Question

18.97 When 1.000 g of gaseous butane, C4H10, is burned at 25°C and 1.00 atm pressure, H2O(/) and CO2(g) are formed with the evolution of 49.50 kJ of heat. a Calculate the molar enthalpy of formation of butane. (Use enthalpy of formation data for H20 and CO2.) b AG of butane is -17.2 kJ/mol. What is AGo for the combustion of 1 mol butane? From a and b, calculate AS° for the combustion of 1 mol butane.
It says the answer is supposed to be A) -126kJ/mol

B) -2745.9kJ

C)-440J/K

Answer 1

Write the reaction as follows: CH,0 + 120, 400,(8)+ 54,0(1) Amount of heat evolved, q = - 49.50 kJ mass of C4H10 = 1.000 g molar mass of C4H10 = 58.0 g/mol mass of C,H, moles of CH = molar mass of C4H10 1.000 g 58.0 g/mol = 0.0172 mol Given, enthalpy of the reaction is, AHO. = — moles of C-H10 49.50 kJ AH* = -0.0172 mol AH = -2878 kJ/mol (a) Calculate the enthalpy of formation of butane as follows: AH = 54H;(H,0) + 4AH; (C0,)-84; (CH,)-12 14; (0) -2878 kJ/mol = 5(-286 kJ/mol)+4(-393.5 kJ/mol) - AH; (C, AH;(C.H.) = (-1430-1574+2878) kJ/mol = -126 kJ/mol Therefore, the required solution is -126 kJ/mol

(b) Calculate the Gibbs free energy of the reaction as follows: AC° = 5AG, (1,0)+486; (00,)-4C;(C.H.)-240;(0) AGⓇ = 5(-237.1kJ/mol+46-394.4 kJ/mol )=(-17.2kJ/mol) – 3 (0) AG° = (-1185.5-1577.6+17.2) kJ/mol AG° = -2745.9 kJ/mol Therefore, the required solution is -2745.9 kJ/mol - - - - - - - - - - - - - - - - - - - - - - - - - - - (C) Write the equation for Gibbs free enrgy as follows: AGⓇ = AH - TAS ...... (1) Temperature, T = 25°C = 25+273 K = 298 K Substitute the correpsonding values in equation (1) and calculate AS. 4.5° = AH'-ACT -2878kJ/mol -(-2745.9 kJ/mol) 298 K AS -132. 1 kJ/mol 1000 J 298 K 1kJ AS° = -442 J/K AS = -440 J/K Therefore, the required solution is -440J/K