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enthalpy question

5. The enthalpy of combustion of ethanol is -1380 kJmol-1. 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 respectively.

This is from the A-level chemistry website! the answer is -265.1 but i can't seem to get it :frown:
My final calculation ends up being -1380 - 1645.1 which is wrong but doing +1380 - 1645.1 seems to get the right answer but i am not too sure how to get the +1380.
Can someone please help?
Original post by Frizzle16523
5. The enthalpy of combustion of ethanol is -1380 kJmol-1. 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 respectively.
This is from the A-level chemistry website! the answer is -265.1 but i can't seem to get it :frown:
My final calculation ends up being -1380 - 1645.1 which is wrong but doing +1380 - 1645.1 seems to get the right answer but i am not too sure how to get the +1380.
Can someone please help?

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Did you create the equations for enthalpy of combustion and formation correctly?
Did you correctly combine them to make formation of ethanol?
Reply 2
Original post by BankaiGintoki
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Did you create the equations for enthalpy of combustion and formation correctly?
Did you correctly combine them to make formation of ethanol?
it makes so much more sense now,thank you πŸ˜€
Reply 3
Original post by BankaiGintoki
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Did you create the equations for enthalpy of combustion and formation correctly?
Did you correctly combine them to make formation of ethanol?

I'm so sorry but do you mind helping me with this one πŸ˜…

"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 respective"

the answer is -227.1. I tried doing the exact same method that you did for this question too but I'm still unable to get the answer.
(I think the different wording in the first half of this question is throwing me off as it says 1368 kjmol-1 of energy is produced. Is this basically the same as the enthalpy of combustion of ethanol? )
(edited 2 months ago)
Original post by Frizzle16523
I'm so sorry but do you mind helping me with this one πŸ˜…
"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 respective"
the answer is -227.1. I tried doing the exact same method that you did for this question too but I'm still unable to get the answer.
(I think the different wording in the first half of this question is throwing me off as it says 1368 kjmol-1 of energy is produced. Is this basically the same as the enthalpy of combustion of ethanol? )

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Reply 5
Original post by BankaiGintoki
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why does it equal -1368 when the energy produced is +1368?
Original post by Frizzle16523
why does it equal -1368 when the energy produced is +1368?

Because it released energy, so is exothermic
Reply 7
Original post by BankaiGintoki
Because it released energy, so is exothermic

Thank you!
Reply 8
Hiya mate, are you sure the answer is -227.1 and not -277.1? I got this by doing (2(-393.7)+3(-285.9))--1368
Original post by pvma
Hiya mate, are you sure the answer is -227.1 and not -277.1? I got this by doing (2(-393.7)+3(-285.9))--1368

hello, the mark scheme said it is -227.1. I re did the question and got that answer πŸ™‚.
Reply 10
Original post by Frizzle16523
hello, the mark scheme said it is -227.1. I re did the question and got that answer πŸ™‚.

I see, which answer did you get sorry?
Original post by pvma
I see, which answer did you get sorry?

-227.1
Reply 12
Original post by Frizzle16523
-227.1

I don't understand how and I can't see a way which you get this, can you explain what you did please as my method works out to be -277.1? Thanks in advance.
Original post by pvma
I don't understand how and I can't see a way which you get this, can you explain what you did please as my method works out to be -277.1? Thanks in advance.

The answer is -277.1
It's probably a typo in the MS

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