Calculatethe r G o , r H o and r S o at 298 K and at 300 K for thecombustion of

Question

Calculatethe rGo, rHoand rSo at 298 K and at 300 K for thecombustion of ethanol.
Combustion ofethanol:
EtOH + 5/2O2 --> 2 CO2 + 5/2H2O
Other dataneeded: fGo EtOH, l =-174.78 kJ/mol fHo EtOH, l = -277.69kJ/mol fGo CO2,g = -394.36 kJ/mol fHo CO2, g = -393.51kJ/mol fGo H2O,g = -228.57 kJ/mol fHo H2O, g = -241.82kJ/mol fGo O2,g = 0 kJ/mol fHo O2,g = 0 kJ/mol
Do i just subtractdelta Gibbs of the products versus the reactants? can i do it thesame way with enthalpy? how do i find deltaentropy?
thanks!
Combustion ofethanol:
EtOH + 5/2O2 --> 2 CO2 + 5/2H2O
Other dataneeded: fGo EtOH, l =-174.78 kJ/mol fHo EtOH, l = -277.69kJ/mol fGo CO2,g = -394.36 kJ/mol fHo CO2, g = -393.51kJ/mol fGo H2O,g = -228.57 kJ/mol fHo H2O, g = -241.82kJ/mol fGo O2,g = 0 kJ/mol fHo O2,g = 0 kJ/mol
Do i just subtractdelta Gibbs of the products versus the reactants? can i do it thesame way with enthalpy? how do i find deltaentropy?
thanks! fGo CO2,g = -394.36 kJ/mol fHo CO2, g = -393.51kJ/mol fGo H2O,g = -228.57 kJ/mol fHo H2O, g = -241.82kJ/mol fGo O2,g = 0 kJ/mol fHo O2,g = 0 kJ/mol
Do i just subtractdelta Gibbs of the products versus the reactants? can i do it thesame way with enthalpy? how do i find deltaentropy?
thanks! fGo H2O,g = -228.57 kJ/mol fHo H2O, g = -241.82kJ/mol fGo O2,g = 0 kJ/mol fHo O2,g = 0 kJ/mol
Do i just subtractdelta Gibbs of the products versus the reactants? can i do it thesame way with enthalpy? how do i find deltaentropy?
thanks! fGo O2,g = 0 kJ/mol fHo O2,g = 0 kJ/mol
Do i just subtractdelta Gibbs of the products versus the reactants? can i do it thesame way with enthalpy? how do i find deltaentropy?
thanks!

Explanation / Answer

we know that G = H-TS now we need to calculate H    H = nHf(products)-nHf( reactants)         = (2*CO2 + 2.5*H2O)- (C2H5OH+2.5*O2)        = (2* -393.51kJ/mol)+(2.5* -241.82 kJ/mol)-(-277.69 kJ/mol)+2.5(0)       = -787.02-604.55 +277.6-0)kJ/mol      = -1113.97 kJ/mol     G = nGf(products)-nGf( reactants)         = (2*CO2 + 2.5*H2O)- (C2H5OH+2.5*O2)       = (2*-394.36 kJ/mol)+2.5*( -228.57 kJ/mol )-(-174.78 kJ/mol ) +2.5(0kJ/mol)      = (-788.72 -571.42 +174.78-0)kJ/mol      = -1185.36 kJ/mol from first equation      S = H-G/T           =(-1113.97 kJ/mol + 1185.36 kJ/mol)/T          = 71.39kJ/mol/ 298K         = 0.2395kJ/mol.K similary we can calculate for 300K            = (2*CO2 + 2.5*H2O)- (C2H5OH+2.5*O2)       = (2*-394.36 kJ/mol)+2.5*( -228.57 kJ/mol )-(-174.78 kJ/mol ) +2.5(0kJ/mol)      = (-788.72 -571.42 +174.78-0)kJ/mol      = -1185.36 kJ/mol from first equation      S = H-G/T           =(-1113.97 kJ/mol + 1185.36 kJ/mol)/T          = 71.39kJ/mol/ 298K         = 0.2395kJ/mol.K similary we can calculate for 300K

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