CHAPTER 6 - Cengage · Web viewIn a blast furnace, solid iron(III) oxide, Fe2O3, is reduced by gaseous carbon monoxide, CO, to liquid iron and gaseous carbon dioxide, CO2. For each

CHAPTER 6

Thermochemistry

CHAPTER TERMS AND DEFINITIONSNumbers in parentheses after definitions give the text sections in which the terms are explained. Starred terms are italicized in the text. Where a term does not fall directly under a text section heading, additional information is given for you to locate it.

thermodynamics* science of the relationships between heat and other forms of energy (6.1, introductory section)

thermochemistry* one area of thermodynamics; study of the quantity of heat absorbed or evolved by chemical reactions (6.1, introductory section)

energy potential or capacity to move matter (6.1)

kinetic energy (Ek) energy associated with an object by virtue of its motion (6.1)

joule (J) SI unit of energy, kg ∙ m2/s2 (6.1)

watt* measure of quantity of energy used per unit time; 1 J/s (6.1)

calorie (cal) non-SI unit of energy commonly used by chemists; originally defined as the amount of energy required to raise the temperature of one gram of water by one degree Celsius; 1 cal = 4.184 J (exact definition) (6.1)

potential energy (Ep) an object’s energy because of its position in a field of force (6.1)

internal energy (U) sum of kinetic and potential energies of the particles making up a substance (6.1)

Etot* total energy of a substance; sum of the kinetic, potential, and internal energies of the substance (6.1)

law of conservation of energy energy may be converted from one form to another, but the total quantity of energy remains constant (6.1)

thermodynamic system (or system) substance or mixture of substances under study in which a physical or chemical change occurs (6.2)

surroundings everything in the vicinity of a thermodynamic system (6.2)

heat (q) energy that flows between system and surroundings because of a difference in temperature between the thermodynamic system and its surroundings (6.2)

thermal equilibrium* state in which energy does not flow as heat between system and surroundings; temperature equality (6.2)

heat of reaction value of q required, at a given temperature, to return a system to the given temperature at the completion of the reaction (6.2)

exothermic process chemical reaction or physical change in which heat is evolved (q is negative) (6.2)

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128 Chapter 6: Thermochemistry

endothermic process chemical reaction or physical change in which heat is absorbed (q is positive) (6.2)

qp* heat of reaction at constant pressure (6.3)

enthalpy (H) extensive property of a substance used to obtain the heat absorbed or evolved in a chemical reaction (6.3)

state function property of a system that depends only on its present state, which is determined by variables such as temperature and pressure and is independent of any previous history of the system (6.3)

enthalpy of reaction (ΔH) change in enthalpy for a reaction at a given temperature and pressure; equals the heat of reaction at constant pressure (6.3)

enthalpy diagram* pictorial representation of the enthalpy change for a reaction (6.3)

pressure–volume work* energy required by a system to change volume against the constant pressure of the atmosphere (6.3)

thermochemical equation chemical equation for a reaction (including phase labels) in which the equation is given a molar interpretation, and the enthalpy of reaction for these molar amounts is written directly after the equation (6.4)

heat capacity (C) quantity of heat needed to raise the temperature of the sample of substance one degree Celsius (or one kelvin) (6.6)

specific heat capacity (specific heat) quantity of heat required to raise the temperature of one gram of a substance by one degree Celsius (or one kelvin) at constant pressure (6.6)

calorimeter device used to measure the heat absorbed or evolved during a physical or chemical change (6.6)

bomb calorimeter* calorimeter used for reactions involving gases (6.6)

Hess’s law of heat summation for a chemical equation that can be written as the sum of two or more steps, the enthalpy change for the overall equation equals the sum of the enthalpy changes for the individual steps (6.7)

standard state standard thermodynamic conditions chosen for substances when listing or comparing thermodynamic data: 1 atm pressure and the specified temperature (usually 25°C) (6.8)

standard enthalpy of reaction (ΔH°)* enthalpy change for a reaction in which reactants in their standard states yield products in their standard states (6.8)

allotrope one of two or more distinct forms of an element in the same physical state (6.8)

reference form stablest form (physical state and allotrope) of the element under standard thermodynamic conditions (6.8)

standard enthalpy of formation (standard heat of formation) ( ) enthalpy change for the formation of one mole of a substance in its standard state from its elements in their reference forms and in their standard states (6.8)

fuel* any substance that is burned or similarly reacted to provide heat and other forms of energy (6.9)

CHAPTER DIAGNOSTIC TEST


Chapter 6: Thermochemistry 129

1. How much heat is produced when 8.95 g C2H5OH is burned in a constant-pressure system? The equation and enthalpy of reaction are

C2H5OH(l) + 3O2(g) 2CO2(g) + 3H2O(l) ΔH = 1.367 103 kJ

2. Calculate the enthalpy change at 298 K for the reaction

Ni(s) + 4CO(g) Ni(CO)4(g)

Heats of formation at 298 K are for CO = 110.5 kJ/mol

for Ni(CO)4 = 605 kJ/mol

3. Calculate ΔH° at 298 K for the reaction C (graphite) + CO2(g) 2CO(g)

using the following ΔH° data at 298 K:

H2(g) + CO(g) C (graphite) + H2O(g) ΔH° = 131.38 kJFeO(s) + H2(g) Fe(s) + H2O(g) ΔH° = 24.69 kJFeO(s) + CO(g) Fe(s) + CO2(g) ΔH° = 16.32 kJ

4. What is the kinetic energy of an oxygen molecule traveling at a speed of 479 m/s in a tank at 21°C? (Hint: Use Avogadro’s number and the molar mass of O2 to get the actual mass of an oxygen molecule.)

5. For the reaction H2S(g) + 4H2O2(l) H2SO4(l) + 4H2O(l), ΔH° is 1.186 103kJ. The enthalpy change per mole of H2O2 is

a. 1.301 102 kJ.

b. 4.742 103 kJ.

c. 2.965 102 kJ.

d. 4.741 103 kJ.

e. none of the above.

6. Indicate whether each of the following statements is true or false. If a statement is false, change it so that it is true.

a. The enthalpy of reaction is independent of the exact state of the reactants or products. True/False:________________________________________________________________

_________________________________________________________________________.

b. If the enthalpy of reaction for N2(g) + O2(g) 2NO(g) is 180.5 kJ, then the enthalpy of reaction for ½N2(g) + ½O2(g) NO(g) is 90.25 kJ. True/False:

_________________________________________________________________________

_________________________________________________________________________.

c. A calorimeter is a useful apparatus for determining heats of reaction. True/False:

_________________________________________________________________________

_________________________________________________________________________.



d. For reactions involving gases and carried out at constant pressure, ΔH = qp. True/False:

_________________________________________________________________________

_________________________________________________________________________.

e. Hess’s law permits the calculation of ΔH values for reactions from values for reactants and products and the calculation of ΔH values for hypothetical reactions. True/False: _______________________________________________________________

_________________________________________________________________________.

f. The value for an element is always zero. True/False: __________________________

_________________________________________________________________________.

7. Use heat of formation data in Appendix C in the text to calculate the enthalpy of the transition from the liquid to the gaseous state for 1 mole of HCN. Report the answer in kilojoules and kilocalories.

8. Determine the enthalpy of ionization for Cs(g) using all the thermodynamic data below. The equation is Cs(g) Cs+(g) + e(g).

F(g)+ e(g) F(g) ΔH = 336 kJ F2(g) 2F(g) ΔH = 158 kJ Cs(s) Cs(g) ΔH = 78 kJ

Cs(s) + ½F2(g) CsF(s) ΔH = 555 kJ CsF(s) Cs+(g) + F(g) ΔH = 757 kJ

9. When 2.89 g N2H4(g) is combusted in a constant-pressure calorimeter containing exactly 1000 g of water, a temperature increase of 6.68°C is observed. The heat capacity of the calorimeter is 1.00 kJ/°C, and the specific heat of water is 4.184J/(g∙°C). All products are gaseous. Determine the enthalpy change per mole of N2H4 combusted.

10. A 17.9-g sample of an unknown metal was heated to 48.31°C. It was then added to 28.05 g of water in an insulated cup. The water temperature rose from 21.04 to 23.98°C. What is the specific heat of the metal?

11. Ethane gas, C2H6, burns in oxygen to form carbon dioxide gas, CO2, and gaseous water. For each mole of ethane burned, 1.60 kJ of heat is evolved at constant pressure. Write the thermochemical equation for this reaction, including labels for the states of all reactants and products.

ANSWERS TO CHAPTER DIAGNOSTIC TESTIf you missed an answer, study the text section and problem-solving skill (PS Sk.) given in parentheses after the answer.

1. 2.66 102 kJ (6.5, PS Sk. 4)

2. 163 kJ (6.8, PS Sk. 9)

3. 172.39 kJ (6.7, PS Sk. 7)

4. 6.10 1021 J/molecule (6.1, PS Sk. 1)

5. e (6.4, PS Sk. 3)



6.

a. False. The enthalpy of reaction depends on the exact state of the reactants or products. (6.4)

b. True. (6.4, PS Sk. 3)

c. True. (6.6)

d. True. (6.3)

e. True. (6.7)

f. False. The value for an element is always zero when the state of the element is the form of the element that exists at standard conditions. (6.8)

7. 3.0 101 kJ; 7.2 kcal (6.8, PS Sk. 8)

8. 381 kJ (6.7, PS Sk. 7)

9. 383 kJ (6.6, PS Sk. 6)

10. 0.792 J/(g ∙ °C) (6.6, PS Sk. 5)

11. C2H6(g) + O2(g) 2CO2(g) + 3H2O(g), ΔH = 1.60 kJ (6.4, PS Sk. 2)

SUMMARY OF CHAPTER TOPICSStudents find thermochemistry one of the more difficult topics in chemistry. This is said not to scare you but to assure you that if you find yourself really scratching your head, you are not alone. It is also said to let you know that this material is going to take a great deal of time and study to master. Your text presents the subject quite well, but you probably will need to read it over several times before the concepts begin to make sense. One of the biggest stumbling blocks for students is the arithmetic signs (+ and ) that go with almost every term. If you memorize the sign conventions stressed in the text, you will find things considerably easier. If you are stuck in your thinking or in an exercise or problem, go back and review the sign conventions to see if your error is there.

6.1 Energy and Its Units

Learning Objectives Define energy, kinetic energy, potential energy, and internal energy.

Define the SI unit of energy joule, as well as the common unit of energy calorie.

Calculate the kinetic energy of a moving object. (Example 6.1)

State the law of conservation of energy.

Problem-Solving Skill 1. Calculating kinetic energy. Given the mass and speed of an object, calculate the kinetic energy

(Example 6.1).

Energy has many forms. Some of these are electromagnetic energy (such as light, heat, and x rays), electrical energy, sound energy, gravitational potential energy, elastic potential energy, and chemical potential energy. The forms of energy we will be concerned with in this course are electromagnetic energy, electrical energy, and chemical potential energy.



Exercise 6.1An electron, whose mass is 9.11 1031 kg, is accelerated by a positive charge to a speed of 5.0 106 m/s. What is the kinetic energy of the electron in joules? in calories?

Known: Ek = ½mv2; 1 J = 1 kg ∙ m2/s2; 1 cal = 4.184 J

Solution:

Ek = 9.11 10–31 kg (5.0 106 m/s)2

= 1.1 10–17 J

Ek = 1.14 10–17 J = 2.7 10–18 cal

6.2 Heat of ReactionHeat is a difficult term to understand. The word heat makes a good verb but a poor noun. Heat is energy in transit from a hotter object to a colder one. Objects do not possess heat. They possess energy that can be transferred as heat. Once the energy arrives at its destination, it is absorbed and is no longer called heat.

In an exothermic reaction, the reactants are always at a higher state of enthalpy than are the products. Heat is given off, and the heat of reaction is negative (). In an endothermic reaction, it is just the reverse. The reactants are at a lower state of enthalpy than are the products. Energy must be added to the reactants to get the products. Thus the heat of the reaction is positive (+). The following diagrams illustrate these concepts.

Exothermic Reaction Endothermic Reaction

Learning Objectives Define a thermodynamic system and its surroundings.

Define heat and heat of reaction.

Distinguish between an exothermic process and an endothermic process.

Exercise 6.2Ammonia burns in the presence of a platinum catalyst to give nitric oxide, NO.

4NH3(g) + 5O2(g) 4NO(g) + 6H2O(l)

In an experiment, 4 mol NH3 is burned and evolves 1170 kJ of heat. Is the reaction endothermic or exothermic? What is the value of q?

Wanted: whether reaction is endothermic or exothermic; value of q

Given: 1170 kJ of heat evolves.



Known: definitions of two terms; sign is + if heat is absorbed.

Solution: Reaction is exothermic; q = 1170 kJ.

6.3 Enthalpy and Enthalpy Change

Learning Objectives Define enthalpy and enthalpy of reaction.

Explain how the terms enthalpy of reaction and heat of reaction are related.

Explain how enthalpy and internal energy are related.

6.4 Thermochemical Equations

Learning Objectives Define a thermochemical equation.

Write a thermochemical equation given pertinent information. (Example 6.2)

Learn the two rules for manipulating (reversing and multiplying) thermochemical equations.

Manipulate a thermochemical equation using these rules. (Example 6.3)

Problem-Solving Skills2. Writing thermochemical equations. Given a chemical equation, states of substances, and the

quantity of heat absorbed or evolved for molar amounts, write the thermochemical equation (Example 6.2).

3. Manipulating thermochemical equations. Given a thermochemical equation, write the thermochemical equation for different multiples of the coefficients or for the reverse reaction (Example 6.3).

Exercise 6.3A propellant for rockets is obtained by mixing the liquids hydrazine, N2H4, and dinitrogen tetroxide, N2O4. These compounds react to give gaseous nitrogen, N2, and water vapor, evolving 1049 kJ of heat at constant pressure when 1mol N2O4 reacts. Write the thermochemical equation for this reaction.

Solution: The equation and heat of reaction are

2N2H4(l) + N2O4(l) 3N2(g) + 4H2O(g) ΔH = 1049 kJ

Exercise 6.4(a) Write the thermochemical equation for the reaction described in Exercise 6.3 for the case involving 1 mol N2H4. (b) Write the thermochemical equation for the reverse of the reaction described in Exercise 6.3.

a. Known: Each coefficient would be divided by 2, as would ΔHrxn.

Solution: N2H4(l) + N2O4(l) N2(g) + 2H2O(g) ΔH = 5.245 102 kJ

b. Solution: 3N2(g) + 4H2O(g) 2N2H4(l) + N2O4(l) ΔH = 1.049 103 kJ

Note that ΔH in this case is positive.



6.5 Applying Stoichiometry to Heats of Reaction

Learning Objective Calculate the heat absorbed or evolved from a reaction given its enthalpy of reaction and

mass of a reactant or product. (Example 6.4)

Problem-Solving Skill 4. Calculating the heat of reaction from the stoichiometry. Given the value of ΔH for a chemical

equation, calculate the heat of reaction for a given mass of reactant or product (Example 6.4).

Exercise 6.5How much heat evolves when 10.0 g of hydrazine reacts according to the reaction described in Exercise 6.3?

Wanted: ΔH per 10.0 g hydrazine

Given: 10.0 g hydrazine; reaction from Exercise 6.3

Known: ΔHrxn = 1.049 103 kJ per 2 mol hydrazine; formula = N2H4 = 32.0 g/mol.

Solution: Find the moles of N2H4 and then kilojoules:

10.0 g N2H4 = 1.64 102 kJ

A Chemist Looks at: Lucifers and Other Matches

Questions for Study1. What element must have been present in the mixture used in the first friction match?

2. What was the purpose of the glue in the early white phosphorus match?

3. Write the chemical reaction that occurs when the “strike anywhere” match ignites.

4. How do safety matches differ from “strike anywhere” matches?

Answers to Questions for Study1. Sulfur, because the ignition produced SO2.

2. The glue held the match mixture together and protected the white phosphorus from air.

3. P4S3(s) + 8O2(g) P4O10(s) + 3SO2(g)

4. With the safety match, the striking surface contains the red phosphorus, and the match head contains the oxidizing agent. In the “strike anywhere” match, the head contains both the phosphorus and the oxidizing agent that ignite from the friction of rubbing the match head against a surface.



6.6 Measuring Heats of Reaction

Learning Objectives Define heat capacity and specific heat.

Relate the heat absorbed or evolved to the specific heat, mass, and temperature change.

Calculate using this relation between heat and specific heat. (Example 6.5)

Define calorimeter.

Calculate the enthalpy of reaction from calorimetric data (its temperature change and heat capacity). (Example 6.6)

Problem-Solving Skills5. Relating heat and specific heat. Given any three of the quantities q, s, m, and Δt, calculate the

fourth one (Example 6.5).

6. Calculating ΔH from calorimetric data. Given the amounts of reactants and the temperature change of a calorimeter of specified heat capacity, calculate the heat of reaction (Example 6.6).

In calculations in this section, the temperature used is in °C. Note that whether you use the Celsius or Kelvin scale, the number of units of temperature change is the same because both scales use the same-sized unit.

When you calculate the temperature difference (Δt), be sure you always take tfinal tinitial (tf ti). This way the sign of the heat or enthalpy always will come out correctly.

Exercise 6.6Iron metal has a specific heat of 0.449 J/(g ∙ °C). How much heat is transferred to a 5.00-g piece of iron, initially at 20.0°C, when it is placed in a pot of boiling water? Assume that the temperature of the water is 100.0°C and that the water remains at this temperature, which is the final temperature of the iron.

Wanted: heat transferred (J)

Given: 5.00 g of iron; tinitial = 20.0°C; tfinal = 100.0°C; specific heat of iron is 0.449 J/(g ∙ °C).

Known: Heat transferred = specific heat mass temperature change

Solution: Heat transferred = 5.00 g 80.0°C = 1.80 102 J

Exercise 6.7Suppose that 33 mL of 1.20 M HCl is added to 42 mL of a solution containing excess sodium hydroxide, NaOH, in a coffee-cup calorimeter. The solution temperature, originally 25.0°C, rises to 31.8°C. Give the enthalpy change ΔH for the reaction

HCl(aq) + NaOH(aq) NaCl(aq) + H2O(l)

Express the answer as a thermochemical equation. For simplicity, assume that the heat capacity and the density of the final solution in the cup are those of water. (In more accurate work, these values must be determined.) Also assume that the total volume of the solution equals the sum of the volumes of HCl(aq) and NaOH(aq).

Wanted: ΔHrxn



Given: 33 mL 1.20 M HCl + 42 mL NaOH = 75 mL; ti = 25.0°C; tf = 31.8°C. Use density and C of water for calorimeter (cal).

Known: Mass = density volume; ΔHrxn = qrxn = CcalΔt; dwater = 1.0 g/mL; specific heat of water = 4.184 J/(g ∙ °C), so use 4.18 J/(g ∙ °C) for solution.

Solution: First, find Ccal as follows, using C = specific heat mass:

Mass soln = 75 mL = 75 g

C = 75 g = 3.14 102 J/°C

Then find ΔHrxn for 33 mL of 1.20 M HCl:

ΔHrxn = Ccal Δt = 3.14 102 (31.8 25.0) °C = 2.14 kJ

To find ΔHrxn, determine the moles HCl in 33 mL of 1.20 M HCl:

0.033 L soln = 0.0396 mol HCl

ΔHrxn = = 54 kJ/mol HCl

The thermochemical equation is

HCl(aq) + NaOH(aq) NaCl(aq) + H2O(l) ΔH = 54 kJ

6.7 Hess’s Law

Learning Objectives State Hess’s law of heat summation.

Apply Hess’s law to obtain the enthalpy change for one reaction from the enthalpy changes of a number of other reactions. (Example 6.7)

Problem-Solving Skill 7. Applying Hess’s law. Given a set of reactions with enthalpy changes, calculate ΔH for a reaction

obtained from these other reactions by using Hess’s law (Example 6.7).

It is important to understand that the enthalpies (H) of the reactants or products describe the state of a given reaction system, just as volumes (V) and temperatures (T) do. This means that the values of H depend only on the state of the chemical system, not on the history of how it got to that state. Thus we can use Hess’s law, combining as many reactions as needed, to come out with the specific ΔH of the reaction wanted. The physical states of the chemical species in the reaction are important to the energy of reaction. Be sure that you use the correct symbols and cancel them properly. Remember that the value given for an enthalpy change (ΔH) of a reaction is specific for that written reaction: for the physical states of each species and for the amounts of materials specified by the reaction.

Exercise 6.8Manganese metal can be obtained by reaction of manganese dioxide with aluminum.

4Al(s) + 3MnO2(s) 2Al2O3(s) + 3Mn(s)



What is ΔH for this reaction? Use the following data:

2Al(s) + O2(g) Al2O3(s) ΔH = 1676 kJ

Mn(s) + O2(g) MnO2(s) ΔH = 521 kJ

Wanted: ΔHrxn

Given: reactions and enthalpies

Known: Use Hess’s law. Reversing an equation changes the sign of ΔH; if you multiply coefficients by a factor, ΔHrxn must be multiplied by the same factor.

Solution: Write equation (1), multiplying coefficients and ΔH by 2; reverse equation (2), changing the sign of ΔH, and multiply the coefficients and ΔH by 3. Then add the equations and ΔH.

(1) 4Al(s) + 3O2(g) 2Al2O3(s) ΔH = 3352 kJ(2) 3MnO2(s) 3Mn(s) + 3O2(g) ΔH = 1563 kJ

4Al(s) + 3MnO2(s) 2Al2O3(s) + 3Mn(s) ΔH = 1789 kJ

6.8 Standard Enthalpies of Formation

Learning Objectives Define standard state and reference form.

Define standard enthalpy of formation.

Calculate the heat of a phase transition using standard enthalpies of formation for the different phases. (Example 6.8)

Calculate the heat (enthalpy) of reaction from the standard enthalpies of formation of the substances in the reaction. (Example 6.9)

Problem-Solving Skills8. Calculating the heat of phase transition from standard enthalpies of formation. Given a table

of standard enthalpies of formation, calculate the heat of phase transition (Example 6.8).

9. Calculating the enthalpy of reaction from standard enthalpies of formation. Given a table of standard enthalpies of formation, calculate the enthalpy of reaction (Example 6.9).

The summarizing statement to remember for calculating enthalpy changes for reactions is “the enthalpy change for a reaction at standard conditions is the sum of the enthalpies of formation of the products minus the sum of the enthalpies of formation of the reactants.” It is written

ΔH o = ΣnΔ (products) – ΣmΔ (reactants)

where the Greek letter Σ (capital sigma) indicates summation. Memorize this equation. Later, you will see other summation statements in similar form. As your text cautions, pay particular attention to signs. This will be your greatest source of error.



Exercise 6.9Calculate the heat of vaporization Δ of water using standard enthalpies of formation (text Table 6.2).

Known: ΔH° = ΣnΔ (products) – ΣmΔ (reactants)

Solution: Write the reaction. Then write the Δ values under each formula and subtract as indicated above.

H2O(l) H2O(g)

285.8 241.8 (kJ)

ΔH° = Δ = Δ [H2O(g)] Δ [H2O(l)]

= 241.8 kJ – (285.8 kJ)

= 44.0 kJ per mole of water vaporized

Exercise 6.10Calculate the enthalpy change for the following reaction:

3NO2(g) + H2O(l) 2HNO3(aq) + NO(g)

Use standard enthalpies of formation.


Solution: Write Δ under each species in the equation.

3NO2(g) + H2O(l) 2HNO3(aq) + NO(g)

3 33.1 285.8 2 207.4 90.3 (kJ)

Then use the ΔH equation:

ΔH° = ΣnΔ (products) – ΣmΔ (reactants)

= [2(207.4) + 90.3] – [3(33.1) + (–285.8)] kJ

= –138 kJ

You have now seen two methods for calculating the enthalpy change of a reaction: (1) by using Hess’s law and enthalpies of other reactions and (2) by using heats of formation of reactants and products.

Exercise 6.11Calculate the standard enthalpy change for the reaction of an aqueous solution of barium hydroxide, Ba(OH)2, with an aqueous solution of ammonium nitrate, NH4NO3, at 25°C. (Figure 6.1 illustrated this reaction using solids instead of solutions.) The complete ionic equation is

Ba2+(aq) + 2OH–(aq) + 2NH4+(aq) + 2NO3

–(aq) 2NH3(g) + 2H2O(l) + Ba2+(aq) + 2NO3–(aq)


Solution: Write Δ under each species in the net ionic equation:

2OH–(aq) + 2NH4+(aq) 2NH3(g) + 2H2O(l)

2 230.0 2 132.5 2 45.9 2 285.8 (kJ)



Then use the ΔH° equation:

ΔH° = (2{Δ [NH3(g)]} + {Δ [H2O(l)]})

– (2{Δ [OH–(aq)]} + {Δ [NH4+(aq)]})

= [2(45.9) + 2(285.8)] – [2(230.0) + 2(132.5)] kJ

= [(663.4) – (725.0)] kJ = 61.6 kJ

6.9 Fuels—Foods, Commercial Fuels, and Rocket Fuels

Learning Objectives Define fuel.

Describe the three needs of the body that are fulfilled by foods.

Give the approximate average values quoted (per gram) for the heat values (heats of combustion) for fats and for carbohydrates.

List the three major fossil fuels.

Describe the processes of coal gasification and coal liquefaction.

Describe some fuel–oxidizer systems used in rockets.

ADDITIONAL PROBLEMS1. When butane gas, C4H10, burns in oxygen gas, O2, to form carbon dioxide gas, CO2, and gaseous

water, 2845 kJ of thermal energy is evolved per mole of butane burned.

a. Write the thermochemical equation for the reaction using whole-number coefficients.

b. Calculate the thermal energy evolved when 25.0 g of butane is burned.

2. A 100.0-g sample of water at 25.30°C was placed in an insulated cup. Then 45.00 g of lead pellets at 100.00°C was added. The final temperature of the water was 34.34°C. What is the specific heat of lead?

3. A 23.3-g sample of copper at 75.7°C is added to 100.0 mL of benzene at 20.0°C. Assuming that no thermal energy is used in evaporation, calculate the final temperature of the benzene. [The specific heat of copper is 0.389 J/(g ∙ °C), the specific heat of benzene is 1.70 J/(g ∙ °C), and the density of benzene is 0.879 g/mL.] How would this value change if we took into account the energy that actually would go toward the increased evaporation?

4. A 0.875-g sample of anthracite coal was burned in a bomb calorimeter. The temperature rose from 22.50 to 23.80°C. The heat capacity of the calorimeter was found in another experiment to be 20.5 kJ/°C.

a. What was the heat evolved by the reaction?

b. What is the energy released on burning 1 metric ton (exactly 1000 kg) of this type of coal?



5. POCl3 is formed according to the following reaction:

PCl5(g) + H2O(g) POCl3(g) + 2HCl(g)

The enthalpy of reaction is 126 kJ per mole of POCl3 formed. Write the thermochemical equation for the reverse reaction, doubling the coefficients.

6. What is the ratio of speeds of H2 and N2 molecules at 125°C?

7. Write the equation that represents the heat of formation of each of the following species. Include states of matter.

a. Sb2O3(s)

b. C6H12O6(s)

c. N2H4(l)

d. SF6(g)

e. Ca(NO3)2(s)

8. Calculate ΔH° for the reaction

2NO(g) + 2CO(g) N2(g) + 2CO2(g)

given that Δ [NO(g)] = 90.3 kJ/mol, Δ [CO(g)] = 110.5 kJ/mol, and Δ [CO2(g)] = 393.5 kJ/mol.

9. Using Δ data given in text Table 6.2, calculate the amount of heat released from the combustion of a mixture of 15.0 g C5H12(l), n-pentane, and 66.6 g O2(g).

Δ [C5H12(l)] = 173.2 kJ/mol

10. Calculate the number of grams of natural gas at 25.0°C that are required to heat 1.00kg H2O from 25.0 to 100.0°C. Assume that the natural gas is 80.0% CH4 and 20.0% C2H6 by mass.

Δ [CH4(g)] = 74.9 kJ/mol Δ [C2H6(g)] = 84.7 kJ/mol,

and the specific heat of liquid water is 4.184 J/(g ∙ °C).

ANSWERS TO ADDITIONAL PROBLEMSIf you missed an answer, study the text section and problem-solving skill (PS Sk.) given in parentheses after the answer.

1.

a. 2C4H10(g) + 13O2(g) 8CO2(g) + 10H2O(g)

ΔH = 5.690 103 kJ (6.4, PS Sk. 2)

b. 25.0 g C4H10 = 1.22 103 kJ (6.5, PS Sk. 4)

2. ΔHlead = ΔHwater

Final temperature of both is the same.

–(45.0 g) (specific heat) (34.34 100.00)°C =

(100.0 g) [4.184 (J/g ∙ °C)] (34.34 – 25.30) °C



Specific heat = = 1.28 J/(g ∙ °C) (6.6, PS Sk. 5)

3. ΔHcopper = ΔHbenzene

–(23.3 g) [0.389 J/(g ∙ °C)] (tf 75.7°C)

= (100.0 mL 0.879 g/mL) [1.70 J/(g ∙ °C)] (tf 20.0°C)9.064tf + 686.1°C = 149.4tf 2989°C

tf = 23.2°C

The final temperature would be lower because not all the energy would go to raising the temperature of the benzene. (6.6)

4.

a. ΔHrxn = ΔHcal

ΔHcal = 20.5 kJ/°C (23.80 – 22.50) °C = 26.6 kJ

ΔHrxn = 26.6 kJ

b. 106 g = 3.04 107 kJ (6.6, PS Sk. 6)

5. 4HCl(g) + 2POCl3(g) 2H2O(g) + 2PCl5(g) ΔH = 252 kJ (6.4, PS Sk. 3)

6. The kinetic energies of gases are the same at the same temperature, so

Ek (N2) = Ek (H2)

=

=

and since we are interested in the mass ratio, we can use molar masses. Thus

= = = 3.72 (6.1)

7.

a. 2Sb(s) + O2(g) Sb2O3(s)

b. 6C (graphite) + 6H2(g) + 3O2(g) C6H12O6(s)

c. N2(g) + 2H2(g) N2H4(l)

d. S(s)+ 3F2(g) SF6(g)

e. Ca(s) + N2(g) + 3O2(g) Ca(NO3)2(s) (6.8)

8. ΔH° = ΣnΔ (products) – ΣmΔ (reactants)

= [0 + 2(393.5)] [2(90.3) + 2(110.5)]

= 787.0 (180.6 221.0) = 746.6 kJ (6.8, PS Sk. 9)



9. The balanced equation for the combustion of n-pentane is

C5H12(l) + 8O2(g) 5CO2(g) + 6H2O(l)

The calculation of Δ is

ΔH° = 5(393.5 kJ) + 6(285.8 kJ) (173.2 kJ) 8(0 kJ) = 3509.1 kJ

Find the limiting reagent:

15.0 g C5H12 = 1.040 mol CO2

66.6 g O2 = 1.301 mol CO2

C5H12 is the limiting reagent. The amount of heat released in the combustion of 0.208 mol C5H12 is

15.0 g C5H12 = 7.30 102 kJ (6.8, PS Sk. 4, 9)

10. Calculate the heats of combustion of CH4 and C2H6.

2O2(g) + CH4(g) CO2(g) + 2H2O(l)

ΔH°= 393.5 kJ + 2(285.8 kJ) (74.9 kJ) = 890.2 kJ

O2(g) + C2H6(g) 2CO2(g) + 3H2O(l)

ΔH° = 2(393.5 kJ) + 3(285.8 kJ) (84.68 kJ) = 1559.7 kJ

The amount of heat required to raise the temperature of 1.00 kg of liquid water

(100.0 25.0)°C = 75.0°C is

75.0°C 1.00 kg 4.184 J/(g ∙ °C) 1000 g/kg 1 kJ/1000 J = 3.138 102 kJ

The heat released from the combustion of 1 g of natural gas (0.800 g CH4 and 0.200 g C2H6) is

+ = 54.87 kJ/g gas

The amount of natural gas needed to provide 313.8 kJ of heat is

313.8 kJ = 5.72 g gas (6.6, 6.8, PS Sk. 5, 6, 9)

CHAPTER POST-TEST1. Calculate the heat produced when 3.76 g CO at 298 K reacts with an excess of Ni to give Ni(CO)4

in a constant-pressure system.

Ni(s) + 4CO(g) Ni(CO)4(g) ΔH° = 163 kJ



2. Calculate the enthalpy change at 298 K for the reaction

CaO(s) + 2HCl(g) CaCl2(s) + H2O(l)

from the following heats of formation at 298 K:

Δ ΔCaO 635.1 kJ/mol CaCl2 795.0 kJ/molHCl 92.31 kJ/mol H2O 258.8 kJ/mol

3. Determine Δ for CH4(g) from the following ΔH° data and corresponding reactions at 298 K:

C (graphite) + O2(g) CO2(g) ΔH° = 393.55 kJ CH4(g) + 2O2(g) CO2(g) + 2H2O(l) ΔH° = 890.36 kJ H2(g) + O2(g) H2O(l) ΔH° = 285.85 kJ

4. If we consider a hot kitchen stove as a system, then the transfer of heat from the stove to the kitchen is taken to be ___________________________.

(positive/negative)5. Which of the following is false? For the false statement, change it so it is true.

a. If qp for a reaction is negative, the reaction is endothermic. _________________________

_________________________________________________________________________

_________________________________________________________________________

b. ΔH is the amount of heat released or absorbed by a system at constant pressure.

_________________________________________________________________________

c. If heat is evolved in a reaction, the reaction is exothermic. __________________________

_________________________________________________________________________

d. ΔH = H (products) – H (reactants). _____________________________________________

_________________________________________________________________________

e. Enthalpy is a state function. __________________________________________________

_________________________________________________________________________

6. The enthalpy change for a reaction system open to the atmosphere is

a. dependent on the identity of the reactants and products, and/or

b. zero, and/or

c. negative, and/or

d. qp, and/or

e. positive.



7. The following thermochemical equation for the decomposition of gaseous ammonia, NH3,

NH3(g) N2(g) + H2(g) ΔH = 45.9 kJ

indicates that the formation of gaseous ammonia

a. evolves 45.9 kJ for each mole of ammonia formed, and/or

b. evolves 23 kJ for each mole of nitrogen used, and/or

c. absorbs 45.9 kJ for each mole of ammonia formed, and/or

d. absorbs 23 kJ for each mole of nitrogen used, and/or

e. is an exothermic process.

8. The following reaction is carried out in a bomb calorimeter:

CaO(s) + SO3(g) CaSO4(s)

If 1.00 mol SO3 is reacted with an excess of CaO, 2.000 103 g of water increases in temperature from 17.5 to 55.5°C. The calorimeter has a measured heat capacity of 0.126 kJ/°C. Calculate the heat released in this reaction. [Specific heat of water is 4.184 J/(g ∙ °C).]

9. What is the speed of a CCl4 molecule at 22°C when its kinetic energy is 6.13 10–21 J/molecule?

10. SiH4(g) reacts with gaseous oxygen to form SiO2(s) and liquid water. How much heat is liberated when 5.75 g SiH4 burns in an excess of oxygen at 298 K?

Δ [SiH4(g)] = +34.3 kJ/mol.

11. In a blast furnace, solid iron(III) oxide, Fe2O3, is reduced by gaseous carbon monoxide, CO, to liquid iron and gaseous carbon dioxide, CO2. For each mole of iron oxide reduced, 27.6 kJ of heat is evolved. Write the complete equation for this reaction, including the heat of reaction and labels for the states of all reactants and products.

12. A 48.0-g sample of copper was heated to 98.5°C. It was then added to 105 g of water in an insulated container. The temperature of the water rose from 25.6 to 28.8°C. What is the specific heat of copper?

ANSWERS TO CHAPTER POST-TESTIf you missed an answer, study the text section and problem-solving skill (PS Sk.) given in parentheses after the answer.

1. 5.47 kJ (6.5, PS Sk. 4)

2. 261.1 kJ (6.8, PS Sk. 9)

3. 74.89 kJ/mol (6.7, 6.8, PS Sk. 7)

4. negative (6.2)

5.

a. If qp for a reaction is negative, the reaction is exothermic. (6.2)

6. a and d (6.3)

7. a and e (6.4, PS Sk. 3)

8. 323 kJ released (6.6, PS Sk. 6)



9. 219 m/s (6.1, PS Sk. 1)

10. 272 kJ (6.5, 6.8, PS Sk. 4, 9)

11. Fe2O3(s) + 3CO(g) 2Fe(l) + 3CO2(g) ΔH = 27.6 kJ (6.4, PS Sk. 2)

12. 0.42 J/(g ∙ °C) (6.6, PS Sk. 5, 6)


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CHAPTER 6 - Cengage · Web viewIn a blast furnace, solid iron(III) oxide, Fe2O3, is reduced by gaseous carbon monoxide, CO, to liquid iron and gaseous carbon dioxide, CO2. For each