thermochemistry. copyright © cengage learning. all rights reserved.6 | 2 contents and concepts...

THERMOCHEMISTRY

Copyright © Cengage Learning. All rights reserved. 6 | 2

Contents and Concepts

Understanding Heats of ReactionThe first part of the chapter lays the groundworkfor understanding what we mean by heats of reaction.

1. Energy and Its Units2. Heat of Reaction3. Enthalpy and Enthalpy Changes4. Thermochemical Equations5. Applying Stoichiometry to Heats of Reaction6. Measuring Heats of Reaction


Using Heats of Reaction

Now that we understand the basic properties of heats of reaction and how to measure them, we can explore how to use them.

7. Hess’s Law

8. Standard Enthalpies of Formation

9. Fuels—Foods, Commercial Fuels, and Rocket Fuels


ThermodynamicsThe science of the relationship between heat and other forms of energy.

ThermochemistryAn area of thermodynamics that concerns the study of the heat absorbed or evolved by a chemical reaction.

5

First law of thermodynamics – energy can be converted from one form to another, but cannot be created or destroyed.

Esystem + Esurroundings = 0

or

Esystem = -Esurroundings

C3H8 + 5O2 3CO2 + 4H2O

Exothermic chemical reaction!

Chemical energy lost by combustion = Energy gained by the surroundingssystem surroundings


EnergyThe potential or capacity to move matter.

One form of energy can be converted to another form of energy: electromagnetic, mechanical, electrical, or chemical.

Next, we’ll study kinetic energy, potential energy, and internal energy.


Kinetic Energy, EK

The energy associated with an object by virtue of its motion.

m = mass (kg)

v = velocity (m/s)

2mvE2

1K


The SI unit of energy is the joule, J, pronounced “jewel.”

The calorie is a non-SI unit of energy commonly used by chemists. It was originally defined as the amount of energy required to raise the temperature of one gram of water by one degree Celsius.

The exact definition is given by the equation:

2

2

s

mkgJ

(exact)J4.184cal1

?


A person weighing 75.0 kg (165 lbs) runs a course at 1.78 m/s (4.00 mph). What is the person’s kinetic energy?

2

K s

m1.78kg)(75.0

2

1

E

m = 75.0 kgv = 1.78 m/s

EK = ½ mv2

figures)tsignifican(3

J119s

mkg119

2

2

K

E


Potential Energy, EP

The energy an object has by virtue of its position in a field of force, such as gravitaitonal, electric or magnetic field.

Gravitational potential energy is given by the equation

m = mass (kg)

g = gravitational constant (9.80 m/s2)

h = height (m)

mghE P


Internal Energy, UThe sum of the kinetic and potential energies of the particles making up a substance.


Law of Conservation of EnergyEnergy may be converted from one form to another, but the total quantity of energy remains constant.


Thermodynamic System

The substance under study in which a change occurs is called the thermodynamic system (or just system).

Thermodynamic Surroundings

Everything else in the vicinity is called the thermodynamic surroundings (or just the surroundings).


Heat, qThe energy that flows into or out of a system because of a difference in temperature between the thermodynamic system and its surroundings.

Heat flows spontaneously from a region of higher temperature to a region of lower temperature.

• q is defined as positive if heat is absorbed by the system (heat is added to the system)

• q is defined as negative if heat is evolved by a system (heat is subtracted from the system)


Heat of ReactionThe value of q required to return a system to the given temperature at the completion of the reaction (at a given temperature).


Endothermic ProcessA chemical reaction or process in which heat is absorbed by the system (q is positive). The reaction vessel will feel cool.

Exothermic ProcessA chemical reaction or process in which heat is evolved by the system (q is negative). The reaction vessel will feel warm.

17

Schematic of Exothermic and Endothermic Processes


In an endothermic reaction:

The reaction vessel cools.

Heat is absorbed.

Energy is added to the system.

q is positive.

In an exothermic reaction:

The reaction vessel warms.

Heat is evolved.

Energy is subtracted from the system.

q is negative.


Enthalpy, HAn extensive property of a substance that can be used to obtain the heat absorbed or evolved in a chemical reaction.

Extensive PropertyA property that depends on the amount of substance. Mass and volume are extensive properties.


A state function is a 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.


The altitude of a campsite is a state function.

It is independent of the path taken to reach it.


Enthalpy of Reaction The change in enthalpy for a reaction at a given temperature and pressure:

H = H(products) – H(reactants)

Note: means “change in.”

Enthalpy change is equal to the heat of reaction at constant pressure:

H = qP


The diagram illustrates the enthalpy change for the reaction

2Na(s) + 2H2O(l) 2NaOH(aq) + H2(g)

The reactants are at the top. The products are at the bottom. The products have less enthalpy than the reactants, so enthalpy is evolved as heat. The signs of both q and H are negative.


Thermochemical Equation The thermochemical equation is the 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.

For the reaction of sodium metal with water, the thermochemical equation is:

2Na(s) + 2H2O(l) 2NaOH(aq) + H2(g); H = –368.6 kJ

?


Sulfur, S8, burns in air to produce sulfur dioxide. The reaction evolves 9.31 kJ of heat per gram of sulfur at constant pressure. Write the thermochemical equation for this reaction.


S8(s) + 8O2(g) 8SO2(g)

We first write the balanced chemical equation:

Next, we convert the heat per gram to heat per mole.

Now we can write the thermochemical equation:

S8(s) + 8O2(g) 8SO2(g); H = –2.39 × 103 kJ

Note: The negative sign indicates that heat is evolved; the reaction is exothermic.

kJ102.39Δ

Smol1

Sg256.56

Sg1

kJ9.31Δ

3

8

8

8

H

H


Manipulating a Thermochemical Equation• When the equation is multiplied by a factor, the

value of H must be multiplied by the same factor.

• When a chemical equation is reversed, the sign of H is reversed.


Concept Check 6.3Natural gas consists primarily of methane, CH4. It is used in a process called steam reforming to prepare a gaseous mixture of carbon monoxide and hydrogen for industrial use.CH4(g) + H2O(g) CO(g) + 3H2(g); ΔH = 206 kJThe reverse reaction, the reaction of carbon monoxide and hydrogen, has been explored as a way to prepare methane (synthetic natural gas). Which of the following are exothermic? Of these, which one is the most exothermic?a. CH4(g) + H2O(g) CO(g) + 3H2(g)b. 2CH4(g) + 2H2O(g) 2CO(g) + 6H2(g)c. CO(g) + 3H2(g) CH4(g) + H2O(g)d. 2CO(g) + 6H2(g) 2CH4(g) + 2H2O(g)


a. CH4(g) + H2O(g) CO(g) + 3H2(g)

This reaction is identical to the given reaction. It is endothermic.

H = 206 kJ

CH4(g) + H2O(g) CO(g) + 3H2(g); H = 206 kJ


b. 2 CH4(g) + 2H2O(g) 2CO(g) + 6H2(g)

This reaction is double the given reaction. It is endothermic.

H = 412 kJ

CH4(g) + H2O(g) CO(g) + 3H2(g); H = 206 kJ


c. CO(g) + 3H2(g) CH4(g) + H2O(g)

This reaction is the reverse of the given reaction. It is exothermic.H = -206 kJ

CH4(g) + H2O(g) CO(g) + 3H2(g); H = 206 kJ


d. 2CO(g) + 6H2(g) 2CH4(g) + 2H2O(g)

This reaction is reverse and double the given reaction.

It is exothermic.H = -412 kJ

Equations c and d are exothermic.Equation d is the most exothermic reaction.

CH4(g) + H2O(g) CO(g) + 3H2(g); H = 206 kJ

?


When sulfur burns in air, the following reaction occurs:

S8(s) + 8O2(g) 8SO2(g);

H = – 2.39 x 103 kJ

Write the thermochemical equation for the dissociation of one mole of sulfur dioxide into its elements.


S8(s) + 8O2(g) 8SO2(g); H = –2.39 × 103 kJ

We want SO2 as a reactant, so we reverse the given reaction, changing the sign of H:

8SO2(g) S8(g) + 8O2(g) ; H = +2.39 × 103 kJ

We want only one mole SO2, so now we divide every coefficient and H by 8:

SO2(g) 1/8S8(g) + O2(g) ; H = +299 kJ


Applying Stoichiometry to Heats of Reaction

?


You burn 15.0 g sulfur in air. How much heat evolves from this amount of sulfur? The thermochemical equation is

S8(s) + 8O2(g) 8SO2(g); H = -2.39 × 103 kJ


S8(s) + 8O2(g) 8SO2(g); H = -2.39 103 kJ

Molar mass of S8 = 256.52 g

38

88 8

1mol S 2.39 10 kJ15.0 g S

256.5 g S 1mol Sq

q = –1.40 102 kJ

?


The daily energy requirement for a 20-year-old man weighing 67 kg is 1.3 104 kJ. For a 20-year-old woman weighing 58 kg, the daily requirement is

8.8 103 kJ. If all this energy were to be provided by the combustion of glucose, C6H12O6, how many grams of glucose would have to be consumed by the man and the woman per day?

C6H12O6(s) + 6O2(g) 6CO2(g) + 6H2O(l);H = −2.82 103 kJ


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

H = −2.82 x 103 kJ

4glucose 3

1mol glucose 180.2 g glucose1.3 10 kJ

2.82 10 kJ 1mol glucosem

For a 20-year-old man weighing 67 kg:

= 830 g glucose required(2 significant figures)

3glucose 3

1mol glucose 180.2 g glucose8.8 10 kJ

2.82 10 kJ 1mol glucosem

For a 20-year-old woman weighing 58 kg:

= 560 g glucose required(2 significant figures)


Measuring Heats of ReactionWe will first look at the heat needed to raise the temperature of a substance because this is the basis of our measurements of heats of reaction.


Heat Capacity, C, of a Sample of Substance The quantity of heat needed to raise the temperature of the sample of substance by one degree Celsius (or one Kelvin).

Molar Heat Capacity The heat capacity for one mole of substance.

Specific Heat Capacity, s (or specific heat)The quantity of heat needed to raise the temperature of one gram of substance by one degree Celsius (or one Kelvin) at constant pressure.


The heat required can be found by using the following equations.

Using heat capacity:

q = Ct

Using specific heat capacity:

q = s x m x t


A calorimeter is a device used to measure the heat absorbed or evolved during a physical or chemical change. Two examples are shown below.

?


A piece of zinc weighing 35.8 g was heated from 20.00°C to 28.00°C. How much heat was required? The specific heat of zinc is 0.388 J/(g°C).

m = 35.8 gs = 0.388 J/(g°C)t = 28.00°C – 20.00°C = 8.00°C

q = m s t

C8.00Cg

J0.388g35.8

q

q = 111 J(3 significant figures)

?


Nitromethane, CH3NO2, an organic solvent burns in oxygen according to the following reaction:

CH3NO2(g) + 3/4O2(g) CO2(g) + 3/2H2O(l) + 1/2N2(g)

You place 1.724 g of nitromethane in a calorimeter with oxygen and ignite it. The temperature of the calorimeter increases from 22.23°C to 28.81°C. The heat capacity of the calorimeter was determined to be 3.044 kJ/°C. Write the thermochemical equation for the reaction.


We first find the heat evolved for the 1.724 g of nitromethane, CH3NO2.

Now, covert that to the heat evolved per mole by using the molar mass of nitromethane, 61.04 g.

3 2rxn

3 2 3 2

20.03 kJ 61.04 g CH NO

1.724 g CH NO 1mol CH NOq

H = –709 kJ

kJ 20.03 C22.23C28.81C

kJ3.044

Δ

rxn

calrxn

q

tCq


We can now write the thermochemical equation:

CH3NO2(l) + ¾O2(g) CO2(g) + 3/2H2O(l) + ½N2(g); H = –709 kJ


Hess’s Law of Heat SummationFor 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.


Suppose we want H for the reaction

2C(graphite) + O2(g) 2CO(g)

It is difficult to measure directly.

However, two other reactions are known:

2CO2(g) 2CO(g) + O2(g); H = – 566.0 kJ

C(graphite) + O2(g) CO2(g); H = -393.5 kJ

In order for these to add to give the reaction we want, we must multiply the first reaction by 2.

Note that we also multiply H by 2.


2C(graphite) + O2(g) 2CO(g)

2C(graphite) + 2O2(g) 2CO2(g); H = -787.0 kJ

2CO2(g) 2CO(g) + O2(g); H = – 566.0 kJ

Now we can add the reactions and the H values.

2 C(graphite) + O2(g) 2 CO(g); H = –1353.0 kJ


Hsub = Hfus + Hvap

Concept Check 6.4The heat of fusion (also called heat of melting), ΔHfus, of ice is the enthalpy change for

H2O(s) H2O(l); ΔHfus

Similarly, the heat of vaporization, ΔHvap, of liquid water is the enthalpy change for

H2O(l) H2O(g); ΔHvap

How is the heat of sublimation, ΔHsub, the enthalpy change for the reaction

H2O(s) H2O(g); ΔHsub

related to ΔHfus and ΔHvap?

?


What is the enthalpy of reaction, H, for the reaction of calcium metal with water?

Ca(s) + 2H2O(l) Ca2+(aq) + 2OH−(aq) + H2(g)

This reaction occurs very slowly, so it is impractical to measure H directly. However, the following facts are known:

H+(aq) + OH-(aq) H2O(l); H = –55.9 kJCa(s) + 2H+(aq)

Ca2+(aq) + H2(g); H = –543.0 kJ



First, identify each reactant and product:

H+(aq) + OH-(aq) H2O(l); H = –55.9 kJ

Ca(s) + 2H+(aq) Ca2+(aq) + H2(g); H = –543.0 kJ

Each substance must be on the proper side. Ca(s), Ca2+(aq), and H2(g) are fine.H2O(l) should be a reactant.OH-(aq) should be a product.

Reversing the first reaction and changing the sign of its H accomplishes this.



H2O(l) H+(aq) + OH−(aq); H = +55.9 kJ


The coefficients must match those in the reaction we want.The coefficient on H2O and OH- should be 2. We multiply the first reaction and its H by 2 to accomplish this.



2H2O(l) 2H+(aq) + 2OH−(aq); H = +111.8 kJ


We can now add the equations and their H’s.Note that 2H+(aq) appears as both a

reactant and a product.



2H2O(l) 2H+(aq) + 2OH−(aq); H = +111.8 kJ


Ca(s) + 2H2O(l) Ca2+(aq) + 2OH−(aq) + H2(g); H = –431.2 kJ


Standard Enthalpies of FormationThe term standard state refers to the standard thermodynamic conditions chosen for substances when listing or comparing thermodynamic data: 1 atm pressure and the specified temperature (usually 25°C). These standard conditions are indicated with a degree sign (°).

When reactants in their standard states yield products in their standard states, the enthalpy of reaction is called the standard enthalpy of reaction, H°. (H° is read “delta H zero.”)


Elements can exist in more than one physical state, and some elements exist in more than one distinct form in the same physical state. For example, carbon can exist as graphite or as diamond; oxygen can exist as O2 or as O3 (ozone).

These different forms of an element in the same physical state are called allotropes.

The reference form is the most stable form of the element (both physical state and allotrope).


The standard enthalpy of formation, Hf°, is the enthalpy change for the formation of one mole of the substance from its elements in their reference forms and in their standard states. Hf° for an element in its reference and standard state is zero.

For example, the standard enthalpy of formation for liquid water is the enthalpy change for the reaction

H2(g) + 1/2O2(g) H2O(l)

Hf° = –285.8 kJ

Other Hf° values are given in Table 6.2 and Appendix C.


C(graphite) + 2Cl2(g) CCl4(l); Hf° = –135.4 kJ

We first identify each reactant and product from the reaction we want.

Standard enthalpies of formation can be used to calculate the standard enthalpy for a reaction, H°.

CH4(g) + 4Cl2(g) CCl4(l) + 4HCl(g); H° = ?

Table 6.2 shows the Hf° values:

1/2 H2(g) + 1/2 Cl2(g) HCl(g); Hf° = −92.3 kJ

CH4(g) C(graphite) + 2H2(g); Hf° = +74.9 kJ



Each needs to be on the correct side of the arrow and is. Next, we’ll check coefficients.




1/2 H2(g) + 1/2 Cl2(g) HCl(g); Hf° = −92.3 kJ




Cl2 and HCl need a coefficient of 4. Multiplying the second equation and its H by 4 does this.




1/2 H2(g) + 1/2 Cl2(g) HCl(g); Hf° = −92.3 kJ




Now, we can add the equations.




2H2(g) + 2Cl2(g) 4HCl(g); Hf° = −369.2 kJ







2H2(g) + 2Cl2(g) 4HCl(g); Hf° = −369.2 kJ

CH4(g) + 4Cl2(g) CCl4(g) + 4HCl(g); H° = –429.7 kJ


?


What is the heat of vaporization of methanol, CH3OH, at 25°C and 1 atm?

Use standard enthalpies of formation (Appendix C).


We want H° for the reaction:CH3OH(l) CH3OH(g)

mol

kJ200.7Δ:methanolgaseousFor

mol

kJ238.7Δ:methanolliquidFor

f

f

H

H

reactants

fproducts

freaction ΔΔΔ HnHnH

mol

kJ238.7 mol 1

mol

kJ200.7 mol 1Δ vapH

Hvap= +38.0 kJ

?


Methyl alcohol, CH3OH, is toxic because liver enzymes oxidize it to formaldehyde, HCHO, which can coagulate protein. Calculate Ho for the following reaction:

2CH3OH(aq) + O2(g) 2HCHO(aq) + 2H2O(l)

Standard enthalpies of formation, :ofΔH

CH3OH(aq): −245.9 kJ/molHCHO(aq): −150.2 kJ/molH2O(l): −285.8 kJ/mol


We want H° for the reaction:2CH3OH(aq) + O2(aq) 2HCHO(aq) + 2H2O(l)

reactants

fproducts

freaction ΔΔΔ HnHnH

oreacton

kJ kJΔ 2 mol 150.2 2 mol 285.8

mol mol

kJ kJ2 mol 245.9 1 mol 0

mol mol

H

kJ 491.8kJ 571.6kJ 300.4 Δ o

reaction H

kJ 491.8kJ 872.0Δ oreaction H

kJ 380.2Δ oreaction H


Foods fuels three needs of the body:

• They supply substances for the growth and repair of tissue.

• They supply substances for the synthesis of compounds used in the regulation processes.

• They supply energy.

Foods do this by a combustion process.


For glucose, a carbohydrate:

C6H12O6(s) + 6O2(g) 6CO2(g) + 6H2O(l); Hf° = –2803 kJ

For glycerol trimyristate, a fat:

C45H86O6(s) + 127/2O2(g) 45CO2(g) + 43H2O(l); Hf°= –27,820 kJ

The average value for carbohydrates is 4.0 kcal/g and for fats is 9.0 kcal/g.

thermochemistry. copyright © cengage learning. all rights reserved.6 | 2 contents and concepts...

Documents

mv2copyright cengage

form of energy

internal energy

forms of energy

epthe energy

ekthe energy

gravitational potential

study kinetic energy