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Final_Review

Course: CHEM 6al, Spring 2012
School: UCSB
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Course Hero has millions of student submitted documents similar to the one below including study guides, practice problems, reference materials, practice exams, textbook help and tutor support.

= Review M+ 78, 1 chlorine atom IR: 2921 1450 650 13C NMR: 1H NMR: 63.8 32.1 20.5 3.30, t, 2H 1.59, sextet, 2H 0.89, t, 3H Cl Cl Cl B A C Cl Cl D E M+ = 78, 1 chlorine atom Answer = A IR: 2921 1450 650 13C NMR: 1H NMR: 63.8 32.1 20.5 3.30, t, 2H 1.59, sextet, 2H 0.89, t, 3H Cl Cl Cl B A C Cl Cl D E M+ = 78, 1 chlorine atom IR: 2921 1450 650 13C NMR: 1H NMR: 72.1 25.3 3.84, septet, 1H 1.01, doublet, 6H Cl Cl Cl B A C Cl Cl D E M+ = 78, 1 chlorine atom Answer = B IR: 2921 1450 650 13C NMR: 1H NMR: 72.1 25.3 3.84, septet, 1H 1.01, doublet, 6H Cl Cl Cl B A C Cl Cl D E M+ = 74, 1 oxygen atom IR: 35003300 br 2974 1202 13C NMR: 1H NMR: 72.1 25.3 2.0, singlet, 1H 1.21, 9H singlet, OH OH OH B A OH D C O E M+ = 74, 1 oxygen atom Answer = D IR: 35003300 br 2974 1202 13C NMR: 1H NMR: 72.1 25.3 2.0, singlet, 1H 1.21, singlet, 9H OH OH OH B A OH D C O E M+ = 74, 1 oxygen atom IR: 35003300 br 2974 1202 13C NMR: 1H NMR: 64.6 32.1 31.7 22.8 3.74, sextet, 1H 2.3, singlet, 1H 1.47, quintet, 2H 1.12, doublet, 3H 0.96, triplet, 3H OH OH OH B A OH D C O E M+ = 74, 1 oxygen atom Answer = B IR: 35003300 br 2974 1202 13C NMR: 1H NMR: 64.6 32.1 31.7 22.8 3.74, sextet, 1H 2.3, singlet, 1H 1.47, quintet, 2H 1.12, doublet, 3H 0.96, triplet, 3H OH OH OH B A OH D C O E UCLA Website has many good problems and solutions http://www.chem.ucla.edu/~webspectra/

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UCSB - CHEM - 6al

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UCSB - CHEM - 6al

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UCSB - CHEM - 113b

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UCSB - CHEM - 113A

Chemistry 113A Fall 2011Lecture: TR 11AM 12:15PM1701 Theater and DanceInstructor: Prof. Steve Buratto4148B Chemistryburatto@chem.ucsb.eduOffice Hours: MTW 9:00 - 10:00 AM (or by appointment)Course TA: Taylor Judkinstjudkins@chem.ucsb.eduOffice Ho

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UCSB - CHEM - 113A

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UCSB - CHEM - 113A

or for infinitesimal changes work of expansion/compression (aka pV work)closed system (const. n) reversible implies equilibrium transformation and weuse the pressure inside the cylinder=If we know the equation of state we can integrateand solve for

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UCSB - CHEM - 113A

Heat Capacityandconstandconst Enthalpy (H)and for an ideal gas the change in internal energy heat capacities are extensive (i.e. depend on n) intensive heat capacities (per mole quantity) are writtenand for ideal gases we can show that the ra

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UCSB - CHEM - 113A

Adiabatic means no heat (q=0) dU = dwad =-pdVIf we have an ideal gas dU = CVdT = dwadThus, for an ideal gas CVdT = -pdVAs seen from the graph,p, V and T all changeSince we have a closed system(constant n), the variables p, Vand T are interdependen

Lecture6

UCSB - CHEM - 113A

are the variables for a van der Waals gas state functions?dp is exact and p is a state function(so are V and T, proof left up tostudents) now consider the exact differential for an ideal gas (from Thurs. 10/6/11) for a van der Waals gas these equa

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UCSB - CHEM - 113A

similar to what we did for the internal energy we examinethe quantity (H/p)T for an ideal gas from Tuesdays lecture we showed that for an ideal gas using the product rule for differentiationfor an ideal gas using the definition of enthalpyfor an id

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UCSB - CHEM - 113A

nd2nd Law of Thermodynamics (Kelvin-Planck formulation)It is impossible for a system to undergo a cyclic processwhose sole effects are the flow of heat into the systemfrom a heat reservoir and the performance of anequivalent amount of work by the sys

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UCSB - CHEM - 113A

Consider the system and the surroundings.andFor any real transformation (i.e. spontaneous process)This is our first condition for equilibrium (and spontaneity)!We had shown earlier that for a reversible transformationSince U is a state function the a

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UCSB - CHEM - 113A

A = U - TSThis is our second condition for equilibrium (and spontaneity)!Note the function A depends only on the system variables.If we consider our transformation at constant T (alreadyassumed) and constant V in the limit of only pV work (dV = 0 dw

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UCSB - CHEM - 113A

Consider the pressure dependence firstandconstant Tandintegratep = 1 barideal gasConsider the temperature dependence of G as followsGibbs-Helmholtz EquationWe can create a more useful expression by using the chain ruleset up to integrateConside

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UCSB - CHEM - 113A

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UCSB - CHEM - 142A

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UCSB - CHEM - 150

Chem 150First Midterm Exam_(3 points)Name (Please Print)_(2 points)Perm #1I. Multiple Choice (2.5 pts each)1.The equilibrium constant of a reaction depends on:(a) The concentrations of the reactants(b) The concentrations of the products(c) The

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UCSB - CHEM - 113C

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UCSB - CHEM - 113C

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UCSB - CHEM - 113C

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UCSB - CHEM - 113C

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UCSB - CHEM - 113C

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UCSB - CHEM - 113C

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UCSB - MCBD - 1B

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Chapter 9