Standard enthalpies of formation are obtained from thermodynamic tables as: C 2

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 ?Ho, then combine the equations to give the required process, and evaluate the associated ?Ho

Explanation / Answer

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

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