Answer to Question #260379 in General Chemistry for Bestie

Question

The standard enthalpies of formation, at 25.00 oC, of methanol (CH4O (l)), water (H2O (l)), and carbon dioxide (CO2 (g)) are respectively -238.7 kJ / mol, - 285.8 kJ / mol, and -393.5 kJ / mol. Calculate the change in the entropy of the surroundings (in J / K) when burning 15.8 g of methanol under a constant pressure of 1,000 atm at 25.00 oC (NB, combustion is the reaction of a substance with oxygen molecular to produce water and carbon dioxide).

Answer 1

CH3OH+32O2→CO2+2H2O

To find the entropy of the surroundings, we will find the enthalpy of the system as

ΔHsys=ΔHf(CO2)+2×ΔHf(H2O)−32×ΔHf(O2)−ΔHf(CH3OH)ΔHsys=−393.5 kJ/mol+2×−285.8 kJ/mol−32×0 kJ/mol−(−238.7 kJ/mol)ΔHsys=−726.4 kJ/mol