Chem 161-2010 Lecture 12(1)

Chem 161-2010 Lecture 12(1) - CHEMISTRY 161-2010 LECTURE 12...

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CHEMISTRY 161-2010 LECTURE 12 ANNOUNCEMENTS E-MAIL ATTENDANCE EXAMS Chem 161-2010 Lecture 12 1
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PLAN FOR TODAY : • STANDARD ENTHALPIES OF FORMATION • HISTORY OF THE STRUCTURE OF THE ATOM Chem 161-2010 Lecture 12 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 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 161-2010 Lecture 12 3
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Chem 161-2010 Lecture 12 4
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Enthalpy of Formation – Simple Enthalpy of Formation = ΔH o f ET: Two ways already studied for ∆H (1) Direct measurement (calorimetry), (2) Hess’ Law. Third way: ΔH o f Simply look it up. ET: Reference analogy for ΔH o f is measuring heights of mountains vs. sea level ET: Define standard state: Room temperature (298 o K) and 1 atmosphere pressure Standard enthalpy of formation of a compound: the change in enthalpy that accompanies the formation of one mole of a compound, from its elements, with all substances in their standard states. e.g., C( s ) + 2H 2 ( g ) + ½ O 2 ( g ) → CH 3 OH( l ) ∆H o f = -219 kJ/mol o Note that we never really know the “H” of a substance; we only know the difference (change) in enthalpy between two substances; hence “ΔH”. o ΔH o f is for one mole of compound/product o ΔH o f is from its elements • all elements in standard state*, e.g., oxygen = O 2 (g); mercury = Hg( l ); sodium = Na(s) • ΔH o f of elements = 0 * working definition, not actual definition Chem 161-2010 Lecture 12 5
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23 Chem 161-2005 Hourly Exam II Chapter 6 - Thermodynamics Enthalpy of formation For which one of the following equations does H correspond to
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This document was uploaded on 11/02/2011 for the course GEN CHEM 162 at Rutgers.

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Chem 161-2010 Lecture 12(1) - CHEMISTRY 161-2010 LECTURE 12...

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