# How to solve enthalpy of formation when energy produced is given

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Thread starter 1 month ago
#1
This is the 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 respectively.

Im confused as first I worked out the sum of the products and i got an answer of -1645.1kJmol-1. Im not able to figure out what the enthalpy of formation of the reactants would be as from the equation it needs to total to 1368 kJmol-1
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1 month ago
#2
Have you tried a Hess's cycle?
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Thread starter 1 month ago
#3
(Original post by Bookworm_88)
Have you tried a Hess's cycle?
I have tried to but im kinda confused on how to go about it
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1 month ago
#4
(Original post by Fa123123)
I have tried to but im kinda confused on how to go about it
Have you written the balanced equation for combustion of ethanol?
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Thread starter 1 month ago
#5
(Original post by Hellllpppp)
Have you written the balanced equation for combustion of ethanol?
Yes I did it as C2H5OH + 6 1/2 O2 --> 2CO2 + 3H2O
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1 month ago
#6
(Original post by Fa123123)
Yes I did it as C2H5OH + 6 1/2 O2 --> 2CO2 + 3H2O
Is that 6 x 1/2 for CO2

It won’t make a difference with the calculations of O2 is wrong but yeah it’s C2H5OH + 3 O2 --> 2CO2 + 3H2O

Do you remember the definition for enthalpy of formation?
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Thread starter 1 month ago
#7
(Original post by Hellllpppp)
Is that 6 x 1/2 for CO2

It won’t make a difference with the calculations of O2 is wrong but yeah it’s C2H5OH + 3 O2 --> 2CO2 + 3H2O

Do you remember the definition for enthalpy of formation?
yes and yes Standard enthalpy change of formation (ΔHfo) is the energy change when 1 mole of substance made from its elements in their standard state.
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1 month ago
#8
(Original post by Fa123123)
yes and yes Standard enthalpy change of formation (ΔHfo) is the energy change when 1 mole of substance made from its elements in their standard state.
Yeah so at the bottom of the Hess’ cycle put the elements in their standard state then draw arrows from the elements to the equation.
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Thread starter 1 month ago
#9
(Original post by Hellllpppp)
Yeah so at the bottom of the Hess’ cycle put the elements in their standard state then draw arrows from the elements to the equation.
This is what ive done
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1 month ago
#10
As it’s combustion the enthalpy change is always negative

I haven’t got a calculator on me so I’ll assume the calculation is right.

Follow the arrows and use Hess’ law
Last edited by Hellllpppp; 1 month ago
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1 month ago
#11
(Original post by Fa123123)
This is what ive done
Provided the calculation for formation of co2 and h2o is correct and assuming you mean -1368, formation of ethanol from its constituent elements is -286.1 kj/mol. You work this out by (-1654.1 + 1368) = -286.1
Last edited by gogrizz123; 1 month ago
1
Thread starter 1 month ago
#12
(Original post by Hellllpppp)
As it’s combustion the enthalpy change is always negative

I haven’t got a calculator on me so I’ll assume the calculation is right.

Follow the arrows and use Hess’ law
yeah but i still need to figure out the sum of reactants as thats what the question is asking so how would i do that?
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1 month ago
#13
(Original post by Fa123123)
yeah but i still need to figure out the sum of reactants as thats what the question is asking so how would i do that?
What’s the enthalpy of formation of oxygen?
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Thread starter 1 month ago
#14
(Original post by gogrizz123)
Provided the calculation for formation of co2 and h2o is correct and assuming you mean -1368, formation of ethanol from its constituent elements is -286.1 kj/mol
thank you so much, how did you get to this?
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1 month ago
#15
(Original post by Fa123123)
thank you so much, how did you get to this?
▲hF for ethanol = ▲Hf(co2+h20) - ▲H combustion of ethanol (but remember we are going against the arrow so you have to reverse the sign e.g turn a negative into a positive)
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1 month ago
#16
(Original post by Fa123123)
thank you so much, how did you get to this?
The enthalpy change is independent of the route taken the up arrow is -1645.1 by your maths and the arrow to the left is 1368

-1645.1 + 1368 = 1368 - 1645.1
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1 month ago
#17
(Original post by Fa123123)
I have tried to but im kinda confused on how to go about it
Hav you tried posting on the chem stuy group?
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