Chemistry

CHEMISTRYTHIRD EDITIONGilbert | Kirss | Foster | Davies

© 2012 by W. W. Norton & Company

CHAPTER 5

Thermochemistry


Chapter Outline

5.1 Energy: Basic Concepts and Definitions» Work, Potential Energy, and Kinetic Energy» Kinetic Energy and Potential Energy at the

Molecular Level 5.2 Systems, Surroundings, and Energy Transfer 5.3 Enthalpy and Enthalpy Changes 5.4 Heating Curves and Heat Capacity 5.5 Calorimetry: Measuring Heat Capacity

and Calorimeter Constants 5.6 Enthalpies of Formation/Enthalpies of Reaction 5.7 Fuel Values and Food Values 5.8 Hess’s Law


Definitions

Thermodynamics:

• Transformation of energy from one form to another.

Thermochemistry:

• Energy in the form of heat consumed or produced by chemical reactions.

• 2H2(g) + O2(g) → 2H2O(l) + energy

Energy: Heat vs Work


Definitions (cont.)

Heat:• Energy transferred between objects because of a

difference in their temperatures.

Work: w = F × d• Work (w) is done when a force (F) moves an

object through a distance (d).


Work and Energy


Two Types of Energy

Potential: due to position or composition:

• Can be converted to work: PE = m × g × h

» m = mass, g = force of gravity, and h = vertical distance

• Chemical energy = a form of potential energy

Kinetic: due to motion of the object

• KE = 1/2 mu2 (m = mass, u = velocity)


Potential Energy: A State Function

Depends only on the difference between initial and final state of the system.• Independent of path between states.


The Nature of Energy

Law of Conservation of Energy:• Energy can be neither created nor destroyed.

• Can be converted from one form to another.

» Potential energy → kinetic energy

» Chemical energy → heat


Kinetic energy at the molecular level depends on:• Mass and velocity of the particle (KE = ½ mu2).

• Temperature» As T increases, molecular motion and KE increase.

Potential energy at molecular level:• Electrostatic attractions:

Where Eel is electrostatic energy, Q1 and Q2 are charges

separated by distance, d.

Energy at the Molecular Level

d

Q2 × Q1Eel


Electrostatic Potential Energy


Chapter Outline

5.1 Energy: Basic Concepts and Definitions 5.2 Systems, Surroundings, and Energy Transfer

» Isolated, Closed, and Open Systems» Exothermic and Endothermic Processes» Energy Units and P-V Work

5.3 Enthalpy and Enthalpy Changes 5.4 Heating Curves and Heat Capacity 5.5 Calorimetry: Measuring Heat Capacity



Terms Describing Energy Transfer

System: The part of the universe that is the focus of a thermodynamic study.

• Isolated / open / closed

Surroundings: Everything in the universe that is not part of the system.

Universe = System + Surroundings


Examples


Heat Flow

Exothermic process: Heat flows out of system to surroundings (q < 0).

Endothermic process: Heat flows into system from surroundings (q > 0).


Phase Changes and Heat FlowE

xothermi

cE

ndot

herm

ic


Energy and Phase Changes

Absorbed heat increases kinetic energy of molecules.

Loss of kinetic energy caused by release of heat by molecules.


Internal Energy

Internal energy of a system = sum of all KE and PE of all components of the system.

• Different types of molecular motion contribute to overall internal energy: (a) translational, (b) rotational, and (c) vibrational.


First Law of Thermodynamics

First Law of Thermodynamics = Law of Conservation of Energy!

Energy of the universe is constant!

• Universe = system + surroundings, so…

• energy gained or lost by a system must equal the energy lost or gained by surroundings.


Units of Energy

Calorie (cal):• The amount of heat necessary to raise the

temperature of 1 g of water 1°C.

Joule (J):• The SI unit of energy.

• 4.184 J = 1 cal.

Energy = heat and/or work (same units!).


Energy Flow Diagrams


Change in Internal Energy

ΔE = q + w:

• ΔE = change in system’s internal energy

• q = heat

• w = work

Work:

• w = −PΔV

» where P = pressure, ΔV = change in volume.

• Work done by the system is energy lost by the system (added to the surroundings), hence the negative sign.


Problem: Calculation of Work

Calculate the work in L∙atm and joules associated with the expansion of a gas in a cylinder from 54 L to 72 L at a constant external pressure of 18 atm. (Note: 1 L∙atm = 101.32 J)

• Collect and Organize: • Analyze:• Solve:• Think about It:


Chapter Outline

5.1 Energy: Basic Concepts and Definitions 5.2 Systems, Surroundings, and Energy Transfer 5.3 Enthalpy and Enthalpy Changes 5.4 Heating Curves and Heat Capacity 5.5 Calorimetry: Measuring Heat Capacity



Enthalpy and Change in Enthalpy

Enthalpy (H) = E + PV

Change in Enthalpy (ΔH) = ΔE + PΔV

ΔH = change in enthalpy; energy flows as heat at constant pressure:

ΔH = qP = ΔE + PΔV

ΔH > 0, Endothermic; ΔH < 0, Exothermic

Add subscripts to indicate ΔH for specific process or part of the universe; e.g., ΔHvap,ΔHrxn,ΔHsys.


Chapter Outline

5.1 Energy: Basic Concepts and Definitions 5.2 Systems, Surroundings, and Energy Transfer 5.3 Enthalpy and Enthalpy Changes 5.4 Heating Curves and Heat Capacity

» How does heat flow translate into changes in temperature or phase changes?

5.5 Calorimetry: Measuring Heat Capacity and Calorimeter Constants

5.6 Enthalpies of Formation/Enthalpies of Reaction 5.7 Fuel Values and Food Values 5.8 Hess’s Law


Heating Curves

Heat in → kinetic energy


Heating Curves

Heat in → phase change


Heat Capacities

Molar heat capacity (cp) is the heat required to raise the temperature of 1 mole of a substance by 1°C at constant pressure.

• q = ncpΔT (cp = J / (mol∙°C)

Specific heat (cs) is the heat required to raise the temperature of 1 gram of a substance by 1°C at constant pressure.

• q = ncsΔT (cs = J / (g∙°C)

Heat capacity (Cp) is the quantity of heat needed to raise the temperature of some specific object by 1°C at constant pressure.


Molar Heat Capacities


Heating Curves

heat in, q = ncp(g)ΔT

heat in, q = ncp(l)ΔT

heat in, q = ncp(s)ΔT


Calculating Energy Through a Change in State

Molar heat of fusion:

• ΔHfus = heat needed to convert 1 mole of a solid at its melting point to 1 mole of liquid.

• q = nΔHfus

Molar heat of vaporization:

• ΔHvap = heat needed to convert 1 mole of a liquid at its boiling point to 1 mole of vapor.

• q = nΔHvap


Heating Curves

heat in, q = nΔHvap

heat in, q = nΔHfus


Problem: Specific Heat Capacity

During a strenuous workout, a student generates 2000 kJ of heat energy. What mass of water would have to evaporate from the student’s skin to dissipate this much heat?

• Collect and Organize:• Analyze:• Solve:• Think about It:


Problem: Heat Gain vs Heat Loss

Exactly 10 mL of water at 25°C was added to a hot iron skillet. All of the water was converted into steam at 100°C. If the mass of the pan was 1.20 kg and the molar heat capacity of iron is 25.19 J/mol∙°C, what was the temperature change of the skillet?



Chapter Outline


and Calorimeter Constants» Determining Molar Heat Capacity and Specific Heat» Measuring Calorimeter Constants

5.6 Enthalpies of Formation/Enthalpies of Reaction 5.7 Fuel Values and Food Values 5.8 Hess’s Law


Calorimetry

Calorimetry = measurement of heat.

• Typically, measuring change in heat that accompanies a physical change or chemical process.

A calorimeter = device used to measure the absorption or release of heat by a physical or chemical process.

• A closed system! −qsystem = qcalorimeter


(a) 23.5 g of Al beads are heated to 100.0°C in boiling water.

(heat transferred from water to Al until the final temp of Al beads = 100.0°C)

Measuring Heat Capacity


Measuring Heat Capacity (cont.)

(b) The Al beads (at 100°C) are placed in a Styrofoam box calorimeter containing 103.0 g of water at 23.0°C.

(c) Heat is transferred from Al beads to water until they reach a thermal equilibrium at 26.0°C; heat lost by Al = heat gained by H2O.

−qaluminum = qwater

mAlcs(Al)ΔTal = mwatercs(water)ΔTwater


Measuring Heats of Reaction (ΔHrxn)

A bomb calorimeter is a constant-volume device used to measure the heat of a combustion reaction.

Heat produced by reaction = heat gained by calorimeter

ΔH = −qcal = −CcalΔT


Chapter Outline


and Calorimeter Constants 5.6 Enthalpies of Formation/Enthalpies of Reaction

» Standard Enthalpies of Formation» A Practical Application

5.7 Fuel Values and Food Values 5.8 Hess’s Law


Heat of Reaction

Heat of reaction:

• Also known as enthalpy of reaction (ΔHrxn).

• The heat absorbed or released by a chemical reaction.

Table 5.2B: Some Standard Enthalpies of Combustion (ΔHcomb°)


A Specific Enthalpy

The standard enthalpy of formation, ΔHf°:

• The enthalpy change of the formation reaction.

• A formation reaction is the process of forming 1 mole of a substance in its standard state from its component elements in their standard states.

• For example, formation reaction for water:

» H2(g) + ½ O2(g) → H2O(l)

» ΔHrxn = ΔHf°(H2O)

(The standard state of a substance is its most stable form under 1 bar pressure and at 25°C.)


Methods of Determining ΔHrxn

1. From calorimetry experiments:

• ΔHrxn = −Ccal ΔT

2. From enthalpies of formation:

• ΔHrxn° = ΣnpΔHf°(products) − ΣnrΔHf°(reactants)

• ΔHfvalues for substances in Appendix 4.

§ Using Hess’s Law (Section 5.8).


Problem: Using ΔHf°to Find ΔHrxn°

Use Table 5.2 to calculate an approximate enthalpy of reaction for

CH4(g) + 2O2(g) → CO2(g) + 2H2O(l).



Problem 2: Using ΔHf°to Find ΔHrxn°

One step in the production of nitric acid is the combustion of ammonia. Use data in Appendix 4 to calculate the enthalpy of this reaction:

4NH3(g) + 5O2(g) → 4NO(g) + 6H2O(g)



Chapter Outline


and Calorimeter Constants 5.6 Enthalpies of Formation/Enthalpies of Reaction 5.7 Fuel Values and Food Values

» Fuel Values vs. Fuel Densities» Food Value

5.8 Hess’s Law


Fuel Values

Amount of energy (in kJ/g) from combustion reaction.

• CH4(g) + 2O2(g) → CO2(g) + 2H2O(g)

• ΔHcomb = 802.3 kJ/mol

• Fuel value = (802.3 kJ/mol)∙(1 mol/16.04 g) = 50.02 kJ/g

Fuel Density:

• For liquid fuels, energy released in kJ/L.


Fuel Values (cont.)

CompoundMolecular Formula

Fuel Value (kJ/g)*

Methane CH4 50.0

Ethane C2H6 47.6

Propane C3H8 46.3

Butane C4H10 45.8

* Based on the formation of H2O (g)


Food Values

Amount of energy produced when food is burned completely: • Determined by bomb calorimetry. • Nutritional Calorie = 1 kcal = 4.184 kJ.

Food Category Food Value (Cal or kcal)

Food Value(kJ)

Proteins 4.0 16.7

Carbohydrates 4.0 16.7

Fats 9.0 37.7


Chapter Outline



» Calculating ΔHrxn by Combining Reactions


Hess’s Law

Hess’s Law of constant heat of summation:• The ΔH of a reaction that is the sum of two or more

reactions is equal to the sum of the ΔH values of the constituent reactions.

1. CH4(g) + H2O(g) → CO(g) + 3H2(g) ΔH1

2. CO(g) + 3H2(g) + H2O(g) → 4H2(g) + CO2(g) ΔH2

3. CH4(g) + 2H2O(g) → 4H2(g) + CO2(g) ΔH3

ΔH3 = ΔH1 + ΔH2


Hess’s Law (cont.)


Calculations Using Hess’s Law

1. If a reaction is reversed, ΔH sign changes.

N2(g) + O2(g) → 2NO(g) ΔH = 180 kJ

2NO(g) → N2(g) + O2(g) ΔH = −180 kJ

2. If the coefficients of a reaction are multiplied by an integer, ΔH is multiplied by that same integer.

6NO(g) → 3N2(g) + 3O2(g) ΔH = 3(−180 kJ)

ΔH = −540 kJ


Problem: Using Hess’s Law

Using the following data, calculate the enthalpy change for the reaction:

C2H4(g) + H2(g) → C2H6(g)

ΔHrxn

H2(g) + ½O2(g) → H2O(l) −285.8 kJC2H4(g) + 3O2(g) → 2H2O(l) + 2CO2(g) −1411 kJC2H6(g) + 7/2O2(g) → 3H2O(l) + 2CO2(g) −1560 kJ

• Collect and Organize: • Analyze: • Solve:• Think about It:

ChemTour: State Functions and Path Functions


This ChemTour defines and explores the difference between state and path functions using a travel analogy that leads into a discussion of energy, enthalpy, heat, and work.

Click here to launchthis ChemTour

ChemTour: Internal Energy


This ChemTour explores how energy is exchanged between a system and its surroundings as heat and/or work, and how this transfer in turn affects the internal energy (E) of a system. It includes Practice Exercises.


ChemTour: Pressure–Volume Work


An animated ChemTour of an internal combustion engine shows how a system undergoing an exothermic reaction can do work on its surroundings; students can explore the relationship among pressure, volume, and work. It includes Practice Exercises.


ChemTour: Heating Curves


In this ChemTour, students use interactive heating curve diagrams to explore phase changes, heat of fusion, and heat of vaporization. Macroscopic views of ice melting and water boiling are shown in sync with the appropriate sections of the heating curve. It includes Practice Exercises.


ChemTour: Calorimetry


This ChemTour demonstrates how a bomb calorimeter works and walks students through the equations used to solve calorimetry problems. It includes an interactive experiment.


ChemTour: Hess’s Law


This ChemTour explains Hess’s law of constant heat of summation using animated sample problems and step-by-step descriptions. It includes Practice Exercises.



Sample Exercise 5.1

Describe the flow of heat during the purification of water by distillation (Figure 5.13), identify the steps in the process as either endothermic or exothermic, and give the sign of q associated with each step. Consider the water being purified to be the system.


Sample Exercise 5.1 (cont.)

Collect and Organize: Since the water is the system, we must evaluate how the water gains or loses energy during distillation.

Analyze: In distillation, energy flows in three steps: (1) liquid water is heated to the boiling point and (2) vaporizes. (3) The vapors are cooled and condense as they pass through the condenser.



Solve:a. Energy flows from the surroundings (hot plate) to heat the impure water (the system) to its boiling point and then to vaporize it. Therefore, processes 1 and 2 are endothermic. The sign of q is positive for both. b. Because energy flows from the system (water vapor) into the surroundings (condenser walls), process 3 is exothermic. Therefore, the sign of q is negative.



Think about It: “Endothermic” means that energy is transferred from the surroundings into the system, the water in the distillation flask. When the water vapor is cooled in the condenser, energy flows from the vapor as it is converted from a gas to a liquid; the process is exothermic.


Sample Exercise 5.2

Figure 5.19 shows a simplified version of a piston and cylinder in an engine. Suppose combustion of fuel injected into the cylinder produces 155 J of energy. The hot gases in the cylinder expand, pushing the piston down. In doing so, the gases do 93 J of P–V work on the piston. If the system is the gases in the cylinder, what is the change in internal energy of the system?

Collect and Organize: The change in internal energy is related to the work done by a system or on a system and the heat gained or lost by the system (Equation 5.5). First we have to decide whether q and w are positive or negative according to the sign convention shown in Figure 5.18.



Analyze: The system (gases in the cylinder) absorbs energy, so q > 0, and the system does work on the surroundings (the piston), so w < 0.

Solve:

ΔE = q + w = (155 J) + (-93 J) = 62 J

Think about It: More energy enters the system (155 J) than leaves it (93 J), so a positive value of ΔE is reasonable.


Sample Exercise 5.3

A tank of compressed helium is used to inflate balloons for sale at a carnival on a day when the atmospheric pressure is 1.01 atm. If each balloon is inflated from an initial volume of 0.0 L to a final volume of 4.8 L, how much P–V work is done by 100 balloons on the surrounding atmosphere when they are inflated? The atmospheric pressure remains constant during the filling process.



Collect and Organize: Each of 100 balloons goes from empty (V = 0.0 L) to 4.8 L, which means ΔV = 4.8 L, and the atmospheric pressure P is constant at 1.01 atm. The identity of the gas used to fill the balloons doesn’t matter, because under normal conditions all gases behave the same way, regardless of their identity. Our task is to determine how much P–V work is done by the 100 balloons on the air that surrounds them.



Analyze: We focus on the work done on the atmosphere (the surroundings) by the system; the balloons and the helium they contain are the system.

Solve: The volume change (ΔV) as all the balloons are inflated is

100 balloons × 4.8 L/balloon = 480 LThe work (w) done by our system (the balloons) as they inflate against an external pressure of 1.01 atm is

Because the work is done by the system on its surroundings, the work is negative from the point of view of the system:



Think about It: The internal energy of the balloons (the system) decreases when they are inflated. Air injected into a balloon cools when the balloon expands and does work on the surroundings. The work is done by the system.


Sample Exercise 5.4

Between periods of a hockey game, an ice-refinishing machine spreads 855 L of water across the surface of a hockey rink.

a. If the system is the water, what is the sign of ΔHsys as the water freezes?

b. To freeze this volume of water at 0°C, what is the value of ΔHsys? The density of water is 1.00 g/mL.

Collect and Organize: The water from the ice-refinishing machine is identified as the system, so we can determine the sign of ΔH by determining whether heat transfer is into or out of the water. We are told the volume of the water and its density, so we can calculate the mass of water involved in the change of state from liquid water to ice.



Analyze: (a) Because the freezing takes place at constant pressure, ΔHsys = qP. (b) To calculate the amount of energy lost from the water as it freezes, we must convert 855 L of water into moles of water, because the conversion factor between the quantity of energy removed and the quantity of water that freezes is the enthalpy of solidification of water, ΔHsolid = -6.01 kJ/mol.



Solve:

a. For the water (the system) to freeze, energy must be removed from it. Therefore, ΔHsys must be a negative value.

b. We convert the volume of water into moles:

and then calculate ΔHsys:



Think about It: ΔH changes with pressure but the magnitude of the change is negligible for pressures near normal atmospheric pressure. The magnitude of the answer to part b is reasonable given the large amount of water that is frozen to refinish the rink’s ice.


Sample Exercise 5.5

Calculate the amount of energy required to raise the temperature of 237 g of solid ice from 0.0°C to 80.0°C. The molar heat of fusion (ΔHfus) of ice is 6.01 kJ/mol. The molar heat capacity of liquid water is 75.3 J/(mol . °C).

Collect and Organize: This problem refers to a process symbolized by segments and in Figure 5.21. The ice must first be melted, and we are given the molar heat of fusion of ice. The amount of energy required to heat the water formed when all the ice has been melted can be determined using the molar heat capacity of liquid water, and Equation 5.8. We can calculate the temperature change (ΔT) of the water from the initial and final temperatures.



Analyze: Four mathematical steps are required: (1) calculate the number of moles of ice; (2) determine the amount of energy required to melt the ice (3) calculate the amount of energy required to raise the liquid water temperature from 0.0°C to 80.0°C; (4) add the results of steps (2) and (3). We can calculate the number of moles in 237 g of water using the molar mass of water (M =18.02 g/mol). The water increases in temperature from 0.0°C to 80.0°C.

Solve:1. Calculate the number of moles of water:



2. Determine the amount of energy needed to melt the ice (Equation 5.9):

3. Determine the amount of energy needed to warm the water (Equation 5.8):

4. Add the results of parts 2 and 3:



Think about It: Using the definitions of molar heat of fusion and molar heat capacity, we can solve this exercise without referring back to Equations 5.8 and 5.9. Molar heat of fusion defines the amount of energy needed to melt 1 mole of ice. Multiplying that value (6.01 kJ/mol) by the number of moles (13.2 mol) gives us the energy required to melt the given amount of ice. By the same token, the molar heat capacity of water [75.3 J/(mol . °C )] defines the amount of energy needed to raise the temperature of 1 mole of liquid water by 1 °C.We know the number of moles (13.2 mol), and we know the number of degrees by which we want to raise the temperature (80.0 °C - 0.00°C = 80.0°C ); multiplying those factors together gives us the heat needed to raise the temperature of the water.


Sample Exercise 5.6

If you add 250.0 g of ice initially at -18.0°C to 237 g (1 cup) of freshly brewed tea initially at 100.0°C and the ice melts, what is the final temperature of the tea? Assume that the mixture is an isolated system (in an ideal insulated container) and that tea has the same molar heat capacity, density, and molar mass as water.

Collect and Organize: We know the mass of tea, its initial temperature, and the molar heat capacity of tea, for which we just use the cP value for water. The amount of energy released when the tea is cooled will be the same as the amount of energy gained by the ice and, once it is melted, the amount gained by the water that is formed as it warms to the final temperature. We know the amount of ice, its initial temperature, and the heat of fusion of ice. Our task is to find the final temperature of the tea ice–water mixture.



Analyze: Before solving this problem, we need to think about the changes that take place when the tea and ice come into contact. Assuming the ice melts completely, three heat transfers account for the energy lost by the tea:

1. q1: raising the temperature of the ice to 0.0°C

2. q2: melting the ice.

3. q3: bringing the mixture to the final temperature; the temperature of the water from the melted ice rises and the temperature of the tea (Tinitial = 100.0°C ) falls to the final temperature of the mixture (Tfinal = ?).

The energy gained by the ice equals the energy lost by the tea:

qice = -qtea

Based on our analysis, we know that

qice + q1 + q2 + q3



Solve: The energy lost by the tea as it cools from 100.0°C to Tfinal is, from Equation 5.8,

The transfer of this heat is responsible for the changes in the ice. We can treat the heat transfer from the hot tea to the ice in terms of the three processes we identified in Analyze. In step 1, the ice is warmed from -18.0°C (Tinitial) to 0.0°C (Tfinal in this step):



In step 2, the ice melts, requiring the absorption of energy:

In step 3, the water from the ice, initially at 0.0°C, warms to the final temperature (where ΔT = Tfinal - Tinitial = Tfinal - 0.0°C ):



The sum of the quantities of energy absorbed by the ice and the water from it during steps 1 through 3 must balance the energy lost by the tea:

Expressing all values in kilojoules:



and rearranging the terms to solve for Tfinal, we have

Think about It: This calculation was carried out assuming the system (the tea plus the ice) to be isolated and the vessel to be a perfect insulator. Our answer, therefore, is an “ideal” answer and reflects the coldest temperature we can expect the tea to reach. In the real world, the ice would absorb some energy from the surroundings (the container and the air) and the tea would lose some energy to the surroundings, so the final temperature of the beverage could be different from the ideal value we calculated.


Sample Exercise 5.7

Before we can determine the enthalpy change of a reaction run in a calorimeter, we must determine the heat capacity (the calorimeter constant) of the calorimeter. What is the calorimeter constant of a bomb calorimeter if burning 1.000 g of benzoic acid in it causes the temperature of the calorimeter to rise by 7.248°C ? The heat of combustion of benzoic acid is ΔHcomb = -26.38 kJ/g.

Collect and Organize: We are asked to find the calorimeter constant of a calorimeter. We are given data describing how much the temperature of the calorimeter rises when a known amount of benzoic acid is burned in it, and we are given the heat of combustion of benzoic acid.



Analyze: We need to determine the amount of energy required to raise the temperature of the calorimeter by 1°C. The heat capacity of a calorimeter can be calculated using Equation 5.13 and the knowledge that the combustion of 1.000 g of benzoic acid produces 26.38 kJ of heat.



Solve:

This calorimeter can now be used to determine the heat of combustion of any combustible material.



Think about It: The calorimeter constant is determined for a specific calorimeter. If anything changes—if the thermometer breaks and has to be replaced, or if the calorimeter loses any of the water it contains—a new constant must be determined.


Sample Exercise 5.8

Which of the following reactions are formation reactions at 25°C? For those that are not, explain why not.



Collect and Organize: We are given four balanced chemical equations and information about the standard states of specific reactants and products.

Analyze: For a reaction to be a formation reaction, it must meet the criteria that it produces 1 mole of a substance from its component elements in their standard states. Each reaction must therefore be evaluated for the quantity of product and for the state of each reactant and product.



Solve:a. The reaction shows 1 mole of water vapor formed from its constituent elements in their standard states. This is the formation reaction for H2O(g), and its enthalpy of reaction is correctly symbolized ΔH°f.

b. The reaction shows 1 mole of liquids methanol formed from its constituent elements in their standard states. This reaction therefore is a formation reaction, and its change in enthalpy is correctly symbolized ΔH°f.



c. This is not a formation reaction because the reactants are not elements in their standard states and because more than one product is formed. The change in enthalpy of this reaction would be symbolized ΔHcomb.

d. This is not a formation reaction because the product is 4 moles of POCl3. Note, however, that all of the constituent elements are in their standard states and that only one compound is formed. We could therefore call the heat of reaction (ΔHrxn) for this reaction a standard heat of reaction, which we indicate by adding a superscript degree to the symbol: ΔH°rxn. We could easily convert this into a formation reaction by dividing all the coefficients and the standard heat of reaction (ΔH°rxn) by 4. Once we do this, ΔH°rxn = ΔH°f.



Think about It: Just because we can write formation reactions for substances like methanol does not mean that anyone would ever use that reaction to make methanol. Remember that formation reactions are defined to provide a standard against which other reactions can be compared when their thermochemistry is evaluated.


Sample Exercise 5.9

Using the appropriate values from Table 5.2, calculate ΔH°rxn for the combustion of the fuel propane (C3H8) in air.

Collect and Organize: The reactants are propane and elemental oxygen, O2, and the products are carbon dioxide and water. Even though combustion is an exothermic reaction, we assume that water is produced as liquid, i.e., H2O(ℓ), so that all products and reactants are in their standard states. The heats of formation of propane, carbon dioxide, and H2O(ℓ) are given in Table 5.2. The heat offormation of the element O2 in its standard state is zero. Our task is to use the balanced equation and the data from Table 5.2 to calculate the heat of combustion.



Analyze: Equation 5.17 defines the relation between heats of formation of reactants and products and the standard enthalpy of reaction. We also need the balanced equation for combustion of propane:



Solve: Inserting ΔH°f values for the products [CO2(g) and H2O(ℓ)] and reactants [C3H8(g) and O2(g)] from Table 5.2 and the coefficients in the balanced chemical equation into Equation 5.17, we get



Think about It: The result of the calculation has a large negative value, which means that the combustion reaction is highly exothermic, as expected for a hydrocarbon fuel.


Sample Exercise 5.10

Most automobiles run on either gasoline or diesel fuel. Although both fuels are mixtures, the energy content in gasoline can be approximated by considering it to have the formula C9H20 (d = 0.718 g/mL), while diesel fuel may be considered to be C14H30 (d = 0.763 g/mL). Using these two formulas, compare (a) fuel value per gram of each fuel and (b) fuel density per liter of each fuel.



Collect and Organize: We are given representative chemical formulas for gasoline (C9H20) and diesel fuel (C14H30) and can calculate the molar mass for each fuel. Heat of combustion data for C9H20 and C14H30 are given in Table 5.2. Our task is to use these data and the respective molar masses to obtain the fuel values. We can then use the given density of each fuel to convert the fuel value from kJ/g to kJ/L.



Analyze: The enthalpies of combustion in Table 5.2 are given in kilojoules per mole, so to answer this question in terms of grams and liters, we need molar masses to convert moles to grams in part a and for part b we need density (d = m/V, grams per milliliter) to convert grams to liters. Taking C9H20 and C14H30 as the average molecules in regular gasoline and diesel fuel, respectively, we can determine their molar masses. Then the fuel value can be calculated from the heats of combustion. Using the densities given in the problem, we can calculate fuel densities in kJ/L.



Solve:a. Fuel values

b. Fuel densities



Think about It: We wouldn’t expect the fuel values of gasoline and diesel fuel to be very different, or otherwise there would be a strong preference for gasoline-fueled cars over diesel or vice versa. The values are similar to the fuel values of methane and propane calculated in the text, so these answers seem reasonable. In the United States, fuel for cars and trucks is bought by the gallon and its consumption is rated in miles per gallon, whereas in most countries it is bought by the liter (1 U.S. gal = 3.785 L). Because of the way we buy automobile fuels, it makes sense to compare fuel densities rather than fuel values. On either basis, diesel is a slightly inferior fuel compared to gasoline.



Glucose (C6H12O6) is a simple sugar formed by photosynthesis in plants. The complete combustion of 0.5763 g of glucose in a calorimeter (Ccalorimeter = 6.20 kJ/°C ) raises the temperature of the calorimeter by 1.45°C . What is the food value of glucose in Calories per gram?

Collect and Organize: We are asked to determine the food value of glucose, which means the energy given off when 1 g is burned. We have the mass of glucose burned and the calorimeter constant. We can relate the energy given off by the glucose to the energy gained by the calorimeter (Equation 5.15) and the energy given off by the glucose to the temperature change and the calorimeter constant (Equation 5.12).



Analyze: We use the data from the calorimetry experiment to determine how much heat in kilojoules is given off when the stated amount of glucose is burned. We can convert that quantity into Calories by using the conversion factor 1 Cal = 4.184 kJ.



Solve:

To convert this quantity of energy to a food value, we divide by the sample mass:

and then convert into Calories:



Think about It: One gram of a doughnut, which is mostly carbohydrates, and 1 g of glucose have about the same food value (19 kJ/g and 15.6 kJ/g, respectively), so the answer to this Sample Exercise seems reasonable.



Hydrocarbons burned in a limited supply of air may not burn completely, and CO(g) may be generated. One reason furnaces and hot-water heaters fueled by natural gas need to be vented is that incomplete combustion can produce toxic carbon monoxide:



Collect and Organize: We are given two reactions (B and C) with thermochemical data and a third (A) for which we are asked to find ΔH °comb. All the reactants and products of reaction A are present in reactions B and/or C.

Analyze: We can manipulate the equations for reactions B and C so that they sum to give the equation for which ΔH °comb is unknown. Then we can calculate this unknown value by applying Hess’s law. The reaction of interest (A) has methane on the reactant side. Because reaction B also has methane as a reactant, we can use B as written. Reaction A has CO as a product. Reaction C involves CO as a reactant, so we have to reverse C in order to get CO on the product side. Once we reverse C, we must change the sign of ΔH°comb. If the coefficients as given do not allow us to sum the two reactions to yield reaction A, we can multiply one or both reactions by other factors.



Solve: We start with reaction B as written and add the reverse of reaction C, remembering to change the sign of (C) ΔH°comb:

Because methane has a coefficient of 2 in reaction A, we multiply all the terms in reaction B, including ΔH°comb (B), by 2:



Because the carbon monoxide in reaction A has a coefficient of 2, we do not need to multiply reaction C by any factor. Now we add (2 × B) to the reverse of C and cancel out common terms:



Think about It: We used Hess’s law to calculate the heat of combustion of methane to make CO, which is impossible to achieve in an experiment because any CO produced can react with O2 to give CO2, which results in a mixture of CO and CO2 as a product. The answer of -1038 kJ is a little less negative than the heat of combustion for the production of CO2 from methane, so the answer is reasonable.


Clicker Questions Note

Note to Instructors

The following in-class questions are also applicable to Chapter 12, The Chemistry of Solids.


Clicker Question: Isothermal Expansion of Ideal Gas

An ideal gas in a sealed piston is allowed to expand isothermally (at a constant temperature) against a pressure of 1 atm. In what direction, if at all, does heat flow for this process? (Hint: E = q − PΔV)

A) into the system B) out of the system C) heat does not flow


Clicker Question: Isothermal Expansion of Ideal Gas

Consider the following arguments for each answer and vote:

A. When the gas expands isothermally, it does work without a decrease in its energy, so heat must flow into the system.

B. During the expansion, the gas pressure decreases, thereby releasing heat to the surroundings.

C. The fact that the process is isothermal means that heat does not flow.


Clicker Question: Adiabatic Compression of an Ideal Gas

An ideal gas in an insulated piston is compressed adiabatically (q = 0) by its surroundings. What can be said of the change in the temperature (ΔT) of the gas for this process?

A) ΔT > 0 B) ΔT = 0 C) ΔT < 0


Clicker Question: Adiabatic Compression of an Ideal Gas


A. The surroundings are doing work on the system, and no heat is flowing. Therefore, ΔE > 0 and ΔT > 0.

B. The volume of the gas decreases, but the pressure increases to keep the product of the pressure and volume constant. Therefore, the temperature is also constant.

C. The gas is being compressed to a more ordered state, which corresponds to a lower temperature.


Clicker Question: ΔT of a Released Rubber Band

Consider a stretched rubber band that is suddenly released. What can be said of the change in the temperature (ΔT) of the rubber band for this process?

A) ΔT > 0 B) ΔT = 0 C) ΔT < 0


Clicker Question: ΔT of a Released Rubber Band


A. The stretched rubber band is at a higher energy state than the unstretched rubber band. Releasing the stretched rubber band causes the energy to be released.

B. Because the recoil of the rubber band is rapid, this process is essentially adiabatic. Therefore, the temperature of the rubber band will not change.

C. As the rubber band contracts, it does work and its energy decreases, resulting in a decrease in its temperature.


Clicker Question: Specific Heat Capacity of Al and Fe

A 1.0 gram block of Al (cs = 0.9 J∙°C − 1·g − 1) at 100°C and a 1.0 gram block of Fe (cs = 0.4 J∙°C − 1∙g − 1) at 0 °C are added to 10 mL of water (cs = 4.2 J∙°C − 1∙g − 1) at 50°C. What will be the final temperature of the water?

A) < 50°C B) 50°C C) > 50°C


Clicker Question: Specific Heat Capacity of Al and Fe


A. The specific heat capacity of Fe(s) is smaller than that of Al(s), so heat from both the Al(s) and the water will be required to warm the Fe(s).

B. The average initial temperature of the three components is 50°C. Therefore, the final temperature of the water will be 50°C.

C. The specific heat capacity of Al(s) is greater than that of Fe(s), so the Al block at 100°C will heat the water more than the Fe block will cool it.


Clicker Question: Equilibrium Constantof Cyclooctatetraene

Cyclooctatetraene, C8H8, can undergo a transformation between two possible states, A and B, by rearranging its 4 double bonds.

Which of the following graphs depicts the dependence

of the equilibrium constant (K) on temperature for the

conversion from state A to state B?

A) B) C)


Clicker Question: Equilibrium Constantof Cyclooctatetraene


A. The intermediate is at a higher energy state than either states A or B, so at high temperatures, the reaction will favor the intermediate and K will decrease.

B. The enthalpies of formation for states A and B are equal, so ΔH° = 0 and K is not temperature dependent.

C. At high temperatures, the conversion from state A to state B will occur at a much faster rate, thus increasing the value of K.


Clicker Question: Enthalpies of N2, NH3, and N2O

The reaction of N2(g) and O2(g) to form N2O(g) is an endothermic process. The reaction of N2(g) and H2(g) to form NH3(g) is an exothermic process. Given this information, which of the following species has the lowest enthalpy of formation?

A) N2(g) B) NH3(g) C) N2O(g)


Clicker Question: Enthalpies of N2, NH3, and N2O


A. N2(g) is the elemental form of nitrogen, which by definition will have ΔHf° = 0, the lowest possible enthalpy of formation.

B. Since the formation reaction for N2O(g) is endothermic, it has a higher ΔHf° than N2(g), O2(g), and H2(g); NH3(g) has a negative ΔHf°, less than N2(g), O2(g), and H2(g).

C. The formation of an N—N double bond and an N—O double bond, as found in N2O, releases more energy than does the creation of 3 N—H bonds to form NH3(g).


Clicker Question: Polymerization of Ethylene

Which of the following is true of ΔH° for the polymerization of ethylene to form polyethylene? Note: the C—C single bond enthalpy is ~350 kJ/mole and the C—C double bond enthalpy is ~600 kJ/mole.

A) ΔH° > 0 B) ΔH° = 0 C) ΔH° < 0


Clicker Question: Polymerization of Ethylene


A. The energy required to break double bonds is more than the energy released by forming new single bonds.

B. The total number of C—C bonds (if we count double bonds twice) does not change with polymerization. Therefore, there can be no change in ΔH°.

C. For each C-C double bond that breaks (~600 kJ/mole), two single bonds form (2 × ~350 = ~700 kJ/mole).

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