Name: Student Number: PRINT family name first UNIVERSITY OF VICTORIA EXAMINATIONS DECEMBER Chemistry 245 Introductory Physical Chemistry

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1 Name: Stuent Number: PRIN family name first UNIERSIY OF ICORI EXMINIONS DECEMER 2015 Chemistry 245 Introuctory Physical Chemistry Duration: 3 hours Instructor: M. G. Moffitt Print your name an stuent number in the saces above. his exam consists of a total of 11 ages (incluing cover sheet an formula sheet). Stuents must count the ages in this exam before beginning to write, an reort any iscreancy to the invigilator. ll questions must be answere in en on the exam aer in the sace rovie. Numerical answers must be written with the inicate units an aroriate significant figures. Inicate if you want the back of any ages to be marke. Do NO use your own scratch aer. Shar EL-510R, EL-510RN, or EL-510RN calculators only may be use. You may carefully remove the formula sheet from the back of the exam only when instructe by the invigilator. Do not han back the formula sheet. Maximum score = 110 oints. Question 1 Mark / 15 Question 2 Mark / 15 Question 3 Mark / 15 Question 4 Mark / 15 Question 5 Mark / 20 Question 6 Mark / 10 Question 7 Mark / 8 Question 8 Mark / 12 OL / 110

2 1. (15 oints) 2.00 mol of CO (a linear gas) is confine by a iston in a thermally-insulate cyliner at an initial ressure an temerature of 1.00 bar an 25.0 C, resectively. he gas is then comresse reversibly an aiabatically until the final ressure is 10.2 bar. Calculate U,, an S, q an w. ssume erfect gas behaviour. U = kj (6 oints) = kj (3 oints) S = J K -1 (2 oints) q = kj (2 oints) w = kj (2 oints) 2

3 2. (15 oints) g of ethanol, C 2 5 O (l), unergoes combustion in a bomb calorimeter at exactly 25 C (calorimeter constant C = kj K -1 ) resulting in a measure temerature increase in the calorimeter of C. Given the constant-ressure heat caacities below (assume to be constant with resect to temerature), calculate the enthaly of combustion for ethanol at exactly 75 C. C,m [C 2 5 O (l)] = J K -1 mol -1 C,m [CO 2 (g)] = J K -1 mol -1 C,m [ 2 O (l)] = J K -1 mol -1 C,m [O 2 (g)] = J K -1 mol -1 c (75 C) = kj mol -1 3

4 3. (15 oints) 4.52 mol of liqui CO 2 is confine by a iston at a temerature of K an an initial ressure of bar; at this ressure an temerature, CO 2 liqui an vaour are in equilibrium an the enthaly of vaorization is kj mol -1. he iston rests on the surface of the liqui so that only liqui is resent in the initial state. hen, the iston is rawn back isothermally until only CO 2 vaour is resent at a final ressure of bar. Calculate S for the isothermal rocess, assuming CO 2 vaour behaves as a erfect gas. S = J K -1 4

5 4. (15 oints) Dr. Fizz Kem has been given a solution to analyze. he start-u comany Freeze-Not has eveloe the solution as very smelly tye of anti-freeze. Phew, what in blazes is in this?! Dr. Kem asks Ms. Kay F., Freeze-Not founer an resient. It s an aqueous solution, Ms. Kay F. resons. ut I can t tell you the solute or its concentration rorietary knowlege. I can only tell you a few things that our scientists have learne so far. First, the solute is ilute, an it s not a olymer, so ieal-ilute behaviour is followe. Secon, the solution has an osmotic ressure of 122 kpa at 298 K. We nee you to etermine its freezing oint! Dr. Fizz Kem suenly remembers that her freezing oint aaratus is broken, so she whis out an enveloe, en, an her trusty Shar EL-510RN calculator to o some crafty calculations. nything else you can tell me? she asks Ms. F. n aqueous solution means the solvent is water. I know that. Sorry. Oh the ensity of the solution is the same as that of ure water, which is g ml -1. hat hels. hanks. n the freezing oint constant for water is K f = 1.86 K kg mol -1, as Ms. Kay F. hat hels too...wait Got it! cries Dr. Kem. ere s your answer What is the freezing oint of Freeze-Not s solution? f = C 5

6 5. (20 oints) Consiering the soliliqui hase iagram (right) for mixtures of the elements hoshorous (P) an sulfur (S), answer the questions below in the saces rovie. Use a ruler when require an reort comosition information to two significant figures an temerature information to three significant figures. (a) Ientify the three soli comouns of hoshorous an sulfur by inicating the missing stoichiometric coefficient b in each case, in orer of lowest to highest sulfur content. Comoun 1, P 4 S b : b = (2 oints) Comoun 2, P 4 S b : b = (2 oints) Comoun 3, P 4 S b : b = (2 oints) 6

7 5. (cont.) (b) What is the melting oint of the soli comoun containing the most S (Comoun 3)? f = C (2 oint) (c) Describe the system as Comoun 3 melts, by stating the nature (soli, liqui, or soli comoun) an comosition (mole fraction x S ) of all hases resent. Only fill in the chart for the number of hases that are resent, leaving unnecessary saces blank. (4 oints) Phase 1 Phase 2 Phase 3 i.) nature ii.) comosition x S = x S = x S = () liqui mixture of comosition x S = 0.30 is coole from 300C to 100C. Describe the system after cooling, by stating the nature (soli, liqui, or soli comoun), comosition (mole fraction x S ), an relative amounts (% of the hase) of all hases resent. Only fill in the chart for the number of hases that are resent, leaving unnecessary saces blank. (6 oints) Phase 1 Phase 2 Phase 3 i.) nature ii.) comosition x S = x S = x S = iii.) rel. amount % Phase 1 = % % Phase 2 = % % Phase 3 = % (e) he eutectic comosition containing the least S is x S =. (2 oint) 7

8 6. (10 oints) Calculate the vaour ressure (in kpa) of water at exactly 56C, given va = kj mol -1. va. ressure = kpa 8

9 7. (8 oints) Starting with the funamental equation U = S, erive the following general thermoynamic relation:. he following efinitions for the exansion coefficient,, an comressibility,, aly: Clearly write all stes, stating any assumtions. an. 9

10 8. (12 oints) Calculate the equilibrium constant at K for the reaction: C 3 8 (g) + C 4 (g) C 4 10 (g) + 2 (g), given the following thermoynamic ata at K: f [C 3 8 (g)] = kj mol -1 ; f [C 4 (g)] = kj mol -1 ; f [C 4 10 (g)] = kj mol -1 ; S m [C 3 8 (g)] = J K -1 mol -1 ; S m [C 4 (g)] = J K -1 mol -1 ; S m [C 4 10 (g)] = J K -1 mol -1 ; S m [ 2 (g)] = J K -1 mol -1. K = END OF EXM 10

11 11 Useful Constants an Formulas R (gas constant) = J K -1 mol -1 = L bar K -1 mol -1 = L atm K -1 mol -1 k (oltzmann constant) = x J K -1 N (vogaro constant) = x mol -1 1 atm = bar = 760 orr = kpa 1 L atm = J 1 L bar = 100 J 1L = 1 m 3 = 1000 ml = 1000 cm 3 0C = K tomic Masses: N = g mol -1 = g mol -1 C = g mol -1 e = g mol -1 O = g mol -1 r = g mol -1 2 n a nb nr * 1 1 * ln va R nr ',, J J n n G w q U ln R w ex ln * x R 2 P C F C q * x U K b S G b b b K q S rev f f b K W k S ln R f f i i gh trs trs 2 1,m r 1 r 2 r C 2 c M R M R c β β α α l n l n G G, r K R G ln r