Question

11. The standard enthalpy of combustion of ethyl ethanoate (CH3COOC2H5, liq.) is -2331 kJ mol at 298 K, and its standard molar entropy is 259.4 JK mol. Calculate the standard Gibbs energy for formation of the compound at 298 K. (The standard enthalpies of formation of CO2(g) and H2O(l) are -393.51 kJ and -285.83 k), respectively, at 298 K.)

is this correct or do i just plug in the values of the standard enthalpy of combustion and standard molar entropy given in the begining of the question into the gibbs equation

Answer 1

Dear Student, Kindly refer to the complete solution. Feel free to ask if you face difficulty in understanding any step of the solution attached, so that we can assist you in a better way.

Dear student, Actually, the standard enthalpy of combustion of ethyl ethanoate (hoogts us - 2231 kJ. molt but in the question we are provided with - 2331 kjmet thee we will consider - 2331 kJ molt as the data provided Pel ethyl ethanoate, DH = - 2331 kJ moll formation of ethyl ethanoate : (s) + 445 +0g (9) hy cool, Hgle) Fel combustion :- ery cooca H5 W + 509&g) > 460, ) + 4HOCO Standard Gibbs energy for the formation of molecule is given is a AGº Aphº - TAFS° - 0 Using the expression for standard enthalpy of combustion, we can calculate the " standard enthalpy of formation AH ScHº = { 44*° (004.g) + 444°C40,)} | - SAH (chloogH l + ANO P.T.O."

ACH°C CH. COOCHE, 1) = 944,4°(cy_g) + 4 AFH°C H30,4)} - Acto AH (0₂, 8)) = 0 OfHº (cf. Cooghs, l) = +(-393.51K5) + 4(-a85.83K) – (-2331) AfH = - 386.36 kJ molt To calculate standared entropy of formations 4,3° = E Sm (producte) - Esem (Reactant) - Son (CAL LOGHs, e) - (45 m (Ce) + 4 sm (h. g) + sm (02, g)} 259.4 J K molt -54(5.74Jkt mort 1 4 (130.685 k" mal) + 205. 14Jk mot AS = -491.42Jkt molt AFS = -0.4914 KJ kmal

4 Gº AFHOP T Sesº = -386-36 KJ mol - 2981 (-0.4914 KI kimer' segº = -239.9 KJ molt # Remember if we used H=2231 kJ/onal we will get AG= - 339.9 kJ molt Best wishes !!