Question

Standard enthalpies of formation are obtained from thermodynamic tables as:

C₂H₅OH(l) -227 kJ/mol
CO₂(g)    -390 kJ/mol
H₂O(l)    -285 kJ/mol.

Calculate the enthalpy change of the reaction

C₂H₅OH(l) + 3 O₂ ? 2 CO₂(g) + 3 H₂O

Follow the procedures based on Hess's Law:
First write down the reactions corresponding to the enthalpies of formation you have been given, reverse the equations if necessary, remembering to change the sign of ?H^o, then combine the equations to give the required process, and evaluate the associated ?H^o

Answer 1

Formation of C2H5OH

2C(s) + 3H2(g) +1/2O2(g) = C2H5OH(l)

Reverse the equation

C2H5OH(l) = 2C(s) + 3H2(g) +1/2O2(g)....... Eq1

H1 = +227 kJ/mol

Formation of CO2

C(s) + O2(g) = CO2(g)

Multiply by 2

2C(s) + 2O2(g) = 2CO2(g)........... Eq2

H2 = - 390*2 = - 780 kJ/mol

Formation of H2O

H2(g) + 1/2O2(g) = H2O (l)

Multiply by 3

3H2(g) + 3/2O2(g) = 3H2O (l)........... Eq3

H3 = -285*3 = - 855 kJ/mol

From the Hess law

Add eq1 eq2 and eq3

We get

C2H5OH(l) + 3 O2 ? 2 CO2(g) + 3 H2O

enthalpy change

H = H1 + H2 + H3

= 227 - 780 - 855

= - 1408 kJ/mol