Chemistry, Student Solutions Manual Chapter 3

Chapter 3 Energy and Its Conservation

Solutions to Problems in Chapter 3

3.1 Conservation of energy states that energy is neither created nor destroyed. It can only be transformed from one form into another:

(a) Total energy, which is conserved, is the sum of potential, kinetic, and thermal energies. On the tree, an apple possesses some gravitational potential energy. As an apple falls, that potential energy is converted into kinetic energy of motion.

(b) When an apple hits the ground, the impact transfers energy to molecules in the earth and in the apple. As a result, there is a slight increase in temperature; the kinetic energy of the apple has been converted into thermal energy.

3.3 Speed and kinetic energy are related through the equation, :

3.5 Electrical energy can be calculated using Equation 3-1: .

When distance is expressed in picometres (1 pm = 10–12 m) and charges are in electronic units, the constant in the equation is k = 2.31 × 10–16 J pm. Attractive energies are negative because when close together, the charges have lower potential energy than when far apart. The charges on the ions in this example are +1 and –1:

3.7 Energy comes in various forms: radiant, kinetic, potential, and thermal:

(a) Radiant energy is consumed and thermal energy is produced.

(b) Chemical potential energy is consumed and kinetic energy is produced.

3.9 Speed and kinetic energy are related through the equation, vThe joule has units kg m2 s–2, so mass must be expressed in kg and v has units m/s:

mneutron = 1.6749 × 10–27 kg

3.11 Calculate the temperature change resulting from an energy input using C values from Table 3-1 of your textbook and Equation 3-2:

3.13 Calculate an energy change accompanying a temperature change using C values from Table 3-1 of your textbook and Equation 3-2:

3.15 When two objects at different temperatures are placed in contact, energy flows from the warmer to the cooler object until the two objects are at the same temperature. Let that temperature be T. The energy lost by the warmer object equals the energy gained by the cooler object:

Substitute and solve for T:

3.17 Calculate the energy change for a temperature change using Equation 3-2 and C values from Table 3-1 of your textbook:

3.19 First determine the energy difference (DE) between chicken and beef:

A 55-kg person walking 6.0 km/h consumes1090 kJ/h:

Therefore, to consume the additional energy a person must walk

3.21 (a) In a combustion reaction, the products are CO2 and H2O:

C7H6O2 + O2 ® CO2 + H2O (unbalanced)

Follow standard procedures to balance the equation. Give CO2 a coefficient of 7 to balance C, H2O a coefficient of 3 to balance H:

C7H6O2 + O2 ® 7 CO2 + 3 H2O

7 C + 6 H + 4 O ® 7 C + 6 H + 17 O

Balance O by giving O2 a coefficient of 15/2, and then multiply by 2 to clear fractions:

2 C7H6O2 + 15 O2 ® 14 CO2 + 6 H2O

(b) Find energy per mole from the energy released by 1.350 g using the molar mass (122.1 g/mol):

3.23 In Table 3-2, we find the following bond energies for H–X bonds where X is an element in Group 16:

Bond / H–O / H–S / H–Te
Energy (kJ/mol) / 460 / 365 / 240

There is a trend: moving down the column, the bond energy decreases. From this trend, we predict that the H–Se bond has an energy between H–S and H–Te, around 300 kJ/mol (the experimental value is 315 kJ), and the H–Po bond will be weaker than H–Te, perhaps around 200 kJ/mol (no experimental value is available).

3.25 To estimate the energy change in a reaction, list the number of bonds of each type in reactants and products, and subtract the sum of average bond energies for products from the sum of average bond energies for reactants, using values from Table 3-2 of your textbook:

Reactants / Products
Bond / No. / Energy (kJ/mol) / Bond / No. / Energy (kJ/mol)
N=N / 1 / 945 / N–N / 1 / 160
O=O / 2 / 495 / N–O / 2 / 200
N=0 / 2 / 605

3.27 Figure 3-13 in your textbook shows three paths for a reaction. One is the actual path. A second is decomposition of reactants into elements and then recombination of the elements into products. A third is breakage of all bonds in the reactants into atoms of the elements, and then reaction of the atoms to form all bonds in the products:

3.29 To estimate the energy change in a reaction, list the number of bonds of each type in the reagents, and subtract the sum of average bond energies for products from the sum of average bond energies for reactants, using values from Table 3-2 of your textbook:

2 H2 + O2 ® 2 H2O

Reactants / Products
Bond / No. / Energy (kJ/mol) / Bond / No. / Energy (kJ/mol)
H–H / 2 / 435 / O–H / 4 / 460
O=O / 1 / 495

3 H2 + N2 ® 2 NH3

Reactants / Products
Bond / No. / Energy (kJ/mol) / Bond / No. / Energy (kJ/mol)
H–H / 3 / 435 / N–H / 6 / 390
N≡N / 1 / 945

3 H2 + CO ® CH4 + H2O

Reactants / Products
Bond / No. / Energy (kJ/mol) / Bond / No. / Energy (kJ/mol)
H–H / 3 / 435 / C–H / 4 / 415
C≡O / 1 / 1070 / O–H / 2 / 460

3.31 Calculate the heat accompanying a temperature change in a calorimeter using Equation 3-6 and the heat capacity C of the calorimeter:

The problem statement tells us to assume that Ccal @ Cwater:

The heat gained by the calorimeter is heat lost by the solution process, giving

qsolution = -1.4 × 103 J

3.33 The heat capacity of a calorimeter can be found from energy and temperature data using Equation 3-6,

Here, the heat released by the combustion process is absorbed by the calorimeter:

3.35 Calculate the energy released during combustion from the temperature increase and the total heat capacity of the calorimeter. Then convert to molar energy using molar mass:

3.37 Expansion work can be calculated using Equation 3-8, The external pressure opposing inflation of a balloon is atmospheric pressure, 1.00 atm:

Vf = 2.5 L, Vi = 0.0 L

DV = 2.5 L – 0.0 L = 2.5 L = 2.5 × 10–3 m3

3.39 Standard enthalpy changes are calculated from Equation 3-12 using standard enthalpies of formation, which can be found in Appendix D of your textbook:

3.41 Reaction energies are related to reaction enthalpies (Problem 3.39) through Equation 3-11:

DHreaction @ DEreaction + RTD(n)gases

Since temperature does not change, we can rearrange the last term:

DEreaction @ DHreaction – RTD(ngases)

Begin by calculating the change in the number of moles of gases, and then use the rearranged equation to determine DEreaction:

(a) Dngases = 2 – (3 + 1) = –2 mol

(b) Dngases = 3 + 1 –2 = 2 mol

3.43 A formation reaction has elements in their standard states as reactants and 1 mole of a single substance as the product:

(a) 3 K (s) + P (s) + 2 O2 (g) ® K3PO4 (s)

(b) 2 C (graphite) + 2 H2 (g) + O2 (g) ® CH3CO2H (l)

(d) 2 Al (s) + 3/2 O2 (g) ® Al2O3 (s)

3.45 Standard enthalpy changes are calculated from Equation 3-12 using standard enthalpies of formation, which can be found in Appendix D of your textbook:

3.47 Figure 3-21 shows that the energy needs per capita in the computer age are 1 × 106 kJ/day. To determine the annual energy usage of Canada, multiply this need by the total population and the number of days in a year:

Annual energy usage = (1 × 106 kJ/person-day)(365 days/yr)(33 212 696 persons)

=1 × 1016 kJ

3.49 Use a ruler to measure the energy consumption for each source for the year 1975 on Figure 3-22. Here are the values:

Source / Amount (1018 kJ) / %
Wood / 1.7 / 2.2
Hydro / 3.4 / 4.3
Nuclear / 2.1 / 2.7
Coal / 13.8 / 17.5
Gas / 20.7 / 26.2
Petroleum / 37.2 / 47.1
Total / 78.9 / 100

Use these values to construct a pie chart, which appears on the right.

3.51 Because energy is a state function, the energy released (∆E) when 1 g of gasoline burns is the same regardless of the conditions. The energy released can be set equal to the work done (w) plus the heat transferred (q). Work is done when an automobile accelerates, but no work is done when an automobile idles:

Heat released (idle) = Heat released (acceleration) + Work done (acceleration). Thus, more heat must be removed under idling conditions (this is part of the reason why automobiles tend to overheat in traffic jams).

3.53 The difference between DHreaction and DEreaction can be estimated using Equation 3-11:

3.55 To work a problem involving heat transfers, it is useful to set up a block diagram illustrating the process. In this problem, a copper block transfers energy to ice:

Thus, qice = –qCu

Substituting gives

nice DHfus = – nCuCDT

Here are the data needed for the calculation:

Substitute and solve for the amount of ice that melts:

Finally, convert to mass:

3.57 (a) In a combustion reaction, a substance reacts with O2 to form CO2 and H2O:

C6H12O6 + O2 ® CO2 + H2O

Balance the equation using standard procedures. Give CO2 a coefficient of 6 and H2O a coefficient of 6 to balance C and H:

C6H12O6 + O2 ® 6 CO2 + 6 H2O

6 C + 12 H + 8 O ® 6 C + 12 H + 18 O

Give O2 a coefficient of 6 to balance O:

C6H12O6 (s) + 6 O2 (g) ® 6 CO2 (g) + 6 H2O (l)

(b) To determine the molar heat of combustion, multiply the heat released (negative, indicating an exothermic reaction) in burning 1 g by the molar mass:

(c) Use the molar heat of combustion along with Equation 3-12 and data from Appendix D to determine the heat of formation of glucose:

3.59 To predict stability from bond energies, tabulate the number of bonds of each type and add up their contributions. Expand the structures in order to count bonds of each type:

We can ignore the 5 C–H bonds and 1 C–O bond that the two compounds have in common, leaving O–H and C–C in ethanol vs. C–H and C–O in diethyl ether: Ethanol: 460 kJ/mol (O–H) + 345 kJ/mol (C–C) = 805 kJ/mol

Diethyl ether: 415 kJ/mol (C–H) + 360 kJ/mol (C–O) = 775 kJ/mol

Based on average bond energies, ethanol is more stable by 30 kJ/mol.

3.61 When an airplane accelerates down a runway, its engines burn fuel, converting chemical energy into kinetic energy of motion. During takeoff, some kinetic energy is converted into gravitational potential energy. The net transformation is conversion of stored chemical energy of the airplane’s fuel into kinetic energy and gravitational potential energy of the airplane (much of the chemical energy also is “wasted” as heat that is transferred to the air passing through the engines).

3.63 A molar heat capacity can be calculated from a temperature change using q = nCDT:

3.65 A calculation using average bond energies requires a list of all bonds in the reagents. To obtain an estimate for reaction energy per mole of material, use Equation 3-5:

Ethane reagents: / H3C–CH3 / 7/2 O2 / 2 CO2 / 3 H2O
Bonds: / 1 C–C, 6 C–H / 7/2 (1 O=O) / 2 (2 C=0) / 3 (2 O–H)
Energies (kJ/mol): / 345, 415 / 495 / 800 / 460
Ethane reagents: / H2C–CH2 / 3 O2 / 2 CO2 / 2 H2O
Bonds: / 1 C–C, 4 C–H / 3 (1 O=O) / 2 (2 C=0) / 2 (2 O–H)
Energies (kJ/mol): / 615, 415 / 495 / 800 / 460
Ethane reagents: / HCºCH2 / 5/2 O2 / 2 CO2 / H2O
Bonds: / 1 CºC, 4 C-H / 5/2 (1 O=O) / 2 (2 C=0) / 2 O–H
Energies (kJ/mol): / 835, 415 / 495 / 800 / 460

These estimates indicate that acetylene releases slightly more energy per unit mass than ethane or ethylene, but experimental values show that ethane releases the most. Remember that average bond energy calculations provide estimates, not exact values.

3.67 Kinetic energy is given by , expressed in SI units: