Question

Brazilians are quite familiar with fueling their automobiles with ethanol, a fermentation product from sugarcane.
C2H5OH(l) +3O2(g) = 2CO2(g)+3H2O(g)
Calculate the standard molar enthalpy for the complete combustion of liquid ethanol (C2H5OH) using the standard enthalpies of formation of the reactants and products. ansswer in kj

Accepted Answer

HESS'S LAW STATE :ΔH0(REACTION)=\SUM ΔHF0(PRODUCTS) - \SUM
ΔHF0(REACTANTS)WHERE  ΔHF IS AN ENTHALPY OF FORMATION, THE 0 -
SUPERSCRIPT INDICATES STANDARD STATE VALUES.

∑ ΔHF0(REACTANTS)= ΔHF0(C2H5OH) + ΔHF0(O2)THE ENTHALPY OF
FORMATION OF A SIMPLE SUBSTANCE IS 0.\IMPLIES ∑ ΔHF0(REACTANTS)=
ΔHF0(C2H5OH)ΔHF0(C2H5OH)=– 277,6 KJ/MOLE

∑ ΔHF0(PRODUCTS)= 2 * ΔHF0(CO2) + 3 * ΔHF0(H2O)ΔHF0(CO2)=–
393,5 KJ/MOLEΔHF0(H2O)(GASE)= –241.8 KJ/MOLE∑ ΔHF0(PRODUCTS) =
-1512,4 KJ/MOLEΔH0(REACTION)=-1234,8 KJ/MOLE

[REFERENCE (TABLE) VALUES ΔHF0(PRODUCTS) ]

Question #99586

Expert's answer

Hess's law state :ΔH⁰_(reaction)= $\sum$ ΔH_f⁰_(products) - $\sum$ ΔH_f⁰_(reactants)where ΔH_f is an enthalpy of formation, the ⁰- superscript indicates standard state values.

∑ ΔH_f⁰_(reactants)= ΔH_f⁰(C₂H₅OH)+ ΔH_f⁰(O₂)The enthalpy of formation of a simple substance is 0. $\implies$ ∑ ΔH_f⁰_(reactants)= ΔH_f⁰(C₂H₅OH)ΔH_f⁰(C₂H₅OH)=– 277,6 kJ/mole

∑ ΔH_f⁰_(products)= 2 * ΔH_f⁰(CO₂) + 3 * ΔH_f⁰(H₂O)ΔH_f⁰(CO₂)=– 393,5 kJ/moleΔH_f⁰(H₂O)_(gase)= –241.8 kJ/mole∑ ΔH_f⁰_(products)= -1512,4 kJ/moleΔH⁰_(reaction)=-1234,8 kJ/mole

[reference (table) values ΔH_f⁰_(products) ]

Question #99586

Expert's answer

Hess's law state :ΔH0(reaction)=∑\sum∑ ΔHf0(products) - ∑\sum∑ ΔHf0(reactants)where ΔHf is an enthalpy of formation, the 0 - superscript indicates standard state values.

∑ ΔHf0(reactants)= ΔHf0(C2H5OH) + ΔHf0(O2)The enthalpy of formation of a simple substance is 0. ⟹ \implies⟹ ∑ ΔHf0(reactants)= ΔHf0(C2H5OH)ΔHf0(C2H5OH)=– 277,6 kJ/mole

∑ ΔHf0(products)= 2 * ΔHf0(CO2) + 3 * ΔHf0(H2O)ΔHf0(CO2)=– 393,5 kJ/moleΔHf0(H2O)(gase)= –241.8 kJ/mole∑ ΔHf0(products) = -1512,4 kJ/moleΔH0(reaction)=-1234,8 kJ/mole

[reference (table) values ΔHf0(products) ]

Hess's law state :ΔH⁰_(reaction)= $\sum$ ΔH_f⁰_(products) - $\sum$ ΔH_f⁰_(reactants)where ΔH_f is an enthalpy of formation, the ⁰- superscript indicates standard state values.

∑ ΔH_f⁰_(reactants)= ΔH_f⁰(C₂H₅OH)+ ΔH_f⁰(O₂)The enthalpy of formation of a simple substance is 0. $\implies$ ∑ ΔH_f⁰_(reactants)= ΔH_f⁰(C₂H₅OH)ΔH_f⁰(C₂H₅OH)=– 277,6 kJ/mole

∑ ΔH_f⁰_(products)= 2 * ΔH_f⁰(CO₂) + 3 * ΔH_f⁰(H₂O)ΔH_f⁰(CO₂)=– 393,5 kJ/moleΔH_f⁰(H₂O)_(gase)= –241.8 kJ/mole∑ ΔH_f⁰_(products)= -1512,4 kJ/moleΔH⁰_(reaction)=-1234,8 kJ/mole

[reference (table) values ΔH_f⁰_(products) ]