Unit 7 Chap 15 B and L notes.ppt

Chemical Equilibrium

Lab Feedback

• Dissolving is Physical not chemical

change, no new bonds form!!

• Overall heat in a dissolution depends on

three factors, not broken and formed

bonds

• q

= -q

must flip sign of heat term b/c

surrounding are what was measured.

Equilibrium

is a state in which there are

no observable changes as time goes by.

In stoichiometry we talk about theoretical yields, and the many reasons actual yields may be lower.

Another critical reason actual yields may be lower is the reversibility of chemical reactions: some

reactions may produce only 70% of the product you may calculate they ought to produce.

Chemical Equilibrium Terms

Chemical Equilibrium Terms

Reversible Reaction Definition:

A chemical reaction in which the products can react to re-form the reactants

Chemical Equilibrium Definition:

When the rate of the forward reaction equals the rate of the reverse reaction and the concentration of products and reactants remains unchanged after

equilibrium is met.

2HgO(s)  2Hg(l) + O₂(g)

Arrows going both directions (  ) indicates equilibrium in a chemical

Equilibrium Example:

Graphically?

What does this graph tell us?

x and y-axis?

Concept of Dynamic

Equilibrium:

1. When rxns begin

reactant conc high

and prod conc low.

2. As rxn proceeds,

react conc dec and

prod conc inc. (rxn

moves forward)

Which is react?

Which is prod?

Reactant

Dynamic Equilibrium

form, rxn begins to

reverse and prod makes reactants. (reverse)

2. Eventually, rate of react making prod and prod making react are equal. This is when dynamic equilibrium is reached. *This is point on graph

N₂O₄ (g) 2NO₂ (g)

Start with NO₂ Start with N₂O₄ Start with NO₂ & N₂O₄ equilibrium

equilibrium

equilibrium

14.1

The Concept of Equilibrium

The Concept of Equilibrium

• Consider colorless frozen N2O4. At room temperature, it

decomposes to brown NO₂:

N₂O₄(g)  2NO₂(g).

• At some time, the color stops changing and we have a mixture of N₂O₄ and NO₂.

• Chemical equilibrium is the point at which the rate of the forward reaction is equal to the rate of the reverse

reaction.

At that point, the concentrations of all species are

Using the collision model:

•as the amount of NO

builds up, there

is a chance that two NO

molecules will

collide to form N

O

.

•At the beginning of the reaction,

there is no NO

so the reverse

reaction (2NO

(

g

)



N

O

(

g

)) does not

If rateforward = ratereverse then kforward[reactants]m = k

reverse[products]n

= = K the equilibrium constant

k_forward k_reverse

[products]n

[reactants]m

•values of m and n are those of the coefficients in the balanced chemical equation.

•Note that this is equilibrium, not kinetics.

•The rates of the forward and reverse reactions are equal, NOT the concentrations of reactants and products.

Law of Mass Action

Law of Mass Action

For the reaction

For the reaction:

Where

Where

K

K

is the equilibrium constant,

is the equilibrium constant,

and is

and is

unitless!

unitless!

Writing an Equilibrium Expression

Writing an Equilibrium Expression

2NO

2NO

(g)

(g)





2NO(g) + O

2NO(g) + O

(g)

(g)

K =

???

Write the equilibrium expression for the reaction:

Product Favored Equilibrium

Product Favored Equilibrium

Large values for K signify the reaction is

“product favored” (Large = K > 1)

When equilibrium is achieved, most

Reactant Favored Equilibrium

Reactant Favored Equilibrium

Small values for K signify the reaction is

“reactant favored” (Small K = K < 1)

Summarizing Significance of

Equilibrium Constant, Kc

1. K < 1

Reverse reaction is favored; forward

reaction does not proceed very far.

2. K = 1

Neither direction is favored; forward rxn

proceeds about half way.

3. K > 1

Example:

Write the equilibrium expression for K

for

the following reactions:

1.H

(g)

+ I

(g)

2HI

(g)

2.3H

(g)

+ N

(g)

2NH

(g)

Calculate the Equilibrium

Constant, K

Example:

H

(g)

+ I

(g)

2HI

(g)

1.If the equilibrium concentrations are:

[H

] = 2.0 M

[I

] = 1.0 M

[HI] = 4.0 M

N

(g)

+3H

(g)



2NH

(g)

2. If the equilibrium concentrations are:

[N

]=0.5M

[H

]=1.0 M

[NH

]=0.5 M

*What does the value of K_c indicate for each rxn?

Types of Equilibrium Rxns:

Homogeneous equilibrium

applies to

reactions in which all reacting species

are

in the same phase.

Heterogeneous equilibrium

applies to

reactions in which reactants and products

are in different phases.

CO (g) + Cl₂ (g) COCl₂ (g)

The position of a heterogeneous equilibrium

does not depend on the amounts of pure solids or liquids present.

Write the equilibrium expression for the reaction:

PCl₅(s)  PCl₃(l) + Cl₂(g)

Pure

solid Pure liquid

The concentration of

solids

and

pure

liquids

are not included in the

expression for the equilibrium

constant.

P_CO2 = K_p

CaCO₃ (s) CaO (s) + CO₂ (g)

P_CO2 does not depend on the amount of CaCO₃ or CaO

Calculate the equilibrium

constant, K

for the following:

CaCO

CaO

+ CO

At equilibrium, in a 2.0 L flask at 250_{C, 6.75 g of}

Relationships between K

and the

Chemical Equations Examples

1. Given the equilibrium reaction and concentrations below, calculate the value of K_c.

2NO

2NO₂₂(g) (g)  2NO(g) + O _{2NO(g) + O22}(g)_(g)

[0.25M] [0.12M] [0.023M][0.25M] [0.12M] [0.023M]

2. Given the equilibrium reaction and concentrations below, calculate the value of K_c.

2NO(g) + O

2NO(g) + O₂₂(g) (g)  2NO 2NO₂₂(g) (g)

[0.12M] [0.023M] [0.25M] [0.12M] [0.023M] [0.25M]

3. Given the equilibrium reaction and concentrations below, calculate the value of K_c.

NO

NO₂₂(g) (g)  NO(g) + ½O _{NO(g) + ½O22}(g)_(g)

[0.25M] [0.12M] [0.023M][0.25M] [0.12M] [0.023M]

Relationships between K

and

the Chemical Equations

1. If you reverse the equation, invert the

equilibrium constant.

2. If you multiply the coefficients in the equation

by a factor, raise the equilibrium constant to

the same factor.

3. If you add two or more individual chemical

equations to obtain an overall equation,

multiply the corresponding equilibrium

Example (reciprocal):

Example (reciprocal):

The equilibrium expression for a reaction is

The equilibrium expression for a reaction is

the reciprocal for a reaction written in

the reciprocal for a reaction written in

reverse

reverse

2NO

2NO

(g)

(g)





2NO(g) + O

2NO(g) + O

(g

(g

)

)

2NO(g) + O

N₂O₄ (g) 2NO₂ (g)

= 4.63 x 10-3

K = [NO2]2 [N₂O₄]

2NO₂ (g) N₂O₄ (g)

K = [N2O4] [NO₂]2

‘

K = 216

When the equation for a reversible reaction is written in the opposite direction, the equilibrium constant

Example (Differ by a factor):

Example (Differ by a factor):

When the balanced equation for a reaction

When the balanced equation for a reaction

is multiplied by a factor

is multiplied by a factor nn, the equilibrium , the equilibrium expression for the new reaction is the

expression for the new reaction is the

original expression, raised to the

original expression, raised to the n_nthth power. _power.

2NO

2NO

(g)

(g)





2NO(g) + O

2NO(g) + O

(g

(g

)

)

NO

Example (Putting them together) :

Example (Putting them together) :

Consider the chemical equation and equilibrium

constant for the following:

N

(g)

+3H

(g)



2NH

(g)

K

= 5.6 x 10

Calculate the equilibrium constant for the

following rxn at 25°C.

NH

(g)



1/2 N

(g)

+3/2 H

(g)

K

’ = ?

Understanding reactions involving N₂ and O₂, the most abundant gases in air, is essential for solving problems dealing with atmospheric pollution.

Here is a reaction sequence between N₂ and O₂ to form nitrogen dioxide, a toxic pollutant that contributes to photochemical smog.

(1) N₂(g) + O₂(g) 2NO(g) K_c1 = 4.3 x 10-25

(2) 2NO(g) + O₂(g) 2NO₂(g) K_c2 = 6.4 x 109

Example (Multiply Kc’s together):

SOLUTION:

Writing the Overall Equilibrium Constant for a series:

N₂(g) + 2O₂(g) 2NO₂(g)

K_c = [NO2]

2

[N₂][O₂]2

K_c1 = [NO]2 [N₂][O₂] K_c2 = [NO₂]2

[NO]2[O 2]

[NO]2

[N₂][O₂]

K_c1x K_c2 = [NO2]2

[NO]2[O 2]

= [NO2]2

[N₂] [O₂] 2

(2) 2NO(g) + O₂(g) 2NO₂(g) (1) N₂(g) + O₂(g) 2NO(g)

K_c = K_c1 x K_c2 = (4.3 x 10-25) x (6.4 x 109) = 2.8 x 10-15

(b) Calculate the K_c for the overall reaction.

Equilibrium with Gases

• Equilibria involving gases can be described using pressures instead of concentrations

• If pressure data is used, equilibrium constant is expressed as K_p instead of K_c

Example: Write the equilibrium constant expression for the equilibrium below:

N₂(g) + 3H₂(g)  _2NH3(g)

Equilibrium with Gases

How can you relate K_c and K_p?

N

(g) + 3H

(g)



2NH

(g)

RT

P

C

CRT

P

RT

V

n

P

nRT

PV

=



=

×



=



=

Equilibrium with Gases

N

(g) + 3H

(g)



2NH

(g)

( )

( )( )

1

1

1

÷

ø

ö

ç

è

æ

×

÷

ø

ö

ç

è

æ

÷

ø

ö

ç

è

æ

×

×

=

÷÷

ø

ö

çç

è

æ

×

÷÷

ø

ö

çç

è

æ

÷÷

ø

ö

çç

è

æ

=

RT

RT

RT

P

P

P

RT

P

RT

P

RT

P

K

2

)

(

RT

K

K

C

=

P

×

K or K_c = equilibrium constant in terms of molar concentrations K_p = equilibrium constant in terms of partial pressures

Equations you Need to know:

For Pressures instead of concentrations:

K

=

P

P

Reactants

K

=

K

(RT)

To convert between K

and K

:

R = Gas constant (0.0821)

T = Temperature (Kelvin)

When will Kp = Kc?

K

= K

(RT) can cancel out if total of gaseous

coefficients are same on each side b/c

∆n = 0.

K

=

K

(RT)

The equilibrium concentrations for the reaction

between carbon monoxide and molecular chlorine to form COCl₂ (g) at 740C are [CO] = 0.012 M, [Cl

2] = 0.054

M, and [COCl₂] = 0.14 M. Calculate the equilibrium constants K_c and K_p.

CO (g) + Cl₂ (g) COCl₂ (g)

K_c = [COCl2]

[CO][Cl₂] =

0.14

0.012 x 0.054 = 220

K_p = K_c(RT)n

n = 1 – 2 = -1 R = 0.0821 T = 273 + 74 = 347 K

The equilibrium constant K_p for the reaction

is 158 at 1000K. What is the equilibrium pressure of O₂ if the P_NO2 = 0.400 atm and P_NO = 0.270 atm?

2NO₂ (g) 2NO (g) + O₂ (g)

K_p = PNO PO2

2

P_NO2 ₂

P_{O2 = Kp} PNO2 ₂

P_NO2

Consider the following equilibrium at 295 K:

The partial pressure of each gas is 0.265 atm. Calculate K_p and K_c for the reaction?

NH₄HS (s) NH₃ (g) + H₂S (g)

K_p = P_NH

3PH2S = 0.265 x 0.265 = 0.0702

K_p = K_c(RT)n

K_c = K_p(RT)-n

n = 2 – 0 = 2 T = 295 K

K_c = 0.0702 x (0.0821 x 295)-2 = 1.20 x 10-4

AP Sample Question

2X(

g

) + Y(

g

) 3Z(

⇄

g

)

The reaction mixture represented above is at

equilibrium at 298 K, and the molar concentrations

are [X] = 2.0

M

, [Y] = 0.5

M

, and [Z] = 4.0

M

. What

is the value of the equilibrium constant for the

reaction at 298 K ?

A.0.50

In an experiment, X(g) and Y(g) were combined in a rigid container at constant temperature and allowed to react as shown in the equation above. The table provides the data

collected during the experiment. Based on the data, when did the reaction likely achieve an equilibrium?

The particle diagram above illustrates the changes that take place when X(g) and Y(g) are placed inside a rigid container at constant temperature. What is a

characteristic of a system at equilibrium that is best represented by the particle diagram?

The particle diagram shows that after 200s there are no

A sample of N₂O₄(g) is placed into an evacuated container at 373K and allowed to undergo the reversible reaction N₂O₄(g) ⇄2NO₂(g). The

concentration of each species is measured over time, and the data are used to make the graph shown above. When is equilibrium is first

reached? Provide a correct explanation.

At 60 seconds, because [NO₂] and [N₂O₄] remain constant,

Solving for Equilibrium Concentration

Solving for Equilibrium Concentration

Consider this reaction at some temperature: H2O(g) + CO(g)  H2(g) + CO2(g) K = 2.0

Assume you start with 8 moles of H₂O and

6 moles of CO in a 1.0 L container. What

concentration of H₂O, CO, H₂, and CO₂ are

present at equilibrium?

Here, we learn about ““ICEICE”” – the most

Solving for Equilibrium Concentration

Solving for Equilibrium Concentration

H₂O(g) + CO(g)  H₂(g) + CO₂(g) K = 2.0

Step #1: We write the law of mass action for the reaction:

2 2

2

[

][

]

2.0

[

][

]

H

CO

H O CO

Solving for Equilibrium Concentration

Solving for Equilibrium Concentration

H₂O(g) + CO(g)  H₂(g) + CO₂(g)

I

Initial:

C

Change:_hange:

E

Equilibrium:_quilibrium:

Step #2: We “ICE” the problem by setting

up an “ICE” chart. Start by filling in the

Initial concentrations. Reactant conc are

usually given. Initially, products have no concentration.

6 0 0

Solving for Equilibrium Concentration

Solving for Equilibrium Concentration

H₂O(g) + CO(g)  H₂(g) + CO₂(g)

I

Initial:

C

Change:_hange:

E

Equilibrium:_quilibrium:

Step #3: Fill in the change portion of the chart. It is based on the stoichiometry of the reaction (moles of each substance.)

Reactants are “-” b/c they decrease prod are “+” b/c they increase.

6 0 0

-x -x +x +x

Solving for Equilibrium Concentration

Solving for Equilibrium Concentration

H₂O(g) + CO(g)  H₂(g) + CO₂(g)

I

Initial:

C

Change:_hange:

E

Equilibrium:_quilibrium:

Step #4: Write the equilibrium

concentrations of each substance. Subtract the initial from the change for reactants. Products, it is just the change.

6 0 0

-x -x +x +x

8-x 6-x x x

Solving for Equilibrium Concentration

Solving for Equilibrium Concentration

E

Equilibrium:quilibrium: 8-x8-x 6-x6-x xx xx

Step #5: We plug equilibrium concentrations into our equilibrium expression, and solve for x

H₂O(g) + CO(g)  H₂(g) + CO₂(g)

2.0

=

(

x

)(

x

)

(8

-

x

)(6

-

x

)

4

Solving for Equilibrium Concentration

Solving for Equilibrium Concentration

Step #6: Substitute x into our equilibrium

concentrations to find the actual concentrations H₂O(g) + CO(g)  H₂(g) + CO₂(g)

E

Equilibrium:_quilibrium: 8-x_8-x 6-x_6-x x_x x_x

E

Equilibrium:_quilibrium: 8-4=4_8-4=4 6-4=2_6-4=2 4₄ 4₄

4

Quadratic Formula

ax

2

+

bx

+

c

=

0

x

= -

b

±

b

2

-

4

ac

2

a

*If you cannot solve your quadratic by factoring, use the quadratic formula!

Another ICE box

Consider the reaction:

N

(g)

+ O

(g)



2NO

(g),

K

= 0.10 (at 2000C)

A reaction mixture at 2000°C initially contains

[N

] = 0.200 M and [O

] = 0.200 M. Find the

equilibrium concentrations of both the reactants

and products at this temperature.

And Another ICE Box:

Consider the reaction:

2 IBr

(g)



Br

(g)

+ I

(g)

K

= 8.5x 10

at 150 °C

A reaction initially contains IBr at a partial

pressure of 0.017 atm at 150 °C. Calculate the

equilibrium partial pressures of IBr, Br

and I

.

What else can ICE do?

Consider the decomposition of water:

2 H

O

(l)



2

H

(g)

+ O

(g)

After the reaction has come to

equilibrium, the total pressure is 2.10 atm.

Calculate Kp and the equilibrium partial

pressures of H

and O

.

What else can ICE do?

At a particular temperature, 12.0 mol of SO

is placed into a 3.0 L rigid container, and the

SO

dissociates by the following reaction:

2 SO

(g)



2 SO

(g)

+ O

(g)

At equilibrium, 3.0 mol of SO

is present.

Calculate the equilibrium constant, K for this

reaction.

AP Sample Question

COCl₂(g) decomposes according to the equation above. When pure COCl₂(g) is injected into a rigid, previously evacuated flask at 690 K, the pressure in the flask is

initially 1.0 atm. After the reaction reaches equilibrium at 690 K, the total pressure in the flask is 1.2 atm. What is the value of K_p for the reaction at 690 K?

A mixture of NO₂(g) and N₂O₄(g) is placed in a glass tube and allowed to reach equilibrium at 70°C, as represented above.

If P_N2O4 is 1.33 atm when the system is at equilibrium at 70°C, what is P_NO2 ?

The Reaction Quotient

The Reaction Quotient

For some time, t, when the system is not at

equilibrium, the reaction quotient, Q takes the

place of K, the equilibrium constant, in the

law of mass action. (Formula to write one is exactly the same!)

Q

=

[

C

]

[

D

]

[

A

]

[

B

]

Significance of the Reaction Quotient

Significance of the Reaction Quotient

 If _Q = _K, the system is at equilibrium

 If _{Q > K}, the system shifts to the left, consuming products and forming reactants until equilibrium is achieved

Relationships:

Equilibrium Constant, Kc or Kp

Reaction Quotient, Q

Direction

K < 1 K < Q Reverse, Make Reactants K > 1 K > Q Forward, Make

Products

Example:

At 1000 K, K

= 0.338 for the reaction:

2 SO

(g)



2 SO

(g)

+ O

(g)

Calculate the value of Q

and the direction in which

the reaction proceeds toward equilibrium if the

initial partial pressures are:

P

= 0.16atm P

= 0.41atm P

= 2.5atm

Q

= 16

Recall:

Another Q vs. K

Consider the reaction and its equilibrium

constant, K

:

A reaction mixture contains [NO

] = 0.0255 M and

[N

O

] = 0.0331 M. Determine the direction in

which the reaction will proceed to re-establish

equilibrium.

Q

= 1.96 x 10

, reaction proceeds left

Another Example:

CO

(g)

+ 2 H

(g)



CH

OH

(g)

K

= 14.5

A 1.00 L reaction vessel initially contains 1.2

mol CO

(g)

, 1.5 mol H

(g)

, and 2.0 mol

CH

OH

(g).

How will the total pressure change

as the system approaches equilibrium at

AP Exam Style Q

A 1.0 mol sample of X(g) and a 1.0 mol sample of Q(g) are introduced into an evacuated, rigid 10.0 L container and allowed to reach equilibrium at 50°C according to the equation above. At equilibrium, which of the following is true about the concentrations of the gases?

At a certain point in time, a 1.00 L rigid reaction vessel contains 1.5 mol of PCl₃(g), 1.0 mol of Cl₂(g), and 2.5

mol of PCl₅(g). Which of the following describes how the measured pressure in the reaction vessel will change

and why it will change that way as the reaction system approaches equilibrium at constant temperature?

A 0.10 mol sample of each of the four species in the

reaction represented above is injected into a rigid,

previously evacuated 1.0 L container. Which of the

following species will have the highest concentration

when the system reaches equilibrium?

LeChatelier

LeChatelier

’

’

s Principle

s Principle

When a system at equilibrium is placed

When a system at equilibrium is placed

under stress, the system will undergo a

under stress, the system will undergo a

change in such a way as to relieve that

change in such a way as to relieve that

stress. Types of stress:

stress. Types of stress:

1.

1.concentration, 2. temperature, 3. pressure, concentration, 2. temperature, 3. pressure, 4. volume 5. catalyst

4. volume 5. catalyst

Translated:

Translated: The system undergoes a _{The system undergoes a}

temporary shift in order to restore

temporary shift in order to restore

equilibrium.

1. Concentration

• If a substance is

increased, system will shift to use up what was added.

• If a substance is

removed, system will shift to produce what was taken.

2. Temperature

• Treat like concentration. • Find heat term

• If Inc Temp, rxn shifts to use up heat

• If dec Temp, rxn shift to make more heat

• Only stress that will affect the K_c value.

• Exo, K_c will dec. if add heat • Endo, K_c will inc. if add heat

• Ex.

3. Pressure

• Only for gases!

• If press is inc, rxn will shift to side that

makes less mols of gas

• If press is dec, rxn will shift to side that

makes more mols of gas

4. Volume

• Similar to pressure and will only affect gas.

• Inc volume, shift to side with more moles of gas • Dec volume, shift to

side with less moles of gas

5. Catalyst

• Catalysts do not affect equilibrium

concentrations, shifts, or K_c values at all.

• They speed up

chemical reactions in both the forward and reverse directions

Example:

N

+ 3 H



2 NH

+ heat

Which way will the rxn shift if:

Inc conc of NH

? ______

Dec conc of H

? ______

Inc temp? ______

Dec press? ______

Inc volume?

______

Le Châtelier’s Principle Summary of Stress

Change Shift Equilibrium Change Equilibrium Constant

Concentration yes no

Pressure yes no

Volume yes no

Temperature yes yes

Which of the following changes alone would cause a

decrease in the value of K_eq for the reaction represented above?

A.Decreasing the temperature B.Increasing the temperature

A mixture of NO₂(g) and N₂O₄(g) is placed in a glass tube and allowed to reach equilibrium at 70°C, as represented above.

Which of the following statements best helps to explain why the contents of the tube containing the equilibrium mixture turned a lighter color when the tube was placed into an ice bath?

A. The forward reaction is exothermic. B. The forward reaction is endothermic.

After the equilibrium represented above is established, some pure O₂(g) is injected into the reaction vessel at constant temperature. After equilibrium is reestablished, which of the following has a lower value compared to its value at the original equilibrium?

A.K_eq for the reaction

B.The total pressure in the reaction vessel

The synthesis of CH₃OH(g) from CO(g) and H₂(g) is represented by the equation above. The value of K_c for the reaction at 483 K is 14.5.

Which of the following explains the effect on the equilibrium constant, K_c, when the temperature of the reaction system is increased to 650 K?

A. K_c will increase because the activation energy of the forward reaction increases more than that of the reverse reaction.

B. K_c will increase because there are more reactant molecules than product molecules.

Chemistry In Action

Life at High Altitudes and Hemoglobin Production

K_c = [HbO2] [Hb][O₂]

Hb (aq) + 4 O₂ (g) Hb(O₂)₄ (aq)

Le Chatlier also explains shifts in this system.

High altitudes mean low pressure,

causing a shift to the left of the equation, less oxygenated Hb in the blood.

(shortness of breath)

Delivery of O₂ to body tissues.