Combustion! One mole of either carbon monoxide or benzene are completely combust
ID: 479361 • Letter: C
Question
Combustion! One mole of either carbon monoxide or benzene are completely combusted with oxygen at constant temperature and pressure (298 K and 1 atm) to generate CO_2 and H_2O. Assume all substances are ideal gases for calculating volume changes. Write out balanced combustion reactions for each reaction. Calculate the change in entropy for the system for each reaction, using the table, below Use the enthalpies of formation to calculate the heat lost or gained during this reaction. Use your result in part c) to calculate the change in entropy of the surroundings for each reaction. Also calculate delta U, delta A, delta G for the system for each reaction. Which of the four thermodynamic potentials (delta U, delta H, delta A, delta G) is suited for an experiment at constant T, P when evaluating if the reaction is spontaneous? What if the experiment is conducted in a rigid container at constant temperature?Explanation / Answer
CO+ C2H6 +4O2-----> 3CO2 + 3H2O
moles of CO and C2H6= 1 moles
moles of oxygen requried =4 moles of CO2 = 3 moles and moles of H2O= 3 mol
enthalpy balance, enthalpy change= 3* enthalpy of formationof CO2 +3* enthalpy of formation of H2O- { enthalpy of CO*1 +1* enthalpy of C2h6+4* enthalpy of formatino of O2}
=3* (-393.5)+3* (-241.8) - {(-110.5) +82.9+4*0} =-1878.3 Kj. 3,3, 1,1,4 are the coefficients of CO2, H2O, CO,C2H6 and O2 repsectively.
enthalpy change ,deltaH= deltU+d*(PV)= deltaU+deltan*RT
deltan= change in number of moles of products- reactants = 3+3-(1+1+4)= 0
hence change in internal energy =chane in enthalpy
delatS = 3*(213.7)+3*188.8-{ 197.7+269.3+4*205.1}=-79.9 J/K
deltaG= deltaH - TdeltaS = -1878,3+*298*79.9/1000=-1854.49 Kj
A= U-TS, deltaA= deltaU- T*deltaS= -1878.3+298*79.9/1000=-1854.49 Kj
deltaG is -ve suggests spontaneous process.