Equilibrium
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Equilibrium. For equilibrium to occur:. System must be closed. Temperature must be constant. Reactions must be reversible (do not go to completion). H 2 (g) + Cl 2 (g) 2HCl(g) + energy No visible change…… A dynamic equilibrium exists.

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Equilibrium

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Equilibrium

Equilibrium


Equilibrium

For equilibrium to occur:

  • System must be closed.

  • Temperature must be constant.

  • Reactions must be reversible (do not go to completion).

  • H2(g) + Cl2(g) 2HCl(g) + energy

  • No visible change…… A dynamic equilibrium exists.

  • The rate of forward rx. = the rate of the reverse rx.

  • Homogeneous Equilibria: all gaseous or aqueous phases.


Equilibrium in n 2 o 4 g q 2 no 2 g

Equilibrium in N2O4(g) + q 2 NO2 (g)

  • Initially, [N2O4] large.

Rate f = Rate r

  • It decreases rapidly then more slowly.

  • The [NO2] starts at zero; increases rapidly.

concentration

N2O4

NO2

  • Both eventually, plateau…no change

  • Equilibrium reached.

Time

Teq


Equilibrium

N2O4(g) + q 2 NO2 (g)

What happened here?

concentration

Here?

Time

Teq


Predicting changes

Predicting Changes

Le Chatelier’s Principle: when a stress is applied to a system at equilibrium…. The reaction will “shift” in a direction to minimize the stress.

2H2(g) + O2(g) 2H2O(g) + energy

“shift” = rxn speeds up in a particular direction.

E.g. shifts right…. Forward rxn speeds up. More products “appear”. We say the reaction is “favored” to the right or the “position of the equilibrium” shifts right.

  • Types of stress: change in concentrations; change in volume; in temperature…


2h 2 g o 2 g 2h 2 o g energy

2H2(g) + O2(g) 2H2O(g) + energy

[H2]

Make more

[H2O]

[O2]

Use it up!

[H2O]

[H2]

[O2]

Temp

Make more pressure

Volume

[H2O]

[O2]

[H2]

What would a catalyst do to the equilibrium position?

Nothing! It would simply be reached sooner.

What would happen if we added He gas ?

Nothing! It does not affect the partial press of other gases.


N 2 o 4 g q 2 no 2 g

N2O4(g) + q 2 NO2 (g)

colorless

orange

Describe what happens when your instructor removes the tube from the freezer, containing the system described above. Explain your observation(s) using LeChatelier’s Principle, and all of the appropriate terminology.


Quantitative aspect of equilibrium

Quantitative Aspect of Equilibrium

  • Measurements in the N2O4 - NO2 system @ 100C

If we divide Equilibrium product values by reactant, each factor raised to the correct power (explained in a moment)…. What do you notice????


Equilibrium

Quantitative Aspect of Equilibrium

N2O4(g) + q 2 NO2(g)


Mass action expression

Mass Action Expression

Given the general equation:

aA(aq)* + bB(aq)cC(aq) + dD(aq)

*also if g = gas

[C]c [D]d

[A]a [B]b

K eq

equilibrium constant expression

@ constant Temp. ( Keq can also be Kp , Kc , Ksp , Ka , Kb , etc.)

Product(s)

Right

K eq =

0r

Reactant(s)

Left


Equilibrium

What does the K value tell us?

Prod.

large

  • If K is > 1

At Equilibrium… there is more product than reactant!

Reactant

small

  • If K is < 1

Prod.

At Equilibrium… there

is more reactant than product!

small

Reactant

Large

  • If K is = 1 [Products] = [Reactants]

  • If K is verysmall…..then practically no product

  • is formed.


Equilibrium

Eg. Weak Acid:

HC2H3O2 (+ H2O) H+ (aq) + C2H3O2-(aq)

[H+]

[C2H3O2- ]

= 1.8 x 10-5

Ka =

[HC2H3O2]

What is there more of……reactant or product?

Reactant !

HC2H3O2 only 3% ionizes. There is almost no product present.

This is why it is a weak acid. Strong acids 100% ionize….Ka is LARGE.

Insoluble salts: PbI2 (s) (+ H2O) Pb+2 (aq) + 2 I-(aq) Ksp = [Pb+2][I-]2 = 8.4 x 10-9 What does this value tell us?

This solid is very insoluble; mostly solid PbI2 present.


Equilibrium

Keqis used to calculate concentrations of species at equilibrium.

Given: N2(g) + O2(g) = 2NO(g). At 25°C the

Kc = 1.0 x 10-30. [N2] =0.040 & [O2] = 0.010.

What is the concentration of NO?

[NO]2

M.A.E. : Kc =

= 1.0 x 10-30

[N2] [O2]

Small K value…….very little product!

[NO]2 = [N2] [O2] • 1.0 x 10-30

= (0.040) (0.010) (1.0 x 10-30) = 4.0 x 10-34

[NO] = 2.0 x 10-17mol/L


Equilibrium

For the system: CO2 + H2 = CO + H2O Kc = 0.64 @ 900°C.

The initial concentrations of reactants are both 0.100mol/L. When the system reaches equilibrium what are the concentrations of reactants and products?

Orig. Conc (mol/L) Change in Conc. Equilib. Conc.

CO2

H2

CO

H2O

0.100

0.100

0.000

0.000

-x

0.100-x

-x 0.100-x

+x

+x

x

x

[CO] [H2O]

= x2

Kc = 0.64 =

[CO2] [H2]

(0.100-x)2

x = 0.044; [CO] = [H2O] = 0.044mol/L

[CO2] = [H2] = 0.100 - 0.044 = 0.056 mol/L


Equilibrium

Heterogeneous equilibrium

  • Reactions in which one or more of the substances involved is a pure liquid or solid.

CO2(g) + H2(g) CO(g) + H2O (l)

  • Experimentally, we find, that the position of the equilibrium is independent of the amount of solid or liquid. Adding or removing a liquid or a solid has no effect on the equilibrium.

  • We do not need to include terms for solids or liquids in the expression for Keq. They are totally ignored.

[CO]

Thus, Keq =

[CO2] [H2]


Equilibrium

Solubility & K: (read pgs.597-612)

  • Calculating solubility from Ksp

PbSO4(s) = Pb+2(aq) + SO4-2 (aq)

Given, PbSO4 , Ksp= 1.6 x 10-8. Calculate its solubility, [Pb+2] & [SO4-2].

,

Ksp = [Pb+2][SO4-2] = 1.6 x 10-8

x2 = 1.6 x 10-8 (since x = [Pb+2] & [SO4-2])

x = 1.3 x 10-4. [Pb+2] & [SO4-2] = 1.3 x 10-4M


Equilibrium

Try this problem:

Mg(OH)2(s) = Mg+2(aq) + 2OH -1(aq)

Ksp = 1.5 x 10 -11 . Calculate the [Mg+2], [OH-1], and the [Mg(OH)2] at equilibrium.

If X = [Mg+2]

[OH-1] =

2x

So… how do we set up the mass action expression?

(x) (2x)2 =

1.5 x 10-11

x = 1.6 x10 -4 = [Mg+2]; [OH-1] = 3.2 x10 -4

[Mg(OH)2] = 1.6 x10 -4


Equilibrium

More Problems:

1. Bromine(I) chloride gas is formed in an endothermic reaction. At 400°C, after the reaction reaches equilibrium, the mixture contained 0.82M BrCl, 0.20M Br2(g)and 0.48M Cl2(g).

a. write the equation for this reaction:

1 Cl2 (g) + 1Br2(g) + q = 2 BrCl (g)

b. Write the equilibrium Expression, the MAE

[BrCl]2

= Keq

[Cl2][Br2]

[.82]2

= 7.0

c. Calculate the value for Keq:

[.20] [.48]

d. What direction is favored?


Equilibrium

2. Will a precipitate form when the following two solutions are combined? (assume volumes are additive.)

0.0025 M Pb(NO3)2 + 0.000036 M NaI

The Ksp for PbI2 = 8.4 x 10-9

[Pb+2] = 2.5 x 10-3 , [I-] = 3.6 x 10-5

Reaction Quotient: (2.5 x 10-3) (3.6 x 10-5)2 = 3.24 x 10-12

Since the rx. Quotient < the Ksp , No ppt. Forms!


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