# Chemistry AS Enthalpy Change Question Help!!

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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.

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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#2

(Original post by

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.

**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.

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(Original post by

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'.

**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'.

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#4

(Original post by

Yes they are enthalpy of formation values, thank you! Could you tell me roughly how to do Q2 please? Thanks

**Gwenog_quidditch**)Yes they are enthalpy of formation values, thank you! Could you tell me roughly how to do Q2 please? Thanks

If you need help with that I'll go into more detail.

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#5

**Gwenog_quidditch**)

Yes they are enthalpy of formation values, thank you! Could you tell me roughly how to do Q2 please? Thanks

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#6

(Original post by

is the answer for part a 1368?

**rumana101**)is the answer for part a 1368?

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(Original post by

is the answer for part a 1368?

**rumana101**)is the answer for part a 1368?

Products - reactants

(-394)+(286) - 278

= -680 - 278 = -958kj ???

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#8

(Original post by

I got -958

Products - reactants

(-394)+(286) - 278

= -680 - 278 = -958kj ???

**Gwenog_quidditch**)I got -958

Products - reactants

(-394)+(286) - 278

= -680 - 278 = -958kj ???

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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(Original post by

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.

**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.

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**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.

(I got something like 0.0087719... so I think that this is probably wrong )

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#11

(Original post by

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 )

**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 )

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(Original post by

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.

**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.

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#13

(Original post by

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

**Gwenog_quidditch**)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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#14

**Gwenog_quidditch**)

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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(Original post by

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

**EricPiphany**)Yes, looks right. So octane releases more energy per gram.

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#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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(Original post by

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

**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

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#18

(Original post by

Thank you so much!

**Gwenog_quidditch**)Thank you so much!

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#19

(Original post by

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

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

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#20

(Original post by

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.

**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.

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