Week 3   thermodynamics
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Week 3 thermodynamics

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Week 3   thermodynamics Week 3 thermodynamics Document Transcript

  • ENVE 501 – Week 3 1
  • Quiz 1Imagine you can follow a molecule of copper from the source to the mouth of the Mississippi River. Assume that the average concentration of dissolved copper remains constant at 5 µg/L. The total dissolved solids, however, increases significantly from source to mouth.Would you expect the activity to increase, decrease, or remain the same? Why?From the perspective of a fish, what is more important, the activity or the concentration of copper? Briefly explain your answer.2
  • Quiz 1 - Solution• Activity will decrease • Ionic strength will increase as the total dissolved solids (TDS) concentration increases • Activity coefficient will decrease as the ionic strength increases ai = γi × [i]• Activity is more important • Toxicity involves biochemical reactions • Reactions depend on chemical activity 3
  • Teams - I Jennings Diane Jin Weidong Knudsen Westen Gobi Mu Binbin Temino Boes Regina Xu Limeimei Dai Wei Douce JordanKalahari Glick Nicole Huang Jinjin Jiang Yuanzan 4
  • Teams - II Mirza Muhammad Sharpe JessicaAtacama Short Bradley Wang Chen Wang Zehua Gong Qijian Izadmehr Mahsa Meyer KristineSahara Mocny William She Congwei Wang Xiaolang 5
  • Teams - III Azu Laud Hernandez Alcides Huo YechenSonoran Pujari Aniket Ramos Tiffanie Zhang Yue Cao Jicong Lin KehsunMojave Tadeusz Bobak Wen Jiahong Young Meghan 6
  • Team assignment for next week• Use a 9” by 9” slide in Landscape format• Introduce your consulting firm • Names • Photographs(? ) • History • Specialization • Company logo…. 7
  • Details posted on Blackboard• No more than four slides• Email to me by 8 PM on September 14• Use Blackboard tools to collaborate• Select one team member to present next week• Students at remote sites should consider an electronic presentation 8
  • Chapter 3 –Thermodynamics 9
  • Key ConceptsEntropy pp 62-64; 70-74Gibbs (free) energy pp 80-93; 98-104Thermodynamics & chemical systems pp 104-125 • Page numbers are from Chapter 2 of our text. • The handout provides a more comprehensive description. 10
  • Laws of thermodynamics - IZeroth law of thermodynamics • Systems in thermal equilibrium have the same temperature.First law of thermodynamics • Energy is conserved. 11
  • Laws of thermodynamics - IISecond law of thermodynamics • Nature is not symmetric; the quality of energy decreases. • A process will occur spontaneously if it increases the total entropy of the universeThird law of thermodynamics • Entropy = 0 for a perfect crystal at T = 0. 12
  • An alternative look at the lawsWe cant winWe are sure to loseWe cant get out of the game (Garrett Hardin from an unknown source) 13
  • Entropy and lifeLife is an open or continuous system that is able to decrease its internal entropy at the expense of free energy taken in from the environment and subsequently rejected in a degraded form. • Lovelock, J. (2000) GAIA. A new look at life on Earth. Oxford University Press, New York. 14
  • Entropy and deathEntropy is a measure of a systems thermal disorder. It implies the predestined and inevitable run-down and death of the Universe. • Lovelock, J. (2000) GAIA. A new look at life on Earth. Oxford University Press, New York. 15
  • Fundamental thermodynamic propertiesGibbs (free) energy = G • Energy available to do useful workEnthalpy = H • Total energy (molecular + mechanical work)Entropy = S • A measure of disorder 16
  • Changes in G, H, and S are related∆Gi = ∆H i − T ∆Si 17
  • What do the symbols mean?Subscript identifies the compoundGi ≡ Gibbs energy associated with total iThe over-bar means per molGi ≡ energy per mol of iThe superscript ‘o’ refers to the standard state oGi ≡ standard molar Gibbs energy of i, also the standard Gibbs energy of formation 18
  • Thermodynamic properties are relativeBy definition, the Gibbs energy of a pure element under standard conditions is 0 • For example (T = 25°C, P = 1 bar) GH2 = 0; pure hydrogen gas GCu = 0; pure solid copperFor dissolved ions (T = 25°C, P = 1 bar) GH = 0; dissolved hydrogen ion + • Other ions defined relative to this reference 19
  • Corrections for different activity 20
  • 21 + RT ln ci + RT ln γ i ActivityEffect on Gibbs energy coefficient Concentration o Gi = Gi Standard Gibbs energy of formation
  • Example What is the Gibbs energy per mole of atmospheric oxygen? Actual Gibbs energy = Standard Gibbs energy + Corrections for activity o Gi = Gi + RT ln ci + RT ln γ i 22
  • 23
  • Example What is the Gibbs energy per mole of atmospheric oxygen? 24
  • Example Substituting 25
  • So what?Change in Gibbs energy tells us if a reaction can occur.Reaction is possible: • If the change in the Gibbs energy of reaction is negative 26
  • ExampleFind molar Gibbs energy of reaction for: 1.0 mole hydrogen gas + 0.5 mole oxygen gas = 1.0 mole water as a gasAssume 1.0 bar and T = 25° C H2(g) + ½ O2(g) ↔ H2O(g) 27
  • 28
  • Example Substitute 29
  • -228.57 kJ per mol of what?Reaction involves: 1 mol H2 + 0.5 mol O2 + 1 mol H2OPer mole of stoichiometric reaction.For example: aA+bB ↔ cC+dD 1.0 mole of stoichiometric reaction: a mol A + b mol B → c mol C + d mol D Stoichiometry is important 30
  • For the reaction: aA+bB↔ cC+dD G r , products = G r ,C + G r , D o o = G r ,C + RT ln aC + G r , D + RT ln aD G r ,reactants = G r , A + G r , B o o = Gr, A + RT ln a A + G r , B + RT ln aB 31
  • o o o o  aC aD  ∆ Gr = GC + GD − GA − GB+ RT ln    a A aB  o  aC aD  = ∆ Gr + RT ln    a A aB   γ c [ C ] c γ d [ D] d  C   o D∆Gr = ∆Gr + RT ln    γ A [ A] γ B [ B ] a a b b    32
  • For dilute solutionsAs γi → 1.0, NOTE: From here on we often assume a dilute solution [ C ] c [ D] d    o∆Gr = ∆Gr + RT ln    [ A] [ B ] a b    o = ∆Gr + RT ln Q Q = reaction quotient 33
  • Example H2(g) + ½ O2(g) ↔ H2O (g • Average atmospheric partial pressures pH2 = 6×10-7 bar pO2 = 0.21 bar pH2O = 0.02 bar • Gibbs energy under conditions different from the standard state is : Gi = G + RT ln ai i o 34
  • 35
  • ExampleUnder standard conditions: RT = 8.314 (J/mol.K) × 298 (K) RT = 2.48 (kJ/mol) ai = γi × pi 36
  • Example 37
  • Substituting for each term 38
  • Quiz 2• Match the reaction described in the first column with the dimensions of the rate constant in the final column• You might not need all the choices for k 39
  • Interpreting values of ∆Gr ∆Gr < 0reaction can proceed in forward direction ∆Gr > 0reaction can proceed in reverse direction ∆Gr = 0forward and reverse reaction rates are the same; reaction is at equilibrium 40
  • A special reaction quotient when ∆ Gr = 0 o ∆Gr = − RT ln K eq Keq = equilibrium constant    [ C ] [ D]  o ∆Gr c d K eq = exp − =  RT  [ A] [ B ] a b   41
  • Example How much oxygen should be dissolved in Lake Michigan when T = 25°C? Reaction is: O2(g) ↔ O2(aq) Gibbs energy for the reaction as written is: o  [ O2 ]    ∆Gr = ∆Gr + RT ln    pO2    What are the dimensions of the term in brackets? 42
  • 43
  • Example Recall the “hidden” terms:  mol  cO2   Numerator is:  L   mol  1.0   L  Denominator is: pO2 ( bar ) 1.0( bar ) Change in standard state Gibbs energy for this reaction is: o ∆Gr = 16.32 - 0 = 16.32 (kJ/mol) 44
  • Example  ∆ Gro    K eq = exp  −   RT     16.32 × 103    −3 = exp  −  = 1.38 × 10  8.314 × 298    Solving for equilibrium dissolved oxygen: [O2] = 1.38×10-3 × 0.21 [O2] = 2.9×10-4 (mol/L) ⇒ 9.3 mg/L Dimensions follow from “hidden” terms. 45
  • Example Estimate Lake Michigan dissolved oxygen concentration for T = 1°C. It looks as if :     o ∆Gr K eq = exp −   RT    o Knowing ∆Gr Can we substitute T = 1°C? 46
  • Example No!Superscript ‘o’ refers to the reference state.These values apply only for T = 25°C 47
  • Example Equilibrium constant at T = 1°C estimated from the van’t Hoff equation.  ∆H r  1 1    o  K eq ,T2 = K eq ,T1 exp   − ÷  R  T1 T2     −3  −11.71 × 103  1  1  K eq ,T2 = 1.38 × 10 exp   − ÷  8.314   298 274    = 2.09 ×10−3 [O2] = 2.09×10-3 × 0.21 [O2] = 4.39×10-4 (mol/L) ⇒ 14 mg/L 48
  • Example Compare available energy Aerobic versus anaerobic process Degradation of acetic acid Aerobic process C2H4O2(aq) + 2 O2(aq) = 2 H2CO3 (aq) ∆Gr = 2×(-623.2) – (-396.6) - 2×(16.32) ∆Gr = -882.44 kJ/mol 49
  • Example Anaerobic process C2H4O2(aq) + H2O(l) = CH4(aq) + H2CO3(aq) ∆Gr = (-34.39)+(-623.2)-(-396.6)-(-237.18) ∆Gr = -23.81 kJ/mol Relative to anaerobic processes, aerobic process release more energy 50
  • Example Temperature affects equilibrium: Dissolution of O2 ∆H r = -11.71 kJ/mol Precipitation of CaCO3(s) Ca2+(aq) + CO32-(aq) = CaCO3(s) ∆H r= (-1207.4) - (-542.83) – (-677.1) ∆H r = 12.53 kJ/mol 51
  • Example Sign of ∆H is important O2 solubility increases with decreasing temperature Implications for fish? CaCO3(s) solubility decreases with decreasing temperature Implications for water softening? 52
  • Example Precipitation of FeCO3(s) Find changes in: Gibbs energy Enthalpy Entropy Fe2+ + CO32- = FeCO3(s) 53
  • 54
  • Example ∆Gr = -666.7 – (-78.87 – 527.9) = -59.93 kJ/mol ∆H r = -737.0 – (-89.10 – 677.1) = 29.2 kJ/mol ∆Sr = 105 – (-138 – 56.9) = 299.90 J/mol.K Entropy increases for formation of this solid. Why? 55
  • Summary I ∆Gi = ∆H i − T ∆Si oGi = Gi + RT ln ci + RT ln γ i∆Gr = Gproducts − Greactants 56
  • Summary II   o ∆Gr { C} { D}  c d K eq = exp − =   RT  { A} a { B} b   ∆H r  1 1    o K eq ,T1 = K eq ,T2 exp   − ÷  R  T1 T2     57
  • Some time for teams or Maxwell’s Demon 58
  • 0.5 N2 + O2 ⇔ NO2 + heatThe expression above shows an exothermic reaction (heat is generated) wherein nitrogen gas is oxidized to NO2.What effect would each of the following have on the reaction? • Decrease the temperature. • Increase the volume. • Decrease the O2 concentration. • Add a catalyst. 59
  • Solutions…Decrease the temperature. Equilibrium shifts to the right to increase the temperature.Increase the volume. Equilibrium shifts to the left to increase the pressure.Decrease the O2 concentration. Equilibrium shifts to the left to increase O2.Add a catalyst. A catalyst will affect the rate of the reaction, but not the equilibrium. 60