ATAR Notes: Forum

VCE Stuff => VCE Science => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Chemistry => Topic started by: 98.40_for_sure on July 29, 2010, 10:04:41 pm

Title: Le Cheteliar's Graphs
Post by: 98.40_for_sure on July 29, 2010, 10:04:41 pm
Lets assume that the equation is 3A + 2B -> C
And you dilute it with water

Which graph is correct? Is the dilution the same for each concentration line, or do you have to draw them different lengths according to mole ratio?

CHEERS  :knuppel2:
Title: Re: Le Cheteliar's Graphs
Post by: Studyinghard on July 29, 2010, 10:19:23 pm
im pretty sure its the second graph. work the with mole ratios
Title: Re: Le Cheteliar's Graphs
Post by: stonecold on July 29, 2010, 10:20:47 pm
yeah the second graph.  dilutions should always go further down for higher concentrations.

it is not correct to drop them all by the same amount and apparently examiners don't like it...
Title: Re: Le Cheteliar's Graphs
Post by: 98.40_for_sure on July 29, 2010, 10:21:03 pm
From what i've seen in textbooks and notes, they just dilute them all the same

But a few checkpoints questions make a fuss about it, so unsure...
Title: Re: Le Cheteliar's Graphs
Post by: stonecold on July 29, 2010, 10:24:28 pm
let me give you a scenario.

-dilute something that is 1.0 M by half, new conc is 0.5 M
-dilute something that is 0.2 M by half, new conc is 0.1 M

see how the higher the initial conc, the more the dilution will drop the conc?
Title: Re: Le Cheteliar's Graphs
Post by: 98.40_for_sure on July 29, 2010, 10:27:37 pm
But isn't this mole ratio, not concentration ratio
Title: Re: Le Cheteliar's Graphs
Post by: 8039 on August 02, 2010, 11:10:12 pm
A and B would lower rapidly then slowly reach equilibrium and C would rapidly lower, then slightly rise (not as high as original tho, obviously). Isn't that right?

Is this a time/concentration graph? then yeah lines would be equal length. Reactants on top in any order and product at bottom. The graphs look the same to me?