Equilibrium
Group Work.1
- A Zen Moment
In silence contemplate the beaker of water for 2 minutes.
During the first minute, allow your mind to go where it wants.
During the second minute, become aware–think–about what is in the beaker at the molecular level.
Today's Reaction System: copper chloride dihydrate CuCl2(H2O)2
Species identities:
turquoise solid: CuCl2(H2O)2
green (in solution): [CuCl(H2O)5]+ ion
sky-blue (in solution): [Cu(H2O)6]2+ ion
a- Add 6 drops water to the turquoise solid. What happens? Write a balanced equation for the reaction.
CuCl2(H2O)2(s) + 3H2O <==> [CuCl(H2O)5]+ + Cl-
color changes from turquoise to green; some solid remains
b- Add one full pipette of water. What happens? Write a balanced equation for the reaction.
[CuCl(H2O)5]+ + H2O <==> [Cu(H2O)6]2+ + Cl-
color changes from green to sky blue
c -Add the vial of white powder, LiCl. What happens? Suggest the reaction. Write an equation describing system.
[Cu(H2O)6]2+ + LiCl <==> [CuCl(H2O)5]+ + H2O + Li+
color changes from sky blue to green
d- What is the net equilibrium reaction between the sky blue species and the green species?
(i.e, the reaction in step c. Ignore all spectator ions, such as Li+)
[Cu(H2O)6]2+ + Cl- <==> [CuCl(H2O)5]+ + H2O
e-Write Kc for the reaction in step d. Kc = [Cu Cl (H2O)5][H2O]/[Cu(H2O)6][Cl-]
f- Note that pure liquids do not appear in Kc expressions. (More precisely, their "concentration" is defined to be unity or 1. ) Now re-write Kc taking this into account. Kc = [CuCl(H2O)5]/[Cu(H2O)6][Cl-]
g- What information is needed to calculate Kc?
Concentration of all species @ equilibrium: [CuCl(H2O)5] = 0.8 M, [Cu(H2O)6] = 0.4 M, [Cl-] = 7.1 M
h- Calculate Kc
Kc = 0.8/0.4*7.1 = 0.28
i- Add two full pipettes of water. What happens? Suggest the reaction.
[CuCl(H2O)5]+ + H2O <==> [Cu(H2O)6]2+ + Cl-
color changes from green to sky blue
j - Explain the results observed in steps c and i using the equilibrium reaction you wrote in step d.
Adding water stresses the reactant side of the reaction, which pushes the equilibrium towards the products. This must be true since Kc is a constant, so while the individual concentrations change, the ratio of the three species in the equilibrium expression must be constant.
Now extend what you've learned:
k -What do you expect to happen if you added 10 pipettes of water to the turquoise solid in the first step?
It would turn sky blue due to the formation of [Cu(H2O)6]2+
l - Write the equilibrium expression for the reaction in step k.
CuCl2(H2O)2(s) + 4H2O <==> [Cu(H2O)6]2+ + 2Cl-
m -Show how you could add two reactions above (step a and step b) to get the overall reaction in step l.
CuCl2(H2O)2(s) + 3H2O <==> [CuCl(H2O)5]+ + Cl- (1)
[CuCl(H2O)5]+ + H2O <==> [Cu(H2O)6]2+ + Cl- (2)
CuCl2(H2O)2(s) + 4H2O <==> [Cu(H2O)6]2+ + 2Cl-
n- Show how the Kc you can write for the equation in step l can be written as a product of Kc' s.
Kcoverall = Kc1*Kc2 = ([Cu(H2O)6][Cl-]/[CuCl(H2O)5]) * ([Cl-][CuCl(H2O)5])
[Cu(H2O)6] [Cl-]2
- Compare the microscopic and the macroscopic descriptions of the system provided.
- Is nothing happening in a reaction that appears to have stopped?
- Do all reactions go forward? Do all reactions go to completion? Do all reactions "go"?
- Prove why the pressure of CO2 above _any_ mixture of CaO and calcium carbonate is always the same.