DOC PREVIEW
Wright CHM 1220 - CHM 1220_Exam1_Fall2017_answers

This preview shows page 1 out of 3 pages.

Save
View full document
View full document
Premium Document
Do you want full access? Go Premium and unlock all 3 pages.
Access to all documents
Download any document
Ad free experience
Premium Document
Do you want full access? Go Premium and unlock all 3 pages.
Access to all documents
Download any document
Ad free experience

Unformatted text preview:

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) BExam #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 CCl4? 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? 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, CHCl3. 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) P1 = 213 torr T1 = 40 °C = 313 K T2 = 80 °C = 353 K ΔHvap = ? P2 = 836 torr Relating vapor pressures at different temperatures, need to use Clausius-Clapyeron equation: Don’t know ΔHvap, need to use data to calculate: Plug-in values, solve for ΔHvap: 12vap2111lnTTRHPPK 1331K 5331KmolJ314.8 torr368 torr132lnvapHmolJ1014.34vapHNow with ΔHvap known, can calculate vapor pressure at new temperature: 12vap2111lnTTRHPPK 13311KmolJ314.8molJ1014.3 torr607 torr213ln24TC 76.9K 350 10858.21 K 1331110368.323224TTT  514.0 torr175 torr428 torr175 torr 0531octaneheptaneoctanetotalnecyclopentaoctaneheptaneheptaneoctanetotaloctaneheptaneoctaneheptaneheptanetotaloctaneheptaneheptaneheptanetotalPPPPPPPPPPPPPPPoctaneoctaneheptaneheptanetotalPPP1octaneheptaneheptaneoctane1P1 = 213 torr T1 = 40 °C = 313 K T2 = ? ΔHvap = 3.14×104 J/mol P2 = 760 torr Raoult’s law describes the vapor pressure of a mixture of volatile components: Definition of mole fraction: Substitute into first expression and solve for mole fraction of heptane: Relationship for boiling point elevation: Relationship for molar mass: miKTb bg/mol 55.163smellraspberry mol 108.264smellraspberry g 0.135mol g4-MMUse molality as a conversion factor to get to mol raspberry smell smellraspberry mol 10264.8CHCl kg 1smellraspberry mol 03305.0g 1000kg 1CHCl g 0.35433Can solve for molality to get eventually to molecular weight/fomula,   smellraspberry 03305.0C/ .6331C 70.61C 1.826bmmiKTmbTo find molecular formula from empirical formula, figure out ratio between them: the two: g/mol 55.163smellraspberry mol 108.264smellraspberry g 0.135mol g4-MM g/mol 82 1(16) 6(1) 5(12)OHC of 65MMActual MM is twice, so multiple empirical formula by 2: 21210OHC16. 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. 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: 4 AlCl3(g) + 9 O2(g) → 2 Al2O3(s) + 12 ClO(g) 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. Substance Gf °(kJ/mol) AlCl3(g)  O2(g) 0 Al2O3(s) –1034 ClO(g) 75 0surrsysuniv SSSbeakerbenzeneqq vapbenzeneHnq vapbenzeneSnS for a spontaneous process: benzene = system; beaker = surroundings: Assume from First Law of Thermodynamics: Again, benzene is undergoing a phase change, for melting phase change: beakerbeakerbeakerTqS So, beakerbenzeneunivSSS So, beakervapunivSSnS But, Then, beakerbeakervapunivTqSnS vapbenzeneHnq Substituting into expression for ∆Suniv, beakervapvapunivTHnSnSPlug-in values,     KJ 63.4K 373molJ108.30mol 1KmolJ 2.87mol 13univS∆Suniv > 0, so yes benzene does evaporate spontaneously at 100 °C )reactants()products(ofofrxn GnGnG     kJ 700 0molkJ467mol 4 molkJ75mol 12molkJ1034mol 2rxnGNeed to calculate ΔGrxn: A reaction that has ΔGrxn 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


View Full Document

Wright CHM 1220 - CHM 1220_Exam1_Fall2017_answers

Documents in this Course
Load more
Download CHM 1220_Exam1_Fall2017_answers
Our administrator received your request to download this document. We will send you the file to your email shortly.
Loading Unlocking...
Login

Join to view CHM 1220_Exam1_Fall2017_answers and access 3M+ class-specific study document.

or
We will never post anything without your permission.
Don't have an account?
Sign Up

Join to view CHM 1220_Exam1_Fall2017_answers 2 2 and access 3M+ class-specific study document.

or

By creating an account you agree to our Privacy Policy and Terms Of Use

Already a member?