ATAR Notes: Forum
VCE Stuff => VCE Science => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Chemistry => Topic started by: Xavier1234 on November 08, 2010, 02:24:21 am
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Im not great at chem but i thought if you increase the volume of the vessel of a gaseous system, the reaction will favour the production of MORE particles so as to partially compensate for that decreased pressure.
so the qtn goes like,
CO(g) + 2H2(g) ↔ CH3 OH(g)
if you increase the vol of the vessel at constant temp what happens to the:
•mass of CO at equi I would have said increase since favour ←
•mass of H2 at equi I would have said increase since favour ←
•effecct on mass of methanol at equi I woulda said decrease since favour ←
but the solutions suggests the complete opposite. can anyone explain why? please... :buck2: :-\
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X.x can u provide the full question?
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MHS 08 exam n' sols attached :police:
Edit: its question 1a of extended response :coolsmiley:
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lol the exam solutions are wrong. That's the only conclusion I can come up with (can't find any tricks in it sorry).
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lol the exam solutions are wrong. That's the only conclusion I can come up with (can't find any tricks in it sorry).
whew thank you so much.
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You will understand it better if you draw the concentration vs time graph.
A decrease in pressure causes all of the substances to decrease in concentration (an immediate drop). However, when the system regains equilibrium (by moving in the net backward direction) it will never 'exceed' its original concentration. (It asymptotically approaches it) Therefore, the net effect on the mass of CO and H2 will be a decrease. But I can't explain why ch3oh increases....
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You will understand it better if you draw the concentration vs time graph.
A decrease in pressure causes all of the substances to decrease in concentration (an immediate drop). However, when the system regains equilibrium (by moving in the net backward direction) it will never 'exceed' its original concentration. (It asymptotically approaches it) Therefore, the net effect on the mass of CO and H2 will be a decrease. But I can't explain why ch3oh increases....
oh right. i get it now. silly me. i was getting confused by basic things. thanks!
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is it me or does that question have strange wording? do they want the effects of the changes made at an established equilibrium or the effects after equilibrium is established after change
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lol ok I just did this exam (fml it is hard =.=) and I think multiple choice Q10 answer should be D, instead of A. And I also don't get why Q15 answer is C. And also, wat is this "rate determining step", I've never heard of that before.
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You will understand it better if you draw the concentration vs time graph.
A decrease in pressure causes all of the substances to decrease in concentration (an immediate drop). However, when the system regains equilibrium (by moving in the net backward direction) it will never 'exceed' its original concentration. (It asymptotically approaches it) Therefore, the net effect on the mass of CO and H2 will be a decrease. But I can't explain why ch3oh increases....
this is mass not concentration. I agree with Xavier.
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You will understand it better if you draw the concentration vs time graph.
A decrease in pressure causes all of the substances to decrease in concentration (an immediate drop). However, when the system regains equilibrium (by moving in the net backward direction) it will never 'exceed' its original concentration. (It asymptotically approaches it) Therefore, the net effect on the mass of CO and H2 will be a decrease. But I can't explain why ch3oh increases....
this is mass not concentration. I agree with Xavier.
Martoman can you please elaborate? im abit confused now.
lol ok I just did this exam (fml it is hard =.=) and I think multiple choice Q10 answer should be D, instead of A. And I also don't get why Q15 answer is C. And also, wat is this "rate determining step", I've never heard of that before.
that exam was indeed difficult. the rate determining step is always the reaction that takes up the most energy to take the reactants to form an activated complex, and then reform as products. the q15 mc, yeah i didnt get it either.
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I haven't done the exam but conceptually speaking
say you have N2O4 -> 2NO2
You decrease the volume. Meaning you increase the pressure.
Now overall the concentration has increased.
So irrespective, The concentration of NO2 say is increased. But the number of mol, the amount that is in transit left and right in the reactions, decreases at the new equlibrium as a net back reaction is happening.
Say you had 10 mol of NO2 in a 10 L container. At the end you'd have say 8 mol but now in a 2 L container. Concentration has increased yes, from 1M to 4M but the mass has decreased.
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:2funny: i was right after all. thanks martoman! ;D :D
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Hello, why can't Sodium Hydroxide and Ethanoic acod be mixed to create a buffer solution? I would have thought that adding sodium Hydroxide and sodium ethanoate would be purely basic, and so could not be a buffer? This is question 6, multiple choice.
Having not done the question, Sodium Hydroxide is a strong base so you cannot use that for a buffer solution... Has to be weak acid and base...
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lol ok I just did this exam (fml it is hard =.=) and I think multiple choice Q10 answer should be D, instead of A. And I also don't get why Q15 answer is C. And also, wat is this "rate determining step", I've never heard of that before.
yeah the q10 got me thinking too. since temp is decreased, Ka should be smaller, hence weaker acid, and an increase in pH. pure water=neutral so that bit's right. hmm. fml got 16/20 on mc.
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Mm I chose A for that one cause I was thinking that sodium hydroxide totally neutralizes the ethanoic acid, so overall you get a strong base solution, which is not a buffer.
Hello, why can't Sodium Hydroxide and Ethanoic acod be mixed to create a buffer solution? I would have thought that adding sodium Hydroxide and sodium ethanoate would be purely basic, and so could not be a buffer? This is question 6, multiple choice.
Having not done the question, Sodium Hydroxide is a strong base so you cannot use that for a buffer solution... Has to be weak acid and base...
=/ one of the other choices is NaOH and CH3COONa, and apparently that acts as a buffer
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I dunno how that works...
If you add base to that solution though I don't know where the H+ ions come from to neutralise it - and I thought that buffer solutions are meant to be able to neutralise or partly neutralise a solution of acid or base...