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March 29, 2024, 04:03:14 pm

Author Topic: VCE Chemistry Question Thread  (Read 2313586 times)  Share 

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sweetcheeks

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Re: VCE Chemistry Question Thread
« Reply #6075 on: February 19, 2017, 10:37:47 pm »
+2
For this http://m.imgur.com/a/uwc5t

Is my working correct? The book has my answer but with an increase of factor of 1000 for both answer.

Also, for the combustion of butanol, is it in a liquid state? The answer wrote a gaseous state..

Thanks

Seems like the book made a mistake in the first part not converting J to kJ. They likely used the answer from part A in their next answer, resulting in it being 1000x larger than it should be.

As for the combustion of butanol, referring to the data booklet for the state they list both 1-propanol and 2-propanol as liquids in the molar enthalpy section. From this it is most likely that 1-butanol will also be in a liquid state. This area is a bit complicated and VCAA will generally accept either terms. In reality the molar enthalpy will be higher for the combustion of liquid butanol compared to gaseous butanol.

sweetcheeks

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Re: VCE Chemistry Question Thread
« Reply #6076 on: February 19, 2017, 10:39:24 pm »
+1
Also with combustion, they'll often tell you the state but if I remember correctly C1-4 was gas, 5-10 liquid and above solid (someone may want to confirm this though). Usually it'll be given to you though.

This is true for alkanes but not alkanols.

peanut

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Re: VCE Chemistry Question Thread
« Reply #6077 on: February 19, 2017, 11:06:04 pm »
0
Why is enthalpy change expressed in kJ mol^-1? Shouldn't it just be kJ? I ask this because when you calculate enthalpy change, you use delta H = heat of combustion * moles of fuel. This should give you units of kJ, not kJmol^-1?

peanut

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Re: VCE Chemistry Question Thread
« Reply #6078 on: February 19, 2017, 11:08:33 pm »
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Similarly, why do you double the enthalpy change when you double a thermochemical equation (to balance it, for example). Shouldn't the units of kJmol^-1 already take into account the new mole amount? Doesn't doubling the enthalpy change quadruple the enthalpy change?

sweetcheeks

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Re: VCE Chemistry Question Thread
« Reply #6079 on: February 20, 2017, 06:01:13 am »
+2
Why is enthalpy change expressed in kJ mol^-1? Shouldn't it just be kJ? I ask this because when you calculate enthalpy change, you use delta H = heat of combustion * moles of fuel. This should give you units of kJ, not kJmol^-1?

Thermochemistry can be quite tricky to understand at the start (better to be getting it done now than at the end of the year, you have time to clear up any difficulties).

The equation used is actually change in energy/moles = delta h. The units will be xkJ/ymol resulting in the enthalpy change is kJ/mol.

There needs to be a reference. kJ would be worthless in giving us any useful information. Enthalpy change can be viewed as how much energy is released when 6.02 x 1023 of the reaction occurs (one mole). We have the combustion of octane 2C8H18 + 25O2 --> 16CO2 + 18H2O delta h= -10000kJ/mole. What this means is that when 2 atoms of octane react with 25 atoms of oxygen producing 16 atoms of CO2 and 18 atoms of water, occurring 6.02 x 1023 times, the total energy released (or energy difference between bonds) will be -10000kJ. Therefore the delta h of the reaction is going to be -10000kJ/mol, as for every 1 mole of a reaction occurring -10000kJ of energy will be released.

Similarly, why do you double the enthalpy change when you double a thermochemical equation (to balance it, for example). Shouldn't the units of kJmol^-1 already take into account the new mole amount? Doesn't doubling the enthalpy change quadruple the enthalpy change?
I've covered most of this above about it being per 1 mole of the reaction occurring. The molar enthalpy of octane is -5000kJ. What this means is that 1 mole of octane undergoing complete combustion; This reaction occurring C8H18 + 12.5O2 --> 8CO2 + 9H2O 6.02 x 1023 times will release 5000kJ.

When we write the proper thermochemical equation everything has doubled. So wouldn't it be fair to say that twice as much energy is going to be output?

Compare; C8H18 + 12.5O2 --> 8CO2 + 9H2O to 2C8H18 + 25O2 --> 16CO2 + 18H2O
occurring 6.02 x 1023 times, there being twice as many reactants means that the energy change is going to be 2x greater.

deStudent

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Re: VCE Chemistry Question Thread
« Reply #6080 on: February 22, 2017, 09:26:57 pm »
0
For this question http://m.imgur.com/a/Adl83
The book didn't have a similar example question and I couldn't find one.

I want to double check for a,b and c. Is this how I'd set it out? In particular the chemical equations I wrote, for example since for (a) they gave us mole of silver, therefore my chemical equation should only consider silver as it makes it clear what the mole ratio is?

for (d), my method of working above breaks down here. I'm not sure how I can relate NaCl to hydrogen molecules?

Shadowxo

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Re: VCE Chemistry Question Thread
« Reply #6081 on: February 23, 2017, 12:38:35 am »
+2
For this question http://m.imgur.com/a/Adl83
The book didn't have a similar example question and I couldn't find one.

I want to double check for a,b and c. Is this how I'd set it out? In particular the chemical equations I wrote, for example since for (a) they gave us mole of silver, therefore my chemical equation should only consider silver as it makes it clear what the mole ratio is?

for (d), my method of working above breaks down here. I'm not sure how I can relate NaCl to hydrogen molecules?

I'm a bit rusty on this, but your equations seem to be fine.
For d), the same process can be used but the relevant equations would be
2Cl-(aq) -> Cl2(g) +2e- and
2H+(aq) + 2e- -> H2(g)
The electrons from the first equation can be used for the second.
So since you need 1 mol of H2, you need 2 mols of e-, and therefore 2 Faradays
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deStudent

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Re: VCE Chemistry Question Thread
« Reply #6082 on: February 23, 2017, 09:34:07 pm »
0
I'm a bit rusty on this, but your equations seem to be fine.
For d), the same process can be used but the relevant equations would be
2Cl-(aq) -> Cl2(g) +2e- and
2H+(aq) + 2e- -> H2(g)
The electrons from the first equation can be used for the second.
So since you need 1 mol of H2, you need 2 mols of e-, and therefore 2 Faradays
thanks
For the reductant, why isn't it 2H2O -> O2 + 4H + 4e-? Isn't this the stronger reductant?

----
http://m.imgur.com/a/k62Ve
The main problem I'm having here is that, how do I relate this unknown metallic ionic compound to the amount of electrons gained in this reaction?

After I obtain that I can just use Q = n(e-) * F, yes?

Thanks

Shadowxo

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Re: VCE Chemistry Question Thread
« Reply #6083 on: February 23, 2017, 10:47:51 pm »
+2
thanks
For the reductant, why isn't it 2H2O -> O2 + 4H + 4e-? Isn't this the stronger reductant?

----
http://m.imgur.com/a/k62Ve
The main problem I'm having here is that, how do I relate this unknown metallic ionic compound to the amount of electrons gained in this reaction?

After I obtain that I can just use Q = n(e-) * F, yes?

Thanks

It may be, but if I remember correctly, Cl is a strange one as both Cl- and H2O can be oxidised at the same time (someone correct me if I'm wrong). But since charge is independent of voltage, it shouldn't matter for that question anyway. But yes, H2O is the stronger reductant in this case.

For that question, you have charge and the number of molecules
n(e-) = Q/F = 9.33*10-3
n(e-)/n(element) = 9.33*10-3/(4.67*10-3) = 2.00
Therefore 2 e- per molecule therefore the charge is 2+.
Or a shorter way (shortcut, I wouldn't do this on exam/SAC)
f=Q/n = 900/(4.67*10-3), divide this by F (96500) and you get 2 faradays
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deStudent

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Re: VCE Chemistry Question Thread
« Reply #6084 on: February 24, 2017, 12:16:22 am »
0
It may be, but if I remember correctly, Cl is a strange one as both Cl- and H2O can be oxidised at the same time (someone correct me if I'm wrong). But since charge is independent of voltage, it shouldn't matter for that question anyway. But yes, H2O is the stronger reductant in this case.

For that question, you have charge and the number of molecules
n(e-) = Q/F = 9.33*10-3
n(e-)/n(element) = 9.33*10-3/(4.67*10-3) = 2.00
Therefore 2 e- per molecule therefore the charge is 2+.
Or a shorter way (shortcut, I wouldn't do this on exam/SAC)
f=Q/n = 900/(4.67*10-3), divide this by F (96500) and you get 2 faradays
Thanks!

I accidentally did it with the shorter way before I saw this. No clue how I got there because I was mucking around with some values haha.

peanut

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Re: VCE Chemistry Question Thread
« Reply #6085 on: February 25, 2017, 01:42:18 pm »
0
Is there any good way of memorising the solubility table for ions and the 7 equations for acids?

sweetiepi

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Re: VCE Chemistry Question Thread
« Reply #6086 on: February 25, 2017, 01:46:16 pm »
+1
Is there any good way of memorising the solubility table for ions and the 7 equations for acids?
There are multiple ways, but imo flashcards seemed the best way to remember them (from experience here). :)
(You could always do a mind-map or keep writing them all down, but that seems rather dull after a while)

Either way, it's mostly rote learning, haha :)
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Sine

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Re: VCE Chemistry Question Thread
« Reply #6087 on: February 25, 2017, 01:53:10 pm »
+1
Is there any good way of memorising the solubility table for ions and the 7 equations for acids?
Some schools may tell you to memorise the solubility table but afaik it isn't 100% necessary (unless it is for your sacs). VCAA questions usually make it very clear what is soluble. As insanipi said learning the solubility table is rote-learning.

Equations for acids if you know what an acid does (and which species they are ) it becomes quite easy. At the start you may be learning a list of equations but it's better to understand what exactly is happening.(i.e where the H+ is coming from and where it is going) Doing this makes it rather intuitive and helps out when writing equations for reactions that you aren't familiar with.

plsbegentle

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Re: VCE Chemistry Question Thread
« Reply #6088 on: February 25, 2017, 03:46:10 pm »
0
What are the trends between the number of carbon atoms and deltaH value?
Thanks
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Syndicate

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Re: VCE Chemistry Question Thread
« Reply #6089 on: February 25, 2017, 04:36:05 pm »
+2
What are the trends between the number of carbon atoms and deltaH value?
Thanks

The more the amount of carbon in a molecule, the more the energy is released (so the delta H value will be more negative). It's all related to bonds being broken (which requires input of energy) and made (which releases energy). The more the amount of carbon atoms in a hydrocarbon chain, the more C=O bonds are created, which therefore releases more energy (CO2 molecules).
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