CHEM 105 HOUR EXAM III 28-OCT-99. = -163 kj/mole determine H f 0 for Ni(CO) 4 (g) = -260 kj/mole determine H f 0 for Cr(CO) 6 (g)

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1 CHEM 15 HOUR EXAM III 28-OCT-99 NAME (please print) 1. a. given: Ni (s) + 4 CO (g) = Ni(CO) 4 (g) H Rxn = -163 k/mole determine H f for Ni(CO) 4 (g) b. given: Cr (s) + 6 CO (g) = Cr(CO) 6 (g) H Rxn = -26 k/mole determine H f for Cr(CO) 6 (g) c. determine H Rxn for the reaction: Cr (s) + Ni(CO) 4 (g) + 2 CO (g) = Cr(CO) 6 (g) + Ni (s) 2. When 36. grams of water in the solid state at -1..C is mixed with 6. grams of water in the gas state at 15.C, what is the final temperature and state of the mixture? 3. Balance the following equation for a reaction presumed to take place in acidic media: OsO 4 (s) + C 3 H 7 O 2 (liq) = OsO (aq) + CO 2 (g) 4. Balance the following equation for a reaction presumed to take place in alkaline media: [ Sn 4 (OH) 1 ] 2 - (aq) + [ I(OH) 6 ] 1 - (aq) = [ Sn 4 (OH) 18 ] 2 - (aq) + I 2 O (aq) 5. Suppose grams of an organic compound of formula, C 3 H 7 O 2, is combusted in a bomb calorimeter. Suppose further that the calorimeter was determined in advance to have a heat capacity of 12.4 k /. Suppose further yet that before this particular combustion the temperature of the calorimeter was C, and that after the combustion the maximum temperature recorded was 2.13.C. As a final supposition, suppose that the combustion was carried out so that q v is the same as q p. Determine the enthalpy of combustion for this reaction. 6. Determine the enthalpy of formation for the organic compound of formula, C 3 H 7 O 2, that was combusted in a bomb calorimeter in problem 5 above. 7. A compound containing carbon, hydrogen, nitrogen and chlorine is decomposed under conditions of constant temperature and pressure, such that individual volumes of product gases can be measured. When 2 ml of the compound is decomposed, the individual volumes of the gas products are: 1. L carbon dioxide,.9 L water, 3 ml nitrogen, and 2 ml of chlorine. a. Determine the formula of the initial compound. b. Write a balanced chemical equation for the decomposition of the initial compound.

2 8. Suppose at an atmospheric pressure of 75 tore, the per cent by mass of various gases present in air are as follows: nitrogen 79 %, oxygen 19 %, argon 1.3 %, and that balance is made up of water vapor. a. Determine the moll fraction of each gas present in air. b. Determine the partial pressure of the: I. nitrogen gas. ii. argon gas. c. Suppose at the temperature of this sampling, the maximum possible vapor pressure of water is 14 tore. What is the per cent humidity of this air mass? 9. A gram sample of an organic liquid is placed in an evacuated 25 ml flask, sealed, and vaporized at 22.C. Under these conditions the vapor exerts a pressure of 11 tore. a. Determine the molecular weight of the organic substance. b. This organic liquid is known to be a binary compound composed of carbon and chlorine. In a separate experiment the compound is analyzed and found to contain per cent carbon by weight. i. Determine the empirical formula of the compound. ii. Determine the molecular formula of the compound. 1. For the following oxidation-reduction reaction occurring in acidic media: MnO (aq) + C 2 O (aq) = MN 2 + (aq) + CO 2 (g) Suppose a student weighed out.1952 g of solid sodium oxalate, NA 2 C 2 O 4, and reacted it according to the procedure used in Monday's laboratory exercise. Given an initial burette reading of.38 ml, and a final burette reading of ml, determine the molarity of the permanganate solution. ANSWERS 1. a. 163 k / mole = [ H {Ni(CO )} ] [ 4 H {CO} ] f 4 f H f Ni(CO) 4 = (-11) = - 63 k/mole b. 26 k/ mole = [ H {Cr(CO )} ] [ 6 H {CO} ] f 6 f H f Cr(CO) 6 = (-11) = - 92 k/mole c. reverse the equation shown for preparation of Ni(CO) 4, and add it to the equation shown for preparation of Cr(CO) 6 H Rxn = = -97 k / mole

3 2. a. work with water in the solid state: i. warm to.c q = ( 2.3 )(36 g)( - (-1) ) = ii. melt to liquid state at freezing point temperature q = 36 g 1 mole 6 18 g 1 mole = + 12, iii. total energy required for steps i and ii = = b. work with water in the gas state: i. cool gas to 1.C q = ( 2.2 )(6 g)(1-15 ) = - 61 ii. condense to liquid state at boiling point temperature q = 6 g 1 mole g 1 mole = iii. cool liquid to.c q = ( )(6 g)( - 1 ) = iv. total energy liberated in steps i, ii, and iii = = c. bring parts a. and b. together the energy involved in bringing = 42 g water from their initial conditions, to a common condition of liquid at.c, is algebraically add a. iii to b. iv = = this excess heat energy was removed from the 42 g water (i.e., the system) and placed in the "care" of the surroundings. It now needs to be returned (i.e., ADDED) to the system, and in so doing the system will gain in heat energy = ( )(42 g)(tf - ) Tf = = 19.4.C (4.184)(42) 3. 2 H OsO e 1 - = OsO H 2 O 4 H 2 O + C 3 H 7 O 2 = 3 CO e H 1 +

4 4. [ Sn 4 (OH) 1 ] OH 1 - = [ Sn 4 (OH) 18 ] e 1-2 [ I(OH) 6 ] e 1 - = I 2 O OH H 2 O 12.4 k = k 5. a. the calorimeter q(cal) = ( ) b. the compound q(cmpd) = - q(cal) q(cmpd) = k for g c. MW cmpd = 3(12) + 7(1) + 2(16) = 75 g/mole H Rxn = 75 g 1 mole k g = k / mole 6. balanced thermochemical equation for the combustion: C 3 H 7 O /2 O 2 = 3 CO 2 + 7/2 H 2 O H Rxn = k / mole (could multiply X 2) 2 C 3 H 7 O O 2 = 6 CO H 2 O H Rxn = k / mole o o o [ f 2 f 2 ] [ f ] H = 6 H (CO ) + 7 H (H O) 2 H (C H O ) Rxn -313 = [ 6(-393) + 7(-285) ] - [ 2 H f ( C 3 H 7 O 2 ) ] H f ( C 3 H 7 O 2 ) = - [ ] / 2 = k / mole 7. Under conditions of constant temperature and pressure, VOLUME is directly related to MOLES. SUBSTANCE VOLUME MOLES (divide all by 2 ml) compound 2 ml 1. ATOMS CO 2 1 ml 5. of CO 2 5 carbon H 2 O 9 ml 4.5 of H 2 O 9 hydrogen N 2 3 ml 1.5 of N 2 3 nitrogen Cl 2 2 ml 1. of Cl 2 2 chlorine

5 a. molecular formula C 5 H 9 N 3 Cl 2 b. decomposition reaction was a combustion reaction, b/c oxides of carbon and hydrogen are formed. C 5 H 9 N 3 Cl /2 O 2 = 5 CO 2 + 9/2 H 2 O + 3/2 N 2 + Cl 2 8. a. MASSES present in air sample: (for convenience, use grams in place of per cents) 79 grams N 2 19 grams O grams Ar.7 grams H 2 O convert all to moles mole N 2 = 79 / 28 = 2.82 mole O 2 = 19 / 32 =.594 mole Ar = 1.3 / 4 =.325 mole H 2 O =.7 / 18 =.389 total moles = = mol fraction N 2 = 2.82 / =.89 mol fraction O 2 =.594 / =.17 mol fraction Ar =.325 / =.932 mole fraction H 2 O =.389 / =.112 b. partial pressure = mol fraction x total pressure partial pressure N 2 = (.89)(75 torr) = 67 torr partial pressure Ar = (.932)(5 torr) = 7 torr c. partial pressure H 2 O = (.112)(75 torr) = 8.4 torr equilibrium vapor pressure H 2 O = 14 torr relative pre cent humidity = 8.4 / 14 = 6 % 9. a. VOL =.25 L TEMP = = 493 K PRESSURE = 11 / 76 atm g L atm 1.329? g/mole K mol ( atm )(.25 L) =.82 ( 493 K) MW = 237 g/mole b. i. mass carbon =.2526 (237) = 6 6/12 = 5 carbon atoms mass chlorine = = / 35.5 = 5 chlorine atoms empirical formula is: C Cl in a one : one ratio ii. molecular formula is: C 5 Cl 5

6 1. balance the equation: 2 [ 8 H MnO e 1 - = Mn H 2 O ] 5 [ C 2 O = 2 CO e 1 - ] multipliers of 2 and 5 give the CONVERSION FACTOR for this reaction as: every TWO moles MnO require FIVE moles C 2 O mole MnO moles MnO = moles C 2 O mole C O 2 4 (? M MnO 4 1 )( L) =.1952 g 134 g/mole [.4 ] [ MnO 4 1- ] =.216 molar