This question is perhaps a bit basic for this group. I’m in year 12 and
I would like to know the explanation for this chemical equilibrium
observation.
Say we have a solution in equilibrium:
2CrO4 2-(aq) + 2H3O+(aq) <–> Cr2O7 2-(aq) + 3H2O(l)
Say that the solution currently has an orange colour because the
solution is mostly filled with dichromate ions. If water is then added
to the solution, what happens, and why?
I have done the experiment and I’m pretty sure that the solution
becomes more yellow in colour due to the equilibrium position being
shifted to left. The reason I’m confused is because our syllabus
(http://www.curriculum.wa.edu.au/pages/syllabus_manuals/volumes/VII_sc…)
says:
"3.22 State and apply that changes in the mass of a solid, the volume
of a liquid and the presence of a catalyst have no effect on the
relative proportions of products to reactants at equilibrium."
Adding water (a liquid) should have no effect on the position of
equilibrium according to this statement. My guess is that adding water
dilutes the concentration of the ions and therefore the position of
equilibrium could possibly change. Could someone please help out by
explaining what is happening? Thanks very much.
Regards,
James Midolo
In article <1160104018.250180.7…@h48g2000cwc.googlegroups.com>,
"James Midolo" <james_…@hotmail.com> wrote:
>This question is perhaps a bit basic for this group. I’m in year 12 and
>I would like to know the explanation for this chemical equilibrium
>observation.
>Say we have a solution in equilibrium:
> 2CrO4 2-(aq) + 2H3O+(aq) <–> Cr2O7 2-(aq) + 3H2O(l)
>Say that the solution currently has an orange colour because the
>solution is mostly filled with dichromate ions. If water is then added
>to the solution, what happens, and why?
>I have done the experiment and I’m pretty sure that the solution
>becomes more yellow in colour due to the equilibrium position being
>shifted to left. The reason I’m confused is because our syllabus
Yes. If you assume everything was originally at 1 M, your K would be one.
Now assume you double the volume so everything is at 0.5 M. Calculate Q.
It’s not equal to K, so the equilibrium has been disturbed and the system has
to shift to return to equilibrium.
You have 1 mol of aqueous "stuff" on the right and 4 moles on the left.
Diluting thus decreases the concentration of the left more than the right, so
the system shifts left to restore equil.
It’s analogous to a system with gases and you change the volume of the
container. If there’s a different no. of moles of gas on the 2 sides of the
equil, one side will change pressure more than the other, throwing it out of
equil.
As Lyod says, adding the water has DISTURBED the equilibrium. It will
eventually upon its own return to equilibrium in time. By adding
either a product or reactant you will disturb a reaction at equilibrium
and the shift will follow Le Chatelier’s Principle. According to your
syllabus it is somewhat a confusing statement but is not incorrect. By
adding those mentioned substances we all agree the equilibrium is
disturbed and at this point the relative proportions of products to
reactants will have changed, hence Q doesn’t equal K. But once the
reaction returns to equilibrium it will now have the same relative
proportions of products to reactants. The syllabus statement saves
itself in saying at the very end "at equilibrium". In my opinion it is
a poorly written objective that as shown will only institute confusion
from the students.
Jeremy