1. Your Turn!
Which of the following represents a decrease in the potential energy of the system?
A book is raised six feet above the floor
A ball rolls downhill
Two electrons come close together
A spring is stretched completely
Two atomic nuclei approach each other

2. Your Turn!
Which of the following represents an increase in the potential energy of the system?
Na  Na+ + e–
A periodic table falls off the wall
Ca2+ + 2e–  Ca
A firecracker explodes
Gasoline is burned and creates CO2 and H2O

3. A bag of chocolate candy has 220 Cal. How much energy is this in kJ?

4. A bag of chocolate candy has 220 Cal. How much energy is this in kJ?
Answer: a

5. Your Turn! Which statement about kinetic energy (KE) is true?
Atoms and molecules in gases, liquids and solids possess KE since they are in constant motion.
At the same temperature, gases, liquids and solids all have different KE distributions.
Molecules in gases are in constant motion, while molecules in liquids and solids are not.
Molecules in gases and liquids are in constant motion, while molecules in solids are not.
As the temperature increases, molecules move more slowly.

6. Your Turn! A closed system can __________ include the surroundings
absorb energy and mass
not change its temperature
not absorb or lose energy and mass
absorb or lose energy, but not mass

7. Learning Check: Heat Capacity
A cup of water is used in an experiment. Its heat capacity is known to be 720 J/˚C. How much heat will it absorb if the experimental temperature changed from 19.2 ˚C to 23.5 ˚C?
q = 3.1 × 103 J

8. Learning Check: Heat Capacity
If it requires J to raise the temperature of g of water by 1.00 C, calculate the heat capacity of 1.00 g of water.
4.18 J/°C

9. Your Turn!
What is the heat capacity of 300. g of an object if it requires J to raise the temperature of the object by 2.00˚ C?
4.18 J/˚C
418 J/˚C
837 J/˚C
1.26 × 103 J/˚C
2.51 × 103 J/°C
1255 J/°C

10. Using Specific Heat
How much heat is required to raise g H2O from 18.2 C to 84.3 C ?
q = m x s x Dt
q = g x J/g.oC x (84.3oC – 18.2oC)
q = 4,682 J

11. Learning Check: Specific Heat
Calculate the specific heat of a metal if it takes 235 J to raise the temperature of a g sample by 2.53 °C.
The symbol q is used to mean “quantity of heat”.

12. Your Turn!
The specific heat of copper metal is J/(g ˚C). How many J of heat are necessary to raise the temperature of a 1.42 kg block of copper from 25.0 ˚C to 88.5 ˚C?
547 J
1.37 × 104 J
3.47 × 104 J
34.7 J
4.74 × 104 J

13. Specific Heat Calculation
How much heat energy must you lose from a 250. mL cup of coffee for the temperature to fall from 65.0 ˚C to 37.0 ˚C? (Assume density of coffee = 1.00 g/mL, scoffee = swater = 4.18 J/g˚C)
q = s  m  t
t = – 65.0 ˚ C = – 28.0 ˚C
q = 4.18 J/g˚C  250. mL  1.00 g/mL  (– 28.0 ˚C)
q = (–29.3  103 J) = –29.3 kJ

14. Using Specific Heat
If a 38.6 g piece of gold absorbs 297 J of heat, what will the final temperature of the gold be if the initial temperature is 24.5 ˚C? The specific heat of gold is J/g˚C.
Need to find tfinal t = tf – ti
First use q = s  m  t to calculate t
Next calculate tfinal
59.6 °C = tf – 24.5 °C
tf = 59.6 °C °C = °C
= ˚ C

15. Exothermic Reaction Example:
A reaction where products have less chemical energy than reactants
The reaction releases heat to the surroundings
Some chemical (PE) energy is converted to kinetic energy
Heat leaves the system; q is negative ( – )
Heat energy is a product
Reaction gets warmer, the temperature increases
Example:
CH4(g) + 2O2(g)  CO2(g) + 2H2O(g) + heat

16. Endothermic Reaction
A reaction where products have more chemical energy than reactants
The reaction absorbs heat from its surroundings
Some kinetic energy is converted into chemical (PE) energy
Heat is added to the system; q is positive (+)
Heat energy is a reactant
Reaction becomes colder,
temperature decreases
Example: Photosynthesis
6CO2(g) + 6H2O(g) + solar energy  C6H12O6(s) + 6O2(g)

17. Bond Strength
Measured by how much energy is needed to break a bond or how much energy is released when a bond is formed.
A larger amount of energy equals a stronger bond
Weak bonds require less energy to break than do strong bonds
A key to understanding reaction energies
Example: If a reaction has
Weak bonds in the reactants and
Stronger bonds in the products
Heat released (the system loses energy as it becomes more stable)

18. Learning Check: –1,256 kJ Consider the following reaction:
2C2H2(g) + 5O2(g) → 4CO2(g) + 2H2O(g)
ΔE ° = –2511 kJ
The reactants (acetylene and oxygen) have 2511 kJ more energy than products. How many kJ are released for 1 mol C2H2?
–1,256 kJ

19. Learning Check:
Given the equation below, how many kJ are required for 44 g CO2 (MM = g/mol) to react with H2O?
6CO2(g) + 6H2O → C6H12O6(s) + 6O2(g) ΔH˚reaction = 2816 kJ
If 100. kJ are provided, what mass of CO2 can be converted to glucose?
470 kJ
9.4 g

20. Your Turn! Based on the reaction
CH4(g) + 4Cl2(g)  CCl4(g) + 4HCl(g)
H˚reaction = – 434 kJ/mol CH4
What energy change occurs when 1.2 moles of methane reacts?
–3.6 × 102 kJ
5.2 × 102 kJ
–4.3 × 102 kJ
3.6 × 102 kJ
–5.2 × 102 kJ
H = –434 kJ/mol × 1.2 mol
H = –520.8 kJ = × 102 kJ

21. Your Turn! Rxn. 1: C3H8(g) + 5O2(g) → 3 CO2(g) + 4 H2O(g)
ΔH °rxn= –2043 kJ
Rxn. 2: C3H8(g) + 5O2(g) → 3 CO2(g) + 4 H2O(l )
ΔH °rxn = –2219 kJ
Why does rxn. 2 release more heat than rxn. 1?
It shouldn’t, and the DH values should be equal
More reactants must have been used in rxn 2
In rxn 1 some of the heat of the reaction is used up converting liquid water to gas, so less heat can be given off.
Liquids always have lower temperatures than gases

22. Example: Calculate Horxn for C (s, graphite)  C (s, diamond)
Given C (s, gr) + O2(g)  CO2(g) H°rxn = –394 kJ
C (s, dia) + O2(g)  CO2(g) H°rxn = –396 kJ
To get desired equation, must reverse second equation and add resulting equations
C(s, gr) + O2(g)  CO2(g) H°rxn = –394 kJ
CO2(g)  C(s, dia) + O2(g) H°rxn = –(–396 kJ)
C(s, gr) + O2(g) + CO2(g)  C(s, dia) + O2(g) +CO2(g)
H° = –394 kJ kJ = + 2 kJ
–1[ ]

23. Your Turn!
The standard enthalpy of formation of sulfur dioxide is kJ. What is DHf for the formation of g of sulfur dioxide in its standard state from its elements in their standard states?
A kJ
B kJ
C. -4,759 kJ
D kJ
E kJ
= ̶ 74.3 kJ

24. Example: Calculate Horxn Using Hf°
Calculate H°rxn using Hf° data for the reaction
SO3(g)  SO2(g) + ½O2(g)
Multiply each Hf° (in kJ/mol) by the number of moles in the equation
Add the Hf° (in kJ/mol) multiplied by the number of moles in the equation of each product
Subtract the Hf° (in kJ/mol) multiplied by the number of moles in the equation of each reactant
H°rxn has units of kJ
H°f has units of kJ/mol
H°rxn = 99 kJ

25. Determine the for the combustion of ammonia4
NH3 (g) + O2 (g) N2 (g) H2O (g) 3 2 6
(kJ)
4 NH3 (g)
4 x (-46.19)
3 O2 (g)
3 x 0
2 N2 (g)
2 x 0
6 H2O (g)
6 x (-241.8)
= pdts – rcts = [( )-( )]
= kJ

26. Ethanol is used as an additive in many fuels today
Ethanol is used as an additive in many fuels today. What is ΔHºrxn (kJ) for the combustion of ethanol? 2 C2H5OH (l ) + 6 O2 (g) → 4 CO2 (g) + 6 H2O (l )
ΔHºf (kJ/ mol)
C2H5OH (l)
–277.6
CO2 (g)
–393.5
H2O (g)
–241.8
H2O (l)
–285.8
– 401.7
–2469
+ 2734
– 2734

27. Ethanol is used as an additive in many fuels today
Ethanol is used as an additive in many fuels today. What is ΔHºrxn (kJ) for the combustion of ethanol? 2 C2H5OH (l ) + 6 O2 (g) → 4 CO2 (g) + 6 H2O (l )
ΔHºf (kJ/ mol)
C2H5OH (l)
–277.6
CO2 (g)
–393.5
H2O (g)
–241.8
H2O (l)
–285.8
– 401.7
–2469
+ 2734
– 2734
Answer: e

28. DH°reaction = S n DHf°(products) − S n DHf°(reactants)
Calculate the DHrxn for 2 C2H2(g) + 5 O2(g) ® 4 CO2(g) + 2 H2O(l) DHf°C2H2 = kJ, DHf°H2O = −285.8 kJ, DHf°CO2 = −393.5 kJ
DH°reaction = S n DHf°(products) − S n DHf°(reactants)
DHrxn = [(4•DHf°CO2 + 2•DHf°H2O) – (2•DHf°C2H2 + 5•DHf°O2)]
DHrxn = [(4•(−393.5) + 2•(−285.8)) – (2•(+227.4) + 5•(0))]
DHrxn = − kJ

29. Your Turn! Which of the following is not a form of kinetic energy?
A pencil rolls across a desk
A pencil is sharpened
A pencil is heated
A pencil rests on a desk
A pencil falls to the floor

30. For a system that gains thermal energy and does work on the surroundings,
heat (q) and work (w) are both positive.
heat (q) is positive and work (w) is negative.
heat (q) is negative and work (w) is positive.
heat (q) and work (w) are both negative.
Answer: b

31. For a system that gains thermal energy and does work on the surroundings,
heat (q) and work (w) are both positive.
heat (q) is positive and work (w) is negative.
heat (q) is negative and work (w) is positive.
heat (q) and work (w) are both negative.
Answer: b

32. Calculate the internal energy, ΔE, for a system that does 422 J of work and loses 227 J of energy as heat.
+ 649 J
– 649 J
0 J
– 195 J
+ 195 J

33. Calculate the internal energy, ΔE, for a system that does 422 J of work and loses 227 J of energy as heat.
+ 649 J
– 649 J
0 J
– 195 J
+ 195 J
Answer: b

34. A piece of metal at 85 °C is added to water at 25 °C, the final temperature of the both metal and water is 30 °C. Which of the following is true?
Heat lost by the metal > heat gained by water
Heat gained by water > heat lost by the metal
Heat lost by the metal > heat lost by the water
Heat lost by the metal = heat gained by water
More information is required

35. A piece of metal at 85 °C is added to water at 25 °C, the final temperature of the both metal and water is 30 °C. Which of the following is true?
Heat lost by the metal > heat gained by water
Heat gained by water > heat lost by the metal
Heat lost by the metal > heat lost by the water
Heat lost by the metal = heat gained by water
More information is required

36. Learning Check: Heat Capacity
A cup of water is used in an experiment. Its heat capacity is known to be 720 J/˚C. How much heat will it absorb if the experimental temperature changed from 19.2 ˚C to 23.5 ˚C?
q = 3.1 × 103 J

37. Learning Check: Heat Capacity
If it requires J to raise the temperature of g of water by 1.00 C, calculate the heat capacity of g of water.
4.18 J/°C

38. Your Turn!
What is the heat capacity of 300. g of an object if it requires J to raise the temperature of the object by 2.00˚ C?
4.18 J/˚C
418 J/˚C
837 J/˚C
1.26 × 103 J/˚C
2.51 × 103 J/°C
1255 J/°C

39. Your Turn!
A copper mug has a heat capacity of 77.5 J/˚C. After adding hot water to the mug, the temperature of the mug changed from 77.0 ˚F to 185 ˚F. How much heat did the mug absorb from the water?
4.65 × 103 J
1.29 J
8.37 × 103 J
5.97 × 103 J
1.43 × 104 J
Note units: temps. must be in oC
ti = 25.0 oC, tf = 85.0 oC,
So Dt = 85.0 oC – 25.0 oC = 60.0 oC
= 4.65 × 103 J

40. Learning Check 4.18 J/g °C 4.18 J/g °C
Calculate the specific heat of water if the heat capacity of 100. g of water is 418 J/°C.
What is the specific heat of water if heat capacity of g of water is 4.18 J/°C?
Thus, heat capacity is independent of amount of substance – an intensive property!
4.18 J/g °C
4.18 J/g °C

41. Your Turn!
The specific heat of silver J g–1 °C–1. What is the heat capacity of a 100. g sample of silver?
0.235 J/°C
2.35 J/°C
23.5 J/°C
235 J/°C
2.35 × 103 J/°C

42. Your Turn!
A cast iron skillet is moved from a hot oven to a sink full of water. Which of the following is false?
The water heats
The skillet cools
The heat transfer for the skillet has a negative (–) sign
The heat transfer for the skillet has the same sign as the heat transfer for the water

43. 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 Note: Cs is same as s
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

44. Specific Heat Capacity
If 50 g samples of all substances in Table 6.4 absorbed 100 J of energy, which would show the greatest temperature change?
DT = q /Cs so DT 1/Cs
If 50 g samples of all substances in Table 6.4 cooled by 75 oC, which would release the most energy?
Delta T = q/mCs ; Delta T is proportional to 1/Cs
Answer Top Question – Glass
Answer Bottom Question - Water

45. Example 6. 3: A 32. 5-g cube of aluminum initially at 45
Example 6.3: 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? Cs, Al = J/g•ºC, Cs, H2O = 4.18 J/g•ºC
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

46. A hot piece of metal weighing 350. 0 g is heated to 100. 0 °C
A hot piece of metal weighing g is heated to °C. It is then placed into a coffee cup calorimeter containing g of water at 22.4 °C. The water warms and the copper cools until the final temperature is 35.2 °C. Calculate the specific heat of the metal and identify the metal.
metal: g, T1 = °C, T2 = 35.2 °C
H2O: g, T1 = 22.4 °C, T2 = 35.2°C, Cs = 4.18 J/g °C
FIND: Cs , metal, J/gºC
Cs= 0.377m; presumably copper
Set up m ∙ Cs ∙ delta T for the water and the metal. Solve for the specific heat of the metal.
m, Cs, DT q

47. A 3. 54 g piece of aluminum is heated to 96
A 3.54 g piece of aluminum is heated to 96.2 ºC and allowed to cool to room temperature, 22.5 ºC. Calculate the heat (in kJ) associated with the cooling process. The specific heat of aluminum is J/ g · K.
– 236
+ 236
– 0.236
– 0.638
q = m c delta T
The metal (system) loses heat so the sign is negative

48. A 3. 54 g piece of aluminum is heated to 96
A 3.54 g piece of aluminum is heated to 96.2 ºC and allowed to cool to room temperature, 22.5 ºC. Calculate the heat (in kJ) associated with the cooling process. The specific heat of aluminum is J/ g · K.
– 236
+ 236
– 0.236
– 0.638
Answer: d

49. A 1. 5 g iron nail is heated to 95
A 1.5 g iron nail is heated to 95.0 °C and placed into a beaker of water. Calculate the heat gained by the water if the final equilibrium temperature is 57.8 ° C. The specific heat capacity of iron = J/ g ° C, and the specific heat capacity of water = 4.18 J/ g ° C.
6.0 J
25 J
–25 J
39 J
– 39 J
m c delta T for iron = m c delta T for water

50. A 1. 5 g iron nail is heated to 95
A 1.5 g iron nail is heated to 95.0 °C and placed into a beaker of water. Calculate the heat gained by the water if the final equilibrium temperature is 57.8 ° C. The specific heat capacity of iron = J/ g ° C, and the specific heat capacity of water = 4.18 J/ g ° C.
6.0 J
25 J
–25 J
39 J
– 39 J
Answer: b

51. Identical amounts of heat are applied to 50 g blocks of lead, silver, and copper, all at an initial temperature of 25 °C. Which block will have the largest increase in temperature?
Lead
Silver
Copper
None; all will be at the same temperature.

52. Identical amounts of heat are applied to 50 g blocks of lead, silver, and copper, all at an initial temperature of 25 °C. Which block will have the largest increase in temperature?
Lead
Silver
Copper
None; all will be at the same temperature.
Answer: c
q = m c delta T. Which has the smaller value for c? c and delta T are inversely proportional given the same amounts of heat and masses.
Lead should be the correct answer.

53. 50.0 g of water at 22 °C is mixed with 125 g of water initially at 36 ° C. What is the final temperature of the water after mixing, assuming no heat is lost to the surroundings?
29 °C
42 ° C
33 ° C
31 ° C
36 ° C

54. 50.0 g of water at 22 °C is mixed with 125 g of water initially at 36 ° C. What is the final temperature of the water after mixing, assuming no heat is lost to the surroundings?
29 °C
42 ° C
33 ° C
31 ° C
36 ° C
Answer: c
Use m c delta T. c is the same in both instances. Use T(final) – T(initial) for delta T.

55. Calorimeter Problem
When g of olive oil is completely burned in pure oxygen in a bomb calorimeter, the temperature of the water bath increases from ˚C to ˚C.
a) How many Calories are in olive oil, per gram? The heat capacity of the calorimeter is kJ/˚C.
t = ˚C – ˚C = ˚C
qabsorbed by calorimeter = Ct = kJ/°C × ˚C
= kJ
Calculate heat absorbed by calorimeter
Convert this to heat released by combustion of olive oil
Convert heat to Cal/gram
qreleased by oil = – qcalorimeter = – kJ
–8.740 Cal/g oil

56. Calorimeter Problem (cont)
Olive oil is almost pure glyceryl trioleate, C57H104O6. The equation for its combustion is
C57H104O6(l) + 80O2(g)  57CO2(g) + 52H2O
What is E for the combustion of one mole of glyceryl trioleate (MM = g/mol)? Assume the olive oil burned in part a) was pure glyceryl trioleate.
E = qV = –3.238 × 104 kJ/mol oil

57. Coffee Cup Calorimetry
NaOH and HCl undergo rapid and exothermic reaction when you mix 50.0 mL of 1.00 M HCl and 50.0 mL of 1.00 M NaOH. The initial t = 25.5 °C and final t = 32.2 °C. What is H in kJ/mole of HCl? Assume for these solutions s = J g–1°C–1. Density: 1.00 M HCl = 1.02 g mL–1; 1.00 M NaOH = 1.04 g mL–1.
NaOH(aq) + HCl(aq)  NaCl(aq) + H2O(aq)
qabsorbed by solution = mass  s  t
massHCl = 50.0 mL  1.02 g/mL = 51.0 g
massNaOH = 50.0 mL  1.04 g/mL = 52.0 g
massfinal solution = 51.0 g g = g
t = (32.2 – 25.5) °C = 6.7 °C
1. Determine how much heat is absorbed by the calorimeter.

58. Coffee Cup Calorimetry
qcal = g  J g–1 °C–1  6.7 °C = 2890 J
Rounds to qcal = 2.9  103 J = 2.9 kJ
qrxn = –qcalorimeter = –2.9 kJ
= mol HCl
Heat evolved per mol HCl =
2. Heat evolved by reaction
= -58 kJ/mol

59. Your Turn!
A g sample of solid is transferred from boiling water (t = 99.8 ˚C) to 152 g water at 22.5 ˚C in a coffee cup. The temperature of the water rose to 24.3 ˚C. Calculate the specific heat of the solid.
–1.1 × 103 J g–1 ˚C–1
1.1 × 103 J g–1 ˚C–1
1.0 J g–1 ˚C–1
0.35 J g–1 ˚C–1
0.25 J g–1 ˚C–1
q = m × s × t
Determine how much heat is absorbed by the calorimeter.
Calculate the specific heat of the solid.
= 1.1 × 103 J
qsample = – qwater = – 1.1 × 103 J

60. How much heat is evolved in the complete combustion of 13
How much heat is evolved in the complete combustion of 13.2 kg of C3H8(g) ?
C3H8(g) + 5 O2(g) → 3 CO2(g) + 4 H2O(g) DH = −2044 kJ
MM = g/mol
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 number is reasonable because the amount of C3H8 is much more than 1 mole

61. What is DHrxn/mol Mg for the reaction: Mg(s) + 2 HCl(aq) → MgCl2(aq) + H2(g) if g Mg completely reacts in mL of solution and changes the temperature from 25.6 °C to 32.8 °C?
Given:
Find:
6.499 x 10−3 mol Mg, 1.00 x 102 g, DT = 7.2 °C,,Cs = 4.18 J/g∙°C
DH, kJ/mol
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

62. In a coffee cup calorimetry experiment, 50. 0 mL of 0
In a coffee cup calorimetry experiment, 50.0 mL of M NaOH is added to 50.0 mL of M HCl. The temperature of the solution increased by 14.9 °C. Calculate ΔH for the reaction. Use 1.00 g/ mL as the density of the solution and 4.18 J/ g ° C as the specific heat capacity.
1.25 kJ/ mol
6.23 kJ/ mol
31.0 kJ/ mol
1250 kJ/ mol
6230 kJ/ mol

63. In a coffee cup calorimetry experiment, 50. 0 mL of 0
In a coffee cup calorimetry experiment, 50.0 mL of M NaOH is added to 50.0 mL of M HCl. The temperature of the solution increased by 14.9 °C. Calculate ΔH for the reaction. Use 1.00 g/ mL as the density of the solution and 4.18 J/ g ° C as the specific heat capacity.
1.25 kJ/ mol
6.23 kJ/ mol
31.0 kJ/ mol
1250 kJ/ mol
6230 kJ/ mol
Answer: d

64. qwater = mwaterswaterΔtwater
The specific heat of water is J / goC. In an isolated system, a 12.0 g sample of hot metal at oC is dropped into 25.0 g of water at oC. The temperature of the water rises to oC to reach equilibrium. What is the specific heat of this metal?
qwater = mwaterswaterΔtwater
qwater = (25.0 g)(4.184 J / goC)(23.34 oC – oC)
qwater = J
qwater = - qmetal
qmet = J
smetal = qmetal / mmetalΔtmetal
smetal= J / (12.0g ×(23.34 oC – oC) )
smetal = J / goC 68

65. Your Turn! Which of the following is an endothermic process?
Na+ + e–  Na
wood burning
CH4(g)  C(g) + 4H(g)
a bomb exploding
water condensing
Attraction of charges releases heat
Produces heat (i.e., heat released)
Breaking bonds requires heat
Produces heat (i.e., heat released)
Heat must be released

66. Which of the following statements about enthalpy is FALSE?
The value of ΔH for a chemical reaction is the amount of heat absorbed or evolved in the reaction under conditions of constant pressure.
An endothermic reaction has a positive ΔH and absorbs heat from the surroundings. An endothermic reaction feels cold to the touch.
An exothermic reaction has a negative ΔH and gives off heat to the surroundings. An exothermic reaction feels warm to the touch.
Statements (a), (b), and (c) are all false statements.
Statements (a), (b), and (c) are all true statements.

67. Which of the following statements about enthalpy is FALSE?
The value of ΔH for a chemical reaction is the amount of heat absorbed or evolved in the reaction under conditions of constant pressure.
An endothermic reaction has a positive ΔH and absorbs heat from the surroundings. An endothermic reaction feels cold to the touch.
An exothermic reaction has a negative ΔH and gives off heat to the surroundings. An exothermic reaction feels warm to the touch.
Statements (a), (b), and (c) are all false statements.
Statements (a), (b), and (c) are all true statements.
Answer: e

68. Cu(s) + CuCl2(s)  2 CuCl(s) DH° = ? kJ
Given the following equations, Find DHrxn for
Cu(s) + CuCl2(s)  2 CuCl(s) DH° = ? kJ
Cu(s) + Cl2(g)  CuCl2(s) DH° = −206 kJ
2 Cu(s) + Cl2(g)  2 CuCl(s) DH° = − 36 kJ
Cu(s) + Cl2(g)  CuCl2(s) DH° = −206 kJ
2 Cu(s) + Cl2(g)  2 CuCl(s) DH° = − 36 kJ

69. Cu(s) + CuCl2(s)  2 CuCl(s) DH° = ? kJ
Given the following equations, find DHrxn for
Cu(s) + CuCl2(s)  2 CuCl(s) DH° = ? kJ
Cu(s) + Cl2(g)  CuCl2(s) DH° = −206 kJ
2 Cu(s) + Cl2(g)  2 CuCl(s) DH° = − 36 kJ
CuCl2(s)  Cu(s) + Cl2(g) DH° = +206 kJ
2 Cu(s) + Cl2(g)  2 CuCl(s) DH° = kJ
Cu(s) + CuCl2(s)  2 CuCl(s) DH° = kJ

70. Given the following equations, Find DHrxn for 3 NO2(g) + H2O(l)  2 HNO3(aq) + NO(g)
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
2 NO2(g)  2 NO(g) + O2(g) DH° = +116 kJ
2 N2(g) + 5 O2(g) + 2 H2O(l)  4 HNO3(aq) DH° = −256 kJ
2 NO(g)  N2(g) + O2(g) DH° = −183 kJ
X 3/2
3 NO2(g)  3 NO(g) O2(g) DH° = (+174 kJ)
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
X 1/2

71. Use Hess’s Law to determine ΔH for the following target reaction
Use Hess’s Law to determine ΔH for the following target reaction. ½ N2 (g) + ½ O2 (g) → NO (g) ΔH = kJ NO (g) + ½ Cl2 (g) → NOCl (g) ΔH = –38.6 kJ 2 NOCl (g) → N2 (g) + O2 (g) + Cl2 (g) ΔH = ?
–51.7 kJ
51.7 kJ
–103.4 kJ
103.4 kJ
142.0 kJ

72. Use Hess’s Law to determine ΔH for the following target reaction
Use Hess’s Law to determine ΔH for the following target reaction. ½ N2 (g) + ½ O2 (g) → NO (g) ΔH = kJ NO (g) + ½ Cl2 (g) → NOCl (g) ΔH = –38.6 kJ 2 NOCl (g) → N2 (g) + O2 (g) + Cl2 (g) ΔH = ?
–51.7 kJ
51.7 kJ
–103.4 kJ
103.4 kJ
142.0 kJ
Answer: c

73. Which of the following has a negative value of ΔH?
The combustion of gasoline
The melting of snow
The evaporation of ethanol
The sublimation of CO2 at room temperature
All of the above have positive ΔH values.

74. Which of the following has a negative value of ΔH?
The combustion of gasoline
The melting of snow
The evaporation of ethanol
The sublimation of CO2 at room temperature
All of the above have positive ΔH values.
Answer: a

75. Standard Enthalpies of Formation

76. Learning Check Calculate H for this reaction using Hf° data.
2Fe(s) + 6H2O(l)  2Fe(OH)3(s) + 3H2(g)
Hf° – –
H°rxn = 2×Hf°(Fe(OH)3, s) + 3×Hf°(H2, g) – 2× Hf°(Fe, s) – 6×Hf°(H2O, l )
H°rxn = 2 mol× (– kJ/mol) + 3×0 – 2×0 – 6 mol× (–285.8 kJ/mol)
H°rxn = –1393 kJ kJ
H°rxn = kJ

77. Learning Check Calculate H°rxn for this reaction using Hf° data.
CO2(g) + 2H2O(l )  2O2(g) + CH4(g)
Hf° – – – 74.8
H°rxn = 2×Hf°(O2, g) + Hf°(CH4, g) –Hf°(CO2, g) – 2× Hf°(H2O, l )
H°rxn = 2 × mol × (–74.8 kJ/mol) – 1 mol × (–393.5 kJ/mol) – 2 mol × (–285.8 kJ/mol)
H°rxn = –74.8 kJ kJ kJ
H°rxn = kJ

78. Your Turn!
Calculate Hf° for FeO(s) using the information below. Hf° values are shown below each substance.
Fe3O4(s) + CO(g)  3FeO(s) + CO2(g) Ho=21.9 kJ
kJ kJ ?? – kJ
A kJ
B kJ
C kJ
D J
E kJ

79. Your Turn! sol’n
Important
H°rxn = [3Hf° (FeO, s) + Hf°(O2, g)] – [Hf°(Fe3O4, s) + Hf°(CO, g)]
+21.9 kJ = [3Hf° (FeO, s) kJ)] – [ kJ kJ)]
+21.9 kJ = [3Hf° (FeO, s) kJ]
kJ = 3Hf° (FeO, s)
kJ = Hf° (FeO, s)

80. Learning Check
Calculate H°rxn using Hf° for the reaction 4NH3(g) + 7O2(g)  4NO2(g) + 6H2O(l )
H°rxn = [136 – ] kJ
H°rxn = –1395 kJ