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Chapter 13: EquilibriumPowerPoint Presentation

Chapter 13: Equilibrium

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Chapter 13: Equilibrium

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Chapter 13: Equilibrium

AP Chemistry

- The state in which there are no observable changes as time goes by
- Rate of fwd rxn = Rate of rev rxn
- [R] and [P] are constant
- Examples: acetone, NaCl(s), NO2 tubes

K

E

K

E

K

E

Label the region in which the study of kinetics is concerned

Label the region that indicates the region has reached equilibrium

- Both Reactants and Products are Present
- Rate of Forward Rxn = Rate of Reverse Rxn
- Closed System
- No macroscopic changes (Pressure, Temp, color, etc.)

- A lit Bunsen burner
- An Alka Seltzer tablet in water
- Vapor pressure in a sealed flask
- A saturated solution of PbCl2
- Student population of OHS over a year
- Salt dissolving in water

N2O4 2NO2

aA + bB cC + dD

No units!!

If K = 1

If K >> 1

If K << 1

Concentration of products = concentration of Reactants

Concentration of products > concentration of Reactants

“Equilibrium lies far to the right”

Concentration of products < concentration of Reactants

“Equilibrium lies far to the left”

All species are in the same phase (i.e. all gases)

N2O4 (g) 2 NO2(g)

But…with gases

Kc ≠ Kp

CO +2Cl COCl2

The equilibrium concentrations for the rxn between carbon monoxide & molecular chlorine to form COCl2 (g) at 74 oC are [CO] = 0.012 M, [Cl]=0.054 M, and [COCl2] = 0.14 M. Calculate Kc and Kp.

CaCO3 (s) CaO (s) + CO2 (g)

Reactants and products are in different phases

How are the 2 pictures different from each other?

How are they the same?

dPCO2

CaCO3 (s) CaO (s) + CO2 (g)

Kc = [CO2]

Kp = PCO2

Solids and liquids do not affect K; they have a constant concentration (i.e. density)

Problem: “Hard Water” contains lots of Ca2+

Solution: “Soften” water by replacing Ca2+ with Na+

N2O4(g) 2NO2(g)

2NO2(g) N2O4(g)

‘

K =

K =

= 4.63 x 10-3

[NO2]2

[N2O4]

[N2O4]

[NO2]2

1

K

=

Original rxn

Reverse rxn

= 216

When the equation for a reversible reaction is written in the opposite direction, the equilibrium constant becomes the reciprocal of the original equilibrium constant.

N2O4(g) 2NO2(g)

½N2O4(g) NO2(g)

2N2O4(g) 4NO2(g)

K =

K =

K =

= 4.63 x 10-3

= K1/2

= K2

[NO2]2

[NO2]

[NO2]4

[N2O4]1/2

[N2O4]

[N2O4]2

Original rxn

New rxn

= (4.63 X 10-3)1/2

New rxn

= (4.63 X 10-3)2

aA + bB cD + dD

initial concentrations

Substitute of the reactants and products into the equilibrium constant (Kc) expression.

Reactants Products

- IF
- Qc > Kc
- system proceeds from right to left to reach equilibrium
- (reverse reaction is favored)

- Qc = Kc
- system is at equilibrium
- Qc < Kc
- system proceeds from left to right to reach equilibrium
- (forward reaction is favored)

[Br]2

[Br2]

Kc =

Br2 (g) 2 Br (g)

= 1.1 x 10-2

Kc =

(0.0120 + 2x)2

0.163 - x

At a given temp, (Kc) = 1.10 x 10-2 for the reaction Br2(g)2 Br (g)

If initially [Br2] i = 0.163 M and [Br] i = 0.0120 M, calculate the concentrations of these species at equilibrium.

Are we at equilibrium?????

ICE

Let x be the change in concentration of Br2

Initial (M)

0.163

0.0120

Change (M)

-x

+2x

Equilibrium (M)

0.163 - x

0.0120 + 2x

Solve for x

-b ± b2 – 4ac

x =

2a

Br2 (g) 2Br (g)

Initial (M)

0.163

0.0120

Change (M)

-x

+2x

Equilibrium (M)

0.163 - x

0.0120 + 2x

= 1.10 x 10-2

Kc =

(0.0120 + 2x)2

0.163 - x

ax2 + bx + c =0

4x2 + 0.0590x - 0.00165 = 0

Recall the quadratic equation gives 2 solutions! We must determine which is correct to find the equilibrium concentrations

x = 0.0142

x = -0.0290

[Br2] eq = 0.148 M

[Br] eq= 0.0404 M

[CO]2[O2]

2CO2 (g) 2 CO (g) + O2 (g)

[CO2]2

Kc =

(2x)2 (x)

= 2.0 x 10-6

Kc =

(0.4 - 2x)2

At a particular temperature, K = 2.0 X 10-6 for the reaction:

2CO2 (g) 2CO(g) + O2 (g)

If 2.0 mol of CO2 is initially placed into a 5.0 L vessel, calculate the equilibrium concentrations of all species

Are we at equilibrium?????

0

0

0.40

Initial (M)

+2x

+x

-2x

Change (M)

x

Equilibrium (M)

0.4 - 2x

2x

Solve for x

2CO2 (g) 2 CO (g) + O2 (g)

(2x)2 (x)

4x3

(4x3)

4x3

= 2.0 x 10-6

= 2.0 x 10-6

= 2.0 x 10-6

= 2.0 x 10-6

Kc =

Kc =

Kc =

Kc =

(0.4 - 2x)2

(0.4)2

(0.4 - 2x)2

(0.4 - 2x)2

LOOK! K IS REALLY SMALL

YUCK!!!

0

0

0.40

Initial (M)

+2x

+x

-2x

Change (M)

x

Equilibrium (M)

0.4 - 2x

2x

≈0.4

x = 4.3 x 10-3M = [O2]eq

[CO2]eq= 0.40 M

[CO] eq= 8.6 X 10-3M

MATH

“0: Solver”

0= AX2 + BX + C

ENTER

After X = , type 0

“Solve”

“QUIT”

“X,t,theta,n” button

ENTER

2CO2 (g) 2 CO (g) + O2 (g)

0

0

0.40

Initial (M)

+2x

+x

-2x

Change (M)

x

Equilibrium (M)

0.4 - 2x

2x

≈0.4

0.40 – 2x =

0.40 – 2(4.3 X 10-3) =

0.3914

K is very small (K < 10-3)

K is very big (K > 103)

Inventor of acetylene torch

Professor of Industrial Chemistry and Metallurgy

Instrumental in the development of cement and Plaster of Paris

When a stress is applied

to a system at equilibrium,

the system will respond

to partially undo the stress

Add Reactant, Add Product, Remove Reactant, Remove Product, Add Heat, Increase Pressure,…

Haber process

N2 + 3 H2 2 NH3 + energy

produced

used

produced

used

Use energy

Produce NH3

N2 (g) + 3H2 (g) <===> 2NH3 (g) + heat

Problem: The forward Rxn is slow at Room Temp,

BUT also exothermic.

Adding heat favors the REVERSE Rxn!

Le Chatelier’s Solution:

Use an iron catalyst, at apressure of 200 ATM!

Accidental contamination of the Rxn chamber with air results in an huge explosion blowing fragments of steel through the lab floor and ceiling. Le Chatelier abandons the project, and less than five years later, Haber and Claude were successful in producing ammonia on a commercial scale using a similar process.

"I let the discovery of the ammonia synthesis slip through my hands. It was the greatest blunder of my scientific career."

produced

used

produced

used

2 H+ + 2 CrO42-Cr2O72- + H2O

[Cr2O72-]

[CrO42-]

[H2O]

Use H+

ORANGE

(Add H+)

[H+]

[CrO42-]

[Cr2O72-]

[H2O]

(Use H+)

Produce H+

YELLOW

Reaches equilibrium faster---no change!

H+

H+

H+

Na+

Add H+

X= CrO4-2

O= Cr2O7-2

x x xx x x

x x xox xx

x x x xx xxxx

o x x x x

x x x x

ox o oo o

o o ooo o

o o x o oo oo

oo o ooo

oo

Add OH-

produced

used

produced

used

2 NO2N2O4 + energy

[NO2]

[N2O4]

DARKER

Use Heat

Produce Heat

LIGHTER

Darker; then

lighter

Decrease Pr.

Lighter; then

Darker

Increase Pr.

Ca5 (PO4)3OH (s) 5 Ca+2(aq) + 3 PO4–3 (aq) + OH- (aq)

N2(g) + 3H2(g) 2NH3(g)

Equilibrium shifts left to offset stress

Add

NH3

Le Châtelier’s Principle

If an external stress is applied to a system at equilibrium, the system adjusts in such a way that the stress is partially offset as the system reaches a new equilibrium position.

- Changes in Concentration

14.5

aA + bB cC + dD

Le Châtelier’s Principle

- Changes in Concentration continued

Change

Shifts the Equilibrium

Increase concentration of product(s)

left

Decrease concentration of product(s)

right

Increase concentration of reactant(s)

right

Decrease concentration of reactant(s)

left

14.5

A (g) + B (g) C (g)

Le Châtelier’s Principle

- Changes in Volume and Pressure

Change

Shifts the Equilibrium

Increase pressure

Side with fewest moles of gas

Decrease pressure

Side with most moles of gas

Increase volume

Side with most moles of gas

Decrease volume

Side with fewest moles of gas

14.5

Le Châtelier’s Principle

- Changes in Temperature

Change

Exothermic Rx

Endothermic Rx

Increase temperature

K decreases

K increases

Decrease temperature

K increases

K decreases

- Adding a Catalyst
- does not change K
- does not shift the position of an equilibrium system
- system will reach equilibrium sooner

14.5

Change Equilibrium

Constant

Le Châtelier’s Principle

Change

Shift Equilibrium

Concentration

yes

no

Pressure

yes

no

Volume

yes

no

Temperature

yes

yes

Catalyst

no

no

14.5