Exercises 2, 4, 16, 18, 20, 26, 28, 34, 36, 46, 50, 52, 54, 58, 60, 62, 72, 84

2. (a) As a tennis ball moves upward, the potential energy increases due to an increase in the gravitational force, which increases as altitude increases. Since the potential energy increases, the kinetic energy must decrease since the total energy of the tennis ball is constant.

(b) The potential energy of the tennis ball increases as the ball moves higher.

(c) The potential energy of gravity is given by the expression "mgh" where m is mass, g is the gravitational acceleration constant, and h is height. If the mass of the ball is doubled and is thrown with the same energy as before, then the height must be halved to come up with the same value of potential energy. Therefore, the heavier ball will go to half the height that the tennis ball did.

4. Kinetic energy is expressed by the formula: . In order to get energies in joules, the mass and velocity must be expressed in S. I. units (kg for mass and m/s for velocity).

(a) Converting 950 lb to kg (1 kg = 2.2046 lb)

Converting the velocity of 68 mph to m/s:

The kinetic energy is equal to:

(b) Convert 34 mph to m/s:

The kinetic energy is equal to:

The kinetic energy is reduced by a factor of 4. You may avoid the math in part (b) by recognizing that 34 mph is half of the original speed of 68 mph. If the velocity decreases by a factor of 2, then the kinetic energy decreases by a factor of 2² or 4.

(c) The major portion of the kinetic energy goes into the friction caused by the brakes rubbing on the wheel rims.

16. (a) When a positively charged particle moves in a circle at a fixed distance from a negatively charge particle, there is an attractive electrostatic force between the positively charged particle and the negatively charged particle. This force is dependent upon the distance between the particles. Since there is no change in distance between the particles, no work is being performed on the positively charged particle.

(b) There is an attractive magnetic force between the iron nail and the magnet whose strength depends on the distance between the nail and the magnet. Since you are pulling off the nail from the magnet, work is being performed to counterbalance the force of the magnet pulling on the nail.

18. (a) The first law of thermodynamics is expressed by the equation: ∆E = q + w.

(b) In thermodynamics, we are generally interested only in changes in internal energy and energy flows, not the total internal energy of an object. Therefore, we don't have to measure the internal energy in applying the first law.

(c) q is negative when heat flows out of the system into the surroundings. w is negative when work is performed by the system on the surroundings.

20. (a) Since heat flows into the balloon, this process is endothermic and q = 900 J. Since the balloon performs work, w = –422 J. The change in internal energy is equal to:

∆E = q + w = 900 J + (–422 J) = 478 J

(b) Since the water loses 3140 J of heat, this process is exothermic and q = –3140 J. No work is mentioned so assume it is equal to zero. The change of internal energy is equal to:

∆E = q + w = –3140 J + 0 J = –3140 J

(c) Since the reaction releases heat, this process is exothermic and q = –8.65 kJ. Since no work is performed: ∆E = q = –8.65 kJ.

26. (a) The change in potential energy when a book is transferred from a table to a shelf depends only on the difference in height between the table and the shelf. It is not dependent on how the book travels from table to shelf.

(b) The heat releases when a sugar cube is oxidized to CO₂ and H₂O depends on how the sugar cube is oxidized. Heat depends on how the process is performed.

(c) The work accomplished in burning a gallon of gasoline depends on how the work is being produced by the combustion of the gallon of gasoline. Work depends on how the process is performed.

28. (a) The enthalpy change is equal to the amount of heat transferred into or out of the system under conditions of constant pressure.

(b) The change in a state function is equal to the difference between the initial and final value of the state function. It does not depend on how you get from the initial to the final state of the system.

(c) If a system absorbs heat, then the value of q is positive. At constant pressure, the value of q is equal to ∆H. Since q is positive, ∆H is also positive, which means enthalpy will increase.

34. (a) Because ∆H is positive, heat is absorbed during the course of this reaction.

(b) Convert 1.60 kg of CH₃OH into number of moles of CH₃OH:

Convert moles of CH₃OH into amount of heat transferred using the thermochemical equation:

(c) Convert the enthalpy change of 64.7 kJ into number of moles of H₂ produced using the thermochemical equation:

Convert moles of H₂ into grams of H₂:

(d) If you reverse the direction of the reaction, the sign on ∆H reverses. Therefore, ∆H = –90.7 kJ for the reverse reaction.

Convert 32.0 g of CO into moles of CO:

Convert moles of CO into the heat released:

The negative sign indicates heat is released.

36. (a) Convert 4.34 mol of O₂ into ∆H:

(b) Convert 200.8 g of KCl into moles of KCl:

Convert moles of KCl into ∆H:

(c) Generally, a reaction that proceeds spontaneously in one direction will not proceed spontaneously in the reverse direction. Therefore, you would not expect KCl to react with O₂ to form KClO₃.

46. (a) The specific heat of water is 1.00 cal/g·°C or 4.18 J/g·°C. Converting the specific heat into the molar heat capacity:

(b) The heat capacity of 8.42 mol of liquid water is equal to:

(c) Convert 2.56 kg of water into moles of water:

Using the molar heat capacity of water, the heat needed is equal to:

50. The specific heat of pure water is 4.18 J/g·°C. The mass of the solution is equal to 63.9 g (60.0 g water + 3.88 g NH₄NO₃). The heat of reaction is equal to:

Convert 3.88 g of NH₄NO₃ into moles of NH₄NO₃:

∆H in units of kJ/mol is equal to:

52. (a) The balanced reaction for the combustion of C₆H₅OH is:

C₆H₅OH(s) + 7 O₂(g) ¾ ¾ ® 6 CO₂(g) + 3 H₂O(l)

(b) Calculate the heat of combustion for 1.800 g of phenol:

The heat of combustion per gram of phenol is equal to:

Convert 1.800 g of phenol into moles of phenol:

The heat of combustion per mole of phenol is equal to:

54. (a) Calculate the heat of combustion of 1.640 g of benzoic acid:

The total heat capacity of the calorimeter is the heat of combustion divided by the temperature change:

(b) Calculate the heat of combustion of 1.320 g of the substance:

The heat of combustion per gram of the substance is equal to:

(c) If some water is lost when changing samples, the total heat capacity of the calorimeter would decrease since there is less water to absorb the heat.

58. (a) The reaction Y ¾ ¾ ® Z is the sum of the following two reactions:

Y ¾ ¾ ® X ∆H = –(–35 kJ) = 35 kJ

X ¾ ¾ ® Z ∆H = 90 kJ

Y ¾ ¾ ® Z ∆H = 125 kJ

According to Hess's law, the enthalpy change for a reaction that is the sum of two thermochemical equations is the sum of the enthalpy changes for the two reactions, or 125 kJ.

(b) The enthalpy diagram is:

(c) Two reactions must be performed at the same temperature for Hess's Law to be valid. Therefore, if the first reaction is performed at 25°C and the second reaction at 240°C, then you couldn't add them.

60. The reaction:

3 H₂(g) + O₃(g) ¾ ¾ ® 3 H₂O(g)

is the sum of 3/2 times the first thermochemical reaction plus ½ times the reverse of the second thermochemical reaction:

3 H₂(g) + O₂(g) ¾ ¾ ® 3 H₂O(g) ∆H = (–483.6 kJ) = –725.4 kJ

O₃(g) ¾ ¾ ® O₂(g) ∆H = –(284.6 kJ) = –142.3 kJ

3 H₂(g) + O₃(g) ¾ ¾ ® 3 H₂O(g) ∆H = –867.7 kJ

62. The reaction

N₂O(g) + NO₂(g) ¾ ¾ ® 3 NO(g)

is equal to the sum of ½ the third thermochemical equation plus ½ times the reverse of the second thermochemical equation plus the first thermochemical equation:

N₂(g) + O₂(g) ¾ ¾ ® 2 NO(g) ∆H = 180.7 kJ

NO₂(g) ¾ ¾ ® NO(g) + O₂(g) ∆H = –(–113.1 kJ) = 56.6 kJ

N₂O(g) ¾ ¾ ® N₂(g) + O₂(g) ∆H = (–163.2 kJ) = –81.6 kJ

N₂O(g) + NO₂(g) ¾ ¾ ® 3 NO(g) ∆H = 155.7 kJ

72. (a) N₂O₄(g) + 4 H₂(g) ¾ ¾ ® N₂(g) + 4 H₂O(g)

∆H = (N₂(g)) + 4 (H₂O(g)) – (N₂O₄(g)) – 4 (H₂(g))

= 0 kJ + 4(–241.82 kJ) – (9.66 kJ) – 4(0 kJ) = –967.28 kJ – 9.66 kJ = –976.94 kJ

(b) 2 KOH(s) + CO₂(g) ¾ ¾ ® K₂CO₃(s) + H₂O(g)

∆H = (K₂CO₃(s)) + (H₂O(g)) – 2 (KOH(s)) – (CO₂(g))

= –1150.18 kJ + (–241.82 kJ) – 2(–424.7 kJ) – (–393.5 kJ)

= –1150.18 kJ – 241.82 kJ + 849.4 kJ + 393.5 kJ = –149.1 kJ

(c) SO₂(g) + 2 H₂S(g) ¾ ¾ ® S₈(s) + 2 H₂O(g)

∆H = (S₈(s)) + 2(H₂O(g)) – (SO₂(g)) – 2(H₂S(g))

=(0 kJ) + 2(–241.82 kJ) – (–269.9 kJ) – 2(–20.71 kJ)

= –483.64 kJ + 269.9 kJ + 20.71 kJ = –193.0 kJ

(d) Fe₂O₃(s) + 6 HCl(g) ¾ ¾ ® 2 FeCl₃(s) + 3 H₂O(g)

∆H = 2(FeCl₃(s)) + 3(H₂O(g)) – (Fe₂O₃(s)) – 6(HCl(g))

= 2 (–400 kJ) + 3(–241.82 kJ) – (–822.16 kJ) – 6(–92.30 kJ)

= –800 kJ – 725.46 kJ + 822.16 kJ + 553.8 kJ = –150 kJ

84. Calculate the mass of 177 mL of wine given its density of 1.0 g/mL:

Use the mass percent alcohol to calculate the mass of ethanol in the glass of wine:

Convert the mass of ethanol into number of moles:

Multiply by the heat of combustion to get the heat released:

Convert to kcals:

A dietetic calorie is equal to a kcal so a glass of wine contains 133.5 dietetic calories.