TODAY

• A new chapter: Equilibrium

• Writing equilibrium expressions

• observing LeChatelier’s Principle

• Calculating equilibrium constants, K

TODAY

• A new wrinkle on Equilibrium:

Q

Q, from the Q continuum, Star Trek ,The Next GenerationNo, not that Q ..

TODAY

• A new chapter: Equilibrium

• Writing equilibrium expressions

• observing LeChatelier’s Principle

• Calculating equilibrium constants, K

TODAY

• A quick recap of Monday’s concepts

•Magnitudes of Keq

•Q: the reaction quotient

• Q: how to use it

A quick recap on equilibrium expressions:

Keq [reagents]

[products]

[Cu(H2O)6]2+ + Cl- [CuCl(H2O)5]+ + H2OFor this reaction:

But if reaction goes BOTH to right and to left, which are reagents and which are products?

By convention”reagents are species to the left of arrowproducts are species to the right of arrow

A quick recap on equilibrium behavior:

[Cu(H2O)6]2+ + Cl- [CuCl(H2O)5]+ + H2OYou observed this chemical system:

When you added excess Cl- :[Cu(H2O)6]2+ + Cl- [CuCl(H2O)5]+ + H2O

When you added excess H2O:[Cu(H2O)6]2+ + Cl- [CuCl(H2O)5]+ + H2O

The series of reactions observed:

3. [Cu(H2O)6]2+ + Cl- [CuCl(H2O)5]2+ + H2O

To calculate Keq [Cu(H2O)62+][Cl]

[CuCl(H2O)52+]

1. CuCl2(H2O)2 + 3H2O [CuCl(H2O)5]+ + Cl-

2. [CuCl(H2O)5]+ + H2O [Cu(H2O)6]2+ + Cl-

Need Keqconcentrations 0.40M x 7.1 M

0.80 M0.28,NO UNITS!

Pure liquids and solids don’t appear in Keq expression:

Combining reactions equals multiplying K’s

Net CuCl2(H2O)2 + 4H2O [Cu(H2O)6]2+ + 2 Cl-

K1eq x K2

eq CuCl2(H2O)2

[CuCl(H2O)52+ [Cl]

(a) CuCl2(H2O)2 + 3H2O [CuCl(H2O)5]+ + Cl-

(b) [CuCl(H2O)5]+ + H2O [Cu(H2O)6]2+ + Cl-

K1eq

Kneteq

K2eq

[Cu(H2O)62+][Cl]

[CuCl(H2O)52+

][Cu(H2O)6

2+][Cl]2

CuCl2(H2O)2 Kneteq K1

eq x K2eq

Are Your Eyes Misleading You?What is in the graduated cylinder?

Abs

wavelength 700 nm400 nm

[Cu(H2O)62+]

[CuCl(H2O)5+]

Visible electronicspectra

mixture can appear GREEN

Are Your Eyes Misleading You?What is in the graduated cylinder?

Keq [Cu(H2O)6

2+][Cl]

[CuCl(H2O)5+]

Recall equilibrium concentrations:

0.280.8 M

0.4 M x 7.1M

Abs

wavelength 700 nm400 nm

[Cu(H2O)62+]

[CuCl(H2O)5+]

mixture can appear GREEN

Visible electronicspectra

If Keq ~ 1, product concentrations are similar to reagent concentrations

How large must Keq be for reaction to be “complete”?Consider this reaction:

Keq 2.0 x 108

[Ni(H2O)6]2+ + 6 NH3 [Ni(NH3)6]2+ + 6H2O

[Ni(en)3]2+ + 6 NH3[Ni(NH3)6]2+ + 3 “en”

Keq 7.3 x 109

What is Keq for this reaction: ?

K1 = 2.0 x 108[Ni(H2O)6]2+ + 6 NH3 [Ni(NH3)6]2+ + 6H2O[Ni(en)3]2+ + 6 NH3[Ni(NH3)6]2+ + 3 “en” K2 = 7.3 x 109

[Ni(H2O)6]2+ + 3 “en” [Ni(en)3]2+ + 6 H2O

[Ni(H2O)6]2+ + 3 “en” [Ni(en)3]2+ + 6 H2O

K = K1 x K2 = (7.3 x 109)(2.0 x 108) = 1.5 x 1018

K = K1 x K2

What happens if the concentrations are equal:

[Cu(H2O)6]2+ + Cl- [CuCl(H2O)5]+ + H2O

[Cu(H2O)62+][Cl]

[CuCl(H2O)5+]

But you know Keq = 0.28 ≠ 1.25: what does this mean?

You investigate by calculation:

1.25

[CuCl(H2O)5+]= 0.8 M, [Cu(H2O)6

2+]= 0.8 M, [Cl] = 0.8M

for this reaction:

0.8 M x 0.8 M0.8 M

It’s not at equilibrium!!

So under these concentration conditions:

[Cu(H2O)6]2+ + Cl- [CuCl(H2O)5]+ + H2O

[Cu(H2O)62+][Cl]

[CuCl(H2O)5+]

How will reaction system species behave?

1.25 > 0. 28 = Keq

[CuCl(H2O)5+]= 0.8 M, [Cu(H2O)6

2+]= 0.8 M, [Cl] = 0.8M

This tells you one definite thing:there’s too much in numerator, or,there’s too much product

[CuCl(H2O)5+] decreases, [Cu(H2O)6

2+] increases, [Cl] increases

This is Q!!! Ratio of Concentrations under Non-Equilibrium conditions

aA + bB cC + dD

Q[reagents]

[products]

Q [A]a[B]b

[C]c[D]d

Q: the Reaction Quotient

The reaction quotient Q can be determined for any set of concentrations

Possible outcomes

[A]a[B]b1. Q [C]c[D]d

Keq

2. Q [A]a[B]b

[C]c[D]d

> Keq

3. Q [A]a[B]b

[C]c[D]d

< Keq

Example problems to be used with reaction:[Cu(H2O)6]2+ + Cl- [CuCl(H2O)5]+ + H2O

Keq = 0.28A. [CuCl(H2O)5

+]= 0 M, [Cu(H2O)62+]= 0.4 M, [Cl] = 0.4 M

B. [CuCl(H2O)5+]= 1 M, [Cu(H2O)6

2+]= 1 M, [Cl] = 0.5 M

C. [CuCl(H2O)5+]= 0.01 M, [Cu(H2O)6

2+]= 0.01 M, [Cl] = 0.01 M

Compare this result with earlier equimolar at 0.8M !!

Calculations, calculations, calculations,

Many types:1. Calculating K from equil. concentrations

2. Calc’g Keq from initial and changed concentrations

3. Calc’g final concentrations from initial, change and Keq

4. Calculating a new K from adding 2 reactions

The ICEbox method

For these types:2. Calc’g Keq from initial and changed concentrations

3. Calc’g final concentrations from initial, change and Keq

Lots of K’s

Kc - equilibrium constant in concentrations

Kp - equilibrium constant in partial pressures

Ka - equilibrium constant for H+ dissociation

Kb - equilibrium constant for OH- formation

Ksp - equilibrium constant for solubility

Not to be confused with:k – rate constants

Keq – a general equilibrium constant

Kf - formation constant for metal complexes

The K “Zoo”

KcKa Kblittle kKp

Keq KfKsp