“A” students work (without solutions manual) ~ 10 problems/night

“A” students work(without solutions manual)~ 10 problems/night.

Alanah FitchFlanner Hall 402508-3119afitch@luc.edu

Office Hours W – F 2-3 pm

Module #11Thermochemistry

FITCH Rules

G1: Suzuki is SuccessG2. Slow me down G3. Scientific Knowledge is ReferentialG4. Watch out for Red HerringsG5. Chemists are LazyC1. It’s all about chargeC2. Everybody wants to “be like Mike”C3. Size MattersC4. Still Waters Run DeepC5. Alpha Dogs eat first

E kq q

r ror k

Set up like others

1. Want to understand how moleculesform and interact.

2. Need a measurement tool and scaling3. Q – measure of the energy within and

1. Between molecules.

Energy:

Or transfer heat, q

E E E q wfina l in itia l

Energy ConsumedDepends onBoth workAnd heat

capacity to do work

Marie the Jewess, 300 Jabir ibn Hawan, 721-815

Galen, 170 Abbe Jean Picard1620-1682

Galileo Galili1564-1642

Daniel Fahrenheit1686-1737

Evangelista Torricelli1608-1647

Isaac Newton1643-1727

Robert Boyle, 1627-1691

Blaise Pascal1623-1662

Anders Celsius1701-1744

Charles Augustin Coulomb 1735-1806 John Dalton

1766-1844 B. P. Emile Clapeyron1799-1864

Jacques Charles 1778-1850 Germain Henri Hess

1802-1850

Fitch Rule G3: Science is Referential

William ThompsonLord Kelvin, 1824-1907

James Maxwell1831-1879

Johannes D.Van der Waals1837-1923

Justus von Liebig (1803-1873

1825-1898Johann Balmer

James Joule (1818-1889)

Johannes Rydberg1854-1919

Rudolph Clausius1822-1888

Thomas Graham1805-1869

Heinrich R. Hertz, 1857-1894

Max Planck1858-1947

J. J. Thomson1856-1940

Linus Pauling1901-1994

Werner Karl Heisenberg1901-1976

Wolfgang Pauli1900-1958

Count Alessandro Guiseppe Antonio Anastasio Volta, 1747-1827

Georg Simon Ohm 1789-1854

Henri Louis LeChatlier1850-1936

Svante Arrehenius1859-1927

Francois-Marie Raoult1830-1901

William Henry1775-1836physician

Gilbert Newton Lewis1875-1946

Fritz Haber1868-1934

Michael Faraday1791-1867

Luigi Galvani1737-1798

Walther Nernst1864-1941

Lawrence J. Henderson1878-1942

Amedeo Avogadro1756-1856

J. Willard Gibbs1839-1903

Niels Bohr1885-1962 Erwin Schodinger

1887-1961Louis de Broglie (1892-1987)

Friedrich H. Hund1896-1997

Fritz London1900-1954

An alchemist

Ludwig Boltzman1844-1906

Richard AugustCarl Emil Erlenmeyer1825-1909

Johannes NicolausBronsted1879-1947

Thomas MartinLowry1874-1936

James Watt1736-1819

Dmitri Mendeleev1834-1907

Properties and MeasurementsProperty Unit Reference StateSize m size of earthVolume cm3 mWeight gram mass of 1 cm3 water at specified Temp

(and Pressure)Temperature oC, K boiling, freezing of water (specified

Pressure)1.66053873x10-24g amu (mass of 1C-12 atom)/12quantity mole atomic mass of an element in gramsPressure atm, mm Hg earth’s atmosphere at sea levelEnergy, General

Animal hp horse on tread millheat BTU 1 lb water 1 oF

calorie 1 g water 1 oCKinetic J m, kg, sElectrostatic 1 electrical charge against 1 Velectronic states in atom Energy of electron in vacuum

Electronegativity F Heat flow measurements Reference state?

To set a “heat flow” scale

1. Defined conditions: how experiment is performedopen flask, closed flask, pressure

2. Define the direction of heat flow by giving a positive or negative number

+ heat - heat?

Does the “system (earth)” gain energy?

For image abovesystem (earth) gains heatfrom surroundings (sun)

1. Defined conditions: how experiment is performedopen flask, closed flask, pressure

2. Define the direction of heat flow by giving a positive or negative number

+ heat - heat?

We would get a different answer if we asked“Does the “system (sun)” gain energy?”

For this question: thesystem (sun) loses heatto the surroundings (earth)

First Law of Thermodynamics: Energy is conserved

No universal change in energyJust a transfer of energy

q>0 q<0

E = constant when System AND surroundings considered!

system system

system

surroundings

1. Defined conditions: how experiment is performed open flask, closed flask, pressure

2. Define the direction of heat flow (q) by giving a positive or negative number

q is + when heat flows into the system from the surroundings

q is - when heat flows out of the system into the surroundings

3. Chemical process in the “system” is defined by heat flow

endothermic q>0exothermic q<0

Pressure)1.66053873x10-24g amu (mass of 1C-12 atom)/12quantity mole atomic mass of an element in gramsPressure atm, mm Hg earth’s atmosphere at sea levelEnergy: Thermal BTU 1 lb water 1 oF

calorie 1 g water 1 oCKinetic J 2kg mass moving at 1m/sEnergy, of electrons energy of electron in a vacuum

Electronegativity F Heat Flow into system = +

Z n s H aq Z n aq H g( ) ( ) ( ) ( ) 2 22

Constant Atm.pressure

Chemical reactions involve 1. heat exchange

Heat exchangeAt constantPressure

As a review:

who is oxidized?who is reduced?what is the oxidation number on H2?Who is an oxidizing agent?

1 atm pressure = constant pressure

This means heat flow, q, is enthalpy change

H =enthalpy

Greek: thalpein – to heaten - in

H for (?) heat

SubscriptReminds us thatPressure is constant

Chemical reactions involve 1. heat exchange 2. work

constant Atm. pressure

Z n s H aq Z n aq H g( ) ( ) ( ) ( ) 2 22

Pressure-Volumework

E E E q wfina l in itia l

E q P VP P

Change in volume is typically small for Most reactions

E HP This means we can measure the energy change of A chemical reaction by measuring the heat exchangeAt constant pressure

E H P VP

This is a form of Rule G3

Science is referential

Generally final - initial

five Navy Avengers

disappeared in the Bermuda Triangle on Dec. 5, 194

First example problem will involve methane

We will prove to ourselves that the Pressure-Volume work is a Small contribution to the total energy change

MethaneGas RecoveryAt landfills

3 to 8 standard cubic feet of biogas per pound of manure. The biogas usually contains 60 to 70% methane.

Consider the contribution of volume of gas phase molecules

C H O C O H Og g g l4 2 2 22 2( ) ( ) ( ) ( )

P V n n R Tgas fina l gas in itia l

P V nR T

P V n R T At constant T:

P V m olesL a tm

m ol KK

1 3 0 0 8 2 1 2 9 8.

P V L atm 4 8 9 3 1 6.

P V L atmkJ

L a tmkJ

4 8 9 3 1 6

0 1 0 1 34 9.

2mole change

%&$*! Conversions – ifInterested see slide after next

Consider the contribution of volume change for water in this reaction

C H O C O H Og g g l4 2 2 22 2( ) ( ) ( ) ( )

21 8 1

1 00 0 3 62

3 3m oleH Og

cm w ater

gw ater

cmLl( ) * * * .

P V atm LkJ

L atmkJ

1 0 0 3 6

0 1 0 1 30 0 0 3 6.

Energy in kJMost reactions total (q): ~ 1000 kJPV 1 mole gas ~ 2.5 kJPV 2mole liquid water ~ 0.0036 kJ

Sig fig tells us that PV energy small compared to q

%&$*! Conversions – ifInterested see next slide

By the end ofThis moduleWe will see thisIs “true”

Optional Slide: conversion

a tmx P a

1 0 1 3 2 5 1 0 5.

m sP a

1 0 1 3 2 5 1 0 5 2.

a tmx P a

m sP a

1 0 1 3 2 5 1 0 1 0 1

5 2 3 3

a tmx P a

m sP a

1 0 1 3 2 5 1 0 1 0 1

1 0 1 00 1 0 1 3 2 5

5 2 3 3

0 1 0 1 3 2 5 1 0 1 3 2 5 1 0 1 0 1

1 0 1 0

5 2 3 3

a tm L

m sP a

To set a “heat flow” scale1. Defined conditions: how experiment is performed

Constant Pressure2. But not on the path taken (state property)

Heat flowdependsthe conditions

q H H Hreaction cons tp ressure products reac tstan tan

H= enthalpy

Enthalpy is a state property(measured under constant pressure, but how measured under that constant pressure is not important)

exo therm ic

H O hea t H Os2 2( )

H Hproducts reac ts tan

endotherm ic

H Hproducts reac tss tan

C H O C O H O hea tg g g l4 2 2 22( ) ( ) ( ) ( )

Think of heat as a reactant

Think of heat as a product

Enthalpy is an “extensive” property

Depends upon the amount present

C H O C O H O H kJg g g l4 2 2 22 2 8 9 0( ) ( ) ( ) ( )

890kJ of heat is released when 1 mole of methaneReacts with oxygen

m oleC H g( )

12 1 7 8 0

m oleC Hm oleC H kJ

( )( )

Electronegativity F Heat flow measurements constant pressure, define system vs surroundings

per mole basis (intensive)

2006 Sept Sci. Am: World Wide Petroleum UsageLand People Transport29% of total use

Land Freight 19%Air People and Freight 5%

Non-transportation

U.S. differs from world in distribution of petroleum use

Transportation = 71.8%

57.8% of Transportation=Personal land transport

= 41.5% of total U.S. consumptionData US DOE 2006

Context for the next example problem

3 Non-transportation

Total transportation=53%

H x kJ C H O C O H Og g 1 0 9 1 0 2 2 5 1 6 1 848 1 8 2 2 2. ( ) ( )

H x kJ 1 0 7 1 0 6.

Example: If you drove an automobile 1.50x102 miles at 17.5 miles/gal you consume a certain number of gallons of gasoline. If you burn thatnumber of gallons of gasoline at constant pressure how much heat would be released? Assume the gasoline is pure octane with a density of the octane 0.690 g/mL?

1 5 01

1 7 5.

Strategy: need moles of octane consumed (Golden Bridge)

miles gallons gramsdensitympg

molesMolar mass

1 5 01

3 7 8 5 2.

3 7 8 5 2 1 0 3

3 7 8 5 2 1 0 0 6 9 038 8

. .m i

m L C H

3 7 8 5 2 1 0 0 6 9 0 1

. .m i

m L C H

m oleC H

3 7 8 5 2 1 0 0 6 9 0 1

1 0 9 1 0

. . .m i

m L C H

m oleC H

H kJ 1 0 7 0 2 4 3 9 5 5, , .

3 sig figWe will use part of this problemAgain:

3 7 8 5 2 1 0 0 6 9 0 1

1 0 9 1 0

1 2 4 8 6 138 8

. . . ,L

m L C H

m oleC H

1.Enthalpy is an extensive property (depends upon number of moles)2.Enthalpy change for a reaction is equal in magnitude, but opposite

in sign, to the enthalpy for the reverse reaction

3. Enthalpy change depends upon the state of the reactant and products

H x kJ C H O C O H Og g 1 0 9 1 0 2 2 5 1 6 1 848 1 8 2 2 2. ( ) ( )

H x kJ 1 0 9 1 0 4.

H kJ 4 4

1 6 1 8 2 2 52 2 8 1 8 2C O H O C H Og g( ) ( )

H O H Ol g2 2( ) ( )

ENERGY measurementa) change in temperatureb) some function specific to the material and how

it is organized (bonds)

heat 25 oC15 oC

ENERGY change is a function of a) temperatureb) material E k

ENERGY measurementa) change in temperatureb) some function specific to the material and how

it is organized (bonds)

q C t t

fina l in itia l

C = heat capacity = heat required to raise the temperature of the system 1oC

units: J/oC

q m c t

m = mass c = specific heat of a pure substance

Pure material

http://www.lsbu.ac.uk/water/molecule.html

Electrons on Oxygen sit“out there” causing largeElectrostatic potential Oriented on the electrons

Electron density of waterShape and charge distributionOn water

ch e densityq

C(liquidH2O)=4.18J/g-oC

Liquid water is very strongly organizeddue to the polarity of the molecule, so it has a high specific heat

Material Specific Heats c, (J/g-K)Pb(s) 0.12803Pb(l) 0.16317Cu(s) 0.382Fe(s) 0.446Cl2(g) 0.478C(s) 0.71CO2(g) 0.843NaCl(s) 0.866Al(s) 0.89C6H6(l)1.72H2O(g) 1.87C2H5OH(l) 2.43H2O(l) 4.18

Ability to storeheat in a substanceis variable.

Specific heats, c, of various substances in various physical states

For the same amount of energy, easier to break electrostatic attraction of He compared to water with it’s localized charge

q m c t

Mass of water:

Example: 1.00 cup of water is heated from 25.0 oC to 100.0 oC. Howmany joules were used to heat the water?

t K 7 5

1 0 01

1 0 5 7.

1 0 0. cup 1 0 0

4. cup

1 0 5 7

12 3 6 5 1

.cupq t

t C Co o 1 0 0 2 5 t T Tfina l in itia l

t Co 7 5

q x J 7 4 1 1 0 4.

q 7 4 1 4 8 5 3.

2 3 6 5 1

4 1 87 5.

q kJ 7 4 1.

Example 2: 1 cup of dry soil (specific heat, c = 0.800 J/gK; density =1.28 g/cm3). Calculate the Joules required to raise the temperature of the dry soil from 25oC to 100

1 0 01

1 0 5 7

1 0 1 2 8

12 3 6 5 1 1 2 8 3 0 2 7

.. . .cup

m Lg g

3 0 2 7

0 87 5.

q J kJdry so il 1 8 1 6 4 1 8 1, .

q kJw ater 7 4 1.

Because water has a high heat capacity it takes longer than air or soil to warm up and longer to cool down

LandWater

The temperature change in 24 hours in summer for water is not muchleading to big differences between lake and land

1. Hot air rises over land

2. Cold air from Lake moves to fill in (Lake Breeze)

1. Hot air rises over Lake

2. Cold air moves in from land (Land breeze)

and thus a lake breeze

Heat capacity of large bodiesOf water affect human activity

Example 3: Heat capacity of the metal block of a car combustion engine. Assume that a Prius 1.5 L, 4 cylinder 176.6 kg engine block is made of iron. The specific heat of iron is 440 J/kg-C. If I drive 8 miles twice a day (to work and back) at an average 42 mpg, what fraction of the total available enthalpy in 1 gallon of octane is consumed in heating the engine block from 25oC to 100oC?

q m c t

kg CC kJ

4 4 01 0 0 2 5 5 8 2 8,

H x kJ C H O C O H Og g 1 0 9 1 0 2 2 5 1 6 1 848 1 8 2 2 2. ( ) ( )

Compare to heat available from combustion of 1 gallon of octane (from before

3 7 8 5 2 1 0 0 6 9 0 1

1 0 9 1 0

1 2 4 8 6 138 8

. . . ,L

m L C H

m oleC H

5 8 2 8

2 7 7 3 0m iles

m iles

kJ eng ine hea ting

kJ hea ting

5 8 2 8

1 2 4 8 6 11 0 0 2 2 2 %

Let’s compare to whatI measure

Heating the engine block

3 9 3 1

3 91 0 0 2 0 5 %

Prius 1

4 3 3 4

4 31 0 0 2 0 %

Prius 2

Electronegativity F Heat flow measurements constant pressure, define system vs surroundings

per mole basis (intensive) Calorimetry

q qreaction ca lorim eter

If the temperature of the water rises (heat flow into water)then heat must have been lost from the reaction

Relate reaction heat To the calorimeter heat

Example: When 1.00 g of ammonium nitrate, NH4NO3, is added to 50.0 g of water in a coffee-cup calorimeter, it dissolves:

and the temperature of the water drops from 25.00 to 23.32 oC. Assuming that all the heat absorbed by the reactions comes from the water, calculate q for the reaction system.

N H N O N H N Os aq aq4 3 4 3( ) ( ) ( )

q Jca lorim eter 3 5 1 1 2.A 1 degree change in Celsus Is a 1 degree change in Kelvin

q Jca lorim eter 3 5 1

KKca lorim eter

2 0 9 1 6 8.

gKC Cca lorim eter

5 0 0 4 1 8 2 3 3 2 2 5 0 0. . . . q qreaction ca lorim eter

q m c tca lorim eter

q Jreaction 3 5 1

q q C treaction reaction ca l Heat capacity of the calorimeter

In some problemsYou will Need to determineThis numberIn one step andThen go on

Not all calorimeters can be based on

q q m c treaction ca lorim eter

Example: The reaction between hydrogen and chlorine

can be studied in a bomb calorimeter. It is found that when a 1.00 g sample of H2 reacts completely, the temperature rises from 20.00 to 29.82 oC. Taking the heat capacity of the calorimeter to be 9.33 kJ/oC, calculate the amount of heat evolved in the reaction.

H C l H C lg g g g( ) ( ) ( ) 2 2

q C treaction ca l

CC Creaction o

9 3 3 2 9 8 2 2 0 0 0. . .

q kJreaction 9 1 6.

Heat evolved?Calorimeter heat capacity = 9.33 kJ/oCTinitial =20.00 oCT final = 29.82 oC

Example: Salicyclic acid, C7H6O3, is one of the starting materials in the manufacture of aspirin. When 1.00 g of salicylic acid burns in a bomb calorimeter, the temperature rises to 32.11oC from 28.91 oC. The temperature in the bomb calorimeter increases by 2.48oC when the calorimeter absorbs 9.37 kJ. How much heat is given off when one mole of salicylic acid is burned?

1.00 g of C7H6O3

Tinitial 32.11oCTfinal 29.91oC9.37kJ required to cause 2.48 oC change

q C treaction ca l

Cca l o9 3 7

Creaction o

2 4 83 2 11 2 9 9 1

q kJreaction 8 3 1 2 1.

11 3 8

8 3 1 2 1

13 1 5 8 5 9 87 6 3

7 6 3 7 6 3

m oleC H OgC H O

m oleC H O

gC H OkJ

3 1 6 kJ

Combine

calorimetry

2 2 5 1 6 1 88 1 8 2 2 2C H O C O H Og g ( ) ( )

To get reaction enthalpies

Reaction stoichiometry

N H N O N H N Os aq aq4 3 4 3( ) ( ) ( )

1 g was reacted

Example 2 What is the enthalpy change for the reaction

If exactly 1 g of ammonium nitrate is reacted in a bomb calorimeter made with 50 g of water and the temperature of the water drops from 25.00 oC to 23.32 oC?

q H Jreaction cons tlabpressu retan 3 5 1

q Jreaction 3 5 1

q Jreaction 3 5 1 1 2.

CCreaction o

2 0 9 1 6 8.

q q gJ

g CC Creaction ca lorim eter o

5 0 0 4 1 8 2 3 3 2 2 5 0 0. . . .

2N=28.024H=4.043O=48.00 80.05

8 0 0 5 2 8 1 0 0 2 8 1J

N H N O N H N O H kJs aq aq4 3 4 3 2 8 1( ) ( ) ( ) .

Where did the per/mole go?

The reaction was written as a per/moleEnthalpy is understood as a per/mole of reactant (or as the reaction

is written)

N H N O N H N O H kJs aq aq4 3 4 3 2 8 1( ) ( ) ( ) .

Thermochemical Equation Rules

1. Value of H applies when products and reactants are at same temperature, 25oC unless otherwise specified.

2. Sign of H, indicates whether reaction, when carried out at constant pressure, is exothermic or endothermic

3. ΔH sign changes when reaction is reversed

4. Stoichiometry is important4. Phases of all species must be specified5. Values of H is same regardless of method used to calculate it (Hess’s Law)

When there are noCoefficients it is understood thatIt is “1”

N H N O N H N O H kJaq aq s4 3 4 3 2 8 1( ) ( ) ( ) .

Using a coffee-cup calorimeter, it is found that when an ice cube weighing 24.6 g melts, it absorbs 8.19 kJ of heat. Calculate for the phase change represented by the thermochemical equation

H O H Os l2 2( ) ( )

1 8 0 22

m oleice

1 8 0 21 6 0 02

m olem ole kJ

H OH O

1 8 0 212

m olem ole

H OH O

Example illustrating importance of phases

H O H Og g l2

2 2 2( ) ( ) ( )

An example of several of the rules using Fuel Cells

kJ H O H Og g l

5 7 1 6

22 8 6 2

.( ) ( ) ( )

Fuel cells use the reaction:

Calculate the enthalpy for the equation above given that:

H kJ H O H Ol g g 5 7 1 6 2 22 2 2. ( ) ( ) ( )

Reverse reaction:

H kJ H O H Og g l 5 7 1 6 2 22 2 2. ( ) ( ) ( )

Here we got a number by coming “at it” from an odd direction

Hess’s law

Germain Henri Hess1802-1850 born in Geneva SwitzerlandProfessor of ChemistryAt St. Petersburg Technological Institute

The value of H for a reaction is the same whether it occurs in one step or in a series of steps (enthalpy (constant P, T) is a state function)

Marie the Jewess, 300 Jabir ibn Hawan, 721-815

Galen, 170 Abbe Jean Picard1620-1682

Galileo Galili1564-1642

Daniel Fahrenheit1686-1737

Evangelista Torricelli1608-1647

Isaac Newton1643-1727

Robert Boyle, 1627-1691

Blaise Pascal1623-1662

Anders Celsius1701-1744

Charles Augustin Coulomb 1735-1806 John Dalton

1766-1844 B. P. Emile Clapeyron1799-1864

Jacques Charles 1778-1850 Germain Henri Hess

1802-1850

Fitch Rule G3: Science is Referential

William ThompsonLord Kelvin, 1824-1907

James Maxwell1831-1879

Johannes D.Van der Waals1837-1923

Justus von Liebig (1803-1873

1825-1898Johann Balmer

James Joule (1818-1889)

Johannes Rydberg1854-1919

Rudolph Clausius1822-1888

Thomas Graham1805-1869

Heinrich R. Hertz, 1857-1894

Max Planck1858-1947

J. J. Thomson1856-1940

Linus Pauling1901-1994

Werner Karl Heisenberg1901-1976

Wolfgang Pauli1900-1958

Count Alessandro Guiseppe Antonio Anastasio Volta, 1747-1827

Georg Simon Ohm 1789-1854

Henri Louis LeChatlier1850-1936

Svante Arrehenius1859-1927

Francois-Marie Raoult1830-1901

William Henry1775-1836physician

Gilbert Newton Lewis1875-1946

Fritz Haber1868-1934

Michael Faraday1791-1867

Luigi Galvani1737-1798

Walther Nernst1864-1941

Lawrence J. Henderson1878-1942

Amedeo Avogadro1756-1856

J. Willard Gibbs1839-1903

Niels Bohr1885-1962 Erwin Schodinger

1887-1961Louis de Broglie (1892-1987)

Friedrich H. Hund1896-1997

Fritz London1900-1954

An alchemist

Ludwig Boltzman1844-1906

Richard AugustCarl Emil Erlenmeyer1825-1909

Johannes NicolausBronsted1879-1947

Thomas MartinLowry1874-1936

James Watt1736-1819

Dmitri Mendeleev1834-1907

C O C Os g g( ) ( ) ( ) 1

It is difficult to measure the heat evolved for this reaction because it occurs as the partial burning of carbon in the presence of other reactionsinvolving the complete burning of carbon

H kJ C O C Os g g 3 9 3 5 2 2. ( ) ( ) ( )

To solve rearrange equations to get CO on right hand side

Example of how Hess’s law is useful

CO2(g)

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

This is theNumber we wantBut can’t actually measure

C(s) + O2(g)

Get to the number by an alternativePath (State function!)

H kJ C O O C Og g g 5 6 6 0 2 22 2. ( ) ( ) ( )

Related to 2CO + O2….

H kJ0 11 0 5 . C O C Os g g( ) ( ) ( ) 1

H kJ C O C O

H kJ C O C O O

3 9 3 5

2 8 3 0

( ) ( ) ( )

H kJ C O C O Og g g 2 8 3 0 2

2 2. ( ) ( ) ( )

CO2(g)

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

C(s) + O2(g)

C O C O Og g g

5 6 6 0

.( ) ( ) ( )

H kJ C O C O Og g g 5 6 6 0 2 22 2. ( ) ( ) ( )

H kJ C O O C Og g g 5 6 6 0 2 22 2. ( ) ( ) ( )

H kJ 11 0 5.

H kJ C O C Os g g 3 9 3 5 2 2. ( ) ( ) ( )

2 2O g( )C O C Os g g( ) ( ) ( ) 1

Enthalpies of Formation

Rather than getting the enthalpy for each reaction from a bombcalorimeter use a smaller number of standard reactions from whichHess’s law can be applied to get all the remainder reactions ofinterest

Invoke Rule G5: Chemists are Lazy

Enthalpy associated with standard reaction isenthalpy of formation

which is the enthalpy change when one mole of compound is formed at constant pressure of 1 atm and a fixed temperature, ordinarily 25oC, from the elements in their stable states at that pressure and temperature. STP (Standard Temperature and Pressure)

This allows us to look at enthalpy of compoundsnot reactions which reduces total data which mustBe acquired (Chemists are Lazy!!!)

H kJ N O N OfO

g a tm C g a tm C g a tm C 8 8 21

2 2 1 2 5 2 1 2 5 2 1 2 5( , , ) ( , , ) ( , , )

Standard molar enthalpy of formation of a compound

1 atm pressure25oC

From elements in their stable states at

1 is “understood”

Most Hfo are negative meaning that formation

of the compound from the elements is ordinarily exothermic

Elements in their stable states at 1atm, 25oC have a standardmolar enthalpy of 0

H H HF e sO

F e sO

atm C F e sO

atm C( ) ( ) , ( ) ,

1 2 5 1 2 50

Products (standard state) - Reactants (standard state) = 0

F e F es s

H kJ N O N OfO

2 2 1 2 5 2 1 2 5 2 1 2 5( , , ) ( , , ) ( , , )

Standard molar enthalpy of formation of a compound

1 atm pressure25oC

From elements in their stable states at

Most Hfo are negative meaning that formation

of the compound from the elements is ordinarily exothermic

For aqueous ions, the enthalpy is scaled relative to theproton

Do we detect any patterns?

Fe(s) 0Pb(s) 0Li(s) 0Mg(g) 0Mn(s) 0Hg(l) 0Ni(s) 0N2(g) 0O2(s) 0P4(s) 0K(s) 0Rb(s) 0

kJ/mol

Sc(s) 0Si(s) 0Na(s) 0S(s,rhombic) 0Ti(s) 0Zn(s) 0

kJ/mol

Common non-metalsHave specific formsIn which theyAre standard

Al(s) 0Ba(s) 0Be(s) 0Br2(g) 0Ca(s) 0C(s,graphite) 0Cs(s) 0Cl2(g) 0Cr(s) 0Cu(s) 0F2(g) 0H2(g) 0H+(aq) 0I2(s) 0

H fO kJ/mol

Group 17 with exceptionOf I2 are gases in Stable, standard state

MOST metals are elemental solids (metal) in standard state

electronic states in atom Energy of electron in vacuum

Electronegativity F

Heat flow measurements constant pressure, define system vs surroundingsper mole basis (intensive)

Standard Molar Enthalpy 25 oC, 1 atm, from stable stateHf

o Haq+ =0

Calculation of Ho standard enthalpy change of a reaction

1 atm pressure25oC

H n H n HOf productso

f reac tso , , tan

1. The coefficients of products and reactants in the thermochemical equation must be taken into account

2. Elements in standard states can be omitted because heats of formation are zero

2 22 3 2 3A l F e O F e A l Os s s s( ) ( ) ( ) ( )

H H HOf A l Oo

f F e Oo

s , ,2 3 2 3

H H HOf A l Oo

f F e so , , ( )2 3

2 H H H H HOf A l Oo

f F e so

f F e Oo

f A l so

s , , ( ) , , ( )2 3 2 3

H O 1 6 6 9 8 8 2 2 1 6. . AppendixOJO!

H kJ m olO 1 6 6 9 8 8 2 2 1 6 8 4 7 6 5. . . /

Calculation of Ho standard enthalpy change of hot and cold packs

1 atm pressure25oC

f reac tso , , tan

N H N O N H N Os aq aq4 3 4 3, , ,

H H H HO

of N H N Oo

aq aq s , , ,

, , ,4 3 4 3

AppendixCompound ΔHf

o, kJ/molNH4NO3,s -365.6NH4

+,aq -132.5

NO3-aq -205.0

MgSO4,s -1284.9Mg2+

aq -466.8SO4

2-aq -909.3

Fe,s 0O2,g 0Fe2O3,s -1118.4

H O 1 3 2 5 2 0 5 0 3 6 5 6. . .

m olO 3 3 7 5 3 6 5 6 2 8. .

Compares well to the calorimetry calc. (28.1kJ/mol)!

http://biochempress.com/Files/IECMD_2003/IECMD_2003_027.pdf

http://webmineral.com/data/Hexahydrite.shtml

http://www.lsbu.ac.uk/water/hofmeist.html

H O 4 6 6 8 9 0 9 3 1 2 8 4 9. . .

H H H HO

of M gSOo

a q a q s , , ,

m olO 1 3 7 6 1 1 2 8 4 9 9 1 2, . . .

M gSO sH O

, M gSO M gsH O

, M gSO M g SOsH O

aq aq42

Liquid waterhttp://www.edinformatics.com/math_science/info_water.htm

Products are more stable, lower in the energywell, than reactants

Example: Calculate the Ho for the combustion of one mole of methane CH4 according to the equation

Given the standard enthalpies of formation at 25oC, 1 atm from AppendixkJ/mol

O2(g) 0CO2(g) -393.5H2O(g) -241.8CH4(g) -74.8

C H O C O H Og g g g4 2 2 22 2( ) ( ) ( ) ( )

f reac tso , , tan

H m olH OkJ

m olH Om olC O

m olC OO

2 2 4 1 8 1 3 9 3 52

2 0 1 7 4 82

m olOkJ

m olOm olC H

m olC H.

Example: Calculate the Ho for the combustion of one mole of methane CH4 according to the equation

C H O C O H Og g g g4 2 2 22 2( ) ( ) ( ) ( )

H m olH OkJ

m olH Om olC O

m olC OO

2 2 4 1 8 1 3 9 3 52

2 0 1 7 4 82

m olOkJ

m olOm olC H

m olC H.

H kJ kJ kJO 8 7 7 1 0 7 4 8 8 0 2 3 0. . .

Can also “reverse” the problem (inside out socks)

Sig figs? H kJO 8 0 2 3.

Example: Calculate the standard enthalpy of formation for octane

H x kJ C H O C O H Oog g 1 0 9 1 0 2 2 5 1 6 1 84

8 1 8 2 2 2. ( ) ( )

H H HOf productso

f reac tso , , tan

H x kJ H

Of productso

f reac tso

1 0 9 1 0 4. ,

08 1 8

8 1 82

m olC H xkJ

m olC Hm olO

x x kJ 1 0 9 7 1 0 3.

Given the standard enthalpies of formation at 25oC, 1 atmkJ/mol

O2(g) 0CO2(g) -393.5H2O(g) -241.8CH4(g) -74.8

1 0 9 1 0 1 6 3 9 3 5 1 82 4 1 84

( )( )

x kJ m olC OkJ

m olC Om olH O

m olH Ogg

1 0 9 1 0 6 2 9 6 4 3 5 2 4 2 04. .x kJ kJ kJ x

1 0 9 1 0 8 7 0 4 8 24. .x kJ kJ x

2 1 9 5 2 2. kJ x

Bond Enthalpy

B r B r H kJg g2 2 1 9 3( )

The change in enthalpy when 1 mole of bonds is broken in the gaseousState.

Rule G3: Science is referential!

C l C l H kJg g2 2 2 4 3( ) Which has a stronger bondEnthalpy?

For covalent bonds, bond enthalpies depend on?

Bond length Overall structure of the molecule

Bond Bond Pauling's atomic EnthalpyLength ΔE.N. radii Single Bond

pm (pm) kJ/mol(Average)

Cl-Cl 199 0 99 243Br-Br 228 0 114 193I--I 267 0 133 151

H--F 92 1.8 (37) 64 568H--Cl 127 1 (37) 99 432H--Br 141 0.8 (37) 114 366H--I 161 0.5 (37) 133 298

C--F 135 1.5 77 64 488C--Cl 177 0.7 77 99 330C--Br 194 0.5 77 114 288C--I 214 0.2 77 133 216

Cl-Cl 199 0 99 243Br-Br 228 0 114 193I--I 267 0 133 151

H--F 92 1.8 (37) 64 568H--Cl 127 1 (37) 99 432H--Br 141 0.8 (37) 114 366H--I 161 0.5 (37) 133 298

C--F 135 1.5 77 64 488C--Cl 177 0.7 77 99 330C--Br 194 0.5 77 114 288C--I 214 0.2 77 133 216

C--F 135 1.5 77 64 488C--O 143 1 77 66 360C--N 147 0.5 77 70 308C--C 154 0 77 348

H--H 74 0 -37 436H--O 96 1.3 (37) 66 366H--N 101 0.8 (37) 70 391H--C 109 0.3 (37) 77 413

Bond length is nice, but it doesn’tReally relate to the Periodic tableAND it isn’t the whole story

Electronegativities? Can explain some trends

Cl-Cl 199 0 99 243Br-Br 228 0 114 193I--I 267 0 133 151

H--F 92 1.8 (37) 64 568H--Cl 127 1 (37) 99 432H--Br 141 0.8 (37) 114 366H--I 161 0.5 (37) 133 298

C--F 135 1.5 77 64 488C--Cl 177 0.7 77 99 330C--Br 194 0.5 77 114 288C--I 214 0.2 77 133 216

C--F 135 1.5 77 64 488C--O 143 1 77 66 360C--N 147 0.5 77 70 308C--C 154 0 77 348

H--H 74 0 -37 436H--O 96 1.3 (37) 66 366H--N 101 0.8 (37) 70 391H--C 109 0.3 (37) 77 413

H--H 74 0 -37 436H--C 109 0.3 (37) 77 413H--N 101 0.8 (37) 70 391H--O 96 1.3 (37) 66 366

Cl-Cl 199 0 99 243Br-Br 228 0 114 193I--I 267 0 133 151

H--F 92 1.8 (37) 64 568H--Cl 127 1 (37) 99 432H--Br 141 0.8 (37) 114 366H--I 161 0.5 (37) 133 298

C--F 135 1.5 77 64 488C--Cl 177 0.7 77 99 330C--Br 194 0.5 77 114 288C--I 214 0.2 77 133 216

Cl-Cl 199 0 99 243Br-Br 228 0 114 193I--I 267 0 133 151

H--F 92 1.8 (37) 64 568H--Cl 127 1 (37) 99 432H--Br 141 0.8 (37) 114 366H--I 161 0.5 (37) 133 298

C--F 135 1.5 77 64 488C--Cl 177 0.7 77 99 330C--Br 194 0.5 77 114 288C--I 214 0.2 77 133 216

C--F 135 1.5 77 64 488C--O 143 1 77 66 360C--N 147 0.5 77 70 308C--C 154 0 77 348

H--H 74 0 -37 436H--O 96 1.3 (37) 66 366H--N 101 0.8 (37) 70 391H--C 109 0.3 (37) 77 413

Cl-Cl 199 0 99 243Br-Br 228 0 114 193I--I 267 0 133 151

H--F 92 1.8 (37) 64 568H--Cl 127 1 (37) 99 432H--Br 141 0.8 (37) 114 366H--I 161 0.5 (37) 133 298

C--F 135 1.5 77 64 488C--Cl 177 0.7 77 99 330C--Br 194 0.5 77 114 288C--I 214 0.2 77 133 216

C--F 135 1.5 77 64 488C--O 143 1 77 66 360C--N 147 0.5 77 70 308C--C 154 0 77 348

H--H 74 0 -37 436H--O 96 1.3 (37) 66 366H--N 101 0.8 (37) 70 391H--C 109 0.3 (37) 77 413

H--H 74 0 -37 436H--C 109 0.3 (37) 77 413H--N 101 0.8 (37) 70 391H--O 96 1.3 (37) 66 366

Bond Bond Pauling's atomic EnthalpyLength E.N. radii Single Bond

H--H 74 2.2 37 436C--C 154 2.5 77 348Cl-Cl 199 3.2 99 243S-S 205 2.6 104 226

Br-Br 228 3 114 193N-N 145 3 70 159I--I 267 2.7 133 151O-O 148 3.5 66 145

Bottom line – atomic radii ΔE.N. seem to “best” determine bond enthalpies

Bond enthalpies increase Single <Double < TripleBut not by multiples of the single bond

Bond Enthalpy (kJ/mol)X-X X=X X=X

C-C 347 612 820 measured694 1041 calc.

N-N 159 418 941 measured318 477 calc.

C-N 293 615 890 measured586 879 calc.

C-O 351 715 1075 measured702 1053 calc.

http://chemviz.ncsa.uiuc.edu/content/doc-resources-bond.html

Your book’s emphasis on bond enthalpies?

Relates to the energyreleased ortaken up

By a reaction

If Bonds of Reactants stronger than bonds products endothermic

We will take up the issue of bond enthalpy when discussingsolids

Examples we have examined about energy so far

H kJ 11 0 5. C O C Os g g( ) ( ) ( ) 1

H kJ C O C Os g g 3 9 3 5 2 2. ( ) ( ) ( )

Fossil fuel burning (coal)

H x kJ C H O C O H Og g 1 0 9 1 0 2 2 5 1 6 1 848 1 8 2 2 2. ( ) ( )

Fossil fuel burning (octane)

H kJ C H O C O H Og g g l 8 9 0 2 24 2 2 2( ) ( ) ( ) ( )

Fossil fuel burning (methane)

Hydrogen Fuel Cell

H kJ H O H Og g l 2 8 6 2

2 2 2( ) ( ) ( )

Energy Density Pure octane

FITCH Rules

G1: Suzuki is SuccessG2. Slow me down G3. Scientific Knowledge is ReferentialG4. Watch out for Red HerringsG5. Chemists are LazyC1. It’s all about chargeC2. Everybody wants to “be like Mike”C3. Size MattersC4. Still Waters Run DeepC5. Alpha Dogs eat first

E kq q

Summary Slides

A B hea t C q endo therm ichea t to system

A B hea t C q exo therm ichea t to surround in gs

q cm t t cm tfina l in itia l

c is a measure of the intermolecular interactions; very large forwater = measure of the polarity of the water molecule

q H entha lpych ange H Hcons tp ressure products reac tstan tan

H H O H O

fu sion H O s l

vaporiza tion H O l g

, ( ) ( )

Says something aboutintermolecular interactions (polarity!)

kJ H O H Og g l

5 7 1 6

22 8 6 2

.( ) ( ) ( )

H kJ H O H Ol g g 5 7 1 6 2 22 2 2. ( ) ( ) ( )

H kJ H O H Og g l 5 7 1 6 2 22 2 2. ( ) ( ) ( )

An example of Hess’s Law

H H HOf productso

f reac tso , , tan

H kJ N O N OfO

2 2 1 2 5 2 1 2 5 2 1 2 5( , , ) ( , , ) ( , , )

An example of a reaction of standard molar enthalpy of formation

H E P V

H E P V P Vproducts reac ts

electronic states in atom Energy of electron in vacuum

Electronegativity F

Heat flow measurements constant pressure, define system vs surroundingsper mole basis (intensive)

Standard Molar Enthalpy 25 oC, 1 atm, from stable stateHf

o Haq+ =0

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