Physical Chemistry I For Biochemists Chem340 Lecture 14

Physical Chemistry I forBiochemistsChem340Lecture 14 (2/14/11)Y hit k IIshiiYoshitakahii Ch4.1-4.4Announcement Exam 1 will be returned on Friday.pat the web site. This HW5 is uploadedshould be easier. Exam 2 will be held 3/18 Right beforeSpring Break ( 4 weeks from now)1

Ch 4.1 Energy Stored in Chemical Bonds isRelated or Taken Up in Chemical Reactions Significant amount of internal energy orenthalpy is stored in the form of chemicalbond.1/2N2(g) q(472 J) N(atom, g)1/2H2(g) q(218 J) H(atom, g)Q. What is the formation energy for NH3? 2

Which of H and U Should We Use toCharacterize Chemical Reactions? Since most of reactions are carried out at aconstant ppressure,, we use H rather than U tomeasure the change in the energy associated withchemical reaction.Constant Volume UConstant Pressure HFor an ideal gas, H (U PV) U (nRT) U RT n(see Example 4.2 p69; Sec 4.5)(if T const)Standard Partial Molar Enthalpy(Raff p105)For a molecule A, dHA CpAdT ( H/ P)dP( H/ P)dP H A dnd A H is the partial molar enthalpy of A. H A n A T ,P At the standard condition (T 298.15 K, P 1bar)dHA H A dnA HA H A nAIf we have more than one type of molecule, H n AH A nB H B nC HC . n X H XX3

Reaction Enthalpy using PartialMolar Enthalpy For a reaction that takes place at constant P and T,the heat flow needed for a chemical reaction is Hreaction. AA BB XX YY Hreaction X H X Y H Y - A H A- B H B H reaction iPr oduct H i kRe ac tan t H ki Xk A In general, it is difficult to find partial molarenthalpy. We use formation enthalpy.Standard Formation Enthalpy Consider a reaction from pure elements that are moststable at 298.15 K and 1 bar, which results in theformation of one mole of a molecule of your interestinterest. C(s, graphite) O2(g) CO2(g) q(393.5 kJ) 1/2N2(g) 3/2H3 NH3 (g) q(45.9 kJ)Note, the only product should be the molecule of your interest. AAssume thatth t q isi theth heath t generatedt d forf theth reactionti whenh P 1 bar and T 298.15K (Standard State). The energy isdescribed as a standard formation enthalpy Hf0 for CO2(g)and NH3(g) as Hf0 for CO2 393.5 kJmol-1 Hf0 for NH3 45.9 kJmol-14

Standard Formation Enthalpy Listed inAppendix B Table 4.1You can tell which compound is more stable!Reaction Enthalpy usingStandard Formation Enthalpy For a reaction that takes place at constant P and T,the heat flow needed for a chemical reaction is Hreaction. AA BB XX YY0000 Hreaction [Q1]X HfX Y HfY - A HfA - B HfB In general, H reaction iPr oduct H f0,i kRe ac tan t H f0,kori X H reaction i Hk A0f ,ii(Note use I 0 for products and I 0 for reactants)5

Reaction Enthalpy from StandardFormation Enthalpy (Example) C(s, graphite) O2(g) CO2(g) q(393.5 kJ) H reaction H 0f (CO2 ( g )) H 0f (O2 ( g )) H 0f (C ( s, graphite)) H 0f (CO2 ( g )) 393.5 kJ 1/2N2(g) 3/2H2 NH3 (g) q(45.9 kJ) H reaction H 0f ( NH 3 ( g )) (1 / 2) H 0f ( N 2 ( g )) (3 / 2) H 0f ( H 2 ( g )) H 0f ( NH 3 ( g )) 45.9 kJ Hf0 for CO2 393.5 kJmol-1 Hf0 for NH3 45.9 kJmol-1Sample Question Write reactions to obtain standard formation enthalpyof (a) O2(g), (b) Hg(l), (c) Hg(g), (d) CO2(g), (e) H2O(l)(f)C12H22O11(s)(a) O2(g) O2(g) q(0 kJ) Hf0 (O2(g)) 0 kJ(b) Hg (l) Hg(l) q(0 kJ)(c) Hg(l) Hg(g) q(-61.4 kJ) Hf0 (Hg2(g)) 61.4 kJ(d) C(s, graphite) O2(g) CO2(g) q(393.5 kJ)(c) H2(g) 1/2O2(g) H2O(l) q(285.8 kJ)(d) 12C(s) 11H2(g) 5.5O2 C12H22O11(s) q(2226.1 kJ)6

QuizHow much is the reaction enthalpy for the followingreactions at T 293.15K and P 1 bar? H2(g) 1/2O2(g) H2O(l) q(285.8 kJ)- 285.8 kJ Hreaction0 [Q1] 2H2(g) O2(g) 2H2O(l) Hreaction0 -[Q2]285 8 kJ x 2 -571.6285.8-571 6 kJ How much is the standard formation enthalpy forH2O(l)? Hf0 (HO2(l)) -[Q3]285.8 kJSample QuestionEx. 4.1 (p67) Hf0(C12H22O11(s)) -2226 kJmol-1 Hf0(CO2(aq)) -412.9412 9 kJmol-11 Hf0(H2O (l)) -258.8 kJmol-1 How much is the reaction enthalpy for the followingreaction?C12H22O11(s) 12O2(g) 12CO2(aq) 11H2O(l) Hreaction XHfX0 YHfY0 - AHfA0 - BHfB0 12Hf0(CO2(aq)) 11Hf0(H2O (l)) Hf0(C12H22O11(s)) Hf0(O2(g)) 12x(-412.9kJ) 11x (-258.8kJ) (-2226 kJ) -5873kJ7

Hess’s Law is Based on EnthalpyBeing a State Function How can we Hf0 calculate from known reactionenthalpy Hreaction? For exampleis the reaction to obtain Hf0(urea(s)), but thisreaction does not happen readily. Instead, we useQ. How to obtain Hf0(urea(s))?Solution8

Alternative SolutionAxBxCxDx(B)(D/2)(C)(7/2A B C/2 D) (-A) -A 1Bond EnthalpySample Question (Example 4.2) The average bond enthalpy of the O-Hbond in water is defined as one-half of theenthalpy change for the reaction H2O.H2O(g) q(927.0 kJ) 2H(g) O(g)What is the bond enthalpy obtained from theg reactions for N2 and H2?followingN2(g) q(944 J) 2N(atom, g)H2(g) q(436 J) 2H(atom, g)Q. How about the bond for NH3(g)?9

Q. What is the average bond energy for NH3? 4.4 The Temperature Dependenceof Reaction EnthalpiesSuppose we would like to perform a reactionatt a higherhi h ttemperaturetthanth 298298.15K,15K thethenthalpy can be calculated fromT H0T H0298.15 K C P (T ' )dT '298.15 KFor a reactionT H0reaction ,T H0reaction , 298 .15 K CP(T ' )dT '298 .15 K C P (T ' ) j C p, j (T ' )10

Molar mass of sucrose 342 gTwo Type of Reactions andReaction EnthalpyExothermic Reaction: Heat is generatedafterft a reaction.ti(T iis iincreased).d)A B X Y q( 0) H 0Endothermic Reaction: Heat is needed for areaction (T is decreased).reaction.decreased)A B q X Y H 011

Bond EnthalpySample Question (Example 4.2) The average bond enthalpy of the O-Hbond in water is defined as one-half of theenthalpy change for the reaction H2O.H2O(g) q(927.0 kJ) 2H(g) O(g)What is the bond enthalpy obtained from theg reactions for N2 and H2?followingN2(g) q(944 J) 2N(atom, g)H2(g) q(436 J) 2H(atom, g)Q. How about the bond for NH3(g)?Q. What is the average bond energy for NH3? 12

Bond Energy: U for Reactions for aBond Formation (P69) Hbond0 Hreaction0/2 Hreaction0 2 Hf0(H(g)) Hf0(O(g)) Hf0(H2O(g)) 218.0 x 2 kJ 249.2 kJ (-241.8 kJ) 927 kJ Ureaction0 Hreaction0 – (PV) Hreaction0 – RT (n) 927 kJ – 2RT 922kJQ. What is n? n 24.4 The Temperature Dependenceof Reaction EnthalpiesSuppose we would like to perform a reactionatt a higherhi h ttemperaturetthanth 298298.15K,15K thethenthalpy can be calculated fromT H0T H0298.15 K C P (T ' )dT '298.15 KFor a reactionT H0reaction ,T H0reaction , 298 .15 K CP(T ' )dT '298 .15 K C P (T ' ) j C p, j (T ' )13

HW5 Q1 Using dHA CpAdT ( H/ P)T, ndP ( H/ n)T, PdnA (A B, X, or Y), prove that the reaction enthalpy Hreaction(T)is given by eq. (4.27) in the text. Specify the path of (P,T n) you would like to useT,use. Assume a reaction 2B C X 3Y.T H0reaction ,T H0reaction , 298 .15 K CP(T ' )dT '298 .15 K((1)) 2B(T)( ) C(T) ( ) 2B(298K)() C(298K)() p,B[Q1] H(1) (C Cp,C )(298 K – T) -(Cp, B Cp, C)( T)0[Q2](2) B(298K) X(298K) Y(298K) H(2) Hreaction,298K(3) X(298 K) Y(298 K) X(T) 3Y(T) H(3) (C[Q3]p, X Cp,Y)( T)4.5 The Experimental Determination of Uand H for Chemical ReactionsNo heat flow q 0Bomb Calorimeter for Reactioninvolving gas and volatile chemicals Uoverall 0 U overall mS U reactionMSmH 2OCH 2O,m TMH 2O Ccalorimeter TMeasuredFor ideal gas, H U (PV) U RT( n)V Const for the container14

Standard Partial Molar Enthalpy (Raff p105) For a molecule A, dHA CpAdT ( H/ P)dP dnA is the partial molar enthalpy of A. At the standard condition (T 298.15 K, P 1bar) HA A T P A n H H, dHA dnH