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# How do I calculate the enthalpy of formation of ethanol?

I am very confused on how to calculate this. Please show me the process with explanations. Thanks!

Calculate the standard enthalp of formation of ethanol C2H5OH (l) from the heat of combustion of ethanol, which is -1368 kJ/mol, by using tabulated standard enthalpies of formation for CO2 (g) and H20 (l).

### 6 Answers

- denwel33Lv 51 decade agoFavorite Answer
The combustion of C2H5OH is as follows:

C2H5OH (l) + 3O2 (g) -----> 2CO2 (g) + 3H2O (l)

The heat of combustion of ethanol is -1368 kJ/mol.

Heat of formation of CO2 (g) = -393.509 kJ/mol

Heat of formation of H2O (l) = -285.83 kJ/mol

Total enthalpy change = total heat of formation of CO2 reaction + total heat of formation of H2O in reaction

Therefore, total enthalpy change = 2(-393.509) + 3(-285.83) = -787.018 + (-857.49)

= -1644.508 kJ/mol

Obviously, there is a difference between the total enthalpy change of the reaction and the heat of combustion of ethanol. This difference is the enthalpy of formation of the ethanol.

The difference = -1644.508 - (-1368) kJ/mol = -276.51 kJ/mol

Therefore, enthalpy change of formation of ethanol = -276.51 kJ/mol

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- bohlanderLv 43 years ago
Enthalpy Of Combustion Of Ethanol

Source(s): https://shrinks.im/a893j- Login to reply the answers

- brummetLv 43 years ago
Enthalpies Of Formation Table

Source(s): https://shrink.im/a0BBR- Login to reply the answers

- Anonymous5 years ago
RE:

How do I calculate the enthalpy of formation of ethanol?

I am very confused on how to calculate this. Please show me the process with explanations. Thanks!

Calculate the standard enthalp of formation of ethanol C2H5OH (l) from the heat of combustion of ethanol, which is -1368 kJ/mol, by using tabulated standard enthalpies of formation for CO2 (g) and...

Source(s): calculate enthalpy formation ethanol: https://trimurl.im/j36/how-do-i-calculate-the-enth...- Login to reply the answers

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- HPVLv 71 decade ago
The balanced equation for the combustion of ethanol (l) is

C2H5OH(l) + 3O2(g) ==> 2CO2(g) + 3H2O(l) . . .delta H = -1368 kJ

The delta H for any reaction can be calculated as

delta H reaction = sum of delta H of formation of products - sum of delta H of formation of reactants

From tables:

delta H f CO2(g) = -393.5 kJ/mole

delta H f H2O(l) = -285.8 kJ

delta H f O2(g) = 0

delta H reaction = ((2 moles CO2)(-393.5 kJ/mole) + (3 moles H2O)(-285.8 kJ/mole)) - ((1 mole C2H5OH)(delta H f C2H5OH(l))

-1368 kJ = (-787.0 kJ - 857.4 kJ) - (delta H f C2H5OH(l)

-276.4 kJ = delta H f C2H5OH(l)

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- 4 years ago
answer is -1235

- Alexa2 years agoReport
can you explain this?

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