Chapter 6 Thermochemistry

Copyright © 2011 Pearson Education, Inc.

Chapter 6Thermochemistry

Chemistry: A Molecular Approach, 2nd Ed.Nivaldo Tro

Roy KennedyMassachusetts Bay Community College

Wellesley Hills, MA


Chemical Hand Warmers• Most hand warmers work by using the heat

released from the slow oxidation of iron4 Fe(s) + 3 O2(g) → 2 Fe2O3(s)

• The amount your hand temperature rises depends on several factorsthe size of the hand warmerthe size of your glove, etc.mainly, the amount of heat released by the

reaction

2Tro: Chemistry: A Molecular Approach, 2/e


Nature of Energy• Even though chemistry is the study of

matter, energy affects matter• Energy is anything that has the capacity to

do work• Work is a force acting over a distance

Energy = Work = Force x Distance• Heat is the flow of energy caused by a

difference in temperature• Energy can be exchanged between

objects through contactcollisions



Energy, Heat, and Work

• You can think of energy as a quantity an object can possessor collection of objects

• You can think of heat and work as the two different ways that an object can exchange energy with other objectseither out of it, or into it



Classification of Energy

• Kinetic energy is energy of motion or energy that is being transferred

• Thermal energy is the energy associated with temperaturethermal energy is a

form of kinetic energy



Classification of Energy• Potential energy is energy that is stored in an

object, or energy associated with the composition and position of the objectenergy stored in the structure of a compound is

potential



Manifestations of Energy



Some Forms of Energy• Electrical

kinetic energy associated with the flow of electrical charge• Heat or thermal energy

kinetic energy associated with molecular motion• Light or radiant energy

kinetic energy associated with energy transitions in an atom• Nuclear

potential energy in the nucleus of atoms • Chemical

potential energy due to the structure of the atoms, the attachment between atoms, the atoms’ positions relative to each other in the molecule, or the molecules, relative positions in the structure



Conservation of Energy• The Law of Conservation

of Energy states that energy cannot be created nor destroyed

• When energy is transferred between objects, or converted from one form to another, the total amount of energy present at the beginning must be present at the end



System and Surroundings• We define the system as the material or process

within which we are studying the energy changes within

• We define the surroundings as everything else with which the system can exchange energy with

• What we study is the exchange of energy between the system and the surroundings

Surroundings

System

Surroundings

System



Comparing the Amount of Energy in the System and Surroundings During Transfer• Conservation of energy means that the amount

of energy gained or lost by the system has to be equal to the amount of energy lost or gained by the surroundings



• The amount of kinetic energy an object has is directly proportional to its mass and velocity KE = ½mv2

Units of Energy



Units of Energy• joule (J) is the amount of energy needed to

move a 1-kg mass a distance of 1 meter1 J = 1 N∙m = 1 kg∙m2/s2

• calorie (cal) is the amount of energy needed to raise the temperature of one gram of water 1°Ckcal = energy needed to raise 1000 g of water 1°Cfood Calories = kcals

Energy Conversion Factors1 calorie (cal) = 4.184 joules (J) (exact)

1 Calorie (Cal) = 1000 cal = 1 kcal = 4184 J1 kilowatt-hour (kWh) = 3.60 x 106 J



Energy Use



The First Law of ThermodynamicsLaw of Conservation of Energy

• Thermodynamics is the study of energy and its interconversions

• The First Law of Thermodynamics is the Law of Conservation of Energy

• This means that the total amount of energy in the universe is constant

• You can therefore never design a system that will continue to produce energy without some source of energy



Energy Flow and Conservation of Energy

• Conservation of energy requires that the sum of the energy changes in the system and the surroundings must be zeroDEnergyuniverse = 0 = DEnergysystem + DEnergysurroundings

D Is the symbol that is used to mean change final amount – initial amount



Internal Energy• The internal energy is the sum of the kinetic

and potential energies of all of the particles that compose the system

• The change in the internal energy of a system only depends on the amount of energy in the system at the beginning and enda state function is a mathematical function whose

result only depends on the initial and final conditions, not on the process used

DE = Efinal – Einitial

DEreaction = Eproducts − Ereactants



State FunctionYou can travel either of two trails to reach the top of the mountain. One is long and winding, the other is shorter but steep. Regardless of which trail you take, when you reach the top you will be 10,000 ft above the base.

The distance from the base to the peak of the mountain is a state function. It depends only on the difference in elevation between the base and the peak, not on how you arrive there!



Energy Diagrams

19

• Energy diagrams are a “graphical” way of showing the direction of energy flow during a process

Inte

rnal

Ene

rgy

initial

finalenergy added

DE = +

Inte

rnal

Ene

rgy

initial

final

energy removedDE = ─

• If the final condition has a larger amount of internal energy than the initial condition, the change in the internal energy will be +• If the final condition has a smaller amount of internal energy than the initial condition, the change in the internal energy will be ─

Tro: Chemistry: A Molecular Approach, 2/e


Energy Flow• When energy flows out of a

system, it must all flow into the surroundings

• When energy flows out of a system, DEsystem is ─

• When energy flows into the surroundings, DEsurroundings is +

• Therefore:─ DEsystem= DEsurroundings

SurroundingsDE +

SystemDE ─



Energy Flow• When energy flows into a

system, it must all come from the surroundings

• When energy flows into a system, DEsystem is +

• When energy flows out of the surroundings, DEsurroundings is ─

• Therefore:DEsystem= ─ DEsurroundings

SurroundingsDE ─

SystemDE +



energy releasedDErxn = ─

energy absorbedDErxn = +

Energy Flow in a Chemical Reaction• The total amount of internal energy

in 1mol of C(s) and 1 mole of O2(g) is greater than the internal energy in 1 mole of CO2(g) at the same temperature and pressure

• In the reaction C(s) + O2(g) → CO2(g), there will be a net release of energy into the surroundings −DEreaction = DEsurroundings

• In the reaction CO2(g) → C(s) + O2(g), there will be an absorption of energy from the surroundings into the reaction DEreaction = − DEsurroundings

Inte

rnal

Ene

rgy

CO2(g)

C(s), O2(g)

Surroundings

SystemC + O2 → CO2

SystemCO2 →C + O2



Energy Exchange• Energy is exchanged between the system and

surroundings through heat and workq = heat (thermal) energyw = work energyq and w are NOT state functions, their value depends

on the processDE = q + w



Energy Exchange

• Energy is exchanged between the system and surroundings through either heat exchange or work being done



• The white ball has an initial amount of 5.0 J of kinetic energy

• As it rolls on the table, some of the energy is converted to heat by friction

• The rest of the kinetic energy is transferred to the purple ball by collision

Heat & Work



energy change for the white ball, DE = KEfinal − KEinitial = 0 J − 5.0 J = −5.0 Jkinetic energy transferred to purple ball, w = −4.5 Jkinetic energy lost as heat, q = −0.5 Jq + w = (−0.5 J) + (−4.5 J) = −5.0 J = DE

On a smooth table, most of the kinetic energy is transferred from the white ball to the purple – with a small amount lost through friction

energy change for the white ball, DE = KEfinal − KEinitial = 0 J − 5.0 J = −5.0 Jkinetic energy transferred to purple ball, w = −3.0 Jkinetic energy lost as heat, q = −2.0 Jq + w = (−2.0 J) + (−3.0 J) = −5.0 J = DE

On a rough table, most of the kinetic energy of the white ball is lost through friction – less than half is transferred to the purple ball

26

Heat & Work



Heat, Work, and Internal Energy• In the previous billiard ball example, the DE of the white

ball is the same for both cases, but q and w are not• On the rougher table, the heat loss, q, is greater

q is a more negative number• But on the rougher table, less kinetic energy is

transferred to the purple ball, so the work done by the white ball, w, is lessw is a less negative number

• The DE is a state function and depends only on the velocity of the white ball before and after the collision in both cases it started with 5.0 kJ of kinetic energy and ended

with 0 kJ because it stopped q + w is the same for both tables, even though the values of q

and w are different27Tro: Chemistry: A Molecular Approach, 2/e


Example 6.1: If the burning of the fuel in a potato cannon performs 855 J of work on the potato and

produces 1422 J of heat, what is DE for the burning of the fuel?

the unit is correct, the sign make sense as the fuel should lose energy during the reaction

qpotato = 855 J, wpotato = 1422 J

DEfuel, J

Check:

Solution:

Concept Plan:

Relationships:

Given:

Find:

q, w DE

qsystem = −qsurroundings, wsystem = −wsurroundings, DE = q + w

q, wpotato

q, wfuel



Practice – Reacting 50 mL of H2(g) with 50 mL of C2H4(g) produces 50 mL of C2H6(g) at 1.5 atm. If the reaction

produces 3.1 x 102 J of heat and the decrease in volume requires the surroundings do 7.6 J of work on the gases, what

is the change in internal energy of the gases?



If the reaction produces 3.1 x 102 J of heat and the decrease in volume requires the surroundings do 7.6 J of work on the gases, what is the change in internal energy of the gases?

the units are correct, the sign is reasonable as the amount of heat lost in the reaction is much larger than

the amount of work energy gained

qreaction = −310 J, wsurroundings = −7.6 J

DEgases, J

Check:

Solution:

Concept Plan:

Relationships:

Given:

Find:

q, w DE

qsystem = −qsurroundings, wsystem = −wsurroundings, DE = q + w

wsurrounding

wgases



Heat Exchange• Heat is the exchange of thermal energy

between the system and surroundings• Heat exchange occurs when system and

surroundings have a difference in temperature• Temperature is the measure of the amount of

thermal energy within a sample of matter• Heat flows from matter with high temperature

to matter with low temperature until both objects reach the same temperaturethermal equilibrium



Quantity of Heat Energy Absorbed:Heat Capacity

• When a system absorbs heat, its temperature increases

• The increase in temperature is directly proportional to the amount of heat absorbed

• The proportionality constant is called the heat capacity, Cunits of C are J/°C or J/K

q = C x DT• The larger the heat capacity of the object

being studied, the smaller the temperature rise will be for a given amount of heat



Factors Affecting Heat Capacity

• The heat capacity of an object depends on its amount of matterusually measured by its mass200 g of water requires twice as much heat to raise

its temperature by 1°C as does 100 g of water• The heat capacity of an object depends on the

type of material1000 J of heat energy will raise the temperature of

100 g of sand 12 °C, but only raise the temperature of 100 g of water by 2.4 °C



Specific Heat Capacity• Measure of a substance’s

intrinsic ability to absorb heat• The specific heat capacity is the

amount of heat energy required to raise the temperature of one gram of a substance 1°CCs

units are J/(g∙°C)• The molar heat capacity is the

amount of heat energy required to raise the temperature of one mole of a substance 1°C



Specific Heat of Water• The rather high specific heat of water allows water to

absorb a lot of heat energy without a large increase in its temperature

• The large amount of water absorbing heat from the air keeps beaches cool in the summerwithout water, the Earth’s temperature would be about the

same as the moon’s temperature on the side that is facing the sun (average 107 °C or 225 °F)

• Water is commonly used as a coolant because it can absorb a lot of heat and remove it from the important mechanical parts to keep them from overheating or even melting it can also be used to transfer the heat to something else

because it is a fluid



Quantifying Heat Energy

• The heat capacity of an object is proportional to its mass and the specific heat of the material

• So we can calculate the quantity of heat absorbed by an object if we know the mass, the specific heat, and the temperature change of the objectHeat = (mass) x (specific heat) x (temp. change)

q = (m) x (Cs) x (DT)



Example 6.2: How much heat is absorbed by a copper penny with mass 3.10 g whose temperature rises from

−8.0 °C to 37.0 °C?

q = m ∙ Cs ∙ DTCs = 0.385 J/g•ºC (Table 6.4)

the unit is correct, the sign is reasonable as the penny must absorb heat to make its

temperature rise

T1 = −8.0 °C, T2= 37.0 °C, m = 3.10 g

q, J

Check:• Check

Solution:• Follow the conceptual plan to solve the problem

Conceptual Plan:

Relationships:

• Strategize

Given:

Find:

• Sort Information

Cs m, DT q



Practice – Calculate the amount of heat released when 7.40 g of water cools from 49° to 29 °C

(water’s specific heat is 4.18 J/gºC)



Practice – Calculate the amount of heat released when 7.40 g of water cools from 49° to 29 °C

q = m ∙ Cs ∙ DTCs = 4.18 J/gC (Table 6.4)

T1= 49 °C, T2 = 29 °C, m = 7.40 g

q, J

Check:• Check

Solution:• Follow the concept plan to solve the problem

Conceptual Plan:

Relationships:

• Strategize

Given:

Find:

• Sort Information

Cs m, DT q

39

the unit is correct, the sign is reasonable as the water must release heat to make its

temperature fallTro: Chemistry: A Molecular Approach, 2/e


Heat Transfer & Final Temperature• When two objects at different temperatures are

placed in contact, heat flows from the material at the higher temperature to the material at the lower temperature

• Heat flows until both materials reach the same final temperature

• The amount of heat energy lost by the hot material equals the amount of heat gained by the cold material

qhot = −qcold

mhotCs,hotDThot = −(mcoldCs,coldDTcold)40Tro: Chemistry: A Molecular Approach, 2/e


1. Heat lost by the metal > heat gained by water

2. Heat gained by water > heat lost by the metal

3. Heat lost by the metal > heat lost by the water

4. Heat lost by the metal = heat gained by water

5. More information is required

1. Heat lost by the metal > heat gained by water

2. Heat gained by water > heat lost by the metal

3. Heat lost by the metal > heat lost by the water

4. Heat lost by the metal = heat gained by water

5. More information is required

A piece of metal at 85 °C is added to water at 25 °C, the final temperature of the both metal and water is 30 °C. Which of the

following is true?



q = m ∙ Cs ∙ DTCs, Al = 0.903 J/g•ºC, Cs, H2O = 4.18 J/g•ºC(Table 6.4)

Example 6.3: A 32.5-g cube of aluminum initially at 45.8 °C is submerged into 105.3 g of water at 15.4 °C. What is the final

temperature of both substances at thermal equilibrium?

the unit is correct, the number is reasonable as the final temperature should be between the two initial

temperatures

mAl = 32.5 g, Tal = 45.8 °C, mH20 = 105.3 g, TH2O = 15.4 °C

Tfinal, °C

Check:

Solution:

Conceptual Plan:

Relationships:

Given:

Find:

Cs, Al mAl, Cs, H2O mH2O DTAl = kDTH2O

42

qAl = −qH2O Tfinal



Practice – A hot piece of metal weighing 350.0 g is heated to 100.0 °C. It is then

placed into a coffee cup calorimeter containing 160.0 g of water at 22.4 °C. The water warms and the copper cools until the final temperature is 35.2 °C.

Calculate the specific heat of the metal and identify the metal.


Copyright © 2011 Pearson Education, Inc.44

Practice – Calculate the specific heat and identify the metal from the data

q = m x Cs x DT; qmetal = −qH2O

the units are correct, the number indicates the metal is copper

metal: 350.0 g, T1 = 100.0 °C, T2 = 35.2 °CH2O: 160.0 g, T1 = 22.4 °C, T2 = 35.2°C, Cs = 4.18 J/g °C Cs , metal, J/gºC

Check:

Solution:

Concept Plan:

Relationships:

Given:

Find:

m, Cs, DT q



Pressure –Volume Work• PV work is work caused by a volume change against

an external pressure• When gases expand, DV is +, but the system is doing

work on the surroundings, so wgas is ─• As long as the external pressure is kept constant─ Workgas = External Pressure x Change in Volumegas

w = ─PDV to convert the units to joules use 101.3 J = 1 atm∙L



Example 6.4: If a balloon is inflated from 0.100 L to 1.85 L against an external pressure of 1.00 atm,

how much work is done?

the unit is correct; the sign is reasonable because when a gas expands it does work on the surroundings and loses energy

V1 = 0.100 L, V2 = 1.85 L, P = 1.00 atm

w, J

Check:

Solution:

Conceptual Plan:

Relationships:

Given:

Find:

P, DV w

101.3 J = 1 atm∙L



Practice – A certain process results in a gas system releasing 68.3 kJ of energy. During the process, 15.8 kcal of heat is

released by the system. If the external pressure is kept constant at 1.00 atm and the initial volume of the gas is 10.0 L,

what is the final volume of the gas?(1 cal = 4.18 J, 101.3 J = 1.00 atm•L)



Practice – A certain process results in a gas system releasing 68.3 kJ of energy. During the process, 15.8 kcal of heat is released by the system. If the external pressure is kept constant at 1.00 atm and the initial volume

of the gas is 10.0 L, what is the final volume of the gas?

so both DE and q are −, and DE > q, w must be −; and when w is − the system is expanding, so V2 should be

greater than V1; and it is

DE = −68.3 kJ, q = −15.8 kcal, V1 = 10.0 L, P = 1.00 atm

V2, L

Check:

Solution:

Conceptual Plan:

Relationships:

Given:

Find:

DE = q + w, w = −PDV, 1 kJ = 1000 J, 1 cal = 4.18 J, 101.3 J = 1 atm∙L

q, DE wV2

P, V1

DE = −6.83 x 104 J, q = −6.604 x 104J, V1 = 10.0L, P = 1.00 atm

V2, L



Exchanging Energy BetweenSystem and Surroundings

• Exchange of heat energyq = mass x specific heat x DTemperature

• Exchange of workw = −Pressure x DVolume



Measuring DE, Calorimetry at Constant Volume

• Because DE = q + w, we can determine DE by measuring q and w

• In practice, it is easiest to do a process in such a way that there is no change in volume, so w = 0 at constant volume, DEsystem = qsystem

• In practice, it is not possible to observe the temperature changes of the individual chemicals involved in a reaction – so instead, we measure the temperature change in the surroundings use insulated, controlled surroundings qsystem = −qsurroundings

• The surroundings is called a bomb calorimeter and is usually made of a sealed, insulated container filled with water

qsurroundings = qcalorimeter = ─qsystem50Tro: Chemistry: A Molecular Approach, 2/e


Bomb Calorimeter• Used to measure DE

because it is a constant volume system

• The heat capacity of the calorimeter is the amount of heat absorbed by the calorimeter for each degree rise in temperature and is called the calorimeter constantCcal, kJ/ºC



Example 6.5: When 1.010 g of sugar is burned in a bomb calorimeter, the temperature rises from 24.92 °C to 28.33 °C.

If Ccal = 4.90 kJ/°C, find DE for burning 1 mole

qcal = Ccal x DT = −qrxn

MM C12H22O11 = 342.3 g/mol

the units are correct, the sign is as expected for combustion

1.010 g C12H22O11, T1 = 24.92 °C, T2 = 28.33 °C, Ccal = 4.90 kJ/°C

DErxn, kJ/mol

Check:

Solution:

Conceptual Plan:

Relationships:

Given:

Find:

qrxn, mol DECcal, DT qcal qrxn

2.9506 x 10−3 mol C12H22O11, DT = 3.41°C, Ccal = 4.90 kJ/°C

DErxn, kJ/mol



Practice – When 1.22 g of HC7H5O2 (MM 122.12) is burned in a bomb calorimeter, the temperature rises from 20.27 °C to

22.67 °C. If DE for the reaction is −3.23 x 103 kJ/mol, what is the heat capacity of the calorimeter in kJ/°C?



Practice – When 1.22 g of HC7H5O2 is burned in a bomb calorimeter, the temperature rises from 20.27 °C to 22.67 °C.

If D E for the reaction is −3.23 x 103 kJ/mol, what is the heat capacity of the calorimeter in kJ/°C?

qcal = Ccal x DT = −qrxn

MM HC7H5O2 = 122.12 g/mol

the units and sign are correct

1.22 g HC7H5O2, T1=20.27 °C, T2=22.67 °C, DE= −3.23 x 103 kJ/mol

Ccal, kJ/°C

Check:

Solution:

Conceptual Plan:

Relationships:

Given:

Find:

qcal, DT Ccalmol, DE qrxn qcal

9.990 x 10−3 mol HC7H5O2, DT = 2.40 °C, DE= −3.23 x 103 kJ/mol

Ccal, kJ/°C



Enthalpy• The enthalpy, H, of a system is the sum of the

internal energy of the system and the product of pressure and volumeH is a state function

H = E + PV• The enthalpy change, DH, of a reaction is the

heat evolved in a reaction at constant pressureDHreaction = qreaction at constant pressure

• Usually DH and DE are similar in value, the difference is largest for reactions that produce or use large quantities of gas



Endothermic and Exothermic Reactions• When DH is ─, heat is being released by the system• Reactions that release heat are called exothermic reactions• When DH is +, heat is being absorbed by the system• Reactions that absorb heat are called endothermic reactions• Chemical heat packs contain iron filings that are oxidized in an

exothermic reaction ─ your hands get warm because the released heat of the reaction is transferred to your hands

• Chemical cold packs contain NH4NO3 that dissolves in water in an endothermic process ─ your hands get cold because the pack is absorbing your heat



• For an exothermic reaction, the surrounding’s temperature rises due to release of thermal energy by the reaction

• This extra thermal energy comes from the conversion of some of the chemical potential energy in the reactants into kinetic energy in the form of heat

• During the course of a reaction, old bonds are broken and new bonds are made

• The products of the reaction have less chemical potential energy than the reactants

• The difference in energy is released as heat

Molecular View of Exothermic Reactions



Molecular View of Endothermic Reactions

• In an endothermic reaction, the surrounding’s temperature drops due to absorption of some of its thermal energy by the reaction

• During the course of a reaction, old bonds are broken and new bonds are made

• The products of the reaction have more chemical potential energy than the reactants

• To acquire this extra energy, some of the thermal energy of the surroundings is converted into chemical potential energy stored in the products



Enthalpy of Reaction• The enthalpy change in a chemical reaction is an

extensive property the more reactants you use, the larger the enthalpy

change• By convention, we calculate the enthalpy change

for the number of moles of reactants in the reaction as writtenC3H8(g) + 5 O2(g) → 3 CO2(g) + 4 H2O(g) DH = −2044

kJ



Example 6.7: How much heat is evolved in the complete combustion of 13.2 kg of C3H8(g)?

1 kg = 1000 g, 1 mol C3H8 = −2044 kJ, Molar Mass = 44.09 g/mol

the sign is correct and the number is reasonable because the amount of C3H8 is much more than 1 mole

13.2 kg C3H8,

q, kJ/mol

Check:

Solution:

Conceptual Plan:

Relationships:

Given:

Find:

mol kJkg g



Practice – How much heat is evolved when a 0.483 g diamond is burned?

(DHcombustion = −395.4 kJ/mol C)



Practice – How much heat is evolved when a 0.483 g diamond is burned?

1 mol C = −395.4 kJ, Molar Mass = 12.01 g/mol

the sign is correct and the number is reasonable because the amount of diamond is less than 1 mole

0.483 g C, DH = −395.4 kJ/mol C

q, kJ/mol

Check:

Solution:

Concept Plan:

Relationships:

Given:

Find:

mol kJg



Measuring DHCalorimetry at Constant Pressure

• Reactions done in aqueous solution are at constant pressureopen to the atmosphere

• The calorimeter is often nested foam cups containing the solution

qreaction = ─ qsolution = ─(masssolution x Cs, solution x DT)

DHreaction = qconstant pressure = qreaction to get DHreaction per mol, divide by the

number of moles63Tro: Chemistry: A Molecular Approach, 2/e


Example 6.8: What is DHrxn/mol Mg for the reaction Mg(s) + 2 HCl(aq) → MgCl2(aq) + H2(g) if 0.158 g Mg reacts in

100.0 mL of solution and changes the temperature from 25.6 °C to 32.8 °C?

the sign is correct and the value is reasonable because the reaction is exothermic

0.158 g Mg, 100.0 mL, T1 = 25.6 °C, T2 = 32.8 °C, assume d = 1.00 g/mL, Cs = 4.18 J/g∙°C H, kJ/mol

Check:

Solution:

Conceptual Plan:

Relationships:

Given:

Find:

qrxn, mol DHCs, DT, m qsol’n qrxn

qsol’n = m x Cs, sol’n x DT = −qrxn, MM Mg= 24.31 g/mol

6.499 x 10−3 mol Mg, 1.00 x 102 g, DT = 7.2 °C,,Cs = 4.18 J/g∙°C

DH, kJ/mol



Practice – What will be the final temperature of the solution in a coffee cup calorimeter if a 50.00 mL sample of 0.250 M HCl(aq) is

added to a 50.00 mL sample of 0.250 NaOH(aq). The initial temperature is 19.50 °C and the DHrxn is −57.2 kJ/mol NaOH.

(assume the density of the solution is 1.00 g/mL and the specific heat of the solution is 4.18 J/g∙°C)



qsol’n = m x Cs, sol’n x DT = −qrxn, 0.250 mol NaOH = 1 L

Practice – What will be the final temperature of the solution in a coffee cup calorimeter if a 50.00 mL sample of 0.250 M HCl(aq) is added to a 50.00 mL

sample of 0.250 NaOH(aq). The initial temperature is 19.50 °C.

the sign is correct and the value is reasonable, because the reaction is exothermic we expect the final temperature to be higher than the initial

50.00 mL of 0.250 M HCl, 50.00 mL of 0.250 M NaOH T1 = 19.50 °C, d = 1.00 g/mL, Cs = 4.18 J/g∙°C, D H = −57.2 kJ/molT2, °C

Check:

Solution:

Conceptual Plan:

Relationships:

Given:

Find:

s

nsol'Cm

T

DqHmol rxn Dq rxnnsol' qq

mol rxnqH D

1.25 x 10−2 mol NaOH, 1.00 x 102 g, T1 = 19.50 °C, DH =−57.2 kJ/mol, Cs = 4.18 J/g∙°CT2, °C

qrxn qsol’nDH DTCs, qsol’n, m T2



Relationships Involving DHrxn

• When reaction is multiplied by a factor, DHrxn is multiplied by that factorbecause DHrxn is extensive

C(s) + O2(g) → CO2(g) DH = −393.5 kJ2 C(s) + 2 O2(g) → 2 CO2(g) DH = 2(−393.5 kJ) = −787.0

kJ

• If a reaction is reversed, then the sign of DH is changed

CO2(g) → C(s) + O2(g) DH = +393.5 kJ



Relationships Involving DHrxn

Hess’s Law• If a reaction can be

expressed as a series of steps, then the DHrxn for the overall reaction is the sum of the heats of reaction for each step



[3 NO2(g) 3 NO(g) + 1.5 O2(g)] DH° = (+174 kJ)[1 N2(g) + 2.5 O2(g) + 1 H2O(l) 2 HNO3(aq)] DH° = (−128 kJ)

[2 NO(g) N2(g) + O2(g)] DH° = (−183 kJ)3 NO2(g) + H2O(l) 2 HNO3(aq) + NO(g) DH° = − 137 kJ

Example: Hess’s Law

Given the following information:2 NO(g) + O2(g) 2 NO2(g) DH° = −116 kJ

2 N2(g) + 5 O2(g) + 2 H2O(l) 4 HNO3(aq) DH° = −256 kJN2(g) + O2(g) 2 NO(g) DH° = +183 kJ

Calculate the DH° for the reaction below:3 NO2(g) + H2O(l) 2 HNO3(aq) + NO(g) DH° = ?

[2 NO2(g) 2 NO(g) + O2(g)] x 1.5 DH° = 1.5(+116 kJ)[2 N2(g) + 5 O2(g) + 2 H2O(l) 4 HNO3(aq)] x 0.5 DH° = 0.5(−256 kJ)

[2 NO(g) N2(g) + O2(g)] DH° = −183 kJ



Practice – Hess’s Law

Given the following information:Cu(s) + Cl2(g) CuCl2(s) DH° = −206 kJ2 Cu(s) + Cl2(g) 2 CuCl(s) DH° = − 36 kJ

Calculate the DH° for the reaction below:Cu(s) + CuCl2(s) 2 CuCl(s) DH° = ? kJ



Practice – Hess’s Law

Given the following information: Cu(s) + Cl2(g) CuCl2(s) DH° = −206 kJ2 Cu(s) + Cl2(g) 2 CuCl(s) DH° = − 36 kJ

Calculate the DH° for the reaction below:Cu(s) + CuCl2(s) 2 CuCl(s) DH° = ? kJ

CuCl2(s) Cu(s) + Cl2(g) DH° = +206 kJ2 Cu(s) + Cl2(g) 2 CuCl(s) DH° = − 36 kJCu(s) + CuCl2(s) 2 CuCl(s) DH° = +170. kJ



Standard Conditions• The standard state is the state of a material at a defined

set of conditions pure gas at exactly 1 atm pressure pure solid or liquid in its most stable form at exactly 1 atm pressure

and temperature of interest usually 25 °C

substance in a solution with concentration 1 M• The standard enthalpy change, DH°, is the enthalpy

change when all reactants and products are in their standard states

• The standard enthalpy of formation, DHf°, is the enthalpy change for the reaction forming 1 mole of a pure compound from its constituent elements the elements must be in their standard states the DHf° for a pure element in its standard state = 0 kJ/mol

by definition72Tro: Chemistry: A Molecular Approach, 2/e


Standard Enthalpies of Formation



Formation Reactions

• Reactions of elements in their standard state to form 1 mole of a pure compoundif you are not sure what the standard state of an

element is, find the form in Appendix IIB that has a DHf° = 0

because the definition requires 1 mole of compound be made, the coefficients of the reactants may be fractions



Writing Formation ReactionsWrite the Formation Reaction for CO(g)• The formation reaction is the reaction between the

elements in the compound, which are C and OC + O → CO(g)

• The elements must be in their standard state there are several forms of solid C, but the one with DHf° = 0 is graphite

oxygen’s standard state is the diatomic gasC(s, graphite) + O2(g) → CO(g)

• The equation must be balanced, but the coefficient of the product compound must be 1use whatever coefficient in front of the reactants is

necessary to make the atoms on both sides equal without changing the product coefficient

C(s, graphite) + ½ O2(g) → CO(g)75Tro: Chemistry: A Molecular Approach, 2/e


Write the formation reactions for the following:

CO2(g)

Al2(SO4)3(s)

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

2 Al(s) + 3/8 S8(s, rhombic) + 6 O2(g) Al2(SO4)3(s)



Calculating Standard Enthalpy Change for a Reaction

• Any reaction can be written as the sum of formation reactions (or the reverse of formation reactions) for the reactants and products

• The DH° for the reaction is then the sum of the DHf° for the component reactionsDH°reaction = S n DHf°(products) − S n DHf°(reactants)S means sumn is the coefficient of the reaction



CH4(g) → C(s, graphite) + 2 H2(g) D H° = + 74.6 kJ

2 H2(g) + O2(g) → 2 H2O(g) DH° = −483.6 kJ

CH4(g)+ 2 O2(g)→ CO2(g) + H2O(g)

C(s, graphite) + 2 H2(g) → CH4(g) DHf°= − 74.6 kJ/mol CH4

C(s, graphite) + O2(g) → CO2(g) DHf°= −393.5 kJ/mol CO2

H2(g) + ½ O2(g) → H2O(g) DHf°= −241.8 kJ/mol H2O

CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(g) DH° = −802.5 kJ

DH° = [((−393.5 kJ)+ 2(−241.8 kJ)− ((−74.6 kJ)+ 2(0 kJ))] = −802.5 kJ

78

DH° = [(DHf° CO2(g) + 2∙DHf°H2O(g))− (DHf° CH4(g) + 2∙DHf°O2(g))]



Example: Calculate the enthalpy change in the reaction

2 C2H2(g) + 5 O2(g) 4 CO2(g) + 2 H2O(l)1. Write formation reactions for each compound

and determine the DHf° for each

2 C(s, gr) + H2(g) C2H2(g) DHf° = +227.4 kJ/mol

C(s, gr) + O2(g) CO2(g) DHf° = −393.5 kJ/mol

H2(g) + ½ O2(g) H2O(l)DHf° = −285.8 kJ/mol



2 C2H2(g) 4 C(s) + 2 H2(g) DH° = 2(−227.4) kJ


2 C2H2(g) + 5 O2(g) 4 CO2(g) + 2 H2O(l)

4 C(s) + 4 O2(g) 4CO2(g)DH° = 4(−393.5) kJ

2 H2(g) + O2(g) 2 H2O(l) DH° = 2(−285.8) kJ

2. Arrange equations so they add up to desired reaction

2 C2H2(g) + 5 O2(g) 4 CO2(g) + 2 H2O(l) DH = −2600.4 kJ




2 C2H2(g) + 5 O2(g) 4 CO2(g) + 2 H2O(l)

DH°reaction = S n DHf°(products) − S n DHf°(reactants)

DHrxn = [(4•DHCO2 + 2•DHH2O) – (2•DHC2H2 + 5•DHO2)]

DHrxn = [(4•(−393.5) + 2•(−285.8)) – (2•(+227.4) + 5•(0))]

DHrxn = −2600.4 kJ



Example 6.12: How many kg of octane must be combusted to supply 1.0 x 1011 kJ of energy?

MMoctane = 114.2 g/mol, 1 kg = 1000 g

the units and sign are correct,the large value is expected

1.0 x 1011 kJmass octane, kg

Check:

Solution:

Conceptual Plan:

Relationships:

Given:Find:

DHf°’s DHrxn°Write the balanced equation per mole of octane

kJ mol C8H18 g C8H18 kg C8H18from

above

C8H18(l) + 25/2 O2(g) → 8 CO2(g) + 9 H2O(g)

Look up the DHf°for each material in Appendix IIB

DH°rxn = −5074.1 kJ



Practice – Calculate the DH° for decomposing 10.0 g of limestone, CaCO3, under standard conditions.

CaCO3(s) → CaO(s) + O2(g)



Practice – Calculate the DH° for decomposing 10.0 g of limestone, CaCO3(s) → CaO(s) + O2(g)

MMlimestone = 100.09 g/mol,

the units and sign are correct

10.0 g CaCO3

DH°, kJ

Check:

Solution:

Conceptual Plan:

Relationships:

Given:Find:

DHf°’s DH°rxn per mol CaCO3

g CaCO3 mol CaCO3 kJfrom

above

CaCO3(s) → CaO(s) + O2(g)

DH°rxn = +572.7 kJ



Energy Use and the Environment• In the United States, each person uses over 105 kWh of

energy per year• Most comes from the combustion of fossil fuels

combustible materials that originate from ancient lifeC(s) + O2(g) → CO2(g) DH°rxn = −393.5 kJCH4(g) +2 O2(g) → CO2(g) + 2 H2O(g) DH°rxn = −802.3 kJC8H18(g) +12.5 O2(g) → 8 CO2(g) + 9 H2O(g) DH°rxn = −5074.1 kJ

• Fossil fuels cannot be replenished• At current rates of consumption, oil and natural gas

supplies will be depleted in 50–100 yrs



Energy Consumption

86

• The distribution of energy consumption in the United States

• The increase in energy consumption in the United States



The Effect of Combustion Products on Our Environment

• Because of additives and impurities in the fossil fuel, incomplete combustion, and side reactions, harmful materials are added to the atmosphere when fossil fuels are burned for energy

• Therefore fossil fuel emissions contribute to air pollution, acid rain, and global warming



Global Warming• CO2 is a greenhouse gas

it allows light from the sun to reach the earth, but does not allow the heat (infrared light) reflected off the Earth to escape into outer space it acts like a blanket

• CO2 levels in the atmosphere have been steadily increasing

• Current observations suggest that the average global air temperature has risen 0.6 °C in the past 100 yrs.

• Atmospheric models suggest that the warming effect could worsen if CO2 levels are not curbed

• Some models predict that the result will be more severe storms, more floods and droughts, shifts in agricultural zones, rising sea levels, and changes in habitatsTro: Chemistry: A Molecular Approach, 2/e


• Our greatest unlimited supply of energy is the sun

• New technologies are being developed to capture the energy of sunlightparabolic troughs, solar power towers, and dish

engines concentrate the sun’s light to generate electricity

solar energy used to decompose water into H2(g) and O2(g); the H2 can then be used by fuel cells to generate electricity

H2(g) + ½ O2(g) → H2O(l) DH°rxn = −285.8 kJ • Hydroelectric power• Wind power

90

Renewable Energy


Chapter 6 Thermochemistry

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Transcript of Chapter 6 Thermochemistry