Chapter 5 Written by JoAnne Swanson University of Central Florida Thermodynamics.

Chapter 5Written by

JoAnne Swanson

University of Central Florida

Thermodynamics

Topics Topics Types of energy and units of energy Exothermic vs. endothermic reactions Heat capacity and specific heat Energy transfers in changes of state Calorimetry Enthalpy and entropy Hess’s Law

Energy is defined as the Energy is defined as the capacity to do work.capacity to do work.

There are two categories of energy: Kinetic energy. Potential energy

Kinetic Energy is the energy of motion Potential energy is energy of position,

Chemical energy is a form of Chemical energy is a form of potential energy stored in the potential energy stored in the

structure of a chemical substance structure of a chemical substance due to its composition.due to its composition.

Consider the potential energy of a rock up on a cliff. If one rock is up on a 1000 ft. cliff and another is up on a 10 ft. hill, the rock that is 1000ft. up will hit the ground with much more energy than the other if it falls.

Chemical energy has similar differences in that the structure of one substance may contain much more potential energy than another substance.

Kinetic Energy of an object depends on Kinetic Energy of an object depends on its mass and velocity:its mass and velocity:

EEkk = ½ (mv = ½ (mv22) or mv) or mv22/2/2

mass is in Kg, mass is in Kg,

and Kgand Kg..mm22/s/s22 = 1 Joule, the unit used to = 1 Joule, the unit used to designate energy.designate energy.

From this equation you should see that From this equation you should see that EEk k increases with mass and with velocityincreases with mass and with velocity

Potential Energy depends on gravity, Potential Energy depends on gravity, mass, and position:mass, and position:

EEpp = mgh = (mass, gravity, height) = mgh = (mass, gravity, height)

(gravitational constant = 9.8 m/s(gravitational constant = 9.8 m/s22))

But EBut Epp of submicroscopic substances of submicroscopic substances

(molecules, ions, etc.) (molecules, ions, etc.)

electrical charges between particles.electrical charges between particles.

Electrical charge is designated by the Electrical charge is designated by the symbol ‘Q’ and has the value of an symbol ‘Q’ and has the value of an electron charge = 1.60 x 10electron charge = 1.60 x 10-19-19 C CThe unit for electrical charge is the The unit for electrical charge is the coulomb ( C).coulomb ( C). EEelel = kQ = kQ11QQ22 / d where k=8.99x10 / d where k=8.99x1099JmJm

CC22

This equation shows the electrostatic This equation shows the electrostatic forces of attraction between two particles, forces of attraction between two particles, 1 and 2, are inversely proportional to the 1 and 2, are inversely proportional to the distance between the particles.distance between the particles.

Another energy unit we will use is the Another energy unit we will use is the calorie. 1 c = 1 cal = 4.184 Jcalorie. 1 c = 1 cal = 4.184 J

This differs from the nutritional calorie This differs from the nutritional calorie which is 1C = 1000c = 1Kcalwhich is 1C = 1000c = 1Kcal

Thermochemistry is the study Thermochemistry is the study of heat transfers occurring in of heat transfers occurring in

chemical and physical changes chemical and physical changes of substances.of substances.

Thermal energy random motion of atoms and

molecules.

The Universe is made up of The Universe is made up of the the system and the and the

surroundings

The system is the part of the universe that is of interest to us.

The surroundings is everything outside of the system.

Energy is transferred as heat and work.Energy is transferred as heat and work.

Work = Force x DistanceWork = Force x Distance

= pressure x volume= pressure x volume

It usually pertains to either the It usually pertains to either the electrical work (for instance in a battery) electrical work (for instance in a battery) or mechanical work (like the pressure or mechanical work (like the pressure exerted on or by a gas in a reaction).exerted on or by a gas in a reaction).

The change in Internal Energy of a The change in Internal Energy of a system = system = E = q + wE = q + w

E is the change in internal energyE is the change in internal energy

q = heatq = heat

w = workw = workHeat can flow into the system or out of Heat can flow into the system or out of the system.the system.• sign on qsign on qWork can be done on the system or by Work can be done on the system or by the system.the system.• sign on wsign on w

Exothermic vs. Endothermic Exothermic vs. Endothermic ProcessesProcesses

Any process that gives off heat to the surroundings is an Exothermic process.

bonds are made

When a process absorbs heat from the surroundings it is an Endothermic process. When ice melts it absorbs heat from the surroundings, thus it is an endothermic process.

To measure the change in internal To measure the change in internal energy of a chemical reaction, the energy of a chemical reaction, the reaction is often carried out in a “bomb reaction is often carried out in a “bomb calorimeter”. There is no change in calorimeter”. There is no change in volume in this sealed container, therefore volume in this sealed container, therefore there is no work involved and there is no work involved and

E = qE = qreaction reaction (at constant volume)(at constant volume)

The symbol The symbol HH is used to is used to represent the change in heat represent the change in heat into or out of into or out of the system. It is the system. It is defined as the defined as the change in change in enthalpy. enthalpy.

Enthalpy and Internal Energy are nearly Enthalpy and Internal Energy are nearly equal. Enthalpy takes into account equal. Enthalpy takes into account reactions at reactions at constant pressureconstant pressure where an where an expanding gas does work on the expanding gas does work on the surroundings.surroundings.Remember w = PV, but if w is done by the Remember w = PV, but if w is done by the system, w = -PVsystem, w = -PV

H = E +H = E +PV and PV and H = H = E + E + PV PV

w = - w = - PV PV so, so, H = (q + w) – wH = (q + w) – w

H = q H = q (at constant (at constant pressure)pressure)

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•1 mole gas at 251 mole gas at 25ooCC and 1atm has a and 1atm has a volume of 24.5L =( 24.5 Lvolume of 24.5L =( 24.5 L..atm / mol ) atm / mol ) •1 L1 L..atm = 0.1013 kJ atm = 0.1013 kJ • PV makes little difference in the PV makes little difference in the H H value, and even less difference for liquids value, and even less difference for liquids or solids.or solids.

why?why?

If 1 LIf 1 L..atm = 0.1013 kJ how many kJ are in atm = 0.1013 kJ how many kJ are in the molar volume of a gas at 25the molar volume of a gas at 25ooC?C?

24.5 L24.5 L..atm x atm x 0.1013 kJ0.1013 kJ = 2.5 kJ = 2.5 kJ

1 L1 L..atm atm

see next examplesee next example

Examples: given the values for PV, Examples: given the values for PV, calculate the change in enthalpy for the calculate the change in enthalpy for the combustion of methane where combustion of methane where E = -885 kJE = -885 kJ

PV at 25PV at 25ooCCCHCH44(g)(g) 2.5 KJ/ mol2.5 KJ/ mol

OO22(g)(g) 2.5 KJ/ mol2.5 KJ/ mol

COCO22(g)(g) 2.5 KJ/ mol2.5 KJ/ mol

HH220(l)0(l) 0.0018 KJ/ mol0.0018 KJ/ mol

H = H = E + E + PV PV ((PVPV = = PVPV products – products – PV PV reactants)reactants)

continued………continued………

CHCH44(g) + 2 O(g) + 2 O22 (g) (g) CO CO22 (g) + 2 H (g) + 2 H22O (l)O (l)

PV = PV COPV = PV CO22 + 2PV H + 2PV H22O –[PV CHO –[PV CH44 + 2PV O + 2PV O22

PV = [2.5 KJ + 2(0.0018 KJ)] – [2.5 KJ + 2(2.5 KJ) ] PV = [2.5 KJ + 2(0.0018 KJ)] – [2.5 KJ + 2(2.5 KJ) ]

PV = -5.0 KJPV = -5.0 KJ

H = -885 KJ - 5.0 KJ = -890 KJH = -885 KJ - 5.0 KJ = -890 KJ

Carried out at Carried out at constant pressureconstant pressure, , H = -885 KJH = -885 KJ

Not a big difference when Not a big difference when PV is addedPV is added

EXAMPLE 2:EXAMPLE 2:

a.a. Calculate the kinetic energy in joules, Calculate the kinetic energy in joules, of a 45 g golf ball moving at 61 m/s.of a 45 g golf ball moving at 61 m/s.

b.b. Convert this energy to calories.Convert this energy to calories.

c.c. What happens to the energy when the What happens to the energy when the ball strikes a tree?ball strikes a tree?

Since 1J = 1kg mSince 1J = 1kg m22/s/s22

change g to kg.change g to kg.

45 g = 0.045 kg45 g = 0.045 kg

EEkk = mv = mv2 2 / 2 = / 2 = 0.045 kg x (61m/s)0.045 kg x (61m/s)22 22

= = 84 kg m84 kg m22 = 84 J= 84 Jss22

b.b. 84 J x 84 J x 1cal1cal = 20 cal = 20 cal4.184 J4.184 J

c. When the ball hits the tree, velocity = 0 c. When the ball hits the tree, velocity = 0 and so does Kinetic energy. Kinetic and so does Kinetic energy. Kinetic Energy is transferred as potential energy Energy is transferred as potential energy to the tree and the deformed golf ball.to the tree and the deformed golf ball.

Example 4:Example 4:

Calculate Calculate E when a balloon is inflated E when a balloon is inflated completely by heating it. The volume changes completely by heating it. The volume changes from 4.00 x 10from 4.00 x 1066 L to 4.50 x 10 L to 4.50 x 106 6 L, with the addition L, with the addition of 1.3 x 10of 1.3 x 1088 J of heat. The balloon expands J of heat. The balloon expands against an internal pressure of 1.0 atm. against an internal pressure of 1.0 atm. E = q+w E = q+w

w = - w = - PV because ________________PV because ________________q is positive because _________________q is positive because _________________

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V = 4.50 x 10V = 4.50 x 1066L - 4.00 x 10L - 4.00 x 1066L = 5.00 x 10L = 5.00 x 105 5 LL

-P-PV = 5.00 x 10V = 5.00 x 1055 L x (-1.0 atm) =5.00 x 10 L x (-1.0 atm) =5.00 x 1055 Latm Latm = -5.07 x 10= -5.07 x 107 7 JJE = 1.3 x 10E = 1.3 x 1088J – 5.07 x 10J – 5.07 x 1077J = 7.9 x 10J = 7.9 x 1077 J J

Example 3:Example 3:

Calculate Calculate E when a gas is compressed E when a gas is compressed from 6.0 x 10from 6.0 x 1044L to 5.3 x 10L to 5.3 x 1044L under an L under an external pressure of 1.3 atm. The amount external pressure of 1.3 atm. The amount of heat transferred to the surroundings was of heat transferred to the surroundings was 4.3 x 104.3 x 1022 J. J.

E = w + q, where w = +E = w + q, where w = +PV why?PV why?

V = 5.3 x 10V = 5.3 x 1044L - 6.0 x 10L - 6.0 x 1044L = - 7.0 x 10L = - 7.0 x 1033LL

PV = (1.3 atm)(- 7.0 x 10PV = (1.3 atm)(- 7.0 x 1033L)L)= -9.1 x 10= -9.1 x 1033Latm Latm

-9.1 x 10-9.1 x 103 3 Latm x Latm x 0.1013 kJ0.1013 kJ = -9.2 x 10 = -9.2 x 102 2 KJKJ

LatmLatmIn this case since the gas was In this case since the gas was compressed, w = + and since heat was compressed, w = + and since heat was transferred to the surroundings, q = - .transferred to the surroundings, q = - .E = w – q, = -9.2 x 10E = w – q, = -9.2 x 105 5 J – 4.3 x 10J – 4.3 x 1022 J = J =

= - 9.2043 x 10= - 9.2043 x 1055 J J= - 9.2 x 10= - 9.2 x 1055 J J

The enthalpy of reaction is the The enthalpy of reaction is the differencedifference between the enthalpies between the enthalpies of the of the products and reactantsproducts and reactants..

HH(rxn)(rxn) = = HHff(products) – (products) – ff(reactants)(reactants)

reactantsreactants

productsproducts reactantsreactants

productsproducts

reaction progressionreaction progression reaction progressionreaction progression

Ene

rgy

Ene

rgy

Ene

rgy

Ene

rgy

energy addedenergy added

energy expelledenergy expelled

EEaa

EEaa

EndothermicEndothermic ExothermicExothermic

The diagram illustrates energy given off when bonds are made between H2 and O2 , (a). Energy is absorbed

when bonds are broken in HgO , (b).

The units of energy are The units of energy are Kilojoules and Calories. Kilojoules and Calories. Enthalpy of reaction will be Enthalpy of reaction will be expressed as Kilojoules (Kj).expressed as Kilojoules (Kj).

It is important to note the following It is important to note the following when calculating the enthalpy of when calculating the enthalpy of reaction:reaction:

The enthalpy of a substance in its standard state is equal to zero.

Enthalpy is dependent on the quantity of the substance therefore the enthalpy of a substance must be multiplied by its coefficient in the chemical equation.

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When a rxn is reversed, the sign on the When a rxn is reversed, the sign on the enthalpy of rxn is changed.enthalpy of rxn is changed.

If a rxn is divided through or multiplied If a rxn is divided through or multiplied through by a number, so is the enthalpy through by a number, so is the enthalpy of rxn.of rxn.

the state of matter is important in the state of matter is important in enthalpy calculations. enthalpy calculations.

Enthalpy of Enthalpy of formationformation values of values of compounds are found in tables of compounds are found in tables of thermodynamic data.thermodynamic data.

The enthalpy of The enthalpy of formationformation is defined as the is defined as the energy associated with the energy associated with the formationformation of of 1 mole1 mole of a substance of a substance from its elementsfrom its elements in in their standard statestheir standard states. .

NaNa(s)(s) + 1/2 Cl + 1/2 Cl2 (g)2 (g) NaCl NaCl

The formation equation for sodium chlorideThe formation equation for sodium chloride

Ex. 1

Determine the enthalpy for the following chemical reaction:

COCO2 (g)2 (g) + 2 H + 2 H22O O (l)(l) -----> 2 O -----> 2 O2 (g)2 (g) + CH + CH4 (g)4 (g)

Use appendix C pg. 1041 in your text to find the individual enthalpy values.

CO2 (g) = -393.509, H2O (l) = -285.83

CH4 (g) = -74.81 , O2 (g)= 0

H = Hf(products) – freactants)

H = (0 + -74.81)-(-393.509 + 2(- 285.83))

= -74.81 + 965.17 = 890.36 Kj

CO2 + 2 H2O 2 O2 + CH4

HESS’S LAWHESS’S LAW

ENTHALPY IS A ENTHALPY IS A STATE FUNCTIONSTATE FUNCTION. . ENTHALPY OF A REACTION WILL BE ENTHALPY OF A REACTION WILL BE THE SAME NO MATTER WHAT PATH THE SAME NO MATTER WHAT PATH IS TAKEN TO ARRIVE AT THE IS TAKEN TO ARRIVE AT THE PRODUCTS.PRODUCTS.

This means that if we need to This means that if we need to calculate the enthalpy of a reaction calculate the enthalpy of a reaction for which we do not know the for which we do not know the enthalpies of formation, we can enthalpies of formation, we can algebraically manipulate other algebraically manipulate other reactionsreactions to arrive at the enthalpy for to arrive at the enthalpy for the desired reaction.the desired reaction.

Ex.Ex. Determine the Determine the HHrxnrxn for for

SnSn(s)(s) + 2 Cl + 2 Cl2 (g)2 (g) SnCl SnCl4(l)4(l)

Given the following:Given the following:

SnSn(s)(s) + Cl + Cl2 (g)2 (g) SnCl SnCl2(s)2(s) HH11= -349.8 kJ= -349.8 kJ

SnClSnCl2(s) 2(s) + Cl+ Cl2 (g)2 (g) SnCl SnCl4(l)4(l) HH22= -195.4 kJ= -195.4 kJ

Sn Sn (s) (s) + 2 Cl+ 2 Cl2 (g)2 (g) SnCl SnCl4(l)4(l) HH33= ?= ?

HH11 + + HH22 = = HH3 3 = -545.2 kJ= -545.2 kJ

Ex. Ex. Calculate theCalculate the HHrxnrxn for for

S S (s)(s) + O + O22 (g)(g) SO SO2(g)2(g)

Given Given

S S (s)(s) + 3/2 O + 3/2 O22 (g)(g) SO SO3(g) 3(g) H= -395.2 kJH= -395.2 kJ

SOSO22 (g)(g) + O + O22 (g)(g) 2 SO 2 SO3(g) 3(g) H= -198.2 kJH= -198.2 kJ

NOTICE THAT WE NEED SONOTICE THAT WE NEED SO22 (g)(g) TO BE TO BE

ON THE PRODUCT SIDE. THEREFORE ON THE PRODUCT SIDE. THEREFORE WE MUST REVERSE THE SECOND RXNWE MUST REVERSE THE SECOND RXNCONTINUEDCONTINUED

S S (s)(s) + O + O22 (g)(g) SO SO2(g)2(g)

S S (s)(s) + 3/2 O + 3/2 O22 (g)(g) SO SO3(g) 3(g) H= -395.2 kJH= -395.2 kJ

22 SOSO33 (g)(g) O O22 (g)(g) + 2 SO + 2 SO2(g) 2(g) H= +198.2 kJH= +198.2 kJ

IT IS ALSO NECESSARY TO DIVIDE IT IS ALSO NECESSARY TO DIVIDE EQUATION 2 BY 2, SO THAT WE HAVE EQUATION 2 BY 2, SO THAT WE HAVE 11 SOSO2(g) 2(g) IN THE PRODUCT. WE MUST IN THE PRODUCT. WE MUST

ALSO DIVIDE THE ALSO DIVIDE THE HH+198.2 kJ / 2 = 99.10 kJ+198.2 kJ / 2 = 99.10 kJ

Now add the reactions and the enthalpies.Now add the reactions and the enthalpies.

S S (s)(s) + 3/2 O + 3/2 O22 (g)(g) SO SO3(g) 3(g) H= -395.2 kJH= -395.2 kJ

SOSO33 (g)(g) 1/2 O 1/2 O22 (g)(g) + SO + SO2(g) 2(g) H= +99.1 kJH= +99.1 kJ

S S (s)(s) + O + O22 (g)(g) SO SO2(g)2(g) H= -296.1 kJH= -296.1 kJ

The enthalpy of formation of The enthalpy of formation of a a substance in a chemical reactionsubstance in a chemical reaction can can also be calculated if the also be calculated if the enthalpy of enthalpy of reaction is knownreaction is known and the and the enthalpies of enthalpies of the other substancesthe other substances in the reaction arein the reaction are knownknown. Simply call the unknown . Simply call the unknown enthalpy ‘X’ and plug in all known values enthalpy ‘X’ and plug in all known values to solve for ‘X’.to solve for ‘X’.

example on transparencyexample on transparency

Specific heat and heat CapacitySpecific heat and heat Capacity

Heat capacity Heat capacity is the amount of heat is the amount of heat required to raise the temperature of a required to raise the temperature of a given quantity of a substance by 1given quantity of a substance by 1oo C. C.

Specific Heat Specific Heat is the amount of heat is the amount of heat required to raise the temperature of required to raise the temperature of 1 1 gramgram of a substance by 1 of a substance by 1oo C. J/g C. J/gooCC

Molar heat capacity is J / mol Molar heat capacity is J / mol ooCC

•a substance’s a substance’s ability to absorb heat and ability to absorb heat and to store heatto store heat..

For example, a metal _______________For example, a metal _______________

________________________________________________________________

A liquid like A liquid like waterwater (which contains strong (which contains strong intermolecular forces, hydrogen bonds), intermolecular forces, hydrogen bonds), ____________________________________________________________________________________________________________________________________

The The metalmetal has a has a ______________ ______________ and and the the waterwater has a has a ________________________________..

The specific heat capacity ( C ) of The specific heat capacity ( C ) of water is an important and easy number water is an important and easy number to remember ; C = 1 cal / g to remember ; C = 1 cal / g ooC ,C ,

(or 4.184 J / g(or 4.184 J / gooC )C )Since Since ooC and Kelvin have the same size C and Kelvin have the same size increments, they are interchangeable in increments, they are interchangeable in these equations.these equations.

CalorimetryCalorimetry is a technique used is a technique used in the lab to measure the change in the lab to measure the change in enthalpy of a reaction. One in enthalpy of a reaction. One apparatus used is the apparatus used is the Bomb Bomb Calorimeter.Calorimeter. It is a Heavy It is a Heavy walled, steel container which has walled, steel container which has a known specific heat.a known specific heat.

Types of Calorimetry problems we will cover:Types of Calorimetry problems we will cover:

1.1. simple change in water temperaturesimple change in water temperature

2.2. change in state of waterchange in state of water

3.3. a rxn in a coffee cup calorimetera rxn in a coffee cup calorimeter

4.4. a rxn in a bomb calorimetera rxn in a bomb calorimeter

5.5. a mixture of hot substance with cold a mixture of hot substance with cold substance in a calorimeter.substance in a calorimeter.

Equations needed: Equations needed:

1.1. q = C x m x q = C x m x TT

2.2. q = qq = qiceice + + ff + q + qwaterwater + + HHvv + q + qsteamsteam

3.3. qqrxn rxn ==HHrxnrxn = -q = -qwaterwater

4.4. HHrxn rxn = -[q= -[qwaterwater + q + qbombbomb], ], or = or = -(C-(Ccalcal x x T)T)

5.5. qqcold cold = -q= -qhothot

Some needed values:Some needed values:

specific heat of water(l) = 4.184 J/gKspecific heat of water(l) = 4.184 J/gKspecific heat of water(s) = 2.05 J/gKspecific heat of water(s) = 2.05 J/gKspecific heat of water(g) = 2.01 J/gKspecific heat of water(g) = 2.01 J/gK

HHff = enthalpy of fusion = energy = enthalpy of fusion = energy

involved in a change of state from liquid involved in a change of state from liquid to solid or vice versa = 333 J/gto solid or vice versa = 333 J/g

HHvv = enthalpy of vaporization = energy = enthalpy of vaporization = energy

involved in a change of state from liquid involved in a change of state from liquid to vapor or vice versa = 2444 J/gto vapor or vice versa = 2444 J/g

The heat of the reaction (q) is The heat of the reaction (q) is equal to theequal to the negative of the heat negative of the heat absorbed by theabsorbed by the ((bomb plus the bomb plus the water water surrounding the bomb).surrounding the bomb).

qqrxnrxn = = HHrxnrxn = - (q = - (qbombbomb + q + q waterwater ) )

qq is used to represent the is used to represent the quantityquantity and and direction direction of heat of heat transferred, using a calorimeter.transferred, using a calorimeter.

Necessary Equations:Necessary Equations:

q = C x m x q = C x m x TT

q = (specific heat)(mass)(change in Temp.)q = (specific heat)(mass)(change in Temp.)

q = (J/gK)(g)(q = (J/gK)(g)(T)T)

q(rxn) = - (q of the water + q of the bomb)q(rxn) = - (q of the water + q of the bomb)

Subliminal message.....Subliminal message.....

Wake up !!!Wake up !!!

1. 1. CalorimetryCalorimetryA 466g sample of water is heated from A 466g sample of water is heated from 8.508.50ooC to 74.60C to 74.60ooC. Calculate the amount C. Calculate the amount of heat absorbed by the water.of heat absorbed by the water.

q = (sp. heat)(mass)(q = (sp. heat)(mass)(

q q = (= (4.184 J/g4.184 J/gooC)(466g)(74.60-8.50C)(466g)(74.60-8.50ooC)C)

q = 1.29 x 10q = 1.29 x 105 5 J = 129 KJJ = 129 KJ

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change of state of water example on change of state of water example on transparency.transparency.

coffee cup calorimeter on transparencycoffee cup calorimeter on transparency

Ex. 3Ex. 3

1.435 g of Naphthalene (molar mass 1.435 g of Naphthalene (molar mass =128.2) =128.2) was burned in a bomb was burned in a bomb calorimeter. The temp. rose from calorimeter. The temp. rose from 20.1720.17ooC to 25.84 C to 25.84 ooCC. The mass of . The mass of the the waterwater surrounding the surrounding the calorimeter was calorimeter was 2000.g 2000.g and the heat and the heat capacity of the capacity of the bombbomb was was 1.80 1.80 KJ/KJ/ooCC. Calculate the heat of . Calculate the heat of combustion of Naphthalene.combustion of Naphthalene.

q(rxn) = -(q water + q bomb)q(rxn) = -(q water + q bomb)

q(water) = (2000g)(4.184 J/gq(water) = (2000g)(4.184 J/gooC)(5.67C)(5.67ooC)C)

= 4.74 x 10= 4.74 x 1044

q(bomb) = (1.80 x 10q(bomb) = (1.80 x 1033J / J / ooC)(5.67C)(5.67ooC)C)

= 1.02 x 10= 1.02 x 1044 JJ

q(rxn)q(rxn) = -(4.74 x 10= -(4.74 x 104 4 J + 1.02 x 10J + 1.02 x 104 4 J)J)

= -5.76 x 10= -5.76 x 104 4 JJ

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MORE EXAMPLES WILL BE GIVEN IN MORE EXAMPLES WILL BE GIVEN IN CLASS.CLASS.

SINCE I HAD TO RUSH TO GET SINCE I HAD TO RUSH TO GET THESE NOTES ON LINE, THEY THESE NOTES ON LINE, THEY WERE NOT COMPLETELY EDITED. WERE NOT COMPLETELY EDITED. THERE IS SOME REPETITION, and THERE IS SOME REPETITION, and they may need some corrections.they may need some corrections.