Question

c) The condensation of water vapor d) The formation of CeHe() from its elements in their standard states 7. Consider the following data: C H R) -84.7 -32.9 CO(g) -393.5 -394.4 HaO(I) AH°, (kJ/mole) AG (kJ/mole) -285.9 f -237.2 a) Calculate the entropy change for the combustion of 1.0 mole of C,H (g) at 25°C. b) Are the products more or less ordered than the reactants at 25°C? c) Is the reaction spontaneous at 25°C? d) What is the maximum amount of useful work which can be obtained from the process at 25°C? e) If the process is carried out irreversibly at 25°C, under constant volume, what is the heat flow? f) If the process is carried out irreversibly under constant pressure at 25°C, what is the heat flow? g) What is the heat flow if the process is carried out reversibly at 25°C? h) If the process occurs under constant pressure at 25°C, what is the PAV work?

Answer 1

712 02 (g) 2 CO2 (g) + 3 H2oles QH6 (g) + Dioxn = (9) son products - { on reactants = 2X DG (O2(g) + 3x D G Hrow] - [ OG 2 H6185 1 X DG²0218) AG DG8 = 2xl-394.4)+ 3x (-237. [-32.9+o) Tonio = - 1467.5 KJ 1 mol ) Similarly DH = [2x DH Co2 + 3x DH Ho] - [ DH (246 + 7/24 DH 02] DH'= [2x1-343-5) + 3X1-285.9)-[-84.7 to] (DH - - 1560 kJ / mol ) T = 298 K Now entropy change, as is given by DG = DHTOSO - 1467.5 = - 1560 12988 DSS As = – 0.3104 kJ/mol. R osº = -310-4 I/ mark] on (b) since Dsro - s reactant ) s product It means that products are less orders them! Reactants because gaseous mole of fentents are more 10) since on Ko so sertion is spontaneous (d) wmax = 1on1 = [1467.5 kJ / moi