FITCH Rules “A” students work (without solutions manual) G1: Suzuki is Success ~ 10 problems/night. G2. Slow me down G3. Scientific Knowledge is Referential General Alanah Fitch G4. Watch out for Red Herrings Flanner Hall 402 G5. Chemists are Lazy 508-3119 [email protected] C1. It’s all about charge

Office Hours W – F 2-3 pm C2. Everybody wants to “be like Mike”

⎛qq12 ⎞ qq12⎛ ⎞ C3. Size Matters Ek= ⎜ or⎟ k = ⎜ ⎟ el ⎝ + ⎠ ⎝ ⎠ rr12 d Module #11 Chemistry C4. Still Waters Run Deep Thermochemistry C5. Alpha Dogs eat first

Energy: capacity to do work Galen, 170 Marie the Jewess, 300 Jabir ibn Galileo Galili Evangelista Abbe Jean Picard Daniel Fahrenheit Blaise Pascal Robert Boyle, Isaac Newton Anders Celsius Hawan, 721-815 An alchemist 1564-1642 Torricelli 1620-1682 1686-1737 1623-1662 1627-1691 1643-1727 1701-1744 1608-1647 wFd= ()

Or transfer heat, q Charles Augustin James Watt Luigi Galvani Count Alessandro Amedeo Avogadro John Dalton William Henry Jacques Charles Georg Simon Ohm Michael Faraday Coulomb 1735-1806 1736-1819 1737-1798 Guiseppe 1756-1856 1766-1844 1775-1836 1778-1850 1789-1854 1791-1867 B. P. Emile Germain Henri Hess Antonio Anastasio physician Clapeyron 1802-1850 q Volta, 1747-1827 1799-1864

Thomas Graham William Thompson Justus von Liebig Richard August James Joule Rudolph Clausius 1825-1898 James Maxwell Dmitri Mendeleev Johannes D. 1805-1869 Lord Kelvin, Francois-Marie Energy (1803-1873 Carl Emil (1818-1889) 1822-1888 1824-1907 Johann Balmer 1831-1879 1834-1907 Van der Waals Raoult Erlenmeyer 1837-1923 1830-1901 Consumed 1825-1909 Depends on Both work

Thomas Martin Henri Louis Johannes Rydberg J. J. Thomson Heinrich R. Hertz, Max Planck Svante Arrehenius Walther Nernst Fritz Haber J. Willard Gibbs Lowry And heat Ludwig Boltzman LeChatlier 1854-1919 1856-1940 1857-1894 1858-1947 1859-1927 1864-1941 1868-1934 1874-1936 1839-1903 1844-1906 1850-1936 w Fitch Rule G3: Science is Referential

∆EE=−final E initial qw =+

Johannes Nicolaus Niels Bohr Fritz London Wolfgang Pauli Werner Karl Linus Pauling Erwin Schodinger Friedrich H. Hund Gilbert Newton Lewis Bronsted Lawrence J. Henderson 1885-1962 Louis de Broglie 1900-1954 1900-1958 Heisenberg 1901-1994 1887-1961 1896-1997 1875-1946 1879-1947 1878-1942 (1892-1987) 1901-1976

1 Properties and Measurements To set a “heat flow” scale Property Unit Reference State Size m size of earth Volume cm3 m 1. Defined conditions: how experiment is performed Weight gram mass of 1 cm3 water at specified Temp open flask, closed flask, pressure (and Pressure) Temperature oC, K boiling, freezing of water (specified 2. Define the direction of heat flow by giving Pressure) a positive or negative number 1.66053873x10-24gamu (mass of 1C-12 atom)/12 quantity mole atomic mass of an element in grams Pressure atm, mm Hg earth’s atmosphere at sea level Energy, General Animal hp horse on tread mill heat BTU 1 lb water 1 oF calorie 1 g water 1 oC Kinetic J m, kg, s Electrostatic 1 electrical charge against 1 V Does the “system (earth)” gain energy? electronic states in atom Energy of electron in vacuum + heat For image above Electronegativity F - heat? system (earth) gains heat Heat flow measurements Reference state? from surroundings (sun)

To set a “heat flow” scale First Law of Thermodynamics: Energy is conserved

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

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

We would get a different answer if we asked “Does the “system (sun)” gain energy?” No universal change in energy For this question: the Just a transfer of energy + heat system (sun) loses heat -heat? to the surroundings (earth)

2 To set a “heat flow” scale surroundings 1. Defined conditions: how experiment is performed open flask, closed flask, pressure

system 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 q =? 3. Chemical process in the “system” is defined by heat flow q>0 system q<0 system E = constant when System AND surroundings considered! endothermic q>0 exothermic q<0

++2 Properties and Measurements Chemical reactions involve Zn() s+→2 H ( aq ) Zn ( aq ) H2 ( g +) Property Unit Reference State 1. heat exchange Size m size of earth Volume cm3 m 3 As a review: Weight gram mass of 1 cm water at specified Temp Heat exchange (and Pressure) Constant Temperature oC, K boiling, freezing of water (specified At constant Atm.pressure who is oxidized? Pressure) Pressure who is reduced? -24 1.66053873x10 gamu (mass of 1C-12 atom)/12 what is the oxidation number on H ? quantity mole atomic mass of an element in grams 2 Pressure atm, mm Hg earth’s atmosphere at sea level Who is an oxidizing agent? Energy: Thermal BTU 1 lb water 1 oF calorie 1 g water 1 oC H = Greek: thalpein – to heat Kinetic J 2kg mass moving at 1m/s enthalpy en -in Energy, of electrons energy of electron in a vacuum H for (?) heat Electronegativity F Heat Flow into system = + qH= ∆ P 1 atm pressure = constant pressure

Subscript Reminds us that This means heat flow, q, is enthalpy change Pressure is constant

3 ++2 Chemical reactions involve Zn() s+→2 H ( aq ) Zn ( aq ) H ( g +) 2 ∆EE= final− E initial = qw+ 1. heat exchange 2. work This is a form of Rule G3 ∆EqPP=+(− PV∆ ) Science is referential ( ) Pressure- ∆EHPVP =+∆ − ∆ Generally final - initial Volume constant Atm. pressure ∆EHPV=−∆ ∆ work P ∆V w = −P∆V Change in volume is typically small for Most reactions

∆EHP ≈ ∆ This means we can measure the energy change of A chemical reaction by measuring the heat exchange At constant pressure

3 to 8 standard cubic feet of five Navy Avengers biogas per pound of manure. disappeared in the Bermuda The biogas usually contains 60 Triangle on Dec. 5, 194 to 70% methane.

Methane First example problem will involve methane Gas We will prove to ourselves that the Pressure-Volume work is a Recovery Small contribution to the total energy change At landfills

4 Consider the contribution of volume of gas phase molecules Consider the contribution of volume change for water in this reaction

CH42()gg+→22 O () CO 2 () g+ H 2 O () l CH42()gg+ 22 O ()→ CO 2 () g+ H 2 O () l

PVnRT= 3 At constant T: ⎡18g ⎤ 1cm⎡ water ⎤1L ⎡ ⎤ = []2moleH2 O()l **⎢ ⎥ ⎢ *⎥33⎢ 0 . 036 L⎥ PV()()∆∆= nRT ⎣mol ⎦ 1⎣gwater 10⎦ cm⎣ ⎦

()∆= − %&$*! Conversions – if PV() ngas final n gas initial RT Interested see next slide 01013. kJ PV= [][]1 atm 0. 036 L 0. 0036=kJ − ⎛ Latm ⋅ ⎞ Latm PV()(∆=−1 3 moles 0. 0821⎜ ) 298K ⎟ ⎝ mol K ⋅ ⎠ Energy in kJ Most reactions total (q): ~ 1000 kJ By the end of () %&$*! Conversions – if This module PV∆=−48. 9316 Latm ⋅ Interested see slide after next PV 1 mole gas ~ 2.5 kJ We will see this Is “true” 01013. ⎛kJ ⎞ PV 2mole liquid water ~ 0.0036 kJ PV()(∆=−48. 9316 Latm ⋅ ⎜ ) 49. kJ⎟ =− Latm⎝ ⋅ 2mole⎠ change Sig fig tells us that PV energy small compared to q

Optional Slide: conversion To set a “heat flow” scale 1. Defined conditions: how experiment is performed Constant Pressure 2. But not on the path taken (state property)

kg ⎛ ⎞ 5 ⎛ ⎛ ⎞ ⎞ Heat flow 5 ⎜ ⎜ ⎟ ⎟33 3 ⎜ ⎟ ⎛101325. xPa 10 5 ⎞ ⎜ ⎝2 ⎠ ⎟33⎛ ⎞ ()atm ⎛⎜101325. xPa 10ms⎞⎟ ⎜ ⎝ ⋅ 10⎠ ⎟cm ⎛ 1m ⎞ ⎛ J ⎞ kJ⎜ ⎟ ⎛ ⎞ depends ()atm ⎜⎝ ⎟⎠ ()L ⎜ ⎟ ⎜ ⎟ ⎜ 0101325. ⎟ ⎜kJ ⎟ = ⎝ atm ⎠Pa⎜ ⎟L ⎝ 102cm ⎠ ⎝ 2 ⎠ 10 3 ⎛J ⋅ ⎞ ⎝ ⎠ atm Pa⎜ ⎟L 10 cm kg m ⎜ ⎜ ⎟ ⎟ the conditions ⎝ ⎠ ⎝ ⎝ ⎠ ⎠ s 2

kg ⎧ ⎛ ⎛ ⎞ ⎞ ⎛ ⎞ ⎫ ⎪ 5 ⎜2⎜ 33⎟ ⎟ 3 ⎜ ⎟ ⎪ 0101325..kJ ⎪⎛ 101325xPa 10ms⎞ ⎝ 10⋅ cm⎠ ⎛ 1m ⎞ ⎛ J ⎞ kJ⎜ ⎟ ⎛ ⎞ ⎪ = ⎨⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎬ ()()atm L ⎝ atm Pa⎠ ⎜ L ⎟ ⎝ 102cm ⎠ ⎝ 2 ⎠ 10⎜3 ⎛J ⋅ ⎞ ⎟ ⎝ ⎠ ⎪ ⎜ ⎟ kg m ⎜ ⎜ ⎟ ⎟ ⎪ ⎪ ⎝ ⎠ ⎜ ⎟ ⎪ H= enthalpy ⎩ s 2 ⎝ ⎝ ⎠ ⎠ ⎭

qHHH==∆ − reaction constan tpressure products reactan ts

5 Enthalpy is a state property (measured under constant pressure, but how measured under that Enthalpy is an “extensive” property constant pressure is not important)

qH=>∆ 0 endothermic Depends upon the amount present

HHproducts> reactan ts CH42()gg+→2 O () CO 2 () g+ 2 H 2 O () l ∆ H= − 890 kJ

Think of heat as a reactant 890kJ of heat is released when 1 mole of methane H22 O()s +⎯→ heat⎯ H O Reacts with oxygen q =<∆H 0 − 890kJ exothermic 1moleCH4()g HHproducts< reactan tss ⎛ − 890kJ ⎞ ⎜ ⎟ =− +→ + +Think of heat as a product ⎜ ()2⎟moleCH 1780 kJ CH42()gg2 O () CO 22 () g H O () l heat ⎝ ⎠ 4()g 1moleCH4()g

Properties and Measurements Context for the next example problem Property Unit Reference State 2006 Sept Sci. Am: World Wide Petroleum Usage Size m size of earth Land People Transport 3 Non- Volume cm m 29% of total use Weight gram mass of 1 cm3 water at specified Temp transportation (and Pressure) o Temperature C, K boiling, freezing of water (specified Total transportation Pressure) -24 =53% 1.66053873x10 gamu (mass of 1C-12 atom)/12 Land Freight 19% quantity mole atomic mass of an element in grams Pressure atm, mm Hg earth’s atmosphere at sea level Air People and Freight 5% Energy, General Animal hp horse on tread mill U.S. differs from world in distribution of petroleum use heat BTU 1 lb water 1 oF calorie 1 g water 1 oC 3 Non-transportation Kinetic J m, kg, s Transportation = 71.8% Electrostatic 1 electrical charge against 1 V electronic states in atom Energy of electron in vacuum 57.8% of Transportation= Electronegativity F Personal land transport Heat flow measurements constant pressure, define system vs surroundin = 41.5% of total U.S. consumption per mole basis (intensive) Data US DOE 2006

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 that number 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? 4 ∆HxkJCHOCOHO=−109. 10 2818 25 2 + 16 2()gg → 18 2 () + Strategy: need moles of octane consumed (Golden Bridge) ∆H mpg density Molar mass miles gallons grams moles heat

33 4 ⎛⎛ 11galgal ⎞37852⎞37852.⎛.⎛ LL 10 10 ⎞⎞mLmL⎛⎛ 06900690.. gC⎞gC⎞⎛⎛ H H 1 1moleCmoleC H H⎞⎞⎛⎛ 109. xkJ 10 ⎞⎞ ⎛ − ⎞ ()()150150..mimi ⎜⎜ ⎟⎟⎜⎜ ⎟⎟⎜⎜ ⎟⎟⎜⎜8888 8888⎟⎟⎜⎜ ⎟⎟ ⎜ ⎟ = ⎝⎝17. 5mi ⎠⎠⎝⎝ ⎠⎠⎝⎝ ⎠⎠⎝⎝ ⎠⎠⎝⎝ ⎠⎠ ⎝ ⎠ 1717. 5. 5mimi galgal LL mLCmLC8888 H H 114114gCgC8888 H H 2moleC88 H

∆HkJ= −1,,. 070 243955 We will use part of this problem 6 ∆H = −107. x 10 kJ 3 sig fig Again:

⎛ 37852. L 10⎞ ⎛3 mL 0. 690⎞ ⎛gC H 1moleC⎞ H⎛ 109.,xkJ 104 ⎞ ⎛ − 124 861kJ ⎞ − ⎜ ⎟ ⎜ ⎟ ⎜ 88 88⎟ ⎜ ⎟ ⎜ ⎟= ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ gal L mLC88 H 114 gC88 H 2moleC88 H gal

Rules ENERGY measurement a) change in temperature 1.Enthalpy is an extensive property (depends upon number of moles) b) some function specific to the material and how 2.Enthalpy change for a reaction is equal in magnitude, but opposite it is organized (bonds) in sign, to the enthalpy for the reverse reaction

4 ∆HxkJCHOCOHO=−109. 10 2818 25 2 + 16 2()gg → 18 2 () +

4 +→+ ∆HxkJ=+109. 10 16CO22()gg 18 H O () 2 C 8182 H 25 O

3. Enthalpy change depends upon the state of the reactant and products HO→ HO ∆H =+44kJ 22()lg ( )

7 ENERGY change is a function of a) temperature ENERGY measurement b) material ⎛qq12 ⎞ Ek= ⎜ ⎟ a) change in temperature el ⎝ d ⎠ b) some function specific to the material and how H it is organized (bonds) H H H O Pure material qCt=−()final t initial H O H qmct= ⋅ ⋅ ∆ O H O =∆ O O qCt H H H H O H O m = mass H H H25 H c = specific heat of a pure substance H H

O O

H H

O O C = heat capacity = heat required to raise the temperature of

H H the system 1oC H H units: J/oC 15 oC heat 25 oC

http://www.lsbu.ac.uk/water/molecule.html q charg edensity ≈ Specific heats, c, of various substances in various physical states r

Material Specific Heats c, (J/g-K) Pb(s) 0.12803 Pb(l) 0.16317 Ability to store Cu(s) 0.382 heat in a substance Fe(s) 0.446 is variable.

Cl2(g) 0.478 Electron density of water C(s) 0.71 Shape and charge distribution Electrons on Oxygen sit CO2(g) 0.843 On water “out there” causing large NaCl(s) 0.866 Electrostatic potential o Al(s) 0.89 C(liquidH2O)=4.18J/g- C Oriented on the electrons C6H6(l) 1.72 Liquid water is very strongly organized H2O(g) 1.87 due to the polarity of the molecule, so it C2H5OH(l) 2.43 has a high specific heat H2O(l) 4.18

8 For the same amount of energy, easier to break Example: 1.00 cup of water is heated from 25.0 oC to 100.0 oC. How electrostatic attraction of He compared to water with it’s many joules were used to heat the water? localized charge

qmct=⋅⋅∆ H H Mass of water: He H He O 3 O ⎡ 1 ⎤ ⎡1 10⎤ ⎡ 1 ⎤ He ⎡qt ⎤L⎡ mL⎤ g ⎡ ⎤ O []100. cup ⎢ ⎡ 11qtqt ⎥ ⎢ ⎤ 1L ⎥ ⎢ ⎥ ⎢23651. ⎥g = H [][]100100.100.. cupcupcup⎣ ⎢⎢ ⎦ ⎣⎥⎥⎢ ⎦ ⎣⎥ ⎦ ⎣ ⎦ H [ 4⎣cups]⎣44cupscups1057. 1057⎦⎦.⎣ qt qt L ⎦ 1mL H O H H He ∆tT= − T oo H final initial ∆tCC= 100− 25 O He o

∆tC= 75 ∆t = 75K H O H 418.⎡ J ⎤ qg= 23651. ⎢ 75K⎥ 4

He gK⎣ ⎦ qxJ= 74110. H q = 7414853. qkJ= 741.

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 oC.

3 ⎡ 1qt ⎤ ⎡1L 10⎤mL⎡ 128. g⎤ ⎡ ⎤ []100. cup ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ()23651..⎥gg== 128 302 . 7 ⎣4cups 1057⎦.⎣ qt L⎦ ⎣ 1mL ⎦ ⎣ ⎦ Lag

08.⎡ J ⎤ qg= 302. 7 ⎢ 75K⎥ Water gK⎣ ⎦ Land

qJkJdry soil ==18,. 164 181

qkJwater = 741.

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

9 1. Hot air rises over land

Land 2. Cold air from Lake 1. Hot air rises moves to fill in (Lake over Lake Breeze) Lake

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

Heat capacity of large bodies The temperature change in 24 hours in summer for water is not much Of water affect human activity leading to big differences between lake and land and thus a lake breeze

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? qmct= ⋅ ⋅ ∆ 440⎛ J ⎞ qkg= ()200 ⎜ ()100⎟ 25CkJ−= 5, 828 kg⎝ C ⋅ ⎠

⎛ 42miles ⎞trip⎛ 5,, 828⎞ ⎛kJ engineheating 27 730kJ⎞ heating⎛ ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ = ⎜ ⎟ ⎝ 18gal ⎠miles⎝ ⎠ ⎝ 1trip gal⎠ ⎝ ⎠ Heating the engine block Compare to heat available from combustion of 1 gallon of octane (from before 4 ∆HxkJCHOCOHO=−109. 10 2818 25 2 + 16 2()gg → 18 2 () +

⎛ 37852. L 10⎞ ⎛3 mL 0. 690⎞ ⎛gC H 1moleC⎞ H⎛ 109.,xkJ 104 ⎞ ⎛ − 124 861kJ ⎞ − ⎜ ⎟ ⎜ ⎟ ⎜ 88 88⎟ ⎜ ⎟ ⎜ ⎟ = Prius 1 Prius 2 ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ gal L mLC88 H 114 gC88 H 2moleC88 H gal ⎛ − ⎞ ⎛ − ⎞ ⎛ 5828, kJ ⎞ Let’s compare to what 39 31mpg 43 34mpg ⎜ 100⎟ 22. 2%= ⎜ 100⎟ 205%=. ⎜ 100⎟ 20%= ⎝ 124, 861kJ ⎠ I measure ⎝ 39mpg ⎠ ⎝ 43mpg ⎠

10 Properties and Measurements Property Unit Reference State Relate reaction heat Size m size of earth To the calorimeter heat Volume cm3 m Weight gram mass of 1 cm3 water at specified Temp (and Pressure) Temperature oC, K boiling, freezing of water (specified Pressure) 1.66053873x10-24gamu (mass of 1C-12 atom)/12 quantity mole atomic mass of an element in grams Pressure atm, mm Hg earth’s atmosphere at sea level Energy, General qqreaction= − calorimeter Animal hp horse on tread mill heat BTU 1 lb water 1 oF calorie 1 g water 1 oC If the temperature of the water rises (heat flow into water) Kinetic J m, kg, s then heat must have been lost from the reaction Electrostatic 1 electrical charge against 1 V electronic states in atom Energy of electron in vacuum Electronegativity F Heat flow measurements constant pressure, define system vs surroundin per mole basis (intensive) Calorimetry

Example: When 1.00 g of ammonium nitrate, NH4NO3, is Not all calorimeters can be based on added to 50.0 g of water in a coffee-cup calorimeter, it dissolves: qqreaction= − calorimeter = − mct⋅ ⋅ ∆ +− NH43 NO()saqaq→+ NH 4 ( ) NO 3 ( ) Heat capacity of the calorimeter and the temperature of the water drops from 25.00 to 23.32 qqreaction=− reaction =−{[ Ct cal ] } ∆ oC. Assuming that all the heat absorbed by the reactions comes from the water, calculate q for the reaction system. In some problems You will qJcalorimeter = −351 qmctcalorimeter =⋅⋅∆ Need to determine This number ⎡ J ⎤ oo In one step and qgcalorimeter = []50.. 0 418⎢ [] 2332 .⎥ CC 2500 .− qq= − ⎣ gK ⎦ reaction calorimeter Then go on

qJreaction =−−( 351 ) ⎡ J ⎤ q = 209[] 168.− K calorimeter ⎣⎢ ⎦⎥ qJ= 351 K A 1 degree change in Celsus reaction

qJcalorimeter =−35112. Is a 1 degree change in Kelvin

11 Example: The reaction between hydrogen and chlorine 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 HClHClgg()+→2 () g2 () g 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 can be studied in a bomb calorimeter. It is found that when a 1.00 g of salicylic acid is burned? ⎡937. kJ ⎤ sample of H2 reacts completely, the temperature rises from 20.00 to qreaction=−⎢ o ()32⎥.. 11 29 91− 29.82 oC. Taking the heat capacity of the calorimeter to be 9.33 kJ/oC, 1.00 g of C7H6O3 ⎣248. C ⎦ o calculate the amount of heat evolved in the reaction. Tinitial 32.11 C o Tfinal 29.91 C qkJreaction = −8. 3121 Heat evolved? q 9.37kJ required to cause 2.48 oC change Calorimeter heat capacity = 9.33 kJ/oC qCtreaction=−[] cal ∆ o ⎛ 138gC H O 8⎞ 3121.⎛ − kJ ⎞ Tinitial =20.00 C ()1moleC H O ⎜ 763 ⎟ ⎜ 3158598⎟. =−kJ 763⎝1moleC H O 1gC⎠ ⎝ H O ⎠ o =− ∆ 763 763 T final = 29.82 C qCtreaction[] cal

= −316kJ ⎡ kJ ⎤ oo ??? q =−933...[] 2982CC 2000− reaction⎣⎢ oC ⎦⎥ 937. kJ C = cal248. o C qkJreaction =−916.

Example 2 What is the enthalpy change for the reaction ∆H +− Combine NH43 NO()saqaq→+ NH 4 ( ) NO 3 ( ) 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?

⎡J oo⎤ qqreaction=− calorimeter[] =−50.. 0 g 418⎢ o [] 2332 .CC⎥ 2500 . − gC⎣ ⎦ 2251618CH+→ O CO+ HO 818 2 2()gg 2 () 2N=28.02 ⎡ J ⎤ o 4H=4.04 qreaction=−⎢ 209 o [] 168⎥. − C 1 g was reacted ⎣ C ⎦ 3O=48.00 qJ= 35112. 80.05 reaction ⎡351J ⎤80⎡ .. 05g 28100⎤ J 281kJ ⎢ ⎥ ⎢ ⎥ == ⎣ 1g ⎦ ⎣mol ⎦ mol mol qJreaction = 351 calorimetry Reaction stoichiometry qHJ==∆ 351 reaction constan tlabpressure

To get reaction enthalpies +− NH43 NO()saqaq→+ NH 4 () NO 3 () ∆ H281. kJ =+

12 +− 1NH NO→+11 NH+− NO∆ H281. kJ =+ NH43 NO()saqaq→+ NH 4 () NO 3 () ∆ H281. kJ =+ 43()saqaq 4 () 3 ()

When there are no Where did the per/mole go? Thermochemical Equation Rules Coefficients it is understood that It is “1” The reaction was written as a per/mole 1. Value of )H applies when products and reactants are at same Enthalpy is understood as a per/mole of reactant (or as the reaction temperature, 25oC unless otherwise specified. is written) 2. Sign of )H, indicates whether reaction, when carried out at constant pressure, is exothermic or endothermic 3. ∆H sign changes when reaction is reversed +− NH43()aq+→ NO () aq NH 43 NO () s H281∆. kJ =− 4. Stoichiometry is important 4. Phases of all species must be specified 5. Values of )H is same regardless of method used to calculate it (Hess’s Law)

Example illustrating importance of phases An example of several of the rules using Fuel Cells Using a coffee-cup calorimeter, it is found that when an ice cube Fuel cells use the reaction: 1 weighing 24.6 g melts, it absorbs 8.19 kJ of heat. Calculate HOHO+→ for the phase change represented by the thermochemical equation 2()gg2 22 () () l Calculate the enthalpy for the equation above given that:

HO22()sl→ HO () ∆HkJHOHO=+5716. 2222()lgg 2 ( )→+ ( )

⎡ 819. kJ 18⎤ ⎡. 02g HO ⎤ Reverse reaction: ⎢ 2 ⎥ = ⎢ ⎥ []16001moleHO . kJ ⎣24. 6g ⎦mole⎣⎢ ⎦⎥ 2 ice HO2 ∆HkJHOHO=−5716. 222()gg+ ()→ 2 2 () l

scale

−5716. kJ 1 ∆H = 286=−kJ H O + H O → 2 2()gg2 22 () () l Here we got a number by coming “at it” from an odd direction

13 Hess’s law

Galen, 170 Marie the Jewess, 300 Jabir ibn Galileo Galili Evangelista Abbe Jean Picard Daniel Fahrenheit Blaise Pascal Robert Boyle, Isaac Newton Anders Celsius Hawan, 721-815 An alchemist 1564-1642 Torricelli 1620-1682 1686-1737 1623-1662 1627-1691 1643-1727 1701-1744 The value of )H for a reaction is the same whether it occurs in one 1608-1647 step or in a series of steps (enthalpy (constant P, T) is a state function) Charles Augustin James Watt Luigi Galvani Count Alessandro Amedeo Avogadro John Dalton William Henry Jacques Charles Georg Simon Ohm Michael Faraday Coulomb 1735-1806 1736-1819 1737-1798 Guiseppe 1756-1856 1766-1844 1775-1836 1778-1850 1789-1854 1791-1867 B. P. Emile Germain Henri Hess Antonio Anastasio physician Clapeyron 1802-1850 Volta, 1747-1827 1799-1864

Thomas Graham William Thompson Justus von Liebig Richard August James Joule Rudolph Clausius 1825-1898 James Maxwell Dmitri Mendeleev Johannes D. 1805-1869 Lord Kelvin, Francois-Marie (1803-1873 Carl Emil (1818-1889) 1822-1888 Johann Balmer 1831-1879 1834-1907 Van der Waals 1824-1907 Raoult Erlenmeyer 1837-1923 1830-1901 1825-1909

Thomas Martin Henri Louis Johannes Rydberg J. J. Thomson Heinrich R. Hertz, Max Planck Svante Arrehenius Walther Nernst Fritz Haber J. Willard Gibbs Lowry Ludwig Boltzman LeChatlier 1854-1919 1856-1940 1857-1894 1858-1947 1859-1927 1864-1941 1868-1934 1839-1903 1874-1936 1844-1906 1850-1936 Germain Henri Hess Fitch Rule G3: Science is Referential 1802-1850 born in Geneva Switzerland Professor of Chemistry

Johannes Nicolaus Niels Bohr Fritz London Wolfgang Pauli Werner Karl Linus Pauling Erwin Schodinger Friedrich H. Hund Gilbert Newton Lewis Bronsted Lawrence J. Henderson 1885-1962 Louis de Broglie 1900-1954 1900-1958 Heisenberg 1901-1994 At St. Petersburg Technological Institute 1887-1961 (1892-1987) 1896-1997 1875-1946 1879-1947 1878-1942 1901-1976

Example of how Hess’s law is useful ∆HkJCOOCO= −566. 0 2()gg+ 22 ()→ 2 () g 1 CO()sg+→2 2 () CO () g ∆HkJCOCOO= +566. 0 222()ggg→ 2 ()+ ()

It is difficult to measure the heat evolved for this reaction because it +566. 0kJ 1 ∆H = CO CO→+ O occurs as the partial burning of carbon in the presence of other reactions 2 2()ggg ()2 2 () involving the complete burning of carbon 1 C(s) + O2(g) ∆HkJCOCOO=+2830. → + This is the 2()ggg ()2 2 () CO +1/2O Number we want (g) 2(g) ∆HkJCOCO= −3935. + → But can’t actually measure 0 ()sg22 () () g ∆HkJCOCO=−3935. ()sg22 ( ) + ( g → ) Related to 2CO + O2…. Get to the number by an alternative ∆ 0 =+11 → + CO ∆H HkJCOCOO= −1105.2830kJ. CO+→22O()ggg CO () () 2(g) Path (State function!) ()sg22 22() ()g () g

To solve rearrange equations to get CO on right hand side 0 1 C(s) + O2(g) ∆H =−1105. kJ CO()sg+→2 2 () CO () g ∆HkJCOCO=−3935. + → ()sg22 () () g CO(g) +1/2O2(g)

∆HkJCOOCO=−566. 0 2()gg22 () + 2 () g → CO2(g)

14 Enthalpies of Formation Invoke Rule G5: Chemists are Lazy 1 is “understood” Rather than getting the enthalpy for each reaction from a bomb O 1 1 calorimeter use a smaller number of standard reactions from which ∆HkJNf =−882 2 21(,g atm , 25 C ) O 21 (, g atm ,+ 25 C ) NO 21 (, g atm , 25 → C ) Hess’s law can be applied to get all the remainder reactions of interest Standard molar enthalpy of formation of a compound 1 atm pressure Enthalpy associated with standard reaction is From elements in their stable states at 25oC enthalpy of formation which is the enthalpy change when one mole of compound is o formed at constant pressure of 1 atm and a fixed temperature, Most )Hf are negative meaning that formation ordinarily 25oC, from the elements in their stable states at that of the compound from the elements is ordinarily pressure and temperature. STP (Standard Temperature and exothermic Pressure) Elements in their stable states at 1atm, 25oC have a standard This allows us to look at enthalpy of compounds molar enthalpy of 0 not reactions which reduces total data which must Why? Be acquired (Chemists are Lazy!!!)

Elements in their stable states at 1atm, 25oC have a standard O 1 molar enthalpy of 0 ∆HkJNf =−882 2 21(,g atm , 25 C ) O 21 (, g atm ,+ 25 C ) NO 21 (, g atm , 25 → C ) Standard molar enthalpy of formation of a compound Fess→ Fe From elements in their stable states at 1 atm pressure O O O o ∆∆HHFe() s =−=Fe () s HFe() s ∆0 25 C 125atm, C 125atm, C o Most )Hf are negative meaning that formation Products (standard state) - Reactants (standard state) = 0 of the compound from the elements is ordinarily exothermic

For aqueous ions, the enthalpy is scaled relative to the proton

O + ∆HHf aq = 0

15 Do we detect any patterns? Properties and Measurements kJ/mol Property Unit Reference State O O O kJ/mol ∆H f kJ/mol ∆H Size m size of earth ∆H f f Fe(s) 0 Sc(s) 0 Volume cm3 m Al(s) 0 3 Pb(s) 0 Si(s) 0 Weight gram mass of 1 cm water at specified Temp Ba(s) 0 (and Pressure) Be(s) 0 Li(s) 0 Na(s) 0 Temperature oC, K boiling, freezing of water (specified Br (g) 0 Mg(g) 0 S(s,rhombic) 0 Pressure) 2 Mn(s) 0 Ti(s) 0 1.66053873x10-24gamu (mass of 1C-12 atom)/12 Ca(s) 0 quantity mole atomic mass of an element in grams C(s,graphite) 0 Hg(l) 0 Zn(s) 0 Ni(s) 0 Pressure atm, mm Hg earth’s atmosphere at sea level Cs(s) 0 Energy, General N (g) 0 Cl (g) 0 2 Common non-metals electronic states in atom Energy of electron in vacuum 2 O (s) 0 Cr(s) 0 2 Have specific forms Electronegativity F P (s) 0 Cu(s) 0 4 In which they K(s) 0 Heat flow measurements constant pressure, define system vs surroundings F (g) 0 Are standard per mole basis (intensive) 2 Rb(s) 0 H2(g) 0 Standard Molar Enthalpy 25 oC, 1 atm, from stable state H+(aq) 0 MOST metals are Group 17 with exception o + )Hf Haq =0 I2(s) 0 elemental solids (metal) in Of I2 are gases in standard state Stable, standard state

Calculation of )Ho standard enthalpy change of a reaction Calculation of )Ho standard enthalpy change of hot and cold packs O o o O o o 1 atm pressure ∆∆HnHnH=− ∆ 1 atm pressure ∆∆HnHnH=−∑∑f,,tan products freac ∆ ts ∑ f,,tan products ∑ freac ts 25oC 25oC NH NO→+ NH+− NO 1. The coefficients of products and reactants in the 43,,,saqaq 4 3 thermochemical equation must be taken into account

22Al()ssss+→ Fe23 O () Fe ()+ Al 23 O () O o o o ∆∆HH=+−+− H ∆{ HfNHNO, ∆} {}fNH,,43,,aqfNO aq 43, s O o o o o ∆∆HH=+222 H ∆ H − ∆ H + ∆ {}{}{}fAlO,,,()23 fFes , ()fFeO , 23s fAls , () Appendix o O o o Compound ∆Hf , kJ/mol ∆∆HH=− H ∆ O fAlO,,23fFeO 23s ∆H =−{}132.. 5 +−2050{ 3656 −− .} NH4NO3,s -365.6 + 2. Elements in standard states can be omitted because heats of NH4 ,aq -132.5 NO - -205.0 formation are zero 3 aq OJO! O kJ MgSO4,s -1284.9 O Appendix ∆H =−{}{}337.. 5 −− 3656 28 =+ 2+ ∆H =−1669.. 8() − 82216 − Mg aq -466.8 mol 2- SO4 aq -909.3 O Fe,s 0 ∆HkJmol=−1669.. 8 82216 + 847 =− ./ 65 Compares well to the calorimetry calc. (28.1kJ/mol)! O2,g 0 Fe2O3,s -1118.4

16 HO 2 +−+ 2 2 o MgSO4,,,s ⎯→⎯⎯⎯⎯+ Mgaq SO4 aq Example: Calculate the )H for the combustion of one mole of methane CH according to the equation O o o o 4 ∆∆HH=+−2+− H ∆2 { Hf, MgSO ∆} {}fMg,,,aqfSO4 aq 4, s CH42()gg+ 22 O ()→ CO 2 () g+ H 2 O () g O ∆H =−{}4668.. +−909 3{ 1284 −− . 9} Given the standard enthalpies of formation at 25 oC, 1 atm from Appendix kJ/mol O kJ ∆ =−{} −− =− H {}1,. 3761 1284 . 9 912 . O2(g) 0 O o o mol ∆∆HnHnH=− ∆ CO -393.5 ∑ f,,tan products ∑ freac ts Products are more stable, lower in the energy 2(g) H2O(g) -241.8 http://webmineral.com/data/Hexahydrite.shtml well, than reactants CH4(g) -74.8

O ⎡ ⎛ kJ ⎞ kJ⎛ ⎞ ⎤ ∆HmolHO=−⎢()2 2418⎜ ..() 1molCO⎟ +− 3935 ⎜ ⎟ ⎥ 2 ⎝ ⎠ 2 ⎝ ⎠ ⎣ molH2 O molCO2 ⎦

http://www.edinformatics.com/math_science/info_water.htm kJ kJ Liquid water ⎡ ⎛ ⎞ ⎛ ⎞ ⎤ − ⎢()20molO2 ⎜ () 1molCH⎟ +−4 748. ⎜ ⎟ ⎥ ⎣ molO⎝ ⎠ molCH⎝ ⎠ ⎦ http://www.lsbu.ac.uk/water/hofmeist.html 2 24

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

Example: Calculate the )Ho for the combustion of one mole of Example: Calculate the standard enthalpy of formation for octane

methane CH4 according to the equation o 4 ∆HxkJCHOCOHO=−109. 10 2818 25 2 + 16 2()gg → 18 2 () + CH+→22 O CO+ H O 42()gg () 2 () g 2 () g Given the standard enthalpies of formation at 25 oC, 1 atm

kJ/mol O o o O ⎡ ⎛ kJ ⎞ kJ⎛ ⎞ ⎤ ∆∆HH=− H ∆ ∆HmolHO=−⎢()2 2418⎜ ..() 1molCO⎟ +− 3935 ⎜ ⎟ ⎥ O 0 ∑ fproducts,,tan∑ freac ts 2 ⎝ ⎠ 2 ⎝ ⎠ 2(g) ⎣ molH2 O molCO2 ⎦ CO2(g) -393.5 O 4 o ∆∆HxkJH=−109. 10 ∑ = f, products ⎡ ⎛ kJ ⎞ ⎛kJ ⎞ ⎤ H2O(g) -241.8 − ⎢()20molO ⎜ () 1molCH⎟ +− 748. ⎜ ⎟ ⎥ − o ∆ 2 ⎝ ⎠ 4 ⎝ ⎠ CH -74.8 ⎣ molO2 molCH24 ⎦ 4(g) ∑ H freac,tan ts

4 ⎡ ⎛ kJ ⎞ 2418. kJ ⎛ − ⎞ ⎤ O −=109..xkJmolCO 10 16⎢ 3935⎜ − 18molH O⎟ + ⎜ ⎟ ⎥ ∆ =−[][] −− =− ()2 ()g ⎜ ()2 ⎟()g ⎜ ⎟ HkJkJkJ87710... 74 8 802 30 ⎣⎢ molCO⎝ 2 ()g ⎠ molH2 O()g ⎝ ⎠ ⎦⎥ O Sig figs? ∆HkJ=−802. 3 ⎡ ⎛ kJ ⎞ 0kJ ⎛ ⎞ ⎤ − ⎢()225molC H x ⎜ ()molO⎟ + ⎜ ⎟ ⎥ 818 ⎝ ⎠ 2 ⎝ ⎠ ⎣ molC818 H molO2 ⎦

4 −=−+−−+109..xkJ 10[]() 6296 kJ( 43524 kJx) [ 2 0] Can also “reverse” the problem (inside out socks) 21952. kJ = − 2 x

4 3 − 109..x 10kJ=− 8704 8 kJ − 2 x x = −1097. x 10 kJ

17 For covalent bonds, bond enthalpies depend on? Bond Enthalpy BondBond BondBondPauling's Pauling's atomicatomic Enthalpy LengthLength ∆∆E.N.E.N. radiiradii SingleSingle Bond Bond length pmpm (pm)(pm) kJ/molkJ/mol (Average)(Average) The change in enthalpy when 1 mole of bonds is broken in the gaseous Overall structure of the molecule Cl-ClCl-Cl 199 199099 0 99 243243 Br-BrBr-Br 228 22800114 0 114114 193193 State. I--II--I 26726700133 0 133133 151151

H--FH--F 92921.8 1.8 (37)(37) 6464 568568 Rule G3: Science is referential! H--ClH--Cl 127 1271 1 (37)(37) 9999 432432 H--BrH--Br 141 1410.8 0.8 (37)(37) 114114 366366 H--IH--I 1611610.5 0.5 (37)(37) 133133 298298

C--FC--F 1351351.5 1.5 7777 6464 488488 Br2()gg→=+2193 Br H∆ kJ Bond length is nice, but it doesn’t C--ClC--Cl 177 1770.7 0.7 7777 9999 330330 C--BrC--Br 194 1940.5 0.5 7777 114114 288288 Which has a stronger bond Really relate to the Periodic table C--IC--I 2142140.2 0.2 7777 133133 216216 Cl→=+2243 Cl H∆ kJ 2()gg Enthalpy? AND it isn’t the whole story C--FC--F 1351351.5 1.5 77 64 488 C--OC--O 143143177 1 77 66360 C--NC--N 1471470.5 0.5 77 70 308 Electronegativities? C--CC--C 154154077 0 77 348 Can explain some trends H--HH--H 74740-37 0 -37 436 H--OH--O 96961.3 1.3 (37)(37) 66 366 H--NH--N 1011010.8 0.8 (37)(37) 70 391 H--CH--C 1091090.3 0.3 (37)(37) 77 413

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

Bond Bond Pauling's atomic Enthalpy Bond enthalpies increase Single

H--H 74 0 -37 436 H--C 109 0.3 (37) 77 413 H--N 101 0.8 (37) 70 391 http://chemviz.ncsa.uiuc.edu/content/doc-resources-bond.html H--O 96 1.3 (37) 66 366

18 Examples we have examined about energy so far Your book’s emphasis on bond enthalpies? Hydrogen Fuel Cell Relates to the energy 1 ∆ =− + → released HkJHOHO286 2()gg2 22 () () l or Fossil fuel burning (coal) taken up 1 By a reaction ∆H = −1105. kJ CO()sg+→2 2 () CO () g ∆HkJCOCO= −3935. + → If Bonds of Reactants stronger than bonds products endothermic ()sg22 () () g Fossil fuel burning (methane) We will take up the issue of bond enthalpy when discussing solids ∆HkJCHOCOHO= −89042()gg+ 2 ()→ 2 () g+ 2 2 () l

Fossil fuel burning (octane) 4 ∆HxkJCHOCOHO=−109. 10 2818 25 2 + 16 2()gg → 18 2 () +

Energy Density Pure octane FITCH Rules G1: Suzuki is Success G2. Slow me down G3. Scientific Knowledge is Referential

General G4. Watch out for Red Herrings G5. Chemists are Lazy C1. It’s all about charge C2. Everybody wants to “be like Mike”

⎛qq12 ⎞ C3. Size Matters Ek= ⎜ ⎟ el ⎝ + ⎠ rr12 Chemistry C4. Still Waters Run Deep C5. Alpha Dogs eat first

19 Summary Slides “A” students work (without solutions manual) A+ B+→ heat C+ q endothermic ~ 10 problems/night. heat to system A+→ B heat C + q − exothermic heat to surroundings Alanah Fitch Flanner Hall 402 508-3119 [email protected] qcmt=−=∆[]final t initial cmt

Office Hours W – F 2-3 pm c is a measure of the intermolecular interactions; very large for water = measure of the polarity of the water molecule

q==∆ H enthalpychange H = H − constan tpressure products reactan ts

∆HHOHO− Says something about fusion,()() H2 O22 s l intermolecular ∆HHOHO− interactions (polarity!) vaporization,()() H2 O22 l g

∆HkJHOHO=+5716. 2222()lgg 2 ( )→+ ( ) Properties and Measurements Property Unit Reference State ∆HkJHOHO=−5716. 222()gg+ ()→ 2 2 () l Size m size of earth Volume cm3 m −5716. kJ 1 Weight gram mass of 1 cm3 water at specified Temp ∆H = 286=−kJ H O + H O → 2 2()gg2 22 () () l (and Pressure) Temperature oC, K boiling, freezing of water (specified Pressure) An example of Hess’s Law 1.66053873x10-24gamu (mass of 1C-12 atom)/12 An example of a reaction of standard molar enthalpy of formation quantity mole atomic mass of an element in grams Pressure atm, mm Hg earth’s atmosphere at sea level Energy, General electronic states in atom Energy of electron in vacuum O 1 ∆HkJNf =−882 2 21(,g atm , 25 C ) O 21 (, g atm +, 25 C ) NO 21 (, g atm , 25 → C ) Electronegativity F Heat flow measurements constant pressure, define system vs surroundings O o o ∆∆HH=−∑∑fproducts,,tan Hfreac ∆ ts per mole basis (intensive) Standard Molar Enthalpy 25 oC, 1 atm, from stable state )H o H + =0 ∆∆∆HE=+( PV) f aq ∆∆=+ − H E() PVproducts PV () reactan ts

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

Alanah Fitch Flanner Hall 402 508-3119 [email protected]

Office Hours W – F 2-3 pm

21