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# stuck on equation Watch

1. Question "When ethanol burns in oxygen under standard conditions, carbon dioxide, water and 1368 kJmol-1 of energy are produced. Calculate the enthalpy of formation of ethanol, given that the enthalpies of formation of carbon dioxide and water are -393.7 and -285.9 kJmol-1 respectivel"

Hi , Im stuck on to how I get to the answer on this question

I understand the answer by using the mark scheme is -227.1kjmol-1

but I cant seem to get to that , can anyone help please?
2. Are you sure the energy produced is +1368? Because I only get the final answer answer if the energy produced is (minus)1368 kJmol-1. If so, I would happily help you
3. C2H5OH + 3O2===>2CO2 + 3H2O dont forget to times the H2O and CO2 enthalpies as there is more than 1 mole of it

2 moles of CO2 so -393.7x2= -787.4
3 moles of H2O so -285.9x3= -857.7
-787.4+-857.7= -1645.1

then

-1645.1--1368= -277.1

Not sure if this is any help but this is how I would get to the answer

Edit: dont forget to make sure the equation is correctly balanced
C2H5OH + 3O2===>2CO2 + 3H2O dont forget to times the H2O and CO2 enthalpies as there is more than 1 mole of it

2 moles of CO2 so -393.7x2= -787.4
3 moles of H2O so -285.9x3= -857.7
-787.4+-857.7= -1645.1

then

-1645.1+1368= -277.1

Not sure if this is any help but this is how I would get to the answer
But for formation you have to do products minus reactants so if it is plus 1368 then your answer would be -3013.1 because + and - become -?
5. (Original post by Sameer599)
Are you sure the energy produced is +1368? Because I only get the final answer answer if the energy produced is (minus)1368 kJmol-1. If so, I would happily help you
The energy produced, i.e. energy released by the reaction, was +1368 kJmol-1. The enthalpy change was thus -1368 kJmol-1. The calculation will come out fine.
6. (Original post by Sameer599)
But for formation you have to do products minus reactants so if it is plus 1368 then your answer would be -3013.1 because + and - become -?

The energy produced, i.e. energy released by the reaction, was +1368 kJmol-1. The enthalpy change was thus -1368 kJmol-1. The calculation will come out fine.
this guys right
7. (Original post by Sameer599)
But for formation you have to do products minus reactants so if it is plus 1368 then your answer would be -3013.1 because + and - become -?
cheers this i definitely how to do it i confused myself by overcomplicating the question and trying to use vector diagrams.
8. Spent an hour on trying to get to the answer LOL wish I posted on here and saved the time, thanks very much for your help, I will know better next time.

Cheers

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Updated: April 3, 2013
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