Chem_208_S_2007_Prelim_I_Review

# Chem_208_S_2007_Prelim_I_Review - S Opalka Prelim I Review...

This preview shows pages 1–4. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: S. Opalka - Prelim I Review Session - Cornell University - Spring 2007 Part I: Heat, Temperature Change, and Heat Capacity. 1. Folks that use solar energy to heat their homes often store heat in beds of rocks. Assuming the homeowner stored heat in marble, which has a speciﬁc heat capacity of 0.880J/gK, how much heat would 100. kg of rocks store if their temperature increased by 100°C? 13*? QB: S x. m '2‘ AT - e at”. to: w- we» I- gt: >1“Ito Q Cb: <3\% 7(\OE' ‘5' \s. QUOSOU‘D-eck 2. A key component in rocket fuel is methylhydrazine (CH6N2). Upon combustion N2 (g g, 5 (a c WW CO2 (g), and H20 (1) are produced. You want to determme the heat of the reactlon when 4.32 g of methylhydrazine is combusted in a bomb calorimeter. You notice that a temperature changes by 10.43°C. Assume the heat capacity of the calorimeter is 7.832kJ/°C. When-.5 :. RCA‘QULLQ.) 4% 50109:?) "rm-'5 a=M1th '4‘" acolteﬁ A (““60 (In “an 2) (10\$;chth Men-k chxn -:> QoﬂWQ+ k-o mmv 0.1:wa __<5\L.q\<.5 ,2 L\0.\5C"\bbt :'%,'=r‘a. x\01§ L‘"7o';c3c\'\ewz_ '\ rm\ (3.9“, ”2. “CA CH'JIK S. Opaika - Prelim 1 Review Session - Cornell University - Spring 2007 Part II: Hess’s Law 3. Given the data: H; (g) + F2(g) 9 2HF (g) AH = —537 k] C (s) + 2F2 (g) 9 CF4 (g) AH = -680 kJ 2c (5) = 2H2 (g) 9 C2H4 (g) AH = +523 kJ use Hess’s law to calculate AH for the reaction: C2H4 (g) + 6F2(g) 9 2CF4 (g) + 4HF (g) -—-. Wag 4» 2.2a.L (933 --—-‘> LH-H? AH:~B’_>‘-=t waxzziom, {(-3 @4263 + #411ch ___~5 ace, R! (:3) AHtﬂoao K's a r-Ieeo \q Clut‘ (:9 Ms W?) a [am Add ‘= - 53“; k3 mun-mummy - 1~\\¢\ 2C ‘03 KS Part III: Using Enthalpies of Reaction 5. You notice that the appendix in your book is missing the standard enthalpy of formation value for CaC03 (5). You, however, are know the standard enthalpy change for the following reaction: CaCO3 (s) 9 CaO (s) + €02 (g) is 1178.1 k]. Given that AH°f(Ca0) = -635.5 kJ and AH°f (C02) = -393.5KJ. What is the standard entlﬁfé of fOTagolfO/rscﬂgﬁog‘w — 2 m INAOQ (.CQCJLLMIV‘\\ w“ v M" — [1362 (Coal + A“: (to; )1«— AH'HCQwQ nth " men \6 :L- base we sc- - “acne“: K33} - 1m"; (cater); S. Opalka - Prelim 1 Review Session - Cornell University - Spring 2007 Part IV: Obtaining and Manipulating Rate Laws Kinetic data were obtained for the following reaction at 250K: Roe-KG. = h E M01 “ 1 H2]? 2N0(g) + 2H2(g) —) N2 (g) + 2H20 (g) [N01 (M) [H2 (M) Initial Rate (M/s) 6. Determine the rate law for the reaction of interestP - ‘5 9 FVOMQMQ (“DALE 1'. AC [Chmv‘ E0\ZOX = 90““ K \o —: 9. '3 Q _— W rm\ Jet lie/<01“ (Gaol? : \\'3.?>=(\o‘3 E Ff0m®cn~d® c0313 __ J1 £0101“ C0401" -.-_ went K\o"3 ‘ a“ -:. '-\ fodﬁ‘ [040 '\ E0“ ? \t'l's‘K‘Q’?’ 7. What is the overall reactlon or er? 01 r\ "-'- '1 l omega“ occlet = '3- title-£5: C069 1 h [mil [“73 . . 3; 8. What is the rate constant for this react1on (Wlth units)? Rah-JALUOELEH'LX‘ .M ‘ Aflb M‘Ls-‘2JL _ *- _ 7.. Lifesaving: )REQt‘O “1 [0J0 3 l. :3- LC “5" :7. )K La'sna‘jg: J1(o.00\ M53 10. What would the rate be when [N0] = 0.043M and [H2] = 0.23M? RDA: :. L1 W‘s“ ENC)? 1:an ‘ my. :5.\ snot” .MIS 11. Assume that this reaction has an activation energy of 25 kJ/mol. Assuming Arrhenius behavior, calculate the rate constant if the temperature is increased by 50°C. 1353 4. vwcﬁ -:_ 15000: \$1., oh': We- E“ NIT“): m‘ ~ Ks "d O kZSO ‘1 AC- ERIE-T140 A- - J. \ - — “ESL '* U _. _. Eq at; Eq/ 2 \ “D a a Two Trio 7’00 _ e. 2: *19K —§_o, -..L .L __ A \L (Two- Two jam; - “use: 8. __ A \ A. ‘ - 26000 '3 {end "" \Q "' 1‘0; xix : \‘Q H’ZS' e q\3\\\-slmo\~K 31E: S. Opal/ca - Prelim 1 Review Session - Cornell University - Spring 2007 Part V: Reaction Mechanisms Your friend suggests the following mechanism for a reaction she is studying: Step 1: A +B—> <w;_£‘/(g) Step 2;»6" —2.é>éP 12. What is the overall reaction she is studying? Pats—av P 13. What are the intermediates in the reaction? C, 14. Use the steady state approximation to determine the rate law of this reaction. Aﬂ— 4; cock 0‘; Qosmiocx &[d* rcxNQ- 0"; dwkuc-an d [(-1 =0 Gist thsl— A Emma-— Aqua «- )2 [61: Autumn = slam + M. Ed’s 41¢»?ng :EQUM + M Jh—\+ A21 cits-7‘1 ._ halal-— h3.k.[A1LB‘S dt Iii—k *1“. 15. If i told you the ﬁrst step were fast and reversible and the second step was slow what rate law would you derive for the above mechanism using the simple equilibrium approximation? For Q‘wsﬁ- “bile-Q h‘EA-XLE‘L 2- 1MB“: 5‘53 1w M dL'PI : Alta—3 = In in. muss—1 (At A. 16. What assumptions would you make to have the rate law obtained using the steady state approximation equal the one obtained using the simple equilibrium approximation? mm; o? Qoxmdt: tame, oi} E‘QUQKQQ J9“ 3‘)» A: AW 2—.» W 1h». M" A \ ...
View Full Document

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern