anactualmess
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The enthalpy of combustion of ethanol is -1371 kJmol-1. The density of ethanol is 0.789 g cm -3.
Calculate the heat energy released in kJ when 0.500dm^3 of ethanol is burned.

I am not really sure what the question is asking. What is the difference between heat energy released when ethanol is burned and the enthalpy of combustion?

What exactly am I supposed to be doing? I have looked at the mark scheme but am still confused.

Thank you!
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username3249896
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Enthalpy of combustion is the amount of energy released when one mole of the compound is completely burnt in oxygen.

"The enthalpy of combustion of ethanol is -1371 kJmol-1" means that 1371kJ is released (due to the negative sign) when one mole of ethanol is completely combusted (hence kJmol-1).

For this question, you should first find the mass of ethanol burnt, then the number of moles of ethanol burnt. Then find the heat released
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anactualmess
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(Original post by BobbJo)
Enthalpy of combustion is the amount of energy released when one mole of the compound is completely burnt in oxygen.

"The enthalpy of combustion of ethanol is -1371 kJmol-1" means that 1371kJ is released (due to the negative sign) when one mole of ethanol is completely combusted (hence kJmol-1).

For this question, you should first find the mass of ethanol burnt, then the number of moles of ethanol burnt. Then find the heat released
Thank you for reminding me of the definition of enthalpy of combustion, that cleared up a lot

I have calculated that 8.58 moles of ethanol are burnt.

How do I now find the heat released?
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Yatayyat
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I’ve tried the question myself.

Here is a picture of my working. Does this make sense to you?

Although I may be wrong too...

Name:  0874A511-4689-46FE-B565-A03BAE7D7F17.jpg.jpeg
Views: 103
Size:  13.7 KB

What was the answer anyway?
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anactualmess
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(Original post by Yatayyat)
I’ve tried the question myself.

Here is a picture of my working. Does this make sense to you?

Although I may be wrong too...

Name:  0874A511-4689-46FE-B565-A03BAE7D7F17.jpg.jpeg
Views: 103
Size:  13.7 KB

What was the answer anyway?
Thank you so much! Yeah according to the mark scheme your answer is correct.

Thanks for including the direct relationship between q and and enthalpy change.

Just one more question, would you be able to define the difference between q and delta H?
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Yatayyat
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(Original post by anactualmess)
Thank you so much! Yeah according to the mark scheme your answer is correct.

Thanks for including the direct relationship between q and and enthalpy change.

Just one more question, would you be able to define the difference between q and delta H?
Q is the energy that is given out when ethanol is burnt for a specified mass and this has units in joules

Delta H is the enthalpy change and hence the amount of energy given out for every one individual mole of ethanol that is burnt. Units are kJ per mole.

Hope that makes sense. Let me know if I still need to clarify anything.
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anactualmess
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(Original post by Yatayyat)
Q is the energy that is given out when ethanol is burnt for a specified mass and this has units in joules

Delta H is the enthalpy change and hence the amount of energy given out for every one individual mole of ethanol that is burnt. Units are kJ per mole.

Hope that makes sense. Let me know if I still need to clarify anything.
Thank you very much.

And yeah one more question sorry, why is the answer automatically in kJ without the need for conversion from J to kJ? I would have thought that since q is measured in joules that the answer would initially be in joules and then need to be converted?
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Yatayyat
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(Original post by anactualmess)
Thank you very much.

And yeah one more question sorry, why is the answer automatically in kJ without the need for conversion from J to kJ? I would have thought that since q is measured in joules that the answer would initially be in joules and then need to be converted?
Yes sorry. You get an answer of joules initially when you use the the q=mc delta t. Then convert it back to kilojoules by dividing by 1000.

And then with the conversion with J to KJ you divide it again with mol to get enthalpy change.

You’re right! that was my mistake. Apologies.
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