Group Work.7
- A Zen Moment
In silence contemplate the bottle for 2 minutes.
During the first minute, allow your mind to go where it wants.
During
the second minute, become awareÑthinkÑabout why the bottle is so blue.
Molecular Push-Me Pull-You
The goal for this group work is to observe microscopic dynamics, i.e.
molecular changes, driven by macroscopic stasis, i.e., equilibrium.
NOTE: it is critical
that you follow the directions precisely to see the intended changes.
1. Dump the vial A containing the red-pink crystals into the 25 mL Erlenmeyer flask marked A that contains water. Make a careful note of its color and appearance. Swirl the mixture until the solid dissolves. Note the color.
2. Dump the vial B containing the red-pink crystals into a second 25 mL Erlenmeyer flask marked B that contains 2-propanol (isopropanol) as solvent. Swirl the mixture until the solid dissolves. Note the color.
3. Is the color of the dissolved solid in B a surprise?
Yeah!
It's blue, not red-pink!
Make a hypothesis to account for what you have observed.
ItÕs different than from the bottle: a
reaction occurred in propanol, (not water?!)
4. Which solution is more intensely colored? Keep this in mind ...
5. Now, carefully take your propanol solution in Erlenmeyer B and pipette it into four test tubes 1-4 so that each is about half-full and has approximately the same volume.
6. (a) To test tube 1, add one drop of water (from vial marked H2O) with a pipette. Continue adding drops of water until the color is ÒlavenderÓ or ÒlilacÓ, a pale purple hue.
(b) Repeat this with test tube 2.
(c) Repeat this with test tube 3. Then, add a full pipette full of water to the third pipette. What color is the
solution now? Does it look familiar?
ItÕs pale pinkÉ looks like the
CoCl2 in water in Erlenmeyer A
WhatÕs going on
here??!!
Now itÕs time to introduce the characters n this chemical drama!
The solid is cobalt (II) chloride hexahydrate, whose molecular formula is usually written as CoCl2 6H2O.
This formula does NOT correctly convey the species which predominates. When dissolved in water, the solid CoCl2 6H2O is actually the complex ion [Co(H2O)6]2+ with two chloride counterions, better written as [Co(H2O)6]Cl2. When dissolved in other solvents, a different complex can form whose formula is [Co(Cl)4]2-.
To get these characters clearly in mind, draw the structures of the cobalt complexes below.
octahedral [Co(H2O)6]2+ tetrahedral
[Co(Cl)4]2-.
The important chemical reactions involving these two cobalt complexes are:
(a) CoCl2 6H2O (s) + water --> [Co(H2O)6]2+ (aq) + 2 Cl - (aq)
(b) [Co(H2O)6]2+ + 4 Cl - --> [Co(Cl)4]2- + 6H2O
8. Referring to your notes, identify the characteristic colors of the two Co complexes, [Co(H2O)6]2+ and [Co(Cl)4]2-. Which complex predominates in aqueous solution and which predominates in propanol?
[Co(H2O)6]2+ is pink in water; [Co(Cl)4]2- is blue in propanol
9. Referring to the reactions above, is it possible for 100% of the [Co(Cl)4]2- to be formed in propanol? Why not?
No! CoCl2 6H2O has only 2
Cl- per Co; so could only form ~ 50% of [Co(Cl)4]2- is blue in propanol.
The counter cation for [Co(Cl)4]2- is [Co(H2O)6]2+
So why is the propanol solution so blue? Suggest an
explanation for why the propanol solution is blue (not lavender). Blue is
a more intense color and masks the pink.
10. (a) What happens when water is added to the test tubes 1 and 2? What reaction occurs? What species are present?
More [Co(H2O)6]2+ is formed as water is added and replaces chloride ligands (reaction b ,
backwards).
Both [Co(H2O)6]2+ and [Co(Cl)4]2- are present to make lavender color.
(b) What happened in test tubes 3?
With even more water, now there is mainly [Co(H2O)6]2+ ; no blue species remains.
11. Now take test tube 1 and place it in the ice bath provided (see dr.b. or PLI ).
Take test tube 2 and place it in the hot water bath. What do you observe?
Propose reactions to account for your observations.
The cold bath turns the lavender
tube pink: reaction (b) goes
backward.
The hot bath turns the lavender
tube blue: reaction (b) goes
forward.
12. Remember Le ChatelierÕs Principle from CHEM103? How does it apply to the changes in steps 7-10?
As water is added, reaction shifts
backwards to form [Co(H2O)6]2+ to use up added water.
13. Since no reagent or solvent additions are made in step 11, what must be the cause of the observed change?
14. Consider reaction (b) above: what is the entropy change DSrxn when the reaction goes forward? Why?
Entropy increases because
there are now more molecules (7 vs 5), DS > 0
What is the entropy change DSrxn when the reaction goes backwards? Why?
Entropy
decreases because there are now fewer molecules, DS < 0
15. What must the enthalpy change DHrxn be for the backward reaction if it can occur spontaneously?
Since DS < 0, then DH must be < 0 and negative
enough to compensate for - DS in order to make DG < 0
What must the enthalpy change DHrxn be for the forward reaction?
DH must be > 0 if the reaction goes the opposite direction.
16. Finally, compare these enthalpy changes DHrxn and the behavior of the solutions when chilled or heated.
What is the relationship between DHrxn and the temperature and the way the species in solution change?
- Adding heat makes reaction go forward ˆ endothermic direction, makes blue [Co(Cl)4]2-
- Removing heat (cooling) heat makes reaction go backward ˆ exothermic direction, makes pink Co(H2O)6]2+