CHAPTER 9 | Thermochemistry: Energy Changes in Chemical Reactions

9.1. Collect and Organize From the depiction of a diesel engine piston (Figure P9.1), we are to describe how the internal energy of the trapped gases changes when the piston moves up.

Analyze Here, the system we are considering is the gases. As the piston moves up, the gases in the cylinder are compressed.

Solve Upon compression the molecules are squeezed together and the change in volume on the system is negative. Therefore, work is done on the system and w = –PΔV, where ΔV is negative, so w is positive and the internal energy change E = q + w is positive. Here, the internal energy of the gases in the cylinder increases.

Think About It If work is done by the system, in this situation where the gases expand, the internal energy of the gases in the cylinder decreases.

9.2. Collect and Organize From the depiction of a diesel engine piston in which the fuel has been ignited and the cylinder is pushed down (Figure P9.2), we are to describe how the internal energy of the trapped gases changes and assign signs to the quantities E, q, and w.

Analyze Here, the system we are considering is the gases in the cylinder. As the gases in the cylinder expand, the piston moves down. Upon ignition, the system releases heat to the surroundings.

Solve Upon expansion, the change in volume of the system is positive. Therefore, work is done by the system on the surroundings. In this expansion, ΔV is positive, so w is negative because w = –PΔV. Because heat is transferred from the ignited gases to the surroundings, q is negative. The internal energy change, E = q + w, is therefore negative and the internal energy of the gases in the cylinder decreases.

Think About It When the piston moves up to compress the gases in the cylinder, work is done on the system by the surroundings and the surroundings heat up the sample, so w and q are both positive for that situation.

9.3. Collect and Organize All the molecules shown in Figure P9.3 are hydrocarbons with either five or six carbon atoms. On the basis of differences in their structures, we are to predict which has the highest and which has the lowest fuel value.

Analyze For hydrocarbons, as the hydrogen-to-carbon ratio decreases the fuel value also decreases. Therefore, we will distinguish between the hydrocarbon in this group that has the most and the least fuel value on the basis of the hydrogen-to-carbon ratio.

Solve The hydrogen-to-carbon ratios of those hydrocarbons are as follows: (a) C6H14 H:C = 2.33:1 (b) C6H12 H:C = 2:1 (c) C5H12 H:C = 2.4:1 (d) C6H14 H:C = 2.33:1 From those ratios, (c) C5H12 has the highest fuel value and (b) C6H12 has the lowest fuel value.

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Think About It Although structurally different, (a) and (d) have the same H:C ratio and therefore the same fuel value.

9.4. Collect and Organize The diagram in Figure P9.4 shows the reaction between N2 and H2 in a closed vessel that is maintained by the piston at constant pressure while being held at constant temperature. The diagram shows that as the reaction proceeds, the volume of the reaction decreases. We are to determine whether work is done by or on the system, calculate the enthalpy of the reaction by using standard enthalpy data, and predict the direction of heat flow in the reaction.

Analyze The balanced reaction to prepare 1 mol of ammonia from nitrogen and hydrogen is 1 3 /2 N2(g) + /2 H2(g) → NH3(g) The fact that the volume of the reaction vessel is smaller after the reaction will help us determine the sign of work for the reaction (the system). Once we calculate the enthalpy of the reaction in part b, we will be able to determine the direction of heat flow for the reaction.

Solve (a) Because the volume of the vessel is less after the reaction (ΔV is negative), the reaction was compressed, and therefore work was done on the system by the surroundings. (b) The enthalpy of the reaction under standard conditions is ° ° 1 ° 3 ° ΔH rxn = ⎣⎡1 mol (ΔH f )NH3(g)⎦⎤ – ⎣⎡ 2 mol(ΔH f )N2 (g) + 2 mol(ΔH f )H2 (g)⎦⎤ 1 3 = [1 mole × –46.1 kJ/mol]−[ 2 mol × 0 kJ/mol + 2 mol × 0 kJ/mol)] = −46.1 kJ (c) The negative value of the enthalpy for that reaction means that the reaction is exothermic; heat flows out from the reaction to the surroundings.

Think About It The reverse reaction would be endothermic and would have an increase in volume of the reaction vessel.

9.5. Collect and Organize For the reaction depicted in Figure P9.4, we are to assign signs to the values ΔE, q, and w and then determine the percent yield of the reaction if each molecule represents 1 mol.

Analyze From our answer to Problem 9.4, we know that the reaction is exothermic and is compressed in going from reactants to products. From the diagram we have 3 mol of N2 and 9 mol of H2 as reactants that form 4 mol of NH3, with 3 mol of H2 and 1 mol of N2 remaining at the end of the reaction.

Solve Because the reaction volume is less at the end of the reaction, the sign of w is positive, showing that work was done on the system. Because the reaction is exothermic, the sign of q is negative, showing flow of heat from the system to the surroundings. The change in E for the reaction is negative if the heat released by the reaction is greater than the work done on the system by the surroundings by decrease of volume. For the percent yield we consider the system in comparison with the balanced equation

N2(g) + 3 H2(g) → 2 NH3(g)

The 3 mol of N2 present in the reaction mixture would be expected to use 9 mol of hydrogen (which we have exactly) and yield 6 mol of ammonia. Only 4 mol of ammonia is produced, so the percent yield of the reaction is experimental yield 4 mol NH ×100 = 3 ×100 = 67% theoretical yield 6 mol NH3

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Think About It A 100% yield for this reaction would have used up all the nitrogen and the hydrogen initially present in the reaction mixture.

9.6. Collect and Organize ° The enthalpy diagram in Figure P9.6 shows the relative enthalpies of formation (∆Hf, kilojoules per mole) for the elements C, H2, and O2 along with that of C2H2, CO, H2O, and CO2. From the diagram we are to explain why all the elements have the same standard enthalpy of formation, explain why C2H2 is an “endothermic compound,” and predict which reaction has the more negative enthalpy change: acetylene reacting with oxygen to form carbon monoxide, or acetylene reacting with oxygen to form carbon dioxide.

Analyze ° ° All the elements have ∆Hf = 0. The ∆Hf is positive for C2H2 and negative for CO, H2O, and CO2. For reactions, the enthalpy is calculated by subtracting the enthalpy of the reactants from the enthalpy of the products.

Solve ° (a) The elements are all on the same horizontal line (∆Hf = 0) because the enthalpy of formation of an element in its standard state is defined as zero. (b) C2H2 is an endothermic compound because the enthalpy of its formation is positive. Therefore, energy would have to be added to the elements for C2H2 to form. (c) The only differences in the two reactions of acetylene with oxygen, 2 C H (gg ) 3 O ( ) 4 CO( g ) 2 H O( g ) 22+ 2→+ 2 2 C H (g) 5 O (g) 4 CO (g) 2 H O(g) 2 2 + 2 → 2 + 2 are the amounts of oxygen required for the reaction and the product formed (CO or CO2). The enthalpy values for the reactions are calculated by subtracting the sum of the enthalpies of formation of the reactants (multiplied by the number of moles in the balanced equation for each product) from the sum of enthalpies of formation of the products (again multiplied by their molar amounts from the balanced equation).

For the first reaction: ΔH  = ⎡4 mol (ΔH  )CO(g) + 2 mol (ΔH  )H O(g)⎤ – ⎡2 mol (ΔH  )C H (g) + 3 mol (ΔH  )O (g)⎤ rxn ⎣ f f 2 ⎦ ⎣ f 2 2 f 2 ⎦ For the second reaction: ΔH  = ⎡4 mol (ΔH  )CO (g) + 2 mol (ΔH  )H O(g)⎤ – ⎡2 mol (ΔH  )C H (g) + 5 mol (ΔH  )O (g)⎤ rxn ⎣ f 2 f 2 ⎦ ⎣ f 2 2 f 2 ⎦ Because the enthalpy of formation of oxygen is zero, the amount of oxygen used in the second reaction does not contribute to any enthalpy of reaction difference between those two reactions. However, the enthalpy of formation of CO2 is more negative than that of CO. That makes the enthalpy of the second reaction, in which the combustion of acetylene forms carbon dioxide, more negative.

Think About It If you consider the magnitudes of the enthalpies of the two combustion reactions from Figure P9.6 you can estimate the enthalpy of the combustion reaction to form CO as approximately –1400 kJ and the reaction to form CO2 as approximately –2480 kJ.

9.7. Collect and Organize The diagram in Figure P9.7 shows that the volume of the cylinder on the product side is greater than that on the reactant side. We are considering that reaction at constant temperature and pressure, and we are to write a balanced equation for the reaction, calculate the enthalpy of reaction for the formation of 1 mol of CO, and determine the heat flow in the reaction in converting reactants to products.

Analyze From the color scheme for the elements shown on the inside back cover of the textbook, the reactants are CH4 and H2O reacting to form CO and H2. The reactant has four molecules of CH4 and four molecules of H2O. Those react to form 4 molecules of CO and 12 molecules of H2. The enthalpy of the reaction is calculated by subtracting the sum of the enthalpies of formation of the reactants (multiplied by the number of moles in the 4 | Chapter 9

balanced equation for each product) from the sum of enthalpies of formation of the products (again multiplied by their molar amounts from the balanced equation).

Solve (a) The balanced equation is

4 CH4 (g) + 4 H2O(g) → 4 CO(g) +12 H2 or, more simply, CH (g) H O(g) CO(g) 3 H 4 + 2 → + 2 (b) ΔH ! = ⎡1 mol (ΔH ! )CO(g) + 3 mol (ΔH ! )H (g)⎤ – ⎡1 mol (ΔH ! )CH (g) + 1 mol (ΔH ! )H O(g) ⎤ rxn ⎣ f f 2 ⎦ ⎣ f 4 f 2 ⎦ ° ΔH rxn = [1 mol CO × –110.5 kJ/mol + 3 mol H2 × 0.0 kJ/mol]−

[1 mol CH4 × –74.8 kJ/mol + 1 mol H2O × –241.8 kJ/mol +] = 206.1 kJ/mol

(c) Because that reaction is endothermic, heat flows into the reaction mixture to the surroundings.

Think About It The volume of the cylinder increases because the reaction yields more product molecules than the number of reactant molecules and the temperature and the pressure of the system are held constant.

9.8. Collect and Organize With reference to the representations in Figure P9.8 we are to answer questions comparing particulate images with phase changes, compare endothermic and exothermic processes, consider the bond breaking in the dissolution of ammonium nitrate, and chose an image that shows an element of compound in its standard state.

Analyze Representations [A] and [I] both show carbon dioxide (solid and gaseous states, respectively), with image [H] showing the sublimation of solid CO2. Representations [C] and [G] both show nitrogen (gaseous and liquid states, respectively), with image [B] showing the vaporization of liquid N2. Representations [D] and [F] compare the enthalpy change for endothermic and exothermic processes. Representation [E] shows the solid structure of ammonium nitrate, a constituent of instant ice packs.

Solve (a) [C] shows gaseous nitrogen and [G] shows liquid nitrogen. (b) [A] shows solid CO2 and [I] shows gaseous CO2. (c) [D] shows an exothermic process; neither photo [B] or [H] shows an exothermic process. (d) [F] shows an endothermic process; both photo [B] and [H] show an endothermic process as well as the dissolution of ammonium nitrate for a cold pack shown in photo [E]. (e) When NH4NO3 dissolves in water, the ionic bonds between the ammonium cations and the nitrate anions break. (f) Particulate images [C], [E], and [I] show molecules or elements in their standard states (25˚C, 1 atm).

Think About It The reverse processes of CO2 deposition (gas to solid) and N2 condensing (gas to liquid) would be exothermic.

9.9. Collect and Organize Energy and work must be related since from our everyday experience we know that doing work takes energy.

Analyze In this context, energy is defined as the capacity to do work. Work is defined as moving an object against a force over some distance. Energy is also thought to be a fundamental component of the universe. The Big Bang theory postulates that all matter originated from a burst of energy, and Albert Einstein proposed that m = E/c2 (mass equals energy divided by the speed of light squared).

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Solve Energy is needed to do work, and doing work uses energy.

Think About It A system with high energy has the potential to do a lot of work.

9.10. Collect and Organize Both potential energy (PE) and kinetic energy (KE) are forms of energy, and therefore they both have the potential to do work. We are to interpret those forms of energy as they exist in molecules.

Analyze PE is the energy an object has as a result of its position or its structure. KE is the energy of an object resulting from its motion.

Solve PE in molecules arises from the strength of the chemical bonds in the molecules or from the interactions between molecules, whereas KE is energy of motion, which depends on a molecule’s mass (m) and speed (u).

Think About It A molecule moving fast has more kinetic energy than the same molecule moving more slowly. Also, a more massive molecule moving at the same speed has more kinetic energy than a less massive molecule.

9.11. Collect, Organize, and Analyze We are to explain what is meant by a state function.

Solve The value of a state function is independent of the path taken in reaching a particular state; only the initial and final values are important.

Think About It Examples of state functions are internal energy and enthalpy, but not work and heat.

9.12. Collect, Organize, and Analyze To answer whether potential energy and kinetic energy are state functions, we need to recognize that both quantities are forms of energy.

Solve Energy is a state function; therefore, both potential energy and kinetic energy are state functions.

Think About It All forms of energy, including electrical, chemical, and mechanical, must be state functions.

9.13. Collect and Organize We are asked whether two particles attract or repel each other if their potential energy increases when they are moved away from each other.

Analyze The basic definition of potential energy is the energy of position from PE = mgh. But potential energy is also present at the molecular level in the form of stored energy in the chemical bonds or the attraction of one molecule to another.

Solve Because the potential energy increases as those particles are moved farther apart, those particles attract each other.

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Think About It You can think of it another way. As the molecules attract and move closer to each other, their potential energy decreases (they enter into a lower potential energy state).

9.14. Collect and Organize Kinetic energy (KE) is defined as the energy of motion.

Analyze The equation for KE is 1 2 KE = 2 mu

Solve Because the macroscopic ice cube is not moving, u = 0, and therefore the kinetic energy is zero. However, we must look closer. The molecules of water within the ice cube are moving (as long as the ice cube is above absolute zero), so the molecules of H2O do indeed have kinetic energy.

Think About It In chemistry, we often have to look closer and view matter at the molecular or atomic scale.

9.15. Collect and Organize Internal energy is the sum of the potential and kinetic energies of the components of a system.

Analyze To increase kinetic energy, increase the motion of the molecules in the gas. To increase the potential energy, compress the gas to increase its pressure. It then can do work when it expands to its original volume.

Solve We can increase the motion of the gas molecules by raising the temperature. To increase the pressure, we can compress the gas (decreasing the volume).

Think About It Absolute internal energy is difficult to determine for a system, but changes in internal energy are easy to measure.

9.16. Collect and Organize For the popping of popcorn we are to describe the signs of q, w, and ΔE.

Analyze The system we are considering is a kernel of popping corn. We should determine the signs of q, w, and ΔE from that point of view.

Solve The kernel absorbs heat to pop, so q is positive. The kernel expands upon popping, thereby doing work on the surroundings, so w is negative. The popping kernel releases energy upon popping, so the sign of ΔE is negative.

Think About It The unpopped kernel has a higher potential energy than that of the popped kernel.

9.17. Collect, Organize, and Analyze We consider how pressure–volume (P–V) work can have energy units (joules).

Solve P–V work (which would, at first glance, have units such as liter ⋅ atmospheres) has energy units because work is done by expending energy. P–V work is the energy expended or released in changing the volume against an outside pressure, so it may have units of energy. Thermochemistry | 7

Think About It To convert liter ⋅ atmospheres to joules, we multiply by 101.325. If, however, pressure is measured using pascals, the SI unit for pressure, the units for P–V work are pascal ⋅ L, This can easily be converted to joules since Pa = kg ⋅ m–2 ⋅ s–2 and J = kg ⋅ m2 ⋅ s–2. We simply would need to multiply by the volume expressed in m3.

9.18. Collect and Organize In the equation ΔE = q –PΔV we are to explain why a negative sign is in front of the PΔV term.

Analyze The PΔV term describes the work done on or by the system on the surroundings.

Solve When work is done on a system, the change in volume is negative because the final volume is less than the initial volume. That corresponds to positive work done on the system, and so the negative sign gives the correct sign for w from the point of view of the system.

Think About It Remember that the sign of qsystem is also determined from the point of view of the system. If the system releases heat to the surroundings, the sign is negative; if the system absorbs heat from the surroundings, the sign is positive.

9.19. Collect and Organize For the processes listed, we are asked to identify which are exothermic and which are endothermic.

Analyze An exothermic process transfers energy from the system to the surroundings. Thus, something feels warm or hot in an exothermic process. An endothermic process transfers energy from the surroundings to the system. Thus, something feels cool or cold in an endothermic process.

Solve (a) When molten aluminum solidifies, energy in the form of heat is released to the surroundings to cool the metal, so that process is exothermic. (b) When rubbing alcohol evaporates from the skin, energy in the form of heat is absorbed by the alcohol from the skin (the surroundings) to evaporate the alcohol, so that process is endothermic. (c) When fog forms, energy in the form of heat is released by the water vapor in the air to condense, so that process is exothermic.

Think About It The opposite processes are the reverse: melting aluminum is endothermic, condensing alcohol is exothermic, and evaporating fog is endothermic.

9.20. Collect and Organize For the processes listed, we are asked to identify which are exothermic and which are endothermic. All the processes describe phase changes for water.

Analyze To determine whether the process is exothermic or endothermic, we need to examine the energy transfer for the process. If energy is transferred from the system to the surroundings, it is exothermic. If the energy is transferred from the surroundings to the system, it is endothermic.

Solve (a) When water freezes to form ice cubes, energy is removed from the water; energy is transferred from the system (water) to the surroundings (the freezer). That process is exothermic. (b) When water sublimes from ice cubes, energy must be transferred into the water to cause it to change from the solid phase to the gaseous phase. That process is endothermic. 8 | Chapter 9

(c) Dew forms on the grass because atmospheric water vapor cools down (condensation). Energy then must be transferred from the water vapor (the system) to the surroundings. That process is exothermic.

Think About It The reverse of a phase change that is exothermic is endothermic. For example, the freezing of water is exothermic, but the melting of ice is endothermic (we need energy from the surroundings to melt ice).

9.21. Collect and Organize Internal energy is defined as ∆E = q + w where q = energy transferred and w = work done on a system (–P∆V). Commonly, the energy transfer is caused by a temperature difference, so q is called “heat.”

Analyze When a process releases energy, the internal energy decreases and q is negative. The reverse is true (∆E increases and q is positive) if the surroundings transfer energy to the system. If the volume of a system increases (∆V is positive), then work is negative and the internal energy of the system decreases. The reverse is true (∆E increases and w is positive) when the volume of the system decreases (for example, a gas is compressed).

Solve When a liquid vaporizes at its boiling point, energy is absorbed from the surroundings. Thus, q is positive. The volume increases (∆V > 0), so work is done by the system on the surroundings and the sign of w is negative. More energy is transferred to the liquid than work done, so the sign of q + w is positive; therefore, ∆E > 0.

Think About It Remember to focus on work and energy transfer from the system’s point of view. Doing so will help you be clear about the sign conventions as you learn more about thermochemistry.

9.22. Collect and Organize Internal energy is defined as ∆E = q + w where q = energy transferred and w = work done on a system (–P∆V).

Analyze When a gas expands, ∆V is positive and therefore work done is negative. Because no energy flow occurs for the system described, q = 0.

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Solve The equation for internal energy for an expanding gas with no energy flow is ∆E = q + w = 0 + (–P∆V) Because ∆V is positive, ∆E is negative. Therefore, the internal energy of a gas decreases when it expands.

Think About It In that situation all the change in internal energy is due to work being done by the system (the gas) on the surroundings.

9.23. Collect and Organize The work being done is due to an expansion of the gas from 250.0 mL to 750.0 mL.

Analyze Work is expressed as –P∆V. Here P is constant at 1.00 atm and the volume change is 500.0 mL, or 0.5000 L. We are to express work in both liter ⋅ atmospheres and joules. We can convert from one unit to another by using 101.325 J/L ⋅ atm.

Solve w = –P∆V = (1.00 atm)(0.5000 L) = –0.500 L ⋅ atm In joules, that work is 101.325 J –0.500 L ⋅atm × = –50.7 J L ⋅atm

Think About It Because work was done by the system on the surroundings, the sign of work is negative.

9.24. Collect and Organize We are asked to calculate the final volume of the system given the work done by the system, the pressure, and the starting volume of the system.

Analyze Work is defined as –P∆V, with ∆V = Vf – Vi. For this problem, Vi = 68 mL, or 0.068 L, and P = 1.01 atm. To perform the calculation with consistent units, we use the conversion 101.325 J/L ⋅ atm to convert the work in joules to units of liter ⋅ atmospheres. Because work is done by the system, w = –150.0 J.

Solve 1 L ⋅atm –150.0 J × = –1.01 atm × (V – 0.068 L) 101.325 J f

1.53 L = Vf

Think About It The final volume should be greater than the initial volume because work was done by the system on the surroundings.

9.25. Collect and Organize For each part, we are to calculate internal energy (∆E) from energy and work values.

Analyze The formula for internal energy from energy transferred as heat and work is ∆E = q + w We need only add the values of q and w given.

Solve (a) ∆E = 100 J + (–50 J) = 50 J 10 | Chapter 9

(b) ∆E = 6.2 × 103 J + 0.7 L ⋅ atm = 6200 J + (0.7 L ⋅ atm × 101.325 J/ L ⋅ atm) = 6300 J, or 6.3 kJ (c) ∆E = –615 kJ + (–3.25 kWh) = –615 kJ + (–3.25 kWh × 3600 kJ/kWh) = –12,300 kJ

Think About It When adding q and w, be sure to add values with consistent units.

9.26. Collect and Organize The change in internal energy for a system is ∆E = q + w

Analyze The system absorbs energy from its surroundings, so q is positive. Because the system does work on the surroundings, w is negative.

Solve ∆E = q + w = 726 kJ + (–526 kJ) = 200 kJ

Think About It Be careful to always define the signs for energy and work from the point of view of the system.

9.27. Collect and Organize The change in internal energy for a system is ∆E = q + w

Analyze The system releases energy to its surroundings, so q is negative. Because the system does work on the surroundings, w is negative.

Solve ∆E = q + w = –210.0 kJ + (–65.5 kJ) = –275.5 kJ

Think About It Be careful to always define the signs for q and w from the point of view of the system.

9.28. Collect and Organize The change in internal energy for a system is ∆E = q + w

Analyze The system produces energy, so q is negative. Because no work is done, w = 0.

Solve ∆E = q + w = –90.7 kJ + 0 kJ = –90.7 kJ

Think About It When no work is done either by the system or on the system, the change in internal energy is equal to the change in heat energy transferred because of a temperature difference, q.

9.29. Collect and Organize Work will be done by a system on the surroundings when the volume of the system shown in Figure P9.29 increases. For each reaction we are to determine for which one work is being done by the system on the surroundings and then determine the sign of w for that reaction.

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Analyze The volume of gas is proportional to the number of moles of gas (n) at constant temperature and pressure. Because both temperature and pressure are constant for each reaction, if n increases in going from reactants to products, the volume of the system increases and work is done by the system on the surroundings.

Solve (a) In that reaction, 3 mol of gaseous reactants forms 3 mol of gaseous products. The ∆n for the reaction is 0, so the reaction does not do work on the surroundings. (b) In that reaction, 6 mol of gaseous reactants forms 7 mol of gaseous products. The ∆n for the reaction is +1, so the reaction does work on the surroundings. (c) In that reaction, 3 mol of gaseous reactants forms 2 mol of gaseous products. The ∆n for the reaction is –1, so the reaction does not do work on the surroundings.

Reaction (b) does work on the surroundings. When work is done on the surroundings, the change in volume of the system is positive. Because w = –PΔV, the sign of w is negative for ΔV that is positive.

Think About It Reaction c has +w (from –P∆V, where ∆V = Vf – Vi and Vi > Vf, so ∆V is negative). Therefore, the surroundings did work on the system. The system was compressed.

9.30. Collect and Organize For each reaction listed we are to determine the direction the piston in Figure 9.29 will move. If the volume of the system increases, that is, if work is done by the system on the surroundings, the piston will move up. The piston will move down if the volume of the system decreases.

Analyze The volume of gas is proportional to the number of moles of gas (n) at constant temperature and pressure. Because both temperature and pressure are constant for each reaction, if n increases in going from reactants to products, the volume of the system increases and work is done by the system on the surroundings.

Solve (a) In that reaction, 4 mol of gaseous reactants forms 2 mol of gaseous products. The ∆n for the reaction is –2. The volume decreases and the piston moves down into the cylinder and the sign of w is positive.

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(b) In that reaction, 1 mol of a gaseous reactant forms 1 mol of a gaseous product. The ∆n for the reaction is 0. The volume stays the same and the piston does not move and the value of w is zero.

(c) In that reaction, 4 mol of gaseous reactants forms 5 mol of gaseous products. The ∆n for the reaction is +2. The volume increases and the piston moves up the cylinder and the sign of w is negative.

Think About It Only gaseous reactants and products affect the volume in the reaction.

9.31. Collect and Organize We are asked to define a change in enthalpy.

Analyze We can refer to the mathematical expression for the change in enthalpy (∆H) to answer this question. ∆H = ∆E + P∆V

Solve By the equation above, a change in enthalpy is the sum of the change of internal energy and the product of the system’s pressure and change in volume and is the heat released or absorbed by a system at constant pressure.

Think About It Because ∆E = q + w, internal energy changes may involve changes in energy or work or both.

9.32. Collect and Organize Enthalpy and internal energy are related to each other through the definition of the change in enthalpy, as the energy that is transferred at constant pressure with only P–V work being done.

Analyze The mathematical definitions of enthalpy change (∆H) and internal energy change (∆E) are ∆H = qP ∆H = ∆E + P∆V

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Solve Consequently, the difference between the change in energy and the change in enthalpy is ∆E – ∆H = –P∆V

Think About It Enthalpy is energy transferred as heat at constant pressure with only P–V work being done.

9.33. Collect and Organize We are asked why we assign a negative sign to ∆H for an exothermic process.

Analyze In an exothermic process, the system transfers energy to the surroundings.

Solve If the system transfers energy to the surroundings, its energy will be less after the process than at the start of the process, and q is negative. If the pressure is constant and only P–V work is done, then this q is ∆H, so ∆H must also be negative.

Think About It The signs of thermodynamic quantities are always assigned from the point of view of the system. That approach can be confusing because we are often observing the process from the point of view of the surroundings.

9.34. Collect and Organize Enthalpy for any process can be defined for either a forward or a reverse reaction.

Analyze Endothermic reactions have positive values of ΔH, whereas exothermic reactions have negative values of ∆H.

Solve When an exothermic reaction is reversed, the reaction becomes endothermic. The sign of ∆H changes from negative to positive. Because energy (here in the form of heat) is conserved, the magnitude of the enthalpy change for the reverse reaction is the same as the enthalpy change for the forward reaction.

Think About It A reaction that requires an input of energy to form products produces energy when the reaction is reversed.

9.35. Collect and Organize A drainpipe gets hot when Drano is added. We are asked what the sign for ∆H is for that process.

Analyze Enthalpy is related to energy transferred as heat (q) by the equation ∆H = qP

Solve Because energy is released in the reaction between the Drano and the clog in the pipe, q is released from the system (the Drano) to the surroundings (the drainpipe), so q is negative and therefore ∆H is also negative.

Think About It All exothermic reactions have a negative ∆H.

9.36. Collect and Organize The cold pack gets cold when ammonium nitrate mixes with the water. We are to determine the sign of ∆H.

Analyze Enthalpy is related to energy (q) by the equation ∆H = qP 14 | Chapter 9

Solve Because energy is transferred into the system (the cold pack) from the surroundings, q is positive for the system, so ∆H is positive.

Think About It All endothermic reactions have a positive ∆H.

9.37. Collect and Organize If O2 is the stable form of oxygen under standard conditions, it must have lower energy than other forms. The reaction describes breaking the oxygen–oxygen bond in O2. We are asked what the sign of ∆H is.

Analyze If a species is at low energy, then it must require an input of energy to take it out of its most stable form. Breaking the oxygen–oxygen bond, therefore, must require an input of energy.

Solve The process of breaking the oxygen–oxygen bond is endothermic, and the sign of ∆H will be positive.

Think About It Breaking of chemical bonds is always an endothermic process.

9.38. Collect and Organize The formation of plaster of Paris from gypsum requires heating to 150˚C to drive off the water.

Analyze Because energy input is required for that reaction, q is positive.

Solve Because ∆H = qP, when qP is positive, ∆H is also positive.

Think About It Reactions that require energy input are endothermic reactions.

9.39. Collect and Organize Compression of H2 gas will give solid H2. To predict the sign of the enthalpy change of this transformation, we have to consider the enthalpy of the phase change from gas to solid.

Analyze To solidify substances, we must cool them to reduce their molecular motion.

Solve To solidify hydrogen gas, we would remove energy from the gas, so the sign of q from the point of view of the system would be negative. Because ∆H = qP, the sign of ∆H is also negative.

Think About It The reverse reaction, in which H2(s) sublimes into H2(g), is endothermic.

9.40. Collect and Organize The reaction of vinegar with baking soda inflates a balloon through the evolution of CO2 gas.

Analyze The volume of the balloon increases as CO2 is produced in the reaction. The work done by the system is defined by the equation w = –P∆V = –P(Vf – Vi) Thermochemistry | 15

Solve For the expanding balloon, Vf > Vi, so ∆V is positive. Therefore, –P∆V is a negative quantity and work is negative. Therefore, work was done by the system on the surroundings.

Think About It Compressing a gas gives a positive value for work, which means that the surroundings have done work on the system to increase its internal energy.

9.41. Collect and Organize We look at the definitions and units of heat capacity and specific heat to differentiate those two terms.

Analyze Heat capacity is the amount of energy needed to raise the temperature of an object by 1˚C. Specific heat is the amount of energy needed to raise the temperature of 1 g of a substance by 1˚C.

Solve The difference in the terms lies in the specificity. Heat capacity does not take into account how much of a substance is present; it is defined for a given object. Specific heat is specified for 1 g of the substance.

Think About It Specific heat for a substance is characteristic of that substance.

9.42. Collect and Organize We compare the heat capacity and the molar heat capacity of 1 L of water versus 1 m3 of water.

Analyze Heat capacity is the amount of energy needed to raise the temperature of a substance by 1˚C. Molar heat capacity is the amount of energy needed to raise the temperature of 1 mol of a substance by 1˚C. Therefore, heat capacity is an extrinsic property, whereas molar heat capacity is an intrinsic property.

Solve Because the heat capacity depends on the size or mass of a material, 1 m3 of water will have more heat capacity than the smaller amount of water, 1 L. However, since the substance in both of those volumes is water, they have the same molar heat capacity (heat capacity on a per-mole basis).

Think About It Specific heat values (also an intrinsic property, and is on a per-mass basis) or molar heat capacity (on a per- mole basis) are better means than heat capacity to compare different materials because the values for substances do not depend on their size. 16 | Chapter 9

9.43. Collect and Organize We consider why the heat of vaporization of water is so much greater than its heat of fusion.

Analyze The enthalpy of fusion is the energy required to melt a solid substance. The enthalpy of vaporization is the energy required to vaporize the substance from a liquid to a gaseous state.

Solve Melting and vaporizing are different processes, so we would not expect them to be the same. By melting ice we are giving the water molecules enough energy to temporarily break the hydrogen bonds between individual water molecules that are then re-formed with a neighboring water molecule. In vaporization, however, the strong hydrogen bonds between the water molecules are completely broken. Thus, vaporizing water requires much more energy.

Think About It Usually the enthalpy of vaporization is more endothermic than the enthalpy of fusion because in the phase change from liquid to gas, individual molecules must be completely separated from one another.

9.44. Collect and Organize In comparing the heating curves for chloroform, water, and ethanol, we are asked to rank the liquids in order of their molar enthalpies of vaporization.

Analyze We are given the boiling points of each liquid (chloroform, 142˚C; water, 100˚C; ethanol, 78˚C). The segment on the heating curves corresponding to vaporization is BC and those segments indicate the liquids’ boiling point and the length of the BC segment is directly related to the molar enthalpy of the liquid.

Solve The yellow heating curve has a boiling point of 78˚C and is assigned to ethanol. The blue heating curve has a boiling point of 100˚C and is assigned to water. The red heating curve has a boiling point of 142˚C and is assigned to chloroform. The longer the line segment, the greater the enthalpy of vaporization; water has the longest line segment, and chloroform has the shortest line segment. Therefore, the liquids in order of increasing molar enthalpy of vaporization are: chloroform < ethanol < water.

Think About It Recall that the temperature of a substance does not change when undergoing a phase transition.

9.45. Collect and Organize Referring to Figure P9.44 we are to explain why the line segments AB have different slopes.

Thermochemistry | 17

Analyze Line segment AB in all three of the heating curves shows the temperature change of the liquid substance upon adding energy. In other words, it shows the heating of the liquids before the boiling point shown by line segment BC.

Solve The slope of the lines AB correlate to the amount of heat needed to raise the temperature of the substance. They have different slopes because the liquids have different molar heat capacities.

Think About It A steeper slope indicates that the substance has a lower heat capacity. In order of increasing molar heat capacity of the liquids, the order for these substances is blue (water), yellow (ethanol), and red (chloroform). That ranking is in agreement with the accepted molar heat capacities: water, 75.28 J/(mol⋅K); ethanol, 112.4 J/(mol⋅K); chloroform, 114.25 J/(mol⋅K).

9.46. Collect and Organize For two metals of equal mass that are heated with the same quantity of heat, but with one having a larger heat capacity than the other, we are to compare their final temperatures.

Analyze The heat capacity of a substance is the heat required to increase the temperature of 1 g of the substance 1˚C.

Solve No. The higher the heat capacity, the more heat would be required to raise the temperature of the substance. Therefore, the metal with the higher heat capacity will have a lower temperature for the same quantity of heat added than the metal with the lower heat capacity.

Think About It The difference in temperature will be related to the difference in the heat capacities of the metals. The change in temperature of the metal with a heat capacity double that of another metal will have only half of the temperature increase of the other metal.

9.47. Collect and Organize To compare the advantage of water-cooled engines over air-cooled engines, we have to compare the heat capacity of water with that of air.

Analyze The specific heat capacity of water (4.18 J/g⋅˚C) is higher than that of air (1.01 J/g⋅˚C) at typical room conditions.

Solve Water’s high heat capacity compared with that of air means that water carries away more energy from the engine for every Celsius degree rise in temperature, so water is a good choice to cool automobile engines.

Think About It On a volume basis, water has an even higher heat capacity than air does.

9.48. Collect and Organize By comparing the thermal conductivities and molar heat capacities for water and liquid sodium, we are to state why liquid sodium is a better coolant than water for nuclear reactors.

Analyze Sodium has a much higher thermal conductivity (a measure of how fast energy is transferred through the material) but a lower heat capacity than water.

18 | Chapter 9

Solve The higher thermal conductivity of sodium means that thermal energy can be removed from the reactor core of a nuclear power plant faster than if water were used. That makes up for the lower heat capacity of sodium, which indicates that sodium does not absorb as much energy as water would with the same increase in temperature.

Think About It Sodium, however, needs to be heated to ~100˚C to be liquid (that energy can be supplied by the reactor itself) and is a highly reactive metal, so it presents some safety issues.

9.49. Collect and Organize The amount of energy needed to raise the temperature of a substance depends on the substance’s heat capacity and the change in temperature of the substance.

Analyze The energy required to raise the temperature of a substance is related to the molar heat capacity by the equation

q = ncP∆T

where q = energy, n = moles of the substance, cP = molar heat capacity of the substance, and ∆T = difference in temperature, Tf – Ti.

Solve ⎛ 1 mol ⎞ 75.3 J ⎛ ⎞   q required = 100.0 g × × ⎜  ⎟ × 100.0 C – 30.0 C = 29,300 J, or 29.3 kJ ⎝⎜ 18.02 g⎠⎟ ⎝ mol⋅ C⎠ ( )

Think About It If the water were cooled from 100˚C to 30˚C, the sign of ∆T would be negative, and therefore q would be negative, showing that the system was cooled.

9.50. Collect and Organize We need to determine the final temperature of water (1.33 kg starting at 30˚C) that has absorbed a certain amount of energy (290.0 kJ, or 2.900 × 105 J) and determine whether that water will be boiling at an altitude where the boiling point is 93˚C.

Analyze The energy required to raise the temperature of a substance is related to the specific heat capacity by the equation q = ncP∆T

where q = energy, n = moles of substance, cP = molar heat capacity of the substance, and ∆T = difference in temperature, Tf – Ti.

Solve 4.18 J 2.900 ×105 J = 1330 g × × (T – 30 C) g ⋅ C f   52.2 C = Tf – 30 C 82 C = T f If the boiling point of water on the mountain is 93˚C, then the 290 kJ of energy supplied by the climber’s stove is not sufficient to boil the water because that amount of heat energy will raise the temperature of the water to only 82˚C.

Think About It Alternatively, we could solve that by asking how much energy is required to boil the water and then compare that energy with the energy supplied by the camper’s stove. Thermochemistry | 19

4.18 J q = 1330 g × × (93C – 30 C) = 3.5 ×10 J, or 350 kJ g C ⋅ Because that is more than the energy supplied by the stove, the water does not boil.

9.51. Collect and Organize A heating curve plots temperature as a function of energy added to a substance.

Analyze Methanol at –100˚C is a solid. As energy is added, the solid methanol increases in temperature until it reaches its melting point, at which point the added energy does not change the temperature of the methanol until all the solid methanol has melted. Only when methanol is entirely liquid will added energy increase the temperature of the liquid until the boiling point of methanol is reached. During boiling, the temperature of the methanol does not change. Once the methanol is converted to gaseous form, added energy increases the temperature of the gaseous methanol. Relevant equations for finding the energy required for each step are as follows: q = ncP∆T for energy added to the solid, liquid, and gas phases, where n = 1 mol, cP = heat capacity for that phase, and ∆T = temperature change for that phase q = n∆Hfus energy for melting q = n∆Hvap energy for vaporization

Solve

Step in Heating Curve Ti, ˚C Tf, ˚C q Total q Heating methanol ice –100 –94 q = 1 mol × 48.7 J/mol ⋅ ˚C × 6˚C = 0.292 kJ 0.29 kJ Melting methanol ice –94 –94 qfus = 1 mol × 3.18 kJ/mol = 3.18 kJ 3.47 kJ Warming liquid methanol –94 65 q = 1 mol × 81.1 J/mol ⋅ ˚C × 159˚C = 12.9 kJ 16.4 kJ Boiling liquid methanol 65 65 qvap = 1 mol × 35.3 kJ/mol = 35.3 kJ 51.7 kJ Heating methanol gas 65 100 q = 1 mol × 43.9 J/mol ⋅ ˚C × 35˚C = 1.54 kJ 53.2 kJ

150

100 (˚C)

T 50

0 0 10 20 30 40 50 60 -50 Heat, q (kJ) Temperature, Temperature, -100

-150

Think About It Methanol has a wide temperature range as a liquid, making it useful as a solvent.

9.52. Collect and Organize A heating curve plots temperature as a function of energy added to a substance.

Analyze Acetic acid at 16˚C is a solid just about to melt. As energy is added, the solid acetic acid melts and its temperature does not change until all of the solid is melted. Only when acetic acid is entirely liquid does added energy increase the temperature of the liquid until the boiling point of acetic acid is reached. During boiling, the temperature of the acetic acid does not change. Once all the acetic acid is converted to gaseous form, added energy increases the temperature of the gaseous acetic acid. Relevant equations for finding the energy required for each step are as follows: q = ncP∆T for energy added to the solid, liquid, and gas phases, where n = 1 mol, cP = heat capacity for that phase, and ∆T = temperature change for that phase 20 | Chapter 9

q = n∆Hfus energy for melting q = n∆Hvap energy for vaporization

Solve

Step in Heating Curve T , i T , ˚C q Total q ˚C f

Melting acetic acid ice 16 16 qfus = 1.5 mol × 11.7 kJ/mol = 18 kJ 18 kJ Warming liquid acetic acid 16 118 q = 1.5 mol × 0.1231 kJ/mol ⋅ ˚C × 102˚C = 19 kJ 37 kJ Boiling liquid acetic acid 118 118 qvap = 1.5 mol × 23.7 kJ/mol = 36 kJ 73 kJ Heating acetic acid gas 118 130.0 q = 1.5 mol × 0.0634 kJ/mol ⋅ ˚C × 12˚C = 1.1 kJ 74 kJ

140 120 100 , (˚C) T 80 60 40 Temperature, Temperature, 20 0 0 10 20 30 40 50 60 70 80 Heat, q (kJ)

Think About It The temperature does not change throughout a phase change. The system uses all the added energy to melt the solid or boil the liquid.

9.53. Collect and Organize Sweating helps to cool athletes. During a workout, an athlete generates 233 kJ of energy. We are to calculate how much water the athlete would lose if all the energy is used to evaporate water.

Analyze The energy generated by the athlete must vaporize the water, so the pertinent equation is q = n∆Hvap where ∆Hvap for water is 40.7 kJ/mol.

Solve The amount of water in moles that 233 kJ of energy would vaporize is 233 kJ = n × 40.7 kJ/mol

n = 5.72 mol H2O Converting that to mass, 18.02 g H O 5.72 mol × 2 = 103 g H O 1 mol 2

Think About It Since the density of water is 1.00 g/mL, the athlete would use 886 mL of water to dissipate the energy. To rehydrate, the athlete should drink about 1 L of water.

9.54. Collect and Organize The final temperature of the four metals of equal masses when the same amount of heat energy is added to them is determined from their relative specific heats.

Thermochemistry | 21

Analyze The specific heat of these metals as listed in Table 9.3 in the textbook are as follows: titanium, 0.523 J/(g ⋅ ˚C); iron, 0.45 J/(g ⋅ ˚C); silver, 0.233 J/(g ⋅ ˚C); and lead, 0.129 J/(g ⋅ ˚C).

Solve The lower the specific heat, the less energy needed to raise the temperature of that metal to the same temperature. If the same amount of energy is added to all the metal, as here, then the one with the highest final temperature will be the one with the lowest specific heat capacity because its temperature will rise higher when the same amount of energy is added. Therefore, lead will have the highest final temperature.

Think About It Likewise, the metal with the lowest final temperature will be the one with the highest specific heat capacity: titanium.

9.55. Collect and Organize When the water is converted into steam, the skillet must lose energy to heat and boil the water away. We are asked to calculate the change in the temperature of the pan with a mass of 1.20 kg when all 10.0 mL water initially at 25˚C is converted into steam at 100.0˚C.

Analyze The equation describing the energy exchange between the skillet and the water is

qwater gained = –qskillet lost where q = nc ΔT + nΔH water gained P vap qskillet lost = ncP ΔT Solve

qwater gained = –qskillet lost ⎡ 1 mol 75.3 J ⎤ ⎡ 1 mol 40,670 J ⎤ 10.0 g (100.0 !C – 25.0 !C) 10.0 g ⎢ × × ! × ⎥ + ⎢ × × ⎥ ⎣ 18.02 g mol⋅ C ⎦ ⎣ 18.02 g mol ⎦ ⎛ 1 mol Fe⎞ 25.19 J = –⎜1200 g × ⎟ × ! × ΔT ⎝ 55.845 ⎠ mol⋅ C ΔT = –47.5 !C

Think About It Because the iron skillet has a moderately high molar heat capacity and contains much iron (in terms of moles), boiling the water into steam does not completely cool the skillet to room temperature.

9.56. Collect and Organize When the hot 20.0 g pieces of iron and gold are dropped into water, the metals cool down and the water heats up until they are both at thermal equilibrium; that is, they will have the same final temperature.

Analyze The energy lost by the two metals equals the energy gained by the water. The energy equations necessary are qmetal = ncP∆T since we are given the molar heat capacities of the metals, and qwater = ncP∆T

given that 1.00 L = 1000 g and cP = 75.3 J/mol ⋅ ˚C and the molar mass of water is 18.02 g/mol.

Solve –qlost by metals = qgained by water –qlost by iron + –qlost by gold = qgained by water 22 | Chapter 9

We are given molar heat capacities for the metals and told that the specific heat capacity of water is 75.3 J/mol ⋅ ˚C: –(n c T n c T ) nc T Fe P,Fe Δ Fe + Au P,Au Δ Au = P Δ H2O

– ⎡nFecP,Fe Tf – Ti + nAucP,Au Tf – Ti ⎤ = ncP Tf – Ti ⎣ ( )Fe ( )Au ⎦ ( )H2O The metals (initially at 100.0˚C) reach the same temperature along with the water (initially at 20.0˚C), so that equation becomes ⎡   ⎤  – nFecP,Fe(Tf –100.0 C) + nAucP,Au(Tf –100.0 C) = ncP(Tf – 20.0 C) ⎣ Fe Au ⎦ H2O Rearranging that equation and solving, – n c + n c T –100.0 C = nc T – 20.0 C ( Fe P,Fe Au P,Au )( f ) P,H2O( f ) H2O T –100.0 C nc ( f ) P,H2O  = (Tf – 20.0 C) –(nFecP,Fe + nAucP,Au ) 1 mol 75.3 J  1000 g × ×  (Tf –100.0 C) 18.02 g mol ⋅ C = T – 20.0 C ⎡⎛ 1 mol Fe 25.19 J ⎞ ⎛ 1 mol Au 25.41 J ⎞ ⎤ ( f ) – 20.0 g Fe 20.0 g Au ⎢⎜ × ×  ⎟ + ⎜ × ×  ⎟ ⎥ ⎣⎝ 55.845 g mol ⋅ C⎠ ⎝ 196.967 g mol ⋅ C⎠ ⎦  3  (Tf –100.0 C) 4.18 ×10 J/ C  =  = –360.3 (Tf – 20.0 C) –11.6 J/ C   Tf –100.0 C = –360.3(Tf – 20.0 C)  Tf –100.0 C = –360.3Tf + 7206

361.3Tf = 7306 T = 20.2 C f

Think About It That small increase in the temperature of the water goes along with everyday experience. The high molar heat capacity of water, coupled with the large quantity of water available to cool the metals, means that the metals cool to nearly the initial temperature of the water.

9.57. Collect and Organize To know why the calorimeter constant (that is, the heat capacity of the calorimeter) is important, we need to define the system and surroundings for the calorimetry experiment.

Analyze The system in a calorimetry experiment is defined as the substance for which, for example, the heat capacity is being measured. The calorimeter is everything but the system (that is, the calorimeter is the surroundings).

Solve Because energy is transferred between the system and the surroundings, the heat capacity of the calorimeter (the surroundings) is important because we need to know how much energy (generated or absorbed by the system) is required to change the temperature of the surroundings (the calorimeter) to calculate the heat capacity or final temperature of the system in an experiment.

Think About It Calorimeter constants vary from calorimeter to calorimeter.

9.58. Collect and Organize An endothermic reaction has energy flow from the surroundings (the calorimeter) to the system.

Thermochemistry | 23

Analyze Endothermic reactions show a temperature drop when the temperature of the calorimeter is monitored. Energy lost by the surroundings (the calorimeter) is equal to the energy gained by the system.

Solve Yes, an endothermic reaction can be used to measure the calorimeter constant as long as the signs are consistent with energy flow.

Think About It The sign for energy for the calorimeter here is negative because energy is lost from the calorimeter to the system.

9.59. Collect, Organize, and Analyze The calorimeter constant is the heat capacity of the surroundings. In replacing water in a calorimeter with another liquid, we are changing the surroundings.

Solve The heat capacity of the new liquid is different from that of water. The liquid is part of the calorimeter and therefore part of the surroundings. Yes, the calorimeter constant must be redetermined.

Think About It If the system is expected to transfer a lot of energy to the calorimeter, then using a liquid with a higher heat capacity than water might be necessary.

9.60. Collect and Organize A small amount of material releases a small amount of energy upon combustion.

Analyze To detect a small amount of energy, we would want to use a calorimeter that readily registers a change in temperature.

Solve A calorimeter with a small heat capacity would be best since it takes less energy to change the temperature than with a large-heat-capacity calorimeter.

Think About It If, however, we combusted a large mass of material, a calorimeter with a large heat capacity is the best choice because the calorimeter would be able to absorb the large amount of energy given off in the combustion reaction with a reasonable temperature rise.

9.61. Collect and Organize Benzoic acid is often used to determine calorimeter constants. As mentioned in the text, when 1 g of it combusts, 26.38 kJ is released to the surroundings.

Analyze The calorimeter constant is defined as q C = calorimeter ΔT In the combustion of benzoic acid for this calorimeter, we use 5.000 g of benzoic acid and get a temperature change of 16.397˚C.

Solve 26.38 kJ × 5.000 g g benzoic acid kJ C = = 8.044 calorimeter 16.397! C ! C 24 | Chapter 9

Think About It Be sure to account for how many grams of benzoic acid is used in measuring the calorimeter constant.

9.62. Collect and Organize Benzoic acid is often used to determine calorimeter constants. When 1 g of it combusts, 26.38 kJ is released to the surroundings.

Analyze The calorimeter constant is defined as q C = calorimeter ΔT In the combustion of benzoic acid for that calorimeter, we use 4.663 g of benzoic acid and get a temperature change of 7.149˚C.

Solve 26.38 kJ × 4.663 g g benzoic acid kJ C = =17.21 calorimeter 7.149 ! C ! C

Think About It Be sure to account for how many grams of benzoic acid is used in measuring the calorimeter constant.

9.63. Collect and Organize In a bomb calorimeter qsystem = ∆Ecomb, but since the P–V work is usually small, ∆Ecomb ≈ ∆Hcomb. So we may assume that ∆Hcomb = –qcalorimeter since qrxn = –qcalorimeter.

Analyze We can find ∆Hcomb through

ΔH comb = –qcalorimeter = –Ccalorimeter ΔT

The ∆Hcomb we find is for the combustion of 1.200 g of cinnamaldehyde. To find ∆Hcomb in terms of kilojoules per mole, we need to divide the calculated ∆Hcomb by the moles of cinnamaldehyde (C9H8O).

Solve   ΔH comb = –3.640 kJ/ C ×12.79 C = –46.56 kJ –46.56 kJ kJ molar ΔH = = –5129 comb ⎛ 1 mol ⎞ mol 1.200 g × ⎝⎜ 132.2 g⎠⎟

Think About It Expressing the enthalpies of reactions in terms of molar enthalpies allows us to compare reactions on a per- mole basis.

9.64. Collect and Organize In a bomb calorimeter qsystem = ∆Ecomb, but since the P–V work is usually small, ∆Ecomb ≈ ∆Hcomb. So we may assume that ∆Hcomb = –qcalorimeter since qrxn = –qcalorimeter.

Analyze We can find ∆Hcomb through

ΔH comb = –qcalorimeter = –Ccalorimeter ΔT

The ∆Hcomb we find is for the combustion of 1.608 g of cymene. To find ∆Hcomb in terms of kilojoules per mole, we need to divide the calculated ∆Hcomb by the moles of cymene (C10H14).

Thermochemistry | 25

Solve ! ! ΔH comb = −3.640 kJ/ C ×19.35 C = –70.43 kJ –70.43 kJ kJ molar ΔH = = –5878 comb ⎛ 1 mol ⎞ mol 1.608 g × ⎝⎜ 134.2 g⎠⎟

Think About It Expressing the enthalpies of reactions in terms of molar enthalpies allows us to compare the enthalpy for reactions on a per-mole basis.

9.65. Collect and Organize We are asked to work backward from the molar enthalpy of combustion of dimethyl phthalate to the final temperature of the calorimeter.

Analyze First, we have to calculate ∆Hcomb from the molar heat of combustion and the grams (which we convert into moles) of dimethyl phthalate (C10H10O4). We can then find Tf by rearranging the equation for ∆Hcomb.

ΔH comb = –Ccalorimeter ΔT

ΔH comb ΔT = = Tf – Ti Ccal

Solve

The ΔH comb for 1.00 g of dimethyl phthalate is 1 mol 4685 kJ 1.00 g××= 24.13 kJ 194.19 g 1 mol The change in the temperature of the calorimeter is

24.13 kJ  ΔTf =  = 3.072 C 7.854 kJ/ C

Think About It Although the molar ∆Hcomb is large for dimethyl phthalate, that experiment combusts so little dimethyl phthalate that the change in temperature is small.

9.66. Collect and Organize We are asked to work backward from the molar enthalpy of combustion of anethole to the final temperature of the calorimeter.

Analyze First, we have to calculate ∆Hcomb from the molar heat of combustion and the grams (which we convert into moles) of anethole (C10H12O). We can find Tf by rearranging the equation for ∆Hcomb.

ΔH comb = Ccalorimeter ΔT

ΔH comb ΔT = = Tf – Ti Ccal

Solve The ∆Hcomb for 0.950 g of anethole is 1 mol 5541 kJ 0.950 g × × = 35.52 kJ 148.20 g 1 mol

35.52 kJ  ΔTf =  = 4.523 C 7.854 kJ/ C 26 | Chapter 9

Think About It Although the molar ∆Hcomb is large for anethole, that experiment combusts so little anethole that the change in temperature is small.

9.67. Collect and Organize For the reaction of 0.243 g of Mg with 100.0 mL of 1.00 M HCl, we are to calculate the enthalpy of the reaction, knowing the temperature change of the solution and given the specific heat and density of the solution.

Analyze All the energy from the reaction is transferred to the solution. We can find that heat from the equation

qsolution = mcPΔT Because all that heat came from the reaction of 0.243 g of Mg, the enthalpy of the reaction is q ΔH = rxn rxn mol of Mg

Solve The energy transferred in the reaction is

⎛ 1.01 g⎞ 4.18 J ! ! 3 q = 100.0 mL solution × × × 33.4 C − 22.4 C = 4.640 ×10 J solution ⎝⎜ ⎠⎟ ! ( ) mL g ⋅ C The enthalpy of the reaction per mole of Mg is 4.640 ×103 J ΔH = = 4.64 ×105 J/mol, or 464 kJ/mol rxn ⎛ mol ⎞ 0.243 g Mg × ⎝⎜ 24.305 g⎠⎟ Because that reaction is exothermic, the enthalpy of reaction is –464 kJ/mol.

Think About It Always be sure to determine the sign of enthalpy by considering whether the reaction is exothermic or endothermic.

9.68. Collect and Organize For the dissolution of 4.00 g of NH4NO3 in 96.0 g of water, we are to calculate the enthalpy of the reaction, knowing the temperature change of the solution. We can assume here that the density of the final solution is close to that of pure water (1.00 g/mL) and that the heat capacity of the solution is that of water also (4.18 J/g ⋅ ˚C).

Analyze All the energy from the reaction is transferred to the solution. We can find that heat from the equation

qsolution = mcPΔT

Where m is the total mass of the solution (96.0 g water + 4.00 g NH4NO3). Because all that heat came from the dissolution of 4.00 g of NH4NO3, the enthalpy of the reaction is qrxn ΔH rxn = mol of NH4NO3

Solve The energy transferred in the reaction is

⎛ 1.00 g⎞ 4.18 J ! 3 q = 100.0 mL solution × × × 3.07 C = 1.280 ×10 J solution ⎝⎜ ⎠⎟ ! ( ) mL g ⋅ C The enthalpy of the reaction per mole of NH4NO3 is 1.280 ×103 J ΔH = = 2.56 ×104 J/mol, or 25.6 kJ/mol rxn ⎛ mol ⎞ 4.00 g NH NO × ⎝⎜ 4 3 80.04 g⎠⎟ Because that reaction is endothermic, the enthalpy of reaction is +24.6 kJ/mol. Thermochemistry | 27

Think About It Always be sure to determine the sign of enthalpy by considering whether the reaction is exothermic or endothermic.

9.69. Collect and Organize Compare Hess’s law with the law of conservation of energy.

Analyze The law of conservation of energy states that energy cannot be created or destroyed; it can be converted from one form into another. Hess’s law states that the enthalpy change for a reaction can be obtained by summing the enthalpies of constituent reactions.

Solve When we apply Hess’s law, all the energy is accounted for in the reaction; energy is neither created nor destroyed.

Think About It Hess’s law makes it easy for us to calculate energy changes for chemical reactions from those for other chemical reactions.

9.70. Collect and Organize We consider whether Hess’s law would remain valid if enthalpy were not a state function.

Analyze Hess’s law states that the enthalpy change for a reaction can be obtained by summing the enthalpies of constituent reactions. The value of a state function is independent of the path taken in reaching a particular state; only the initial and final values are important.

Solve If enthalpy was not a state function, then the path of the changes would matter and we could not simply add them through Hess’s law.

Think About It Heat, q, and work, w, are not state functions.

9.71. Collect and Organize ° To calculate ∆Hf for SO2 from the equations given, we use Hess’s law.

Analyze ° The equation for ∆Hf of SO2 has S and O2 as the reactants and SO2 as the product. The sum of the other two ° reactions must add up to the overall ∆Hf . In both reactions, SO3 is produced. If we reverse the first equation and add the second equation, the overall reaction will consist of S and O2 as the reactants and SO2 as the product. ° When the first reaction is reversed, the ∆Hrxn will change from exothermic to endothermic.

Solve  2 SO3(g) → 2 SO2 (g) + O2 (g) ΔH rxn = 196 kJ 1  4 S8 (s) + 3 O2 (g) → 2 SO3(g) ΔH rxn = −790 kJ 1 S (s) + 2 O (g) → 2 SO (g) ΔH  = −594 kJ 4 8 2 2 rxn ° That will be twice that of the ∆Hf for the formation reaction (for 1 mol of SO2 formed): 1 8 S()82sg+ O()→ SO() 2 g ° Therefore, ∆Hf = –297 kJ/mol.

28 | Chapter 9

Think About It ° Remember that enthalpy is stoichiometric. If 5 mol of SO2 was formed, then ∆Hf = –1485 kJ.

9.72. Collect and Organize To calculate the enthalpy change for a reaction, we use Hess’s law to add chemical reactions and their associated enthalpy changes.

Analyze If we reverse the first reaction, ClO(g) will be a product, as it is in the overall desired equation. That exothermic reaction will now be endothermic. Adding the second equation to that equation gives us the overall equation.

Solve  Cl(g) + 2 O2 (g) → ClO(g) + O3(g) ΔH rxn = 29.90 kJ  2 O3(g) → 3 O2 (g) ΔH rxn = 24.18 kJ Cl(g) + O (g) → ClO(g) + O (g) ΔH  = 54.08 kJ 3 2 rxn

Think About It The overall reaction is endothermic, meaning that energy must be supplied by the surroundings to form ClO(g).

9.73. Collect and Organize The two equations when added in the appropriate way describe the conversion of α-spodumene into β- spodumene.

Analyze Because α-spodumene is the reactant in the conversion, we have to reverse the first equation. That reaction will then be endothermic. Hess’s law then states that we can add the equations and their corresponding ∆H values to give the overall equation and its enthalpy.

Solve 3  2 α-LiAlSi2O6 (s) → Li2O(s) + 2 Al(s) + 4 SiO2 (s) + 2 O2 (g) ΔH rxn = 1870.6 kJ 3  Li2O(s) + 2 Al(s) + 4 SiO2 (s) + 2 O2 (g) → 2 β-LiAlSi2O6 (s) ΔH rxn = −1814.6 kJ 2 α-LiAlSi O (s) → 2 β-LiAlSi O (s) ΔH  = 56.0 kJ 2 6 2 6 rxn This is for the conversion of 2 mol of the α form into the β form. For 1 mol: α-LiAlSi O (s) → β-LiAlSi O (s) ΔH  = 28.0 kJ/mol 2 6 2 6 rxn

Think About It Because the conversion of α-spodumene into β-spodumene is endothermic, we can say that the α form is more stable (by enthalpy) than the β form.

9.74. Collect and Organize The overall process of converting diamond into graphite is Cdiamond(s) → Cgraphite(s)

Analyze Because diamond is a reactant and graphite the product, the first and third reactions do not need to be reversed. We want to cancel CO2 on the product side and CO on the reactant side, which can be achieved using the second reaction.

Thermochemistry | 29

Solve ! Cdiamond (s) + O2 (g) → CO2 (g) ΔH rxn = –395.4 kJ ! 2 CO2 (g) → 2 CO(g) + O2 (g) ΔH rxn = 566.0 kJ ! 2 CO(g) → Cgraphite (s) + CO2 (g) ΔH rxn = –172.5 kJ C (s) → C (s) ΔH ! = –1.9 kJ diamond graphite rxn

Think About It The fact that the enthalpy of conversion of graphite to diamond is exothermic, and therefore a favorable process, means that diamond is unstable with respect to graphite.

9.75. Collect and Organize The two chemical reactions must add up to the overall reaction in which NOCl decomposes to nitrogen, oxygen, and chlorine. We can use Hess’s law to find the enthalpy of decomposition of NOCl.

Analyze NOCl must be on the reactant side, so the second reaction must be reversed. That reaction will then be endothermic (ΔH˚rxn = 38.6 kJ). The first reaction also has to be reversed because we need to have N2 and O2 on ° the product side of the overall equation. That will be an exothermic reaction (∆Hrxn = –90.3 kJ).

Solve 1 1  NO(g) → 2 N2 (g) + 2 O2 (g) ΔH rxn = –90.3 kJ 1  NOCl(g) → NO(g) + 2 Cl2 (g) ΔH rxn = 38.6 kJ NOCl(g) → 1 N (g) + 1 O (g) + 1 Cl (g) ΔH  = –51.7 kJ 2 2 2 2 2 2 rxn ° Multiplying that chemical reaction by 2, we obtain the ΔHrxn for the desired overall equation: ° 2 NOCl(g) → N2 (g) + O2 (g) + Cl2 (g) ΔH rxn = –103 kJ

Think About It That reaction, because it is exothermic, releases energy in the decomposition of NOCl into its constituent elements.

9.76. Collect and Organize Here, we are asked to combine four chemical equations to calculate the enthalpy of formation of CH4 from CO and H2.

Analyze The second equation must be reversed so that CO is a reactant in the overall equation. That results in an endothermic reactant.

Solve ! 3×[C(graphite) + 2 H2 (g) → CH4 (g)] ΔH = –74.8 kJ × 3 1 ! 3×[CO(g) → C(graphite) + 2 O2 (g)] ΔH = 110.5 kJ × 3 1 ! CO(g) + 2 O2 (g) → CO2 (g) ΔH = –283.0 kJ 1 ! 2 ×[H2 (g) + 2 O2 (g) → H2O(g)] ΔH = –285.8 kJ × 3 8 H (g) + 4 CO(g) → 3 CH (g) + 2 H O(g) + CO (g) ΔH ! = –747.5 kJ 2 4 2 2 rxn

Think About It The enthalpy of formation of methane in that process is exothermic. Therefore, the products (CH4, H2O, and CO2) are more stable (lower in energy) than the reactants H2 and CO.

30 | Chapter 9

9.77. Collect and Organize We are asked to explain why the standard heat of formation of carbon monoxide as a gas is difficult to measure experimentally.

Analyze Carbon monoxide is formed from the combustion of carbon. Another product, however, also forms when carbon reacts with oxygen: CO2.

Solve The enthalpy of formation of CO is difficult to measure because the competing reaction to form CO2 is favorable, so isolating just the enthalpy due to the formation of carbon monoxide is hard.

Think About It The enthalpy of formation of carbon dioxide is more exothermic (–412.9 kJ/mol) and therefore more favorable by enthalpy than the formation of carbon monoxide (–393.5 kJ/mol).

9.78. Collect and Organize We are to explain how using enthalpies of formation to determine the enthalpy of a reaction is an application of Hess’s law.

Analyze Hess’s law states that the enthalpy change for a reaction can be obtained by summing the enthalpies of constituent reactions. The enthalpy of formation is a reaction in which a substance is formed from the elements in their standard states.

Solve Because formation reactions are also reactions, we can add them to find the enthalpy of a reaction, just as we use Hess’s law.

Think About It Hess’s law is valid only because enthalpy is a state function.

9.79. Collect and Organize The standard enthalpy of formation is the energy absorbed or evolved when 1 mol of a substance is formed from the elements, all in their standard states. Here we compare the enthalpy of formation of O2 to that of O3.

Analyze Standard conditions are 1 atm and some specified temperature. Both ozone (O3) and elemental oxygen (O2) exist under those conditions.

Solve Because ozone and elemental oxygen are different forms of oxygen, their standard enthalpies of formation are ° different. From Appendix 4, ∆Hf for O2 is 0 kJ/mol (because it is an element in its most stable form under ° standard conditions) and ∆Hf for O3 is 142.7 kJ/mol.

Think About It ° Because ∆Hf for O3 is more positive than that for O2, ozone is less stable than O2.

9.80. Collect, Organize, and Analyze The standard enthalpies of formation of the elements in their standard states are all assigned a value of zero. Standard conditions are defined as 1 atm and some specified temperature.

Solve The standard enthalpies of formation of elements from the elements themselves are assigned to be zero because we are most interested in changes in enthalpy, so we can set the zero point for enthalpy where it is convenient.

Thermochemistry | 31

Think About It ° ° When using ∆Hf in further calculations, remember that ∆Hf of an element in its most stable form is zero under standard conditions.

9.81. Collect and Organize In calculating an estimate of the enthalpy change for a reaction by using bond energies, we are to explain why we need to know the stoichiometry of the reaction.

Analyze Bond breaking is endothermic, and bond formation is exothermic. The stoichiometry of the reaction tells us how many bonds break and how many form.

Solve We must account for all the bonds that break and all the bonds that form in the reaction. To do so we must start with a balanced chemical reaction. If we miss a bond that breaks, our calculated enthalpy of reaction would be too negative. If we miss a bond that forms, our calculated enthalpy of reaction would be too positive.

Think About It Before using an equation to calculate enthalpy change, whether from enthalpies of formation or from bond energies, always start with a balanced chemical equation.

9.82. Collect and Organize We are to explain why we must know the structures of the reactants and products to use bond energies to estimate the enthalpy for a reaction.

Analyze If we know the structure of the molecules involved in the reaction, then we know exactly which type of bonds are being broken and formed.

Solve Because different bonds have different bond energies, we must know the type of bond involved. We cannot assume connectivity of the atoms or the type of bonds involved. We might not realize that we are breaking a double bond instead of a single bond, for example.

Think About It Luckily, we have Lewis structures to help us know what type of bonds are in a molecule and which bonds would be broken and which ones would be formed.

9.83. Collect and Organize We are to explain why the phase of the reactants is important in enthalpy calculations, even in those involving bond energies—in particular, why the reactants and products should all be in the gas phase when we use bond energies to estimate the enthalpy of a reaction.

Analyze The bond energy is the enthalpy change required to break 1 mol of bonds in a substance in the gas phase.

Solve If the compounds are in the solid or liquid phase, interactions between molecules may slightly change the bond energy for a given bond.

Think About It Having bond energy data tabulated only for the gaseous phase ensures that the measured energies are for the bonds breaking only and do not include any intermolecular interactions. Later, you will learn more about intermolecular forces. Some can be very strong (ion–ion forces), and others are quite weak (van der Waals forces).

32 | Chapter 9

9.84. Collect and Organize We are to compare the bond energies of products (CO2) to that of reactants (C and O2) to explain why the enthalpy of formation of CO2 (C + O2 → CO2) is exothermic.

Analyze The reaction in terms of Lewis structures is C + O O O C O

The enthalpy of the reaction as estimated using bond energies is given as ΔH  = ΔH + ΔH rxn ∑ bond breaking ∑ bond forming where ΔHbond breaking and ΔHbond forming are average bond energies for the bonds in the reactants and products, respectively. The bond energy values are always positive because the bond energy is defined as the energy required to break the bond.

Solve If the energy required to break 2 mol of C O bonds is greater (more positive) than the sum of the heat of sublimation of carbon (as a measure of the “bond energy” for solid carbon) and the bond energy of the O O bond, then the ∆Hrxn is negative because a larger negative enthalpy (from the bond energy of the product, CO2) is added to a smaller positive bond energy of the reactants (C + O2).

Think About It The enthalpy for a reaction is exothermic when the strength of the bonds in the products is greater than the strength of the bonds in the reactants.

9.85. Collect and Organize The heat of formation is reflected in a reaction when (1) 1 mol of the substance is produced, (2) the substance is produced under standard-state conditions, and (3) it is produced from the substance’s constituent elements in their standard state.

Analyze ° Each reaction must meet all the criteria for its ∆Hrxn to be classified as a heat of formation.

Solve ° (a) One mole of CO is produced from elemental carbon and oxygen in their standard states, so ∆Hrxn for that ° reaction represents ∆Hf. (b) Because 2 mol of Mn3O4 is produced from 6 mol of MnO4 (which is not an element) and 1 mol of O2, that ° reaction does not represent ∆Hf. 3 (c) Because Na2CO3 is produced from the reaction of 2 mol of Na, 1 mol of C, and /2 mol of O2 (all elements ° in their standard states), that reaction does represent ∆Hf. (d) Two moles of CH4 is produced from 2 mol of elemental carbon and 4 mol of hydrogen each in their ° standard states. That at first appears to represent ∆Hf, but the equation does not represent the product in ° 1 mol of product, CH4, and therefore that reaction does not represent ∆Hf.

Think About It Heat of formation must involve the reaction of the elements to form compounds. Remember, though, that some elements, such as O2, H2, and N2, are diatomic in their elemental state.

9.86. Collect and Organize The heat of formation is reflected in a reaction when (1) 1 mol of the substance is produced, (2) the substance is produced under standard-state conditions, and (3) it is produced from the substance’s constituent elements in their standard state.

Analyze ° Each reaction must meet all the criteria for its ∆Hrxn to be classified as a heat of formation.

Thermochemistry | 33

Solve (a) Because two substances are produced from elemental nitrogen and oxygen and more than 1 mol of the ° products is produced, that reaction does not represent ∆Hf. ° (b) Because 2 mol of NO is produced from the elements, that reaction does not represent ∆Hf. ° (c) Because the N2O4 in that reaction is not produced from the elements, that reaction does not represent ∆Hf. ° (d) Because 2 mol of NO2 is produced from the elements, that reaction does not represent ∆Hf.

Think About It Heat of formation must involve the reaction of the elements to form compounds. Remember, though, that some elements, such as O2, H2, and N2, are diatomic in their elemental state.

9.87. Collect and Organize The heat of a reaction can be computed by finding the difference between the sum of the heats of formation of the products and the sum of the heats of formation of the reactants.

Analyze We have to take into account the moles of products formed and the moles of reactants used as well, because enthalpy is a stoichiometric quantity. ΔH  = n ΔH  – n ΔH  rxn ∑ products f,products ∑ reactants f,reactants ° Values for ∆Hf for the reactants and products are found in Appendix 4.

Solve ! ΔH rxn = ⎣⎡(1 mol CH4 )(−74.8 kJ/mol) + (2 mol H2O)(−285.8 kJ/mol)⎦⎤

– ⎣⎡(4 mol H2 )(0 kJ/mol) + (1 mol CO2 )(−393.5 kJ/mol)⎦⎤ ΔH ! = –252.9 kJ rxn

Think About It ° Be careful to note and find the appropriate ∆Hf for a compound that may exist in different phases. For example, ° ° ∆Hf of H2O(g) = –241.8 kJ/mol, but ∆Hf of H2O(ℓ) = –285.8 kJ/mol.

9.88. Collect and Organize The enthalpy of a reaction can be computed by finding the difference between the sum of the enthalpies of formation of the products and the sum of the enthalpies of formation of the reactants.

Analyze We have to take into account the moles of products formed and the moles of reactants used as well, since enthalpy is a stoichiometric quantity. ΔH  = n ΔH  – n ΔH  rxn ∑ products f,products ∑ reactants f,reactants ° ° Values for ∆Hf for the reactants and products are found in Appendix 4; the ∆�! for methylamine is given in the problem as –22.97 kJ/mol.

Solve  ΔH rxn = ⎣⎡(3 mol CH4 )(−74.8 kJ/mol) + (1 mol CO2 )(−393.5 kJ/mol) + (4 mol NH3 )(−46.1 kJ/mol)⎦⎤

– ⎣⎡(4 mol CH3NH2 )(−22.97 kJ/mol) + (2 mol H2O)(−285.8 kJ/mol)⎦⎤ ΔH  = –138.8 kJ rxn

Think About It ° Be careful to note and find the appropriate ∆Hf for a compound that may exist in different phases. For example, ° ° ∆Hf of H2O(g) = –241.8 kJ/mol, but ∆Hf of H2O(ℓ) = –285.8 kJ/mol.

34 | Chapter 9

9.89. Collect and Organize ° To calculate ∆Hrxn for the decomposition of NH4NO3 to N2O and H2O vapor, we need the balanced equation because the enthalpy of the reaction depends on the moles of reactants consumed and moles of products formed in the reaction.

Analyze ° From the balanced chemical equation and the values of ∆Hf of the reactants and products, we use ΔH  = n ΔH  – n ΔH  rxn ∑ products f,products ∑ reactants f,reactants Because the reaction is run at 250–300˚C, the water product is in the gaseous phase.

Solve The balanced equation for that reaction is

NH4NO3(s) → N2O(g) + 2 H2O(g) ° We use the coefficients in the equation for ∆Hrxn:  ΔH rxn = ⎣⎡(1 mol N2O)(82.1 kJ/mol) + (2 mol H2O)(−241.8 kJ/mol)⎦⎤

– ⎣⎡(1 mol NH4NO3 )(−365.6 kJ/mol)⎦⎤ ΔH  = –35.9 kJ rxn

Think About It The reaction is exothermic, releasing 36 kJ for every mole of NH4NO3 decomposed.

9.90. Collect and Organize We are given the balanced chemical equation for the decomposition of TNT. The enthalpy change from Problem 9.89 for the decomposition of ammonium nitrate is –35.9 kJ for 1 mol of NH4NO3. We are asked to determine the amount of TNT needed to equal the enthalpy change for that decomposition reaction.

Analyze We know from the given information that the decomposition of 1 mol of TNT releases 10,153 kJ (exothermic). To calculate how much TNT is required to equal the enthalpy released by the NH4NO3 reaction, we need to divide the enthalpy of the NH4NO3 reaction by the enthalpy of the TNT reaction. Then, we can convert moles of TNT to mass of TNT by using the molar mass of TNT (C7H5N3O6, 227.13 g/mol).

Solve Equivalent mole of TNT to equal enthalpy release of the NH4NO3 reaction: 1 mol TNT –35.9 kJ × = 3.54 ×10−3 mol TNT –10,153 kJ 227.13 g The mass of TNT = 3.54 ×10−3 mol × = 0.803 g TNT 1 mol

Think About It Clearly, TNT is a much more powerful explosive.

9.91. Collect and Organize We are given the balanced chemical equation for the explosive reaction of fuel oil with ammonium nitrate in the presence of oxygen.

Analyze ° To calculate ΔHrxn we use ΔH  = n ΔH  – n ΔH  rxn ∑ products f,products ∑ reactants f,reactants

Thermochemistry | 35

Solve ! ΔH rxn = ⎣⎡(3 mol N2 )(0.0 kJ/mol) + (17 mol H2O)(−241.8 kJ/mol) + (10 mol CO2 )(−393.5 kJ/mol)⎦⎤

– ⎣⎡(3 mol NH4NO3 )(−365.6 kJ/mol) + (1 mol C10H22 )(249.7 kJ/mol) + (14 mol O2 )(0.0 kJ/mol)⎦⎤ ΔH ! = –7198.5 kJ rxn

Think About It That is a very exothermic reaction that occurs very fast and is therefore explosive.

9.92. Collect and Organize For the decomposition reaction of ammonium nitrate, NH4NO3, to give N2, O2, and H2O, we are to write the balanced chemical equation and calculate the enthalpy of decomposition, ΔH˚rxn, from the heat of formation values in Appendix 4.

Analyze ° To calculate ΔHrxn we use ΔH  = n ΔH  – n ΔH  rxn ∑ products f,products ∑ reactants f,reactants

Solve The balanced decomposition reaction is

2 NH4NO3(s) → 2 N2(g) + O2(g) + 4 H2O(g) The enthalpy for that decomposition reaction under standard conditions is  ΔH rxn = ⎣⎡(2 mol N2 )(0.0 kJ/mol) + (1 mol O2 )(0.0 kJ/mol) + (4 mol H2O)(−241.8 kJ/mol)⎦⎤

– ⎣⎡(2 mol NH4NO3 )(−365.6 kJ/mol)⎦⎤ ΔH  = –236.0 kJ rxn The enthalpy calculated above is for the decomposition of 2 mol of ammonium nitrate. The molar enthalpy for that reaction would be –118.0 kJ/mol.

Think About It That decomposition reaction of ammonium nitrate is more exothermic than when ammonium nitrate decomposes to N2O and H2O (Problem 9.89).

9.93. Collect and Organize We can use average bond energy data in Table A4.1 to estimate the enthalpy of three reactions.

Analyze The enthalpy of a reaction as estimated from bond energies is given as HH + H Δrxn=∑∑ΔΔ bond breaking bond forming where ∆Hbond breaking and ∆Hbond forming are average bond energies for the bonds in the reactants and products, respectively. In computing ∆Hbond breaking and ∆Hbond forming we have to take into account the number (or moles) of a particular type of bond that breaks or forms. We must also keep in mind that bond breaking requires energy (+∆H), whereas bond formation releases energy (–∆H).

Solve (a) ∆Hrxn = [(2 × 945 kJ/mol) + (3 × 436 kJ/mol)] + [–(6 × 391 kJ/mol)] = 852 kJ N N + 3 H H 2 H N H

H

36 | Chapter 9

(b) ∆Hrxn = [(1 × 945 kJ/mol) + (2 × 436 kJ/mol)] + [–(1 × 163 kJ/mol) – (4 × 391 kJ/mol)] = 90 kJ H H N N + 2 H H N N

H H (c) ∆Hrxn = [(2 × 945 kJ/mol) + (1 × 498 kJ/mol)] + [–(2 × 945 kJ/mol) – (2 × 201 kJ/mol)] = 96 kJ 2 N N + O O 2 N N O

Think About It We have to use the “best” Lewis structure for these calculations of reaction enthalpy. In part c, an alternative form (but contributing less to the bonding by formal charge arguments) for N2O would be 2 N N O O 2 N N O

∆Hrxn = [(2 × 945 kJ/mol) + (1 × 498 kJ/mol)] + [– (2 × 418 kJ/mol) – (2 × 607 kJ/mol)] = 338 kJ

9.94. Collect and Organize We can use bond energy data in Table A4.1 to estimate the enthalpy of three reactions.

Analyze The enthalpy of a reaction as estimated from bond energies is given as HH + H Δrxn=∑∑ΔΔ bond breaking bond forming where ∆Hbond breaking and ∆Hbond forming are average bond energies for the bonds in the reactants and products, respectively. In computing ∆Hbond breaking and ∆Hbond forming we have to take into account the number (or moles) of a particular type of bond that breaks or forms. We must also keep in mind that bond breaking requires energy (+∆H), whereas bond formation releases energy (–∆H).

Solve (a) ∆Hrxn = [(2 × 799 kJ/mol) + (1 × 436 kJ/mol)] + [–(2 × 463 kJ/mol) – (1 × 1072 kJ/mol)] = 36 kJ O C O + H H H O H + C O

(b) ∆Hrxn = [(1 × 945 kJ/mol) + (1 × 498 kJ/mol)] + [–(2 × 607 kJ/mol)] = 229 kJ N N + O O 2 N O

(c) ∆Hrxn = [(1 × 719 kJ/mol) + (2 × 799 kJ/mol)] + [–(2 × 1072 kJ/mol)] = 173 kJ C + O C O 2 C O

Think About It All those reactions are predicted to be endothermic. The energy required to break the bonds in the reactants is greater than the energy released to form the bonds of the products.

9.95. Collect and Organize We can use bond energies to compare the reaction enthalpies of the two reactions ! C2H6 + ! O2 → 2 CO + 3 H2O ! C2H6 + ! O2 → 2 CO2 + 3 H2O

Analyze From the Lewis structures and the average bond energy values in Table A4.1, we can estimate the reaction enthalpies. The enthalpy of a reaction as estimated from bond energies is given as HH + H Δrxn=∑∑ΔΔ bond breaking bond forming where ∆Hbond breaking and ∆Hbond forming are average bond energies for the bonds in the reactants and products, respectively.

Thermochemistry | 37

Solve For the incomplete combustion of C2H6 to CO: H H

H C C H + 5/2 O O 2 C O + 3 H O H

H H ΔHrxn = [(6 × 413 kJ/mol) + (1 × 348 kJ/mol) + (5/2 × 498 kJ/mol)] + [–(2 × 1072 kJ/mol) – (6 × 463 kJ/mol)] ∆Hrxn = –851 kJ/mol For the complete combustion of C2H6 to CO2: H H

H C C H + 7/2 O O 2 O C O + 3 H O H

H H ∆Hrxn = [(6 × 413 kJ/mol) + (1 × 348 kJ/mol) + (7/2 × 498 kJ/mol)] + [–(4 × 799 kJ/mol) – (6 × 463 kJ/mol)] ∆Hrxn = –1405 kJ/mol The complete combustion reaction releases 554 kJ more energy than the incomplete combustion reaction.

Think About It Although weaker C O bonds are formed in the complete combustion reaction, two such bonds are formed in CO2, which outweighs the strong C O bond in carbon monoxide in the incomplete combustion reaction.

9.96. Collect and Organize We can use bond energies to compare the reaction enthalpies of the two reactions

C(s) + O2(g) → CO2(g) ! C(s) + ! O2(g) → CO(g)

Analyze From the Lewis structures and the average bond energy values in Table A4.1, we can calculate the reaction enthalpies. The enthalpy of a reaction as estimated from bond energies is given as HH + H Δrxn=∑∑ΔΔ bond breaking bond forming where ∆Hbond breaking and ∆Hbond forming are average bond energies for the bonds in the reactants and products, respectively. We are not given the “bond energy” (really the heat of sublimation) of C(s), but because that value is the same for both reactions it cancels out in finding the difference between the enthalpies of the two reactions.

Solve For the complete combustion of C to CO2: C + O O O C O

∆Hrxn = [x + (1 × 498 kJ/mol)] + [–(2 × 799 kJ/mol)] ∆Hrxn = –1100 + x kJ/mol For the incomplete combustion of C to CO: C + 1/2 O O O C

1 ∆Hrxn = [x + ( 2 × 498 kJ/mol)] + [–(1 × 1072 kJ/mol)]

∆Hrxn = –823 + x kJ/mol The difference in ∆Hrxn values is (–1100 + x) – (–823 + x) = –277 kJ/mol. The reaction that produces CO2 releases 277 kJ/mol more heat than the reaction that produces CO.

Think About It Although weaker C O bonds are formed in the complete combustion reaction, two such bonds are formed in CO2, which outweighs the strong C O bond in carbon monoxide in the incomplete combustion reaction.

38 | Chapter 9

9.97. Collect and Organize For the reaction of ammonia with oxygen to give water and nitrogen dioxide, we can use the bond energies of the N—H (391 kJ/mol), O O (498 kJ/mol), N O (607 kJ/mol), N—O (201 kJ/mol), and O—H (463 kJ/mol) bonds to estimate the enthalpy of the reaction.

Analyze To be sure which bonds are breaking and which bonds are being formed, drawing the Lewis structures of each of the products and reactants will help us. The enthalpy of a reaction as estimated from bond energies is given as HH + H Δrxn=∑∑ΔΔ bond breaking bond forming where ∆Hbond breaking and ∆Hbond forming are average bond energies for the bonds in the reactants and products, respectively.

Solve 4 H N H + 7 O O 4 O N O + 6 H O H

H ∆Hrxn = [(12 × 391 kJ/mol) + (7 × 498 kJ/mol)] + [–(4 × 201 kJ/mol) – (4 × 607 kJ/mol) – (12 × 463 kJ/mol)] ∆Hrxn = –610 kJ

Think About It The bonding in NO2 is not strictly one single bond and one double bond, as shown by the resonance structures O N O O N O

For calculations using bond energies, however, we can use one of the “frozen” resonance structures to assign bond energy values.

9.98. Collect and Organize Given the enthalpy of the reaction for the combustion of H2S to give SO2 and H2O, we are to use the average bond energies of the S—H (347 kJ/mol), O O (498 kJ/mol), and O—H (463 kJ/mol) bonds to estimate the bond energy for S O.

Analyze To be sure which bonds are breaking and which bonds are being formed, drawing the Lewis structures of each of the products and reactants will help us. The enthalpy of a reaction as estimated from bond energies is given as HH + H Δrxn=∑∑ΔΔ bond breaking bond forming where ∆Hbond breaking and ∆Hbond forming are average bond energies for the bonds in the reactants and products, respectively.

Solve 2 H S H + 3 O O 2 O S O + 6 H O H

∆Hrxn = –1036 kJ = [(4 × 347 kJ/mol) + (3 × 498 kJ/mol)] + [–(4x) – (4 × 463 kJ/mol)] x = 516 kJ/mol

Think About It The bond energy for S O calculated here compares favorably with the average value of 523 kJ/mol in Table A4.1.

9.99. Collect and Organize We are asked to describe how the strength of ion–dipole interactions between the solvent and the cations and anions of an ionic salt determine whether the heat of solution of that ionic salt is exothermic or endothermic.

Thermochemistry | 39

Analyze The heat of solution of an ionic salt involves the breaking of ionic bonds in the salt (endothermic) and the formation of ion–dipole interactions (exothermic) between the ions and the polar solvent molecules.

Solve When the ion–dipole forces between the ions and the solvent molecules are stronger than the ion–ion forces between the cations and anions in the salt, the heat of solution will be exothermic. If, however, they are weaker than the ion–ion forces, the heat of solution will be endothermic.

Think About It Whether the heat of solution is exothermic or endothermic does not determine whether a salt will dissolve. For example, we might expect that an endothermic heat of solution would mean that the salt does not dissolve well. However, dissolving sodium chloride is endothermic, but certainly the salt dissolves well in water.

9.100. Collect and Organize Given that sodium hydroxide dissolves better in hot water than in cold water, we are asked to explain how the compound’s dissolution is nonetheless an exothermic process.

Analyze Because the process is exothermic we know that the dissolution releases heat. A greater solubility means that more of the solid dissolves in hot water than in cold water.

Solve Enthalpy and solubility are two very different things, and one does not necessarily correlate with the other.

Think About It We will see in a later chapter in the textbook on thermodynamics that we must take entropy into account as well as enthalpy in order to predict whether a process or a reaction is spontaneous.

9.101. Collect, Organize, and Analyze We are asked to define and describe the term fuel value.

Solve We often are concerned about the energy a fuel provides per mass; fuel values give the energy per gram that a fuel releases upon burning.

Think About It For some fuels, such as gasoline, thinking of energy per volume (liter or gallon) might be more convenient.

9.102. Collect, Organize, and Analyze Units of fuel values depend on the convenient units for measurement of energy released per mass of fuel.

Solve Common units for fuel values are kilojoules per gram, but we could also express the energy in joules, BTUs (British thermal units), or other energy units, and the mass in kilograms, pounds, tons, or other mass units.

Think About It When comparing fuel values, be sure that the energy units for the fuels being compared are consistent.

9.103. Collect and Organize For this question we compare the fuel values (without making any calculations) of CH4 and H2 on a per-mole and a per-gram basis.

Analyze The balanced equations for the combustion reactions are

CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(g) 40 | Chapter 9

! H2(g) + ! O2(g) → H2O(g) The heats of formation for these reactants and products are CH4(g) = –74.8 kJ/mol O2(g) = 0.0 kJ/mol CO2(g) = –393.5 kJ/mol H2O(g) = –241.8 kJ/mol H2(g) = 0.0 kJ/mol

Solve (a) In examining the two reactions and the associated heats of formation, we can see easily that the combustion of 1 mol of CH4 gives off more energy than the combustion of 1 mol of H2. (b) However, because 1 g is 1/16 mol of CH4, whereas 1 g is 1/2 mol of H2 on a per-gram basis, H2 releases more energy upon combustion per gram.

Think About It The high-energy combustion reaction of hydrogen on a per-gram basis makes it an attractive fuel source to develop.

9.104. Collect and Organize We consider the usefulness of the terms fuel density and fuel value.

Analyze Fuel value is expressed in energy per gram (kilojoules per gram), whereas fuel density is expressed as energy per volume (kilojoules per liter).

Solve The fuel density value would be more useful for liquid fuels because it is energy per volume. We could have a substance with a lower fuel value, but if it takes up less volume it can pack into a tank and be transported efficiently. We would then get more energy per tank.

Think About It If we could produce a low-density fuel such as hydrogen on the spot for use, that would be a better fuel to use, but when we carry fuel in tanks, looking for fuels with high fuel density makes sense.

9.105. Collect and Organize We are to use a Born–Haber cycle along with values for the ionization energy of K, electron affinity of Cl, sublimation energy of K, the bond energy of Cl2, and the enthalpy of formation of KCl to determine the lattice energy for KCl.

Analyze A Born–Haber cycle considers the formation of a compound from its elements as a sequence of steps. It uses Hess’s law to calculate the value of a step with a missing thermodynamic value or the overall heat of formation for that compound. For ionic salts such as KCl, the Born–Haber cycle adds the enthalpies of sublimation of the metal, the ionization energy of the metal, the bond energy of the diatomic halogen, the electron affinity of the halogen atom, and the lattice energy of the cation and anion to give the heat of formation of the salt. In this particular case

! 1 ΔHf, KCl = ΔHsub, K + BECl + IE1, K + EACl +U KCl 2 2 Solving for UKCl gives

1 ! −U KCl = ΔHsub, K + BECl + IE1, K + EACl − ΔHf, KCl 2 2

Thermochemistry | 41

Solve ⎛ 1 ⎞ –U KCl = 89 kJ/mol + × 243 kJ/mol + 419 kJ/mol + (–349 kJ/mol) − (–436.5 kJ/mol) = 717 kJ/mol ⎝⎜ 2 ⎠⎟ U = −717 kJ/mol KCl

Think About It Lattice energy for an ionic solid is defined as the energy change when gaseous cations and anions for a solid. Because of the attraction of the oppositely charged ions, that is always exothermic for ionic solids.

9.106. Collect and Organize We can use a Born–Haber cycle along with values for the ionization energy of Na, electron affinity of O, sublimation energy of Na, the bond energy of O2, and the enthalpy of formation of Na2O to determine the lattice energy for Na2O.

Analyze A Born–Haber cycle considers the formation of a compound from its elements as a sequence of steps. It uses Hess’s law to calculate the value of a step with a missing thermodynamic value or the overall heat of formation for that compound. For ionic salts such as Na2O, the Born–Haber cycle adds the enthalpies of sublimation of the metal, the ionization energy of the metal, the bond energy of the diatomic halogen, the electron affinity of the halogen atom, and the lattice energy of the cation and anion to give the heat of formation of the salt. In this particular case 1 H ! (2 H ) BE (2 IE ) EA U Δ f, Na O = × Δ sub, Na + O + × 1, Na + O,2e− + Na O 2 2 2 2 Solving for UKCl gives 1 U (2 H ) BE (2 IE ) EA H ! − Na O = × Δ sub, Na + O + × 1, Na + O,2e− − Δ f, Na O 2 2 2 2

Solve ⎛ 1 ⎞ −U Na O = (2 ×109 kJ/mol) + × 498 kJ/mol + (2 × 495 kJ/mol) + 603 kJ/mol − (–416 kJ/mol) = 2480 kJ/mol 2 ⎝⎜ 2 ⎠⎟

U Na O = −2480 kJ/mol 2

Think About It It may have surprised you that the electron affinity for two electrons on oxygen is positive (endothermic). Although the electron affinity for one electron is indeed, as expected, negative, adding the second electron to an already negative ion, O–, is not favorable and is therefore highly endothermic.

9.107. Collect and Organize To determine the amount of water that could be heated from 20.0˚C to 45.0˚C when 1.00 pound of propane is burned, we use the fuel value of propane and the heat capacity equation for water.

Analyze First, to compute the energy (q, in kilojoules) that 1.00 pound of propane generates, we multiply the fuel value (46.35 kJ/g) by the mass of propane (after converting pounds to grams, using 1 lb = 453.6 g). Then we use the molar heat capacity equation q = ncP∆T where cP = 75.3 J/mol ⋅ ˚C, ∆T = 25.0˚C, to find the moles (n) and finally mass (m) of water that can be heated 25.0˚C by 1.00 pound of propane.

Solve Energy generated by propane: 453.6 g 46.35 kJ 1.00 lb×× = 2.10 × 104 kJ 1 lb g 42 | Chapter 9

Mass of water heated 25.0˚C by this q: 1000 J J 2.10 ×104 kJ × = n × 75.3 × 25.0 ! C 1 kJ mol ⋅ ! C n =1.11 × 104 mol 18.02 g That is 1.11×104 mol× = 2.01×105 g, or 201 kg mol

Think About It Propane’s relatively high fuel value makes it an efficient fuel.

9.108. Collect and Organize We can use the heat capacity equation to find the energy the eight runners produce. We have to convert ounces to grams for the calculation.

Analyze We have to first melt the ice (molar enthalpy of fusion = 6.01 kJ/mol), then heat the water from 0˚C to 100˚C (molar heat capacity = 75.3 J/mol ⋅ ˚C), and then after it reaches 100˚C, boil it all away (molar enthalpy of vaporization = 40.67 kJ/mol). The total energy needed is 6.01 kJ 0.0753 kJ 40.67 kJ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ q = ⎜ n × ⎟ + ⎜ n ×  × ΔT⎟ + ⎜ n × ⎟ ⎝ mol ⎠ ⎝ mol ⋅ C ⎠ ⎝ mol ⎠ Here, ∆T is 100˚C (from melted water to nearly boiling water) and the given mass of 128 ounces has to be converted into grams (28.35 g = 1 ounce) and then to moles (18.02 g/mol).

Solve 28.35 g 1 mol mol water in 1 gallon = 128 ounces × × = 201 mol 1 ounce 18.02 g

⎛ 6.01 kJ⎞ ⎛ 0.0753 kJ  ⎞ ⎛ 40.67 kJ⎞ q = ⎜ 201 mol × ⎟ + ⎜ 201 mol ×  ×100 C⎟ + ⎜ 201 mol × ⎟ = 10, 900 kJ ⎝ mol ⎠ ⎝ mol ⋅ C ⎠ ⎝ mol ⎠

Think About It That amounts to 10,090 kJ ÷ 8 runners, or 1360 kJ generated by each runner in the race in 10 s.

9.109. Collect and Organize Once we compute the fuel value of C5H12, we use it to calculate the energy released when 1.00 kg of C5H12 is burned and how much C5H12 is needed to heat 1.00 kg of water by 70.0˚C.

Analyze ° (a) The fuel value can be calculated by dividing the absolute value of the given ∆Hcomb (–3535 kJ/mol) by the molar mass of C5H12 (72.15 g/mol). (b) The energy released when 1.00 kg of C5H12 is burned can be found by multiplying the mass in grams by the fuel value. (c) The molar heat equation is q = ncP∆T

where moles (n) can be determined from m = 1.00 kg (1000 g) water, cP = 75.3 J/mol ⋅ ˚C, and ∆T = 70.0˚C. That gives us the energy (in joules) required to heat the water. The amount of C5H12 needed to heat the water can then be calculated by dividing the q value by the fuel value for C5H12.

Solve 3535 kJ 1 mol (a) Fuel value of C H = × = 49.00 kJ/g 5 12 mol 72.15 g Thermochemistry | 43

49.00 kJ (b) Heat released by 1.00 kg C H = 1000 g × = 4.90 ×104 kJ 5 12 g (c) Energy needed to raise 1.00 kg water from 20.0˚C to 90.0˚C: 1 mol 75.3 J q = 1000 g × × × 70.0°C = 2.925 ×105 J, or 293 kJ 18.02 g mol ⋅  C 1 g Mass of C5H12 needed to generate that energy = 292.5 kJ × = 5.97 g 49.00 kJ

Think About It Only about 6 g of fuel is required from the camper’s stove to heat the water. The white gas has a relatively high fuel value.

9.110. Collect and Organize Once we compute the fuel value of C6H14, we use it to calculate the energy released when 1.00 kg of C6H14 is burned and how much C6H14 is needed to heat 1.00 kg of water by 60.0˚C. Finally, we use the fuel values from this and Problem 9.103 to calculate how many grams of white gas, which is a mixture of C5H12 and C6H14, is needed to heat 1.00 kg of water by 60.0˚C.

Analyze ° (a) The fuel value can be calculated by dividing the absolute value of the given ∆Hcomb (–4163 kJ/mol) by the molar mass of C6H14 (86.18 g/mol). (b) The energy released when 1.00 kg of C6H14 is burned can be found by multiplying the mass in grams by the fuel value. (c) The molar heat equation is q = ncP∆T

where moles n can be determined from m = 1.00 kg (1000 g) water, cP = 75.3 J/mol ⋅ ˚C, and ∆T = 60.0˚C. That gives us the energy (in joules) required to heat the water. The amount of C6H14 needed to heat the water can then be calculated by dividing the q value by the fuel value for C6H14. (d) The energy required to raise the water’s temperature by 60.0˚C is calculated in part c. The fuel value of the mixture is a weighted average based on the percentages of the C5 and C6 hydrocarbons. The mass of the mixture required then is obtained by dividing q required by the weighted average fuel value. From Problem 9.109, the fuel value of C5H12 is 48.99 kJ/g.

Solve 4163 kJ 1 mol (a) Fuel value of C H=×= 48.31 kJ/g 614 mol 86.18 g 48.31 kJ (b) Heat released by 1.00 kg CH= 1000 g×= 4.831× 104 kJ 614 g (c) Energy needed to raise 1.00 kg water from 25.0˚C to 85.0˚C: 1 mol 75.3 J q = 1000 g × × × 60.0  C = 2.51×105 J, or 251 kJ 18.02 g mol ⋅  C 1 g Mass of C6H14 needed to generate that energy =251 kJ×= 5.20 g 48.31 kJ (d) Weighted average fuel value= 0.75× 48.31 kJ g +0.25× 48.99 kJ g = 48.48 kJ g 1 g Mass of fuel mixture required =251 kJ×= 5.18 g 48.48 kJ

44 | Chapter 9

Think About It Because the fuel values for C5 and C6 hydrocarbon fuels are so close, the mass of fuel required to heat the water does not change significantly. Slightly less (by 0.02 g) of the fuel mixture is needed than for C6H14 because C5H12 has a better fuel value.

9.111. Collect and Organize For two preparations of HCl(g) we are to use the heat of reaction to determine whether we should cool or heat those reactions.

Analyze If a reaction is exothermic, it would make sense to cool the reaction (not heat it) so that it does not boil or cause an explosion. If a reaction is endothermic it would make sense to gently heat the reaction to provide the necessary energy for the reaction to proceed.

Solve The heats of those two reactions are

H2(g) + Cl2(g) → 2HCl(g) ∆Hrxn = [(2 mol HCl × –92.3 kJ/mol)] – [(1 mol H2 × 0.0 kJ/mol) + (1 mol Cl2 × 0.0 kJ/mol)] = –184.6 kJ

2NaCl(s) + H2SO4(,) → 2HCl(g) + Na2SO4(s) ∆Hrxn = [(2 mol HCl × –92.3 kJ/mol) + (1 mol Na2SO4 × –1387.1 kJ/mol)] – [(2 mol NaCl × –411.2 kJ/mol) + (1 mol H2SO4 × –814.0 kJ/mol)] = 64.7 kJ The first reaction using hydrogen and chlorine gases is exothermic, so we should cool that reaction; the second reaction using sodium chloride and sulfuric acid is endothermic, so we should heat that reaction.

Think About It This problem illustrates that whether the formation of a particular product is endothermic or exothermic depends greatly on the process.

9.112. Collect and Organize In this problem, we calculate the minutes needed to burn off 563 kilocalories if a walker burns 4.70 kilocalories per minute.

Analyze To determine the minutes needed to walk at a pace of 4.70 kcal/min to burn off 563 kcal, all we need to do is simply divide the kilocalories needed to burn by the rate those Calories are burned.

Solve 563 kcal minutes needed to burn off Calories = =120 min, or 2 h 4.70 kcal/min

Think About It We did not have to convert between Calories and kcal for this problem because those units would have canceled in doing the division.

9.113. Collect and Organize When a sodium hydroxide solution is mixed with a sulfuric acid solution, the solution gets hot. We are asked to find the ∆Hrxn for the reaction.

Analyze (a) That is a neutralization reaction that produces water and sodium sulfate. (b) To determine the molar ratio (to see whether any reactant is in excess), we multiply the volume of reactant by its concentration. (c) The water absorbs all the energy generated by the reaction, causing the temperature to rise. The energy then is q = mcS∆T Thermochemistry | 45

where m = total mass of the solution (100 g + 50.0 g), cS is the specific heat of water (4.184 J/g · ˚C), and ∆T is the change in temperature (Tf – Ti).

Solve (a) 2 NaOH(aq) + H2SO4(aq) → 2 H2O () + Na2SO4(aq) (b) Stoichiometry of the reaction: 1.0 mol mol NaOH = 100.0 mL×= 0.10 mol NaOH 1000 mL 1.0 mol mol H SO = 50.0 mL×= 0.050 mol H SO 241000 mL 24 From the balanced equation, 0.050 mol H2SO4 would require 0.100 mol of NaOH. Therefore, neither NaOH nor H2SO4 is left over after the reaction and 0.10 mol of H2O is produced in the reaction. 4.184 J (c) q = mc ΔT = 150 g × × 31.4 !C – 22.3!C = 5.7 ×103 J, or 5.7 kJ S g !C ( ) ⋅ That is for the reaction of 0.10 mol of H2O that is produced in that particular reaction. For 1 mol of H2O, therefore, 5.71 kJ = 57 kJ/mol H2O 0.10 mol H2O

Because the reaction transfers energy (the mixture gets hot), that reaction is exothermic and ∆Hrxn = –57 kJ/mol H2O.

Think About It We can assign the sign of ∆H at the end of our calculation. If temperature rises, the reaction is exothermic and the sign of enthalpy is negative.

9.114. Collect and Organize When NaOH and H2SO4 solutions are mixed, an exothermic reaction occurs, which heats the solution. We are asked to determine the final temperature of the solution, Tf, and compare that with the result of Problem 9.113.

Analyze From Problem 9.113, we know that the molar ratio of H2SO4 to NaOH in the balanced equation is 1:2. We must first find out how much H2SO4 reacts by computing the moles of H2SO4 and NaOH. The q for the reaction then may be found by multiplying ∆Hrxn (–57.1 kJ/mol H2O) by the moles of H2O produced in the reaction. Then ∆T can be found from q = mcS∆T where m = total mass of the solution (165 g) and cS = 4.184 J/g ⋅ ˚C.

Solve Molar amounts of the reactants: 1.0 mol mol H SO= 65.0 mL×= 0.065 mol 24 1000 mL 1.0 mol mol NaOH= 100.0 mL×= 0.10 mol 1000 mL The limiting reactant here is NaOH, which would produce 0.10 mol of H2O. Because ∆Hrxn = –57.1 kJ/mol H2O, for that reaction –57.1 kJ q = × 0.10 mol = –5.7 kJ, or – 5700 J mol H2O 4.184 J q = 5700 J = 165 g × × ΔT g ⋅ C  ΔT = 8.3 C That increase in temperature is slightly less than that measured in Problem 9.113.

46 | Chapter 9

Think About It

The ΔHrxn in Problem 9.113 was defined per mole of H2O, so we needed to focus on that quantity in computing the value of q.

9.115. Collect and Organize When a hot sample of copper is dropped into water, energy is transferred from the metal to the water until thermal equilibrium is established.

Analyze The energy lost by the copper sample is

qlost = mCu × cS × (Tf – Ti )

where mCu = 7.25 g, cs = 0.385 J/g ⋅ ˚C, and Ti = 100.1˚C. The energy gained by the water is q n c T – T gained = H2O × P × ( f i ) where n = 50.0 g/(18.02 g/mol), c = 75.3 J/mol ˚C, and T = 25.0˚C. H2O P ⋅ i The law of conservation of energy gives –qlost,Cu = qgained,water

Solve

–qlost,Cu = qgained,water 0.385 J 1 mol 75.3 J –7.25 g × T −100.1C = 50.0 g × × × T – 25.0 C g ⋅ C ( f ) 18.02 g mol ⋅ C ( f ) J J –2.79  × Tf + 279 J = 209  × Tf − 5223 J C C J –211.8 × T = –5502 J C f T = 26.0 C f Think About It Because the heat capacity of the copper is so small and the heat capacity of water is so large, very little increase in the temperature of water occurs: only 1˚C.

9.116. Collect and Organize We are asked to relate the enthalpy of reaction to the balanced chemical equation.

Analyze We should be able to balance the reaction between FeO [iron(II) oxide] and O2 by inspection. Once balanced, we can use the stoichiometric relationships to determine that ∆Hrxn on the basis of ∆Hrxn for 1 mol of Fe3O4 being –318 kJ.

Solve 1 (a) 3 FeO(sg ) + 2 O234 ( ) → Fe O ( s ) Eliminating fractional coefficients gives

6 FeO(sg ) + O234 ( ) → 2 Fe O ( s ) (b) If ∆Hrxn = –318 kJ for 1 mol of Fe3O4, then for 2 mol of Fe3O4 in the balanced equation, ∆Hrxn = –636 kJ.

Think About It Remember that enthalpy is stoichiometric. The more magnetite you produce in that reaction, the greater the energy released.

9.117. Collect and Organize We are asked to calculate the standard heat of combustion of acetylene and determine its fuel value.

Thermochemistry | 47

Analyze (a) To calculate the molar enthalpy of combustion, we need to write a balanced chemical equation and then use ° the following equation to calculate ∆Hrxn ΔH  = n ΔH  – n ΔH  rxn ∑ products f,products ∑ reactants f,reactants (b) The fuel value is the energy released per gram of the fuel, so we will divide our answer in part a by the

molar mass of acetylene.

Solve (a) The balanced chemical equation for the combustion of 1 mol of C2H2 is 5 C22 H (gg )+2 O 2 ( ) → 2 CO 2 ( gg )+ H 2 O( ) The standard enthalpy for burning 1 mol of C2H2 is  ΔH rxn = ⎣⎡(2 mol CO2 × –393.5 kJ/mol) + (1 mol H2O × –241.8 kJ/mol)⎦⎤ 5 − ⎣⎡(1 mol C2H2 × 226.7 kJ/mol) + ( 2 mol O2 × 0.0 kJ/mol)⎦⎤ ΔH  = –1255.5 kJ/mol C H rxn 2 2 (b) The fuel value of acetylene is 1255.5 kJ 1 mol × = 48.22 kJ/g mol 26.037 g

Think About It Some other examples of endothermic compounds: Inorganic compounds = B2H6, CS2, HI, N2O, O3 Organic compounds = C2H4, C6H6, (CH3)2C== C(CH3)2

9.118. Collect and Organize We cannot compute the enthalpy for a reaction without the balanced chemical equation.

Analyze ° The chemical equation can be balanced by inspection. The ∆Hrxn can be computed from ΔH  = n ΔH  – n ΔH  rxn ∑ products f,products ∑ reactants f,reactants Solve The balanced chemical equation for the reaction of iron(II) oxide with oxygen to form iron(III) oxide is 1 2 FeO(sg ) + 2 O223 ( ) → Fe O ( s )

4 FeO(sg ) + O223 ( ) → 2 Fe O ( s ) ° We use the coefficients in the equation for ∆Hrxn:  ΔH rxn = [2 mol Fe2O3 × –824.2 kJ/mol]

− ⎣⎡(4 mol FeO × –271.9 kJ/mol) + (1 mol O2 × 0.0 kJ/mol)⎦⎤ ΔH  = –560.8 kJ rxn ° That enthalpy change is for the production of 2 mol of Fe2O3. For 1 mol, ∆Hrxn = –280.4 kJ.

Think About It Remember that the enthalpy for a particular reaction depends on the form of the balanced chemical equation.

9.119. Collect and Organize We are asked to determine, using average bond energy value, whether the formation of polyethylene from ethylene is endothermic, is exothermic, or has no change in enthalpy.

Analyze The enthalpy of a reaction as estimated from bond energies is given as HH + H Δrxn=∑∑ΔΔ bond breaking bond forming 48 | Chapter 9

where ∆Hbond breaking and ∆Hbond forming are average bond energies for the bonds in the reactants and products, respectively. In computing ∆Hbond breaking and ∆Hbond forming we have to take into account the number (or moles) of a particular type of bond that breaks or forms. We must also keep in mind that bond breaking requires energy (+∆H), whereas bond formation releases energy (–∆H).

Solve The number of C–H bonds does not change, so we need not consider those in the calculation, but rather we can focus on the change in the carbon–carbon bonds.

For the formation of polyethylene from ethylene, the heat of reaction is

n H2C CH2 CH2 CH2 n ΔHrxn = [(n × 614 kJ/mol)]+ [–(2n × 348 kJ/mol)] where we will consider n = 1 ∆Hrxn = –82 kJ/mol That reaction is exothermic.

Think About It That finding may be unexpected in that we are breaking a double bond, but remember: From that double bond we are forming two single bonds to each carbon atom. Two single bonds added together is stronger than a double bond.

9.120. Collect and Organize Given the balanced overall thermochemical equation in which methanol is oxidized to formic acid, we are to determine whether the reaction is exothermic or endothermic by checking the sign of the enthalpy of the reaction. We also are asked to calculate the energy change when 60.0 g of methanol is oxidized and then predict the magnitude of the enthalpy change for the conversion of methanol to formaldehyde, the first step in the reaction.

Analyze ° (a) ∆Hrxn is the enthalpy change under standard conditions to form 1 mol of the product (HCOOH). ΔH  = n ΔH  – n ΔH  rxn ∑ products f,products ∑ reactants f,reactants If energy (in kilojoules) is a reactant in the thermochemical equation, the reaction is endothermic; if energy is a product in the thermochemical equation, the reaction is exothermic. (b) The energy change in the thermochemical equation is for 2 mol of methanol consumed. We multiply that relationship by the moles of CH3OH in 60.0 g. (c) When methanol converts to formic acid in a two-step process, the intermediate formaldehyde has only one oxygen atom, whereas formic acid has two.

Solve (a) That reaction is exothermic because the enthalpy given is negative, showing that energy as heat is produced (as a “product”) in the thermochemical equation. (b) The moles of methanol in 60.0 g is 1 mol 60.0 g×= 1.87 mol 32.04 g For that number of moles, the enthalpy is −1019.5 kJ 1.87 mol × = −953 kJ 2 mol (c) We can expect that the first step that converts methanol to formaldehyde, where only one oxygen atom is present, would be less exothermic than the overall conversion of methanol to formic acid, so the enthalpy of reaction for the formation of formaldehyde from methanol is expected to be smaller (less exothermic) than the overall enthalpy of the reaction to form formic acid.

Thermochemistry | 49

Think About It That reaction, because it is exothermic, is favored by enthalpy.

9.121. Collect and Organize ° From the balanced equation and the ∆�!"#we are to determine whether energy is consumed or produced in the reaction and how much energy is involved when 60.0 g of CH3OH is used in the reaction.

Analyze (a) We can balance that reaction by inspection. For this reaction we have to also realize that oxygen, O2, is a product. (b) We are given that the reaction requires 164 kJ/mol (of methanol) of energy, so that is an endothermic reaction. (c) To calculate the energy needed to transform 60.0 g of methanol, we need only determine the moles of CH3OH in 60.0 g and then multiply that result by the energy required in the reaction for 1 mol of CH3OH (164 kJ/mol).

Solve 1 (a) CH3OH(g) + N2 (g) → HCN(g) + NH3(g) + 2 O2 (g) (b) Because that is an endothermic reaction, the energy is a reactant: 1 164 kJ++ CH32 OH(gg ) N ( ) → HCN( g )++ NH 32 ( g )2 O ( g ) (c) The moles of CH3OH used is 1 mol 60.0 g×= 1.87 mol 32.04 g The enthalpy for the reaction using that amount of CH3OH is 164 kJ 1.87 mol×= 307 kJ 1 mol Think About It The energy required or evolved in a reaction is an extensive property; it depends on the amount of reactants involved in the reaction.

9.122. Collect and Organize We combine three reactions to give the overall reaction and calculate the enthalpy for the formation of NiSO3 from nickel, sulfur, and oxygen.

Analyze Because NiSO3 is a product in the overall reaction, the first reaction will have to be reversed. Because S, O2, and Ni are reactants, neither reaction 2 nor reaction 3 needs to be changed.

Solve  NiO(s) + SO2 (g) → NiSO3(s) ΔH rxn = –156 kJ 1 S (s) + O (g) → SO (g) ΔH  = –297 kJ 8 8 2 2 rxn 1  Ni(s) + 2 O2 (g) → NiO(s) ΔH rxn = –241 kJ Ni(s) + 1 S (s) + 3 O (g) → NiSO (s) ΔH  = –694 kJ 8 8 2 2 3 rxn

Multiplying by 2 gives the desired equations: 2 Ni(s) + 1 S (s) + 3 O (g) → NiSO (s) ΔH ! = –1388 kJ 4 8 2 3 rxn

Think About It All the reactions added together are exothermic. The overall reaction, therefore, must also be exothermic. The overall enthalpy of reaction is also the enthalpy of formation of NiSO3.

50 | Chapter 9

9.123. Collect and Organize We combine three reactions to give the overall reaction for the formation of PbCO3.

Analyze Because PbCO3 is a product in the overall equation, reaction 3 has to be reversed. Neither reaction 1 nor reaction 2 needs to be changed.

Solve 1  Pb(s) + 2 O2 (g) → PbO(s) ΔH rxn = –219 kJ C(s) + O (g) → CO (g) ΔH  = –394 kJ 2 2 rxn  PbO(s) + CO2 (g) → PbCO3(s) ΔH rxn = –86 kJ Pb(s) + 3 O (g) + C(s) → PbCO (s) ΔH  = –699 kJ 2 2 3 rxn That is for 1 mol of C(s). For 2 mol of carbon in the target equation, the energy evolved in that reaction is −699 kJ 2 mol × = −1398 kJ mol

Think About It All the reactions added together are exothermic. The overall reaction, therefore, must also be exothermic. The overall enthalpy of reaction is also the enthalpy of formation of PbCO3.

9.124. Collect and Organize ° We are to use the ∆Hf for the reactants and products to calculate the enthalpy of reaction for the combustion of 1 mol of ethanol.

Analyze The balanced equation for the combustion of ethanol is

C2H5OH(ℓ) + 3 O2(g) → 2 CO2(g) + 3 H2O(g) To calculate the enthalpy of the combustion, use the equation ΔH  = n ΔH  – n ΔH  comb ∑ products f,products ∑ reactants f,reactants

Solve ! ΔH comb = [(2 mol CO2 × –393.5 kJ/mol) + (3 mol H2O × –241.8 kJ/mol)]

– [(1 mol C2H5OH × –277.7 kJ/mol) + (3 mol O2 × 0 kJ/mol)] ΔH ! = −1234.7 kJ comb

Think About It That value is for 1 mol of ethanol combusted because that is how we wrote the balanced reaction.

9.125. Collect and Organize ° ° Using the ∆Hf values for the reactants and products, we are asked to calculate the ∆Hrxn for the decomposition of 1 mol of NaHCO3.

Analyze The overall balanced decomposition reaction is 2 NaHCO3(s) → Na2CO3(s) + CO2(g) + H2O(g)

To calculate the enthalpy of the combustion, use the equation ΔH  = n ΔH  – n ΔH  rxn ∑ products f,products ∑ reactants f,reactants

Thermochemistry | 51

Solve  ΔH rxn = [(1 mol CO2 × –393.5 kJ/mol) + (1 mol H2O × –241.8 kJ/mol) + (1 mol Na2CO3 × –1130.7 kJ/mol)]

– [(2 mol NaHCO3 × –950.8 kJ/mol)] ΔH  = 135.6 kJ rxn For the formation of 1 mol of NaHCO3(s):

 135.6 kJ ΔH f = = 67.8 kJ/mol 2 mol

Think About It ° Remember that enthalpy is stoichiometric. You may have to adjust the ∆Hrxn according to what the problem asks for. Here the enthalpy of decomposition of 1 mol of NaHCO3 was specified.

9.126. Collect and Organize By multiplying the molar mass of various elements by their corresponding specific heats, we investigate the validity of the law of Dulong and Petit.

Analyze Let’s set up a table of the metals in Table 9.3 and compute the value of the multiplication of their molar mass and their specific heat values. We can then determine the average value of those results to predict the specific heat values for nickel and platinum.

Solve (a) The results for the metals in Table 9.3 are shown in the following table:

Metal Molar Mass (g/mol) Specific Heat (J/(g ⋅ ˚C) Molar Mass × Specific Heat Al 26.982 0.897 24.2 Cr 51.996 0.449 23.3 Cu 63.546 0.385 24.5 Au 196.97 0.129 25.4 Fe 55.845 0.450 25.1 Pb 207.2 0.129 26.7 Ag 196.97 0.233 25.1 Sn 118.71 0.227 32.1 Ti 47.867 0.523 25.0 Zn 65.38 0.387 25.3

Using the calculation for Au to determine the units gives 197.0 g 0.129 J × = 25.4 J/mol ⋅ C 1 mol g C ⋅ The average value for the fourth column values is 25.7. Because the values are all close to one another, the data for those metals support this law.

(b) The predicted value of the specific heat of nickel is 25.7 J/(mol ⋅˚C) = 0.438 J/g ⋅˚C 58.693 g/mol The predicted value of the specific heat of platinum is 25.7 J/(mol ⋅˚C) = 0.132 J/g ⋅˚C 195.08 g/mol

Think About It The specific heat of nickel listed in several reference tables is 0.445 J/g ⋅ ˚C, and the specific heat of platinum is listed as 0.133 J/g ⋅ ˚C. Our prediction for those metals on the basis of the Dulong and Petit law is quite good.

52 | Chapter 9

9.127. Collect and Organize ° ° Using ∆Hf values for urea, carbon dioxide, water, and ammonia, we calculate ∆Hrxn for the conversion of urea into ammonia.

Analyze ° The ∆Hrxn can be found from the enthalpy of formation values by using ΔH  = n ΔH  – n ΔH  rxn ∑ products f,products ∑ reactants f,reactants

Solve  ΔH rxn = [(1 mol CO2 × –412.9 kJ/mol) + (2 mol NH3 × –80.3 kJ/mol)]

– [(1 mol urea × –319.2 kJ/mol) + (1 mol H2O × –285.8 kJ/mol)] ΔH  = 31.5 kJ rxn

Think About It The overall reaction is slightly endothermic.

9.128. Collect and Organize We are to compare the fuel value and the fuel density of propane as a liquid and as a gas.

Analyze The fuel value is expressed in energy per gram (kilojoules per gram), whereas the fuel density is expressed as energy per volume (kilojoules per liter).

Solve The fuel value is not affected by the physical state of the propane because it is on a per-gram basis. However, because the fuel density is on a per-volume basis, liquid propane will have a higher fuel density than gaseous propane.

Think About It For purchasing and transport we are more likely to use fuels that are “denser” rather than those that might have a higher “value.”

9.129. Collect and Organize We are given balanced chemical reactions for the combustion of two fuels, dimethylhydrazine and hydrogen, along with their enthalpy of combustion values to help us determine which fuel releases more energy per pound.

Analyze The ∆H values are given on a per-mole-of-fuel basis. We can convert those to a per-gram basis by dividing the absolute value of the enthalpy of the combustion by the molar mass of the fuel. That energy per gram of fuel can then be converted into energy per pound by using the conversion 1 lb = 453.6 g.

Solve For dimethylhydrazine: 1694 kJ 1 mol 453.6 g ×× =12,790 kJ/lb mol 60.10 g 1 lb For hydrogen: 286 kJ 1 mol 453.6 g ×× =64,200 kJ/lb mol 2.02 g 1 lb Pound for pound, therefore, hydrogen is a better fuel.

Think About It Even though the combustion of hydrogen is less exothermic on a per-mole basis, its use as a fuel on a per- weight basis yields more energy than dimethylhydrazine. Thermochemistry | 53

9.130. Collect and Organize Given the combustion reactions in an automobile engine to produce NO and NO2, we are to add the equations ° and use Hess’s law to calculate ∆Hcomb for the overall reaction and determine whether that reaction is exothermic or endothermic.

Analyze Because 2 mol of NO is produced in the first reaction and is consumed in the second reaction, NO cancels and the reactions can be directly added.

Solve Adding the reactions gives ! N2 (g) + O2 (g) → 2 NO(g) ΔH comb = +180 kJ ! 2 NO(g) + O2 (g) → 2 NO2 (g) ΔH comb = –112 kJ N (g) + 2 O (g) → 2 NO (g) ΔH ! = 68 kJ 2 2 2 comb That reaction is endothermic under standard conditions.

Think About It Automobile engines, of course, run hot, so energy is available to produce NO2 as a product of combustion.