Previously in Chem104

• 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 Generation

No, 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]+ + H2O

For 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]+ + H2O

You watched:

You added excess Cl- :

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

You added excess H2O:

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

A quick recap on equilibrium expressions:

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

Keq [reagents]

[products]

[CuCl(H2O)5]+ + H2O

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

[CuCl(H2O)5+]

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

You watched:

You wrote:

0.28

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

2+]= 0.4 M, [Cl] =7.1M

You know:

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 these reactions:

Keq 2.0 x 108

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

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

Keq 7.3 x 109

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

Consider this reaction:

Keq 2.0 x 108

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

What is Keq for this reaction: ?

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

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

[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

This Keq seems huge! What is ratio of rgt. Ni to prdt. Ni species

[Ni(en)3]2+ + 6 H2OK = 1.5 x 1018

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 ≠ 0.8: 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 M

0.8 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