Question

(Hint: Once again we can use the same heats of formation process on enthalpy or entropy. Because they are both state functions). Calculate 45 px and AH , for the following reaction at 25°C using the information provided. N,C) + O2(g) + 2NO() Molecule s(molx) $H; (kl/mol) N (B) 0,(6) NOIR) 191.6 205.2 210.8

Answer 1

Ans: - → at 25 °C find AS, Mom and AH, AS, = 191.6 J/mol.K AS = 205.2 J/mol.K AS O = 210.8 J/mol.K N2(g) + O2(g) + 2NO(g) now entropy change for the wholereactionis AS = Asproducer – E 4Spectant = [2450]-[AS, +ASE] = [2x210.8J/mol.K]-[191.6 J/mol.K + 205.2J/mol.K] =[421.6J/mol.K]-[396.8J/mol.K] AS.Mn = 24.8J/mol.K this is per 2 mole NO for per mole NO it is 12.4J/mol.K AHP, = 0 kJ / mol AHO, = 0 kJ / mol AHO =91.3 kJ/mol N2(g) + O2(g) + 2N0g) now entropy change for the wholereaction is AH, = product - AH pectant = [2AH0]-[AH , +AH ] = [2x91.3 kJ / mol]-[0 kJ/mol +0 kJ / mol] = [182.6 kJ / mol]-[0 kJ / mol] AHM = 182.6 kJ / mol—this is per 2 mole NO for per mole NO it is AH, n = 91.3 kJ / mol|