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# PRINCIPLES OF CHEMISTRY II CH 302

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This 35 page Class Notes was uploaded by Brady Spinka on Monday September 7, 2015. The Class Notes belongs to CH 302 at University of Texas at Austin taught by Stephen McCord in Fall. Since its upload, it has received 20 views. For similar materials see /class/181862/ch-302-university-of-texas-at-austin in Chemistry at University of Texas at Austin.

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ye xy755 7 H12 Electrochem 1 7 mccord 7 51625 1 This print out should have 21 questions Multiple choice questions may continue on the next column or page 7 nd all choices before answering 001 100 points Chlorine bromine and iodine are good 1 oxidizing agents correct 2 reducing agents 3 bases Explanation 002 100 points When the Ags l AgCls l Cl aq electrode acts as a cathode the reaction is 1 Agaq 6 a Ags 2 Ags Cl aq a AgCls 6 3 AgCls 6 a Ags Cl aq cor rect 4 Ags e Agaq 6 5 2AgCls 2 6 e 2Agaq Clgg Explanation 003 100 points Which metal will dissolve in hydrochloric acid 1 Fe correct 2 Ag 3 Au 4 Pt 5 All will dissolve Explanation 004 100 points Identify the galvanic cell for the redox re action between sodium dichromate and mer curyl nitrate in an acidic solution L 11W l Hg aq H CI3aq l CNS 2 Hglt gt 1Hg2012s 10118101 y Haq CrgO aq Cr aq l Pts 3 HM l Hg2 ltan H HWW a CM 07 31 5 Crglwq 1 PWS 4 Pts l Hg aq Hg2aq H Haq Crg O aq Cr3 aq l Pts correct Explanation At the cathode CrgO aq 14Haq 6 6 a 2 Cr3aq 7 Hgow At the anode Hg3aq a 2Hg2 2 6 Equate the 6 CrgO aq 717 14Haq 717 26 a 2 Cr3aq 7 Hgow 3 Hggaq e 2Hg2 26 Add the balanced half reactions CI203 aq 3 Hg3aq 14 Haq 2 Cr3aq 6 Hg2 7 Hgow The cell diagram is MS 1 Hg aq Hg2aq H WW1 CI2O aq Cr3aq 1 PW 005 100 points Copper is plated on zinc by immersing a piece of zinc into a solution containing copperll ions In the plating reaction copperll ions 1 lose two electrons and are reduced 2 gain two electrons and are oxidized 3 lose two electrons and are oxidized 4 gain two electrons and are reduced cor rect Explanation For this question Cu2aq a Cus ye xy755 7 H12 Electrochem l 7 mccord 7 51625 2 For this to happen7 the charge must be bal anced by the Cu2 reacting With 2 electrons Cu2 2e e Cus This is a reduction reaction represented by the gain of electrons 006 100 points What takes place at the negatively charged terminal of a voltaic cell H combustion Equot reduction 00 the net redox reaction 4 oxidation correct Explanation 007 100 points A copper Wire dipped in a solution contain ing AgCN ions acquires a silver coating because 1 Ag oxidizes Cu to Cu2 in a spontaneous process until silver metal completely covers the copper correct 2 Cu2 is a stronger oxidizing agent than Ag 3 H is reduced by Cu to H2 and Ag ions are reduced by H2 to metallic silver 4 Ag is a stronger reducing agent than Cu Explanation AgCN provides Ag ions in solution AgoN Ag 2 CN Ag 16 a Ag E0 07994 v Cu a 0112 2 e7 E0 70337 v Sum E2611 04624 v Since E3611 is positive7 the process is spon taneous Note that Cu is a stronger reducing agent than Ag and Ag is a stronger oxidizing agent than Cu2 Remember reducing agents get oxidized and oxidizing agents get reduced 008 100 points Consider the following reaction taking place in a voltaic cell Cus 2Agaq a Cu2aq 2Ags 1 Cu is reduced at the cathode and Ag is oxidized at the anode 2 Cu is oxidized at the cathode and Ag is reduced at the anode 3 Cu is oxidized at the anode and Ag is reduced at the cathode correct 4 Cu is reduced at the anode and Ag is oxidized at the cathode Explanation Oxidation is loss of electrons and occurs at the anode Which is What happens to the copper metal Cu Reduction is gain of elec trons and occurs at the cathode Which is What happens to the silver ion Ag 009 100 points The standard potential of the Cu2 l Cu elec trode is 034 V and the standard potential of the cell Ags l AgClS l 017aq H 0112aq l 0118 is 012 V What is the standard potential of the AgCl l AgCl electrode 1 022 V correct 2 046 V 3 7 046 V 4 7 022 V 5 024 V ye xy755 7 H12 Electrochem l 7 mccord 7 51625 3 Explanation 010 100 points Which species will oxidize Cr2 but not WW 1 V3 correct 2 P104 3 03 in acidic medium 4 Zn2 5 Fe2 Explanation 011 100 points Consider the cell ZIIltSgt l Zn2aq H Cl aq l Ag01s l Ags Calculate E0 1 098 V correct 2 7 120 V 3 7 054 V 4 054 V 5 120 V Explanation 012 100 points Consider the half reactions 0112 25 a Cu E0 034 V Hg 2eie2Hg E0080V Pd2 2 e a Pd E0 099 V Au33eieAu E0142V Using the redox couples to establish a voltaic cell which cell reaction would be non spontaneous 1 Hg Cu e Cu2 2 Hg 2 2Au3 3Cue2Au3Cu2 3 Pd2 2Hg a Pd Hg 4 3 HgJr 2 Au a 6 Hg 2 Au3 correct 5 Pd2 Cu a Pd 0112 Explanation 013 100 points Consider the following electrode reactions Pe3 1a a Pe2 E0 0771 V lg2e e2l E00535V What would be E3611 for the spontaneous reaction 1 1007V 2 71306 V 3 0236 V correct 4 70236 V 5 1306 V Explanation The anode is the place where oxidation oc curs while the cathode is the place where reduction occurs So looking at the overall re action Fe3 is reduced to form Fe2 because it gains electrons So Fe3 l Fe2 is the reac tion at the cathode 2l has to lose 26 to form lg so it has to be oxidized and this has to occur at the anode Eget Egathode 7 E0 anode 077170535 0236 V 014 100 points Consider two hypothetical metals X and Y which can exist as metal ions X2 and Y3 respectively The standard reduction poten tials are x2 2 e a x 72375 V V3 35 a Y 2873V What is the numerical value for the stan dard cell potential for the reaction 3Xs 2Y3aq a 3X2aq 2Ys ye xy755 7 H12 Electrochem 1 7 mccord 7 51625 4 at 25 C Correct answer 5248 V Explanation 015 100 points Consider the half reactions and the balanced equation for the cell CNS 1 0910101 11121131001 lAuS What is its standard potential Correct answer 231 V Explanation The two half reactions7 written as reduc tions7 are Au3aq 3 6 a Aus Cr2aq 717 2 6 a Crs Equate the 6 2 Au3aq 717 36 a Aus E 140 V 3 Crs a Cr2aq 26 7E 091 V Add the balanced half reactions 3 Crs 717 2 Au3aq a 2 Aus 3 Cr2aq E 140 V E 7091 V E 114 V 7 4091 V 231 V 0 Gel 016 100 points Consider the cell Zns l Zn2aq H F62aq l Fes at standard conditions Calculate the value of AG for the reaction that occurs when current is drawn from this cell 1 7 62 kJ mol 1 correct 2 62 kJ mol 1 3 i 31 kJ mol 1 4 i 230 kJ mol 1 5 230 M A molil Explanation 017 100 points The standard voltage of the cell Pt 1H2g l Haq H 017aq 1Ag01s 1Ags is 022 V at 25 C Calculate the equilibrium constant for the reaction 2Ag01s H2g V 2 Ags 2 Haq 2 Cl aq 1 52 gtlt 103 2 74 3 17 gtlt 103 4 27 gtlt 107 correct 5 37 Explanation 018 100 points The free energy of formation of liquid water at 25 C is 72373 kJmol The standard potential for the H2 1H10 M H QHQO 1 024H10 M cell is 1 226 V 2 123 V correct 3 000 V 4 492 V Explanation AG 72373 kJmol 96485 J T 25 C Note 1 Faraday mol Free energy of formation of liquid water is for one mole of water7 so rewrite the cell as two balanced half cell reations as below H2e2H2f 1 HQOgt OQ2H267 ye xy755 7 H12 Electrochem 1 7 mccord 7 51625 The number of moles of 6 involved is 2 3 increases AG 7nFE 72373 kJmol 2 H101 6 Explanatlon 96485 J 39 7 X E 021 part 3 0f 3 100 p01nt5 V H101 5 What is the voltmeter reading 7237300 J mol 2 mo 1 340V V mol 6 122972 V 2 160 V correct 019 part 1 0f 3 100 points 339 050 V The galvanic cell below uses the standard 4 250 V half cells Mg2 l Mg and Zn2 l Zn7 and a salt bridge containing KClaq The voltmeter 5 430V gives a positive voltage reading Explanation Salt Bridge Identify A and write the half reaction that occurs in that compartment 1 Zns ZnS ZD2a l 1 2 5 2 Mgs Mgs a Mg2aq 2 6 cor rect 3 Zns Zn2aq 2 6 a Zns 4 Mgs Mg2aq 2 6 a Mgs Explanation 020 part 2 0f 3 100 points What happens to the size of the electrode A during the operation of the cell 1 decreases correct 2 No change munoz mmm3898 7 Homework 14 7 holcombe 7 51160 1 This print out should have 21 questions Multiple choice questions may continue on the next column or page 7 nd all choices before answering 001 100 points The reaction 2NO2g 2N0g 02g is postulated to occur via the mechanism N02g NO2g N0g N03g slow N03g a NOg 02g fast What is an intermediate in this reaction 1 N03g correct 2 NOg 3 N02g 4 ON7N03g 5 02g Explanation 002 100 points Consider the mechanism N02 F2 7gt NOQF F F N02 NOQF What is the rate law k1 slow kg 7 fast 1 rate k2 N02 2 rate k1 N02 F2 correct 3 rate k1 NOQF 4 rate k1 k2 N022 5 rate k2 N022 Explanation 003 100 points Consider the multistep reaction that has the overall reaction 2ABe2CD What would be the observed rate expression Mechanism 2AeC1 1BeCD slow fast 1 Rate k A2 B 2 Rate k A 3 Rate k 1 El 4 Rate k A2 correct 5 Rate k A 1 B Explanation The rst step is the slowest step the rate determining step7 so the rate law can be written based on the stoichiometry of the re actant 004 100 points A given reaction has an activation energy of 145 kJmol This reaction is normally run at room temperature 25 C and takes about an hour to get to completion At what temper ature should the reaction be run so that it is running twice as fast as at room temperature 1 50 C 2 65 C correct 3 135 C 4 31 C 5 77 C 6 98 C Explanation Use the Arrhenius equation For the k7s just put in 2 for kg and 1 for k1 and use the temperatures as appropriate T1 is room tem perature7 T2 is the unknown temperature Twice the rate at the same concentrations means double the value of k so that you can munoz mmm3898 7 Homework 14 7 holcombe 7 51160 2 use any k values that are in the ratio 2 l or 1 2 for that matter7 as long as you get the temperatures in the corresponding order Also remember to use Kelvin for temperature7 not Celsius degrees 005 100 points Suppose for some reaction the rate constant doubles in going from T 50 C to 70 C What is Ea for this reaction 1 8 kJmol 2 32 kJmol correct 3 24 kJmol 4 l6 kJmol 5 40 kJmol Explanation T1 50 C 7 273 323 K T2 70 C 273 343 K h i i k1 R T1 T2 k2 R lnk71 l 1 if 72 8314 JmolK ln 2 l l k22k1 Ea 323 K 7 343 K 335 kJmol 006 100 points Note the frequency factor in the Arrhenius equation is occasionally referred to as 77Arrhe nius factor77 What is the temperature that would pro duce a rate constant of 6 X 105 sec 1 if the activation energy is 40 kJ mol and the Arrhe nius factor is 62gtlt1012 1 305 K 2 298 K correct 3 260 K 4 25 K 5 273 K Explanation Ea 40000 M k 6 x 105 sec 1 A 6 gtlt 1012 k A67EaRT A EaRT i 6 k Ea 71 A RT n k T 7 E 7 A R1 7 n 1 40000 Jmol i 8314 Jmol K 62 gtlt 1012 ln 6 gtlt 105 298 K 007 100 points Increasing the temperature of a reaction causes 7 in the reaction rate 1 no change 2 an increase correct 3 a decrease 4 You need to know AH for the reaction Explanation At increased temperature more heat is pro vided to the reaction components7 so the num ber of effective collisions increase 008 100 points The graph describes the energy pro le of a reaction munoz mmm3898 7 Homework 14 7 holcornbe 7 51160 3 400 300 Energy kJ 50A Time What are the values for AH and Ea7 respec tively7 for the reaction in the direction writ ten 1 250 kJ7100 id 2 7250 kJ7 7100 id 3 250 kJ7 350 kJ correct 4 7250 kJ7100 id 5 7250 kJ7 350 kJ Explanation Time AH300 kJ750 kJ 250 kJ Ea 400 M 7 50 kJ 350 kJ 009 part 1 0f 4 100 points Consider the following potential energy dia gram gt Potential energy r Reaction progress gt Which arrow represents the potential en ergy of the reactants 11 2f 36 a a correct 5 c 6d Explanation 010 part 2 0f 4 100 points Which arrow represents the potential energy of the products 1 a 2 f correct 3b 56 6d Explanation 011 part 3 0f 4 100 points Which arrow represents the activation en ergy munoz mmm3898 7 Homework 14 7 holcornbe 7 51160 4 4d 5a 6 b correct Explanation 012 part 4 0f 4 100 points Which arrow represents the heat of reaction enthalpy change 1 0 correct 2 f 3 d 4 e 51 6a Explanation 013 100 points If Substance A is a catalyst7 Which equation best represents What happens in a chemical reaction 1XYZ2AXYZ22A 2XYZ2AXYZ2 3XYZ2 A7XY Z2 Ac0rrect 4XYZ22AXYZ2A Explanation 014 100 points What is the activation energy for a reaction if its rate constant is found to triple When the temperature is raised from 300 K to 310 K 1 4187400 Jrnol 2 20300 Jrnol 3 847900 Jrnol correct 4 1957600 Jrnol 5 No other choice is Within 3 percent Explanation T1300K T2310K k2 3 k1 1mg 77 iii k1 3 T1 T2 Rln Ea 1 1 1 T1 T2 The rate constant tripled7 so k2 3 k1 8314 Jrnol K ln3 1 1 300 K 7 310 K 84944 Jrnol 015 100 points Consider the reaction NOBrg a NOg Br2g A plot of NOBrli1 VS tirne gives a straight line With a slope of 200 M71 s7 The order of the reaction and the rate constant7 respec tively7 are 1 rst order and 200 s71 2 rst order and 0241 s71 3 second order and 200 M71s 1 cor rect 4 second order and 166 M71 s71 5 second order and 0500 M71 s71 Explanation munoz mmm3898 7 Homework 14 7 holcombe 7 51160 5 016 100 points A non steroidal anti inflammatory drug is me tabolized with a rst order rate constant of 325 dayil What is the half life for the metabolism reaction 1 0213 day correct 2 225 day 3 163 day 4 0308 day Explanation 017 100 points The decomposition of cyclobutane is a rst order reaction At a certain temperature7 the half life for this reaction is 137 seconds What fraction of a sample of cyclobutane would be left after 685 seconds at this temperature 1 00625 2 00156 3 00312 correct 4 0125 5 025 Explanation Isl21 t685s a 1 7 ln 2 t12 W a kt cth cyclobutanelt 7512 7 685 s ln2 7 137 s 34657 cyclobutanelo 7 34657 cyclobutanelt 7 32 cy clobut ane t 1 7 003125 cyclobutanelo 32 018 100 points Consider the reaction 1 H202aq gt 5 02aq 14205 if an aqueous system initially has a H202 of 016 M and two days later has a H202 of 002 M 7 what is the half life of H202aq 1 12 hours 2 16 hours correct 3 24 hours 4 not enough information 5 48 hours Explanation 016 057 002 x 3 half lives two days 48 hours 1512 16 hours 019 100 points The following graph is for a rst order reac tion Which are the appropriate units for the axes 1 xv axis temperature y axis Ea 1 2 xv axis t y axis lnA 1 me 1 1 y aXis 7 3 x axis temperature k nguyen vtn285 7 H01 Fundamentals 7 mccord 7 50965 1 This print out should have 16 questions Multiple choice questions may continue on the next column or page 7 nd all choices before answering 001 100 points Which compound has the wrong chemical for mula 1 Ba3PO42 2 CaOH correct 3 NH4QSO4 4 MgOH2 Explanation The calcium ion is Ca2 the hydroxide ion is OH Two OH are needed to balance the charge of each Ca27 so the formula is CaOH2 The magnesium ion is Mg2g the hydroxide ion is OH Two OH are needed to balance the charge of each Mgng7 so the formula is MgOH2 The barium ion is Ba2g the phosphate ion is P027 Two P027 are needed to balance the charge of every three Ba2 This gives a total anion charge of 76 and a total cation charge of 6 The formula is Ba3PO42 The ammonium ion is NHZI the sulfate ion is 803 Two NH4 are needed to balance the charge of each SOZ 7 so the formula is NH4QSO4 002 100 points Which one has the greatest number of atoms 1 305 moles of helium 2 305 moles of argon 3 All have the same number of atoms 4 305 moles of CH4 correct 5 305 moles of water Explanation For 305 moles of water 602 gtlt 1023mOleC 7 atoms 305 mol H20 gtlt lmol 3 atoms 1 molecule 551 X 1024 atoms For 305 moles of CH4 602 gtlt 1023 molec atoms 305 mol CH4 gtlt lmol 5 atoms 1 molecule 918 gtlt 1024 atoms For 305 moles of helium 602 gtlt 1023 atoms 7 atoms 305 mol He gtlt 1 mol 184 gtlt 1024 atoms For 35 moles of argon 602 x 1023 atoms 7 atoms 305 mol Ar gtlt lmol 184 gtlt 1024 atoms 003 100 points If 1000 grams of copper Cu completely re acts with 250 grams of oxygen7 how much copperH oxide CuO will form from 1400 grams of copper and excess oxygen Note CuO is the only product of this reaction 1 1600g 2 2000 g 3 350 g 4 l500g 5 1750 g correct Explanation mCuini1000 g mCu n 1400 g If 100 g copper and 25 g oxygen react com pletely with each other7 there must be 125 g of product formed law of conservation of mass This product is CuO m02 250 g nguyen vtn285 7 H01 Fundamentals 7 mccord 7 50965 2 Now we have a ratio for every 100 g of Cu reacted 125 g of CuO will be produced assuming there is enough oxygen We use this ratio to nd the mass of CuO that could be formed from 140 g of Cu and excess oxygen We set our known ratio 100 g Cu 125 g CuO equal to our experimental ratio 140 g Cu x g CuO and solve for the unknown 100 g Cu 7 140 g Cu 125 g CuO 7 an 140 g Cu 125 g CuO 100 g Cu 175 g CuO 004 100 points What is the volume of 156 kg of a compound whose molar mass is 8186 gmole and whose density is 412 gmL 1 0464 mL 2 379 mL correct 3 311 mL 4 3110 mL 5 783 mL 6 64300 mL Explanation m 156 kg density 412 gmL MW 8186 gmol 1000g X 1mL 1kg 412g V 156kg gtlt 379mL 005 100 points Consider the reaction 4 Fes 3 02g 7 2 Fe203s H 125 g of iron111 oxide rust are pro duced from 874 g of iron how much oxygen gas is needed for this reaction 1 374 g correct 2874 g 3 75 g 4 212 g 5 125 g Explanation H1iron g Inoxide 125 g The balanced equation for the reaction tells us that 4 mol Fe reacts with 3 mol 02 to produce 2 mol F8203 We have two possible starting points We know 125 g F8203 was produced and that 874 g Fe was present at the start of the reaction Choosing the 125 g of F62 03 to start with rst we convert to moles using the molar mass mol F8203 125 g F8203 X 1 mol F8203 1597 g F8203 00783 mol F8203 Now we use the mole ratio from the bal anced equation to nd moles 02 needed to produce 00783 mol F8203 7 1110102 00783 mol F8203 3 mol 02 2 mol F8203 0117 mol 02 We convert from moles to grams 32gOg 7 o 0117 10 g 2 mo 2x11110102 3744 g 02 Starting with 874 g Fe and following the same steps results in the same numerical an swer 006 100 points Upon heating potassium chlorate produces potassium chloride and oxygen 2K0103gt 2KCl3OQ nguyen vtn285 7 H01 Fundamentals 7 mccord 7 50965 3 What mass of oxygen 02 would be pro duced upon thermal decomposition of 25 g of potassium chlorate K0103 with MW 1225 gmol7 1 44g 2 65 g 3 98 g correct 4 49 g 5 33 g Explanation mKClO3 250 g MVVKClO3 1225 gmol The balanced equation for the reaction in dicates that 3 mol 02 are produced for every 2 mol KClOg reacted First we calculate the moles K0103 present 7mol K0103 25 g K0103 1 mol K0103 X 12255 g K0103 0204 mol KClOg Now we use the mole to mole ratio from the balanced equation to nd the moles 02 that could be produced from this amount of K0103 7 mol 02 0204 mol KClOg 3 mol 02 X 2 mol K0103 0306 mol 02 We convert from moles to grams 02 32g02 7 O 0306 10 g 2 mo 2gtlt1molOg 98g 02 007 100 points In the reaction 7 0077 02 7 002 how much oxygen is required to convert 28 g of CO into 0027 1 16 g correct 2 32 g 3 56 g 4 8 g 5 64 g 6 28 g Explanation moo 28 g The balanced equation for the reaction is QCOOQgtQCOQ The coef cients in this equation indicate that 2 mol CO are needed for each mol 02 reacted First we calculate the moles of CO present 7molCO28gCOgtltM 28 g CO 1 mol CO Using the mole ratio from the balanced equa tion7 we nd the moles 02 needed to com pletely react with 1 mol CO 1 mol 02 7molOg1molCOgtltm 05 mol 02 We convert from moles to grams 02 32g02 7 O 05 10 g 2 mo 2gtlt1molOg 16g02 008 100 points Consider the reaction N23H2gt2NH3 How much NH3 can be produced from the reaction of 742 g of N2 and 140 moles of H27 1 562 gtlt 1024 molecules 2 159 gtlt 1024 molecules nguyen vtn285 7 H01 Fundamentals 7 mccord 7 50965 4 3 126 gtlt 1025 molecules 4 319 gtlt 1024 molecules correct 5 169 gtlt 1025 molecules Explanation mN2 742 g 11H2 140 mol First you must determine the limiting reac tant molN2742gN2 gtlt 2 265 mol N2 According to balanced equation we need 3mol H2 1molN2 We have 140 mol H2 528 mol H2 265 moi N2 1 mol N2 Therefore H2 is an excess and N2 is limiting 2 mol NH3 1 mol N2 6022 gtlt 1023 NHg molec 1 mol NH3 molec 319 gtlt 1024 molec NH3 molec NH3 265 mol N2 gtlt 009 100 points For the reaction 7C6H6702 gtCOQHQO 397 grams of C6H6 are allowed to react with 1057 grams of 02 How much 00 will be produced by this reaction Correct answer 1163 grams Explanation mCSH6 397 g m02 1057 g The balanced equation for the reaction is QC6H6 15 02 gt12002 6HQO FW of C6H6 is 781118 gmol giving 05082 mol C6H6 FW of 02 is 319988 gmol giving 3303 mol 02 FW of CO is 440095 gmol 15 mol 02 05082 10 H mo 6 6 X 2molC6H6 3811 mol 02 which is more than what is actually present Therefore the limiting reactant must be 02 3303 10 gtlt 21110100 mo 2 15 moi C6H6 440095 g 002 X 1 mol 002 010 100 points Which of the following describes a chemical change 1163 g 002 1 wood is carved 2 alcohol evaporates 3 water is heated from 0 C to 50 C 4 gasoline burns in an engine correct Explanation A chemical change occurs when one or more substances are consumed at least partially and one or more new substances are formed 011 100 points Which of the following describes a physical change 1 Wood burns 2 Ice melts correct 3 Iron rusts 4 Food is digested in the stomach 5 Plants decompose in a compost pile Explanation A physical change occurs with no change in chemical composition Melting involves no change in composition only a phase change 012 100 points Water is an example of nguyen vtn285 7 H01 Fundamentals 7 mccord 7 50965 5 H None of these 2 a homogeneous mixture 9 a heterogeneous mixture 4 an element 5 a compound correct Explanation The two hydrogen atoms and one oxygen atom making water exempli es two or more atoms bonded together to make a molecule 013 100 points The measurement 32 gtlt 10 3 g could also be written as 1 None of these 2 32 mg correct 3 32 g 4 32 pg 5 32 kg Explanation mg refers to 1073 014 100 points The mole concept is important in chemistry because 1 it allows us to distinguish between ele ments and compounds 2 atoms and molecules are very small and the mole concept allows us to count atoms and molecules by weighing macroscopic amounts of material correct 3 it explains the properties of gases 4 it establishes a standard for reaction stoi chiometry 5 it provides a universally accepted stan dard for mass Explanation The mole concept is important in chemistry because we know that if we weight 6355 g of pure copper7 then we have about a mole of copper atoms 015 100 points How many atoms of hydrogen are contained in 1 mole of methane CH4 1 241 gtlt 1024 atoms correct 2 301 gtlt 1024 atoms 3 4 atoms 4 The correct answer is not given 5 602 gtlt 1023 atoms Explanation n 1 mol Each methane molecule contains 4 hydro gen atoms There are Avogadro7s number of methane molecules in one mole of methane molecules nH 1 mol CH4 X 602 x 1023 molec CH4 1 mol CH4 4 H atoms X 1 molec CH4 241 gtlt 1024 H atoms 016 100 points Which has the greatest number of hydrogen atoms 1 100g of a substance that is 2 H by mass 2 1020 hydrogen atoms 3 5 g of an unknown compound 4 100 g of water 5 20 g of hydrogen gas correct nguyen vtn285 7 H01 Fundamentals 7 mccord 7 50965 Explanation 10 H atoms is much less than 1 mole of H atoms 100g of water is 556 moles of water Which would have 1112 moles of H atoms 5 g of an unknown substance even if it was pure hydrogen could only be 5 moles of H atoms 20 g of hydrogen gas is 10 moles of H2 Which is 20 moles of H atoms 100g of a substance that is 2 by mass hydrogen has 2 g of Hydrogen Which is 2 moles 20 moles of H atoms is the greatest number of atoms wallach baw2266 7 H05 AcidBase 7 mccord 7 51175 This print out should have 26 questions Multiple choice questions may continue on the next column or page 7 nd all choices before answering 001 100 points Which of the following is true in pure water at any temperature 1 pH 70 2 Hgojon j 10 x 10 14 3 pH 70 or greater than 70 4 Hgo 0H7 correct 5 KW decreases with increasing tempera ture Explanation KW is shown to INCREASE with increas ing temperature pH 7 is only true when water is at 24 C H30OH KW which increases with temperature At high temperatures pH can be less than 7 Thus Hgo OH is the only case that is true 002 100 points What is Hgotj when OH 33 gtlt 10 9 M 1 10 gtlt 10 7 M 2 30 gtlt 10 6 M correct 3 33 gtlt 10 5 M 4 33 gtlt 10 9 M 5 66 gtlt 10 5 M Explanation OH 33 gtlt 10 9 M KW H30OH j 1 gtlt 1014 11500 7 if 10 x 1014 W30x10 6M 003 100 points What is OH in a 00050 M HCl solution 1 20 gtlt 10 12 M correct 2 50 gtlt 10 3 M 3 10 M 4 66 gtlt 10 5 M 510 gtlt 10 7 M Explanation 0H7 00050 M Since HCl is a strong acid it completely dissociates and H7L is 00050 M H017 H 01 KW HOH 1 x 10 14 KW 011 7 Ht 1 gtlt 1014 2 gtlt 10 12 M 00050 004 100 points What is the conjugate acid of NO 1 Nogg 2 NH3 3 N027 4 H 5 HN03 correct 6 OH Explanation Since the question asks for the conjugate acid we can assume NO is acting as a base This means that it is a proton acceptor To form the conjugate acid it accepts a H making HNOg wallach baW2266 7 H05 AcidBase 7 rnccord 7 51175 2 005 100 points According to the Bronsted Lowry Theory of acids and bases7 a base is 1 a proton acceptor correct 2 a proton donor 3 an electron acceptor 4 a substance Which When dissolved in water yields OH 5 an electron donor Explanation Bases are de ned by the Arrhenius the ory as substances Which in water produce hy droxide ions7 by the Bronsted Lowry theory as substances Which accept protons7 and by the Lewis theory as substances Which provide electron pairs 006 100 points Which is NOT a conjugate acid base pair 1 H01 Cl 2 H2804 SOi correct 3 H20 OH 4 H3801L H2804 5 H2 H7 Explanation Except for H2804 and SOZ 7 the members of all of the pairs differ by one proton 007 100 points The hydroniurn ion concentration in a solu tion at pH 10 has What relationship to the hydroniurn ion concentration in a solution at pH 13 1 100 times less than 2 1000 times less than 3 3 times greater than 4 1000 times greater than correct Explanation For pH 10 1er 1 gtlt 1010 For pH 13 MH 1 gtlt 1013 008 100 points Which pH represents a solution With 1000 times higher 0H7 than a solution With pH of 5 1 pH 8 correct 2 pH 7 3 pH 5000 4 pH 2 5 pH 3 6 pH 4 7 pH 6 8 pH 0005 9 pH 1 Explanation pH 5 pOH147pH14759 OH 10pOH 109 M 7 7 3 79 OH LE 1000OH 10 10 M 10 6 M pOHI 7logOHI 6 pH147pOH14768 009 100 points At 250 C7 water solutions Which are neutral have a pH of 1 about 0 wallach baw2266 7 H05 AcidBase 7 mccord 7 51175 3 2 about 14 3 in nity 4 about 7 correct Explanation 010 100 points Calculate the resulting pH if 365 mL of 288 M HN03 is mixed with 335 mL of 110 M CaOH2 solution 1 720 2 0350 correct 3 0460 4 0067 5 146 6 236 Explanation VHNO3 365 mL HNOg 288 M lCja0H2 335 mL 110 M To determine the pH of the nal mixture7 we need to determine how much H or OH is left over after the reaction Remember that for complete neutralization we need H and OH in equal molar amounts H OH H20 First calculate how many moles of HJr and OH we have 288 iHNo 7 moi H 0365 L gtlt X 1 mol H 1 mol HN03 105 moi H 110 mol CaOH2 mol OH 0335 L gtlt 1L X 2 mol OH 1 mol CaOH2 0737 mol OH Now we can see that we have more H than OH so OH will be consumed and H will be left over Let7s nd out by how much 7 moi H 105 moi 4 0737 mol 0313 moi H The next step is to calculate the H z 0313 moi H 7 M H 0700 L 0447 M H pH 7log0447 0350 011 100 points What is the pH of a 006 M BaOH2 aqueous solution 1 0920819 2 980939 3 87 4 130792 5 130792 correct Explanation BaOH2 015 M BaOH2 is a strong base which dissociates in aqueous solution to produce two moles of OH for every mole of BaOH27 so 006 M BaOH2 produces 012 M OH BaOH2 o 1352 2OH mi 006 M 0 M 0 M A 7006 M 006 M 2006 M n 0 M 006 M 012 M pH 147pOH 1477log 012 130792 012 100 points 181 mL of an unknown HCl solution was neu tralized in a titration with 362 mL of 0250 M NaOH What is the molarity of the unknown HCl solution wallach baw2266 7 H05 AcidBase 7 mccord 7 51175 4 1 250 gtlt10 1 M 2 905 gtlt10 2 M 3 164 gtlt 103 M 4 500 gtlt 10 2 M correct 5 800 gtlt10 1 M Explanation VHCl 181 mL NaOH 0250 M The balanced equation for this neutraliza tion reaction is VNaOH 362 mL HCl NaOH NaCl H20 We determine the moles of NaOH used 7 mol NaOH 00362 L soln 0250 mol NaOH 1 L soln 000905 molNaOH From the 11 mole ratio in the balanced chem ical reaction we know we would need 000905 moles of NaOH to neutralize 000905 moles HCl This is the amount of HCl that must have been in the 181 mL sample Molarity is moles solute per liter of solution 000905 mol NaOH 0181 L solution 005 M HCl 7MHCl 013 100 points A 028 M solution of a weak acid is 35 ionized What is the pH of the solution 1 146 2 055 3 525 4 201 correct 5 317 Explanation M028M P35 35 of the 028 M is ionized contributes to pH7 so 35 H 028 M gtlt 7 00098 M 100 pH 7logH 7log00098 200877 014 100 points A 0200 M solution of a weak monoprotic acid HA is found to have a pH of 300 at room temperature What is the ionization constant of this acid 1 530 2 20x10 9 18gtlt10 5 GO 50gtlt10 6 correct a 50gtlt10 3 U 6 10x10 6 7 20gtlt10 5 8 10x10 3 Explanation 015 100 points Assume that ve weak acids7 identi ed only by numbers 17 27 37 47 and 57 have the following ionization constants loniz ation Constant Ka value Acid 10 gtlt 1073 30 gtlt 10 5 26 gtlt 107 40 gtlt 109 73 gtlt 10 11 UVJgtOONgtgt7 wallach baw2266 7 H05 AcidBase 7 mccord 7 51175 5 The anion of which acid is the strongest base 14 21 32 43 5 5 correct Explanation 016 100 points What is the percent ionization for a weak acid HX that is 040 M Ka 40 gtlt107 1 000020 2 010 correct 3 0050 4 20 5 0020 Explanation 017 100 points Like all equilibrium constants KW varies somewhat with temperature Given that KW is 174 gtlt 10 13 at some temperature compute the pH of a neutral aqueous solution at that temperature Correct answer 637973 Explanation KW 174 gtlt 10713 KW H30OH 174 x 10713 In a neutral solution H30 OH H30 174 gtlt 10713 70 gtlt 10 7 pH 7 log H30 637973 018 100 points A 000100 M solution of a weak acid HX is 9 ionized Calculate Ka for the acid Correct answer 89011 gtlt 1076 Explanation 019 100 points The term Ia for the ammonium ion77 de scribes the equilibrium constant for which of the following reactions 1 NHr H20 NH3 H30 correct 2 NH4Clsolid H20 7 NH4 or 3 NH4 OH 7 NHg H20 4 NHg H20 7 NH4 OH 5 NH3 H30 NH4 H20 6 The term is misleading because the am monium ion is not an acid Explanation 020 100 points Hydroxylamine is a weak molecular base with Kb 66 gtlt 1079 What is the pH of a 00500 M solution of hydroxylamine 1 pH 926 correct 2 pH 948 3 pH 893 4 pH 363 5 pH 712 6 pH 1037 7 pH 474 Explanation Hydroxylamine is a weak base so use the equation to calculate weak base OH con centration note that this is the approximate wallach baW2266 7 H05 AcidBase 7 rnccord 7 51175 6 equation Why Because Kb is very small and the concentration is reasonable 0H7 xKb Ob 66 gtlt 10 9 00500 182 gtlt 10 5 After nding 0H77 you can nd pH using either method below A pOH 710g 182 gtlt 1075 474 pH 14 7 474 926 or B K H W l l 0117 7 1 0 X 10714 7 5 52 gtlt10 10 182 gtlt 105 7 pH 710g 552 gtlt 10 10 926 021 part 1 0f 2 100 points Calculate the pH of the solute in an aqueous solution of 0195 M C5H5Naq pyridine if the Kb is 18 gtlt 1079 Correct answer 927265 Explanation 0 0195 M Kb 18 gtlt 10 9 yridine C5H5N 717 H20 061 15ll1 1 0H7 0195 7 0 0 7x 7 26 26 0195 7 x 7 x x Kb C6H5NHOH C5H5N 2 2 18 gtlt 10 9 m 7 m x OH 4019518 gtlt10 9 18735 gtlt 105 molL The pOH is pOH 7log18735 gtlt 1075 472735 and the pH is pH 14 4 472735 927265 022 part 2 0f 2 100 points What is the percentage protonation of the solute Correct answer 000960769 Explanation protonation protonated species 00 d pyn ine 18735 gtlt 105 100 0195 X 000960769 023 100 points What is the pH of a 031 M solution of potas sium generate KR COO Ka for the generic acid R COOH is 27 gtlt 1078 1 6431 2 10660 00 10290 4 3470 5 3340 6 7000 7 7569 8 10380 9 10900 10 10530 correct Explanation MKMCOO 031 M Ka 27 gtlt 10 8 Its a salt of a weak generic acid Get it Generic acid makes generic ions Hal This means you need a Kb for the weak ye xy755 7 H14 Kinetics l7 mccord 7 51625 1 This print out should have 17 questions Multiple choice questions may continue on the next column or page 7 nd all choices before answering 001 100 points The rate of the reaction 2 03 7gt 3 02 is equal to 1 1 t 2 At correc Al03l 2 77 At 3 Al03l 2 At Al02l 4 7 At 7 Al02l 3 At Explanation The rate of a chemical reaction can either be expressed as the rate of disappearance of a reactant as in this case or as the rate of appearance of a product However this rate is multiplied by the inverse of the coef cient of the species in question The nega tive sign here reminds us that as the reaction progresses this species is consumed and de creases in concentration 002 100 points When the reaction 3 N0g e N2Og N02g is proceeding under conditions such that 0015 molL of N20 is being formed each second the rate of the overall reaction is L and the rate of change for NO is L 1 0015 Ms 1 70045 Ms71 correct 2 0015 Ms 170005 Ms 1 3 none of the other answers is correct 4 0030 Ms 170005 Msi1 5 0015 Ms710045 Msf1 Explanation Notice that three moles of NO are required to make one mole of N20 in this reaction so the rate of disappearance of NO which has a negative sign is three times the rate of appearance of N20 003 100 points Consider the irreversible reaction A2Be30 Which of the following correctly expresses the rate of change of B 7 1 2 rate of rxn 2 A rate of rxn 3 72 rate of rxn correct 4 rate of rxn 5 7 rate of rxn 6 7rate of rxn Explanation As B is a reactant and consequently disap pears during the reaction its rate of change will be negative and the inverse of the coef cient is used when writing the rate 1 A03 rate of rxn i 75 W ABl 72 rate of rxn i W 004 100 points Suppose a compound is involved in three different reactions denoted R1 R2 and R3 Tripling the concentration of this reactant in all three reactions causes the rates of reaction to increase by factors of 3 9 and 1 respec tively What is the order of each reaction with respect to this reactant ye xy755 7 H14 Kinetics 17 mccord 7 51625 2 1R11R23R30 2R13R29R3l 3R12R20R3l 4 R11R22 R30c0rrect 5R10R21R32 Explanation Tripling the concentration of the reactant in R1 brings about a tripling 31 increase in the rate of the reaction so the reaction is rst order with respect to the reactant in this case Likewise tripling the concentration of the reactant in R2 brings about a 9 fold 32 increase in the rate of the reaction so the reaction is second order with respect to the reactant here As the rate is not affected by an increase in concentration in R3 we can look at this as a change in the rate of 30 the reaction is said to be zero order with respect to the reactant here 005 100 points The following data were collected for the re action A B a C at a particular temperature Three experiments give Initial Initial Initial rate 111311 A B ACAt M M Mmin 1 01 01 40 gtlt 104 2 02 02 32 x 103 3 01 02 16 gtlt 10 3 What is the rate law expression for this reaction 1 10 4 M Qmm HAHBP 2 04 M 2min 1BA2 3 10 4 M QAmin HBHA 4 04 M72min 1A B2 correct Explanation To determine the order of reaction with respect to B compare trials 1 and 3 rate3 7 initial B3 I ratel 7 initial B 1 16 gtlt10 3 Mmin 7 02 My 40 gtlt 104 Mmin 01 M 4 2 x 2 Thus the reaction is second order with respect to B Likewise to determine the order of reaction with respect to A compare trials 2 and 3 rate3 7 initial A3 y rateg 7 initial A 32 gtlt10 3 Mmin 7 02 M 16 gtlt10 3 Mmin 01 M 2211 y 1 Thus the reaction is rst order with respect to A The rate law will take the form Rate k A B To determine the rate law expression in cluding the value of k the data from any of the trials can be used Using trial 1 as an example Rate k A 132 40 gtlt10 4 Mmin k01M01 M2 k 04 M4 min 1 so the rate law expression here is Rate 04 M72 min 1A B2 006 100 points ye xy755 7 H14 Kinetics 17 mccord 7 51625 3 The following data were obtained in four sep arate experiments measuring the initial rates of the reaction A2B C a products with different concentrations of reactants lnitial lnitial lnitial lnitial Trial A B C rate M M Ms x10 2 x10 2 x10 3 x10 8 1 15 20 12 40 2 30 20 12 80 3 30 20 24 80 4 15 30 36 90 What are the exponents of A B and C in that order in the rate law H 1 2 0 correct 6111 7201 Explanation Rate k AIByCz Rateg MAE 11313 013 Ratei klAl flflllCli y 1 2 z 2 a a a 2 29 Rateg 7 MA 13 CE Ratea k w 031 101 80 30 20 y 24 z 5 w a 1f Z Raw 2 W Ratel kAH 13111011 907 15 30 y 407 15 20 225 15y ln 225 y ln 15 ln 225 y ln 15 007 100 points Consider the data collected for a chemical reaction between compounds A and B that is rst order in A and rst order in B A 13 rate M M Ms 102 005 01 2 7 005 04 3 04 7 08 From the information above for 3 experi ments7 determine the missing concentrations of A and 13 Answers should be in the order of A then 1160 M 040 M 2 020 M 080 M 3 080 M 010 M 4 040 M 020 M 5 040 M 010 M 6 080 M 020 M correct Explanation As we know the reaction is rst order in A and rst order in 137 the rate law has the form ye xy755 7 H14 Kinetics 17 mccord 7 51625 Rate kAB To determine the rate constant k7 consider trial 1 Rate kAB 01 Ms k02 M005 M k 10 M 1 s71 So7 in the second trial7 Rate MA B 04 Ms 10 M71s71A005 M A 020 M And in the third trial7 Rate kA B 08 Ms 10 M71s7104 MB B 020 M 008 100 points Consider a chemical reaction between com pounds A and B that is rst order in A lnitial lnitial lnitial Trial A B rate M M Ms 1 01 01 10 gtlt10 4 2 02 02 80 gtlt 10 4 480 gtlt 10 4 3 03 7 What is the missing concentration of B for trial 3 1 09 M 2 03 M 3 01 M 4 04 M correct 5 02M Explanation Rate k A BY Rateg i k A2 f Ratel k B 80X10 4 7 02 02 10 gtlt10 4 01 1 8227 8 i 4 22 29 2 262 Rateg i kA3 B Ratel 2 1 480 gtlt 104 i 03 B 10 gtlt 104 7 01 012 48gtlt10 4 01 01 10gtlt10 4 03 04 Bla 009 100 points Which of the following does not affect the rate of a reaction 1 the temperature of the reactants 2 the presence of a catalyst 3 the value of Ea 4 the value of AHIXD correct Explanation AH is a state function it is independent of the path 010 part 1 0f 2 100 points Dr McCord7s helpful hint here You7ll need to double the rate constant here because of that coef cient of 77277 in front of the CQF4 Another way of saying it is that the rate con stant given in the problem is for the OVER ALL reaction and CQF4 must be reacting TWICE as fast a 2 The rate constant for the following equation is 00315 M 1 s71 We start with 0261 mol CQF4 in a 400 liter container7 with no C4Fg initially present ye xy755 7 H14 Kinetics 17 mccord 7 51625 5 2 CQF4 gt C4Fg What will be the concentration of 02134 after 100 hour Correct answer 000413 M Explanation k 00315 M71s71 t 100 hour 0261 1 02mm 006525 M Use the second order integrated rate law 1 1 7 m i akt 1 1 W 7 m i l kt 1 7 1 W 7 006525 M 2 00315 M 1 s l gtlt 3600 s A 000413 M 011 part 2 0f 2 100 points What will be the concentration of C4Fg after 100 hour Note that this is the PRODUCT of the reaction Correct answer 00306 M Explanation The concentration of C4Fg is equal to HALF the amount of CQF4 that was LOST in the reaction Since 000413 M 02134 remains 006525 M 7 000413 M is what reacted Take 1 5 of that and get the answer of 00306 M 012 100 points A certain reaction is rst order in reactant A and second order in reactant B If the concen tration of both reactants are doubled what happens to the reaction rate 1 The reaction rate increases eight fold correct 2 The reaction rate increases two fold 3 The reaction rate remains the same 4 The reaction rate increases sixteen fold 5 The reaction rate increases four fold Explanation 1f the reaction is rst order in reactant A and second order in reactant B the rate law would be written as Rate k A2 B So doubling the concentration of A will increase the rate twofold and doubling the concentration of B will increase the rate four fold 22 Thus the reaction rate will in crease by a factor of 2 gtlt 4 or eight fold 013 100 points A reaction A a products is observed to obey rst order kinetics Which of the following plots should give a straight line 7 1nA VS t correct 8 A VS2 15 Explanation ye xy755 7 H14 Kinetics 17 mccord 7 51625 The rst order integrated rate equation is n a 1 w M lnA0 7 lnA akt lnA 7akt 717 lnA0 Since a k and lnAO are constants7 this equa tion is in the form y man 717 b7 Where y ln A and x t A straight line is produced by y mar b 014 100 points A rst order reaction7 Where A 7 pro ducts has a rate constant of 156 gtlt 107 s71 At some time7 a concentration of 106 gtlt 10 6 M of species A is introduced into the reactor How long does it take for the concentration of A to fall by a factor of 8 1 t 88 nsec 88 gtlt 1078 sec 2 t 133 nsec 133 gtlt 10 7 sec correct 3 t 55 nsec 55 gtlt 10 9 sec 4 t 352 nsec 352 gtlt10 7 sec 5 t 44 nsec 44 gtlt 1078 sec Explanation A10 106 gtlt 106 M A 3 We 8 51 k15gtlt107s 1 The concentration of A falls by a factor of 87 so the nal concentration is 1 g106 gtlt 106 M 1325 gtlt10 7 M We lni akt Al 106 gtlt 106 7 ln 71156 gtlt 10 t 208 156 gtlt 1072 ti 208 7 156 gtlt 107 133 gtlt 10 7 seconds 015 100 points Consider the rst order reaction A 5 products Where 25 of A disappears in 24 seconds What is the half life of this reaction 1 1512 S 2 1512 S 31512 58 s correct 4 1512 S 5 1512 S 6 1512 S Explanation a 1 t 24 s A10 100 Alt 75 11 11 59 1199 gtlt 10 2 sf1 12 tlQDT58S 016 100 points 908r is one of the harmful radionuclides re sulting from nuclear ssion explosions 1t decays by a rst order process With a half life of 28 years HOW many years would it take for 9999 of a given sample released in the atmosphere to disintegrate Correct answer 372 years Explanation Since 9999 disintegrates7 001 remains 1 1 1512 y ye xy755 7 H14 Kinetics 17 mccord 7 51625 Ab 100 Ah 001 k 1 In t12 114 am M ifit t12 4 t12 ifio ti 11112 In 28y 100 71D 7 1H2 001 372y 017 100 points The rst order reaction X e products has Ea 4 kcaimol and a half life of 252 sec at 300 K What is the half life at 361 K Correct answer 810899 s Explanation Ea 4 kcaimol 7512 252 s T1300K T2361K 7512 Olt ak 1 7512 OC in M in Q E iii 1532 k1 R T1 T2 1 252 i 4000 calmoi n 1512 71987ca1m01K X 1 i 1 300K 361K f 5113387 310766 t12 t i 252 12 7 310766 810899 s munoz mmm3898 7 Homework 14 7 holcombe 7 51160 1 This print out should have 21 questions Multiple choice questions may continue on the next column or page 7 nd all choices before answering 001 100 points The reaction 2NO2g 2N0g 02g is postulated to occur via the mechanism N02g NO2g N0g N03g slow N03g a NOg 02g fast What is an intermediate in this reaction 1 N03g correct 2 NOg 3 N02g 4 ON7N03g 5 02g Explanation 002 100 points Consider the mechanism N02 F2 7gt NOQF F F N02 NOQF What is the rate law k1 slow kg 7 fast 1 rate k2 N02 2 rate k1 N02 F2 correct 3 rate k1 NOQF 4 rate k1 k2 N022 5 rate k2 N022 Explanation 003 100 points Consider the multistep reaction that has the overall reaction 2ABe2CD What would be the observed rate expression Mechanism 2AeC1 1BeCD slow fast 1 Rate k A2 B 2 Rate k A 3 Rate k 1 El 4 Rate k A2 correct 5 Rate k A 1 B Explanation The rst step is the slowest step the rate determining step7 so the rate law can be written based on the stoichiometry of the re actant 004 100 points A given reaction has an activation energy of 145 kJmol This reaction is normally run at room temperature 25 C and takes about an hour to get to completion At what temper ature should the reaction be run so that it is running twice as fast as at room temperature 1 50 C 2 65 C correct 3 135 C 4 31 C 5 77 C 6 98 C Explanation Use the Arrhenius equation For the k7s just put in 2 for kg and 1 for k1 and use the temperatures as appropriate T1 is room tem perature7 T2 is the unknown temperature Twice the rate at the same concentrations means double the value of k so that you can munoz mmm3898 7 Homework 14 7 holcombe 7 51160 2 use any k values that are in the ratio 2 l or 1 2 for that matter7 as long as you get the temperatures in the corresponding order Also remember to use Kelvin for temperature7 not Celsius degrees 005 100 points Suppose for some reaction the rate constant doubles in going from T 50 C to 70 C What is Ea for this reaction 1 8 kJmol 2 32 kJmol correct 3 24 kJmol 4 l6 kJmol 5 40 kJmol Explanation T1 50 C 7 273 323 K T2 70 C 273 343 K h i i k1 R T1 T2 k2 R lnk71 l 1 if 72 8314 JmolK ln 2 l l k22k1 Ea 323 K 7 343 K 335 kJmol 006 100 points Note the frequency factor in the Arrhenius equation is occasionally referred to as 77Arrhe nius factor77 What is the temperature that would pro duce a rate constant of 6 X 105 sec 1 if the activation energy is 40 kJ mol and the Arrhe nius factor is 62gtlt1012 1 305 K 2 298 K correct 3 260 K 4 25 K 5 273 K Explanation Ea 40000 M k 6 x 105 sec 1 A 6 gtlt 1012 k A67EaRT A EaRT i 6 k Ea 71 A RT n k T 7 E 7 A R1 7 n 1 40000 Jmol i 8314 Jmol K 62 gtlt 1012 ln 6 gtlt 105 298 K 007 100 points Increasing the temperature of a reaction causes 7 in the reaction rate 1 no change 2 an increase correct 3 a decrease 4 You need to know AH for the reaction Explanation At increased temperature more heat is pro vided to the reaction components7 so the num ber of effective collisions increase 008 100 points The graph describes the energy pro le of a reaction munoz mmm3898 7 Homework 14 7 holcornbe 7 51160 3 400 300 Energy kJ 50A Time What are the values for AH and Ea7 respec tively7 for the reaction in the direction writ ten 1 250 kJ7100 id 2 7250 kJ7 7100 id 3 250 kJ7 350 kJ correct 4 7250 kJ7100 id 5 7250 kJ7 350 kJ Explanation Time AH300 kJ750 kJ 250 kJ Ea 400 M 7 50 kJ 350 kJ 009 part 1 0f 4 100 points Consider the following potential energy dia gram gt Potential energy r Reaction progress gt Which arrow represents the potential en ergy of the reactants 11 2f 36 a a correct 5 c 6d Explanation 010 part 2 0f 4 100 points Which arrow represents the potential energy of the products 1 a 2 f correct 3b 56 6d Explanation 011 part 3 0f 4 100 points Which arrow represents the activation en ergy munoz mmm3898 7 Homework 14 7 holcornbe 7 51160 4 4d 5a 6 b correct Explanation 012 part 4 0f 4 100 points Which arrow represents the heat of reaction enthalpy change 1 0 correct 2 f 3 d 4 e 51 6a Explanation 013 100 points If Substance A is a catalyst7 Which equation best represents What happens in a chemical reaction 1XYZ2AXYZ22A 2XYZ2AXYZ2 3XYZ2 A7XY Z2 Ac0rrect 4XYZ22AXYZ2A Explanation 014 100 points What is the activation energy for a reaction if its rate constant is found to triple When the temperature is raised from 300 K to 310 K 1 4187400 Jrnol 2 20300 Jrnol 3 847900 Jrnol correct 4 1957600 Jrnol 5 No other choice is Within 3 percent Explanation T1300K T2310K k2 3 k1 1mg 77 iii k1 3 T1 T2 Rln Ea 1 1 1 T1 T2 The rate constant tripled7 so k2 3 k1 8314 Jrnol K ln3 1 1 300 K 7 310 K 84944 Jrnol 015 100 points Consider the reaction NOBrg a NOg Br2g A plot of NOBrli1 VS tirne gives a straight line With a slope of 200 M71 s7 The order of the reaction and the rate constant7 respec tively7 are 1 rst order and 200 s71 2 rst order and 0241 s71 3 second order and 200 M71s 1 cor rect 4 second order and 166 M71 s71 5 second order and 0500 M71 s71 Explanation munoz mmm3898 7 Homework 14 7 holcombe 7 51160 5 016 100 points A non steroidal anti inflammatory drug is me tabolized with a rst order rate constant of 325 dayil What is the half life for the metabolism reaction 1 0213 day correct 2 225 day 3 163 day 4 0308 day Explanation 017 100 points The decomposition of cyclobutane is a rst order reaction At a certain temperature7 the half life for this reaction is 137 seconds What fraction of a sample of cyclobutane would be left after 685 seconds at this temperature 1 00625 2 00156 3 00312 correct 4 0125 5 025 Explanation Isl21 t685s a 1 7 ln 2 t12 W a kt cth cyclobutanelt 7512 7 685 s ln2 7 137 s 34657 cyclobutanelo 7 34657 cyclobutanelt 7 32 cy clobut ane t 1 7 003125 cyclobutanelo 32 018 100 points Consider the reaction 1 H202aq gt 5 02aq 14205 if an aqueous system initially has a H202 of 016 M and two days later has a H202 of 002 M 7 what is the half life of H202aq 1 12 hours 2 16 hours correct 3 24 hours 4 not enough information 5 48 hours Explanation 016 057 002 x 3 half lives two days 48 hours 1512 16 hours 019 100 points The following graph is for a rst order reac tion Which are the appropriate units for the axes 1 xv axis temperature y axis Ea 1 2 xv axis t y axis lnA 1 me 1 1 y aXis 7 3 x axis temperature k munoz mmm3898 7 Homework 14 7 holcombe 7 51160 6 xv axis y axis A 021 100 points 4 tlme Which of the following does not affect the rate 5 xv axis time y axis k Of a Ieacuon 6 1 1 the value of Ea x axrs time7 y aXis W 2 the value of AHIXD correct 4 xv axis time y aXis lnA correct 8 wads time yams Ab 3 the presence of a catalyst 9 wads time yams Ea 4 the temperature of the reactants 1 Explanation 10 wads yams Ea AH is a state function it is independent of temperature the path Explanation 020 100 points The rst order rate constant for the conver sion A a B is 00693 minil How many grams of A would be left after 1 hour if the reaction started With 80 grams of A at time 0 1 100 grams 2 125 grams correct 3 50 grams 4 400 grams 5 Cannot be determined from the given information Explanation t 1 hour 60 min k 00693 min 1 a 1 Thus A ln a kt Al ln A0 7 lnA a kt ln A0 7 a kt lnA lnA ln80 g i 1 00693 minil 60 min 0224 A 60224 125 grams

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