Trial exams discussion

[mystikal]

can sum1 upload the neap 08 exam so i can do it too 

[jaja]

go to the forum "unit four practise exams" 9 topics down from this one

[IntoTheNewWorld]

can sum1 upload the neap 08 exam so i can do it too  

I believe NEAP has been uploaded in this thread

http://vcenotes.com/forum/index.php/topic,6208.0.html

Just look down the page a bit

[jaja]

lol.. didnt noe how to copy that!

[kendraaaaa]

Hey guys just a couple of queries from the IARTV 08 exam, any help would be greatly appreciated.



Why is it that the concentration of oxygen increases here? I assumed that since the volume is halved (increase in pressure) the system shifts to the right which decreases the concentration of oxygen?  



Where do we get this equation from? How is it known that the oxygen is the other product?



And finally, in this question they ignored the coefficient of the butane.



To find delta H wouldn't you use the following calculations:



Where 22070 is the energy found in a), 0.014 is the mol, and 2 is the coeffcient of the butane.

Thank you.

[StringFever]

Hey guys just a couple of queries from the IARTV 08 exam, any help would be greatly appreciated.

(Image removed from quote.)

Why is it that the concentration of oxygen increases here? I assumed that since the volume is halved (increase in pressure) the system shifts to the right which decreases the concentration of oxygen?  

(Image removed from quote.)

Where do we get this equation from? How is it known that the oxygen is the other product?

(Image removed from quote.)

And finally, in this question they ignored the coefficient of the butane.



To find delta H wouldn't you use the following calculations:



Where 22070 is the energy found in a), 0.014 is the mol, and 2 is the coeffcient of the butane.

Thank you.

Hi there,

With your oxygen question - you have to bear in mind of the overall changes. What I mean by that is, by halving the pressure, you're inevitably going to get more number of mole per volume - regardless of what Le Chatelier's Principal (which in itself deals with amount (number of mole) rather than concentration per se).

An easier way to consider it is to think of the equation n = CV.
We know that C= n/V.
Therefore, if n remains constant for all intensive purposes and we halve the volume, our concentration increases.

Also for the fuel cell, think of it as a combustion reaction. If you've been given the half-equation of methanol/ethanol, the other half equation would be for oxygen because that is required for combustion.

Also, for your last question (although I'm a bit sketchy on it), I'm pretty sure you use a balance equation as you did to figure out molar enthalpy.

[methodsboy]

^can someone please upload the IARTV 08 exam?

[chem-nerd]

Question 2
yes, the reaction does move forward reducing the amount of O₂ present. However, the volume has halved hence increasing the concentration (c = n/V). The reduction in the amount of O₂ present, due to the system partially opposing the change, would not be by as much as half.

Question 4
You can assume that the fuel cell reactants will always be O₂ and something else (eg H₂, methanol, ethanol etc) Because the question said it was an acidic fuel cell, you needed the O₂ equation with H⁺

Question 7
No, you don't need to worry about the coefficient

[StringFever]

Question 7
No, you don't need to worry about the coefficient

Why is that? I thought molar ratios were important to enthalpy calculations?!

[lacoste]

Because its kJ per mol

the equation is not balanced

[kendraaaaa]

Thanks a lot guys 



edit 1 IARTV 08 trial exam for methodsboy

edit 2 problem solved

[chem-nerd]

Question 7
No, you don't need to worry about the coefficient

Why is that? I thought molar ratios were important to enthalpy calculations?!

In this question, you've calculated n(butane) from the mass change and calculated the E released from the mass of water and change in temperature. Enthalpy, in kJ/mol, can then be simply calculated by E/n.

If you include the coefficient from the balanced equation, you would be calculating the enthalpy in kJ per 2 mol

[StringFever]

Has anyone done Question 6 in STAV 2008? The one where they ask you to write the balanced half-equation for the reactions at the cathode and anode.

My query about that is that the solutions suggest that H+ is reduced at the cathode instead of Cu2+; but I thought reduction happened to the strongest oxidant, and unless I'm blind, I'm pretty sure Cu2+ is a better oxidant than H+?

Any thoughts or suggestions would be appreciated!

[thisongaintgottaname]

umm.. because Cu metal is present, but not Cu2+?

this is what comes from my memory, i don't have it with me, so i may be wrong

[StringFever]

umm.. because Cu metal is present, but not Cu2+?

this is what comes from my memory, i don't have it with me, so i may be wrong

Hahahahaha that's so awkward didn't read it properly!