Equiliberium and Concentration question (1 Viewer)

mitchwong650

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Yeah i agree, wouldnt a decrease in pressure result in all substances being reduced by an EQUAL amount? because all gases are 24.79L in STP and each gas CO, Cl2 and COCl2 were all the same amount of moles (1:1:1 ratio) .'. they would all decrease the same amount, what the question was saying was that each was different, hence they it cant be reduction in pressure.
well i had second thoughts about pressure, so i wrote what this guy did:

so for the last part where they all dropped is it valid to say that an amount of gases was removed from the system

but yeah.... who knows.
 

Michaelmoo

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I don't think pressure was reduced in the last one. If it was, then what all concentrations would've reduced by equal amounts.

Also quick check- who here wrote something about the plateau-ing being equilibrium?
I thought that too. But then I thought about it and decreasing pressure doesn't mean concentrations are reduced by equal amounts. Consider this analogy. I have a system with 40 balls:

20 red balls
10 blue balls
10 green balls

I want to double the pressure (half the volume). The ratio of balls should still stay the same, i.e. 2:1:1. In doing so, the number of balls will now be:

10 red
5 blue
5 green

The concentration of red balls has decreased by 10/40 = 25%. The concentration of blue and green balls has droped by 5/40 = 12.5%. In other words, the higher the relative concentration, the greater the drop in concentration (upon disturbing the pressure of the system).

So a pressure decrease does explain a different drop in the concentration of gasses.
 

Drewx

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Was the reaction endothermic in the forward direction?
 

mirror-match

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I thought that too. But then I thought about it and decreasing pressure doesn't mean concentrations are reduced by equal amounts. Consider this analogy. I have a system with 40 balls:

20 red balls
10 blue balls
10 green balls

I want to double the pressure (half the volume). The ratio of balls should still stay the same, i.e. 2:1:1. In doing so, the number of balls will now be:

10 red
5 blue
5 green

The concentration of red balls has decreased by 10/40 = 25%. The concentration of blue and green balls has droped by 5/40 = 12.5%. In other words, the higher the relative concentration, the greater the drop in concentration (upon disturbing the pressure of the system).

So a pressure decrease does explain a different drop in the concentration of gasses.
I disagree, considering that all of the substances in equilibrium are a gas. All of them are at the have the same amount of moles in the space and the same amount of volume (1:1:1 ratio + STP) as a result the pressure should decrease them all by the SAME amount. The only thing that explains a variable change is some sort of dilution, which is what i put as what happened at 14 minutes, but know that i think about it, HOW CAN YOU DILUTE A GAS WITH WATER?
 

Michaelmoo

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I disagree, considering that all of the substances in equilibrium are a gas. All of them are at the have the same amount of moles in the space and the same amount of volume (1:1:1 ratio + STP) as a result the pressure should decrease them all by the SAME amount. The only thing that explains a variable change is some sort of dilution, which is what i put as what happened at 14 minutes, but know that i think about it, HOW CAN YOU DILUTE A GAS WITH WATER?
True in this sense. But the graph represents the actual system. If they were all the same concentration, all three lines (on the graph) will be in phase. i.e. All three gasses will have the same line.

But clearly, one line was above another which was above another --> indicating a variation in concentration.
 

shaon0

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The one about the Cl2 gas and the CO was pretty hard :\ i talked about how produts were being removed. Pressure, temp shifting equilibrium.

and how it starts off a a state of equilibrium and is disturbed then shifts bla bla it was all over the place..... im hoping for 4/6..


Other questions were easy tho hoping for a 92% raw :D
Changes were: temperature, concentration and pressure/volume.
 

mirror-match

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True in this sense. But the graph represents the actual system. If they were all the same concentration, all three lines (on the graph) will be in phase. i.e. All three gasses will have the same line.

But clearly, one line was above another which was above another --> indicating a variation in concentration.
I see, well then... Touche young sir, touche. Ahhh well 4/6 :3
 

cdiddy7

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i put:

1st change = temp increase
2nd change = CO concentration decreased manually.
3rd change = pressure reduced
+1
exactly what i put
and i added that it plateaued as it reached equilibrium again
and i quoted what times this applied for
i really hope i got full for it
 

gurmies

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Looking back at it, it IS very simple. At the time however, I couldn't see a single difference between large drops or small ones. I rationalised the situation by saying for the first part, instead of temperature increasing, the pressure slowly dropped (although in exam I just said pressure dropped lol). Did not like this one.
 

Gibbatron

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It was:

increase in temperature, as the system was endothermic, shifting equilibrium to the right
Removal of CO, shifting equlibrium to the right
Decrease in pressure on the system (they all decreased by the same amount, i thought), shifting the equilibrium to the right.
 

P!xel

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so for the last part where they all dropped is it valid to say that an amount of gases was removed from the system
Yes! That's what I wrote too (Although I mentioned removing more products than reactants because the Reactants decreased after the change and the products increased)

Pressure can't affect it because it's 1:1:1 mole ratio. Let's hope me and you are right on this one ;)
 

Gibbatron

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Yes! That's what I wrote too (Although I mentioned removing more products than reactants because the Reactants decreased after the change and the products increased)

Pressure can't affect it because it's 1:1:1 mole ratio. Let's hope me and you are right on this one ;)
Pressure does have a major effect on concentration where gases are concerned. If the pressure was increased, the concentration would increase sharply. It follows that if the pressure were to decrease, then concentration would decrease sharply.

The reason the products increased is because of the molar ratio of the reactants : products was 1:2. This meant as the pressure decreased the equilibrium shifted to the right forming more of the products.
 
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mR sinister

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+1
i wrote most of the correct stuff comparing with use :)
makes me feel alot better since i stuffed up MC

Anyone did anyone find about 6 lines for a 6 marker way too less?
Overall, i think they could have given us much more space to write our answers
 

NoName91

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Point id like to bring up. Note when all parts of the reaction were removed (the vertical drop in everything)... that a far greater concentration of the PRODUCTS were removed (about 0.05 mol L more i think) meaning that equilibrium is forced right to counter the change in concentrations...

?
 

NoName91

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+1
i wrote most of the correct stuff comparing with use :)
makes me feel alot better since i stuffed up MC

Anyone did anyone find about 6 lines for a 6 marker way too less?
Overall, i think they could have given us much more space to write our answers
Yeah I ended up writing like 4 more lines at the top of the page for this question
 

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