Chapter 6 Energy and Chemical Change - Blog FTSL-2 … 6 Energy and Chemical Change Brady and Senese...

Chapter 6Energy and Chemical Change

Brady and Senese

5th Edition

Index

61 An object has energy if it is capable of doing work62 Internal energy is the total energy of an objectrsquos

molecules63 Heat can be determined by measuring temperature

changes 64 Energy is absorbed or released during most chemical

reactions65 Heats of reaction are measured at constant volume or

constant pressure66 Thermochemical equations are chemical equations that

quantitatively include heat67 Thermochemical equations can be combined because

enthalpy is a state function68 Tabulated standard heats of reaction can be used to

predict any heat of reaction using Hessrsquos law

61 An object has energy if it is capable of doing work 3

Energy is the Ability to do Work

bull Energy is the ability to do work (move mass over a distance) or transfer heat

bull Types kinetic and potential Kinetic - the energy of motion

Potential - the stored energy in matter

bull Internal energy (E) - the sum of the kinetic and potential energy for each particle in the system


Kinetic Energy The Energy of Motion

bull KE = frac12mv2

bull Energy can be transferred by moving particles

bull Collision of fast particles with slower particles causes the slow particle to speed up while the fast molecule slows This is why hot water cools in contact with cool air


Potential Energy Depends on Position

bull Potential energy increases when Objects that attract move apart or

Objects that repel move toward each other

bull Stored energy that can be converted to kinetic energy


Your Turn

Which of the following is not a form of kinetic energy

A A pencil rolls across a desk

B A pencil is sharpened

C A pencil is heated

D All are forms of kinetic energy

E None are forms of kinetic energy


Law of Conservation of Energy

bull Energy cannot be created or destroyed but can be transformed from one form of energy to another

bull Also known as the first law of thermodynamics

How does water falling over a waterfall demonstrate this law


Heat and Temperature are Not the Same

bull The temperature of an object is proportional to the average kinetic energy of its particlesmdashthe higher the average kinetic energy the higher the temperature

bull Heat is energy (also called thermal energy) transferred between objects caused by differences in their temperatures until they reach thermalequilibrium


Units of Energy

bull SI unit is the Joule J J = kgm2s2

If the calculated value is greater than 1000 J use the kJ

bull Another unit is the calorie cal cal = 4184 J (exact)

bull Nutritional unit is the Calorie (note capital C) which is one kilocalorie 1 Cal = 1 kcal = 4184 kJ

62 Internal energy is the total energy of an objectrsquos molecules 10

bull 1st Law of Thermodynamics For an isolated system the internal energy (E) is constant

Δ E = Ef - Ei = 0

Δ E = Eproduct- Ereactant= 0

bull We canrsquot measure the internal energy of anything so we measure the changes in energy

bull E is a state function

Internal Energy is Conserved


bull Temperature (T) is proportional to the average kinetic energy of all particle units degC degF K KEaverage= frac12mvaverage

2

bull At a high temperature most molecules are moving at higher average

What is Temperature


State Function

bull A property whose value depends only on the present state of the system not on the method or mechanism used to arrive at that state

bull Position is a state function both train and car travel to the same locations although their paths vary

bull The actual distance traveled does vary with path

New York

Los Angeles

63 Heat can be determined by measuring temperature changes 13

Heat Transfer is a State Function

bull Transfer of heat during a reaction is a state functionbull The route taken to arrive at the products does not

affect the amount of heat that is transferredbull The number of steps does not affect the amount of

heat that is transferred


Heat Transfer q

bull Heat (q) - the transfer of energy from regions of high temperature to regions of lower temperature

bull Units J cal kgmiddotm2s2

A calorie is the amount of energy needed to raise the temperature of 100 g water from 145 to 155 degC

bull A metal spoon at 25 degC is placed in boiling water What happens


Surroundings System Universe

bull System - the reaction or area under studybull Surroundings - the rest of the universebull Open systems can gain or lose mass and energy

across their boundaries ie the human body

bull Closed systems can absorb or release energy but not mass across the boundary ie a light bulb

bull Isolated systems (adiabatic) cannot exchange matter or energy with their surroundings ie a closed Thermos bottle


The Sign Convention

bull Endothermic systems require energy to be added to the system thus the q is (+)

bull Exothermic reactions release energy to the surroundings Their q is (-)

bull Energy changes are measured from the point of view of the system


Your Turn

A cast iron skillet is moved from a hot oven to a sink full of water Which of the following is not true

A The water heats

B The skillet cools

C The heat transfer for the skillet has a (-) sign

D The heat transfer for the skillet is the same as the heat transfer for the water

E None of these are untrue


Heat Capacity and Transfer

bull Heat capacity (C) - the (extensive) ability of an object with constant mass to absorb heat Calorimeter constant

Varies with the sample mass and the identity of the substance

Units J degC-1

bull q = Ctimes ∆t q = heat transferred

C = heat capacity of object

Δt = Change in Temperature (tfinal - tinitial)


Learning Check

A cup of water is used in an experiment Its heat capacity is known to be 720 J degC-1 How much heat will it absorb if the experimental temperature changed from 192 degC to 235 degC

q = Ctimes ∆t

q = 720 J degC-1times (235 - 192 degC)

q = 31 times 103 J


Heat Transfer and Specific Heat

bull Specific heat (s) - The intensive ability of a substance to store heat

C = mtimes s

Units J g-1degC-1 or J g-1 K-1 or J mol-1 K-

1

bull q = mtimes ∆ttimes s

q = heat transferred

m = mass of object

Δt = change in temperature (tfinal - tinitial)


Specific Heats

SubstanceSpecific Heat

J g-1degC-1

(25 degC)

Carbon (graphite) 0711

Copper 0387

Ethyl alcohol 245

Gold 0129

Granite 0803

Iron 04498

Lead 0128

Olive oil 20

Silver 0235

Water (liquid) 418

bull Substances with high specific heats resist temperature changes

bull Note that water has a very high specific heat This is why coastal

temperatures are different from inland temperatures


Learning Check

Calculate the specific heat of a metal if it takes 235 J to raise the temperature of a 3291 g sample by 253 degC

235 J J282

3291 g 253 degC g degC

= times times ∆

= = =times ∆ times

q m s t

qs

m t


The First Law of Thermodynamics Explains Heat Transfer

bull If we monitor the heat transfers (q) of all materials involved and all work processes we can predict that their sum will be zero

bull By monitoring the surroundings we can predict what is happening to our system

bull Heat transfers until thermal equilibrium thus the final temperature is the same for all materials


Learning CheckA 4329 g sample of solid is transferred from boiling water (t = 998 degC) to 152 g water at 225 degC in a coffee cup The twaterrose to 243 degC Calculate the specific heat of the solid

qsample+ qwater+ qcup= 0

qcup is neglected in problem

qsample= -qwater

qsample= mtimes stimes ∆t

qsample= 4329 g times stimes (243 - 998 degC)

qwater= 152 g times 4184 J g-1 degC-1times (243 ndash 225 degC)

4329 g times stimes (243 - 998 degC) = -(152 g times 4184 J g-1degC-1times (243 ndash225 degC))

stimes (-327 times 103) g-1degC-1 = -114 times 103 J

s = 0349 J g-1degC-1


Your Turn

What is the heat capacity of the container if 100 g of water (s = 4184 J g-1degC-1) at 100 degC are added to 100 g of water at 25 degC in the container and the final temperature is 61 degC

A 870 JdegC

B 35 JdegC

C -35 JdegC

D -870 JdegC

E None of these

64 Energy is absorbed or released during most chemical reactions 26

bull Chemical bond - net attractive forces that bind atomic nuclei and electrons together

bull Exothermic reactions form stronger bonds in the product than in the reactant and release energy

bull Endothermic reactions break stronger bonds than they make and require energy

Chemical Potential Energy

65 Heats of reaction are measured at constant volume or constant pressure 27

Work and Pistons

bull Pressure = forcearea

bull If the container volume changes the pressure changes

bull Work = -PtimesΔV Units L bull atm

1 L bull atm = 101 J

bull In expansion ∆V gt 0 and is exothermic

bull Work is done by the system in expansion


How does work relate to reactions

bull Work = Force middot Distance

bull Is most often due to the expansion or contraction of a system due to changing moles of gas

bull Gases push against the atmospheric pressure so Psystem= -Patm

w = -Patm timesΔV

bull The deployment of an airbag is one example of this process

1C3H8(g) + 5O2(g) rarr 3CO2(g) + 4H2O(g)

6 moles of gasrarr 7 moles of gas


Learning Check P-V workEthyl chloride is prepared by reaction of ethylene with HCl How much P-V work (in J) is done if 895 g ethylene and 125 g of HCl are allowed to react at atmospheric pressure and the volume change is -715 L (1 L bull atm = 101 J)

Calculate the work (in kilojoules) done during a synthesis of ammonia in which the volume contracts from 86 L to 43 L at a constant external pressure of 44 atm

w = 19 kJ

w = 722 times 103 J

w = -44 atm times (43 - 86) L = 19 Latm

w = -1atm times -715 L = 715 Latm


Energy can be Transferred as Heat and Work

bull ∆E = q + w

bull Internal energy changes are state functions


Your TurnWhen TNT is combusted in air it is according to the following reaction

4C6H2(NO2)3CH3(s) + 17O2(g) rarr 24CO2(g) + 10H2O(l) + 6N2(g)

The reaction will do work for all of these reasons except

A The moles of gas increase

B The volume of gas increases

C The pressure of the gas increases

D None of these


Calorimetry is Used to Measure Heats ofReaction

bull Heat of reaction - the amount of heat absorbed or released in a chemical reaction

bull Calorimeter - an apparatus used to measure temperature changes in materials surrounding a reaction that result from a chemical reaction

bull From the temperature changes we can calculate the heat of the reaction q qv heat measured under constant volume conditions qp heat measured under constant pressure conditions qp termed enthalpy ∆H


Internal Energy is Measured with a Bomb Calorimeter

bull Used for reactions in which there is change in the number of moles of gas present

bull Measures qv

bull Immovable walls mean that work is zero

bull ΔE = qv


Learning Check Bomb CalorimeterA sample of 500 mg naphthalene (C10H8) is combusted in a bomb calorimeter containing 1000 g of water The temperature of the water increases from 2000 degC to 2437 degC The calorimeter constant is 420 JdegC What is the change in internal energy for the reaction swater= 4184 JgdegC

qcal= 420 Jgtimes (2437 - 2000) degC

qwater= 1000 g times (2437 - 2000) degC times 4184 JgdegC

qv reaction+ qwater+ qcal = 0 by the first law

qv reaction= ∆E = -20 times 104 J


Enthalpy of Combustion

bull When one mole of a fuel substance is reacted with elemental oxygen a combustion reaction can be written

bull Is always negative

bull Learning Check What is the equation associated with the enthalpy of combustion of C6H12O6(s)

C6H12O6(s) + 9O2(g) rarr 6CO2(g) + 6H2O(l)


Your Turn

A 252 mg sample of benzoic acid C6H5CO2H is combusted in a bomb calorimeter containing 814 g water at 2000 degC The reaction increases the temperature of the water to 2170 degC What is the internal energy released by the process

A -711 J

B -285 J

C +711 J

D +285 J

E None of these

swater= 4184 Jg degC

qw + qcal + qv reaction= 0 qcal is ignored by the problemqv reaction= -qw = -814 g times (2170 - 2000) degC times 4184 J g-1degqv reaction= -5789 J

-579 kJ


Enthalpy Change (ΔH)

bull Enthalpy is the heat transferred at constant pressure

ΔH = qp

ΔE = qp - PΔV = ΔH - PΔV

ΔH = ΔHfinal -ΔHinitial

ΔH = ΔHproduct-ΔHreactant


Enthalpy Measured in a Coffee Cup Calorimeter

bull When no change in moles of gas is expected we may use a coffee cup calorimeter

bull The open system allows the pressure to remain constant

bull Thus we measure qp

bull ∆E = qp + w or ∆E = ∆H ndash P∆V

bull Since there is no change in the moles of gas present there is no work

bull Thus we also are measuring ∆E


Learning Check Coffee Cup CalorimetryWhen 500 mL of 0987 M H2SO4 is added to 500 mL of 100 MNaOH at 250 degC in a coffee cup calorimeter the temperature of the aqueous solution increases to 317 degC Calculate heat for the reaction per mole of limiting reactant

Assume that the specific heat of the solution is 418 JgdegC the density is 100 gmL and that the calorimeter itself absorbs a negligible amount of heat

H2SO4(aq) + 2NaOH(aq) rarr 2H2O(l) + Na2SO4(aq)

mol NaOH = 00500 mol is limitingmol H2SO4 = 00494 mol

qsoln = 100 g soln times (317 - 250) degC times 418 JgdegC

qp rxn = -56 times 104 J

qp rxn + qcal + qsoln = 0 thus qp rxn = -qsoln



Your TurnA sample of 5000 mL of 0125 M HCl at 2236 degC is added to a 5000 mL of 0125 M Ca(OH)2 at 2236 degC The calorimeter constant was 72 J g-1degC-1 The temperature of the solution (s = 4184 J g-1degC-1 d = 100 gmL) climbed to 2330 degC Which of the following is not true

A qcal = 677 JB qsolution = 3933 JC qrxn = 4610 JD qrxn = -4610 JE None of these


Calorimetry Overview

bull The equipment used depends on the reaction type

bull If there will be no change in the moles of gas we may use a coffee-cup calorimeter or a closed system Under these circumstances we measure qp

bull If there is a large change in the moles of gas we use a bomb calorimeter to measure qv

66 Thermochemical equations are chemical equations that quantitatively include heat 42

Thermochemical Equations

bull Relate the energy of a reaction to the quantities involved

bull Must be balanced but may use fractional coefficients bull Quantities are presumed to be in molesbull Example

C(s) + O2(g) rarr CO2(g) ∆Hdeg = -3935 kJ


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

ΔH = -2511 kJ

The reactants (acetylene and oxygen) have 2511 kJ more energy than the products How many kJ are released for 1 mol C2H2

Learning Check

1256 kJ


Learning Check

6CO2(g) + 6H2O(l) rarr C6H12O6(s) + 6O2(g) ∆H = 2816 kJ

How many kJ are required for 44 g CO2 (molar mass = 4401 gmol)

If 100 kJ are provided what mass of CO2 can be converted to glucose

938 g

470 kJ


Learning Check Calorimetry of Chemical ReactionsThe meals-ready-to-eat (MRE) in the military can be heated on a flameless heater Assume the reaction in the heater is

Mg(s) + 2H2O(l) rarr Mg(OH)2(s) + H2(g)

ΔH = -353 kJ

What quantity of magnesium is needed to supply the heat required to warm 25 mL of water from 25 to 85 degC Specific heat of water = 4184 J g-1degC-1 Assume the density of the solution is the same as for water at 25 degC 100 g mL-1

qsoln = 25 g times (85 - 25) degC times 4184 J g-1degC-1 = 63times 103 J

masssoln = 25 mL times 100 g mL-1 = 25 g

(63 kJ)(1 mol Mg353 kJ)(243 g mol-1 Mg) = 043 g


Your TurnConsider the thermite reaction The reaction is initiated by the heat released from a fuse or reaction The enthalpy change is -848 kJ mol-1 Fe2O3 at 298 K

2Al(s) + Fe2O3(s) rarr 2Fe(s) + Al2O3(s)

What mass of Fe (molar mass 55847 g mol-1) is made when 500 kJ are released

A 659 g

B 0587 g

C 328 g

D None of these


Learning Check Ethyl Chloride Reaction RevisitedEthyl chloride is prepared by reaction of ethylene with HCl

C2H4(g) + HCl(g) rarr C2H5Cl(g) ΔHdeg = -723 kJ

What is the value of ΔE if 895 g ethylene and 125 g of HCl are allowed to react at atmospheric pressure and the volume change is -715 L

Ethylene = 2805 gmol HCl = 3646 gmol

mol C2H4 319 mol is limitingmol HCl 343 molΔHrxn =319mol times-723 kJmol

=-2306 kJ

w = -1 atm times -715 L

= 715 Latm = 7222 kJΔE= -2306kJ+72kJ= -223 kJ

67 Thermochemical equations can be combined because enthalpy is a state function 48

Enthalpy Diagram


Hessrsquos Law

The overall enthalpy change for a reaction is equal to the sum of the enthalpychangesfor individual steps in the reaction

For example

2Fe(s) + 32O2(g) rarr Fe2O3(s) ∆H = -8222 kJ

Fe2O3(s) + 2Al(s) rarr Al2O3(s) + 2Fe(s) ∆H = -848 kJ

32O2(g) + 2Al(s) rarr Al2O3(s) ∆H = -8222 kJ + -848 kJ

-1670 kJ


Rules for Adding Thermochemical Reactions

1 When an equation is reversedmdashwritten in the opposite directionmdashthe sign of H must also be reversed

2 Formulas canceled from both sides of an equation must be for the substance in identical physical states

3 If all the coefficients of an equation are multiplied or divided by the same factor the value of H must likewise be multiplied or divided by that factor


Strategy for Adding Reactions Together

1 Choose the most complex compound in the equation (1)

2 Choose the equation (2 or 3 orhellip) that contains the compound

3 Write this equation down so that the compound is on the appropriate side of the equation and has an appropriate coefficient for our reaction

4 Look for the next most complex compound


Hessrsquos Law (Cont)

5 Choose an equation that allows you to cancel intermediates and multiply by an appropriate coefficient

6 Add the reactions together and cancel like terms

7 Add the energies together modifying the enthalpy values in the same way that you modified the equation

bull If you reversed an equation change the sign on the enthalpy

bull If you doubled an equation double the energy


Learning Check

How can we calculate the enthalpy change for the reaction 2 H2(g) + N2(g) rarr N2H4(g) using these equations

N2H4(g) + H2(g) rarr 2NH3(g) ΔHdeg = -1878 kJ

3H2(g) + N2(g) rarr 2NH3(g) ΔHdeg = -924 kJ

Reverse the first reaction (and change sign)2NH3(g) rarr N2H4(g) + H2(g) ΔHdeg = +1878 kJ

Add the second reaction (and add the enthalpy)3H2(g) + N2(g) rarr 2NH3(g) ΔHdeg = -924 kJ

2NH3(g) + 3H2(g) + N2(g) rarr N2H4(g) + H2(g) + 2NH3(g)

2H2(g) + N2(g) rarr N2H4(g) (1878 - 924) = +954 kJ

2


Learning CheckCalculate ΔH for 2C(s) + H2(g) rarr C2H2(g) using

bull 2C2H2(g) + 5O2(g) rarr 4CO2(g) + 2H2O(l) ΔHdeg = -25992 kJ

bull C(s) + O2(g) rarr CO2(g) ΔHdeg = -3935 kJ

bull H2O(l) rarr H2(g) + frac12 O2(g) ΔHdeg = +2859 kJ

-frac12(2C2H2(g) + 5O2(g) rarr 4CO2(g) + 2H2O(l) ΔHdeg = -25992)

2CO2(g) + H2O(l) rarr C2H2(g) + 52O2(g) ΔHdeg = 12996

2(C(s) + O2(g) rarr CO2(g) ΔHdeg = -3935 kJ)

2C(s)+ 2O2(g) rarr 2CO2(g) ΔHdeg = -7870 kJ

-1(H2O(l) rarr H2(g) + frac12 O2(g) ΔHdeg = +2859 kJ)

H2(g) + frac12 O2(g) rarrH2O(l) ΔHdeg = -2859 kJ

2C(s) + H2(g) rarr C2H2(g) ΔHdeg = +2267 kJ


Your Turn

What is the energy of the following process6A + 9B + 3D + F rarr 2G

Given that C rarr A + 2B ∆H = 202 kJmol2C + D rarr E + B ∆H = 301 kJmol3E + F rarr 2G ∆H = -801 kJmolA 706 kJB -298 kJC -1110 kJD None of these

68 Tabulated standard heats of reaction can be used to predict any heat of reaction using Hessrsquos law

56

State Matters

bull C3H8(g) + 5O2(g) rarr 3CO2(g) + 4H2O(g) ΔH = -2043 kJ

bull C3H8(g) + 5O2(g) rarr 3CO2(g) + 4H2O(l) ∆H = -2219 kJ

bull Note that there is a difference in energy because the states do not match

bull If H2O(l) rarr H2O(g) ∆H = 44 kJmol

4H2O(l) rarr 4H2O(g) ∆H = 176 kJmol

-2219 + 176 kJ = -2043 kJ


57

Most stable form of the pure substance at

bull 1 atm pressure

bull Stated temperature If temperature is not specified assume 25 degC

bull Solutions are 1 M in concentration

bull Measurements made under standard state conditions have the deg mark ΔHdeg

bull Most ΔH values are given for the most stable form of the compound or element

Standard State


58

Determining the Most Stable State

bull The most stable form of a substance below the melting point is solid above the boiling point is gas between these temperatures is liquid

What is the standard state of GeH4 mp -165 degC bp -885 degC

What is the standard state of GeCl4 mp -495 degC bp 84 degC

gas

liquid


59

Allotropes

bull Are substances that have more than one form in the same physical state

bull You should know which form is the most stable

bull C P O and S all have multiple allotropes Which is the standard state for each C ndash solid graphite

P ndash solid white

O ndash gas O2 S ndash solid rhombic


60

Enthalpy of Formation

bull Enthalpy of formation is the enthalpy change ΔHdegf for the formation of 1 mole of a substance in its standard state from elements in their standard states

bull Note ΔHdegf = 0 for an element in its standard state

bull Learning CheckWhat is the equation that describes the formation of CaCO3(s)

Ca(s) + C(graphite) + 32O2(g) rarr CaCO3(s)


61

Calculating ΔH for Reactions Using ΔHdegdegdegdegf

ΔHdegrxn = [sum of ΔHdegf of all products]

ndash [sum ofΔHdegf of all reactants]

2Fe(s) + 6H2O(l) rarr 2Fe(OH)3(s) + 3H2(g)

0 -2858 -6965 0

CO2(g) + 2H2O(l) rarr 2O2(g) + CH4(g)-3935 -2858 0 -748

ΔHdegrxn = 3218 kJ

ΔHdegrxn = 8903 kJ

Your TurnWhat is the enthalpy for the following reaction

A 965 kJ

B -965 kJ

C 482 kJ

D -482 kJ

E None of these

2H2CO3(aq) + 2OH-(aq) rarr 2H2O(l) + 2HCO3- (aq)

∆Hfordm -69965

kJmol

-2300 kJmol

-2859 kJmol

-69199 kJmol

Index

61 An object has energy if it is capable of doing work62 Internal energy is the total energy of an objectrsquos

molecules63 Heat can be determined by measuring temperature

changes 64 Energy is absorbed or released during most chemical

reactions65 Heats of reaction are measured at constant volume or

constant pressure66 Thermochemical equations are chemical equations that

quantitatively include heat67 Thermochemical equations can be combined because

enthalpy is a state function68 Tabulated standard heats of reaction can be used to

predict any heat of reaction using Hessrsquos law









bull KE = frac12mv2









Your Turn

















Units of Energy







Δ E = Ef - Ei = 0







2


What is Temperature


State Function




New York

Los Angeles







Heat Transfer q












The Sign Convention





Your Turn


A The water heats

B The skillet cools








Units J degC-1





Learning Check


q = Ctimes ∆t


q = 31 times 103 J




C = mtimes s


1



m = mass of object



Specific Heats


J g-1degC-1

(25 degC)


Copper 0387

Ethyl alcohol 245

Gold 0129

Granite 0803

Iron 04498

Lead 0128

Olive oil 20

Silver 0235

Water (liquid) 418





Learning Check


235 J J282


= times times ∆


q m s t

qs

m t










qsample= -qwater






s = 0349 J g-1degC-1


Your Turn


A 870 JdegC

B 35 JdegC

C -35 JdegC

D -870 JdegC

E None of these







Work and Pistons












w = -Patm timesΔV







w = 19 kJ

w = 722 times 103 J

w = -44 atm times (43 - 86) L = 19 Latm




bull ∆E = q + w









D None of these









bull Measures qv


bull ΔE = qv














Your Turn


A -711 J

B -285 J

C +711 J

D +285 J

E None of these



-579 kJ




ΔH = qp




































ΔH = -2511 kJ


Learning Check

1256 kJ


Learning Check




938 g

470 kJ




ΔH = -353 kJ









A 659 g

B 0587 g

C 328 g

D None of these







=-2306 kJ


= 715 Latm = 7222 kJΔE= -2306kJ+72kJ= -223 kJ


Enthalpy Diagram


Hessrsquos Law


For example




-1670 kJ




















Learning Check







2H2(g) + N2(g) rarr N2H4(g) (1878 - 924) = +954 kJ

2














Your Turn




56

State Matters






-2219 + 176 kJ = -2043 kJ


57


bull 1 atm pressure





Standard State


58





gas

liquid


59

Allotropes




P ndash solid white



60







61





0 -2858 -6965 0


ΔHdegrxn = 3218 kJ

ΔHdegrxn = 8903 kJ


A 965 kJ

B -965 kJ

C 482 kJ

D -482 kJ

E None of these


∆Hfordm -69965

kJmol

-2300 kJmol

-2859 kJmol

-69199 kJmol









bull KE = frac12mv2









Your Turn

















Units of Energy







Δ E = Ef - Ei = 0







2


What is Temperature


State Function




New York

Los Angeles







Heat Transfer q












The Sign Convention





Your Turn


A The water heats

B The skillet cools








Units J degC-1





Learning Check


q = Ctimes ∆t


q = 31 times 103 J




C = mtimes s


1



m = mass of object



Specific Heats


J g-1degC-1

(25 degC)


Copper 0387

Ethyl alcohol 245

Gold 0129

Granite 0803

Iron 04498

Lead 0128

Olive oil 20

Silver 0235

Water (liquid) 418





Learning Check


235 J J282


= times times ∆


q m s t

qs

m t










qsample= -qwater






s = 0349 J g-1degC-1


Your Turn


A 870 JdegC

B 35 JdegC

C -35 JdegC

D -870 JdegC

E None of these







Work and Pistons












w = -Patm timesΔV







w = 19 kJ

w = 722 times 103 J

w = -44 atm times (43 - 86) L = 19 Latm




bull ∆E = q + w









D None of these









bull Measures qv


bull ΔE = qv














Your Turn


A -711 J

B -285 J

C +711 J

D +285 J

E None of these



-579 kJ




ΔH = qp




































ΔH = -2511 kJ


Learning Check

1256 kJ


Learning Check




938 g

470 kJ




ΔH = -353 kJ









A 659 g

B 0587 g

C 328 g

D None of these







=-2306 kJ


= 715 Latm = 7222 kJΔE= -2306kJ+72kJ= -223 kJ


Enthalpy Diagram


Hessrsquos Law


For example




-1670 kJ




















Learning Check







2H2(g) + N2(g) rarr N2H4(g) (1878 - 924) = +954 kJ

2














Your Turn




56

State Matters






-2219 + 176 kJ = -2043 kJ


57


bull 1 atm pressure





Standard State


58





gas

liquid


59

Allotropes




P ndash solid white



60







61





0 -2858 -6965 0


ΔHdegrxn = 3218 kJ

ΔHdegrxn = 8903 kJ


A 965 kJ

B -965 kJ

C 482 kJ

D -482 kJ

E None of these


∆Hfordm -69965

kJmol

-2300 kJmol

-2859 kJmol

-69199 kJmol



bull KE = frac12mv2









Your Turn

















Units of Energy







Δ E = Ef - Ei = 0







2


What is Temperature


State Function




New York

Los Angeles







Heat Transfer q












The Sign Convention





Your Turn


A The water heats

B The skillet cools








Units J degC-1





Learning Check


q = Ctimes ∆t


q = 31 times 103 J




C = mtimes s


1



m = mass of object



Specific Heats


J g-1degC-1

(25 degC)


Copper 0387

Ethyl alcohol 245

Gold 0129

Granite 0803

Iron 04498

Lead 0128

Olive oil 20

Silver 0235

Water (liquid) 418





Learning Check


235 J J282


= times times ∆


q m s t

qs

m t










qsample= -qwater






s = 0349 J g-1degC-1


Your Turn


A 870 JdegC

B 35 JdegC

C -35 JdegC

D -870 JdegC

E None of these







Work and Pistons












w = -Patm timesΔV







w = 19 kJ

w = 722 times 103 J

w = -44 atm times (43 - 86) L = 19 Latm




bull ∆E = q + w









D None of these









bull Measures qv


bull ΔE = qv














Your Turn


A -711 J

B -285 J

C +711 J

D +285 J

E None of these



-579 kJ




ΔH = qp




































ΔH = -2511 kJ


Learning Check

1256 kJ


Learning Check




938 g

470 kJ




ΔH = -353 kJ









A 659 g

B 0587 g

C 328 g

D None of these







=-2306 kJ


= 715 Latm = 7222 kJΔE= -2306kJ+72kJ= -223 kJ


Enthalpy Diagram


Hessrsquos Law


For example




-1670 kJ




















Learning Check







2H2(g) + N2(g) rarr N2H4(g) (1878 - 924) = +954 kJ

2














Your Turn




56

State Matters






-2219 + 176 kJ = -2043 kJ


57


bull 1 atm pressure





Standard State


58





gas

liquid


59

Allotropes




P ndash solid white



60







61





0 -2858 -6965 0


ΔHdegrxn = 3218 kJ

ΔHdegrxn = 8903 kJ


A 965 kJ

B -965 kJ

C 482 kJ

D -482 kJ

E None of these


∆Hfordm -69965

kJmol

-2300 kJmol

-2859 kJmol

-69199 kJmol







Your Turn

















Units of Energy







Δ E = Ef - Ei = 0







2


What is Temperature


State Function




New York

Los Angeles







Heat Transfer q












The Sign Convention





Your Turn


A The water heats

B The skillet cools








Units J degC-1





Learning Check


q = Ctimes ∆t


q = 31 times 103 J




C = mtimes s


1



m = mass of object



Specific Heats


J g-1degC-1

(25 degC)


Copper 0387

Ethyl alcohol 245

Gold 0129

Granite 0803

Iron 04498

Lead 0128

Olive oil 20

Silver 0235

Water (liquid) 418





Learning Check


235 J J282


= times times ∆


q m s t

qs

m t










qsample= -qwater






s = 0349 J g-1degC-1


Your Turn


A 870 JdegC

B 35 JdegC

C -35 JdegC

D -870 JdegC

E None of these







Work and Pistons












w = -Patm timesΔV







w = 19 kJ

w = 722 times 103 J

w = -44 atm times (43 - 86) L = 19 Latm




bull ∆E = q + w









D None of these









bull Measures qv


bull ΔE = qv














Your Turn


A -711 J

B -285 J

C +711 J

D +285 J

E None of these



-579 kJ




ΔH = qp




































ΔH = -2511 kJ


Learning Check

1256 kJ


Learning Check




938 g

470 kJ




ΔH = -353 kJ









A 659 g

B 0587 g

C 328 g

D None of these







=-2306 kJ


= 715 Latm = 7222 kJΔE= -2306kJ+72kJ= -223 kJ


Enthalpy Diagram


Hessrsquos Law


For example




-1670 kJ




















Learning Check







2H2(g) + N2(g) rarr N2H4(g) (1878 - 924) = +954 kJ

2














Your Turn




56

State Matters






-2219 + 176 kJ = -2043 kJ


57


bull 1 atm pressure





Standard State


58





gas

liquid


59

Allotropes




P ndash solid white



60







61





0 -2858 -6965 0


ΔHdegrxn = 3218 kJ

ΔHdegrxn = 8903 kJ


A 965 kJ

B -965 kJ

C 482 kJ

D -482 kJ

E None of these


∆Hfordm -69965

kJmol

-2300 kJmol

-2859 kJmol

-69199 kJmol


Your Turn

















Units of Energy







Δ E = Ef - Ei = 0







2


What is Temperature


State Function




New York

Los Angeles







Heat Transfer q












The Sign Convention





Your Turn


A The water heats

B The skillet cools








Units J degC-1





Learning Check


q = Ctimes ∆t


q = 31 times 103 J




C = mtimes s


1



m = mass of object



Specific Heats


J g-1degC-1

(25 degC)


Copper 0387

Ethyl alcohol 245

Gold 0129

Granite 0803

Iron 04498

Lead 0128

Olive oil 20

Silver 0235

Water (liquid) 418





Learning Check


235 J J282


= times times ∆


q m s t

qs

m t










qsample= -qwater






s = 0349 J g-1degC-1


Your Turn


A 870 JdegC

B 35 JdegC

C -35 JdegC

D -870 JdegC

E None of these







Work and Pistons












w = -Patm timesΔV







w = 19 kJ

w = 722 times 103 J

w = -44 atm times (43 - 86) L = 19 Latm




bull ∆E = q + w









D None of these









bull Measures qv


bull ΔE = qv














Your Turn


A -711 J

B -285 J

C +711 J

D +285 J

E None of these



-579 kJ




ΔH = qp




































ΔH = -2511 kJ


Learning Check

1256 kJ


Learning Check




938 g

470 kJ




ΔH = -353 kJ









A 659 g

B 0587 g

C 328 g

D None of these







=-2306 kJ


= 715 Latm = 7222 kJΔE= -2306kJ+72kJ= -223 kJ


Enthalpy Diagram


Hessrsquos Law


For example




-1670 kJ




















Learning Check







2H2(g) + N2(g) rarr N2H4(g) (1878 - 924) = +954 kJ

2














Your Turn




56

State Matters






-2219 + 176 kJ = -2043 kJ


57


bull 1 atm pressure





Standard State


58





gas

liquid


59

Allotropes




P ndash solid white



60







61





0 -2858 -6965 0


ΔHdegrxn = 3218 kJ

ΔHdegrxn = 8903 kJ


A 965 kJ

B -965 kJ

C 482 kJ

D -482 kJ

E None of these


∆Hfordm -69965

kJmol

-2300 kJmol

-2859 kJmol

-69199 kJmol











Units of Energy







Δ E = Ef - Ei = 0







2


What is Temperature


State Function




New York

Los Angeles







Heat Transfer q












The Sign Convention





Your Turn


A The water heats

B The skillet cools








Units J degC-1





Learning Check


q = Ctimes ∆t


q = 31 times 103 J




C = mtimes s


1



m = mass of object



Specific Heats


J g-1degC-1

(25 degC)


Copper 0387

Ethyl alcohol 245

Gold 0129

Granite 0803

Iron 04498

Lead 0128

Olive oil 20

Silver 0235

Water (liquid) 418





Learning Check


235 J J282


= times times ∆


q m s t

qs

m t










qsample= -qwater






s = 0349 J g-1degC-1


Your Turn


A 870 JdegC

B 35 JdegC

C -35 JdegC

D -870 JdegC

E None of these







Work and Pistons












w = -Patm timesΔV







w = 19 kJ

w = 722 times 103 J

w = -44 atm times (43 - 86) L = 19 Latm




bull ∆E = q + w









D None of these









bull Measures qv


bull ΔE = qv














Your Turn


A -711 J

B -285 J

C +711 J

D +285 J

E None of these



-579 kJ




ΔH = qp




































ΔH = -2511 kJ


Learning Check

1256 kJ


Learning Check




938 g

470 kJ




ΔH = -353 kJ









A 659 g

B 0587 g

C 328 g

D None of these







=-2306 kJ


= 715 Latm = 7222 kJΔE= -2306kJ+72kJ= -223 kJ


Enthalpy Diagram


Hessrsquos Law


For example




-1670 kJ




















Learning Check







2H2(g) + N2(g) rarr N2H4(g) (1878 - 924) = +954 kJ

2














Your Turn




56

State Matters






-2219 + 176 kJ = -2043 kJ


57


bull 1 atm pressure





Standard State


58





gas

liquid


59

Allotropes




P ndash solid white



60







61





0 -2858 -6965 0


ΔHdegrxn = 3218 kJ

ΔHdegrxn = 8903 kJ


A 965 kJ

B -965 kJ

C 482 kJ

D -482 kJ

E None of these


∆Hfordm -69965

kJmol

-2300 kJmol

-2859 kJmol

-69199 kJmol






Units of Energy







Δ E = Ef - Ei = 0







2


What is Temperature


State Function




New York

Los Angeles







Heat Transfer q












The Sign Convention





Your Turn


A The water heats

B The skillet cools








Units J degC-1





Learning Check


q = Ctimes ∆t


q = 31 times 103 J




C = mtimes s


1



m = mass of object



Specific Heats


J g-1degC-1

(25 degC)


Copper 0387

Ethyl alcohol 245

Gold 0129

Granite 0803

Iron 04498

Lead 0128

Olive oil 20

Silver 0235

Water (liquid) 418





Learning Check


235 J J282


= times times ∆


q m s t

qs

m t










qsample= -qwater






s = 0349 J g-1degC-1


Your Turn


A 870 JdegC

B 35 JdegC

C -35 JdegC

D -870 JdegC

E None of these







Work and Pistons












w = -Patm timesΔV







w = 19 kJ

w = 722 times 103 J

w = -44 atm times (43 - 86) L = 19 Latm




bull ∆E = q + w









D None of these









bull Measures qv


bull ΔE = qv














Your Turn


A -711 J

B -285 J

C +711 J

D +285 J

E None of these



-579 kJ




ΔH = qp




































ΔH = -2511 kJ


Learning Check

1256 kJ


Learning Check




938 g

470 kJ




ΔH = -353 kJ









A 659 g

B 0587 g

C 328 g

D None of these







=-2306 kJ


= 715 Latm = 7222 kJΔE= -2306kJ+72kJ= -223 kJ


Enthalpy Diagram


Hessrsquos Law


For example




-1670 kJ




















Learning Check







2H2(g) + N2(g) rarr N2H4(g) (1878 - 924) = +954 kJ

2














Your Turn




56

State Matters






-2219 + 176 kJ = -2043 kJ


57


bull 1 atm pressure





Standard State


58





gas

liquid


59

Allotropes




P ndash solid white



60







61





0 -2858 -6965 0


ΔHdegrxn = 3218 kJ

ΔHdegrxn = 8903 kJ


A 965 kJ

B -965 kJ

C 482 kJ

D -482 kJ

E None of these


∆Hfordm -69965

kJmol

-2300 kJmol

-2859 kJmol

-69199 kJmol


Units of Energy







Δ E = Ef - Ei = 0







2


What is Temperature


State Function




New York

Los Angeles







Heat Transfer q












The Sign Convention





Your Turn


A The water heats

B The skillet cools








Units J degC-1





Learning Check


q = Ctimes ∆t


q = 31 times 103 J




C = mtimes s


1



m = mass of object



Specific Heats


J g-1degC-1

(25 degC)


Copper 0387

Ethyl alcohol 245

Gold 0129

Granite 0803

Iron 04498

Lead 0128

Olive oil 20

Silver 0235

Water (liquid) 418





Learning Check


235 J J282


= times times ∆


q m s t

qs

m t










qsample= -qwater






s = 0349 J g-1degC-1


Your Turn


A 870 JdegC

B 35 JdegC

C -35 JdegC

D -870 JdegC

E None of these







Work and Pistons












w = -Patm timesΔV







w = 19 kJ

w = 722 times 103 J

w = -44 atm times (43 - 86) L = 19 Latm




bull ∆E = q + w









D None of these









bull Measures qv


bull ΔE = qv














Your Turn


A -711 J

B -285 J

C +711 J

D +285 J

E None of these



-579 kJ




ΔH = qp




































ΔH = -2511 kJ


Learning Check

1256 kJ


Learning Check




938 g

470 kJ




ΔH = -353 kJ









A 659 g

B 0587 g

C 328 g

D None of these







=-2306 kJ


= 715 Latm = 7222 kJΔE= -2306kJ+72kJ= -223 kJ


Enthalpy Diagram


Hessrsquos Law


For example




-1670 kJ




















Learning Check







2H2(g) + N2(g) rarr N2H4(g) (1878 - 924) = +954 kJ

2














Your Turn




56

State Matters






-2219 + 176 kJ = -2043 kJ


57


bull 1 atm pressure





Standard State


58





gas

liquid


59

Allotropes




P ndash solid white



60







61





0 -2858 -6965 0


ΔHdegrxn = 3218 kJ

ΔHdegrxn = 8903 kJ


A 965 kJ

B -965 kJ

C 482 kJ

D -482 kJ

E None of these


∆Hfordm -69965

kJmol

-2300 kJmol

-2859 kJmol

-69199 kJmol



Δ E = Ef - Ei = 0







2


What is Temperature


State Function




New York

Los Angeles







Heat Transfer q












The Sign Convention





Your Turn


A The water heats

B The skillet cools








Units J degC-1





Learning Check


q = Ctimes ∆t


q = 31 times 103 J




C = mtimes s


1



m = mass of object



Specific Heats


J g-1degC-1

(25 degC)


Copper 0387

Ethyl alcohol 245

Gold 0129

Granite 0803

Iron 04498

Lead 0128

Olive oil 20

Silver 0235

Water (liquid) 418





Learning Check


235 J J282


= times times ∆


q m s t

qs

m t










qsample= -qwater






s = 0349 J g-1degC-1


Your Turn


A 870 JdegC

B 35 JdegC

C -35 JdegC

D -870 JdegC

E None of these







Work and Pistons












w = -Patm timesΔV







w = 19 kJ

w = 722 times 103 J

w = -44 atm times (43 - 86) L = 19 Latm




bull ∆E = q + w









D None of these









bull Measures qv


bull ΔE = qv














Your Turn


A -711 J

B -285 J

C +711 J

D +285 J

E None of these



-579 kJ




ΔH = qp




































ΔH = -2511 kJ


Learning Check

1256 kJ


Learning Check




938 g

470 kJ




ΔH = -353 kJ









A 659 g

B 0587 g

C 328 g

D None of these







=-2306 kJ


= 715 Latm = 7222 kJΔE= -2306kJ+72kJ= -223 kJ


Enthalpy Diagram


Hessrsquos Law


For example




-1670 kJ




















Learning Check







2H2(g) + N2(g) rarr N2H4(g) (1878 - 924) = +954 kJ

2














Your Turn




56

State Matters






-2219 + 176 kJ = -2043 kJ


57


bull 1 atm pressure





Standard State


58





gas

liquid


59

Allotropes




P ndash solid white



60







61





0 -2858 -6965 0


ΔHdegrxn = 3218 kJ

ΔHdegrxn = 8903 kJ


A 965 kJ

B -965 kJ

C 482 kJ

D -482 kJ

E None of these


∆Hfordm -69965

kJmol

-2300 kJmol

-2859 kJmol

-69199 kJmol



2


What is Temperature


State Function




New York

Los Angeles







Heat Transfer q












The Sign Convention





Your Turn


A The water heats

B The skillet cools








Units J degC-1





Learning Check


q = Ctimes ∆t


q = 31 times 103 J




C = mtimes s


1



m = mass of object



Specific Heats


J g-1degC-1

(25 degC)


Copper 0387

Ethyl alcohol 245

Gold 0129

Granite 0803

Iron 04498

Lead 0128

Olive oil 20

Silver 0235

Water (liquid) 418





Learning Check


235 J J282


= times times ∆


q m s t

qs

m t










qsample= -qwater






s = 0349 J g-1degC-1


Your Turn


A 870 JdegC

B 35 JdegC

C -35 JdegC

D -870 JdegC

E None of these







Work and Pistons












w = -Patm timesΔV







w = 19 kJ

w = 722 times 103 J

w = -44 atm times (43 - 86) L = 19 Latm




bull ∆E = q + w









D None of these









bull Measures qv


bull ΔE = qv














Your Turn


A -711 J

B -285 J

C +711 J

D +285 J

E None of these



-579 kJ




ΔH = qp




































ΔH = -2511 kJ


Learning Check

1256 kJ


Learning Check




938 g

470 kJ




ΔH = -353 kJ









A 659 g

B 0587 g

C 328 g

D None of these







=-2306 kJ


= 715 Latm = 7222 kJΔE= -2306kJ+72kJ= -223 kJ


Enthalpy Diagram


Hessrsquos Law


For example




-1670 kJ




















Learning Check







2H2(g) + N2(g) rarr N2H4(g) (1878 - 924) = +954 kJ

2














Your Turn




56

State Matters






-2219 + 176 kJ = -2043 kJ


57


bull 1 atm pressure





Standard State


58





gas

liquid


59

Allotropes




P ndash solid white



60







61





0 -2858 -6965 0


ΔHdegrxn = 3218 kJ

ΔHdegrxn = 8903 kJ


A 965 kJ

B -965 kJ

C 482 kJ

D -482 kJ

E None of these


∆Hfordm -69965

kJmol

-2300 kJmol

-2859 kJmol

-69199 kJmol


State Function




New York

Los Angeles







Heat Transfer q












The Sign Convention





Your Turn


A The water heats

B The skillet cools








Units J degC-1





Learning Check


q = Ctimes ∆t


q = 31 times 103 J




C = mtimes s


1



m = mass of object



Specific Heats


J g-1degC-1

(25 degC)


Copper 0387

Ethyl alcohol 245

Gold 0129

Granite 0803

Iron 04498

Lead 0128

Olive oil 20

Silver 0235

Water (liquid) 418





Learning Check


235 J J282


= times times ∆


q m s t

qs

m t










qsample= -qwater






s = 0349 J g-1degC-1


Your Turn


A 870 JdegC

B 35 JdegC

C -35 JdegC

D -870 JdegC

E None of these







Work and Pistons












w = -Patm timesΔV







w = 19 kJ

w = 722 times 103 J

w = -44 atm times (43 - 86) L = 19 Latm




bull ∆E = q + w









D None of these









bull Measures qv


bull ΔE = qv














Your Turn


A -711 J

B -285 J

C +711 J

D +285 J

E None of these



-579 kJ




ΔH = qp




































ΔH = -2511 kJ


Learning Check

1256 kJ


Learning Check




938 g

470 kJ




ΔH = -353 kJ









A 659 g

B 0587 g

C 328 g

D None of these







=-2306 kJ


= 715 Latm = 7222 kJΔE= -2306kJ+72kJ= -223 kJ


Enthalpy Diagram


Hessrsquos Law


For example




-1670 kJ




















Learning Check







2H2(g) + N2(g) rarr N2H4(g) (1878 - 924) = +954 kJ

2














Your Turn




56

State Matters






-2219 + 176 kJ = -2043 kJ


57


bull 1 atm pressure





Standard State


58





gas

liquid


59

Allotropes




P ndash solid white



60







61





0 -2858 -6965 0


ΔHdegrxn = 3218 kJ

ΔHdegrxn = 8903 kJ


A 965 kJ

B -965 kJ

C 482 kJ

D -482 kJ

E None of these


∆Hfordm -69965

kJmol

-2300 kJmol

-2859 kJmol

-69199 kJmol







Heat Transfer q












The Sign Convention





Your Turn


A The water heats

B The skillet cools








Units J degC-1





Learning Check


q = Ctimes ∆t


q = 31 times 103 J




C = mtimes s


1



m = mass of object



Specific Heats


J g-1degC-1

(25 degC)


Copper 0387

Ethyl alcohol 245

Gold 0129

Granite 0803

Iron 04498

Lead 0128

Olive oil 20

Silver 0235

Water (liquid) 418





Learning Check


235 J J282


= times times ∆


q m s t

qs

m t










qsample= -qwater






s = 0349 J g-1degC-1


Your Turn


A 870 JdegC

B 35 JdegC

C -35 JdegC

D -870 JdegC

E None of these







Work and Pistons












w = -Patm timesΔV







w = 19 kJ

w = 722 times 103 J

w = -44 atm times (43 - 86) L = 19 Latm




bull ∆E = q + w









D None of these









bull Measures qv


bull ΔE = qv














Your Turn


A -711 J

B -285 J

C +711 J

D +285 J

E None of these



-579 kJ




ΔH = qp




































ΔH = -2511 kJ


Learning Check

1256 kJ


Learning Check




938 g

470 kJ




ΔH = -353 kJ









A 659 g

B 0587 g

C 328 g

D None of these







=-2306 kJ


= 715 Latm = 7222 kJΔE= -2306kJ+72kJ= -223 kJ


Enthalpy Diagram


Hessrsquos Law


For example




-1670 kJ




















Learning Check







2H2(g) + N2(g) rarr N2H4(g) (1878 - 924) = +954 kJ

2














Your Turn




56

State Matters






-2219 + 176 kJ = -2043 kJ


57


bull 1 atm pressure





Standard State


58





gas

liquid


59

Allotropes




P ndash solid white



60







61





0 -2858 -6965 0


ΔHdegrxn = 3218 kJ

ΔHdegrxn = 8903 kJ


A 965 kJ

B -965 kJ

C 482 kJ

D -482 kJ

E None of these


∆Hfordm -69965

kJmol

-2300 kJmol

-2859 kJmol

-69199 kJmol


Heat Transfer q












The Sign Convention





Your Turn


A The water heats

B The skillet cools








Units J degC-1





Learning Check


q = Ctimes ∆t


q = 31 times 103 J




C = mtimes s


1



m = mass of object



Specific Heats


J g-1degC-1

(25 degC)


Copper 0387

Ethyl alcohol 245

Gold 0129

Granite 0803

Iron 04498

Lead 0128

Olive oil 20

Silver 0235

Water (liquid) 418





Learning Check


235 J J282


= times times ∆


q m s t

qs

m t










qsample= -qwater






s = 0349 J g-1degC-1


Your Turn


A 870 JdegC

B 35 JdegC

C -35 JdegC

D -870 JdegC

E None of these







Work and Pistons












w = -Patm timesΔV







w = 19 kJ

w = 722 times 103 J

w = -44 atm times (43 - 86) L = 19 Latm




bull ∆E = q + w









D None of these









bull Measures qv


bull ΔE = qv














Your Turn


A -711 J

B -285 J

C +711 J

D +285 J

E None of these



-579 kJ




ΔH = qp




































ΔH = -2511 kJ


Learning Check

1256 kJ


Learning Check




938 g

470 kJ




ΔH = -353 kJ









A 659 g

B 0587 g

C 328 g

D None of these







=-2306 kJ


= 715 Latm = 7222 kJΔE= -2306kJ+72kJ= -223 kJ


Enthalpy Diagram


Hessrsquos Law


For example




-1670 kJ




















Learning Check







2H2(g) + N2(g) rarr N2H4(g) (1878 - 924) = +954 kJ

2














Your Turn




56

State Matters






-2219 + 176 kJ = -2043 kJ


57


bull 1 atm pressure





Standard State


58





gas

liquid


59

Allotropes




P ndash solid white



60







61





0 -2858 -6965 0


ΔHdegrxn = 3218 kJ

ΔHdegrxn = 8903 kJ


A 965 kJ

B -965 kJ

C 482 kJ

D -482 kJ

E None of these


∆Hfordm -69965

kJmol

-2300 kJmol

-2859 kJmol

-69199 kJmol








The Sign Convention





Your Turn


A The water heats

B The skillet cools








Units J degC-1





Learning Check


q = Ctimes ∆t


q = 31 times 103 J




C = mtimes s


1



m = mass of object



Specific Heats


J g-1degC-1

(25 degC)


Copper 0387

Ethyl alcohol 245

Gold 0129

Granite 0803

Iron 04498

Lead 0128

Olive oil 20

Silver 0235

Water (liquid) 418





Learning Check


235 J J282


= times times ∆


q m s t

qs

m t










qsample= -qwater






s = 0349 J g-1degC-1


Your Turn


A 870 JdegC

B 35 JdegC

C -35 JdegC

D -870 JdegC

E None of these







Work and Pistons












w = -Patm timesΔV







w = 19 kJ

w = 722 times 103 J

w = -44 atm times (43 - 86) L = 19 Latm




bull ∆E = q + w









D None of these









bull Measures qv


bull ΔE = qv














Your Turn


A -711 J

B -285 J

C +711 J

D +285 J

E None of these



-579 kJ




ΔH = qp




































ΔH = -2511 kJ


Learning Check

1256 kJ


Learning Check




938 g

470 kJ




ΔH = -353 kJ









A 659 g

B 0587 g

C 328 g

D None of these







=-2306 kJ


= 715 Latm = 7222 kJΔE= -2306kJ+72kJ= -223 kJ


Enthalpy Diagram


Hessrsquos Law


For example




-1670 kJ




















Learning Check







2H2(g) + N2(g) rarr N2H4(g) (1878 - 924) = +954 kJ

2














Your Turn




56

State Matters






-2219 + 176 kJ = -2043 kJ


57


bull 1 atm pressure





Standard State


58





gas

liquid


59

Allotropes




P ndash solid white



60







61





0 -2858 -6965 0


ΔHdegrxn = 3218 kJ

ΔHdegrxn = 8903 kJ


A 965 kJ

B -965 kJ

C 482 kJ

D -482 kJ

E None of these


∆Hfordm -69965

kJmol

-2300 kJmol

-2859 kJmol

-69199 kJmol


The Sign Convention





Your Turn


A The water heats

B The skillet cools








Units J degC-1





Learning Check


q = Ctimes ∆t


q = 31 times 103 J




C = mtimes s


1



m = mass of object



Specific Heats


J g-1degC-1

(25 degC)


Copper 0387

Ethyl alcohol 245

Gold 0129

Granite 0803

Iron 04498

Lead 0128

Olive oil 20

Silver 0235

Water (liquid) 418





Learning Check


235 J J282


= times times ∆


q m s t

qs

m t










qsample= -qwater






s = 0349 J g-1degC-1


Your Turn


A 870 JdegC

B 35 JdegC

C -35 JdegC

D -870 JdegC

E None of these







Work and Pistons












w = -Patm timesΔV







w = 19 kJ

w = 722 times 103 J

w = -44 atm times (43 - 86) L = 19 Latm




bull ∆E = q + w









D None of these









bull Measures qv


bull ΔE = qv














Your Turn


A -711 J

B -285 J

C +711 J

D +285 J

E None of these



-579 kJ




ΔH = qp




































ΔH = -2511 kJ


Learning Check

1256 kJ


Learning Check




938 g

470 kJ




ΔH = -353 kJ









A 659 g

B 0587 g

C 328 g

D None of these







=-2306 kJ


= 715 Latm = 7222 kJΔE= -2306kJ+72kJ= -223 kJ


Enthalpy Diagram


Hessrsquos Law


For example




-1670 kJ




















Learning Check







2H2(g) + N2(g) rarr N2H4(g) (1878 - 924) = +954 kJ

2














Your Turn




56

State Matters






-2219 + 176 kJ = -2043 kJ


57


bull 1 atm pressure





Standard State


58





gas

liquid


59

Allotropes




P ndash solid white



60







61





0 -2858 -6965 0


ΔHdegrxn = 3218 kJ

ΔHdegrxn = 8903 kJ


A 965 kJ

B -965 kJ

C 482 kJ

D -482 kJ

E None of these


∆Hfordm -69965

kJmol

-2300 kJmol

-2859 kJmol

-69199 kJmol


Your Turn


A The water heats

B The skillet cools








Units J degC-1





Learning Check


q = Ctimes ∆t


q = 31 times 103 J




C = mtimes s


1



m = mass of object



Specific Heats


J g-1degC-1

(25 degC)


Copper 0387

Ethyl alcohol 245

Gold 0129

Granite 0803

Iron 04498

Lead 0128

Olive oil 20

Silver 0235

Water (liquid) 418





Learning Check


235 J J282


= times times ∆


q m s t

qs

m t










qsample= -qwater






s = 0349 J g-1degC-1


Your Turn


A 870 JdegC

B 35 JdegC

C -35 JdegC

D -870 JdegC

E None of these







Work and Pistons












w = -Patm timesΔV







w = 19 kJ

w = 722 times 103 J

w = -44 atm times (43 - 86) L = 19 Latm




bull ∆E = q + w









D None of these









bull Measures qv


bull ΔE = qv














Your Turn


A -711 J

B -285 J

C +711 J

D +285 J

E None of these



-579 kJ




ΔH = qp




































ΔH = -2511 kJ


Learning Check

1256 kJ


Learning Check




938 g

470 kJ




ΔH = -353 kJ









A 659 g

B 0587 g

C 328 g

D None of these







=-2306 kJ


= 715 Latm = 7222 kJΔE= -2306kJ+72kJ= -223 kJ


Enthalpy Diagram


Hessrsquos Law


For example




-1670 kJ




















Learning Check







2H2(g) + N2(g) rarr N2H4(g) (1878 - 924) = +954 kJ

2














Your Turn




56

State Matters






-2219 + 176 kJ = -2043 kJ


57


bull 1 atm pressure





Standard State


58





gas

liquid


59

Allotropes




P ndash solid white



60







61





0 -2858 -6965 0


ΔHdegrxn = 3218 kJ

ΔHdegrxn = 8903 kJ


A 965 kJ

B -965 kJ

C 482 kJ

D -482 kJ

E None of these


∆Hfordm -69965

kJmol

-2300 kJmol

-2859 kJmol

-69199 kJmol





Units J degC-1





Learning Check


q = Ctimes ∆t


q = 31 times 103 J




C = mtimes s


1



m = mass of object



Specific Heats


J g-1degC-1

(25 degC)


Copper 0387

Ethyl alcohol 245

Gold 0129

Granite 0803

Iron 04498

Lead 0128

Olive oil 20

Silver 0235

Water (liquid) 418





Learning Check


235 J J282


= times times ∆


q m s t

qs

m t










qsample= -qwater






s = 0349 J g-1degC-1


Your Turn


A 870 JdegC

B 35 JdegC

C -35 JdegC

D -870 JdegC

E None of these







Work and Pistons












w = -Patm timesΔV







w = 19 kJ

w = 722 times 103 J

w = -44 atm times (43 - 86) L = 19 Latm




bull ∆E = q + w









D None of these









bull Measures qv


bull ΔE = qv














Your Turn


A -711 J

B -285 J

C +711 J

D +285 J

E None of these



-579 kJ




ΔH = qp




































ΔH = -2511 kJ


Learning Check

1256 kJ


Learning Check




938 g

470 kJ




ΔH = -353 kJ









A 659 g

B 0587 g

C 328 g

D None of these







=-2306 kJ


= 715 Latm = 7222 kJΔE= -2306kJ+72kJ= -223 kJ


Enthalpy Diagram


Hessrsquos Law


For example




-1670 kJ




















Learning Check







2H2(g) + N2(g) rarr N2H4(g) (1878 - 924) = +954 kJ

2














Your Turn




56

State Matters






-2219 + 176 kJ = -2043 kJ


57


bull 1 atm pressure





Standard State


58





gas

liquid


59

Allotropes




P ndash solid white



60







61





0 -2858 -6965 0


ΔHdegrxn = 3218 kJ

ΔHdegrxn = 8903 kJ


A 965 kJ

B -965 kJ

C 482 kJ

D -482 kJ

E None of these


∆Hfordm -69965

kJmol

-2300 kJmol

-2859 kJmol

-69199 kJmol


Learning Check


q = Ctimes ∆t


q = 31 times 103 J




C = mtimes s


1



m = mass of object



Specific Heats


J g-1degC-1

(25 degC)


Copper 0387

Ethyl alcohol 245

Gold 0129

Granite 0803

Iron 04498

Lead 0128

Olive oil 20

Silver 0235

Water (liquid) 418





Learning Check


235 J J282


= times times ∆


q m s t

qs

m t










qsample= -qwater






s = 0349 J g-1degC-1


Your Turn


A 870 JdegC

B 35 JdegC

C -35 JdegC

D -870 JdegC

E None of these







Work and Pistons












w = -Patm timesΔV







w = 19 kJ

w = 722 times 103 J

w = -44 atm times (43 - 86) L = 19 Latm




bull ∆E = q + w









D None of these









bull Measures qv


bull ΔE = qv














Your Turn


A -711 J

B -285 J

C +711 J

D +285 J

E None of these



-579 kJ




ΔH = qp




































ΔH = -2511 kJ


Learning Check

1256 kJ


Learning Check




938 g

470 kJ




ΔH = -353 kJ









A 659 g

B 0587 g

C 328 g

D None of these







=-2306 kJ


= 715 Latm = 7222 kJΔE= -2306kJ+72kJ= -223 kJ


Enthalpy Diagram


Hessrsquos Law


For example




-1670 kJ




















Learning Check







2H2(g) + N2(g) rarr N2H4(g) (1878 - 924) = +954 kJ

2














Your Turn




56

State Matters






-2219 + 176 kJ = -2043 kJ


57


bull 1 atm pressure





Standard State


58





gas

liquid


59

Allotropes




P ndash solid white



60







61





0 -2858 -6965 0


ΔHdegrxn = 3218 kJ

ΔHdegrxn = 8903 kJ


A 965 kJ

B -965 kJ

C 482 kJ

D -482 kJ

E None of these


∆Hfordm -69965

kJmol

-2300 kJmol

-2859 kJmol

-69199 kJmol




C = mtimes s


1



m = mass of object



Specific Heats


J g-1degC-1

(25 degC)


Copper 0387

Ethyl alcohol 245

Gold 0129

Granite 0803

Iron 04498

Lead 0128

Olive oil 20

Silver 0235

Water (liquid) 418





Learning Check


235 J J282


= times times ∆


q m s t

qs

m t










qsample= -qwater






s = 0349 J g-1degC-1


Your Turn


A 870 JdegC

B 35 JdegC

C -35 JdegC

D -870 JdegC

E None of these







Work and Pistons












w = -Patm timesΔV







w = 19 kJ

w = 722 times 103 J

w = -44 atm times (43 - 86) L = 19 Latm




bull ∆E = q + w









D None of these









bull Measures qv


bull ΔE = qv














Your Turn


A -711 J

B -285 J

C +711 J

D +285 J

E None of these



-579 kJ




ΔH = qp




































ΔH = -2511 kJ


Learning Check

1256 kJ


Learning Check




938 g

470 kJ




ΔH = -353 kJ









A 659 g

B 0587 g

C 328 g

D None of these







=-2306 kJ


= 715 Latm = 7222 kJΔE= -2306kJ+72kJ= -223 kJ


Enthalpy Diagram


Hessrsquos Law


For example




-1670 kJ




















Learning Check







2H2(g) + N2(g) rarr N2H4(g) (1878 - 924) = +954 kJ

2














Your Turn




56

State Matters






-2219 + 176 kJ = -2043 kJ


57


bull 1 atm pressure





Standard State


58





gas

liquid


59

Allotropes




P ndash solid white



60







61





0 -2858 -6965 0


ΔHdegrxn = 3218 kJ

ΔHdegrxn = 8903 kJ


A 965 kJ

B -965 kJ

C 482 kJ

D -482 kJ

E None of these


∆Hfordm -69965

kJmol

-2300 kJmol

-2859 kJmol

-69199 kJmol


Specific Heats


J g-1degC-1

(25 degC)


Copper 0387

Ethyl alcohol 245

Gold 0129

Granite 0803

Iron 04498

Lead 0128

Olive oil 20

Silver 0235

Water (liquid) 418





Learning Check


235 J J282


= times times ∆


q m s t

qs

m t










qsample= -qwater






s = 0349 J g-1degC-1


Your Turn


A 870 JdegC

B 35 JdegC

C -35 JdegC

D -870 JdegC

E None of these







Work and Pistons












w = -Patm timesΔV







w = 19 kJ

w = 722 times 103 J

w = -44 atm times (43 - 86) L = 19 Latm




bull ∆E = q + w









D None of these









bull Measures qv


bull ΔE = qv














Your Turn


A -711 J

B -285 J

C +711 J

D +285 J

E None of these



-579 kJ




ΔH = qp




































ΔH = -2511 kJ


Learning Check

1256 kJ


Learning Check




938 g

470 kJ




ΔH = -353 kJ









A 659 g

B 0587 g

C 328 g

D None of these







=-2306 kJ


= 715 Latm = 7222 kJΔE= -2306kJ+72kJ= -223 kJ


Enthalpy Diagram


Hessrsquos Law


For example




-1670 kJ




















Learning Check







2H2(g) + N2(g) rarr N2H4(g) (1878 - 924) = +954 kJ

2














Your Turn




56

State Matters






-2219 + 176 kJ = -2043 kJ


57


bull 1 atm pressure





Standard State


58





gas

liquid


59

Allotropes




P ndash solid white



60







61





0 -2858 -6965 0


ΔHdegrxn = 3218 kJ

ΔHdegrxn = 8903 kJ


A 965 kJ

B -965 kJ

C 482 kJ

D -482 kJ

E None of these


∆Hfordm -69965

kJmol

-2300 kJmol

-2859 kJmol

-69199 kJmol


Learning Check


235 J J282


= times times ∆


q m s t

qs

m t










qsample= -qwater






s = 0349 J g-1degC-1


Your Turn


A 870 JdegC

B 35 JdegC

C -35 JdegC

D -870 JdegC

E None of these







Work and Pistons












w = -Patm timesΔV







w = 19 kJ

w = 722 times 103 J

w = -44 atm times (43 - 86) L = 19 Latm




bull ∆E = q + w









D None of these









bull Measures qv


bull ΔE = qv














Your Turn


A -711 J

B -285 J

C +711 J

D +285 J

E None of these



-579 kJ




ΔH = qp




































ΔH = -2511 kJ


Learning Check

1256 kJ


Learning Check




938 g

470 kJ




ΔH = -353 kJ









A 659 g

B 0587 g

C 328 g

D None of these







=-2306 kJ


= 715 Latm = 7222 kJΔE= -2306kJ+72kJ= -223 kJ


Enthalpy Diagram


Hessrsquos Law


For example




-1670 kJ




















Learning Check







2H2(g) + N2(g) rarr N2H4(g) (1878 - 924) = +954 kJ

2














Your Turn




56

State Matters






-2219 + 176 kJ = -2043 kJ


57


bull 1 atm pressure





Standard State


58





gas

liquid


59

Allotropes




P ndash solid white



60







61





0 -2858 -6965 0


ΔHdegrxn = 3218 kJ

ΔHdegrxn = 8903 kJ


A 965 kJ

B -965 kJ

C 482 kJ

D -482 kJ

E None of these


∆Hfordm -69965

kJmol

-2300 kJmol

-2859 kJmol

-69199 kJmol










qsample= -qwater






s = 0349 J g-1degC-1


Your Turn


A 870 JdegC

B 35 JdegC

C -35 JdegC

D -870 JdegC

E None of these







Work and Pistons












w = -Patm timesΔV







w = 19 kJ

w = 722 times 103 J

w = -44 atm times (43 - 86) L = 19 Latm




bull ∆E = q + w









D None of these









bull Measures qv


bull ΔE = qv














Your Turn


A -711 J

B -285 J

C +711 J

D +285 J

E None of these



-579 kJ




ΔH = qp




































ΔH = -2511 kJ


Learning Check

1256 kJ


Learning Check




938 g

470 kJ




ΔH = -353 kJ









A 659 g

B 0587 g

C 328 g

D None of these







=-2306 kJ


= 715 Latm = 7222 kJΔE= -2306kJ+72kJ= -223 kJ


Enthalpy Diagram


Hessrsquos Law


For example




-1670 kJ




















Learning Check







2H2(g) + N2(g) rarr N2H4(g) (1878 - 924) = +954 kJ

2














Your Turn




56

State Matters






-2219 + 176 kJ = -2043 kJ


57


bull 1 atm pressure





Standard State


58





gas

liquid


59

Allotropes




P ndash solid white



60







61





0 -2858 -6965 0


ΔHdegrxn = 3218 kJ

ΔHdegrxn = 8903 kJ


A 965 kJ

B -965 kJ

C 482 kJ

D -482 kJ

E None of these


∆Hfordm -69965

kJmol

-2300 kJmol

-2859 kJmol

-69199 kJmol





qsample= -qwater






s = 0349 J g-1degC-1


Your Turn


A 870 JdegC

B 35 JdegC

C -35 JdegC

D -870 JdegC

E None of these







Work and Pistons












w = -Patm timesΔV







w = 19 kJ

w = 722 times 103 J

w = -44 atm times (43 - 86) L = 19 Latm




bull ∆E = q + w









D None of these









bull Measures qv


bull ΔE = qv














Your Turn


A -711 J

B -285 J

C +711 J

D +285 J

E None of these



-579 kJ




ΔH = qp




































ΔH = -2511 kJ


Learning Check

1256 kJ


Learning Check




938 g

470 kJ




ΔH = -353 kJ









A 659 g

B 0587 g

C 328 g

D None of these







=-2306 kJ


= 715 Latm = 7222 kJΔE= -2306kJ+72kJ= -223 kJ


Enthalpy Diagram


Hessrsquos Law


For example




-1670 kJ




















Learning Check







2H2(g) + N2(g) rarr N2H4(g) (1878 - 924) = +954 kJ

2














Your Turn




56

State Matters






-2219 + 176 kJ = -2043 kJ


57


bull 1 atm pressure





Standard State


58





gas

liquid


59

Allotropes




P ndash solid white



60







61





0 -2858 -6965 0


ΔHdegrxn = 3218 kJ

ΔHdegrxn = 8903 kJ


A 965 kJ

B -965 kJ

C 482 kJ

D -482 kJ

E None of these


∆Hfordm -69965

kJmol

-2300 kJmol

-2859 kJmol

-69199 kJmol


Your Turn


A 870 JdegC

B 35 JdegC

C -35 JdegC

D -870 JdegC

E None of these







Work and Pistons












w = -Patm timesΔV







w = 19 kJ

w = 722 times 103 J

w = -44 atm times (43 - 86) L = 19 Latm




bull ∆E = q + w









D None of these









bull Measures qv


bull ΔE = qv














Your Turn


A -711 J

B -285 J

C +711 J

D +285 J

E None of these



-579 kJ




ΔH = qp




































ΔH = -2511 kJ


Learning Check

1256 kJ


Learning Check




938 g

470 kJ




ΔH = -353 kJ









A 659 g

B 0587 g

C 328 g

D None of these







=-2306 kJ


= 715 Latm = 7222 kJΔE= -2306kJ+72kJ= -223 kJ


Enthalpy Diagram


Hessrsquos Law


For example




-1670 kJ




















Learning Check







2H2(g) + N2(g) rarr N2H4(g) (1878 - 924) = +954 kJ

2














Your Turn




56

State Matters






-2219 + 176 kJ = -2043 kJ


57


bull 1 atm pressure





Standard State


58





gas

liquid


59

Allotropes




P ndash solid white



60







61





0 -2858 -6965 0


ΔHdegrxn = 3218 kJ

ΔHdegrxn = 8903 kJ


A 965 kJ

B -965 kJ

C 482 kJ

D -482 kJ

E None of these


∆Hfordm -69965

kJmol

-2300 kJmol

-2859 kJmol

-69199 kJmol







Work and Pistons












w = -Patm timesΔV







w = 19 kJ

w = 722 times 103 J

w = -44 atm times (43 - 86) L = 19 Latm




bull ∆E = q + w









D None of these









bull Measures qv


bull ΔE = qv














Your Turn


A -711 J

B -285 J

C +711 J

D +285 J

E None of these



-579 kJ




ΔH = qp




































ΔH = -2511 kJ


Learning Check

1256 kJ


Learning Check




938 g

470 kJ




ΔH = -353 kJ









A 659 g

B 0587 g

C 328 g

D None of these







=-2306 kJ


= 715 Latm = 7222 kJΔE= -2306kJ+72kJ= -223 kJ


Enthalpy Diagram


Hessrsquos Law


For example




-1670 kJ




















Learning Check







2H2(g) + N2(g) rarr N2H4(g) (1878 - 924) = +954 kJ

2














Your Turn




56

State Matters






-2219 + 176 kJ = -2043 kJ


57


bull 1 atm pressure





Standard State


58





gas

liquid


59

Allotropes




P ndash solid white



60







61





0 -2858 -6965 0


ΔHdegrxn = 3218 kJ

ΔHdegrxn = 8903 kJ


A 965 kJ

B -965 kJ

C 482 kJ

D -482 kJ

E None of these


∆Hfordm -69965

kJmol

-2300 kJmol

-2859 kJmol

-69199 kJmol


Work and Pistons












w = -Patm timesΔV







w = 19 kJ

w = 722 times 103 J

w = -44 atm times (43 - 86) L = 19 Latm




bull ∆E = q + w









D None of these









bull Measures qv


bull ΔE = qv














Your Turn


A -711 J

B -285 J

C +711 J

D +285 J

E None of these



-579 kJ




ΔH = qp




































ΔH = -2511 kJ


Learning Check

1256 kJ


Learning Check




938 g

470 kJ




ΔH = -353 kJ









A 659 g

B 0587 g

C 328 g

D None of these







=-2306 kJ


= 715 Latm = 7222 kJΔE= -2306kJ+72kJ= -223 kJ


Enthalpy Diagram


Hessrsquos Law


For example




-1670 kJ




















Learning Check







2H2(g) + N2(g) rarr N2H4(g) (1878 - 924) = +954 kJ

2














Your Turn




56

State Matters






-2219 + 176 kJ = -2043 kJ


57


bull 1 atm pressure





Standard State


58





gas

liquid


59

Allotropes




P ndash solid white



60







61





0 -2858 -6965 0


ΔHdegrxn = 3218 kJ

ΔHdegrxn = 8903 kJ


A 965 kJ

B -965 kJ

C 482 kJ

D -482 kJ

E None of these


∆Hfordm -69965

kJmol

-2300 kJmol

-2859 kJmol

-69199 kJmol






w = -Patm timesΔV







w = 19 kJ

w = 722 times 103 J

w = -44 atm times (43 - 86) L = 19 Latm




bull ∆E = q + w









D None of these









bull Measures qv


bull ΔE = qv














Your Turn


A -711 J

B -285 J

C +711 J

D +285 J

E None of these



-579 kJ




ΔH = qp




































ΔH = -2511 kJ


Learning Check

1256 kJ


Learning Check




938 g

470 kJ




ΔH = -353 kJ









A 659 g

B 0587 g

C 328 g

D None of these







=-2306 kJ


= 715 Latm = 7222 kJΔE= -2306kJ+72kJ= -223 kJ


Enthalpy Diagram


Hessrsquos Law


For example




-1670 kJ




















Learning Check







2H2(g) + N2(g) rarr N2H4(g) (1878 - 924) = +954 kJ

2














Your Turn




56

State Matters






-2219 + 176 kJ = -2043 kJ


57


bull 1 atm pressure





Standard State


58





gas

liquid


59

Allotropes




P ndash solid white



60







61





0 -2858 -6965 0


ΔHdegrxn = 3218 kJ

ΔHdegrxn = 8903 kJ


A 965 kJ

B -965 kJ

C 482 kJ

D -482 kJ

E None of these


∆Hfordm -69965

kJmol

-2300 kJmol

-2859 kJmol

-69199 kJmol




w = 19 kJ

w = 722 times 103 J

w = -44 atm times (43 - 86) L = 19 Latm




bull ∆E = q + w









D None of these









bull Measures qv


bull ΔE = qv














Your Turn


A -711 J

B -285 J

C +711 J

D +285 J

E None of these



-579 kJ




ΔH = qp




































ΔH = -2511 kJ


Learning Check

1256 kJ


Learning Check




938 g

470 kJ




ΔH = -353 kJ









A 659 g

B 0587 g

C 328 g

D None of these







=-2306 kJ


= 715 Latm = 7222 kJΔE= -2306kJ+72kJ= -223 kJ


Enthalpy Diagram


Hessrsquos Law


For example




-1670 kJ




















Learning Check







2H2(g) + N2(g) rarr N2H4(g) (1878 - 924) = +954 kJ

2














Your Turn




56

State Matters






-2219 + 176 kJ = -2043 kJ


57


bull 1 atm pressure





Standard State


58





gas

liquid


59

Allotropes




P ndash solid white



60







61





0 -2858 -6965 0


ΔHdegrxn = 3218 kJ

ΔHdegrxn = 8903 kJ


A 965 kJ

B -965 kJ

C 482 kJ

D -482 kJ

E None of these


∆Hfordm -69965

kJmol

-2300 kJmol

-2859 kJmol

-69199 kJmol



bull ∆E = q + w









D None of these









bull Measures qv


bull ΔE = qv














Your Turn


A -711 J

B -285 J

C +711 J

D +285 J

E None of these



-579 kJ




ΔH = qp




































ΔH = -2511 kJ


Learning Check

1256 kJ


Learning Check




938 g

470 kJ




ΔH = -353 kJ









A 659 g

B 0587 g

C 328 g

D None of these







=-2306 kJ


= 715 Latm = 7222 kJΔE= -2306kJ+72kJ= -223 kJ


Enthalpy Diagram


Hessrsquos Law


For example




-1670 kJ




















Learning Check







2H2(g) + N2(g) rarr N2H4(g) (1878 - 924) = +954 kJ

2














Your Turn




56

State Matters






-2219 + 176 kJ = -2043 kJ


57


bull 1 atm pressure





Standard State


58





gas

liquid


59

Allotropes




P ndash solid white



60







61





0 -2858 -6965 0


ΔHdegrxn = 3218 kJ

ΔHdegrxn = 8903 kJ


A 965 kJ

B -965 kJ

C 482 kJ

D -482 kJ

E None of these


∆Hfordm -69965

kJmol

-2300 kJmol

-2859 kJmol

-69199 kJmol








D None of these









bull Measures qv


bull ΔE = qv














Your Turn


A -711 J

B -285 J

C +711 J

D +285 J

E None of these



-579 kJ




ΔH = qp




































ΔH = -2511 kJ


Learning Check

1256 kJ


Learning Check




938 g

470 kJ




ΔH = -353 kJ









A 659 g

B 0587 g

C 328 g

D None of these







=-2306 kJ


= 715 Latm = 7222 kJΔE= -2306kJ+72kJ= -223 kJ


Enthalpy Diagram


Hessrsquos Law


For example




-1670 kJ




















Learning Check







2H2(g) + N2(g) rarr N2H4(g) (1878 - 924) = +954 kJ

2














Your Turn




56

State Matters






-2219 + 176 kJ = -2043 kJ


57


bull 1 atm pressure





Standard State


58





gas

liquid


59

Allotropes




P ndash solid white



60







61





0 -2858 -6965 0


ΔHdegrxn = 3218 kJ

ΔHdegrxn = 8903 kJ


A 965 kJ

B -965 kJ

C 482 kJ

D -482 kJ

E None of these


∆Hfordm -69965

kJmol

-2300 kJmol

-2859 kJmol

-69199 kJmol









bull Measures qv


bull ΔE = qv














Your Turn


A -711 J

B -285 J

C +711 J

D +285 J

E None of these



-579 kJ




ΔH = qp




































ΔH = -2511 kJ


Learning Check

1256 kJ


Learning Check




938 g

470 kJ




ΔH = -353 kJ









A 659 g

B 0587 g

C 328 g

D None of these







=-2306 kJ


= 715 Latm = 7222 kJΔE= -2306kJ+72kJ= -223 kJ


Enthalpy Diagram


Hessrsquos Law


For example




-1670 kJ




















Learning Check







2H2(g) + N2(g) rarr N2H4(g) (1878 - 924) = +954 kJ

2














Your Turn




56

State Matters






-2219 + 176 kJ = -2043 kJ


57


bull 1 atm pressure





Standard State


58





gas

liquid


59

Allotropes




P ndash solid white



60







61





0 -2858 -6965 0


ΔHdegrxn = 3218 kJ

ΔHdegrxn = 8903 kJ


A 965 kJ

B -965 kJ

C 482 kJ

D -482 kJ

E None of these


∆Hfordm -69965

kJmol

-2300 kJmol

-2859 kJmol

-69199 kJmol




bull Measures qv


bull ΔE = qv














Your Turn


A -711 J

B -285 J

C +711 J

D +285 J

E None of these



-579 kJ




ΔH = qp




































ΔH = -2511 kJ


Learning Check

1256 kJ


Learning Check




938 g

470 kJ




ΔH = -353 kJ









A 659 g

B 0587 g

C 328 g

D None of these







=-2306 kJ


= 715 Latm = 7222 kJΔE= -2306kJ+72kJ= -223 kJ


Enthalpy Diagram


Hessrsquos Law


For example




-1670 kJ




















Learning Check







2H2(g) + N2(g) rarr N2H4(g) (1878 - 924) = +954 kJ

2














Your Turn




56

State Matters






-2219 + 176 kJ = -2043 kJ


57


bull 1 atm pressure





Standard State


58





gas

liquid


59

Allotropes




P ndash solid white



60







61





0 -2858 -6965 0


ΔHdegrxn = 3218 kJ

ΔHdegrxn = 8903 kJ


A 965 kJ

B -965 kJ

C 482 kJ

D -482 kJ

E None of these


∆Hfordm -69965

kJmol

-2300 kJmol

-2859 kJmol

-69199 kJmol














Your Turn


A -711 J

B -285 J

C +711 J

D +285 J

E None of these



-579 kJ




ΔH = qp




































ΔH = -2511 kJ


Learning Check

1256 kJ


Learning Check




938 g

470 kJ




ΔH = -353 kJ









A 659 g

B 0587 g

C 328 g

D None of these







=-2306 kJ


= 715 Latm = 7222 kJΔE= -2306kJ+72kJ= -223 kJ


Enthalpy Diagram


Hessrsquos Law


For example




-1670 kJ




















Learning Check







2H2(g) + N2(g) rarr N2H4(g) (1878 - 924) = +954 kJ

2














Your Turn




56

State Matters






-2219 + 176 kJ = -2043 kJ


57


bull 1 atm pressure





Standard State


58





gas

liquid


59

Allotropes




P ndash solid white



60







61





0 -2858 -6965 0


ΔHdegrxn = 3218 kJ

ΔHdegrxn = 8903 kJ


A 965 kJ

B -965 kJ

C 482 kJ

D -482 kJ

E None of these


∆Hfordm -69965

kJmol

-2300 kJmol

-2859 kJmol

-69199 kJmol








Your Turn


A -711 J

B -285 J

C +711 J

D +285 J

E None of these



-579 kJ




ΔH = qp




































ΔH = -2511 kJ


Learning Check

1256 kJ


Learning Check




938 g

470 kJ




ΔH = -353 kJ









A 659 g

B 0587 g

C 328 g

D None of these







=-2306 kJ


= 715 Latm = 7222 kJΔE= -2306kJ+72kJ= -223 kJ


Enthalpy Diagram


Hessrsquos Law


For example




-1670 kJ




















Learning Check







2H2(g) + N2(g) rarr N2H4(g) (1878 - 924) = +954 kJ

2














Your Turn




56

State Matters






-2219 + 176 kJ = -2043 kJ


57


bull 1 atm pressure





Standard State


58





gas

liquid


59

Allotropes




P ndash solid white



60







61





0 -2858 -6965 0


ΔHdegrxn = 3218 kJ

ΔHdegrxn = 8903 kJ


A 965 kJ

B -965 kJ

C 482 kJ

D -482 kJ

E None of these


∆Hfordm -69965

kJmol

-2300 kJmol

-2859 kJmol

-69199 kJmol


Your Turn


A -711 J

B -285 J

C +711 J

D +285 J

E None of these



-579 kJ




ΔH = qp




































ΔH = -2511 kJ


Learning Check

1256 kJ


Learning Check




938 g

470 kJ




ΔH = -353 kJ









A 659 g

B 0587 g

C 328 g

D None of these







=-2306 kJ


= 715 Latm = 7222 kJΔE= -2306kJ+72kJ= -223 kJ


Enthalpy Diagram


Hessrsquos Law


For example




-1670 kJ




















Learning Check







2H2(g) + N2(g) rarr N2H4(g) (1878 - 924) = +954 kJ

2














Your Turn




56

State Matters






-2219 + 176 kJ = -2043 kJ


57


bull 1 atm pressure





Standard State


58





gas

liquid


59

Allotropes




P ndash solid white



60







61





0 -2858 -6965 0


ΔHdegrxn = 3218 kJ

ΔHdegrxn = 8903 kJ


A 965 kJ

B -965 kJ

C 482 kJ

D -482 kJ

E None of these


∆Hfordm -69965

kJmol

-2300 kJmol

-2859 kJmol

-69199 kJmol




ΔH = qp




































ΔH = -2511 kJ


Learning Check

1256 kJ


Learning Check




938 g

470 kJ




ΔH = -353 kJ









A 659 g

B 0587 g

C 328 g

D None of these







=-2306 kJ


= 715 Latm = 7222 kJΔE= -2306kJ+72kJ= -223 kJ


Enthalpy Diagram


Hessrsquos Law


For example




-1670 kJ




















Learning Check







2H2(g) + N2(g) rarr N2H4(g) (1878 - 924) = +954 kJ

2














Your Turn




56

State Matters






-2219 + 176 kJ = -2043 kJ


57


bull 1 atm pressure





Standard State


58





gas

liquid


59

Allotropes




P ndash solid white



60







61





0 -2858 -6965 0


ΔHdegrxn = 3218 kJ

ΔHdegrxn = 8903 kJ


A 965 kJ

B -965 kJ

C 482 kJ

D -482 kJ

E None of these


∆Hfordm -69965

kJmol

-2300 kJmol

-2859 kJmol

-69199 kJmol

































ΔH = -2511 kJ


Learning Check

1256 kJ


Learning Check




938 g

470 kJ




ΔH = -353 kJ









A 659 g

B 0587 g

C 328 g

D None of these







=-2306 kJ


= 715 Latm = 7222 kJΔE= -2306kJ+72kJ= -223 kJ


Enthalpy Diagram


Hessrsquos Law


For example




-1670 kJ




















Learning Check







2H2(g) + N2(g) rarr N2H4(g) (1878 - 924) = +954 kJ

2














Your Turn




56

State Matters






-2219 + 176 kJ = -2043 kJ


57


bull 1 atm pressure





Standard State


58





gas

liquid


59

Allotropes




P ndash solid white



60







61





0 -2858 -6965 0


ΔHdegrxn = 3218 kJ

ΔHdegrxn = 8903 kJ


A 965 kJ

B -965 kJ

C 482 kJ

D -482 kJ

E None of these


∆Hfordm -69965

kJmol

-2300 kJmol

-2859 kJmol

-69199 kJmol

























ΔH = -2511 kJ


Learning Check

1256 kJ


Learning Check




938 g

470 kJ




ΔH = -353 kJ









A 659 g

B 0587 g

C 328 g

D None of these







=-2306 kJ


= 715 Latm = 7222 kJΔE= -2306kJ+72kJ= -223 kJ


Enthalpy Diagram


Hessrsquos Law


For example




-1670 kJ




















Learning Check







2H2(g) + N2(g) rarr N2H4(g) (1878 - 924) = +954 kJ

2














Your Turn




56

State Matters






-2219 + 176 kJ = -2043 kJ


57


bull 1 atm pressure





Standard State


58





gas

liquid


59

Allotropes




P ndash solid white



60







61





0 -2858 -6965 0


ΔHdegrxn = 3218 kJ

ΔHdegrxn = 8903 kJ


A 965 kJ

B -965 kJ

C 482 kJ

D -482 kJ

E None of these


∆Hfordm -69965

kJmol

-2300 kJmol

-2859 kJmol

-69199 kJmol
















ΔH = -2511 kJ


Learning Check

1256 kJ


Learning Check




938 g

470 kJ




ΔH = -353 kJ









A 659 g

B 0587 g

C 328 g

D None of these







=-2306 kJ


= 715 Latm = 7222 kJΔE= -2306kJ+72kJ= -223 kJ


Enthalpy Diagram


Hessrsquos Law


For example




-1670 kJ




















Learning Check







2H2(g) + N2(g) rarr N2H4(g) (1878 - 924) = +954 kJ

2














Your Turn




56

State Matters






-2219 + 176 kJ = -2043 kJ


57


bull 1 atm pressure





Standard State


58





gas

liquid


59

Allotropes




P ndash solid white



60







61





0 -2858 -6965 0


ΔHdegrxn = 3218 kJ

ΔHdegrxn = 8903 kJ


A 965 kJ

B -965 kJ

C 482 kJ

D -482 kJ

E None of these


∆Hfordm -69965

kJmol

-2300 kJmol

-2859 kJmol

-69199 kJmol













ΔH = -2511 kJ


Learning Check

1256 kJ


Learning Check




938 g

470 kJ




ΔH = -353 kJ









A 659 g

B 0587 g

C 328 g

D None of these







=-2306 kJ


= 715 Latm = 7222 kJΔE= -2306kJ+72kJ= -223 kJ


Enthalpy Diagram


Hessrsquos Law


For example




-1670 kJ




















Learning Check







2H2(g) + N2(g) rarr N2H4(g) (1878 - 924) = +954 kJ

2














Your Turn




56

State Matters






-2219 + 176 kJ = -2043 kJ


57


bull 1 atm pressure





Standard State


58





gas

liquid


59

Allotropes




P ndash solid white



60







61





0 -2858 -6965 0


ΔHdegrxn = 3218 kJ

ΔHdegrxn = 8903 kJ


A 965 kJ

B -965 kJ

C 482 kJ

D -482 kJ

E None of these


∆Hfordm -69965

kJmol

-2300 kJmol

-2859 kJmol

-69199 kJmol








ΔH = -2511 kJ


Learning Check

1256 kJ


Learning Check




938 g

470 kJ




ΔH = -353 kJ









A 659 g

B 0587 g

C 328 g

D None of these







=-2306 kJ


= 715 Latm = 7222 kJΔE= -2306kJ+72kJ= -223 kJ


Enthalpy Diagram


Hessrsquos Law


For example




-1670 kJ




















Learning Check







2H2(g) + N2(g) rarr N2H4(g) (1878 - 924) = +954 kJ

2














Your Turn




56

State Matters






-2219 + 176 kJ = -2043 kJ


57


bull 1 atm pressure





Standard State


58





gas

liquid


59

Allotropes




P ndash solid white



60







61





0 -2858 -6965 0


ΔHdegrxn = 3218 kJ

ΔHdegrxn = 8903 kJ


A 965 kJ

B -965 kJ

C 482 kJ

D -482 kJ

E None of these


∆Hfordm -69965

kJmol

-2300 kJmol

-2859 kJmol

-69199 kJmol



ΔH = -2511 kJ


Learning Check

1256 kJ


Learning Check




938 g

470 kJ




ΔH = -353 kJ









A 659 g

B 0587 g

C 328 g

D None of these







=-2306 kJ


= 715 Latm = 7222 kJΔE= -2306kJ+72kJ= -223 kJ


Enthalpy Diagram


Hessrsquos Law


For example




-1670 kJ




















Learning Check







2H2(g) + N2(g) rarr N2H4(g) (1878 - 924) = +954 kJ

2














Your Turn




56

State Matters






-2219 + 176 kJ = -2043 kJ


57


bull 1 atm pressure





Standard State


58





gas

liquid


59

Allotropes




P ndash solid white



60







61





0 -2858 -6965 0


ΔHdegrxn = 3218 kJ

ΔHdegrxn = 8903 kJ


A 965 kJ

B -965 kJ

C 482 kJ

D -482 kJ

E None of these


∆Hfordm -69965

kJmol

-2300 kJmol

-2859 kJmol

-69199 kJmol


Learning Check




938 g

470 kJ




ΔH = -353 kJ









A 659 g

B 0587 g

C 328 g

D None of these







=-2306 kJ


= 715 Latm = 7222 kJΔE= -2306kJ+72kJ= -223 kJ


Enthalpy Diagram


Hessrsquos Law


For example




-1670 kJ




















Learning Check







2H2(g) + N2(g) rarr N2H4(g) (1878 - 924) = +954 kJ

2














Your Turn




56

State Matters






-2219 + 176 kJ = -2043 kJ


57


bull 1 atm pressure





Standard State


58





gas

liquid


59

Allotropes




P ndash solid white



60







61





0 -2858 -6965 0


ΔHdegrxn = 3218 kJ

ΔHdegrxn = 8903 kJ


A 965 kJ

B -965 kJ

C 482 kJ

D -482 kJ

E None of these


∆Hfordm -69965

kJmol

-2300 kJmol

-2859 kJmol

-69199 kJmol




ΔH = -353 kJ









A 659 g

B 0587 g

C 328 g

D None of these







=-2306 kJ


= 715 Latm = 7222 kJΔE= -2306kJ+72kJ= -223 kJ


Enthalpy Diagram


Hessrsquos Law


For example




-1670 kJ




















Learning Check







2H2(g) + N2(g) rarr N2H4(g) (1878 - 924) = +954 kJ

2














Your Turn




56

State Matters






-2219 + 176 kJ = -2043 kJ


57


bull 1 atm pressure





Standard State


58





gas

liquid


59

Allotropes




P ndash solid white



60







61





0 -2858 -6965 0


ΔHdegrxn = 3218 kJ

ΔHdegrxn = 8903 kJ


A 965 kJ

B -965 kJ

C 482 kJ

D -482 kJ

E None of these


∆Hfordm -69965

kJmol

-2300 kJmol

-2859 kJmol

-69199 kJmol





A 659 g

B 0587 g

C 328 g

D None of these







=-2306 kJ


= 715 Latm = 7222 kJΔE= -2306kJ+72kJ= -223 kJ


Enthalpy Diagram


Hessrsquos Law


For example




-1670 kJ




















Learning Check







2H2(g) + N2(g) rarr N2H4(g) (1878 - 924) = +954 kJ

2














Your Turn




56

State Matters






-2219 + 176 kJ = -2043 kJ


57


bull 1 atm pressure





Standard State


58





gas

liquid


59

Allotropes




P ndash solid white



60







61





0 -2858 -6965 0


ΔHdegrxn = 3218 kJ

ΔHdegrxn = 8903 kJ


A 965 kJ

B -965 kJ

C 482 kJ

D -482 kJ

E None of these


∆Hfordm -69965

kJmol

-2300 kJmol

-2859 kJmol

-69199 kJmol







=-2306 kJ


= 715 Latm = 7222 kJΔE= -2306kJ+72kJ= -223 kJ


Enthalpy Diagram


Hessrsquos Law


For example




-1670 kJ




















Learning Check







2H2(g) + N2(g) rarr N2H4(g) (1878 - 924) = +954 kJ

2














Your Turn




56

State Matters






-2219 + 176 kJ = -2043 kJ


57


bull 1 atm pressure





Standard State


58





gas

liquid


59

Allotropes




P ndash solid white



60







61





0 -2858 -6965 0


ΔHdegrxn = 3218 kJ

ΔHdegrxn = 8903 kJ


A 965 kJ

B -965 kJ

C 482 kJ

D -482 kJ

E None of these


∆Hfordm -69965

kJmol

-2300 kJmol

-2859 kJmol

-69199 kJmol


Enthalpy Diagram


Hessrsquos Law


For example




-1670 kJ




















Learning Check







2H2(g) + N2(g) rarr N2H4(g) (1878 - 924) = +954 kJ

2














Your Turn




56

State Matters






-2219 + 176 kJ = -2043 kJ


57


bull 1 atm pressure





Standard State


58





gas

liquid


59

Allotropes




P ndash solid white



60







61





0 -2858 -6965 0


ΔHdegrxn = 3218 kJ

ΔHdegrxn = 8903 kJ


A 965 kJ

B -965 kJ

C 482 kJ

D -482 kJ

E None of these


∆Hfordm -69965

kJmol

-2300 kJmol

-2859 kJmol

-69199 kJmol


Hessrsquos Law


For example




-1670 kJ




















Learning Check







2H2(g) + N2(g) rarr N2H4(g) (1878 - 924) = +954 kJ

2














Your Turn




56

State Matters






-2219 + 176 kJ = -2043 kJ


57


bull 1 atm pressure





Standard State


58





gas

liquid


59

Allotropes




P ndash solid white



60







61





0 -2858 -6965 0


ΔHdegrxn = 3218 kJ

ΔHdegrxn = 8903 kJ


A 965 kJ

B -965 kJ

C 482 kJ

D -482 kJ

E None of these


∆Hfordm -69965

kJmol

-2300 kJmol

-2859 kJmol

-69199 kJmol




















Learning Check







2H2(g) + N2(g) rarr N2H4(g) (1878 - 924) = +954 kJ

2














Your Turn




56

State Matters






-2219 + 176 kJ = -2043 kJ


57


bull 1 atm pressure





Standard State


58





gas

liquid


59

Allotropes




P ndash solid white



60







61





0 -2858 -6965 0


ΔHdegrxn = 3218 kJ

ΔHdegrxn = 8903 kJ


A 965 kJ

B -965 kJ

C 482 kJ

D -482 kJ

E None of these


∆Hfordm -69965

kJmol

-2300 kJmol

-2859 kJmol

-69199 kJmol















Learning Check







2H2(g) + N2(g) rarr N2H4(g) (1878 - 924) = +954 kJ

2














Your Turn




56

State Matters






-2219 + 176 kJ = -2043 kJ


57


bull 1 atm pressure





Standard State


58





gas

liquid


59

Allotropes




P ndash solid white



60







61





0 -2858 -6965 0


ΔHdegrxn = 3218 kJ

ΔHdegrxn = 8903 kJ


A 965 kJ

B -965 kJ

C 482 kJ

D -482 kJ

E None of these


∆Hfordm -69965

kJmol

-2300 kJmol

-2859 kJmol

-69199 kJmol









Learning Check







2H2(g) + N2(g) rarr N2H4(g) (1878 - 924) = +954 kJ

2














Your Turn




56

State Matters






-2219 + 176 kJ = -2043 kJ


57


bull 1 atm pressure





Standard State


58





gas

liquid


59

Allotropes




P ndash solid white



60







61





0 -2858 -6965 0


ΔHdegrxn = 3218 kJ

ΔHdegrxn = 8903 kJ


A 965 kJ

B -965 kJ

C 482 kJ

D -482 kJ

E None of these


∆Hfordm -69965

kJmol

-2300 kJmol

-2859 kJmol

-69199 kJmol


Learning Check







2H2(g) + N2(g) rarr N2H4(g) (1878 - 924) = +954 kJ

2














Your Turn




56

State Matters






-2219 + 176 kJ = -2043 kJ


57


bull 1 atm pressure





Standard State


58





gas

liquid


59

Allotropes




P ndash solid white



60







61





0 -2858 -6965 0


ΔHdegrxn = 3218 kJ

ΔHdegrxn = 8903 kJ


A 965 kJ

B -965 kJ

C 482 kJ

D -482 kJ

E None of these


∆Hfordm -69965

kJmol

-2300 kJmol

-2859 kJmol

-69199 kJmol














Your Turn




56

State Matters






-2219 + 176 kJ = -2043 kJ


57


bull 1 atm pressure





Standard State


58





gas

liquid


59

Allotropes




P ndash solid white



60







61





0 -2858 -6965 0


ΔHdegrxn = 3218 kJ

ΔHdegrxn = 8903 kJ


A 965 kJ

B -965 kJ

C 482 kJ

D -482 kJ

E None of these


∆Hfordm -69965

kJmol

-2300 kJmol

-2859 kJmol

-69199 kJmol


Your Turn




56

State Matters






-2219 + 176 kJ = -2043 kJ


57


bull 1 atm pressure





Standard State


58





gas

liquid


59

Allotropes




P ndash solid white



60







61





0 -2858 -6965 0


ΔHdegrxn = 3218 kJ

ΔHdegrxn = 8903 kJ


A 965 kJ

B -965 kJ

C 482 kJ

D -482 kJ

E None of these


∆Hfordm -69965

kJmol

-2300 kJmol

-2859 kJmol

-69199 kJmol


56

State Matters






-2219 + 176 kJ = -2043 kJ


57


bull 1 atm pressure





Standard State


58





gas

liquid


59

Allotropes




P ndash solid white



60







61





0 -2858 -6965 0


ΔHdegrxn = 3218 kJ

ΔHdegrxn = 8903 kJ


A 965 kJ

B -965 kJ

C 482 kJ

D -482 kJ

E None of these


∆Hfordm -69965

kJmol

-2300 kJmol

-2859 kJmol

-69199 kJmol


57


bull 1 atm pressure





Standard State


58





gas

liquid


59

Allotropes




P ndash solid white



60







61





0 -2858 -6965 0


ΔHdegrxn = 3218 kJ

ΔHdegrxn = 8903 kJ


A 965 kJ

B -965 kJ

C 482 kJ

D -482 kJ

E None of these


∆Hfordm -69965

kJmol

-2300 kJmol

-2859 kJmol

-69199 kJmol


58





gas

liquid


59

Allotropes




P ndash solid white



60







61





0 -2858 -6965 0


ΔHdegrxn = 3218 kJ

ΔHdegrxn = 8903 kJ


A 965 kJ

B -965 kJ

C 482 kJ

D -482 kJ

E None of these


∆Hfordm -69965

kJmol

-2300 kJmol

-2859 kJmol

-69199 kJmol


59

Allotropes




P ndash solid white



60







61





0 -2858 -6965 0


ΔHdegrxn = 3218 kJ

ΔHdegrxn = 8903 kJ


A 965 kJ

B -965 kJ

C 482 kJ

D -482 kJ

E None of these


∆Hfordm -69965

kJmol

-2300 kJmol

-2859 kJmol

-69199 kJmol


60







61





0 -2858 -6965 0


ΔHdegrxn = 3218 kJ

ΔHdegrxn = 8903 kJ


A 965 kJ

B -965 kJ

C 482 kJ

D -482 kJ

E None of these


∆Hfordm -69965

kJmol

-2300 kJmol

-2859 kJmol

-69199 kJmol


61





0 -2858 -6965 0


ΔHdegrxn = 3218 kJ

ΔHdegrxn = 8903 kJ


A 965 kJ

B -965 kJ

C 482 kJ

D -482 kJ

E None of these


∆Hfordm -69965

kJmol

-2300 kJmol

-2859 kJmol

-69199 kJmol


A 965 kJ

B -965 kJ

C 482 kJ

D -482 kJ

E None of these


∆Hfordm -69965

kJmol

-2300 kJmol

-2859 kJmol

-69199 kJmol

Date post:	08-Apr-2018
Category:	Documents
Upload:	truongtu
View:	226 times
Download:	4 times

Chapter 6 Energy and Chemical Change - Blog FTSL-2 … 6 Energy and Chemical Change Brady and Senese...

Documents