{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Chapter 5,9 and 10 Notes

Chapter 5,9 and 10 Notes - NOTE To begin CH302 I recommend...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
NOTE: To begin CH302, I recommend strongly you make sure you are comfortable with all the 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 ' 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 (some of these can be measured absolutely, others cannot – we only find the change in them Non - State Functions: DEPENDENT on route: w, q Change in Enthalpy, ' H Definition: Heat flow for process occurring at CONSTANT PRESSURE ' H = qp positive ' H = ENDOTHERMIC negative ' H = EXOTHERMIC Thermochemical Equations balanced chemical reaction plus ' H value for reaction as given : Example: C 5 H 12(l) + 8 O 2(g) o 5 CO 2(g) + 6 H 2 O (l) ' H° = - 3523 kJ "per mole of the reaction" - i.e., the numbers of moles shown in reaction A “°” in the “ ' 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: (NOTE: We’ll use this a lot later on!!) 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 ' H q 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 ' H q f ! The reaction can be reconstructed if needed via above rules REMEMBER ' H q 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 ' ' ¦ ¦ n n Using Hess' Law to Find an Unknown ' If ' H q 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 ' H q rxn for the following reaction: C 3 H 8(g) + 5 O 2(g) o 3 CO 2(g) + 4 H 2 O (l)
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Chapter 5 Highlights Dalton’s Law of Partial Pressures: (see fig 5.9) Mixture of gases in a container: Each gas exerts its own partial pressure as if there were no other gas present .
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

Page1 / 6

Chapter 5,9 and 10 Notes - NOTE To begin CH302 I recommend...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon bookmark
Ask a homework question - tutors are online