“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 Module #11 Thermochemistry.
Download ReportTranscript “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 Module #11 Thermochemistry.
“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 Module #11 Thermochemistry Chemistry General FITCH Rules G1: Suzuki is Success G2. Slow me down G3. Scientific Knowledge is Referential G4. Watch out for Red Herrings G5. Chemists are Lazy C1. It’s all about charge C2. Everybody wants to “be like Mike” qq qq C3. Size Matters or k E k d r r C4. Still Waters Run Deep C5. Alpha Dogs eat first 1 2 el 1 2 1 2 Set up like others 1. Want to understand how molecules form and interact. 2. Need a measurement tool and scaling 3. Q – measure of the energy within and 1. Between molecules. Energy: capacity to do work w F d Or transfer heat, q q w E E final Einitial q w Energy Consumed Depends on Both work And heat Galen, 170 Marie the Jewess, 300 Charles Augustin James Watt Coulomb 1735-1806 1736-1819 Thomas Graham Justus von Liebig 1805-1869 (1803-1873 J. Willard Gibbs 1839-1903 Jabir ibn Hawan, 721-815 Luigi Galvani 1737-1798 Count Alessandro Guiseppe Antonio Anastasio Volta, 1747-1827 Richard August Carl Emil Erlenmeyer 1825-1909 Ludwig Boltzman 1844-1906 Galileo Galili 1564-1642 An alchemist Henri Louis LeChatlier 1850-1936 Amedeo Avogadro 1756-1856 James Joule (1818-1889) Evangelista Torricelli 1608-1647 John Dalton 1766-1844 Rudolph Clausius 1822-1888 Johannes Rydberg 1854-1919 J. J. Thomson 1856-1940 Abbe Jean Picard 1620-1682 William Henry 1775-1836 physician William Thompson Lord Kelvin, 1824-1907 Heinrich R. Hertz, 1857-1894 Daniel Fahrenheit 1686-1737 Jacques Charles 1778-1850 1825-1898 Johann Balmer Max Planck 1858-1947 Blaise Pascal 1623-1662 Robert Boyle, 1627-1691 Georg Simon Ohm 1789-1854 Francois-Marie Raoult 1830-1901 Svante Arrehenius 1859-1927 Michael Faraday 1791-1867 James Maxwell 1831-1879 Walther Nernst 1864-1941 Isaac Newton 1643-1727 B. P. Emile Clapeyron 1799-1864 Dmitri Mendeleev 1834-1907 Fritz Haber 1868-1934 Anders Celsius 1701-1744 Germain Henri Hess 1802-1850 Johannes D. Van der Waals 1837-1923 Thomas Martin Lowry 1874-1936 Fitch Rule G3: Science is Referential Gilbert Newton Lewis 1875-1946 Johannes Nicolaus Bronsted 1879-1947 Lawrence J. Henderson 1878-1942 Niels Bohr 1885-1962 Erwin Schodinger 1887-1961 Louis de Broglie (1892-1987) Friedrich H. Hund 1896-1997 Fritz London 1900-1954 Wolfgang Pauli 1900-1958 Werner Karl Heisenberg 1901-1976 Linus Pauling 1901-1994 Properties and Measurements Property Size Volume Weight Temperature Unit m cm3 gram Reference State size of earth m mass of 1 cm3 water at specified Temp (and Pressure) oC, K boiling, freezing of water (specified Pressure) amu (mass of 1C-12 atom)/12 atomic mass of an element in grams atm, mm Hg earth’s atmosphere at sea level 1.66053873x10-24g quantity mole Pressure Energy, General Animal hp heat BTU calorie Kinetic J Electrostatic electronic states in atom Electronegativity F Heat flow measurements horse on tread mill 1 lb water 1 oF 1 g water 1 oC m, kg, s 1 electrical charge against 1 V Energy of electron in vacuum Reference state? To set a “heat flow” scale 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 Does the “system (earth)” gain energy? + heat - heat? For image above system (earth) gains heat from surroundings (sun) To set a “heat flow” scale 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?” For this question: the + heat system (sun) loses heat - heat? to the surroundings (earth) First Law of Thermodynamics: Energy is conserved No universal change in energy Just a transfer of energy surroundings system q =? q>0 system q<0 system E = constant when System AND surroundings considered! To set a “heat flow” scale 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 3. q is - when heat flows out of the system into the surroundings Chemical process in the “system” is defined by heat flow endothermic q>0 exothermic q<0 Properties and Measurements Property Size Volume Weight Unit m cm3 gram Temperature 1.66053873x10-24g quantity mole Pressure Energy: Thermal Kinetic Reference State size of earth m mass of 1 cm3 water at specified Temp (and Pressure) oC, K boiling, freezing of water (specified Pressure) amu (mass of 1C-12 atom)/12 atomic mass of an element in grams atm, mm Hg earth’s atmosphere at sea level BTU calorie J Energy, of electrons Electronegativity Heat Flow 1 lb water 1 oF 1 g water 1 oC 2kg mass moving at 1m/s energy of electron in a vacuum F into system = + Chemical reactions involve Zn(s) 2 H (aq) Zn 2 (aq) H2 ( g) 1. heat exchange As a review: Heat exchange Constant At constant Atm.pressure Pressure who is oxidized? who is reduced? what is the oxidation number on H2? Who is an oxidizing agent? thalpein – to heat en - in H for (?) heat H = Greek: enthalpy qP H Subscript Reminds us that Pressure is constant 1 atm pressure = constant pressure This means heat flow, q, is enthalpy change Chemical reactions involve Zn(s) 2 H (aq) Zn 2 (aq) H2 ( g) 1. heat exchange 2. work PressureVolume work constant Atm. pressure V w P V E E final Einitial q w E P q P P V E P H P V This is a form of Rule G3 Science is referential Generally final - initial E P H P V Change in volume is typically small for Most reactions EP H This means we can measure the energy change of A chemical reaction by measuring the heat exchange At constant pressure 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 3 to 8 standard cubic feet of biogas per pound of manure. The biogas usually contains 60 to 70% methane. Methane Gas Recovery At landfills Consider the contribution of volume of gas phase molecules CH4( g ) 2O2( g ) CO2( g ) 2 H2 O( l ) PV nRT At constant T: P V n RT P V n gas final n gas initial RT L atm P V 1 3moles 0.0821 298K mol K P V 48.9316 L atm %&$*! Conversions – if Interested see slide after next . kJ 01013 P V 48.9316 L atm 4.9kJ L atm 2mole change Consider the contribution of volume change for water in this reaction CH4( g ) 2O2( g ) CO2( g ) 2 H2 O( l ) 2moleH O 2 (l ) 3 18 g 1 cm water 1L * * * 3 3 0.036 L mol 1gwater 10 cm %&$*! Conversions – if Interested see next slide 01013 . kJ PV 1atm 0.036 L 0.0036kJ L atm Most reactions total (q): PV 1 mole gas PV 2mole liquid water Energy in kJ ~ 1000 kJ ~ 2.5 kJ ~ 0.0036 kJ By the end of This module We will see this Is “true” Sig fig tells us that PV energy small compared to q Optional Slide: conversion kg 5 3 3 3 5 Pa 101325 2 . x 10 10 cm 1m J 101325 . x10 Pa m s kJ atm L . kJ atm 3 0101325 2 2 atm L 10 cm kg m 10 J atm Pa 2 s kg 3 5 3 3 2 0101325 . kJ 101325 . x10 Pa m s 10 cm 1m J kJ 3 2 2 atm L atm Pa L 10 cm kg m 10 J 2 s To set a “heat flow” scale 1. Defined conditions: how experiment is performed Constant Pressure 2. But not on the path taken (state property) Heat flow depends the conditions H= enthalpy qreaction cons tan tpressure H H products Hreac tan ts Enthalpy is a state property (measured under constant pressure, but how measured under that constant pressure is not important) q H 0 endothermic H products Hreac tan ts H2 O( s) heat H2 O Think of heat as a reactant q H 0 exothermic H products Hreac tan tss CH4( g ) 2O2( g ) CO2( g ) H2 O( l ) heat Think of heat as a product Enthalpy is an “extensive” property Depends upon the amount present CH4( g ) 2O2( g ) CO2( g ) 2 H2 O( l ) H 890kJ 890kJ of heat is released when 1 mole of methane Reacts with oxygen 890kJ 1moleCH4 ( g ) 890kJ 2moleCH4 ( g ) 1780kJ 1moleCH4 ( g ) Properties and Measurements Property Size Volume Weight Temperature Unit m cm3 gram Reference State size of earth m mass of 1 cm3 water at specified Temp (and Pressure) oC, K boiling, freezing of water (specified Pressure) amu (mass of 1C-12 atom)/12 atomic mass of an element in grams atm, mm Hg earth’s atmosphere at sea level 1.66053873x10-24g quantity mole Pressure Energy, General Animal hp heat BTU calorie Kinetic J Electrostatic electronic states in atom Electronegativity F Heat flow measurements horse on tread mill 1 lb water 1 oF 1 g water 1 oC m, kg, s 1 electrical charge against 1 V Energy of electron in vacuum constant pressure, define system vs surroundin per mole basis (intensive) Context for the next example problem 2006 Sept Sci. Am: World Wide Petroleum Usage Land People Transport Non29% of total use transportation Total transportation =53% Land Freight 19% Air People and Freight 5% U.S. differs from world in distribution of petroleum use Transportation = 71.8% 3 Non-transportation 57.8% of Transportation= Personal land transport = 41.5% of total U.S. consumption Data US DOE 2006 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? H 109 . x104 kJ 2C8 H18 25O2 16CO2( g) 18H2 O( g) Strategy: need moles of octane consumed (Golden Bridge) miles mpg gallons density grams Molar mass H moles heat 33 4 11gal 37852 00..690 10 690 gC gC H H 1 1 moleC moleC H H . L gal 3 3 . 7852 7852 L 10 mL mL 109 . x 10 kJ 88 88 88 88 150 150..mi mi 17.5mi 17.5mi gal gal LL mLC mLC88H H88 114 114gC gC88H H88 2moleC8 H8 H 1,070,243.955kJ H 107 . x106 kJ 3 sig fig We will use part of this problem Again: 3.7852 L 103 mL 0.690gC8 H8 1moleC8 H8 109 . x104 kJ 124,861kJ gal gal L mLC8 H8 114 gC8 H8 2moleC8 H8 Rules 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 H 109 . x104 kJ H 109 . x104 kJ 2C8 H18 25O2 16CO2( g) 18H2 O( g) 16CO2( g ) 18H2 O( g ) 2C8 H18 25O2 3. Enthalpy change depends upon the state of the reactant and products H 44 kJ H2 O( l ) H2 O( g ) ENERGY measurement a) change in temperature b) some function specific to the material and how it is organized (bonds) ENERGY change is a function of a) temperature b) material E k q1q2 d el H O H H O H O O H H H H 15 oC H 25 H O O H H heat 25 oC ENERGY measurement a) change in temperature b) some function specific to the material and how it is organized (bonds) Pure material q C t final t initial q C t q m c t m = mass c = specific heat of a pure substance C = heat capacity = heat required to raise the temperature of the system 1oC units: J/oC http://www.lsbu.ac.uk/water/molecule.html q ch arg e density r Electron density of water Shape and charge distribution On water C(liquidH2O)=4.18J/g-oC Liquid water is very strongly organized due to the polarity of the molecule, so it has a high specific heat Electrons on Oxygen sit “out there” causing large Electrostatic potential Oriented on the electrons Specific heats, c, of various substances in various physical states Material Pb(s) Pb(l) Cu(s) Fe(s) Cl2(g) C(s) CO2(g) NaCl(s) Al(s) C6H6(l) H2O(g) C2H5OH(l) H2O(l) Specific Heats c, (J/g-K) 0.12803 0.16317 0.382 0.446 0.478 0.71 0.843 0.866 0.89 1.72 1.87 2.43 4.18 Ability to store heat in a substance is variable. For the same amount of energy, easier to break electrostatic attraction of He compared to water with it’s localized charge H H O He O He He H H H He He O H He Example: 1.00 cup of water is heated from 25.0 oC to 100.0 oC. How many joules were used to heat the water? q m c t Mass of water: 3 1qt11qt 1 L 10 mL 1g 1 L qt . 100 cup 236.51g 100 . cup 100 . cup 100 . cup . 1057 L 1mL 44cups qt qt 4cups 1057 . cups t Tfinal Tinitial t 100o C 25o C t 75K t 75o C 418 . J q 23651 . g 75K gK q 74148.53 q 7.41x10 J 4 q 74.1kJ 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. 1qt 1L 103 mL 128 . g . cup 236.51g 128 . 302.7 g 100 . qt L 1mL 4cups 1057 08 . J q 302.7 g 75K gK qdry soil 18,164 J 181 . kJ q water 741 . kJ Lag Water Land Because water has a high heat capacity it takes longer than air or soil to warm up and longer to cool down 1. Hot air rises over land Land 1. Hot air rises over Lake 2. Cold air from Lake moves to fill in (Lake Breeze) Lake 2. Cold air moves in from land (Land breeze) The temperature change in 24 hours in summer for water is not much leading to big differences between lake and land and thus a lake breeze Heat capacity of large bodies Of 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 440 J q 200kg 100 25 C 5,828kJ kg C 42miles trip 5,828kJ engineheating 27,730kJ heating 1trip gal 1gal 8miles Compare to heat available from combustion of 1 gallon of octane (from before H 109 . x104 kJ 2C8 H18 25O2 16CO2( g) 18H2 O( g) 3.7852 L 103 mL 0.690gC8 H8 1moleC8 H8 109 . x104 kJ 124,861kJ gal gal L mLC8 H8 114 gC8 H8 2moleC8 H8 5,828kJ 100 22.2% 124,861kJ Let’s compare to what I measure Heating the engine block Prius 1 39 31mpg . 100 205% 39mpg Prius 2 43 34mpg 100 20% 43mpg Properties and Measurements Property Size Volume Weight Temperature Unit m cm3 gram Reference State size of earth m mass of 1 cm3 water at specified Temp (and Pressure) oC, K boiling, freezing of water (specified Pressure) amu (mass of 1C-12 atom)/12 atomic mass of an element in grams atm, mm Hg earth’s atmosphere at sea level 1.66053873x10-24g quantity mole Pressure Energy, General Animal hp heat BTU calorie Kinetic J Electrostatic electronic states in atom Electronegativity F Heat flow measurements horse on tread mill 1 lb water 1 oF 1 g water 1 oC m, kg, s 1 electrical charge against 1 V Energy of electron in vacuum constant pressure, define system vs surroundin per mole basis (intensive) Calorimetry Relate reaction heat To the calorimeter heat qreaction qcalorimeter If the temperature of the water rises (heat flow into water) then heat must have been lost from the reaction Example: When 1.00 g of ammonium nitrate, NH4NO3, is added to 50.0 g of water in a coffee-cup calorimeter, it dissolves: NH4 NO3( s) NH4(aq) NO3(aq) 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. qcalorimeter m c t qcalorimeter J o o 50.0g 418 . 2332 . C 2500 . C gK qcalorimeter 351J qreaction qcalorimeter qreaction 351J J qcalorimeter 209 168 . K K A 1 degree change in Celsus qcalorimeter 35112 . J Is a 1 degree change in Kelvin qreaction 351J Not all calorimeters can be based on qreaction qcalorimeter m c t Heat capacity of the calorimeter q reaction q reaction Ccal t In some problems You will Need to determine This number In one step and Then go on Example: The reaction between hydrogen and chlorine Hg ( g ) Cl2( g ) 2 HCl( g ) 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. qreaction Ccal t qreaction Heat evolved? q Calorimeter heat capacity = 9.33 kJ/oC Tinitial =20.00 oC T final = 29.82 oC kJ 9.33 o 29.82 o C 20.00o C C qreaction 916 . kJ 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? 9.37kJ 32.11 29.91 qreaction o 1.00 g of C7H6O3 2.48 C Tinitial 32.11oC qreaction 8.3121kJ Tfinal 29.91oC 9.37kJ required to cause 2.48 oC change qreaction Ccal t ??? Ccal 9.37 kJ 2.48 o C 138gC7 H6 O3 8.3121kJ 3158598 . kJ 1moleC H O 1moleC H O 1gC HO 7 6 3 7 6 3 316kJ 7 6 3 Combine 2C8 H18 25O2 16CO2( g ) 18H2 O( g ) calorimetry Reaction stoichiometry To get reaction enthalpies H Example 2 What is the enthalpy change for the reaction NH4 NO3( s) NH4(aq) NO3(aq) 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? qreaction qcalorimeter qreaction qreaction J 209 o 168 . o C C 35112 . J qreaction 351J qreaction J o o 50.0g 418 . 2332 . C 2500 . C o g C cons tan tlabpressure 1 g was reacted 351J 80.05g 28100 J 281 . kJ 1g mol mol mol 2N=28.02 4H=4.04 3O=48.00 80.05 H 351J NH4 NO3( s) NH4(aq) NO3(aq) H 281 . kJ NH4 NO3( s) NH4(aq) NO3(aq) H 281 . kJ Where did the per/mole go? The reaction was written as a per/mole Enthalpy is understood as a per/mole of reactant (or as the reaction is written) 4( aq ) 1 NH 1 NH4 NO3( s) 3( aq ) 1NO Thermochemical Equation Rules H 281 . kJ When there are no Coefficients it is understood that It is “1” 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( aq ) NH 3( aq ) NO NH4 NO3( s) H 281 . 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 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 H2 O( s) H2 O( l ) 819 8.19 kJ 18.02 g H2O 1mole H2O 6.00kJ 24.6 gice mole H2O An example of several of the rules using Fuel Cells Fuel cells use the reaction: H 2 ( g ) O2 ( g ) H 2 O( l ) 1 2 Calculate the enthalpy for the equation above given that: H 5716 . kJ 2 H2 O(l ) 2 H2( g ) O2( g ) Reverse reaction: H 5716 . kJ 2 H2( g ) O2( g ) 2 H2 O( l ) scale 5716 . kJ H 286kJ 2 H2 ( g ) O2 ( g ) H2 O( l ) 1 2 Here we got a number by coming “at it” from an odd direction Hess’s law 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) Germain Henri Hess 1802-1850 born in Geneva Switzerland Professor of Chemistry At St. Petersburg Technological Institute Galen, 170 Marie the Jewess, 300 Charles Augustin James Watt Coulomb 1735-1806 1736-1819 Thomas Graham Justus von Liebig 1805-1869 (1803-1873 J. Willard Gibbs 1839-1903 Jabir ibn Hawan, 721-815 Luigi Galvani 1737-1798 Count Alessandro Guiseppe Antonio Anastasio Volta, 1747-1827 Richard August Carl Emil Erlenmeyer 1825-1909 Ludwig Boltzman 1844-1906 Galileo Galili 1564-1642 An alchemist Henri Louis LeChatlier 1850-1936 Amedeo Avogadro 1756-1856 James Joule (1818-1889) Evangelista Torricelli 1608-1647 John Dalton 1766-1844 Rudolph Clausius 1822-1888 Johannes Rydberg 1854-1919 J. J. Thomson 1856-1940 Abbe Jean Picard 1620-1682 William Henry 1775-1836 physician William Thompson Lord Kelvin, 1824-1907 Heinrich R. Hertz, 1857-1894 Daniel Fahrenheit 1686-1737 Blaise Pascal 1623-1662 Jacques Charles 1778-1850 1825-1898 Johann Balmer Max Planck 1858-1947 Robert Boyle, 1627-1691 Georg Simon Ohm 1789-1854 Francois-Marie Raoult 1830-1901 Svante Arrehenius 1859-1927 Michael Faraday 1791-1867 James Maxwell 1831-1879 Walther Nernst 1864-1941 Isaac Newton 1643-1727 B. P. Emile Clapeyron 1799-1864 Dmitri Mendeleev 1834-1907 Fritz Haber 1868-1934 Anders Celsius 1701-1744 Germain Henri Hess 1802-1850 Johannes D. Van der Waals 1837-1923 Thomas Martin Lowry 1874-1936 Fitch Rule G3: Science is Referential Gilbert Newton Lewis 1875-1946 Johannes Nicolaus Bronsted 1879-1947 Lawrence J. Henderson 1878-1942 Niels Bohr 1885-1962 Erwin Schodinger 1887-1961 Louis de Broglie (1892-1987) Friedrich H. Hund 1896-1997 Fritz London 1900-1954 Wolfgang Pauli 1900-1958 Werner Karl Heisenberg 1901-1976 Linus Pauling 1901-1994 Example of how Hess’s law is useful C( s) O2 ( g ) CO( g ) 1 2 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 reactions involving the complete burning of carbon C(s) + O2(g) This is the Number we want CO(g) +1/2O2(g) But can’t actually measure Related to 2CO + O2…. Get to the number by an alternative Path (State function!) To solve rearrange equations to get CO on right hand side CO2(g) H 3935 . kJ C( s) O2( g ) CO2( g ) H 566.0kJ 2CO( g ) O2( g ) 2CO2( g ) H 566.0kJ 2CO( g ) O2( g ) 2CO2( g ) H 566.0kJ 566.0kJ H 2 2CO2( g ) 2CO( g ) O2( g ) H 283.0kJ CO2 ( g ) CO( g ) O2 ( g ) H 0 3935 . kJ H 3935 . kJ C( s) O2( g ) CO2( g ) C( s) O2 ( g ) CO2 ( g ) 0 H .283 H 110 5kJ.0kJ H 110.5kJ 0 CO2 ( g ) CO( g ) O2 ( g ) 1 2 1 2 C( s) 11 CO CO( g )( g) O2 ( g ) 22 O 2( g 22()(gg)) CO C(s) + O2(g) C( s) O2 ( g ) CO( g ) 1 2 CO(g) +1/2O2(g) CO2(g) Enthalpies of Formation Invoke Rule G5: Chemists are Lazy Rather than getting the enthalpy for each reaction from a bomb calorimeter use a smaller number of standard reactions from which Hess’s law can be applied to get all the remainder reactions of interest Enthalpy associated with standard reaction is enthalpy 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 compounds not reactions which reduces total data which must Be acquired (Chemists are Lazy!!!) 1 is “understood” H Of 882 kJ 1 2 N 2 ( g ,1atm,25C ) O2 ( g ,1atm,25C ) 1NO2 ( g ,1atm,25C ) Standard molar enthalpy of formation of a compound From elements in their stable states at 1 atm pressure 25oC 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 standard molar enthalpy of 0 Why? Elements in their stable states at 1atm, 25oC have a standard molar enthalpy of 0 Fes Fes O O H Fe H ( s) Fe ( s ) 1atm,25C O H Fe ( s) 1atm,25C 0 Products (standard state) - Reactants (standard state) = 0 H Of 882 kJ 1 2 N 2 ( g ,1atm,25C ) O2 ( g ,1atm,25C ) NO2 ( g ,1atm,25C ) Standard molar enthalpy of formation of a compound From elements in their stable states at 1 atm pressure 25oC 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 the proton H Of Haq 0 Do we detect any patterns? H Al(s) Ba(s) Be(s) Br2(g) Ca(s) C(s,graphite) Cs(s) Cl2(g) Cr(s) Cu(s) F2(g) H2(g) H+(aq) I2(s) O f kJ/mol H Of kJ/mol Fe(s) 0 0 Pb(s) 0 0 Li(s) 0 0 Mg(g) 0 0 Mn(s) 0 0 Hg(l) 0 0 Ni(s) 0 0 N2(g) 0 0 O2(s) 0 0 P4(s) 0 0 K(s) 0 0 Rb(s) 0 0 0 MOST metals are 0 elemental solids (metal) in standard state kJ/mol H Of Sc(s) Si(s) Na(s) S(s,rhombic) Ti(s) Zn(s) 0 0 0 0 0 0 Common non-metals Have specific forms In which they Are standard Group 17 with exception Of I2 are gases in Stable, standard state Properties and Measurements Property Size Volume Weight Temperature Unit m cm3 gram Reference State size of earth m mass of 1 cm3 water at specified Temp (and Pressure) oC, K boiling, freezing of water (specified Pressure) amu (mass of 1C-12 atom)/12 atomic mass of an element in grams atm, mm Hg earth’s atmosphere at sea level 1.66053873x10-24g quantity mole Pressure Energy, General electronic states in atom Electronegativity Heat flow measurements Standard Molar Enthalpy Energy of electron in vacuum F constant pressure, define system vs surroundings per mole basis (intensive) 25 oC, 1 atm, from stable state Hfo Haq+ =0 Calculation of Ho standard enthalpy change of a reaction o n H f , products HO o n H f ,reac tan ts 1 atm pressure 25oC 1. The coefficients of products and reactants in the thermochemical equation must be taken into account 2 Al( s) Fe2 O3( s) 2 Fe( s) Al2 O3( s) H O H of , Al2O3 2 H of , Fe ( s) H of , Fe2O3 s 2 H of , Al ( s) H O H of , Al2O3 H of ,Fe2O3s 2. Elements in standard states can be omitted because heats of formation are zero OJO! O Appendix H 1669.8 82216 . H O 1669.8 82216 . 847.65kJ / mol Calculation of Ho standard enthalpy change of hot and cold packs HO n H of , products n H of ,reac tan ts 1 atm pressure 25oC NH4 NO3,s NH4,aq NO3,aq H O H of , NH H of , NO 4 , aq 3 , aq H of , NH4 NO3, s H O 1325 . 2050 . 3656 . kJ H 337.5 365.6 28 mol O Compares well to the calorimetry calc. (28.1kJ/mol)! Appendix Compound NH4NO3,s NH4+,aq NO3-aq MgSO4,s Mg2+aq SO42-aq Fe,s O2,g Fe2O3,s ΔHfo, kJ/mol -365.6 -132.5 -205.0 -1284.9 -466.8 -909.3 0 0 -1118.4 2 2 MgSO4,s Mgaq SO42,aq HO H O H of , Mg 2 H of ,SO2 H of , MgSO4 , s , aq 4 aq H O 4668 . 909.3 1284.9 kJ mol Products are more stable, lower in the energy well, than reactants H 1,3761 . 1284.9 912 . O http://webmineral.com/data/Hexahydrite.shtml http://www.edinformatics.com/math_science/info_water.htm Liquid water http://www.lsbu.ac.uk/water/hofmeist.html http://biochempress.com/Files/IECMD_2003/IECMD_2003_027.pdf Example: Calculate the Ho for the combustion of one mole of methane CH4 according to the equation CH4( g ) 2O2( g ) CO2( g ) 2 H2 O( g ) Given the standard enthalpies of formation at 25oC, 1 atm from Appendix kJ/mol O2(g) 0 O o o H n H n H f , products f ,reac tan ts CO2(g) -393.5 H2O(g) -241.8 CH4(g) -74.8 kJ kJ 1molCO2 3935 H 2molH2 O 2418 . . molH2 O molCO2 O kJ kJ 1molCH4 74.8 2molO2 0 molCH24 molO2 Example: Calculate the Ho for the combustion of one mole of methane CH4 according to the equation CH4( g ) 2O2( g ) CO2( g ) 2 H2 O( g ) kJ kJ 1molCO2 3935 H 2molH2 O 2418 . . molH2 O molCO2 kJ kJ 1molCH4 74.8 2molO2 0 molCH24 molO2 O H O 87710 . kJ 748 . kJ 802.30kJ Sig figs? H O 802.3kJ Can also “reverse” the problem (inside out socks) Example: Calculate the standard enthalpy of formation for octane H o 109 . x104 kJ 2C8 H18 25O2 16CO2( g ) 18H2 O( g ) Given the standard enthalpies of formation at 25oC, 1 atm kJ/mol O o o H H H f , products f ,reac tan ts O2(g) 0 CO2(g) -393.5 H O 109 . x104 kJ H of , products H2O(g) -241.8 CH4(g) -74.8 H of ,reac tan ts 2418 kJ . kJ 18molH2 O( g ) 109 . x10 kJ 16molCO2 ( g ) 3935 . molCO2 ( g ) molH2 O( g ) 0kJ kJ 25molO2 2molC8 H18 x molC8 H18 molO2 4 109 . x104 kJ 6296kJ 4352.4kJ 2 x 0 109 . x104 kJ 8704.8kJ 2 x 2195.2 kJ 2 x x 1097 . x103 kJ Bond Enthalpy The change in enthalpy when 1 mole of bonds is broken in the gaseous State. Rule G3: Science is referential! Br2( g ) 2 Brg Cl2( g ) 2Clg H 193kJ H 243kJ Which has a stronger bond Enthalpy? For covalent bonds, bond enthalpies depend on? Bond Bond Bond length Overall structure of the molecule Bond length is nice, but it doesn’t Really relate to the Periodic table AND it isn’t the whole story Electronegativities? Can explain some trends Bond Bond Length Length pm pm Pauling's Pauling's ΔE.N. ΔE.N. atomic atomic radii (pm) (pm) Cl-Cl Cl-Cl Br-Br Br-Br I--I I--I 199 199 228 228 267 267 00 00 00 99 99 114 114 133 133 Enthalpy Enthalpy Single Bond Bond Single kJ/mol kJ/mol (Average) (Average) 243 243 193 193 151 151 H--F H--F H--Cl H--Cl H--Br H--Br H--I H--I 92 92 127 127 141 141 161 161 1.8 1.8 11 0.8 0.8 0.5 0.5 (37) 64 64 (37) (37) 99 99 (37) (37) 114 (37) 114 (37) 133 133 (37) 568 568 432 432 366 366 298 298 C--F C--F C--Cl C--Cl C--Br C--Br C--I C--I 135 135 177 177 194 194 214 214 1.5 1.5 0.7 0.7 0.5 0.5 0.2 0.2 77 64 64 77 77 99 99 77 77 114 77 114 77 133 133 77 488 488 330 330 288 288 216 216 C--F C--F C--O C--O C--N C--N C--C C--C 135 135 143 143 147 147 154 154 1.5 1.5 11 0.5 0.5 00 77 64 77 77 66 77 77 70 77 77 77 488 360 308 348 348 H--H H--H H--O H--O H--N H--N H--C H--C 74 74 96 96 101 101 109 109 00 1.3 1.3 0.8 0.8 0.3 0.3 -37 -37 (37) 66 (37) 66 (37) 70 70 (37) (37) 77 77 (37) 436 436 366 366 391 391 413 413 H--H H--C H--N H--O 74 109 101 96 0 0.3 0.8 1.3 -37 (37) 77 (37) 70 (37) 66 436 413 391 366 rr aa dd ii ii Bond ?? Bond Bond Length pm Pauling's E.N. atomic radii (pm) H--H C--C Cl-Cl S-S Br-Br N-N I--I O-O 74 154 199 205 228 145 267 148 2.2 2.5 3.2 2.6 3 3 2.7 3.5 37 77 99 104 114 70 133 66 Enthalpy Single Bond kJ/mol (Average) 436 348 243 226 193 159 151 145 Bottom line – atomic radii ΔE.N. seem to “best” determine bond enthalpies Bond Length pm Pauling's ΔE.N. atomic radii (pm) Cl-Cl Br-Br I--I 199 228 267 0 0 0 99 114 133 Enthalpy Enthalpy Single Single Bond Bond kJ/mol kJ/mol (Average) (Average) 243 243 193 193 151 151 H--F H--Cl H--Br H--I 92 127 141 161 1.8 1 0.8 0.5 (37) 64 (37) 99 (37) 114 (37) 133 568 568 432 432 366 366 298 298 C--F C--Cl C--Br C--I 135 177 194 214 1.5 0.7 0.5 0.2 77 64 77 99 77 114 77 133 488 488 330 330 288 288 216 216 C--F C--O C--N C--C C--C 135 143 147 154 154 1.5 1 0.5 0 0 77 64 77 66 77 70 77 77 488 488 360 360 308 308 348 348 H--H H--H H--O H--O H--N H--N H--C H--C 74 74 96 96 101 101 109 109 0 0 1.3 1.3 0.8 0.8 0.3 0.3 -37 -37 (37) (37) 66 66 (37) 70 (37) 70 (37) (37) 77 77 436 436 366 366 391 391 413 413 H--H H--C H--N H--O 74 109 101 96 0 0.3 0.8 1.3 -37 (37) 77 (37) 70 (37) 66 436 413 391 366 rr aa dd ii ii Bond enthalpies increase Single <Double < Triple But not by multiples of the single bond C-C N-N C-N C-O Bond Enthalpy (kJ/mol) X-X X=X 347 612 694 159 418 318 293 615 586 351 715 702 X=X 820 1041 941 477 890 879 1075 1053 measured calc. measured calc. measured calc. measured calc. http://chemviz.ncsa.uiuc.edu/content/doc-resources-bond.html Your book’s emphasis on bond enthalpies? Relates to the energy released or taken up By a reaction If Bonds of Reactants stronger than bonds products endothermic We will take up the issue of bond enthalpy when discussing solids Examples we have examined about energy so far Hydrogen Fuel Cell H 286kJ H 2 ( g ) O2 ( g ) H 2 O( l ) 1 2 Fossil fuel burning (coal) H 110.5kJ C( s) O2 ( g ) CO( g ) H 3935 . kJ C( s) O2( g ) CO2( g ) 1 2 Fossil fuel burning (methane) H 890kJ CH4( g ) 2O2( g ) CO2( g ) 2 H2 O( l ) Fossil fuel burning (octane) H 109 . x104 kJ 2C8 H18 25O2 16CO2( g) 18H2 O( g) Energy Density Pure octane Chemistry General FITCH Rules G1: Suzuki is Success G2. Slow me down G3. Scientific Knowledge is Referential G4. Watch out for Red Herrings G5. Chemists are Lazy C1. It’s all about charge C2. Everybody wants to “be like Mike” qq C3. Size Matters E k r r C4. Still Waters Run Deep C5. Alpha Dogs eat first 1 2 el 1 2 “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 Summary Slides A B heat C q endothermic heat to system A B heat C q exothermic heat to surroundings q cm t final t initial cm t c is a measure of the intermolecular interactions; very large for water = measure of the polarity of the water molecule q cons tan tpressure H enthalpychange H products Hreac tan ts H fusion , H2O H 2 O( s ) H 2 O( l ) H vaporization , H2O H 2 O( l ) H 2 O( g ) Says something about intermolecular interactions (polarity!) H 5716 . kJ 2 H2 O(l ) 2 H2( g ) O2( g ) H 5716 . kJ 2 H2( g ) O2( g ) 2 H2 O( l ) 5716 . kJ H 286kJ 2 H2 ( g ) O2 ( g ) H2 O( l ) 1 2 An example of Hess’s Law An example of a reaction of standard molar enthalpy of formation H 882 kJ O f HO o H f , products 1 2 N 2 ( g ,1atm,25C ) O2 ( g ,1atm,25C ) NO2 ( g ,1atm,25C ) o H f ,reac tan ts H E PV H E PV products PV reac tan ts Properties and Measurements Property Size Volume Weight Temperature Unit m cm3 gram Reference State size of earth m mass of 1 cm3 water at specified Temp (and Pressure) oC, K boiling, freezing of water (specified Pressure) amu (mass of 1C-12 atom)/12 atomic mass of an element in grams atm, mm Hg earth’s atmosphere at sea level 1.66053873x10-24g quantity mole Pressure Energy, General electronic states in atom Electronegativity Heat flow measurements Standard Molar Enthalpy Energy of electron in vacuum F constant pressure, define system vs surroundings per mole basis (intensive) 25 oC, 1 atm, from stable state Hfo Haq+ =0 “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