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HSC 2012-2015 Chemistry Marathon (archive) (4 Viewers)

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HeroicPandas

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re: HSC Chemistry Marathon Archive

Sorry man. Say you have a galvanic cell with no salt bridge. Wouldn't you still have an instantaneous EMF (essentially a voltage) between the two cells until almost immediately this EMF is negated by an electric field as positively charged ions begin to dominate in the anolyte and negatively charged ions begin to dominate in the catholyte? I doubt this is much clearer.

In this example, electricity is conducted through the external wire.
bzz not sure
 
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hayabusaboston

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re: HSC Chemistry Marathon Archive

a) The anode is the magnesium piece of metal and the cathode is the carbon (graphite) rod.
b) A salt bridge is required to complete the circuit in a galvanic cell and to maintain electrical neutrality in the electrolyte solution of each half cell allowing for a continuous flow of current through the galvanic circuit. Aqueous KNO3 would be a suitable solution in which to immerse the salt bridge (made from filter paper) as it will not form precipitates in the electrolyte solutions and therefore allow for electrical neutrality by transferring positive K+ ions to the FeSO4/Fe2(SO4)3 solution to negate the buildup of positive charge due to a decrease in the concentration of positive Fe^(2+)/Fe^(3+) ions and transferring NO3- ions to the MgSO4 solution to negate the buildup of positive charge due to an increase in the concentration of positive Mg^(2+) ions.
c) Not sure.
d) Oxidation half-equation: Mg(s)->Mg^(2+)(aq)+2e
Reduction half-equation(s): Fe(2+)(aq)+2e->Fe(s), Fe^(3+)(aq)+3e->Fe(s)
Observed changes after three months:
- Decolourisation of the FeSO4/Fe2(SO4)3 solution due to the discharge (reduction) of Fe(3+)/Fe(2+) ions from solution
- Decrease in the mass and size of the magnesium electrode (anode) due to the oxidation of magnesium as shown by the oxidation half-equation above.
- Deposit of a brown metal on the carbon rod due to the formation and subsequent corrosion of iron in the presence of water and oxygen.
- Sedimentation of iron hydroxide (insoluble) at the bottom of the container originating from the corrosion of the iron formed in the reduction half-equation.
Corrosion of iron
Oxidation half-equation: Fe(s)->Fe^(2+)(aq)+2e
Reduction half-equation: O2(g)+2H2O(l)+4e->4OH-(aq)
Formation of insoluble Fe(OH)2: Fe^(2+)(aq)+2OH-(aq)->Fe(OH)2(s)

Bro, dat sentece!! faaar too long! I suspect that wouldn't be looked upon favourably by markers, in certain extended questions it might make your response unclear and smashed together into one big block, or thats what markers could potentially say imo.
 

bleakarcher

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re: HSC Chemistry Marathon Archive

Bro, dat sentece!! faaar too long! I suspect that wouldn't be looked upon favourably by markers, in certain extended questions it might make your response unclear and smashed together into one big block, or thats what markers could potentially say imo.
I agree, could have definitely split it up a bit. Thanks for the criticism.
 

HeroicPandas

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re: HSC Chemistry Marathon Archive

Question:

a) What is required for a particular reaction to be in a state of dynamic equilibrium? (3 mark)
b) A manager of a company producing ammonia by the Haber process, has become bankrupt because, sadly a burglar stole the equipment for the Haber process to sustain high pressure and temps (his catalyst has not been stolen), and so his costs are taking over his sales. The reaction vessel is still intact. Luckily a very smart industrial chemist, shows up and suggests the manager to borrow money from the bank and buy: a large amount of argon gas (coincidentally it was cheap) and a machine to inject gas into the reaction vessel.

2 years later, the company has recovered itself and the manager is not bankrupt anymore

With reference to Le Chatelier's Principle, explain why the industrial chemist made that descision to purchase those things. (3 marks)

Note:This is a hypothetical situation
 
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lachy95

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1) if the rates of the forward and reverse reactions are equal, then the reaction is said to be in equilibrium.

2)
B/G:
• argon is an inert gas, it does not react with other chemicals.
• Le chateliers principle states that when a closed system in equilibrium is disturbed, the equilibrium will change to minimise the disturbance.
Points:
• injecting argon gas into the reaction vessel pressurises the reaction mixture, undergoing the following equilibrium:
h2(g) + 3n2(g) <-> 2nh3(g)
This creates a higher pressure of reactants and products, so the equilibrium shifts to favour the forward reaction for a reaction mixture with fewer moles of gas which will occupy less volume as per gay lussacs law
Conc:
• the industrial chemist bought argon to achieve the same yield of products in the absence of high pressure equipment.

Is that enough?

Guess I could also Chuck in gay lussacs law properly, purpose of industry, reason why high temp worked etc?

Also you made this up right?

Sent from my GT-I9300 using Tapatalk 2
 

HeroicPandas

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re: HSC Chemistry Marathon Archive

1) Sorry is was meant to be more marks lol, and yeh thats 1 reason (i wanna see TWO more)

2) Well done, but ur conclusion of "achieve the same yield of products in the absence of high pressure equipment", i dont think its correct, i dont think injecting gas will increase pressure up to 200-250 atmospheres. So yield (using argon) =/= yield (using pressure machines)

yeh thats enough

Gay lassac's law isnt in the syllabus

yeh i made it up
 

lachy95

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Interesting about gay lussacs. Thanks.

For (1) what are the other reasons? Does one have to be spontaneous and the other not? Or one exo and the other endo?



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HeroicPandas

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re: HSC Chemistry Marathon Archive

Interesting about gay lussacs. Thanks.

For (1) what are the other reasons? Does one have to be spontaneous and the other not? Or one exo and the other endo?



Sent from my GT-I9300 using Tapatalk 2
-U need a reversible reaction
-Closed system
-Time
 

albertcamus

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re: HSC Chemistry Marathon Archive

I think that despite the argon gas affecting pressure theoretically, because it is inert, it has no effect on the closed system at all - this was a question in Jacaranda if I remember correctly.
 

someth1ng

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re: HSC Chemistry Marathon Archive

Adding Ar will NOT affect the system (assuming volume stays constant) because what actually causes the system to change is the partial pressures (which is basically a measure of concentration of a gas). Even though adding Ar to the system will cause pressure to increase, it won't actually affect the position of equilibrium.

Basically, the partial pressures of all of them will stay the same but p(Ar) will increase - no effect on equilibrium.

Kp=[p(NH3)^2]/[p(H2)p(N2)^3]

If you add more Ar, nothing in that equation changes. However, if you decrease volume, it does change.
 
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HeroicPandas

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re: HSC Chemistry Marathon Archive

I think that despite the argon gas affecting pressure theoretically, because it is inert, it has no effect on the closed system at all - this was a question in Jacaranda if I remember correctly.
this was not a question from any textbook, it was from my brain

Adding Ar will NOT affect the system (assuming volume stays constant) because what actually causes the system to change is the partial pressures (which is basically a measure of concentration of a gas). Even though adding Ar to the system will cause pressure to increase, it won't actually affect the position of equilibrium.

Basically, the partial pressures of all of them will stay the same but p(Ar) will increase - no effect on equilibrium.

Kp=[p(NH3)^2]/[p(H2)p(N2)^3]

If you add more Ar, nothing in that equation changes. However, if you decrease volume, it does change.
thanks for a graet explanation, i know understand


AND soRRY i mislead u lachy95 lol
 
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lachy95

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Haha that's okay. Point about partial pressures is a good one. Okay I'll try to find a new question but I'm not near a book so if anyone else is, go for it.
 

HeroicPandas

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re: HSC Chemistry Marathon Archive

Question: Why does the neutralisation of any strong acid in an aqueous solution by any 1 strong base always result in a heat of reaction of approx. -57 kJ/mol (HSC question)
 
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albertcamus

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Question: Why does the neutralisation of any strong acid in an aqueous solution by any 1 strong base always result in a heat of reaction of approx. -57 kJ/mol
Because they all have the same proton transfer eqn of H^+ + OH^- ---> H2O

Proton has been transferred from the acid to the base to form H2O.

Add in states for eqn and ye
 
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HeroicPandas

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Because they all have the same proton transfer eqn of H^+ + OH^- ---> H2O

Proton has been transferred from the acid to the base to form H2O.

Add in states for eqn and ye
yep nice, if u may continue this without me, my energy levels have depleted
 

lachy95

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I like that one a lot. Never seen it before.

Do you mean just for tonight heroic? We should keep this going till trials. Good to just discuss hard Qs.
 

HeroicPandas

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I like that one a lot. Never seen it before.

Do you mean just for tonight heroic? We should keep this going till trials. Good to just discuss hard Qs.
When are ur trials? Im out of ideas and i dont want to post up those useless questions that are only require rote learning

 

epz

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re: HSC Chemistry Marathon Archive

To Heroics question above, I don't know the answer. I plan to try and be active in this thread for a little bit as Chemistry is probably one of my weakest subjects. Here is my question:

Amphiprotic substances have the ability to work as buffers in natural systems.
Explain why natural systems require buffers and using a specific example of an amphiprotic substance, show how this can behave as a buffer. Include equations in your answer.

This is from James Ruse 2010 Trials Exam.
Solution is in this image if you are struggling (with marking criteria):
http://pbrd.co/1bQ3fji

I'll try to answer anything I know, and absorb what I don't :D
 
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