Consider the combustion of 3.26 mol of liquidethanol (C2H5OH) to gaseous water and car-bon dioxide. Calculate the enthalpy changefor this reaction. The pertinent enthalpies offormation (in kJ/mol)...

Consider the combustion of 3.26 mol of liquid
ethanol (C2H5OH) to gaseous water and car-
bon dioxide.

Calculate the enthalpy change
for this reaction. The pertinent enthalpies of
formation (in kJ/mol) are
Hf H2O = −241.8 Hf CO2 = −393.5
Hf C2H5OH = −277.7
Answer in units of kJ

Asked on by precious531

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jerichorayel | College Teacher | (Level 2) Senior Educator

Posted on



In order to answer the problem, we need to draw the chemical reaction.

The complete combustion of ethanol will yield water and catrbon dioxide. It can be written as:

C2H5OH    + O2 ---->  CO2  + H2O

by balancing the equation we can have

C2H5OH   +    3O2   ----> 2CO2 + 3H2O


Now we have to get the moles of each using stoichiometry.

Moles C2H5OH = 3.26 moles

Moles CO2=

3.26moles C2H5OH x (2moles CO2/1mole C2H5OH)

               = 6.52 moles CO2


Moles H2O = 

3.26 moles C2H5OH x (3moles H2O/1mole C2H5OH)

                = 9.78 moles H2O


Moles O2  =

3.26 moles C2H5OH x (3moles O2/1mole C2H5OH)

                = 9.78 moles O2


now we can solve for the enthalpy change:


enthalpy change = ( sum of nH of products) - ( sum of nH of reactants)


products: H2O and CO2

reactants: C2H5OH and O2

note: O2 does not participate in enthalpy change


enthalpy change =

= [(-241.8 kj/molx 9.78mol)+(-393.5kj/mol x 6.52 moles)] - [(-277.7kj/mol x 1mole)]

= (-2364.804 + -2565.62) - (-277.7)

= -4652.724 =-4653kJ



hope this helps :)


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