CH302 Notes - NOTE To begin CH302 I recommend strongly you...

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material covered in Chapters 2-4. The first part of these notes are some material from Chapter 5,9 and 10 that should have been covered last semester, and you should make sure you are familiar with them. You should also make sure you are familiar with common thermodynamic terms like exothermic, entropy, etc., and how to interpret them, sign-wise. . e.g., if Δ H is negative, what does that tell you? Chapter 9 Highlights: State Functions: INDEPENDENT of route: P , V , T, U (or E), H, S, G Non - State Functions: DEPENDENT on route: w, q These state functions CAN be measured absolutely: P , V , T, S These state functions CANNOT be measured absolutely: U (or E), H, G We can only find the change in these functions! Change in Enthalpy, Δ H Definition: Heat flow for process occurring at CONSTANT PRESSURE Δ H = qp positive Δ H = ENDOTHERMIC negative Δ H = EXOTHERMIC Thermochemical Equations balanced chemical reaction plus Δ H value for reaction as given : Example: C 5 H 12(l) + 8 O 2(g) 5 CO 2(g) + 6 H 2 O (l) Δ H° = - 3523 kJ "per mole of the reaction" - i.e., the numbers of moles shown in reaction A “°” in the “ Δ H°” indicates a Standard Thermochemical Reaction : ALL products and reactants are in their STANDARD STATES. These are: P = 1.0000 atm and OFTEN (but not always) T = 298.15 K (25°C) Thermochemical standard states of matter: Elements: most stable state at 298.15 K and 1.00 atm (know these!) Pure substances (liquid or solid) pure liquid or solid. G ases (pure) the gas at 1.00 atm G ases (mixture) partial pressure* of the gas of 1.00 atm Aqueous solutions : 1.00 M concentration Standard Molar Enthalpy of Formation Δ H ° f For a very specific balanced chemical reaction : - ONE MOLE of SINGLE PRODUCT in specified state - REACTANTS are ALL ELEMENTS- in STANDARD STATES - Don't need to quote equation, just give Δ H ° f ! The reaction can be reconstructed if needed via above rules REMEMBER Δ H ° f of ALL ELEMENTS in STANDARD STATES is DEFINED as ZERO e.g., C (s,graphite) , Br 2(l) etc. ts coefficien tric stoichiome n H n H n H 0 reactants f 0 products f 0 rxn = Δ Δ = Δ n n Using Hess' Law to Find an Unknown Δ If Δ H ° f values are available: See Fig 9.11. The idea is we “un-form” the reactants into their elements, then “form” the products from these elements. Example: Calculate
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This note was uploaded on 02/28/2010 for the course CH 52440 taught by Professor Sutcliffe during the Spring '10 term at University of Texas.

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CH302 Notes - NOTE To begin CH302 I recommend strongly you...

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