a5s - Chemistry 2000B Spring 2002 Assignment 5 Solutions...

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Unformatted text preview: Chemistry 2000B Spring 2002 Assignment 5 Solutions (a) The reaction is 7CO2 g 3H2 O l ¢ £¡ ¤ ¥¡ 15 O 2 2g ¢ £¡ C6 H5 COOH s ¦ §¡ 1. Therefore 384 8     ¦ ¢¡ ¢ ¦  ¦  ¦ 3227 3 kJ mol  #  ¦ ©  ¦ © ¦ ¦  ©  © ¦ ¦ "© ¨  ¦ ¦ ! q   © ¨ nC6H5 COOH  ¨ 3227 3 kJ mol 1 0043 g 8 2237 10 3 mol 122 123 g mol ¯ nC6 H5 COOH ∆H 8 2237 10 3 mol 26 540 kJ  ¦ 285 830 15 ¯ ∆H f O 2 2 15 0 kJ mol 2  l C6 H5 COOH  ¨ 3 ¯ ∆H f  ¡   393 509 H2 O ¨ ¯ 3 ∆H f  ¡ 7 CO2 ¨ ¯ 7 ∆H f ¡ ¯ ∆H © ©  (b) No. A bomb calorimeter is a constant volume device. The heat produced at constant volume is not necessarily the same as the heat at constant pressure. (a) The density of water is about 1 g mL. Since ginger ale won’t have exactly the same density as water at the same temperature, there is no point using a more accurate value than this. Accordingly, 300 mL of ginger ale weighs about 300 g. If there is no heat exchange with the room, we have  % 94 J K 4 30 C mice 333 J g  ¨     #   ¨ $ ¢% ¨ ¢% © ¦   ¨  &  ¢ § ¨    ( ) '  &  ¢ ¨  ¢ ¦ ¦ &   ¢ © $ 300 g 4 18 J K 1g 1 4 30 C mice 2 06 J K 1g 1 0 5C 11 mice 4 18 J K g 4 0C 35048 J 360 J g mice 97 g ¨  © ¦ ¢ ¢ ¦  © © mice melted ice warming to 4 C ice warming to 0 C ¢% ice melting glass cooling to 4 C ¨ ginger ale cooling to 4 C % $ 0 $ q $ ! (b) Each ice cube has a volume of 2 7 cm 3 20 cm3 . Since the density of ice is 0 917 g cm3 (p 32 of the textbook), this corresponds to 18 g per ice cube. You therefore need five ice cubes to cool your drink. Hopefully, your glass is large enough for this!  ¦ ©  ¦ 2. (c) As your drink begins to cool below room temperature, heat will tend to flow from the surroundings into your drink. It will therefore be necessary to add slighly more ice than calculated above. (a)   ¦ 67 4 kJ mol 67 4 103 J mol 8 314 510 J K 1mol 1 298 15 K 4 5  ¦  & ¦ 3 ¦ ¦ 1011   ©  © ! © 6 51 © ¨ exp 0 kJ mol  © e 137 168 RT Cl2 ¦ ¯ ∆G ¯ ∆G f  ¡ 1 20   ¦  ¡ ¨ 204 6 CO ¦ ¨ K ¯ ∆G f COCl2  ¡ ¯ ∆G f ¨ ¯ ∆G ¡ 3. © ¦ (b) The equilibrium constant is enormous, so the reaction will proceed until just about all of the limiting reagent (chlorine) is used up. Accordingly, the equilibrium pressures of COCl2 and of CO are 0.25 and 0.15 atm, respectively. The equilibrium pressure of chlorine can be calculated from the equilibrium relation: aCOCl2 K aCO aCl2 aCOCl2 0 25 aCl2 2 6 10 12 KaCO 6 51 1011 0 15 ¦  ¦ © ¦   ¦ ¦ &    & ©  © ¦ ! © The equilibrium pressure of chlorine gas is therefore 2 6 12 atm. 10 ¦ 4. The reaction is  6 ¥¡ 1 5 CO 137 168 kJ mol    ¡ 10  ¦ 19 5 103 J mol 8 314 510 J K 1mol 1 298 15 K 4 ¦  3 ¦ ¦ © ¦   ¦ 2 10 ¦ ¦ 3 90  & © ! © © ¦ 2 57 103 0 208  ©  1  103 19 5 kJ mol 4  ¦ 2 57 5 ¯ 5 ∆G f ¦ exp 0 Fe  © ¨ ¦ ¦ 9 9  aCO e RT ¯ ∆G f ¦ ¨ 5 ¯ ∆G ¡¡ 705 3 5 ¨ ! aCO K Fe CO ¡ ¯ ∆G f K . We can calculate K as follows: 7  (a) The reaction becomes spontaneous when Q ¨ © aCO There are at least two ways to solve this problem: ©8 ¡ Q ¯ ∆G 5l ¦ ¢¡ For this reaction, Fe CO  5CO g ¦¡ Fe s ! 9 The reaction therefore becomes spontaneous when PCO 0 208 atm. ¯ ¯ ¯ (b) The reaction becomes spontaneous when ∆G 0. Since ∆G ∆G taneity requires that ¦ 7 ¦ ¨ ¦ K ¦ © 7   ¨ ln Q 7 0 7 ! ! Q © RT ln Q ¯ ∆G RT ¯ e ∆G RT ¢¨ ¯ ∆G RT ln Q, spon- The rest of the calculations are identical to those used in the first approach. 5. Solid barium sulfate dissociates in water: Ba2aq 2 SO4 aq ¢¡ @ ¦¡ 561 kJ mol   1362 2 kJ mol ¦ BaSO4 BaSO4 ¡ © C  ¦ ¨ ¢¡A ¢¡B  ¢  3 ¨ ¡B ¨ 744 00 ¯ ∆G f ¨ 2 SO4 ¯ ∆G f 10 10 ¦¡ ¦ ¦  ¦ ¨ ¨   57 ¯ ∆G f  & ¯ ∆G 298 15 K ln 1 1 ¦ ¨ ©  © © © ¨ © © ¡A Ba2 4 1  RT ln K 8 314 510 J K 1mol 57 kJ mol ¯ ¯ ∆G f Ba2 ∆G f SO2 From this equilibrium con- ¦ © ¯ ∆G f ¦ ¯ Also, ∆G  ¯ ∆G 10 .  The equilibrium constant for this reaction is Ksp 1 1 10 stant, we can calculate the standard free energy of reaction:  6¡ BaSO4 s ¨ ! ...
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