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Exam 2 SG: 5, 6, 7 notes

by: Dkrefft

Exam 2 SG: 5, 6, 7 notes CHEM-1070-40

Marketplace > Tulane University > Chemistry > CHEM-1070-40 > Exam 2 SG 5 6 7 notes
GPA 4.0
General Chemistry I
Lopreore, Courtney

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About this Document

Notes from chapters 5, 6, and 7 all wrapped up in one.
General Chemistry I
Lopreore, Courtney
Study Guide
lopreore, Gen Chem
50 ?




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This 2 page Study Guide was uploaded by Dkrefft on Monday October 19, 2015. The Study Guide belongs to CHEM-1070-40 at Tulane University taught by Lopreore, Courtney in Fall 2015. Since its upload, it has received 62 views. For similar materials see General Chemistry I in Chemistry at Tulane University.


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Date Created: 10/19/15
1012 First law of thermodynamics 0 Internal energy u KEPE 0 Includes translational KE of molecules energy from rotation and vibration energy stored in chemical and into molecular attraction and energy associated with electrons in atoms 0 Heat and work are means by which a system exchanges energy with surroundings Relation between heat q and work w and changes in internal energy AU is dictated by the law of conservation of energy AUqw o AU0 in an isolated system 0 AU system AU surroundings state function a property that has unique value for a specified state of a system ie density of H20 at 20 deg C and 100kPa 099820gmL o AUU2U1 0 AU is not path dependent but work is 2 stage expansion w 124e2 J w2PV the work for each expansion 2 the sum of W 180atm136102L 120atm204136L 061Latm 082Latm 143Latm 143Latm x w 144eZ 1 Latm wlt0 qlt0 heat is given off wgt0 qgt0 heat enters reactants Ui 999 products Ur AU Uf U qrxn w at a constant volume w PAV0 AUqrxn there has to be a volume change for work to be done Constant pressure and volume AU qp w qv 0 Q qp W O AUqp 39 o qp AU PAV these are all state functions Define another state function H enthalpy o HUPV 0 AH Hf Hi UfPFVF Ui PM 0 AH Uf Ui Per PiVi 0 AH AU APV at a constant temperature and pressure Pi Pf 0 AH AU PAV 0 AH qp p being heat flow 0 PAV RTnr hi this only works for gases How much heat is associated with the complete combustion of 100kg of 012H2201 1 AH 565e3 kgmol 1000 g 1 mol 100k ClZH22011 g x 1kg x3423g AH 292565e3 kgmol 165e4 KJ Enthalpy of vaporization the heat required to vaporize a fixed amount of a liquid So it39s liquid to gas fusion is solid to liquid 292m0l 0 Consider one mole of H20 I 99 H20g AH 440 KJ at 298K so the AH is AHvap 0 H20 s 99 H20l AHfus 601 KJ at 27315 K 0 Calculate AH for the process in which 500g of H20 is converted from liquid at 100 deg C to vapor at 250 deg C 2 steps 1 Raising the temperature of liquid from 100 C to 25 C 2 Completely vaporizing at 250 C AH is the sum of those steps 418 1 500g H20 250 100 314K mass x SH C x temp change x conversion 2 Hvap 40 500g x 122 K 30 delta H 122 K 314K 01 12514K 0 AH degrees for standard temp and pressure of reaction 29815K and 1 atm unless told othenNise 0 AH via Hess s law Consider N2 g 02 g 9 2N0g AH18050 KJ To express in terms of 1 mole of N0 divide all by 2 o 12 N2g 12 02g 9 N0g AH 9025KJ o if you need to reverse the reaction change the sign Whatever you do to the coefficients you do to the AH 10I14 Hess s law of constant heat summation 12 N2g 12 029 9 N02g AH 0 We can use 12 N2g 02g 9 N0g 12 02g AH 9025KJ o N0g 12 02g 9 N02 g AH5707 KJ o 12 N2 g 02 g ILlgtL0gJIH1602g 9 N0g44 02 g N02g So 12 N2 g 02 g 9 N02 g AH 332 KJ AHf enthalpy change that occurs in the formation of one mole of a system in standard state from the reference forms of the elements in their standard states The AHf for a pure element is 0 o AH ZVpAHf ZVpAHf 0 you look up AHfin the appendix 0 C2H6 g 7I2 02 g 9 2 C02 g 3 H20 l AH 2 mol C02 x AHfC02g 3 mol H20 l x AHf H20 1 mol C2H6 x AHfC2H6g 72 mol 02 x AHf 02 g C02 g AHf 3935 KJ mol H20 l AHf 2858 KJmol C2H6 g AHf 847 KJmol 02 g AHf 0 2 mol 3935 KJmol 3 mol 2858 KJmol 1 mol 847KJmol 15597 KJ O O


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