#Chem 161-2007 recitation 7th week

#Chem 161-2007 recitation 7th week - CHEMISTRY 161-2007 7TH...

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CHEMISTRY 161-2007 7 TH WEEK RECITATION ANNOUNCEMENTS E-MAIL ATTENDANCE EXAMS Recitation quiz II Wed, Oct. 17 th , Chapter 5.5 – 6.4 Thu, Oct. 18 th , Chapter 5.6 – 6.7 Location of hourly exams II and III Sections 23, 25, 29 (and 17)- Hck 101 Exam II Date: 10/24; 9:40 – 11:00 Location: Sections 23, 25, 29 (and 17)- Hck 101 Review Session: Wednesday, Oct. 17 th , 8-10 PM, Ruth Adams Bldg. 001 Monday, Oct. 22 nd , 8-10 PM, Ruth Adams Bldg. 001 Wednesday’s session will be the first half; Monday’s session will be the second half. At this point I don’t know exactly where the dividing mark will be. Chem 161-2007 recitation 7 th week 1
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PLAN FOR TODAY : CHAPTER 6: • HEAT AND WORK • PROPERTIES OF ENTHALPY • CALORIMETRY AND HEAT CAPACITY • HESS’S LAW • STANDARD ENTHALPIES OF FORMATION Chem 161-2007 recitation 7 th week 2
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CHAPTER 6 RELEVANT EQUATIONS ∆E = ΔU = q + w 1 Latm = 101.3 J w = -PΔV; therefore, ∆E = q –P∆V Heat Capacity = C = q/ΔT molar heat capacity = C/mol = q/(ΔT x mol) Specific Heat Capacity = SHC = SH = C/g = (q/ΔT)/g = q/ΔTg q = SHC x ΔT x g First law of thermodynamics: Energy is neither created nor destroyed. Therefore, heat lost by one = the heat gained by other. -q 1 = +q 2 -SHC 1 x ΔT 1 x g 1 = +SHC 2 x ΔT 2 x g 2 Hess’s Law : ΔH is a state function, so path not important. Rules for changing enthalpy equations: o If the enthalpy equation is reversed, sign of ΔH is reversed. o If the enthalpy equation is halved, quantity of ΔH is halved. o If the enthalpy equation is doubled, quantity of ΔH is doubled. Enthalpy of Formation = ΔH o f o is for one mole of compound/product o is from its elements o all elements in standard state*, e.g., oxygen = O 2(g) ; mercury = Hg (l) ; sodium = Na (s) o ΔH o f of elements = 0 * working definition, not actual definition Enthalpy of reaction : o ΔH o Rx = ∑n p ΔH o f,products - ∑n r ΔH o f,reactants Chem 161-2007 recitation 7 th week 3
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ENTHALP Y OF REACTION ET: Define exothermic and endothermic reactions. HYPOTHETICAL ENDOTHERMIC REACTION 200kJ 300kJ 500kJ 100kJ A + B C + D ∆H =(500+100) – (200+300)=+100kJ 1 st Law of Thermodynamics: Energy can neither be created nor destroyed. Therefore: 200kJ 300kJ 500kJ 100kJ A + B + 100kJ C + D Therefore, positive ∆H corresponds to an endothermic reaction. HYPOTHETICAL EXOTHERMIC REACTION 600kJ 100kJ 300kJ 100kJ E + F G + H ∆H =(300+100) – (600+100)=-300kJ 1 st Law of Thermodynamics: Energy can neither be created nor destroyed. Therefore:
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This note was uploaded on 04/12/2008 for the course CHEM 161 taught by Professor Vacillian during the Fall '08 term at Rutgers.

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#Chem 161-2007 recitation 7th week - CHEMISTRY 161-2007 7TH...

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