# Wright CHM 1220 - CHM 1220_Exam1_Fall2017_answers (3 pages)

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## CHM 1220_Exam1_Fall2017_answers

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- School:
- Wright State University
- Course:
- Chm 1220 - General Chemistry II

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Exam 1 Multiple Choice Answers 1 B 2 B 3 B 4 A 5 C 6 A 7 D 8 C 9 D 10 A 11 extra credit E 12 extra credit B Exam 1 Written Problem Solutions 13 Carbon tetrachloride CCl4 has a vapor pressure of 213 torr at 40 C and 836 torr at 80 C What is the normal boiling point of CCl 4 Relating vapor pressures at different temperatures P H vap 1 1 need to use Clausius Clapyeron equation ln 1 P 2 Don t know Hvap need to use data to calculate P1 213 torr T1 40 C 313 K P2 836 torr T2 80 C 353 K Plug in values solve for Hvap Now with Hvap known can calculate vapor pressure at new temperature R T 2 T1 Hvap H vap 1 213 torr 1 ln 836 torr 8 314 J 353 K 313 K mol K P1 213 torr T1 40 C 313 K P2 760 torr T2 H vap 3 14 104 J mol Hvap 3 14 104 J mol 1 1 3 368 10 4 P1 H vap 1 1 ln T2 313 K R T2 T1 P2 1 4 J 2 858 10 3 1 213 torr 3 14 10 mol 1 T2 ln J 7 60 torr T 3 13 K 8 314 T2 350 K 76 9 C mol K 2 14 Heptane C7H16 100 21 g mol and octane C8H18 114 23 g mol are constituents of gasoline At 80 C the vapor pressure of heptane is 428 torr and the vapor pressure of octane is 175 torr What is the mole fraction of heptane in a mixture of heptane and octane that has a vapor pressure of 305 torr at 80 C Raoult s law describes the vapor pressure Ptotal heptanePheptane octanePoctane of a mixture of volatile components Definition of mole fraction octane 1 heptane heptane octane 1 Substitute into first expression and solve for mole fraction of heptane Ptotal heptanePheptane 1 heptane Poctane Ptotal heptanePheptane Poctane heptanePoctane Ptotal Poctane heptane Pheptane Poctane Ptotal Poctane 305 torr 175 torr cyclopentane 0 514 Pheptane Poctane 428 torr 175 torr 15 The smell of ripe raspberries is due to 4 p hydroxyphenyl 2 butanone which has the empirical formula C5H6O To find its molecular formula you dissolve 0 135 g in 25 0 g chloroform CHCl 3 The boiling point of the solution is 61 82 C What is the molecular formula of the solute normal boiling point for CHCl3 61 70 C Kb 3 63 C m Relationship for boiling point elevation Tb iK b m Can solve for molality to get eventually to molecular weight fomula m Tb 61 82 C 61 70 C 0 03305 m raspberry smell 1 3 63 C m iK b Use molality as a conversion factor to get to mol raspberry smell 35 0 g CHCl3 1 kg 0 03305 mol raspberry smell 8 264 10 4 mol raspberry smell 1000 g 1 kg CHCl3 To find molecular formula from empirical formula Relationship for molar mass MM g 0 135 g raspberry smell 163 55 g mol mol 8 264 10 4 mol raspberry smell figure out ratio between them g 0 135 g raspberry smell g mol MM MM of C5H6O 4 5 12 6 1 1 16 163 5582 g mol mol 8 264 10 mol raspberry smell Actual MM is twice so multiple the two empirical formula by 2 C10H12O2 16 For the vaporization of benzene at its normal boiling point Hvap 30 8 kJ mol and Svap 87 2 J mol K Does 1 mol benzene evaporate spontaneously in a beaker heated to 100 C Calculate Suniv to justify your answer for a spontaneous process S univ Ssys Ssurr 0 benzene system beaker surroundings Assume from First Law of Thermodynamics qbenzene qbeaker Suniv Sbenzene Sbeaker Again benzene is undergoing a phase change qbenzene n H vap for melting phase change So But Sbenzene n Svap So Suniv n Svap Sbeaker Substituting into expression for Suniv q Sbeaker beaker Tbeaker Then q Suniv n S vap beaker Tbeaker qbenzene n H vap Suniv n Svap Plug in values n H vap Tbeaker Suniv 1 mol 87 2 J mol K 1 mol 30 8 103 J mol 373 K 4 63 J K Suniv 0 so yes benzene does evaporate spontaneously at 100 C 17 A rocket fuel would be useless if its oxidation were not spontaneous Although rockets operate under conditions that are far from standard an initial estimation of the potential of a rocket fuel might assess whether its oxidation at the high temperatures reached in a rocket is spontaneous A chemist exploring potential fuels for use in space considered using vaporized aluminum chloride in a reaction for which the unbalanced equation is Substance Gf kJ mol AlCl3 g 4 AlCl3 g 9 O2 g 2 Al2O3 s 12 ClO g O2 g 0 Al2O3 s 1034 ClO g 75 Balance the equation then use the data provided in the table which are for 2000 K to decide whether the fuel is worth further investigating Explain your decision Need to calculate Grxn Grxn n Gfo products n Gfo reactants kJ kJ kJ Grxn 2 mol 1034 12 mol 75 4 mol 467 0 700 kJ mol mol mol A reaction that has Grxn that has such a large positive value is extremely unfavored thermodynamically This particular reaction is nonspontaneous at 2000 K so another reaction perhaps would be better explored

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