74. Metal oxides can be reduced with hydrogen gas to give the metal. H2(g) + ZnO(s) <===> H2O(g) + Zn(s) 1. Compound ΔΗ: kJ/mol Sº J/Komol AGⓇF kJ/mol H2(g) ZnO(s) H2O(g) Zn(s) -348.3 -241.8 130.7 43.6 188.8 41.6 -318.3 -228.6 In which direction does the equilibrium above shift when you raise the temperature? a. left b. right c. no change 75. What is the entropy change for the above reaction? b.-56.1 J/K a. 56.1 J/K e. none of these c. 404.7J/K d. 43.6 J/K 76. Calculate the value of Gº for the above reaction. a. -546.9 kJ b. 546.9 kJ C.-89.7 kJ d. 89.7 kJ e. none of these 77. Is the above reaction predicted to be spontaneous at 25°C? a. yes b.no c. the reaction will be at equilibrium 78. Is the equilibrium constant for the above reaction a. less than 1 b. greater than 1 c. equal to 1
H2(g) + ZnO (s) = H2O(g) + Zn(s) 74) Since enthalpy (AHº) for ZnO = - 348.3 KJ/mol and for H20 = -241.8 KJ/mol, hence reaction is endothermic hence increase in temp causes equilibrium to move in right direction. 75)Entropy change =Entropies of products - Entropies of reactants = [°SH2O + OsZn ] - [OSH2 + "szno] =[41.6+188.8] -[ 130.7 + 43.6] = 56.1 J/Kmol 76) Free energy change EG of products - G of reactants = [OGH2O + OGZn ] - [°GH2 + °GZno] =[-228.6+0] -[ -318.3 +0] = 89.7 KJ/mol 77) At 25 °C = 273 +25 = 298 K AG=AH-TAS = -106.5 - 298 K x 56.1 J/Kmol = -106.50 KJ - 16.717KJ = - 89.78 KJ Negative sign of delta G confirms spontaneous reaction (yes). 78) at 25 C delta G is = A ºG and it is related with K as AG-RT In K, then In K=-A OG/RT =- 89780 J/8.314 x 298K = - 36.23, then log K = - 36.23/ 2.303 = -15.73 K= antilog -15.73 = 1.86, hence greater than one. -
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