Chemistry AS Enthalpy Change Question Help!!

Could anyone please give me some guidance on answering these CH1 questions please? Thank you so much :smile:

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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Reply 1

EricPiphany

Original post by Gwenog_quidditch

Could anyone please give me some guidance on answering these CH1 questions please? Thank you so much :smile:

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'.

Reply 2

Gwenog_quidditch

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 :smile:

Reply 3

EricPiphany

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 :smile:

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.

Reply 4

rumana101

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 :smile:

is the answer for part a 1368?

Reply 5

EricPiphany

Original post by rumana101

is the answer for part a 1368?

Combustion, so it has to be exothermic :wink:

Reply 6

Gwenog_quidditch

Original post by rumana101

is the answer for part a 1368?

I got -958 :s-smilie:

Products - reactants
(-394)+(286) - 278
= -680 - 278 = -958kj ???

Reply 7

EricPiphany

Original post by Gwenog_quidditch

I got -958 :s-smilie:

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)

(edited 8 years ago)

Reply 8

Gwenog_quidditch

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 :wink:

Reply 9

Gwenog_quidditch

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 :s-smilie:

)

Reply 10

EricPiphany

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 :s-smilie:

)

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.

(edited 8 years ago)

Reply 11

Gwenog_quidditch

Original post by EricPiphany

Does -48.35 for octane and -29.7391 for ethanol seem right? (octane- I multiplied 0.008771 by -5512 = -48.35)

Reply 12

EricPiphany

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. :smile:

Reply 13

yorobun

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 ! ^-^ :smile:

Reply 14

Gwenog_quidditch

Original post by EricPiphany

Yes, looks right. So octane releases more energy per gram. :smile:

Thank you

Reply 15

username2280751

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

Reply 16

Gwenog_quidditch

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! :smile:

Reply 17

yorobun

Original post by Gwenog_quidditch

Thank you so much! :smile:

Hey, Could you please go through how you got 2C+ 3H2 + 2O2

Reply 18

username2280751

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.

Reply 19

yorobun

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.