1-25-2008 - ! ! " "($ * + ! ) . ! 5 $ $...

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Unformatted text preview: ! ! " "($ * + ! ) . ! 5 $ $ 67) ! !" %& ( #$ ' !$ ) , - / 0 1+ -! 2 3 - 4 Chemistry 114 General College Chemistry II # $%' & #% ( & Instructor: Dr. Gregory Jursich Email: jursich@uic.edu BioE Office: 233 SEO Office: Phone: 312-413-8641 Office: 3235 SEL Office Phone: 312-996-3178 Lab: 1215 SEL Lab Phone: 312-996-8373 Office Hours: MWF 12:00 12:45 pm and 2:15-2:45 pm or by appointment Office Hour Location: 233 SEO Review from last time.... State properties of "system" Pressure, Temperature, Volume, Composition, Enthalpy H = qp " !% change of energy under constant pressure conditions )* + ! ) ! 8 4 ! Phase Changes: solid liquid Hmelting = Hliq Hsolid liquid gas Hvaporization = Hgas-Hliq Reactions: Hr = Hproducts Hreactants H independent of path & depends on amount Temperature Changes with Adding Heat Phase Changes: Solid -> Liquid -> Gas (A-B) (D-F) (H-J) TBP Hvap TMP Hfus Enthalpy Change with Chemical Changes: Hr = Hproducts Hreactants "independent of path" C(s) + O2(g) CO(g) or C(s) + O2(g) CO2(g) O2(g) CO(g) Enthalpy Change with Chemical Changes: Hr = Hproducts Hreactants "independent of path" Reaction C(s) + O2(g) or CO2(g) -393.5 kJ/mol C(s) + O2(g) CO2(g) CO(g) + O2(g) +283.0 kJ/mol C(s) + O2(g) CO(g) -110.5 kJ/mol H____ CO(g) -110.5 kJ/mol / , * 0 1+ # ! # -! . . & * # !% * # $ $ # %# # - / $ $ $ (1) Sn(s) + Cl2 (g) (2) Sn(s) + 2 Cl2 (g) SnCl2 (s)...... H = - 325.1 kJ/mol SnCl4 (l) ...... H = - 511.3 kJ/mol What is the enthalpy change for the reaction ? (3) SnCl2 (s) + Cl2 (g) SnCl4 (l) ...... H = ? / 0 1+ Reverse rxn 1 and add together: (1) SnCl2(s) Sn(s) + Cl2 (g) ...... H = - (- 325.1 kJ/mol) SnCl4 (l) ...... H = - 511.3 kJ/mol SnCl4 (l) ...... H = -186.2 kJ/mol (2) Sn(s) + 2 Cl2 (g) (3) SnCl2 (s) + Cl2 (g) Note: It's important to state the physical state of compounds / 0 1+ H2O(l) H = -285.8 kJ/mol Consider the reaction: H2(g) + O2(g) If Hvap of H2O is 44.0 kJ/mol, what is H or reaction H2(g) + O2(g) H2O(g) ? / 0 1+ H2(g) + O2(g) H2O(l) H2O(g) H2O(l) H = -285.8 kJ/mol H = +44.0 kJ/mol H2(g) + O2(g) H2O(g) H = -241.8 kJ/mol Reaction is less exothermic because there is more enthalpy in H2O(g) than in H2O(l) / 0 1+ H2O(g) H = -241.8 kJ/mol H2(g) + O2(g) So do we get 241.8 kJ/mol of energy from this reaction in shuttle launch? No. One reason is that the temperature of H2O(g) is very high temperature and temperature of O2(g) and H2(g) is very low Hr = Hproducts Hreactants H of reactants & products depend on temperature H = qp = mcs(T) $ $* 0 $ #$" " $ !% * $ ! * 0 - )1 4 - 2# $ - 3# % -! # # # # 2 4 - - # &* ! 5* 6 # # 8 /(:2 ( 4 9 ! *# ! ; $ #$ ! 7 * - $ $* 0 $ #$" 6 $ #$ # # 8 /(:2 ( 4 9 ! *# ! ; 7 * ! ! * 0 - )1 Hr0 = H0products - H0reactants ! : / -2 24 ) 8/ 29 2 2! - 0 1+ 2 ! 2 -! * # /2) ! 2 ) < $ - 24 ! - 0 ) 2 # $ #$= !# " !* ! # !% * # *# 6 $* # $ #$ & H0f 7 - $ 8 /(: 9 4 8 ; 3 # 2 - ; 3 2 ' Hr0 = c Hf0 (of C) + d Hf0(of D) a Hf0(of A) - b Hf0 (of B) 0 $ #$= !# " !% * # & H0f All H0 s are referenced to standard conditions of the stable elements of products and reactants 0 $ #$= !# " !% * # & H0f Using Hess's Law, H of any reaction can be calculated based on common reference: "stable elements" of reactants and products 0 $ #$= !# " )#1 * >24 !% * # 24 ? 24 @ & H0f # ! <# ' 7 < ' # < #8 ' Hf0 (kJ/mol) %$ )=? > %? > 76$ %7" 6 7> Hr0 = Hf0 (CaO(s)) + Hf0 (CO2) Hf0(CaCO3(s)) Hr0 = (-635.1) + (-393.5) - (-1207.6) = +179.0 kJ/mol 3 - ! 2 -2 @ ! 2! - 9 . 9 . # #; 4 #%4 23 > > 23 > ! > - -, ' 2 - ' # B.E.sreactants B.E.sproducts) (bonds broken) (bonds formed) - /A 8 )' #- / B 8 )' 3 Common reference species ! CH2Cl2(g) + 2 HF(g) Reaction: CCl2F2(g) + 2 H2(g) < C1 8 D Using average bond enthalpies (linked above), estimate the enthalpy change for the following reaction: 4NH3 (g) + 5O2 (g) Hrxn = 4NO(g) + 6H2O (g) BE(bonds formed) BE(bonds broken) - 12 x 391 = 4692 5 x 498 = 2490 -(4 x 607) = - 2428 -(12 x 463) = -5556 -802,000 or -803 kJ/mol ...
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