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VCE Stuff => VCE Science => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Chemistry => Topic started by: hyperblade01 on July 17, 2009, 07:21:24 pm

Title: Equilibrium Questions
Post by: hyperblade01 on July 17, 2009, 07:21:24 pm: $N_2O_4 (g) \rightleftharpoons 2NO_2 (g)$

(Visualise that graph of concentration V time)

If the volume is ~~decreased~~ increased (no amount specified), the lines of $2NO_2$ and $N_2O_4$ drop with the line of $2NO_2$ dropping twice as much. Why is this so?

Prac related:

$Fe^{3+} + SCN^- \rightleftharpoons Fe(SCN)^{2+}$

So a solution of $Fe(SCN)^{2+}$ is added to a test tube containing these ions, the $Fe(SCN)^{2+}$ ion causes the solution to be 'blood red' (but really a ginger brown).

Two seperate test tubes:
$Fe(NO_3)_3$ is added, increasing amount of $Fe^{3+}$ - the solution turns darker and deepens in colour; there is a net forward reaction
$KSCN$ is added, increasing amount of $SCN^-$ - the solution turns darker and deepens in colour; there is a net forward reaction but turns darker than the above

Why is this so? I was told that it doesn't really matter why because the main point of the prac was to observe a change in equilibrium (there were other tests with other colours) but I'm still curious..

Anything to do with dissociation of ions is higher in $KSCN$ hence more ions? Taking a wild guess here..
Title: Re: Equilibrium Questions
Post by: TrueTears on July 17, 2009, 07:48:53 pm: If volume is decreased then concentration increases hence LCP indicates the backwards reaction is favoured.

So assuming the reaction is at equilibrium at the start. There will be a sudden increase in the concentration of both spieces. The $NO_2$ line will increase [vertical straight line on the conc - time graph] twice as much because there are twice as much mol of it. [the mol ratio 1:2]. Since $c = \frac{n}{v}$ , let $\frac{1}{v} = k$ . $c = kn$ . Hence if there is twice as much mol means there will be twice as much concentration of $NO_2$ .

EDIT: Okay if the volume is increased it is the same principle. The drop of $NO_2$ will be twice as much as $N_2O_4$ because there were twice as much mols at the start before the volume change. Once the volume changes (increased) since c is proportional to n, if n is double so is c. Hence it drops twice as much.
Title: Re: Equilibrium Questions
Post by: hyperblade01 on July 17, 2009, 07:52:54 pm: Realised I made a mistake by saying decrease rather than increase..

But I can see that now thanks for that :)
Title: Re: Equilibrium Questions
Post by: lacoste on July 19, 2009, 10:27:14 pm: I did a SAC invovling experiments with those of the above; NO2 and Fe(SCN)2+, it was soooooooooo tricky nd long because I thought that they only had LCP and equilibrium but then there was this hard as acid equilibria qst at the end linking to the ones above and other concepts. all under 1hr

Yeap, TT well explained.

gotta love chem!!
Title: Re: Equilibrium Questions
Post by: hyperblade01 on July 20, 2009, 12:08:57 am: Yea i've got a SAC coming up on LCP and equilibrium and honestly I don't know what they will ask...

It'll probably be related to the prac I recently did and hence my question on why one result was darker than the other - it's the only thing I can't explain why it happened....
Title: Re: Equilibrium Questions
Post by: TrueTears on July 20, 2009, 12:10:44 am: GL with your SAC, I'm sure you'll own it :)
Title: Re: Equilibrium Questions
Post by: lacoste on July 20, 2009, 08:46:19 am: GL mate, you'll be fine!!!!!!