Chapter 6 Thermochemistry. Copyright © Cengage Learning. All rights reserved 2 Capacity to do work...

Chapter 6

Thermochemistry

Copyright © Cengage Learning. All rights reserved 2

• Capacity to do work or to produce heat.• Law of conservation of energy – energy can

be converted from one form to another but can be neither created nor destroyed.– The total energy content of the universe is

constant.

Energy

Energy is the capacity to do work

• Radiant energy comes from the sun and is earth’s primary energy source

• Thermal energy is the energy associated with the random motion of atoms and molecules

• Chemical energy is the energy stored within the bonds of chemical substances

• Nuclear energy is the energy stored within the collection of neutrons and protons in the atom

• Potential energy is the energy available by virtue of an object’s position

Heat is the transfer of thermal energy between two bodies that are at different temperatures.

Energy Changes in Chemical Reactions

Temperature is a measure of the thermal energy.

900C400C

greater thermal energy

Temperature = Thermal Energy


• Potential energy – energy due to position or composition.

• Kinetic energy – energy due to motion of the object and depends on the mass of the object and its velocity.

Energy


• In the initial position, ball A has a higher potential energy than ball B.

Initial Position


• After A has rolled down the hill, the potential energy lost by A has been converted to random motions of the components of the hill (frictional heating) and to the increase in the potential energy of B.

Final Position


• Heat involves the transfer of energy between two objects due to a temperature difference.

• Work – force acting over a distance.• Energy is a state function; work and heat are not

– State Function – property that does not depend in any way on the system’s past or future (only depends on present state).

Energy

Thermodynamics is the scientific study of the interconversion of heat and other kinds of energy.

State functions are properties that are determined by the state of the system, regardless of how that condition was achieved.

Potential energy of hiker 1 and hiker 2 is the same even though they took different paths.

energy, pressure, volume, temperature

E = Efinal - Einitial

P = Pfinal - Pinitial

V = Vfinal - Vinitial

T = Tfinal - Tinitial


• System – part of the universe on which we wish to focus attention.

• Surroundings – include everything else in the universe.

Chemical Energy

Thermochemistry is the study of heat change in chemical reactions.

The system is the specific part of the universe that is of interest in the study.

open

mass & energyExchange:

closed

energy

isolated

nothing


• Endothermic Reaction:– Heat flow is into a system.– Absorb energy from the surroundings.

• Exothermic Reaction:– Energy flows out of the system.

• Energy gained by the surroundings must be equal to the energy lost by the system.

Chemical Energy

Exothermic process is any process that gives off heat – transfers thermal energy from the system to the surroundings.

Endothermic process is any process in which heat has to be supplied to the system from the surroundings.

2H2 (g) + O2 (g) 2H2O (l) + energy

H2O (g) H2O (l) + energy

energy + 2HgO (s) 2Hg (l) + O2 (g)

energy + H2O (s) H2O (l)


Is the freezing of water an endothermic or exothermic process? Explain.

CONCEPT CHECK!CONCEPT CHECK!

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15

Classify each process as exothermic or endothermic. Explain. The system is underlined in each example.

a) Your hand gets cold when you touch ice.

b) The ice gets warmer when you touch it.c) Water boils in a kettle being heated on a

stove.d) Water vapor condenses on a cold pipe.e) Ice cream melts.

Exo

EndoEndo

ExoEndo



For each of the following, define a system and its surroundings and give the direction of energy transfer.

a) Methane is burning in a Bunsen burner in a laboratory.

b) Water drops, sitting on your skin after swimming, evaporate.



Hydrogen gas and oxygen gas react violently to form water. Explain.

– Which is lower in energy: a mixture of hydrogen and oxygen gases, or water?


Thermodynamics

• The study of energy and its interconversions is called thermodynamics.

• Law of conservation of energy is often called the first law of thermodynamics.


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

Thermochemistry• Thermodynamics is the science of the

relationship between heat and other forms of energy.

Thermochemistry is the study of the quantity of heat absorbed or evolved by chemical reactions.

First Law of Thermodynamics• To state the laws of thermodynamics, we

must first understand the internal energy of a system and how you can change it.

Heat is energy that moves into or out of a system because of a temperature difference between system and surroundings.

Work, on the other hand, is the energy exchange that results when a force F moves an object through a distance d; work (w) = Fd

First Law of Thermodynamics

• To state the laws of thermodynamics, we must first understand the internal energy of a system and how you can change it.

Remembering our sign convention.

Work done by the system is negative.Work done on the system is positive.Heat evolved by the system is negative.Heat absorbed by the system is positive.


• Internal energy E of a system is the sum of the kinetic and potential energies of all the “particles” in the system.

• To change the internal energy of a system:ΔE = q + w

q represents heatw represents work

Internal Energy

Work vs Energy Flow


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• Thermodynamic quantities consist of two parts:– Number gives the magnitude of the change.– Sign indicates the direction of the flow.

REMEMBER

Internal Energy

• Sign reflects the system’s point of view.• Endothermic Process:

– q is positive• Exothermic Process:

– q is negative



• Sign reflects the system’s point of view.• System does work on surroundings:

– w is negative• Surroundings do work on the system:

– w is positive

Internal Energy

28

• Work = P × A × Δh = PΔV– P is pressure.– A is area.– Δh is the piston moving a

distance.– ΔV is the change in volume.

Work


• For an expanding gas, ΔV is a positive quantity because the volume is increasing. Thus ΔV and w must have opposite signs:

w = –PΔV • To convert between L·atm and Joules, use 1

L·atm = 101.3 J.

Work

A sample of nitrogen gas expands in volume from 1.6 L to 5.4 L at constant temperature. What is the work done in joules if the gas expands (a) against a vacuum and (b) against a constant pressure of 3.7 atm?

w = -P V

(a) V = 5.4 L – 1.6 L = 3.8 L P = 0 atm

W = -0 atm x 3.8 L = 0 L•atm = 0 joules

(b) V = 5.4 L – 1.6 L = 3.8 L P = 3.7 atm

w = -3.7 atm x 3.8 L = -14.1 L•atm

w = -14.1 L•atm x 101.3 J1L•atm

= -1430 J


Which of the following performs more work?

a) A gas expanding against a pressure of 2 atm from 1.0 L to 4.0 L.

b) A gas expanding against a pressure of 3 atm from 1.0 L to 3.0 L.

They perform the same amount of work.

EXERCISE!EXERCISE!

Change in Enthalpy

• State function• ΔH = q at constant pressure• ΔH = Hproducts – Hreactants


Heat of Reaction

• Heat is defined as the energy that flows into or out of a system because of a difference in temperature between the system and its surroundings.

Heat flows from a region of higher temperature to one of lower temperature; once the temperatures become equal, heat flow stops.

An extensive property is one that depends on the quantity of substance. Enthalpy is a state function, a property of a system that depends only on its present state and is independent of any previous history of the system.

Enthalpy and Enthalpy Change• Enthalpy, denoted H, is an extensive property

of a substance that can be used to obtain the heat absorbed or evolved in a chemical reaction.

Enthalpy (H) is used to quantify the heat flow into or out of a system in a process that occurs at constant pressure.

H = H (products) – H (reactants)

H = heat given off or absorbed during a reaction at constant pressure

Hproducts < Hreactants

H < 0Hproducts > Hreactants

H > 0

Consider the combustion of propane:

C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(l)

ΔH = –2221 kJ

Assume that all of the heat comes from the combustion of propane. Calculate ΔH in which 5.00 g of propane is burned in excess oxygen at constant pressure.

–252 kJ


EXERCISE!EXERCISE!

Thermochemical Equations• A 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.

kJ -91.8H );g(NH2)g(H3)g(N 322

kJ 483.7- H ; )g(OH2)g(O)g(H2 o222

Thermochemical Equations

• In a thermochemical equation it is important to note phase labels because the enthalpy change, H, depends on the phase of the substances.

kJ 571.7- H ; )l(OH2)g(O)g(H2 o222

Applying Stoichiometry and Heats of Reactions

• Consider the reaction of methane, CH4, burning in the presence of oxygen at constant pressure. Given the following equation, how much heat could be obtained by the combustion of 10.0 grams CH4?

4

4

CH mol 1kJ 3.890

g0.16CH mol 1

4CH g 0.10 kJ 556

);l(OH2)g(CO)g(O2)g(CH 2224 kJ -890.3Ho

Calorimetry

• Science of measuring heat• Specific heat capacity:

– The energy required to raise the temperature of one gram of a substance by one degree Celsius.

• Molar heat capacity:– The energy required to raise the temperature

of one mole of substance by one degree Celsius.


The specific heat (s) of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius.

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius.

C = m x s

Heat (q) absorbed or released:

q = m x s x t

q = C x t

t = tfinal - tinitial

Calorimetry

• If two reactants at the same temperature are mixed and the resulting solution gets warmer, this means the reaction taking place is exothermic.

• An endothermic reaction cools the solution.


How much heat is given off when an 869 g iron bar cools from 940C to 50C?

s of Fe = 0.444 J/g • 0C

t = tfinal – tinitial = 50C – 940C = -890C

q = mst = 869 g x 0.444 J/g • 0C x –890C = -34,000 J

A Problem to Consider

• Suppose a piece of iron requires 6.70 J of heat to raise its temperature by one degree Celsius. The quantity of heat required to raise the temperature of the piece of iron from 25.0 oC to 35.0 oC is:

)C 0.25C 0.35()C/J 70.6(TCq ooo

J 0.67q

A Problem to Consider

• Calculate the heat absorbed when the temperature of 15.0 grams of water is raised from 20.0 oC to 50.0 oC. (The specific heat of water is 4.184 J/g.oC.)

Tmsq

)C0.200.50()g0.15()184.4(q oCg

Jo

J1088.1q 3

Heats of Reaction: Calorimetry

• A calorimeter is a device used to measure the heat absorbed or evolved during a physical or chemical change.

The heat absorbed by the calorimeter and its contents is the negative of the heat of reaction.

rxnrcalorimete qq

A Coffee–Cup Calorimeter Made of Two Styrofoam

Cups


Constant-Pressure Calorimetry

No heat enters or leaves!

Calorimetry

• Energy released (heat) = s × m × ΔT

s = specific heat capacity (J/°C·g)m = mass of solution (g)ΔT = change in temperature (°C)


A 100.0 g sample of water at 90°C is added to a 100.0 g sample of water at 10°C.

The final temperature of the water is:a) Between 50°C and 90°Cb) 50°Cc) Between 10°C and 50°C



A 100.0 g sample of water at 90.°C is added to a 500.0 g sample of water at 10.°C.


Calculate the final temperature of the water. 23°C



You have a Styrofoam cup with 50.0 g of water at 10.°C. You add a 50.0 g iron ball at 90. °C to the water. (sH2O = 4.18 J/°C·g and sFe = 0.45 J/°C·g)


Calculate the final temperature of the water.18°C



A Problem to Consider• When 23.6 grams of calcium chloride, CaCl2, was

dissolved in water in a calorimeter, the temperature rose from 25.0 oC to 38.7 oC.

If the heat capacity of the solution and the calorimeter is 1258 J/oC, what is the enthalpy change per mole of calcium chloride?


First, let us calculate the heat absorbed by the calorimeter.

)C 0.25C 7.38()1258(TCq ooCJ

cal o

J1072.1q 4cal

Now we must calculate the heat per mole of calcium chloride.


Calcium chloride has a molecular mass of 111.1 g, so

22

2 CaCl mol 212.0g 1.111

)CaCl mol 1()CaCl g 6.23(

mol/kJ 1.81mol 212.0kJ 2.17

CaCl molq

H2

rxn

Now we can calculate the heat per mole of calcium chloride.

Constant-Volume Calorimetry

No heat enters or leaves!

Chemistry in Action:

Fuel Values of Foods and Other Substances

C6H12O6 (s) + 6O2 (g) 6CO2 (g) + 6H2O (l) H = -2801 kJ/mol

1 cal = 4.184 J

1 Cal = 1000 cal = 4184 J

Substance Hcombustion (kJ/g)

Apple -2

Beef -8

Beer -1.5

Gasoline -34

• In going from a particular set of reactants to a particular set of products, the change in enthalpy is the same whether the reaction takes place in one step or in a series of steps.


• This reaction also can be carried out in two distinct steps, with enthalpy changes designated by ΔH2 and ΔH3.

N2(g) + O2(g) → 2NO(g) ΔH2 = 180 kJ

2NO(g) + O2(g) → 2NO2(g) ΔH3 = – 112 kJ

N2(g) + 2O2(g) → 2NO2(g) ΔH2 + ΔH3 = 68 kJ

ΔH1 = ΔH2 + ΔH3 = 68 kJ

N2(g) + 2O2(g) → 2NO2(g) ΔH1 = 68 kJ


The Principle of Hess’s Law



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Characteristics of Enthalpy Changes

• If a reaction is reversed, the sign of ΔH is also reversed.

• The magnitude of ΔH is directly proportional to the quantities of reactants and products in a reaction. If the coefficients in a balanced reaction are multiplied by an integer, the value of ΔH is multiplied by the same integer.


Example

• Consider the following data:

• Calculate ΔH for the reaction


3 2 2

2 2 2

1 3

2 2NH ( ) N ( ) H ( ) H = 46 kJ

2 H ( ) O ( ) 2 H O( ) H = 484 kJ

g g g

g g g

2 2 2 32 N ( ) 6 H O( ) 3 O ( ) 4 NH ( ) g g g g

Problem-Solving Strategy

• Work backward from the required reaction, using the reactants and products to decide how to manipulate the other given reactions at your disposal.

• Reverse any reactions as needed to give the required reactants and products.

• Multiply reactions to give the correct numbers of reactants and products.


Example

• Reverse the two reactions:

• Desired reaction:


2 2 2 32 N ( ) 6 H O( ) 3 O ( ) 4 NH ( ) g g g g

2 2 3

2 2 2

1 3

2 2N ( ) H ( ) NH ( ) H = 46 kJ

2 H O( ) 2 H ( ) O ( ) H = +484 kJ

g g g

g g g

Example

• Multiply reactions to give the correct numbers of reactants and products:

4( ) 4( ) 3( ) 3( )


2 2 2 32 N ( ) 6 H O( ) 3 O ( ) 4 NH ( ) g g g g

2 2 3

2 2 2

1 3

2 2N ( ) H ( ) NH ( ) H = 46 kJ

2 H O( ) 2 H ( ) O ( ) H = +484 kJ

g g g

g g g

Example

• Final reactions:


ΔH = +1268 kJ


2 2 2 32 N ( ) 6 H O( ) 3 O ( ) 4 NH ( ) g g g g

2 2 3

2 2 2

2 N ( ) 6 H ( ) 4 NH ( ) H = 184 kJ

6 H O( ) 6 H ( ) 3 O ( ) H = +1452 kJ

g g g

g g g

• The following are two important rules for manipulating thermochemical equations:

REVIEW: Thermochemical Equations

When a thermochemical equation is multiplied by any factor, the value of H for the new equation is obtained by multiplying the H in the original equation by that same factor.When a chemical equation is reversed, the value of H is reversed in sign.

H2O (s) H2O (l) H = 6.01 kJ

• The stoichiometric coefficients always refer to the number of moles of a substance


• If you reverse a reaction, the sign of H changes

H2O (l) H2O (s) H = -6.01 kJ

• If you multiply both sides of the equation by a factor n, then H must change by the same factor n.

2H2O (s) 2H2O (l) H = 2 x 6.01 = 12.0 kJ

H2O (s) H2O (l) H = 6.01 kJ

• The physical states of all reactants and products must be specified in thermochemical equations.


H2O (l) H2O (g) H = 44.0 kJ

How much heat is evolved when 266 g of white phosphorus (P4) burn in air?

P4 (s) + 5O2 (g) P4O10 (s) H = -3013 kJ

266 g P4

1 mol P4

123.9 g P4

x3013 kJ1 mol P4

x = 6470 kJ

Standard Enthalpy of Formation (ΔHf°)

• Change in enthalpy that accompanies the formation of one mole of a compound from its elements with all substances in their standard states.


Because there is no way to measure the absolute value of the enthalpy of a substance, must I measure the enthalpy change for every reaction of interest?

Establish an arbitrary scale with the standard enthalpy of formation (H0) as a reference point for all enthalpy expressions.

f

Standard enthalpy of formation (H0) is the heat change that results when one mole of a compound is formed from its elements at a pressure of 1 atm.

f

The standard enthalpy of formation of any element in its most stable form is zero.

H0 (O2) = 0f

H0 (O3) = 142 kJ/molf

H0 (C, graphite) = 0f

H0 (C, diamond) = 1.90 kJ/molf

Conventional Definitions of Standard States

• For a Compound– For a gas, pressure is exactly 1 atm.– For a solution, concentration is exactly 1 M.– Pure substance (liquid or solid)

• For an Element– The form [N2(g), K(s)] in which it exists at 1

atm and 25°C.


A Schematic Diagram of the Energy Changes for the Reaction

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

ΔH°reaction = –(–75 kJ) + 0 + (–394 kJ) + (–572 kJ) = –891 kJ


The standard enthalpy of reaction (H0 ) is the enthalpy of a reaction carried out at 1 atm.

rxn

aA + bB cC + dD

H0rxn dH0 (D)fcH0 (C)f= [ + ] - bH0 (B)faH0 (A)f[ + ]

H0rxn nH0 (products)f= mH0 (reactants)f-

Hess’s Law: When reactants are converted to products, the change in enthalpy is the same whether the reaction takes place in one step or in a series of steps.

(Enthalpy is a state function. It doesn’t matter how you get there, only where you start and end.)

• Hess’s law of heat summation states that for a chemical equation that can be written as the sum of two or more steps, the enthalpy change for the overall equation is the sum of the enthalpy changes for the individual steps.

Hess’s Law

C (graphite) + 1/2O2 (g) CO (g)

CO (g) + 1/2O2 (g) CO2 (g)

C (graphite) + O2 (g) CO2 (g)

Calculate the standard enthalpy of formation of CS2 (l) given that:C(graphite) + O2 (g) CO2 (g) H0 = -393.5 kJrxn

S(rhombic) + O2 (g) SO2 (g) H0 = -296.1 kJrxn

CS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) H0 = -1072 kJrxn

1. Write the enthalpy of formation reaction for CS2

C(graphite) + 2S(rhombic) CS2 (l)

2. Add the given rxns so that the result is the desired rxn.

rxnC(graphite) + O2 (g) CO2 (g) H0 = -393.5 kJ

2S(rhombic) + 2O2 (g) 2SO2 (g) H0 = -296.1x2 kJrxn

CO2(g) + 2SO2 (g) CS2 (l) + 3O2 (g) H0 = +1072 kJrxn+

C(graphite) + 2S(rhombic) CS2 (l)

H0 = -393.5 + (2x-296.1) + 1072 = 86.3 kJrxn

Benzene (C6H6) burns in air to produce carbon dioxide and liquid water. How much heat is released per mole of benzene combusted? The standard enthalpy of formation of benzene is 49.04 kJ/mol.

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

H0rxn nH0 (products)f= mH0 (reactants)f-

H0rxn 6H0 (H2O)f12H0 (CO2)f= [ + ] - 2H0 (C6H6)f[ ]

H0rxn = [ 12x–393.5 + 6x–187.6 ] – [ 2x49.04 ] = -5946 kJ

-5946 kJ2 mol

= - 2973 kJ/mol C6H6

• For example, suppose you are given the following data:

Hess’s Law

kJ -297H );g(SO)g(O)s(S o22

kJ 198H );g(O)g(SO2)g(SO2 o223

Could you use these data to obtain the enthalpy change for the following reaction?

?H );g(SO2)g(O3)s(S2 o32

• If we multiply the first equation by 2 and reverse the second equation, they will sum together to become the third.

Hess’s Law

(2)kJ) -297(H );g(SO2)g(O2)s(S2 o22

(-1)kJ) 198(H );g(SO2)g(O)g(SO2 o322

kJ -792H );g(SO2)g(O3)s(S2 o32

Summary:

• Enthalpy - represents heat energy.

• Change in Enthalpy (Ho) - energy difference between the products and reactants.

• Energy released (exothermic), enthalpy change is negative

– In the combustion of CH4, Ho = -211 kcal

• Energy absorbed (endothoermic), enthalpy change is positive.

– In the decomposition of NH3, Ho=+22 kcal

The enthalpy of solution (Hsoln) is the heat generated or absorbed when a certain amount of solute dissolves in a certain amount of solvent.

Hsoln = Hsoln - Hcomponents

Which substance(s) could be used for melting ice?

Which substance(s) could be used for a cold pack?

The Solution Process for NaCl

Hsoln = Step 1 + Step 2 = 788 – 784 = 4 kJ/mol

So enthalpy explains heat transfer in a reaction…..

• What about predicting if a reaction will occur spontaneously at a given temperature?

Spontaneous and Nonspontaneous Reactions

• Spontaneous reaction - occurs without any external energy input.

• Often, but not always, exothermic reactions are spontaneous.

• Thermodynamics is used to help predict if a reaction will occur.

• Another factor is needed.

Entropy

• The second law of thermodynamics - the universe spontaneously tends toward increasing disorder or randomness.

• Entropy (So) - a measure of the randomness of a chemical system.

• High entropy - highly disordered system

• Low entropy - well organized system

• No such thing as negative entropy.

The SI unit of entropy is joules per Kelvin (J/K) and, like enthalpy, is a state function.

So of a reaction = So(products) - So(reactants)

• A positive So means an increase in disorder for the reaction.

• A negative So means a decrease in disorder for the reaction.

So is positive

So is negative

• If exothermic and positive So…

SPONTANEOUS

• If endothermic and negative So…

NONSPONTANEOUS

• For any other situations, it depends on the relative size of Ho and So.

Free Energy Concept

• The American physicist J. Willard Gibbs introduced the concept of free energy (sometimes called the Gibbs free energy), G, which is a thermodynamic quantity defined by the equation G=H-TS.

This quantity gives a direct criterion for spontaneity of reaction.

Free Energy

• Free energy (Go) - represents the combined contribution of the enthalpy and entropy values for a chemical reaction.

• Predicts spontaneity

• Negative Go…Spontaneous

• Positive Go…Nonspontaneous

Go = Ho - TSo

T in Kelvins

Further summary:

• Thermodynamics determines if a reaction will occur but tells us nothing about the time it will take

• Enthalpy - represents heat energy.• Entropy (So) - a measure of the randomness of a

chemical system.• Free energy (Go) - represents the combined

contribution of the enthalpy and entropy values for a chemical reaction. ∆G predicts spontaneity of a chemical reaction at a given temperature. More in next semester.

Spontaneity and Temperature Change

• All of the four possible choices of signs for Ho and So give different temperature behaviors for Go.

Ho So Go Description

– – – Spontaneous at all T

+ + + Nonspontaneous at all T

– – + or – Spontaneous at low T;

Nonspontaneous at high T

+ + + or – Nonspontaneous at low T;

Spontaneous at high T

Calculation of Go at Various Temperatures

• In this method you assume that Ho and So

are essentially constant with respect to temperature.

You get the value of GTo at any

temperature T by substituting values of Ho and So at 25 oC into the following equation.

oooT STHG

A Problem To Consider

Find the Go for the following reaction at 25 oC and 1000 oC. Relate this to reaction spontaneity.

)g(CO)s(CaO)s(CaCO 23

So: 38.292.9 213.7 J/K

Hfo: -635.1-1206.9 -393.5 kJ

Place below each formula the values of Hf

o and So multiplied by stoichiometric coefficients.




So: 38.292.9 213.7 J/K

Hfo: -635.1-1206.9 -393.5 kJ

You can calculate Ho and So using their respective summation laws.




So: 38.292.9 213.7 J/K

Hfo: -635.1-1206.9 -393.5 kJ

)reactants(Hm)products(HnH of

of

o kJ 3.178kJ)]9.1206()5.3931.635[(

)reactants(Sm)products(SnS ooo




So: 38.292.9 213.7 J/K

Hfo: -635.1-1206.9 -393.5 kJ

K/J 0.159)]9.92()7.2132.38[(

A Problem To ConsiderFind the Go for the following reaction at 25 oC and 1000 oC. Relate this to reaction spontaneity.


So: 38.292.9 213.7 J/K

Hfo: -635.1-1206.9 -393.5 kJ

Now you substitute Ho, So (=0.1590 kJ/K), and T (=298K) into the equation for Gf

o.ooo

T STHG




So: 38.292.9 213.7 J/K

Hfo: -635.1-1206.9 -393.5 kJ


o.)K/kJ 1590.0)(K 298(kJ3.178Go

T



So: 38.292.9 213.7 J/K

Hfo: -635.1-1206.9 -393.5 kJ


o.

kJ 9.130GoT So the reaction is

nonspontaneous at 25 oC.



So: 38.292.9 213.7 J/K

Hfo: -635.1-1206.9 -393.5 kJ

Now we’ll use 1000 oC (1273 K) along with our previous values for Ho and So.

)K/kJ 1590.0)(K 1273(kJ3.178GoT




So: 38.292.9 213.7 J/K

Hfo: -635.1-1206.9 -393.5 kJ

Now we’ll use 1000 oC (1273 K) along with our previous values for Ho and So.

kJ 1.24GoT So the reaction

is spontaneous at 1000 oC.



So: 38.292.9 213.7 J/K

Hfo: -635.1-1206.9 -393.5 kJ

To determine the minimum temperature for spontaneity, we can set Gf

o =0 and solve for T.

o

o

S

HT



So: 38.292.9 213.7 J/K

Hfo: -635.1-1206.9 -393.5 kJ

To determine the minimum temperature for spontaneity, we can set Gf

o =0 and solve for T.

)C 848( K 1121K/kJ 1590.0

kJ 3.178T o



So: 38.292.9 213.7 J/K

Hfo: -635.1-1206.9 -393.5 kJ

Thus, CaCO3 should be thermally stable until its heated to approximately 848 oC.

• Fossil Fuels– Petroleum, Natural Gas, and Coal

• Wood• Hydro• Nuclear


Energy Sources Used in the United States


Fuels• Fossil fuels account for nearly 85% of the

energy usage in the United States.Anthracite, or hard coal, the oldest variety of coal, contains about 80% carbon.Bituminous coal, a younger variety of coal, contains 45% to 65% carbon.Fuel values of coal are measured in BTUs (British Thermal Units).A typical value for coal is 13,200 BTU/lb. 1 BTU = 1054 kJ

Fuels• Natural gas and petroleum account for nearly

three-quarters of the fossil fuels consumed per year.

Purified natural gas is primarily methane, CH4, but also contains small quantities of ethane, C2H6, propane, C3H8, and butane, C4H10.We would expect the fuel value of natural gas to be close to that for the combustion of methane.

)g(OH2)g(CO)g(O2)g(CH 2224 kJ 802H ; o

Fuels• Petroleum is a very complicated mixture of

compounds. Gasoline, obtained from petroleum, contains many different hydrocarbons, one of which is octane, C8H18.

)g(O)l(HC 2225

188

kJ -5,074H );g(OH9)g(CO8 o22

This value of Ho is equivalent to 44.4 kJ/gram.

Fuels• With supplies of petroleum estimated to be 80%

depleted by the year 2030, the gasification of coal has become a possible alternative.

First, coal is converted to carbon monoxide using steam.

)g(H)g(CO)g(OH)s(C 22 The carbon monoxide can then be used to produce a variety of other fuels, such as methane.

)g(OH)g(CH)g(H3)g(CO 242

The Earth’s Atmosphere

• Transparent to visible light from the sun.• Visible light strikes the Earth, and part of it is

changed to infrared radiation.• Infrared radiation from Earth’s surface is

strongly absorbed by CO2, H2O, and other molecules present in smaller amounts in atmosphere.

• Atmosphere traps some of the energy and keeps the Earth warmer than it would otherwise be.


The Earth’s Atmosphere


• Other Energy Alternatives?– Oil shale– Ethanol– Methanol– Seed oil– SOLAR– WIND– TIDAL– OTHER?


Chapter 6 Thermochemistry. Copyright © Cengage Learning. All rights reserved 2 Capacity to do work...

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Transcript of Chapter 6 Thermochemistry. Copyright © Cengage Learning. All rights reserved 2 Capacity to do work...