# Chemistry AS Enthalpy Change Question Help!!

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#1
Could anyone please give me some guidance on answering these CH1 questions please? Thank you so much When ethanol burns in air, it produces oxygen and water.
C2H5OH+3O2-->2CO2+3H2O

1. Use the enthalpy change standard values below to calculate the standard enthalpy change for the combustion of ethanol.
C2H5OH = -278
CO2 = -394
H2O = -286
O2 = 0

2. The standard combustion enthalpy change for octane C8H18 is -5512kjmol-1.
By using this value and your answer to question 1, show that octane gives more energy for every gram of fuel burned than ethanol.
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4 years ago
#2
(Original post by Gwenog_quidditch)
Could anyone please give me some guidance on answering these CH1 questions please? Thank you so much When ethanol burns in air, it produces oxygen and water.
C2H5OH+3O2-->2CO2+3H2O

1. Use the enthalpy change standard values below to calculate the standard enthalpy change for the combustion of ethanol.
C2H5OH = -278
CO2 = -394
H2O = -286
O2 = 0

2. The standard combustion enthalpy change for octane C8H18 is -5512kjmol-1.
By using this value and your answer to question 1, show that octane gives more energy for every gram of fuel burned than ethanol.
Are those enthalpy of formation values you are given? You have to construct a Hess cycle, or use the formula 'sum of products - sum of reactants'.
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#3
(Original post by EricPiphany)
Are those enthalpy of formation values you are given? You have to construct a Hess cycle, or use the formula 'sum of products - sum of reactants'.
Yes they are enthalpy of formation values, thank you! Could you tell me roughly how to do Q2 please? Thanks 0
4 years ago
#4
(Original post by Gwenog_quidditch)
Yes they are enthalpy of formation values, thank you! Could you tell me roughly how to do Q2 please? Thanks Start by calculating how many moles of octane there is in a gram, using the molar mass. Do the same for ethanol.
If you need help with that I'll go into more detail.
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4 years ago
#5
(Original post by Gwenog_quidditch)
Yes they are enthalpy of formation values, thank you! Could you tell me roughly how to do Q2 please? Thanks is the answer for part a 1368?
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4 years ago
#6
(Original post by rumana101)
is the answer for part a 1368?
Combustion, so it has to be exothermic 0
#7
(Original post by rumana101)
is the answer for part a 1368?
I got -958 Products - reactants
(-394)+(286) - 278
= -680 - 278 = -958kj ???
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4 years ago
#8
(Original post by Gwenog_quidditch)
I got -958 Products - reactants
(-394)+(286) - 278
= -680 - 278 = -958kj ???
Multiply each by the balancing numbers in the equation.
For each mole of ethanol, two mole of carbon dioxide and three mole of water are formed.
Also be careful with the signs of the enthalpy changes.
3(-286)+2(-394)-(-278)
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#9
(Original post by EricPiphany)
Start by calculating how many moles of octane there is in a gram, using the molar mass. Do the same for ethanol.
If you need help with that I'll go into more detail.
Thanks, I'll try it 0
#10
(Original post by EricPiphany)
Start by calculating how many moles of octane there is in a gram, using the molar mass. Do the same for ethanol.
If you need help with that I'll go into more detail.
So to find the number of moles of octane in a gram do I divide 1 gram with the molar mass (114)?
(I got something like 0.0087719... so I think that this is probably wrong )
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4 years ago
#11
(Original post by Gwenog_quidditch)
So to find the number of moles of octane in a gram do I divide 1 gram with the molar mass (114)?
(I got something like 0.0087719... so I think that this is probably wrong )
That is correct. That is the moles of octane in one gram. Multiplying this by the energy released per mole (kJ per mole) gives us the energy released (kJ) for octane (for each gram). This is what we need to compare with the kJ per gram for ethanol.
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#12
(Original post by EricPiphany)
That is correct. That is the moles of octane in one gram. Multiplying this by the energy released per mole (kJ per mole) gives us the energy released (kJ) for octane (for each gram). This is what we need to compare with the kJ per gram for ethanol.
Does -48.35 for octane and -29.7391 for ethanol seem right? (octane- I multiplied 0.008771 by -5512 = -48.35)
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4 years ago
#13
(Original post by Gwenog_quidditch)
Does -48.35 for octane and -29.7391 for ethanol seem right? (octane- I multiplied 0.008771 by -5512 = -48.35)
Yes, looks right. So octane releases more energy per gram. 0
4 years ago
#14
(Original post by Gwenog_quidditch)
Does -48.35 for octane and -29.7391 for ethanol seem right? (octane- I multiplied 0.008771 by -5512 = -48.35)
Yeah I got that when I went through it ! ^-^ 0
#15
(Original post by EricPiphany)
Yes, looks right. So octane releases more energy per gram. Thank you 1
4 years ago
#16 Here is how I worked it out. Made notes in red to help you understand! 1 is -1398 and 2 is octane more exothermic by 18kJg-1
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#17
(Original post by RME11) Here is how I worked it out. Made notes in red to help you understand! 1 is -1398 and 2 is octane more exothermic by 18kJg-1
Thank you so much! 0
4 years ago
#18
(Original post by Gwenog_quidditch)
Thank you so much! Hey, Could you please go through how you got 2C+ 3H2 + 2O2
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4 years ago
#19
(Original post by yorobun)
Hey, Could you please go through how you got 2C+ 3H2 + 2O2
That's the intermediate - the constituent elements of each of the reactants in their standard form. This is where the value of enthalpy of formation comes from by definition.
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4 years ago
#20
(Original post by RME11)
That's the intermediate - the constituent elements of each of the reactants in their standard form. This is where the value of enthalpy of formation comes from by definition.
Thankyou! i remember my teacher saying it doesn't matter if you cant do it but I just saw a question on forming the cycle so I will have to talk to him tomorrow.
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