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Exam 2 SG: Notes from Ch. 5, 6, 7

by: Dkrefft

Exam 2 SG: Notes from Ch. 5, 6, 7 CHEM-1070-40

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

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About this Document

Instead of picking through the example problems, I just decided to give all of my notes from 5, 6, and 7 in one little bundle.
General Chemistry I
Lopreore, Courtney
Study Guide
lopreore, Gen Chem, General Chemistry, tulane
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This 9 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 184 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 pter 5 Notes 928 Cha Ions in aqueous solution conduct electricity so they re called electrolytes Strong electrolytes substance completely ionizes in aqueous solution breaks apart NaCl MgCl2 o MgClzaq gt Mgzaq 2610161 Weak electrolytes only partially ionized in solution 0 CH3C00H aq lt gt Haq CHgCOO aq o acetic acid acetate 0 If it s a weak electrolyte draw a double arrow to show that it s reversible Bracket notation 0 Symbol concentration In 00050M HCI HClaq gt Haq Cl aq Haq 00050 M 01 aq 00050M MgCizaqi 00050M MgClzaqf 00M Mg2aqf 00050M Cl aqf 00100M Strong reactions What are Al3 and 5042 in 00165M ofAl2 5003 Write equation for disassociation o 14125043 gt 2Al3 35042 Stoichiometry O O O O O O 3 00165molAlzSO43 2molAl3 3 0 Al 1L 1molAlzSO43 00330 MAI aq 2 00165 molAlzSO43 3mol042 2 o 504 L 1molAlzSO43 00495 M 504 aq precipitation reactions cations and anions combine to form a solid that is insoluble Net ionic equations 0 AgN03aq Na aq gt Ag 3 NaN03 aqso AgI precipitant strong electrolytes Ag N 03A g1 are represented by their separate ions 0 Agaq4 93 aq Nai aq I aq gtAg1s Nai aq Negleaqa get rid of the spectator ions ions that appear on both sides 0 Net ionic equation Agaq I aq gt AgIs Solubility rules 0 Type them here Suppose Precipitant or no reaction A gN03 aq KBraq o Agaq N03 aq K aq Br aq o Agaq W Br aq gt AgBrs l Ki N93 o Agaq Br gt AgBrs Indicate whether a precipitant forms K2504aq FeBr3 aq 0 2K 002 Fe 331 gt FeSO42 KBrso no reaction acid base reactions 0 know the strong acids and bases p 161 0 weak acid has a weak tendency to produce H in a solution 0 Acids produce H in a solution proton donors 930 9999p Bases produce OH in solution proton recievers CH3C00H aq lt gt Haq CH3C00 aq weak HCl gt Haq Cl aq HCi H20 gt 1130 Cl NH3 H20 lt gt NH4 0H H20 1 ltgt Haq 0H aq H OH 10e 7 at 25 degrees Celsius H gt 10e 7 means it s acidic OOOOOOOO Neutralization Reaction 0 Acid base water salt 0 HCl aq NaOH aq gt NaCL aq H20 l 0 Strong They come apart completely in solution and have the ability to cancel H aq eieeiq9 Nai 0H39aq gt Naieeiqaar epeeq H20 1 o H OH gt H20 l to recognize acids v bases ionizable H atoms mean it s an acid HCZH302 or CH3600H Redox reductionoxidation 0 When substances gain 0 atoms oxidation 0 Lose 0 atoms Reduction 0 They always come in pairs When no oxygen is involved 0 The oxidation state of an element increases as electrons are lost 0 Reduction is if the oxidation state of an element decreases Mg02s 4H ZCl gt Mn2 H20 l Cl2g Mn is reduced and Cl is oxidized Zns Cu2aq gt Zn2aq Cu 3 oxidation Zns gt Zn2aq 2e reduction Cu2aq 2e gt Cus Make sure the electrons are equal and on opposite sides so that they cancel out Steps to balancing a redox reaction in an acidic solution Write half reactions For each reaction balance all atoms except for H and O by inspection Balance oxygen by adding H20 Balance the hydrogen by adding H Balance the charges by using electrons Equalize the number of electrons in the redox half reactions by multiplying one or both half reactions by the appropriate integer to make sure the electrons cancel out Add the half reactions and cancel the species common to both sides Check the number of atoms and that the charges balance Balance this in acetic solution 503239aq Mn04aq gt 504239aq Mn2aq Oxi 5503239aq H20 gt 502aq 2H 2e39 Red 2Mn04aq 8H Se gt Mn2aq 4H20 5503239aq 5H20 gt 55042aq 10H 10e 2Mn04aq 16H 10e gt 2Mn2aq 8H20 O O O O O PwNe o remember that the goal is to get the electrons to cancel 0 55032 441429 2Mn04 6H 0e gt 55042 1 9Hi 1 0e 2Mn2 3H2 O we reduced the H and the water and cancelled the electrons 0 55032 6H gt 55042 2Mn2 3H20 0 make sure the atoms and charges balance Fe2aq Mn04aq gt Fe3aq Mn2aq 0 oxidation 5Fe2aq gt Fe3aq e 0 reduction Mn04aq 8H Se gt Mn2aq 4H20 o 5Fe2aq Mn04aq 8H le gt 5Fe3aq Mn2aqe 4H20 o 5Fe2aq Mn04aq 8H gt 5Fe3aq Mn2aq 4H20 Balancing in a basic solution Use the method of balancing for the acidic but don t add yet Add the number of OH ions equal to the number of H ions to both sides of the overall equation On the side containing both H and OH combine them to form H20 Cancel common H20 molecules that appear on both sides In a basic solution you ll never have H in your answer Balance Mn04aq CN gt Mn02s OCN 0 red 2Mn04aq 4HJr 3e gt Mn02s 2H20 o oxi 3CN H20 gt OCN 2H 2e 0 aerate BCN 2H 2Mn04 aq 20H 6e gt 3 OCN 6Hi 6e 2Mn04 1H20 20H39reducing the H and water and cancelling electrons o SCN H20 2Mn0439aq gt 3 OCN 2Mn0439 1 H 29 20H 0 BCN H20 2Mn0439aq gt 3 OCN 2Mn0439 20H Acidic has H at the end Basic has H Chapter 6 Gases 105 Pressureforce per unit area FN o 1 atm 760 torr 760mm Hg 101325 kPa General gas equation where p pressure v volume n number of mols t temp Constant temperature for a fixed amount Pi Vi Pfo E Boyle s law PiVi Pfo where pressure is inversely proportional to volume Charles s law volume is directly proportional to temperature at constant pressure Always use K 27315 C K Vi Vf T7 E Standard temperature and pressure STP 0 deg C and 1 atm 1 mole of gas 22414 L ldeal gas equation R 00821 atmLmolR 1 atm x 22414L quoti Ti o R 1 mglex 27315K 0 PV mM m mass and M molar mass MP o d g E the lower the molar mass the greater the lifting power Gases in chemical reactions Law of combining volumes 0 Consider 2N0g 02 g gt 2N02 g the 2 represents the two moles o 2 mols of NO g 1 mol 02 g 2 mols N02 9 o If gases are compared at the same temperature and pressure we can substitute the mols for volumes so 0 2L N0g 1L 02g gt 2L N02g What volume of N2g measured at 735mm Hg and 26 degrees C is produced when 750g NaN3 is decomposed o 2NaN3s gt 2Nal 2N2g 0 find mols first so 750g NaN3 x 1m0 NaN3 3m0 N2 173 molNz 6501g NaN3 2 mol NaN3 assume ideal gas behavior P 735mm Hg x i 0967 atm 760 mm Hg 173 l 00821Latm 299K 0 VnRT max mom 439L P 0967atm Mixtures of gases 2 equations 2 unknowns The total pressure of the mixture is determined by the total number of mols ntotRT 0 PM at constant 17 and t o Vtot at constant p and t Dalton s Law of Partial Pressures 0 PW PaPb 0 each expands to fill its container and exerts the same pressure as if it were alone 0 In a mixture of n moles of A nb moles of b and so on 1v RT NbRT I V a V a Ptot b Ptot total volume is additive vtot Va Vb V volume A a x 100 N p a O a a a xa Ntot Ptot Vtot o xa xb 1 all components in mixture add to one Collecting a gas over liquid o 2Al S 6HCl aq gt 2AlCl3aq 3H2 g o If 355 mL of H2 g is collected over water at 26 degrees Celsius and 7492 mm Hg how many moles of HC are consumed Vapor pressure of H20 at 26 C 252 mm Hg 0 PM PHzo PHZ 7492 252 7744 mm Hg 1019 atm 0 V 00355 L 1019atm00355L 3 O Ntot 0032226th29915 L473 0 NA2 ntotNH2 1422e 3mol tot o 1422e 3 molH2 x M 284e 3 mol HCl consumed 3 mol H2 Non ideal gas You can account for intramolecular forces of attraction Van der Wal s equation 2 o P v nb nRt a and b vary from molecule to molecule pter 7 Thermochemistry Cha system the part of the universe chosen for study usually small surroundings outside of the system with which the system interacts open system can freely exchange energy and matter with surroundings closed system can exchange energy but not matter isolated system doesn t interact with the system at all No energy no matter Energy the capacity to do work Work is done with a force acts through a distance Kinetic energy energy of a moving object 1 o ek Emu2u veloczty o w mad 1kgm2 32 potential energy stored energy with potential to work thermal kinetic energy associated with random molecular motion Directly proportional to temperature of a system oc N of particles more energy More particles heat energy transferred bw a system and surroundings as a result of change in temperature heat will transfer until the kinetic energy of both are the same The change of temp can change the state of matter q heat depends on quantity and nature substance Calorie heat required to change temp of one gram of degrees C 1 cal 4184J Heat capacity quantity of heat required to raise the temp of one degree Celsius 0 energy gt joule 0 If system is mole it s a molar heat capacity 0 If a system is one gram it s a specific heat capacity 0 At 25 degree specific heat of water 418 Jg deg C 0 Not referring to specific heat No grams in denominator Q m x specific heat x change in temp 0 Heat capacity m x specific heat change in temp tfinal t initial 0 If q is positive heat is absorbed or gained negative heat is evolved or lost Q system q surround 0 Q system q surround Calorimetry O O O 0000 O 0 109 Heats of reaction and calorimetry Chemical energy Q reaction heat of reaction quantity of heat exchanged between a system and surroundings at a constant temp Exothermic produces temperature increase in an isolated system or gives heat to surroundings in nonisolated systems q reaction lt 0 so negative Endothermic temperature decreases in an isolated system or gains heat from surroundings in non isolated system q reaction gt 0 so positive Bomb calorimeter measures heat in a combustion reactions all the surroundings the calorimeter Q reaction q calorimeter q calorimeter q bomb q H20 q thermometer The combustion is the system 0010g C12H22011 Ti 2492 C Tf 2833 C heat capacity of calorimeter 490 kJC q caor 49211 2833 2492 167 K 3 CI TeactiOTl q calorimeter w m 539653 k1 1010g mol mol 0 coffee cup calorimeter O 0 mix reactants generally aqueous in a Styrofoam cup and measure temperature change the cup contents isolated system 0 heat of reaction quantity of heat that would be exchanged with surroundings in restoring calorimeter to its initial temperature 0 q reaction q calor H aq OH 9 H2O liquid 0 00000 0000 Two solutions added to Styrofoam cup 250 mL of 250 M HCI 250 mL of 250 M NaOH tinitial211 C tfinal 378 C determine reaction per mol of H20 formed Density of H20100 gmL specifc heat capacity of water 418 Jg C q calor 50mL x1gmL x 418Jg C 378211 C 350e3j q reaction q calor 35 kj is evolved released 25 molL 0025L 00625 mol q rxn 35 kJ 00625 mol H2O formed 56 kJmol exothermic reaction 0000000000 0 pressure volume work the work involved in the expansion or compression of a gas Pathwork w force x distance 2KI03 s 9 2KCIs 302 9 pressure forcearea work m x ga x change of height x a Pext AV AV gt 0 wlt 0 energy leaves as work expansion AVltO wgt0 energy enters compression 1 bar L 100J 1 L atm101325J Determine W 0225 mol N2 at constant t23 C is allowed to expand by 150L in volume against an external pressure 0750atm W PextAV 0750atm150L101325Jmol 114e2 J of work done by the system


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