Question

Standard enthalpies of formation are obtained from thermodynamic tables as:

C2H5OH(l) -224 kJ/mol

CO2(g) -398 kJ/mol

H2O(l) -281 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

State the answer to four significant figures without decimal point, ±xxxx

Answer 1

Formation of C2H5OH from its elements

2C + 3H2 + 1/2O2 = C2H5OH

Hf1 = - 224 kJ/mol

Reverse the equation

C2H5OH = 2C + 3H2 + 1/2O2.......... Eq1

H1 = 224 kJ/mol

Formation of CO2 from its elements

C + O2 = CO2

Hf2 = - 398 kJ/mol

Multiply by 2

2C + 2O2 = 2CO2.........eq2

H2 = - 2 x 398 = - 796 kJ/mol

Formation of H2O from its elements

H2 + 1/2O2 = H2O

Hf3 = - 281 kJ/mol

Multiply by 3

3H2 + 3/2O2 = 3H2O............eq3

H3 = - 3 x 281 = - 843 kJ/mol

Now add the above equations

Eq1 + Eq2 + Eq3

We get

C2H5OH + 3O2 = 2 CO2 + 3 H2O

From the Hess law

enthalpy change of the reaction

H = H1 + H2 + H3

= 224 - 796 - 843

= - 1415 kJ/mol