10.28.2011 - Lecture 22 Today in Chemistry 260 • ...

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Unformatted text preview: Lecture 22 Today in Chemistry 260 •  •  •  •  •  •  Chemistry 260 Principles of thermochemistry Phase transitions Reaction enthalpies Bond enthalpies Hess s law Formation reactions This Week in Chemistry 260 •  No RQ next lecture. • Kubarych: Reaction spontaneity and entropy Lecture 22 October 28, 2011 Thermochemistry 1 Reversible Adiabatic expansions (Pext = Pi = P) q=0 T2 T1 PVg=const g =Cp/Cv>1 ConcepTest Reversible Adiabatic expansions (work relation) Adiabatic ! q = 0 ΔU = q + w ΔU = w For an ideal gas, ΔU = Cv ΔT Tf ∝ 1/Vf Reversible Adiabatic expansions (work relation) T2 Pressure B. T1 Ti* ( 2V ) !T ad iab at ad iab at Volume Tf c Compare the overall work due to a cycle of reversible expansion and than compression to a cycle with an opposite order of compression and than expansion. Assume doubling the volume and halving and the maximum temperature in both cycles is the same. 1V! 2V !1V vs. 2V!1V!2V 1.  Larger 2.  Smaller. 3.  The same. The actual value depends on the max temp. 4. Zero for both cycles. 4 Back to chemistry: Thermochemistry •  To Adiabatic ! q = 0 ΔU = w ΔU = C v ΔT ΔU = q + w Work done in an infinitesimal step T f* (1V ) T1 B. For an ideal T f ! Vi " monoatomic =# $ # V f $ gas, c = c p / cv ! 1 Ti % & Consider a cycle of reversible expansion/compression T f ( 2V ) = 2! c For expansion to double volume Ti (1V ) T2 Work done in an infinitesimal step ! V " For an ideal = # i $ monoatomic c = c p / cv ! 1 Ti # V f $ % & gas, Consider a cycle of reversible expansion/compression q=0 As Vf ↑, Tf ↓ w = cv dT q=0 Adiabatic ! q = 0 ΔU = w ΔU = C v ΔT ΔU = q + w w = cv dT wi = ! PdV dU = ncv dT = ! PdV nRT ncv dT = ! dV V dT nR ncv =! dV T V Tf Vf dT dV ncv " = !nR " T V Ti Vi Tf Vf V cv ln = ! R ln = R ln i Ti Vi Vf The underlying physics: When the gas expands, it does work, loses energy, and therefore its temperature drops. 3 The internal energy is the only available source of energy for the work w<0 w<0 Work done in an infinitesimal step Pressure w<0 2 c For compression back to half = 2c the volume Tf = Ti compression = "!T inv !T inv f = Ti exp ansion w = cv !T compression = "cv !T exp ansion 5 clarify, • Heating/cooling • Exothermic/endothermic process • Heat flow • Internal energy Heating is an Endothermic process but In an exothermic Reaction the beaker is getting warmer . 6 Volume 1 Lecture 22 Internal Energy, Heat flow and temperature Internal Energy, Heat flow and temperature For simplicity: Assume no work and ideal gas atoms. Chemical reaction: (For simplicity: Assume no work) •  Consider heat flow (for heating or cooling the system) and changes in the internal energy. (do not confuse heating with exothermicity) He gas Translational Energy Electronic Energy Nuclear Energy •  Do not confuse temp. effects (due to the reaction) with changes in the internal energy. (temp. increase is not endothermic) Q<0 •  Kinetic theory of ideal gases: KE is propotional to Temp. (molecules move faster upon heating). •  Heating the gas is endothermic, results with the increase of internal energy. •  If the reaction A+B +HEAT ! C+D, requires heat flowing in to enable the transformation.. Q>0 .. it means that the energy stored in the system (molecules) is increased (endothermic). •  Now consider the possibility of heat flow DUE to a chemical reaction • In this case, the system sucks in energy using heat and will result with temperature falling down. Example Reaction The Role of Enthalpy in Chemistry Enthalpy of reaction/process: change in enthalpy during a reaction/process Change in enthalpy: heat transfer at constant pressure 2Na(s)+2H2O(l) ! 2NaOH(aq) + H2(g) 2 moles of Na(s) + 2 moles 2H2O produces 368.6 kJ of heat and 1 mol H2(g) •  PV work: consider the thermal decomposition of 1.0 mol of CaCO3(s) at 1 atm !U≠q: some energy has been used for expansion work –  Volume of the system increases for the H2(g)   System pushes back the atmosphere THREE PROPERTIES OF ENTHALPY TO KEEP IN MIND: Enthalpy (state function) 9 !H = q p !H = H f " H i !H Cp = !T heat (NOT state function) q>0 10 Physical Processes (phase transition) !H = !U + P !V The enthalpy of phase transitions e.g. vaporization (liquid ! gas), rhombic sulfur ! monoclinic sulfur H2O(l)!H 2O(g ) !Ho = 44 kJ thermochemical equation standard enthalpy of vaporization, !Hovap Phase transitions And Heating curves Enthalpy is a state function: The enthalpy change of an overall process is the sum of the enthalpy changes for the steps (observed or hypothetical) into which it may be divided. !H sub for H2O? !H sub = !H fus + !H vap gas enthalpy So enthalpy is a measure for the change in the energetics of the system including the internal energy corrected by the expansion work w<0 H = U + PV !H = !U + P !V •  At 1 atm, 25°C, – P!V = -2.5 kJ !U = !H – P!V (= q + w) !U = -368.6kJ – 2.5kJ !U = -371.1kJ !Hvap !Hfus 11 !Hsub liquid solid 12 2 Lecture 22 Heating curve: Substances undergoing broad temperature changes ConcepTest: Substances undergoing broad temperature changes 1 mol (18 g) H2O at -20 C is heated to a final temperature of 110 C. What is the change in enthalpy accompanying this process? !H = 55.3 kJ 1 mol (18 g) H2O at -20 C is heated to a final temperature of 110 C. What is the change in enthalpy accompanying this process? !H = 55.3 kJ T 100 Where is the energy going in each part of this diagram? •  During phase transitions ‒ breaking/forming intermolecular forces •  During heating processes ‒ Translational, rotational and vibrtational energy !H = !H vap = (1mol ) 40.67 kJ / mol T = 40.67 kJ 100 !H = q p = nc p !T ( = (1mol ) 35.8 J mol !K ) (10K ) Which line should been drawn longer : !H = !H vap A.  Liquid!Gas (blue) B.  Solid!Liquid (red) !H = q p = nc p !T ( = (1mol ) 35.8 J = 0.358kJ !H = q p = nc p !T !H = !H fus = (1mol ) 6.01kJ / mol ( mol !K 0 ) ( 20K ) = 0.734kJ ( = (1mol ) 75.3 J !H = !H fus = 7.53kJ !H = q p = nc p !T = (1mol ) 37.6 J !H = q p = nc p !T 100 K ) mol !K ) ( ) (10K ) mol !K ) (100K ) = 7.53kJ !H = q p = nc p !T ( = (1mol ) 37.6 J t mol !K ) ( 20K ) = 0.734kJ 13 t Atomic and Molecular Changes 14 How much energy as heat (at constant T and P) must be supplied to 1.0 mol Mg(s) to produce Mg2+(g)? Mg2+(g)+2e-(g) Enthalpy is A state function. Chemical changes Enthalpy (kJ mol-1) 0 = 6.01kJ ( = (1mol ) 75.3 J mol !K = 0.358kJ +1451 !H = +2337 Mg+(g)+e-(g) +738 Mg(g) +148 Mg(s) (second ionization enthalpy) (first ionization enthalpy) (enthalpy of sublimation) Enthalpy is a state function: The enthalpy change of an overall process is the sum of the enthalpy changes for the steps (observed or hypothetical) into which it may be divided. 15 16 Combining Reaction Enthalpies HESS S LAW: Enthalpy is a state function: The enthalpy change of an overall process is the sum of the enthalpy changes for the steps (observed or hypothetical) into which it may be divided. Enthalpy is A state function. Combining several chemical changes C3H 6 ( g ) + H 2 ( g ) !! C3H 8 ( g ) " ! #H A = $124 kJ +5O2 !! C3H8 ( g ) + O 2 (g ) (g) " 3CO 2 (g ) + 4H 2O(" ) ! #H B = $2220 kJ H 2 O(" ) !! H 2 (g ) + " 1 O(g ) 2 ! #H C = 286 kJ 9 C3H 6 ( g ) + O 2 (g ) !! 3CO 2 (g ) + 3H 2O(" ) #H ! = $2058 kJ " 2 ! ! ! !H A + !H B + !H C = !H ! = "2058 kJ !H = 0 for a CYCLIC PROCESS (for any state function X, ΔX=0 for a cyclic process) !HBC 17 !HAB !HCA C B A A !HAB B !HBC !HCA C !HAB + !HBC + !HCA = 0 18 3 Lecture 22 Standard Enthalpies of Formation Small number of reactions give enough information to compute ΔH for many others FOR A STANDARD REACTION S ENTHALPY: o !H rxn = products # ni H o ,i " f reactants i # Formation Reactions: C(graphite) + ½O2(g) + 2H2(g) C(graphite) + ½O2(g) + H2(g) ½O2(g) + H2(g) n j H o, j f j Standard state: Standard thermodynamic conditions chosen for substances when listing or comparing thermodynamic data [1atm, 25 C] (usually) Denoted ! ΔH =ΔH under standard state conditions !H f! standard enthalpy for the formation of one mole of the substance from its constituent elements in their reference states 1 H 2 ( g ) + O 2 ( g ) ! ! H 2 O( ! ) " 2 Reference state: most stable state (phase and allotrope) under the standard state conditions #H f" = $286 kJ Define relative scale for enthalpy, where elemental reference forms have ZERO enthalpy of formation H 2 ( g ) ! H 2 ( g ) ; "H o = 0 kJ f •  Lattice energies difficult to measure Na+(g) + Cl-(g) ! NaCl(s) Max Born & Fritz Haber (1919) Na(s) + "Cl2(g) ! NaCl(s) –  Step 1: Sublimation of Na(s) Na(s) ! Na(g) !H1 = 108 kJ/mol –  Step 2: Dissociation of Cl2(g) " Cl2(g) ! Cl(g) !H2 = Cl—Cl disc. E=120 kJ/mol –  Step 3: Ionization of Na(g) Na(g) ! Na+(g) + e!H3 = IENa = 496 kJ/mol –  Step 4: Formation of Cl ion Cl(g) + e- ! Cl-(g) !H4 = EACl = -349 kJ/mol –  Step 5: Formation of NaCl(s) from ions Na+(g) + Cl-(g) ! NaCl(s) !H5 = -LE Methyl alcohol (CH3OH) ! formaldehyde (HCHO) 2CH3OH(aq) + O2(g) ! 2HCHO(aq)+ 2H2O(l) ΔH = ? ΔHf [CH3OH(aq)]=-245.9kJ ! CH3OH(aq) ;ΔHf =-245.9kJ ! HCHO(aq) ;ΔHf =-150.2kJ ! H2O(l) ;ΔHf =-285.8kJ ΔHf [HCHO(aq)]=-150.2kJ ΔHf [H2O(l)]=-285.8kJ From Formation Reactions: 2x(CH3OH(aq) ! C(gr)+½O2(g)+2H2(g); ΔHf =245.9kJ) 2x(C(gr)+½O2(g)+ H2(g) ! HCHO(aq); ΔHf =-150.2kJ) + 2x( ½O2(g) + H2(g) ! H2O(l); ΔHf =-285.8kJ) 2CH3OH(aq)+2C(gr)+O2(g)+2H2(g)+O2(g)+2H2(g)!2C(gr)+O2(g)+4H2(g)+2HCHO(aq)+2H2O(l) ΔH = 2x(245.9kJ) + 2x(-150.2kJ) + 2x(-285.8kJ) 2CH3OH(aq)+O2(g)!2HCHO(aq)+2H2O(l) ΔH = 2x(245.9kJ) + 2x(-150.2kJ) + 2x(-285.8kJ) ΔH =-2xΔHf [CH3OH(aq)]+2xΔHf [HCHO(aq)]+2xΔHf [H2O(l)] ΔH = - 2xΔHf [CH3OH(aq)] - 2xΔHf [O2(g)] + 2xΔHf [HCHO(aq)] + 2xΔHf [H2O(l)] ΔHrxn =Σn ΔHf (products)‒Σm ΔHf (reactants) 19 Lattice Energies from Born-Haber Cycle •  Standard Enthalpies of Formation (example) 20 Lattice Energies from Born-Haber Cycle Na(s) " Cl2(g) Na(g) Cl(g) + e+(g) + Cl-(g) Na ! Na(g) ;!H1 = 108 kJ/mol ! Cl(g) ;!H2 = 120 kJ/mol ! Na+(g)+e- ;!H3 = 496 kJ/mol ! Cl-(g) ;!H4 = -349 kJ/mol ! NaCl(s) ; !H5 = -LE ________________________________________________________________ Na(s) + " Cl2(g) ! NaCl(s) ; !Hf°[NaCl(s)] !H1+!H2+!H3+!H4+!H5=!Hf°[NaCl(s)] 21 108 kJ/mol +120 kJ/mol +496 kJ/mol -349 kJ/mol –LE = -411 kJ/mol 375 kJ/mol – LE = -411 kJ/mol 22 LE = 786 kJ/mol symbol denotes standard state Typically, 25oC, 1 atm o Bond Dissociation Energies and Enthalpies H 2 ( g ) ! H (g ) + H ( g ) !U bo = !H bo $ P!V Enthalpy is A state function. Bond enthalpies = !H $ P "VH $ VH 2 # % & o b = !H bo $ P " 2VH 2 $ VH 2 # % & ! "H b = 436.4 kJ mol-1 (T = 298 K ) Using ideal gas law "U = "H bo ! PVH 2 o b = 436.4 #10 J $ mol ! (101325 Pa)(22.41#10!3 m3 ) 3 !1 = 434 #103 J $ mol !1 = 434kJ $ mol !1 = !H bo $ PVH 2 Note: D0 is the amount of energy necessary to dissociate the bond, ΔUb Ground vibrational state 23 24 4 Lecture 22 Using Bond Enthalpies Average Bond Enthalpies H 2 O( g ) ! 2 H ( g ) + O( g ) "H ! = 927 kJ mol#1 CH 4 ( g ) + 2O 2 ( g ) ! CO 2 ( g ) + 2 H 2O( g ) The !H for the atomization of water is NOT exactly twice the O-H bond enthalpy OH ( g ) ! H ( g ) + O( g ) "H ! = 428 kJ mol#1 H 2O( g ) ! H ( g ) + OH ( g ) "H = 499 kJ mol (-) Forming bonds releases energy ! ! ! $ 2 " #H b (C=O) + 4 " #H b (OH ) & % ' ! ! 4 ! "H b (CH ) + 2 ! "H b (O=O) 4 H + C + 4O In fact, the O-H bond enthalpy is different for H2O than for OH: ° (+) Breaking bonds takes energy o %H ro = ! 4 $ %H bo ( C # H ) + 2 $ %H b (O = O )" & ' # ! 2 $ %H bo ( C = O ) + 4 $ %H bo (O # H )" & ' -1 The enthalpy of a particular bond depends on the molecule. However, one can define a mean enthalpy value for a particular kind of bond (and use this to estimate a reaction enthalpy) !H b (O-H ) = 463 kJ mol-1 " &H ro = ! 4 # 414 kJ + 2 # 499 kJ mol mol ' % " $ ! 2 # 799 kJ + 4 # 460 kJ mol mol ' % !H ro = "784kJ ΔH 25 Estimate !H for the following reaction using the bond energy table provided (all units are kJ/mol) 2NH3 ! N2 + 3H2 Σ B.E.(Broken) ‒ Σ B.E.(Formed) Compare to: !H ro = "802.2kJ from the standard enthalpies of formation 26 Using Bond Enthalpies •  Can use B.E. to estimate !H Can use the table to generate estimates for reaction enthalpies + ! •  Break 1 C=C bond & 1 Cl-Cl bond •  Form 1 C-C bond, 2 C-Cl bonds !H = # B.E.(Broken) – # B.E. (Formed) !H = B.E.(C=C) + B.E.(Cl-Cl) – B.E.(C-C) – 2xB.E.(C-Cl) !H = 602kJ/mol + 240kJ/mol – 346kJ/mol – 2x327kJ/mol !H = -158 kJ/mol 27 28 5 ...
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This note was uploaded on 01/19/2012 for the course CHEM 260 taught by Professor Staff during the Fall '08 term at University of Michigan.

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