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# Exam 2: Study Guide CHEM-1070-30

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This 5 page Study Guide was uploaded by Nina Kalkus on Saturday October 10, 2015. The Study Guide belongs to CHEM-1070-30 at Tulane University taught by Schmehl, Russell in Fall 2015. Since its upload, it has received 211 views. For similar materials see General Chemistry I in Chemistry at Tulane University.

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Date Created: 10/10/15

Continuing Chapter 7 Notes Speci c Heat heat required to raise temperature of 1 g of a material by 1 degree Celsius water standard 1 calorie per gram per degree centigrade Metric Unit of Energy Joule kg m2s2 1 calorie 4184 Joules Measurement of heat capacity Example Lead pellets 150 g take a beaker of water Bunsen burner underneath put the lead pellets in a test tube boil the water receiving cup thermos with water in it 50 grams and thermometer wait until lead pellets are at temperature of boiling water transfer lead pellets into thermos of water Lead initial temperature 100 degrees centigrade Water thermos initial temperature 220 degrees centigrade Final temperature of water and lead with no further temperature change 288 degrees centigrade Heat absorbed by water in the thermos q speci c heat of watermass of watertemperature change of water qwater 4184 Joulesg degrees centigrade28822 degrees centigrade50 g 1422 J 1400 J heat transferred from lead to water What is the speci c heat of the lead Speci c heat of lead Joules releasedmass leadtemp change of lead 1400 J 150712 degrees centigrade 013 Jgdegrees centigrade In an engine work done on the piston pressure of gas change in volume Energy Forcedistancegt Forceareadistancearea gt Pressurevolume change Pressure ForceArea Change in Energy in combustion AU work heat Example Suppose we combust C3Hg in a chamber where the volume does not changegt No work can be done Bomb calorimeter AUqV v constant volume a AU in open systemwqqV quot In system quot quot First Law of Thermodynamics Energy can neither be created nor destroyed Enthalpy qp AH AU APV AH positive positive heat required for reaction Example 02g gt AHA 2Hgo1 2H2g 02g AHA Enthalpy is an extensive property depends on the amount of material twice as much material twice as much heat AH kJmole Types of Enthalpies AHqu heat of fusion heat required for a solid going to a liquid H20sgt H20l AHqu 60 kJmole AHvap heat of vaporization heat required for liquid going to gas H20lgt H20g AHvap 407 kJmole Cp heat capacity Cp H20s 2 Jgdegree C Cp H20l 418 Jgdegree C Cp H20g 1 Jgdegree C gt quotquot quot not actual value just for example39s purpose Heat released when steam 15g at 150 degrees C is cooled to 25 degrees C moles water 15 g1 mole18g 083 moles water 1 150 degrees Cgt100 degrees C 1 Jgdegree C15 g50 degrees C 750 J 2 Condensation at 100 degrees C 40700 Jmole083 34000 J 3 Liquid at 100 degrees C cools to 25 degrees C 418 Jgdegree C15 g75 degrees C 4700 J Sum total 4700 34000 750 39000 J Heats of Reaction Heats of formation under standard conditions 1 atm pressure 25 degrees centigrade AHf0 Cgr 2H2g gt CH4g 749 kJmole Hess39s Law heats of reactions can be added or subtracted to obtain the heats of new reactions Example COg H2g gt Cgr H20g 131 kJmole Cgr H2g gt CH4 75 kJmole What is AHO for CH4g H20 gt 3H2g COg CH4 gt Cgr 2H2g 75 C gr HZO g gt COg H2 g 131 kJmole CH4g H20 gt 3H2g COg 206 kJmole Example N2 3H2 gt 2NH3 AH1 2N0 2H2 gt N2 2H20 AH2 2H20 gt 2H2 02 AH3 2N0 02 gt 2N02 AH4 Question What is the AH for the following reaction 2N02 4H2 gt N2 4H20 2N02 gt 2N0 02 AH4 2N0 2H2 gt N2 2H20 AH2 2H2 02 gt 2HZO AH3 2N02 4H2 gt N2 4H20 AHQAH3AH4 Reaction releases heat Exothermic Reaction requires heat Endothermic Example H2020 l 2H1ggt 12 l AH rxn 2 AH fproducts 2 AH reactants 2AH fH20 O AH fH202 2AH fHI 2285 kJmole 187 kJmole 226 kJmole 435 kJmole Chapter 5 Test Review Practice Problems 1 Determine the concentration of the ion indicated in each solution a K in 0238 M KNO3 Since you can see that there is only 1 mole of K in every mole of KNO3 the ratio is 11 so the concentration of K in the solution is 023810238 M K b NO339 in 0167 M CaN032 Since there are 2 moles of NO339 in every 1 mole of CaN032 the ratio is 12 so the concentration of NOZ39 in the solution is 01672 0334 M NOs39 c Na in 0201 M Na3PO4 Since there are 3 moles of Na in every 1 mole of Na3PO4 the ratio is 13 so the concentration of Na in the solution is 02013 0627 M Na 2 Balance this equation for redox reactions occurring in acidic solution MnO439 139 Mn2 12s First divide the reaction into half reactions and determine which is the reduction and which is the oxidation I39 gt 12 oxidation MnO439 gt Mn2 reduction Then balance the elements that are not oxygen or hydrogen 2139 gt 14 MnO439 gt Mn2 Next balance the oxygens by adding water wherever needed I39 gt 12 M1104 gt M112 1 Then balance the hydrogens by adding H molecules where needed I39 gt 12 8H MnO439 gt Mn2 4H20 Next balance the total charges on either side of each reaction by adding electrons 2139 gt 12 2e39 5e39 8H MnO439 gt Mn392 4H20 Then multiply each half reaction by a constant so that when added the electrons will cancel hint if one half reaction has 5 electrons added multiply the other half reaction by 5 52139 gt 12 2e39 25e39 8H M1102 gt Mn2 4HZO Finally add the two half reactions together remembering to distribute the constants and cancel out the electrons 16H 2MnO439 10139 gt 512 2Mn2 8H20 3 Balance this equation for redox reactions occurring in basic solution M1102 1 C103 gt M1104 1 Cl First divide the reaction into half reactions and determine which is the reduction and which is the oxidation MnOz gt MnO439 oxidation ClOs39gt Cl39 reduction Then balance the elements that are not oxygen or hydrogen M1103 gt M1104 C103 gt C139 Next balance the oxygens by adding OH wherever needed 20H 1 M1102 gt M1104 C103 gt Cl39 3OH Then balance the hydrogens by adding H molecules where needed 20H l M1102 gt M1104 l 2H 3H C103 gt C139 3OH Next balance the total charges on either side of each reaction by adding electrons 20H MnOz gt MnO439 2H 2e39 8e39 3H ClO339 gt ClA 30H Then multiply each half reaction by a constant so that When added the electrons will cancel hint if one half reaction has 5 electrons added multiply the other half reaction by 5 820H MnOz gt MnO439 2H 2e39 28e39 3H ClO339 gt Cl39 3OH Finally add the two half reactions together remembering to distribute the constants and cancel out the electrons 3H ClO339 16OH39 4Mn02 gt 4M20439 8H C139 3OH

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