Question

The vaporization of compound A is described by the following chemical equation. A(1) A(8) A rap(72.39 °C) = 38.21 kJ/mol Calculate the entropy of vaporization, Aswap, for A(l) at 25.0 °C given that the boiling point of A is 72.39 °C, and the molar heat capacity of A(l) is 119.09 J/mol K). Assume that the molar heat capacity of A(R) is 56.5% of that of A(1). ASP J. mol-K Calculate the standard enthalpy of vaporization. A Hap, for A(1) at 25.0 'C. AHp KJ/mol Calculate the standard Gibbs free energy of vaporization, AG pat 25.0 °C AG vap kJ/mol

Answer 1

A(0) Alg) aniob = 38 Atuab = 38.21 KJ mol Boiling point = 72. 39 °C = 72-39 + 273=345.39 Molas heat capacity = 119.09 J molok Dhuap Boiling point 38.21 kJ/mode - 0.1106 nuo6 KI B . moq - k 345. 39K 1 Kg - 1000 on 110.6 g molok osvap = 110.6 Igor Temprature= 25°c = 298 K Arup As = - Temprature Dvoup 10.6 110. 62 = = 298 K e molok = J = 32-958 kJ = Alvab = 32958.8 mol- mal » sh = ? at T = 25°C = 290k luab= 38.21 h me 6 RJ A Suap = 1105 0.1106 - mol molok AG=AH-TAS - 38-21 Kg - 1298* X 0.11 06 kg ) mol = 5.25 kJ/mole

shvap = - RT en k R = 8.314 J molk R = 8.314x10 ks = 5.25 KJ RT enk 3KJ moe --R7 enk modok = 5.25 KJ .-8.314x102K5 x 456K en molok T= 183.0°C IT=456k = K = 0.250 Far Reaction to be spontanous value of an ehauld be -ve. no Bosh value come we when TAS> Drvop Ah= an-TAS AL TAS > Huap ДИ. - Т DS = > 38.2 kJ/mol 0.1106 kJ - molok = T 7 345. 3k So that tembrague should be Dh become negetive greates tran 345.3 k so and reaction become spontanous. T=345. 3K = 72. 3°C