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HSC 2012-2015 Chemistry Marathon (archive) (1 Viewer)

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Mr_Kap

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re: HSC Chemistry Marathon Archive

So if you were to write your reasoning for why (for example) the equilibrium is on the reactants side, you would say K < 1? That's it? I've seen HSC sample answer say if K > 10^3 then the equilibrium lies to the products side.
If K < 10^-3 it lies WELL to the left, like virtually no product is made.

If K > 10^3 it lies WELL to the right, like virtually its almost at completion.

If K = 1 (Reactants to products is equal at equillibrium)

K <1 means the point of equilibrium is on the reactants side

K >1 means the point of equilibrium is on the products side

I remember being really confused about this when i learnt it. I think there is a thread that I made asking about this. I'll find it.
 

BlueGas

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re: HSC Chemistry Marathon Archive

So is it K>1 or K>10^3?
 

Mr_Kap

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So is it K>1 or K>10^3?
It depends how far to the right the equillibrium lies. Like i said... If K > 10^3 it lies WELL to the right, like virtually its almost at completion.
If K > 1 then it lies to the right. So if K is like 2, then it is SLIGHTLY to the right. If K is like 1001 then it is WELL to the right.
 

BlueGas

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Just a quick bump, how do I find the reaction quotient, Q?
 

kawaiipotato

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re: HSC Chemistry Marathon Archive

It's still [products]\[reactants]
K is a special case of Q only occurring at equilibrium
 

BlueGas

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It's still [products]\[reactants]
K is a special case of Q only occurring at equilibrium
Have there been any trials/HSC question on this reaction quotient?
 

Mr_Kap

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Can someone explain to me the chemistry of the Dry cell?
 

kawaiipotato

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Have there been any trials/HSC question on this reaction quotient?
I think ive seen questions give K and then asking if the current reaction is at equilibrium and you have to calculate Q and say since Q =/= K then ...
 

BlueGas

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I know. Should i do the one in Ahmed's notes? Is it good?
Yeah do the one's in Ahmed's notes, it's only the lead acid cell that looks more "harder" to remember, but nonetheless it won't be a problem as if you know the lead acid cell it'll be easy to further speak about the costs,etc. impacts on society and environment, so yeah stick with Ahmed's chosen electrochemical methods.
 

Mr_Kap

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Yeah do the one's in Ahmed's notes, it's only the lead acid cell that looks more "harder" to remember, but nonetheless it won't be a problem as if you know the lead acid cell it'll be easy to further speak about the costs,etc. impacts on society and environment, so yeah stick with Ahmed's chosen electrochemical methods.
Yeh i looked at it and was like wtf, looks so hard.

I'm usually really efficient at rote learning things but i've dropped off on my efficent recently, and after a few days i'm starting to forget things more than i used to. What works for you when you rote learn things. Need to rote learn the end of my option topic for chem, the second half of option in biology, and the second half of the option in physics.
 
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leehuan

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Q isnt in the syllabus but it helps explain K
Marking criteria in past HSCs have given marks for "finds Q"

Keep in mind that the question won't specifically say to find Q though, pretty much always.
 
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BlueGas

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Yeh i looked at it and was like wtf, looks so hard.

I'm usually really efficient at rote learning things but i've dropped off on my efficent recently, and after a few days i'm starting to forget things more than i used to. What works for you when you rote learn things. Need to rote learn the end of my option topic for chem, the second half of option in biology, and the second half of the option in physics.
Damn I've been feeling the same way too, but what other choice have you got other than just straight memorizing lol. Basically just remember as much as you can throughout the day, next day continue memorizing what you have left and once you've done everything, the next day memorise the whole option module, you'll realise that it'll take you less time than you think because you've already memorized bits of it from before.
 

Ekman

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What was the question?
They gave the value of K, and they gave the initial concentrations of both reactants and products, and asked you to calculate the concentration at equilibrium. You had to calculate Q first with the initial concentrations (because the concentration of the products weren't 0, so you cant assume that the equilibrium will be shifting towards the products side), in order to determine which way the equilibrium will shift, which allows you to know whether the concentration of the reactants and products should increase or decrease.
 
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