Cobaltous Chloride Equilibrium
Date: April 2008
I have checked the archives and found one related entry:
Kris N asked about the cobaltous chloride equilibrium on 18 March,
2003 [1672 Cobalt Chloride], [
http://www.newton.dep.anl.gov/askasci/chem03/chem03002.htm ] but his
question was only partially answered by Vince Calder since the
question was not entirely clear.
My question is about the same
equilibria: 1 mole of hexahydrated cobalt (II) ion reacts with 2
moles of chloride ion to produce in a reversible process 1 mole of
tetrahydrated cobalt (II) chloride and 2 moles of water
[Co(H2O)6]2+(aq) + 2Cl-(aq) <==> Co(H2O)4Cl2(aq) + 2H2O(l) Adding a
source of chloride ion (e.g. hydrochloric acid) drives the
equilibrium forward indicated by the resulting blue-colored
This is predicted by Le Chatelier's Principle and expected
since the increased concentration of chloride ion increases the rate
of the forward reaction. Subsequently adding water to the acidified
equilibrium appears to drive the system in reverse since the color
reverts back to the pink associated with the hydrated cobalt ion. I
require an explanation why this happens. Le Chatelier's Principle
might predict the shift to the left as a response to the addition of
water, a 'reactant' for the 'reverse' reaction. However, this is a
heterogeneous equilibrium, the concentration of the water is not
altered by the addition of more. The reverse reaction should not
increase its rate due to the addition of water. Certainly adding
water lowers the concentration of all the aqueous species, but I
have assumed that this affects the forward and reverse reaction
equally and there should be no shift in equilibrium. Is it a matter
that there are more moles of reactants (for the reaction as written)
and the effect of dilution has a greater impact on the left side
slowing down the forward reaction more that the reverse? Or am I
missing the point altogether and it is a matter of a simple dilution
altering the color intensity (in this case somehow changing it from
blue through purple to pink)?
Note that the system has been observed
at constant temperature. That is to say there has been no
temperature stress applies, although since adding water drives the
system in reverse (the exothermic direction) there is a small
temperature rise as a *result* of the shift.
You need to keep track of which concentrations are constant and which are not.
When water is added to the solution, the concentrations of all reacting
species are lowered, EXCEPT WATER. So the water concentration stays approximately
the same, all the rest are lowered, and the tetrahydrated species replaces
chloride with water, allowing the chloride to dissociate.
Richard Barrans, Ph.D., M.Ed.
Department of Physics and Astronomy
University of Wyoming
You are correct in your statement that the concentration of water is not affected
by adding more water. As such, pure liquids are not included in the expression of
Keq because such systems do not change in concentration over the course of the
reaction and do not, in of themselves, affect rates.
You are also correct in stating that adding more water to the system results in the
further dilution of all aqueous reactants and products. Thus, both the forward and
reverse reaction rates are slowed down. However -as you have thought- the rates are
not slowed down *equally*. The forward reaction rate is slowed down more since it
is dependent on essentially two reactants (one of which is raised to the second
power) while the reverse rate is a function of only one reactant. Consequently
there is a net production of more reactants (as written).
Or if you want to look at it from the perspective of Keq, the numerator has
[Co(H2O)4Cl2] while the denominator has: [Co(H2O)6]2+]*[Cl-]^2. Reducing all the
molarities by the amount of water added would lessen the numerical values of the
denominator more. To require equilibria (set the equation to equal to Keq), there
would need to be a shift toward the production of more reactants as written.
Greg (Roberto Gregorius)
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