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Chemistry 2300 (Loader) Fall 2002 Page 1 of 3 T Chemistry 2300 Quiz 2 November 18 2002

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Chemistry 2300 (Loader) Fall 2002 Page 1 of 3 T Chemistry 2300 Quiz 2 November 18 2002 Chemistry 2300 (Loader) Fall 2002 Page 1 of 3 t Chemistry 2300 Quiz 2 November 18 2002 NAME:___________________________ MUN STUDENT#:____________________ Answer ALL of the questions in the spaces provided. Give your final answer to the correct number of significant digits and show your calculations. State any assumptions that you make. The mark that you obtain for this test will be used in calculating your final grade for the course. 1. Iodine adds to cyclohexene in CCl4 as solvent to form an unstable diiodide: C6H10(in CCl4) + I2(in CCl4) C6H10I2(in CCl4) The equilibrium constant, Keq at 25.0 °C = 20. [4] (a) Calculate DG) for the reaction at 25.0 °C ) 8.314 -1 -1 DG = -RT ln Keq = - 1000 kJ K mol % 298.15 K ln 20 -1 = -7.4259 kJ mol Ans: -7.4 kJ mol-1 [5] (b) Calculate the equilibrium concentration of the diiodide, C6H10I2, in the solution formed by making a solution in which the initial concentrations of both C6H10 and I2 = 0.0100 mol L-1 Let the equilibrium concentration of the diiodide = x mol L-1. C6H10(in CCl4) + I2(in CCl4) C6H10I2(in CCl4) Final concentrations (0.0100-x) (0.0100-x) x mol L-1 x 2 Keq = = 20 so x = 20(0100-x) (0.0100-x)2 solving for x using successive approximation with guess of 0.005 gives x = 1.457 x 10-3 mol L-1. Ans: 1.5 x 10-3 mol L-1 ) ) (c) The equilibrium constant, Keq at 35.0 °C = 13. Calculate DH and DS for the reaction. If the enthalpy change and entropy change to not vary significantly with temperature: Using the van’t Hoff equation, K ) ln eq35 = - DH 1 - 1 Keq25 R 308.15 k 298.15 K 13 DH) 1 1 ln 20 = - 8.314%10-3 kJ K-1 mol-1 308.15 k - 298.15 K ) -1 DH = -32.90 kJ mol Ans: DH) = -33 kJ mol-1 DG = DH - TDS calculating for 25.0 °C -1 -1 ) -7.4259 kJ mol = -32.90 kJ mol - 298.15 K % DS ) -1 -1 -1 -1 DS = -0.08544 kJ K mol or 85.44J K mol Ans : DS) = -85J K-1 mol-1 [6] Chemistry 2300 (Loader) Fall 2002 Page 2 of 3 2. Nickel, Ni, metal can be purified (Mond process) in a 2 steps in which it is first combined with carbon monoxide to give nickel tetracarbonyl, Ni(CO)4(g) in the reaction, Ni(s) + 4 CO(g) t Ni(CO)4(g). o -1 -1 o -1 S298/J K mol DHf,298/kJ mol Ni(s) 30 0 CO(g) 197.9 -110.5 Ni(CO)4(g) 313.4 -633 o [7] (a) Calculate DG f,298 and Keq(25 °C)for the the reaction at 298 K. o -1 -1 DrS = {313.4} - {30 + (4 % 197.9)} = -508.2 J K mol o -1 DrH = {-633} - {0 + [4 % (-110.5)]} = -191.0 kJ mol o -1 -508.2 J K-1mol-1 DrG = -191.0 kJ mol - 298 K % 1000 J kJ-1 o -1 -3 -1 -1 DrG = -39.48 kJ mol = -RTlnKeq = -8.314 % 10 kJ K mol % 298 K % lnKeq 6 Keq = 8.26 % 10 o -1 6 Ans: DrG = -39.5 kJ mol ; Keq = 8.3 % 10 [4] (b) If the Ni(CO)4(g) (boiling point 42 °C) is formed in the industrial process at 80 °C what is the equilibrium constant, Kp(80 °C), at this temperature and give reasons why the higher temperature rather than a lower one is chosen for the reaction. If the enthalpy change can be considered not to vary with temperature o ln K353 = - DH ( 1 - 1 ) so ln K2 = +191.0 1 - 1 then, K298 R T2 T1 8.26%106 8.314%10-3 353 298 gives K353 K = 50.7 Ans: K353 K = 51. Since the equilibrium constant is much smaller at 80 °C the yield of product at equilibrium is much lower that at lower temperatures. The higher temperature is probably chosen to speed up the reaction sufficiently to make it practical on an industrial scale. [4] (c) In the second step the pure nickel is recovered by heating the Ni(CO)4 until it decomposes (the reverse reaction). Calculate the temperature at which the reverse reaction becomes spontaneous. DG = DH - TDS if we assume that the entropy and enthalpy do not vary significantly with temperature then the decomposition reaction will become spontaneous when DG < 0. When DG = 0 = -191.0 kJ mol-1 - T(-0.508.2 kJ mol-1) giving T = 375.98 K. Ans: The reverse reaction becomes spontaneous above 3.8 x 102 K. OR o ln 1 = - DH ( 1 - 1 ) so ln 1 = +191.0 1 - 1 K298 R T2 T1 8.26%106 8.314%10-3 T2 298 T2 = 375.6 K Ans: The reverse reaction is spontaneous above 3.8 x 102 K [5] (d) The yield at equilibrium of Ni(CO)4 from Ni and CO can be improved if the reaction is done at increased pressure. However at 80 °C, the Ni(CO)4 forms the liquid under pressure so the temperature must also be increased. Calculate the maximum pressure that can be applied without the liquefaction of the Ni(CO)4 if the reaction is carried out at 152 °C. The normal boiling point of Ni(CO)4(l) is 42 °C and o -1 the enthalpy of vaporization at the normal boiling point, DHvap, 42 o C is 29.0 kJ mol In this case were are interested in the liquid-gas phase change of the liquid nickel tetracarbonyl. We will calculate the pressure such that the Ni(CO)4(l) is at its boiling point at 152 °C. o P2 DHvap 1 1 P2 129.0 1 1 ln = - ( - ) so ln = -3 - Po(normal bp) R T2 T1 1atm 8.314%10 425 315 P2 = 17.53 atm Ans: The pressure must not exceed 18 atm. Chemistry 2300 (Loader) Fall 2002 Page 3 of 3 3. (a) Bromine can exhibit a variety of different oxidation states. Given the following standard electrode potentials + - - - o 6 H (aq) + BrO3(aq) + 6 e d Br (aq) + 3 H2O(l); E298.15 K = 1.52 V - - o Br2(aq) + 2 e d 2 Br (aq); E298.15 K = 1.09 V calculate the standard electrode potential for the half-reaction + - - 12 H (aq) + 2 BrO3(aq) + 10 e d Br2(aq) + 6 H2O(l) [4] The first two equations can be combined to give the third. The calculate the half-cell potential the free energy change is calculated first and the potential since the electron transfer does not balance. (Multiply the first by 2 and reverse the second and add) + - - - o 12 H (aq) + 2 BrO3(aq) + 12 e d 2 Br (aq) + 6 H2O(l);DG = -12 % 1.52 V % F - - o 2 Br (aq) d Br2(aq) + 2 e ;DG = -2 % -1.09 V % F + - - o (-12%1.52 V%F)+(-2%-1.09 V%F) 12 H (aq) + 2 BrO3(aq) + 10 e d Br2 (aq) + 6 H2 O(l);DG = o (-12%1.52 V%F)+(-2%-1.09 V%F) So E for the third cell = -10 % F = 1.606 V Ans: 1.61 V (b) An electrochemical cell is constructed as shown in the cell diagram: - -1 - Pb PbF2(s)|F (aq, 1.00 mol L )||Cl (aq, a = 1)|AgCl(s)|Ag(s) The measured cell potential Ecell =0.574 V. Given that the standard reduction potentials for - - o AgCl(s) + e d Ag(s) + Cl (aq); E298.15 K = +0.222 V and 2+ - o Pb (aq) + 2 e d Pb(s); E298.15 K = -0.126 V o - - [3] (i) Calculate E298.15 K for the half-cell PbF2(s) + 2 e d Pb(s) + 2 F (aq) For the cell the half-reactions are: - - o AgCl(s) + e d Ag(s) + Cl (aq); E298.15 K = +0.222 V - - o Pb(s) + 2 F (aq) d PbF2 (s) + 2 e ;E298.15 K = ?? V - - o Overall 2 AgCl(s) + Pb(s) + 2 F (aq) d 2 Ag(s) + PbF2 (s) + 2 Cl (aq); E298.15 K = +0.574 V So the unknown ?? V = 0.574 - 0.222 V = 0.352 V so for - - o PbF2(s) + 2 e d Pb(s) + 2 F (aq); E298.15 K = -0.352 V Ans: -0.352 V [4] (ii) Calculate the solubility product for PbF2(s) based on the cell potential. Combining equations from above: - - o PbF2(s) + 2 e d Pb(s) + 2 F (aq);E298.15 K = -0.352 V 2+ - o Pb(s) d Pb (aq) + 2 e ;E298.15 K = 0.126 V gives 2+ - o PbF2 (s) d Pb (aq) + 2 F (aq);E298.15 K = -0.226 V the solubility reaction so we can calculate Ksp. o RT 8.314 J K-1 mol-1%298.15 K E298.15 K = 2F lnKsp so - 0.226 V = 2 % 96485 C mol-1 ln Ksp -8 give Ksp = 2.29 x 10 -8 Ans: Ksp = 2.29 x 10 [4] (c) An excess of zinc metal is added to a 100 mL of a 1.0 mol L-1 solution of copper(II) sulfate. The zinc displaces copper from the solution until equilibrium is reached. Calculate the concentration of copper(II) ion, [Cu2+(aq)] in the final solution. solution at 25.0 °C.
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