A certain process results in a gas system releasing 68.3 kJ of energy. During the process, 15.8 kcal of heat is absorbed by the system. If the external pressure is kept constant at 1.00 atm and the initial volume of the gas is 10.0 L, what is the final volume of the gas?

39. A certain process results in a gas system releasing 68. 3 kJ of energy
A certain process results in a gas system releasing 68.3 kJ of energy. During the process, 15.8 kcal of heat is absorbed by the system. If the external pressure is kept constant at 1.00 atm and the initial volume of the gas is 10.0 L, what is the final volume of the gas?
Given:
Find:
DE = −68.3 kJ, q = kcal, V1 = 10.0 L, P = 1.00 atm
V2, L
Conceptual Plan:
Relationships:
q, DE w V2
P, V1
DE = q + w, w = –PDV, 1 kJ = 1000 J, 1 cal = 4.18 J, J = 1 atm∙L
Solution:
Check:

40. A certain process results in a gas system releasing 68. 3 kJ of energy
A certain process results in a gas system releasing 68.3 kJ of energy. During the process, 15.8 kcal of heat is absorbed by the system. If the external pressure is kept constant at 1.00 atm and the initial volume of the gas is 10.0 L, what is the final volume of the gas?
Given:
Find:
DE = −6.83 × 104 J, q = × 104 J, V1 = 10.0 L, P = 1.00 atm
V2, L
Conceptual Plan:
Relationships:
q, DE w V2
P, V1
DE = q + w, w = –PDV, 1 kJ = 1000 J, 1 cal = 4.18 J, J = 1 atm∙L
Solution:
Check: 40 40

41. A certain process results in a gas system releasing 68. 3 kJ of energy
A certain process results in a gas system releasing 68.3 kJ of energy. During the process, 15.8 kcal of heat is absorbed by the system. If the external pressure is kept constant at 1.00 atm and the initial volume of the gas is 10.0 L, what is the final volume of the gas?
Given:
Find:
DE = −6.83 × 104 J, q = × 104 J, V1 = 10.0 L, P = 1.00 atm
V2, L
Conceptual Plan:
Relationships:
q, DE w V2
P, V1
DE = q + w, w = –PDV, 1 kJ = 1000 J, 1 cal = 4.18 J, J = 1 atm∙L
Solution:
Check: 41 41

42. Since DE is − and q is +, w must be −; and it is.
A certain process results in a gas system releasing 68.3 kJ of energy. During the process, 15.8 kcal of heat is absorbed by the system. If the external pressure is kept constant at 1.00 atm and the initial volume of the gas is 10.0 L, what is the final volume of the gas?
Given:
Find:
DE = −6.83 × 104 J, q = × 104 J, V1 = 10.0 L, P = 1.00 atm
V2, L
Conceptual Plan:
Relationships:
q, DE w V2
P, V1
DE = q + w, w = –PDV, 1 kJ = 1000 J, 1 cal = 4.18 J, J = 1 atm∙L
Solution:
Check:
Since DE is − and q is +, w must be −; and it is.
When w is − the system is expanding, so V2 should be greater than V1; and it is. 42 42

43. Exchanging Energy between System and Surroundings
exchange of heat energy
q = mass × specific heat × DTemperature
exchange of work
w = –Pressure × DVolume

44. Measuring DE, Calorimetry at Constant Volume
Since DE = q + w, we can determine DE by measuring q and w.
In practice, it is easiest to do a process in such a way that there is no change in volume, w = 0.
at constant volume, DEsystem = qsystem
In practice, it is not possible to observe the temperature changes of the individual chemicals involved in a reaction. So instead, we use an insulated, controlled surroundings and measure the temperature change in it.
The surroundings is called a bomb calorimeter and is usually made of a sealed, insulated container filled with water.
qsurroundings = qcalorimeter = –qsystem
–DEreaction = qcal = Ccal × DT

45. Bomb Calorimeter
used to measure DE because it is a constant volume system

46. The units and sign are correct.
Example 6.4 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.
Given:
Find:
1.010 g C12H22O11, T1 = °C, T2 = °C, Ccal = 4.90 kJ/°C
DErxn, kJ/mol
× 10−3 mol C12H22O11, DT = 3.41 °C, Ccal = 4.90 kJ/°C
DErxn, kJ/mol
Conceptual Plan:
Relationships:
Ccal, DT
qcal
qrxn
qrxn, mol DE
qcal = Ccal × DT = –qrxn
MM C12H22O11 = g/mol
Solution:
Check:
The units and sign are correct.

47. Practice—When 1. 22 g of HC7H5O2 (MM 122
Practice—When 1.22 g of HC7H5O2 (MM ) is burned in a bomb calorimeter, the temperature rises from °C to °C. If DE for the reaction is −3.23 × 103 kJ/mol, what is the heat capacity of the calorimeter in kJ/°C?

48. The units and sign are correct.
When 1.22 g of HC7H5O2 is burned in a bomb calorimeter, the temperature rises from °C to °C. If DE for the reaction is −3.23 × 103 kJ/mol, what is the heat capacity of the calorimeter in kJ/°C?
Given:
Find:
1.22 g HC7H5O2, T1 = °C, T2 = °C, DE = –3.23 × 103 kJ/mol
Crxn, kJ/°C
9.990 × 10−3 mol HC7H5O2, DT = 2.40 °C, DE = –3.23 × 103 kJ/mol
Ccal, kJ/°C
Conceptual Plan:
Relationships:
mol, DE
qrxn
qcal
qcal, DT
Ccal
qcal = Ccal x DT = –qrxn
MM HC7H5O2 = g/mol
Solution:
Check:
The units and sign are correct.

49. DHreaction = qreaction at constant pressure
Enthalpy
The enthalpy, H, of a system is the sum of the internal energy of the system and the product of pressure and volume (work done).
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.

50. Endothermic and Exothermic Reactions
When DH is –, heat is being released by the system.
Reactions that release heat are called exothermic reactions.
When DH is +, heat is being absorbed by the system.
Reactions that absorb heat are called endothermic reactions.
Chemical heat packs contain iron filings that are oxidized in an exothermic reaction. Your hands get warm because the released heat of the reaction is absorbed by your hands.
Chemical cold packs contain NH4NO3 that dissolves in water in an endothermic process. Your hands get cold because they are giving away your heat to the reaction.

51. Molecular View of Exothermic Reactions
For an exothermic reaction, the outside temperature rises due to release of thermal energy.
This extra thermal energy comes from the conversion of some of the chemical potential energy in the reactants into kinetic energy in the form of heat.
During the course of a reaction, old bonds are broken and new bonds made.
The products of the reaction have less chemical potential energy than the reactants.
The difference in energy is released as heat.

52. Molecular View of Endothermic Reactions
In an endothermic reaction, the surrounding temperature drops due to absorption of some of its thermal energy.
During the course of a reaction, old bonds are broken and new bonds made.
The products of the reaction have more chemical potential energy than the reactants.
To acquire this extra energy, some of the thermal energy of the surroundings is converted into chemical potential energy stored in the products.

53. 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.
The more reactants you use, the larger the enthalpy change.
By convention, we calculate the enthalpy change for the number of moles of reactants in the reaction as written.
For the reaction:
C3H8(g) + 5 O2(g) → 3 CO2(g) + 4 H2O(g) DH = –2044 kJ

54. The sign is correct and the value is reasonable.
Example 6.6 How much heat is evolved in the complete combustion of 13.2 kg of C3H8(g)?
Given:
Find:
13.2 kg C3H8
q, kJ/mol
1 kg = 1000 g, 1 mol C3H8 = –2044 kJ, Molar Mass = g/mol
Conceptual Plan:
Relationships:
mol kJ kg g
Solution:
Check:
The sign is correct and the value is reasonable.

55. Practice—How much heat is evolved when a 0. 483-g diamond is burned
Practice—How much heat is evolved when a g diamond is burned? (DHcombustion = –395.4 kJ/mol C)

56. How much heat is evolved when a 0.483-g diamond is burned?
Given:
Find:
0.483 g C, DH = –395.4 kJ/mol C
q, kJ/mol
1 mol C = −395.4 kJ, Molar Mass = g/mol
Concept Plan:
Relationships:
mol kJ g
Solution:
Check:
The sign is correct and the value is reasonable.

57. Measuring DH Calorimetry at Constant Pressure
Reactions done in aqueous solution are at constant pressure.
open to the atmosphere
The calorimeter is often nested foam cups containing the solution.
qreaction = –qsolution = –(masssolution × Cs, solution × DT)
DHreaction = qconstant pressure = qreaction
To get DHreaction per mol, divide by the number of moles.

58. The sign is correct and the value is reasonable.
Example 6.7 What is DHrxn/mol Mg for the reaction Mg(s) + 2 HCl(aq) → MgCl2(aq) + H2(g) if g Mg reacting in mL of solution 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
d = 1.00 g/mL, Cs = 4.18 J/g∙°C
DH, kJ/mol
6.499 × 10–3 mol Mg, 1.00 × 102 g, DT = 7.2 °C, Cs = 4.18 J/g∙°C
DH, kJ/mol
Conceptual Plan:
Relationships:
Cs, DT, m
qsol’n
qrxn
qrxn, mol DH
qsol’n = m × Cs, sol’n × DT = –qrxn, MM Mg = g/mol,
Solution:
Check:
The sign is correct and the value is reasonable.

59. Practice—What will be the final temperature of the solution in a coffee cup calorimeter if a mL sample of M HCl(aq) is added to a mL sample of NaOH(aq) if the initial temperature is °C and the DHrxn is −57.2 kJ/mol NaOH? (Assume the density of the solution is 1.00 g/mL and the specific heat of the solution is 4.18 J/g∙°C.)

60. The sign is correct and the value is reasonable.
Practice—What will be the final temperature of the solution in a coffee cup calorimeter if a mL sample of M HCl(aq) is added to a mL sample of NaOH(aq) if the initial temperature is °C?
Given:
Find:
50.00 mL of M HCl, mL of M NaOH T1 = °C,
d = 1.00 g/mL, Cs = 4.18 J/g∙°C, DH = −57.2 kJ/mol
T2, °C
1.25 × 10–2 mol NaOH, 1.00 × 102 g, T1 = °C
DH = –57.2 kJ/mol, Cs = 4.18 J/g∙°C
T2, °C
Conceptual Plan:
Relationships:
qrxn
qsol’n DH DT
Cs, qsol’n, m T2
qsol’n = m × Cs, sol’n × DT = −qrxn, mol NaOH = 1 L
Solution:
Check:
The sign is correct and the value is reasonable.

61. Relationships Involving DHrxn
When a 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 = 2(−393.5 kJ) = −787.0 kJ
If a reaction is reversed, then the sign of DH is reversed.
CO2(g) → C(s) + O2(g) DH = kJ

62. Relationships Involving DHrxn 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.

63. Example—Hess’s Law Given the following information:
2 NO(g) + O2(g)  2 NO2(g) DH° = −116 kJ
2 N2(g) + 5 O2(g) + 2 H2O(l)  4 HNO3(aq) DH° = −256 kJ
N2(g) + O2(g)  2 NO(g) DH° = +183 kJ
Calculate the DH° for the reaction below:
3 NO2(g) + H2O(l)  2 HNO3(aq) + NO(g) DH° = ?
[2 NO2(g)  2 NO(g) + O2(g)] × 1.5 DH° = 1.5(+116 kJ)
[2 N2(g) + 5 O2(g) + 2 H2O(l)  4 HNO3(aq)] × 0.5 DH° = 0.5(−256 kJ)
[2 NO(g)  N2(g) + O2(g)] DH° = −183 kJ
[3 NO2(g)  3 NO(g) O2(g)] DH° = (+174 kJ)
[1 N2(g) O2(g) + 1 H2O(l)  2 HNO3(aq)] DH° = (−128 kJ)
[2 NO(g)  N2(g) + O2(g)] DH° = (−183 kJ)
3 NO2(g) + H2O(l)  2 HNO3(aq) + NO(g) DH° = −137 kJ

64. Practice—Hess’s Law Given the following information:
Cu(s) + Cl2(g)  CuCl2(s) DH° = −206 kJ
2 Cu(s) + Cl2(g)  2 CuCl(s) DH° = −36 kJ
Calculate the DH° for the reaction below:
Cu(s) + CuCl2(s)  2 CuCl(s) DH° = ? kJ

65. Practice—Hess’s Law Given the following information:
Cu(s) + Cl2(g)  CuCl2(s) DH° = −206 kJ
2 Cu(s) + Cl2(g)  2 CuCl(s) DH° = −36 kJ
Calculate the DH° for the reaction below:
Cu(s) + CuCl2(s)  2 CuCl(s) DH° = ? kJ
CuCl2(s)  Cu(s) + Cl2(g) DH° = +206 kJ
2 Cu(s) + Cl2(g)  2 CuCl(s) DH° = −36 kJ
Cu(s) + CuCl2(s)  2 CuCl(s) DH° = kJ

66. 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.
The standard enthalpy of formation, DHf°, is 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.
by definition

67. Formation Reactions
reactions of elements in their standard state to form 1 mole of a pure compound
If you are not sure what the standard state of an element is, find the form in Appendix IIB that has a DHf° = 0.
Since the definition requires 1 mole of compound be made, the coefficients of the reactants may be fractions.

68. Writing Formation Reactions— Write the formation reaction for CO(g).
The formation reaction is the reaction between the elements in the compound, which are C and O.
C + O → CO(g)
The elements must be in their standard state.
There are several forms of solid C, but the one with DHf° = 0 is graphite.
Oxygen’s standard state is the diatomic gas.
C(s, graphite) + O2(g) → CO(g)
The equation must be balanced, but the coefficient of the product compound must be 1.
Use whatever coefficient in front of the reactants is necessary to make the atoms on both sides equal without changing the product coefficient.
C(s, graphite) + ½ O2(g) → CO(g)

69. Write the formation reactions for the following.
CO2(g)
Al2(SO4)3(s)
C(s, graphite) + O2(g)  CO2(g)
2 Al(s) + 3 S(s, rhombic) + 6 O2(g)  Al2(SO4)3(s)

70. 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.

71. CH4(g)+ 2 O2(g)→ CO2(g) + H2O(g)
C(s, graphite) + 2 H2(g) → CH4(g) DHf° = −74.6 kJ/mol CH4
CH4(g) → C(s, graphite) + 2 H2(g) DH° = kJ
DH° = [(DHf° CO2(g) + 2∙DHf°H2O(g))− (DHf° CH4(g) + 2∙DHf°O2(g))]
DH° = [((−393.5 kJ)+ 2(−241.8 kJ) − ((−74.6 kJ) + 2(0 kJ))]
= −802.5 kJ
C(s, graphite) + O2(g) → CO2(g) DHf° = −393.5 kJ/mol CO2
H2(g) + ½ O2(g) → H2O(g) DHf° = −241.8 kJ/mol H2O
2 H2(g) + O2(g) → 2 H2O(g) DH° = −483.6 kJ
CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(g) DH° = −802.5 kJ

72. Example—Calculate the enthalpy change in the reaction
Example—Calculate the enthalpy change in the reaction. 2 C2H2(g) + 5 O2(g) ® 4 CO2(g) + 2 H2O(l)
1. Write formation reactions for each compound and determine the DHf° for each.
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

73. Example—Calculate the enthalpy change in the reaction
Example—Calculate the enthalpy change in the reaction. 2 C2H2(g) + 5 O2(g) ® 4 CO2(g) + 2 H2O(l)
2. Arrange equations so they add up to desired reaction.
2 C2H2(g) ® 4 C(s) + 2 H2(g) DH° = 2(−227.4) kJ
4 C(s) + 4 O2(g) ® 4CO2(g) DH° = 4(−393.5) kJ
2 H2(g) + O2(g) ® 2 H2O(l) DH° = 2(−285.8) kJ
2 C2H2(g) + 5 O2(g) ® 4 CO2(g) + 2 H2O(l) DH = − kJ

74. DH°reaction = S n DHf°(products) – S n DHf°(reactants)
Example—Calculate the enthalpy change in the reaction. 2 C2H2(g) + 5 O2(g) ® 4 CO2(g) + 2 H2O(l)
DH°reaction = S n DHf°(products) – S n DHf°(reactants)
DHrxn = [(4•DHCO2 + 2•DHH2O) – (2•DHC2H2 + 5•DHO2)]
DHrxn = [(4•(–393.5) + 2•(–285.8)) – (2•(+227.4) + 5•(0))]
DHrxn = − kJ

75. The units and sign are correct; the large value is expected.
Example 6.11 How many kg of octane must be combusted to supply 1.0 × 1011 kJ of energy?
Given:
Find:
1.0 × 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.

76. Practice—Calculate the DH° for decomposing 10
Practice—Calculate the DH° for decomposing 10.0 g of limestone, CaCO3, under standard conditions. CaCO3(s) → CaO(s) + O2(g)
Material
DHf°, kJ/mol
CaCO3(s)
−1207.6
O2(g)
CaO(s)
−634.9

77. The units and sign are correct.
Practice—Calculate the DH° for decomposing 10.0 g of limestone, CaCO3(s) → CaO(s) + O2(g).
Given:
Find:
10.0 g CaCO3
DH°, kJ
DH°rxn = kJ
Conceptual Plan:
Relationships:
DHf°’s
DH°rxn per mol CaCO3
MMlimestone = g/mol
g CaCO3
mol CaCO3 kJ
from
above
Solution:
CaCO3(s) → CaO(s) + O2(g)
Check:
The units and sign are correct.