Chem qu (1 Viewer)

[velox]

No idea......

 

[currysauce]

C

adding a solid of anythin to an equilibrium doesn't affect concentrations therefore doesn't disturb


well thats what i've been taught - stupid school

 

[Abtari]

cant even read the question!

 

[velox]

thanks ur correct.

 

[currysauce]

phew!!

and also other things that don't affect equil. are

- addition of liquids
- catalyst
- addition of noble gases

Just for you to know *thumbs up*

 

[Abtari]

why not addition of liquids??? that can affect equilibrium too... correct me if wrong

 

[currysauce]

seriously i couldn't tell u - my teacher told me that list along time ago

and u know what ur prolly right

so i shall remove it if others think what u think

 

[funking_you]

Hi All,

When you add solid sulfate into a solution, this specific compound will dissolve up and you will have sodium and sulfate ions in the solution.
So think you are adding more sulfate ions into the system, and sulfate ions appear in the second reaction. However this increase in concentration of sulfate ions does not affect the reaction, since the second reaction is NOT an equilibrium. Hence there will be no overall changes to the either chemical systems.


cheers,
George

 

[Jumbo Cactuar]

Yes, but sulfate ions and hydrogen sulfate ions exist in equilibria. So adding sulfate ions would decrease the acidity of the solution. (slightly)

 

[Dreamerish*~]

why not addition of liquids??? that can affect equilibrium too... correct me if wrong

Addition of liquids which do not participate in the reaction, that is, do not react with the products or reactants, do not affect the equilibrium.

 

[velox]

Logical too, that's a first.

 

[Jumbo Cactuar]

Consider;

Fe3+ + SCN- <====> FeSCN2+ K=1047

at equilibrium

[FeSCN2+] / [Fe3+][SCN-] = K = 1047

n(FeSCN2+) / n(Fe3+)n(SCN-) * V = 1047

Compare initial 0.01M Fe3+, 0.01 SCN- for V = 0.1 & 0.5L

let the equilibrium amount of FeSCN2+ = y

n(FeSCN2+) / n(Fe3+)n(SCN-) * V = 1047
yV / (n0(Fe3+)-y)(n0(SCN-)-y) = 1047
yV = 1047 (y^2 - 0.02y + 0.0001)
0 = y^2 - 0.02y - yV/1047 + 0.0001
y = 0.01 + V/2094 - sqrt((0.01 + V/2094)^2 - 0.0001)

when V = 0.1L
y = 0.01103
when V = 0.5L
y = 0.01244

change in y = (0.01244 - 0.01103)/0.01103 * 100%
= 12.78%

Or maybe I am mistaken...