CH302 Spring 2006 Worksheet 2 Key (Values for ∆S and ∆H can be found in Appendix 2 in the text.) Part 1. Determining Reaction Spontaneity based on ∆Stotal Reaction at 300K, constant P & T ∆Ssystem ∆H ∆Ssurroungding ∆Stotal Is the rxn spontaneous? NaCl(s) Æ Na(s) + ½ Cl2(g) (51.21 J/K +1/2 * 223.07 J/K ) - 72.13 J/K = 90.615 J/K 411.15 kJ -411.15 kJ/300K = -1370.5 J/K -1279.885 J/K No CaO(s) + H20(l) Æ Ca(OH)2 (aq) 83.39 J/K – (39.75 J/K + 69.91) = -26.27 J/K -986.09 kJ – (-635.09 -285.83) = -65.17kJ 65170 J/300K = 217.2333 J/K 190.963 J/K Yes C6H6(g) Æ 6C(s) + 3H2(g) (3* 130.68 + 6* 5.74 ) – 269.31 = 157.17 J/K -82.9kJ 82900J/300K =276.3333 J/K 433.503 Yes Cu2O(s) + ½ O2(g) Æ 2CuO(s) 2* 42.63 – (93.14 + ½ * 205.14) = -110.45 J/K 2* -157.3-(-168.6) = -146 kJ 146000J/300K = 486.6667 J/K 597.16667 J/K Yes CO2+ 4H2 Æ CH4 + 2H20(g) 186.26 + 188.83 * 2 – (213.74 + 4* 130.68) = -172.54 J/K (-74.81 + 2*-241.82) – (-393.51) = -164.94 kJ 164940 J /300 K = 549.8 277.26 Yes Part 2. Finding an Equilibrium Temperature from ∆S (assume that S and H doesn’t change much with temperature) Reaction/Process ∆Ssystem ∆H Temperature which reaction/process is in equilibrium CH3OH(l) Æ CH3OH(g) 293.81-126.8 = 167.01 J/K -200.66- (-238.86) = 38.2 kJ - 38200 J / - 167.01 J/K = 228.72 K I2(g) Æ I2(s) 116.14 – 260.69 =-144.54 J/K -62.04 kJ 62040 J/ 144.54 J/K = 429.22 K N2(g)+ 3/2H2(g) Æ NH3(g) 192.45 – (191.61 + 3/2 * 130.68 ) = -195.18 J/K -80.29kJ 80 290 J / 195.18 J/K = 411.36 K 3O2(g) Æ 2O3(g) 2* 238.93 – 3* 205.14 = -137.56 J 2* 142.7 = 285.4 kJ - 285400 / - 137.56 = 2074.73 KPart 3. True or False (if false, explain your answer) 1. The Second Law of Thermodynamics says that the entropy of any system is always increasing False. The Second Law states that the entropy of the isolated system (universe) increases for a spontaneous reaction. 2. The universe is an isolated system. True. Don’t confuse the system (which is the reaction) and isolated system (which is usually the universe). 3. A process in which ∆S decreases is not spontaneous False. If ∆S of the surroundings exceeds ∆S of the system then the reaction is spontaneous. This can occur when a reaction is sufficiently exothermic that the heat liberated in the process sufficiently increases the entropy of the surroundings. 4. ∆S equals to -∆H/T when the process happens at constant pressure and temperature. True. Part 4. Determining Stability of a Compound from the Free Energy of Reaction Write out and balance the formation reaction at 300K: ∆H ∆S Free energy ∆G (using the formula with ∆H and ∆S) Is the compound stable at room temperature? H2(g) + ½ O2(g)Æ H20(g) -241.82 kJ 188.83 – (130.68 + ½ 205.14) = -44.42 J -241820 – (-44.42* 300) = -228494 J Yes C + O2Æ CO2 -393.51 kJ 213.74 – (5.740 + 205.14) = 2.86 J/K -393510 – 2.86*300 = -394368 J Yes H2 + I2 Æ 2HI 26.48 kJ 2*206.59- (116.14+ 130.68) = 166.36 J/K 26480 – 166.36 *300 = -23428 J Yes K(s) Æ K(g) 89.24 kJ 160.34-64.18 = 96.16 J/K 89240 – 96.16 *300=60392 J No 4P Æ P4 58.91 kJ 279.93 – 4*41.09 = 109.57 J/K 58910 – 300*109.57 = 26039 J
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