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Does anyone have detailed notes on these two experiments.

Molar heat of combustion as well as the fermentation of glucose into ethanol. I am particularly interested in discussions, in other words how to improve experiment (validity, reliability and accuracy). any help will be appreciated.
 
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heat of combustion question.

what is the hypothesis for this experiment?? I know as carbon chain increases the molar heat of combustion increase, but why is this???
 

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Re: heat of combustion question.

Molar heat of combustion increase with chain length because the longer then chain the more energy you need to put in to break these bonds to combust and therefore will result in a larger molar heat of combustion.
 
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Re: heat of combustion question.

Molar heat of combustion increase with chain length because the longer then chain the more energy you need to put in to break these bonds to combust and therefore will result in a larger molar heat of combustion.
if the chain length increases, doesn't that mean the energy supplied by the spirit burner will not be sufficient enough to combust the substances with the higher amount of C atoms. Hence, less heat will be relesed, since heat of combustion is defined as the heat released per mole of a substance.
 

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Re: heat of combustion question.

Because you need more energy to break the bonds since dispersion forces increase as carbon chain length increases
 
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Re: heat of combustion question.

Because you need more energy to break the bonds since dispersion forces increase as carbon chain length increases
yea but if you need more energy shoudn't the heat of combustion be lower? meaning less heat will be relesed because so much energy is needed and the spirit burner provides a constant supply of energy.
 

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Re: heat of combustion question.

Molar heat of combustion increase with chain length because the longer then chain the more energy you need to put in to break these bonds to combust and therefore will result in a larger molar heat of combustion.
Not quite the energy needed to break the bonds, but the energy of the resulting bonds in comparison to the initial bonds.

The more bonds you can reform the more energy is released, and hence molar heat of combustion increases with chain length for simple hydrocarbons.
 

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Re: heat of combustion question.

yea but if you need more energy shoudn't the heat of combustion be lower? meaning less heat will be relesed because so much energy is needed and the spirit burner provides a constant supply of energy.
Get your concept clear, molar heat of combustion is the amount of energy (heat) needed to burn (break bonds) 1 molar solution 'completely', so the more the dispersion forces because we have long chain length, the more energy required to break em' , therefore more energy released (exothermic) , and therefore high molar heat of combustion :)
 
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Re: heat of combustion question.

Get your concept clear, molar heat of combustion is the amount of energy (heat) needed to burn (break bonds) 1 molar solution 'completely', so the more the dispersion forces because we have long chain length, the more energy required to break em' , therefore more energy released (exothermic) , and therefore high molar heat of combustion :)
I have never read the molar heat of combustion to be the amount of heat needed, but in all textbooks it says released.
 

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Re: heat of combustion question.

I have never read the molar heat of combustion to be the amount of heat needed, but in all textbooks it says released.
Amount of energy, and this energy we provide is in the form of heat ....it could be any form, but in this experiment, it is in the form of heat AND the energy released is also heat since the bonds are being broken and release energy in the form of heat ...remember this reaction is exothermic :)

The explanation can't be simpler than this, I hope you get it now....
 
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Re: heat of combustion question.

so the the ethanol is in a spirit burner, we burn the ethanol and get a specific heat. why is this heat released higher.
and obvious response is diffrent but makes more sense.
 

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Re: heat of combustion question.

I have never read the molar heat of combustion to be the amount of heat needed, but in all textbooks it says released.
Molar heat of combustion is the amount of energy released when a mole of a particular combustible substance is reacted with oxygen to yield:
- All carbon forms carbon dioxide.
- All sulfur forms sulfur dioxide.
- All nitrogen forms molecular nitrogen.
- All hydrogen forms water.
And some other technicalities which aren't that important at this point.

Basically, reforming stronger bonds (H-O mainly - as water; C=O is practically equivalent to 2 x C-C so this isn't that significant) releases an appreciable quantity of the difference in bond energy between the C - H and H - O bonds as thermal energy.

This isn't a perfect explanation by any means, but I hope it highlights the basic gist of things.

_________________________________________________________________________________________

Molar heat of combustion doesn't have anything to do with the heat needed to start the reaction.
 
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Re: heat of combustion question.

Thanks, obvious. so basically this statment is wrong
I think that as the molecules become larger, the enthalpy of combustion will increase. This hypothesis is based on my assumption that as the molecule becomes larger, more bonds are added to a substance. The more the bonds, the harder it is to separate the molecule and the more energy is required to combust the substance.

P.s i din't write this, it is from the web.
 

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Re: heat of combustion question.

Two important things to note:
- Breaking of bonds requires an INPUT of energy.
- Forming of bonds LIBERATES energy.

In small carbon chains, bonds are reformed such that there is a net release of energy. If there is a larger number of carbon atoms, it is logical to expect more energy to be released, even though more energy is needed, an even greater amount of energy is released.

Example
Say, energy in the bonds initially is x and energy in bond finally is y.
Since energy is released, you can conclude that x>y.

The energy released can be seen to be equal to x-y
E(released)=x-y

If there are double the bonds, there will be a total of double the energy in all the bonds.
2x-2y=2E(released)

From this, you can see that a larger energy is released if you double it.

This is a very dodgy way (don't copy it in the exam) but it might help you understand the concept.
 

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Re: heat of combustion question.

Two important things to note:
- Breaking of bonds requires an INPUT of energy.
- Forming of bonds LIBERATES energy.

In small carbon chains, bonds are reformed such that there is a net release of energy. If there is a larger number of carbon atoms, it is logical to expect more energy to be released, even though more energy is needed, an even greater amount of energy is released.

Example
Say, energy in the bonds initially is x and energy in bond finally is y.
Since energy is released, you can conclude that x>y.

The energy released can be seen to be equal to x-y
E(released)=x-y

If there are double the bonds, there will be a total of double the energy in all the bonds.
2x-2y=2E(released)

From this, you can see that a larger energy is released if you double it.

This is a very dodgy way (don't copy it in the exam) but it might help you understand the concept.
This.

generally when you get a large molecule, more bonds are formed during the reaction which therefore releases more energy, resulting in a higher molar heat of combustion.

but i doubt you have to go into this much detail for just a hypothesis, haha.
 

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Re: heat of combustion question.

This.

generally when you get a large molecule, more bonds are formed during the reaction which therefore releases more energy, resulting in a higher molar heat of combustion.

but i doubt you have to go into this much detail for just a hypothesis, haha.
Of course it is overly detailed for a hypothesis at HSC level but the purpose of my post was to try and help the other understand why larger chains will have larger molar heat of combustion.
 

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Re: Does anyone have detailed notes on these two experiments.

Guess I'll be nice.

Here's one I typed up last year sometime.

3.12 Solve problems, plan and perform a first-hand investigation to carry out the fermentation of glucose and monitor mass changes.


Practical: Monitoring Fermentation of Glucose

Aim: To investigate the fermentation of glucose & yeast into ethanol in an incubator.

Equipment:
- 50mL Conical Flask
- 5g Teaspoon yeast
- 10g glucose
- 40mL warm water (300C-40oC)
- Incubator
- Thermometer
- Rubber Tubing
- Retort Stand, Bos and Clamp
- Limewater

Hypothesis: A gas will be given of and the mass will change.

Procedure:
1. Place 5g of dried yeast into conical flask with side-arm.
2. Add 10g of glucose and 40mL of warm water to conical flask with side-arm.
3. Swirl flask to ensure contents are well mixed.
4. Place 150mL of limewater into conical flask (no side-arm)
5. Place cork wsith thermometer into the top of the conical flask with side-arm.
6. Connect apparatus as shown in diagram.
7. Place appartatus in incubator at 37oC
8. Monitor mass of fermentation flask each day for 5 days using electronic balance.
9. Record results in table.

Accuracy:
- Use a control- all ingredients except glucose.
- Use an electronic balance
- Using a data logger

Reliability:
- Repeating experiment and averaging results.

Validity:
- Results were what was expected
- Results were similar to other similar experiments performed
- Valid procedure
- All variables were controlled or monitored
- All parallax & systematic errors were controlled, which reduces random errors.

Test for ethanol:
1. Dehydrate ethanol using concentrated H2SO4 (aq) producing ethylene + water.
2. Add a few drops of Br2(aq) to test for reactive double bond and the Br2 will decolourise!

Results
Initial Day 1 Day 2 Day 3 Day 4 Day 5


Mass of fermentation flask

Errors:
- Fluctuations in temperature:
-yeast thrives at 37oC
- keeping constant temperature maximises the overall yield

- Mass Gain instead of loss:
- Limewater was taken into fermentation flask to fix elevate fermentation flask & to keep the temperature & pressure constant.
Safety:
- Ethanol must be kept away from heat and ignition sources
- Don’t ingest any of the practical
- Safety goggles and lab coat to be worn

Some parts are missing but fill them in with your own info!
 

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