# How do you write an equation for the enthalpy change of combustion of an alcohol

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Thread starter 4 weeks ago
#1
How would you write the equations for the enthalpy change of combustion for methanol and pentan-1-ol . I understand how to calculate the enthalpy change of combustion but how do you write an equation for the specific alcohol?
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4 weeks ago
#2
(Original post by Tigergirl)
How would you write the equations for the enthalpy change of combustion for methanol and pentan-1-ol . I understand how to calculate the enthalpy change of combustion but how do you write an equation for the specific alcohol?
1 mol of the alcohol + as much oxygen as needed for complete combustion (ca be fractional) ==> carbon dioxide + water
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Thread starter 4 weeks ago
#3
(Original post by charco)
1 mol of the alcohol + as much oxygen as needed for complete combustion (ca be fractional) ==> carbon dioxide + water
Thank you, but how do I know how much oxygen was needed for the combustion? I haven't been told
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4 weeks ago
#4
(Original post by Tigergirl)
Thank you, but how do I know how much oxygen was needed for the combustion? I haven't been told
All of the carbon atoms in the alcohol (or any other organic molecule) turns to carbon dioxide.
So if there are 4 carbon atoms they make four carbon dioxide molecules, which requires four oxygen molecules.

All of the hydrogen turns to water.
So if there are four hydrogen atoms in the molecule they turn to two water molecules requiring one oxygen molecule.

Don't forget that alcohols already have one oxygen atom in the structure when you are working out the number of oxygen molecules needed.
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Thread starter 4 weeks ago
#5
(Original post by charco)
All of the carbon atoms in the alcohol (or any other organic molecule) turns to carbon dioxide.
So if there are 4 carbon atoms they make four carbon dioxide molecules, which requires four oxygen molecules.

All of the hydrogen turns to water.
So if there are four hydrogen atoms in the molecule they turn to two water molecules requiring one oxygen molecule.

Don't forget that alcohols already have one oxygen atom in the structure when you are working out the number of oxygen molecules needed.
Ahh I think that makes sense now thank you !

so if it was methanol (CH3OH)

then would the formula be

CH3OH+2O2----->CO2+H2O??

the starting mass of the methanol is 188.15g which gives me around 5.9 mol but I am not sure how to include this into the equation thank you so much for your help so far
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4 weeks ago
#6
(Original post by Tigergirl)
Ahh I think that makes sense now thank you !

so if it was methanol (CH3OH)

then would the formula be

CH3OH+2O2----->CO2+H2O??

the starting mass of the methanol is 188.15g which gives me around 5.9 mol but I am not sure how to include this into the equation thank you so much for your help so far
You have not counted the hydrogen correctly:

Methanol has four hydrogen atoms, so it can make 2H2O
AND
Don't forget the oxygen that's already in methanol when counting up.
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Thread starter 4 weeks ago
#7
oh right

so if two water molecules are formed then would the equation be : 2 CH3OH + 3 O2 ---> 2 CO2 + 4 H2O ?
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4 weeks ago
#8
(Original post by Tigergirl)
oh right

so if two water molecules are formed then would the equation be : 2 CH3OH + 3 O2 ---> 2 CO2 + 4 H2O ?
You did not read the first line of my original post!

1 mol of the alcohol + as much oxygen as needed for complete combustion (can be fractional) ==> carbon dioxide + water
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Thread starter 4 weeks ago
#9
(Original post by charco)
You did not read the first line of my original post!

1 mol of the alcohol + as much oxygen as needed for complete combustion (can be fractional) ==> carbon dioxide + water
Oh this is confusing! so is it CH3OH+3/2O2-->CO2+2H2O

I don't think it can be a fractional number in front though can it?
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4 weeks ago
#10
(Original post by Tigergirl)
Oh this is confusing! so is it CH3OH+3/2O2-->CO2+2H2O

I don't think it can be a fractional number in front though can it?
I already said that the amount of oxygen can be fractional. You are defining the enthalpy of combustion, which HAS to be 1 mol of alcohol burned.

CH3OH + 3/2O2 --> CO2 + 2H2O

Is the correct answer ...
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Thread starter 4 weeks ago
#11
(Original post by charco)
I already said that the amount of oxygen can be fractional. You are defining the enthalpy of combustion, which HAS to be 1 mol of alcohol burned.

CH3OH + 3/2O2 --> CO2 + 2H2O

Is the correct answer ...
Thank you so so much for your help!!!
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