chapter 15 equilibrium - welcome to web.gccaz.eduweb.gccaz.edu/~jaszi38221/2014/summer/chm 152...

5/27/2014

1

Chapter 15 – Equilibrium

Equilibrium in Chemical Reactions

time

Am

ou

nt

of

reac

tan

t/p

rod

uct

Lets’ look back at our hypothetical reaction from the kinetics chapter.

A + 2B → C

[A]

[C]

Why doesn’t the concentration of A ever go to zero?

What if we waited longer?

What is happening here?

In reality, a better was to represent our reaction would be…

A + 2B ⇌ C

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2

Equilibrium – Some Reactions are Reversible

Tells us there is more than one reaction taking place.

N2O4 ⇌ 2NO2

2NO2 → N2O4 N2O4 → 2NO2

⇌ Reaction 1 Reaction 2

and “Forward” reaction “Reverse” reaction

Will there be more reactants or more products at equilibrium?

⇌

Reversible Reactions & Equilibrium

2 Rev worker takes the two pieces, puts back together. Sends back….

However, it takes REV worker twice as long to put the pieces

together….

How will the amounts of products and reactants change over time?

Products Reactants

1 Fwd worker breaks reactant in two, sends pieces down the line….

+

http://en.wikipedia.org/wiki/File:Dinitrogen-tetroxide-3D-vdW.png


http://en.wikipedia.org/wiki/File:Nitrogen-dioxide-3D-vdW.png










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Reactants Products

3 Because the fwd and rev workers are working at different rates,

over time, there will be a build-up of products

4 Since the REV worker is being overwhelmed with more products,

the foremen sends in help!!!

As the workday progresses…..


Reactants

5 Because there are now two REV workers, they together they work

at the same rate as the FWD worker

Eventually…

Products

From this point on, the amounts of reactants and products will stay the same…they are in

dynamic equilibrium


Once dynamic equilibrium is reached, the amounts of reactants and products

DO NOT CHANGE.

The reaction appears to have stopped, but the forward and reverse reaction are taking

place at the same rate.

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4

Notice that

although each trial

started with a

different ratios of

reactant/products,

the “ratio” of

concentrations of

reactants/products

at equilibrium will

be the same

INITIAL FINAL

INITIAL FINAL

Experimental Evidence of Equilibrium

Trial 1

Trial 2

Equilibrium Constant – K

Do not confuse the two kay’s

K k • Upper-case / large K

• Denotes “how far” are reaction

proceeds

• Determine by equilibrium

amounts of reactants / products

• NO UNITS!!

• lower-case / small k

• Related to “how fast” a

reaction proceeds

• Determined based on the

rate law

• UNITS!!
































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Equilibrium Constant – K The quantitative measure of “how “far a reaction proceeds is represented by a constant, K.

K can be thought of as the result of two reactions: the forward reaction and the reverse reaction. Whether we end up with more reactant and more product in the end is determined by the rate of each reaction.

Let’s assume the following reaction is a one-step process. We can determine the rate laws since it is an elementary step:

N2O4⇌ 2NO2

“Forward” reaction “Reverse” reaction

Rate = krev[NO2]2 Rate = kfwd[N2O4]

c

42

22

rev

fwd

]O[N

][NO K

k

k

At equilibrium

Similar rxn rates Fwd faster than rev Rev faster than fwd

Reactant Favored Product Favored Reactant ~ Product

c22rev

42fwd

][NO

]O[NK

k

k mequi l ibriu at

[reactant]

[product]

tcoefficien

tcoefficien

c K

Equilibrium Constant

Re

acta

nt(

s)

Pro

du

ct(s

)

K ≈1 K <<1 K >>1

What does the numerical value of Kc actually mean?

“Reactant Favored” “Product Favored”

K <<<<1 K >>>>1

Negligible Complete

Pro

du

ct(s

)

Re

acta

nt(

s)

mequi l ibriu at [reactant]

[product]

tcoefficien

tcoefficien

c K

Substantial

amount of both

reactants and

products

Mostly reactants,,

very small

amount of

products

Mostly products,

very small

amount of

reactants

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Equilibrium Constant and concentration – Kc

aA + bB ⇌ cC + dD mequi l ibriu at [B][A]

[D][C]

ba

dc

c K

For aqueous species and gases we can use concentration in our constant

Initial (M) Equilibrium (M) Experiment [N2O4] [NO2] [N2O4] [NO2]

1 0.000 0.0200 0.00140 0.0172 2 0.000 0.0300 0.00280 0.0243 3 0.000 0.0400 0.00452 0.0310

]O[N

][NO

42

22

c K 211.0(0.00140M)

(0.0172M)K

2

c

Since

We can determine Kc experimentally by measuring concentrations of reactants and products at equilibrium

Kc

0.211 0.211 0.213

N2O4 ⇌ 2NO2

K values and equilibrium concentration practice

Oxygen gas can be converted into ozone based on the following reaction:

3O2 (g) ⇌ 2O3 (g)

• If the equilibrium concentration of O2 is 0.50M and O3 is

2.2x10-7M, determine the value of the equilibrium constant, Kc

• If the equilibrium concentration of O2 is 0.10M, what would be the

equilibrium concentration of O3?

-8 M

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7

Heterogeneous Equilibria

How do we deal with reactions that have different states of matter within the same reaction? For example:

Fe(OH)3(s) → Fe3+(aq) + 3OH-(aq)

Although a gas can be represented by Kc or Kp, how do we represent solids? What is the concentration of a solid? Well, there is no need to calculate it since it will not change even if you change the amount do we. So, we take it out of the K expression.

Pure liquids and solids are not represented in the equilibrium

equation

3-33c ]][OH[Fe ][Fe(OH) ' K

CO2(g) +C(s) → 2CO(g)

][CO

[CO]

2

2

c K

So for the following equation…

…solid carbon will not be a part of the equilibrium expression

H+(aq) +OH-(aq) → H2O(l)

And for the following equation…

]][H[OH

1

-c K

…liquid H2O will not be a part of the equilibrium expression

][Fe(OH)

]][OH[Fe

3

3-3

c

KA solid or ,liquid does not change

concentration

3-3c ]][OH[Fe K

Equilibrium Constant and Pressure – Kp

Just as we can have an equilibrium constant based on pressure of products and reactants (Kc), we can also use partial pressures when dealing with gases

mequi l ibriu at Reactants)of Pressure (Partial

Products)of Pressure (Partial p K

aA + bB ⇌ cC + dD

mequi l ibriu at )()(

)()(

bB

aA

dD

cC

pPP

PPK Pgas = the partial pressure the gas

exerts in the reaction mixture

So, K’s for the reaction:

2O3⇌ 3O2

23

32

c][O

][OK

2O

3O

p)(

)(

3

2

P

PK

Based on partial pressures Based on concentrations

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The Relationship between Kc and Kp

The relationship between Kc and Kp is determined by the following equation

nKK Δcp (RT)

Mol gas products – mol gas reactants from balanced equation

Kmolatm L0.08206

For example, the following reaction: 2SO2(g) + O2(g) ⇌ 2SO3(g) has a Kc of 4.09x10-2 at 1000.K. To determine Kp:

3)(2

KmolatmL2

p )(1000.K)])[(0.08206(4.09x10 K 4.98x10-4

The reaction: H2(g) + Cl2(g) ⇌ 2HCl(g) has a Kp of 4.0x1031 at 300.K. What is Kc?

2)-(2Kmol

atmLc

31 )(300.K)][(0.08206 )(K )(4.0x10 4.0x1031

When the number of moles of gas are the same on both sides of the equation, Kc = Kp

Relationships between related K’s

aA + bB ⇌ cC + dD ba

dc

1c[B][A]

[D][C] K

Remember that for the reaction….

But what about the reverse reaction?

cC + dD ⇌ aA + bB

Since we have written the equation in reverse, products and reactants are

switched. Thus, K is different. dc

ba

2c[D][C]

[B][A] K

1

1

1

2

1 K

KK

When we deal with reversible reactions, we can write it any direction we choose.

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Relationships between related K’s

aA + bB ⇌ xX ba

x

1c[B][A]

[X] K

xX ⇌ cC

213 KKK

x

c

2c[X]

[C] K

aA + bB ⇌ cC

Let’s say we have two reactions that have a reactant/product in common:

We can add the two reaction together

x

c

ba

x

c21c[X]

C][

[B][A]

[X] KK

ba

c

3[B][A]

[C] K

How do we combine K expressions?

Reaction Quotient and Equilibrium Constant What do we do when we know the concentration of reactants before equilibrium (or before

we mix them to perform a reaction)? Do we have an expression that can quantify this?

K Q

Q

Forward

More Products

Formed

Reverse

More Reactants

Formed

mequi l ibriu at NOT anytime at [B][A]

[D][C]

ba

dc

c Q

Equilibrium Q<K Q>K

ba

dc

c[B][A]

[D][C] Q

Ratio of reactants higher than at

equilibrium ba

dc

c[B][A]

[D][C] Q

Ratio of products higher than at

equilibrium

5/27/2014

10

Le Chatelier’s Principle

When an equilibrium is disturbed, the reaction will proceed in the direction that reestablishes an equilibrium .

Equilibrium mixture of NO2 and N2O4 gives a yellowish color.

More NO2 added

The addition of more NO2 turns the mixture orange (NO2 is

orange in color).

Wait 10 minutes

After 10 minutes, the color returns as before, suggesting that the

equilibrium mixture has returned.

N2O4← 2NO2

Which way did the reaction shift to reach equilibrium?

N2O4 + 2NO2

colorless

Dark orange

N2O4 ⇌ 2NO2 N2O4 ⇌ 2NO2 N2O4 ? 2NO2


A + B ⇌ C + D

What happens when a equilibrium is “disturbed”? Once it is disturbed, it will shift toward reactants or products to return to equilibrium

Increase the amount of a reactant

Q<K

Reaction shifts toward products

A + B → C + D

K is reestablished

A + B ⇌ C + D

A + B ⇌ C + D

Decrease the amount of a reactant

Q>K

Reaction shifts toward reactants

A + B ← C + D

K is reestablished

A + B ⇌ C + D

A + B ⇌ C + D

Increase the amount of a product

Q>K

Reaction shifts toward reactants

A + B ← C + D

K is reestablished

A + B ⇌ C + D

A + B ⇌ C + D

Decrease the amount of a product

Q<K


A + B → C + D

K is reestablished

A + B ⇌ C + D

To make a reaction shift toward products •Decrease the amount of product •Increase the amount of reactant

To make a reaction shift toward reactants •Decrease the amount of reactant •Increase the amount of product

5/27/2014

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Le Chatelier’s Principle Only reactants and/or products that are part of the equilibrium equation, will affect equilibrium.

Fe(OH)3

Fe3+

Fe3+ OH-

OH-

OH-

OH- OH- OH-

OH-

Fe(OH)3

Fe3+

Fe3+ OH-

OH-

OH-

OH-

OH-

OH-

OH-

Add more

Fe(OH)3 Adding more Fe(OH)3 will not cause more ions to dissolve.

No shift in equilibrium

3-3c ]][OH[Fe K

Fe(OH)3 is not part of the equilibrium expression

Ca(CO3)2

CO2 CO2

CO2

CO2

CO2 CO2

CaCO3

CaO

CO2 CO2

CO2

CO2

CO2 CO2

CaCO3

Add more

CaCO3

][CO 2c K

Concentration of Fe(OH)3 ions

does not change

Concentration of CO2 does not

change

Adding more CaCO3 will not cause more CO2 or CaO to be

created. No shift in equilibrium

Fe(OH)3(s) → Fe3+(aq) + 3OH-(aq)

CaCO3(s) → CO2(g) + CaO(s)

Le Chatelier’s Principle – Change in Temp In order to determine the effect of temperature on equilibria, it is convenient to view heat either

as a product or reactant.

A + B ⇌ C + HEAT A + B ⇌ C + HEAT

Reaction shifts toward

reactants


A + B ← C + HEAT

A + B ⇌ C + HEAT

Equilibrium is established w/ smaller K

A + B → C + HEAT

A + B ⇌ C + HEAT

Equilibrium is established w/ larger K

+ A + B ⇌ C HEAT


Equilibrium is established w/ larger K

+ A + B ⇌ C HEAT

Reaction shifts toward

reactants

+ A + B → C HEAT

+ A + B ⇌ C HEAT

Equilibrium is established w/ smaller K

+ A + B ← C HEAT

+ A + B ⇌ C HEAT

Exothermic reactions (-ΔH) Endothermic reactions (+ΔH)

5/27/2014

12


N2O4 ⇌ 2NO2

N2O4 ← 2NO2 N2O4 ⇌ 2NO2

Equilibrium shifts to the side with the fewest moles of gas.

More of that gas (product or reactant depending on the balanced equation) is

produced

2 atm 2 atm 1 atm

Decrease in volume An increase in total pressure will shift a reaction toward the side with the fewest moles of gas.

Ca(CO3)2

CO2 CO2

CO2 CO2

CaCO3

CO2

CO2

CaCO3

CO2 CO2

CO2

CO2

CaCO3

2 atm 2 atm

CO2

CO2

CO2

CO2

CO2

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

The “I.C.E” Table

A certain reaction has a balanced equation of A ⇌ 2B + 3C. After placing 6.00M A in reaction vessel @ 1000.K, the reaction begins. During the reaction, the reactant decreased by1.33M . What are the equilibrium concentrations and what is Kc?

I

E C

nitial

hange

quilibirum

6.00M 0M

-1.33M +(2)1.33

(6.00M – 1.33M) 4.67M

(0M+2.66M) 2.66M

A ⇌ 2B + 3C

0M

+(3)1.33

(0M+3.99M) 3.99M

•We plug everything we know into the table.

•Write the balanced equation

•Since A changed -1.33M, how should the concentration of B and C change in comparison?

Look at the balanced equation

•B should change 2 times as much and C should change 3

times as much a A.

[A]

[C][B] 32

c K 96.24.67

(3.99)(2.66) 32

c K


















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Determining Change from Initial and Equilibrium Values

I

E C

nitial

hange

quilibirum

1.00atm 0atm

-3x +2x

3A ⇌ 2B

0.40atm

Gas “A” is put in a reaction flask and heated to 40.0°C until it changes to gas “B” in a 3-to-2 ratio. Before the reaction, gas “A” had a partial pressure of 1.00atm. At the end of the reaction, after it has cooled gas “A” had a partial pressure of 0.40atm. Determine the

equilibrium pressure of gas “B” and determine Kp for the reaction.


•Plug in what we know

•Since we don’t know the change, we, can substitute with a variable

“x” (temporarily).

•We can actually work backward from equilibrium.

Therefore, we know that -3x = -0.60

x = 0.20 atm

-0.60atm

So 2x = 0.40atm

+0.40atm

0.40atm

Change = Initial – Equilibrium = 1.00atm-0.40atm = 0.60atm

Change = -3x

2.5(0.40)

(0.40)3

2

p K

and

3A

2B

p)(

)(

P

PK

Don’t forget your coefficients

Calculating Equilibrium Concentration using K

I

E C

nitial

hange

quilibirum

A ⇌ B

3.33M 0M

-x +x

3.33M - x 0M + x

A certain reactant “A” rearranges into a product “B” at room temperature in a 1-to-1 ration. If you started with 3.33M “A”, what will be the concentration of both “A” and

“B” at equilibrium if Kc = 1.11x10-1



•Since we don’t know the change, we, can substitute with a variable x.

•Equilibrium is just the initial plus the change

•Substitute equilibrium concentrations

0.111[A]

[B]Kc 0.111

x-3.33

x 0.111x0.370x 0.3701.111x

0.333x [A]eq = 3.33M - x

[B]eq = x

[A]eq = 3.00M

[B]eq = 0.333M

Double-check by working backward

0.1113.00

0.333c K

Since… Then…

5/27/2014

14

Calculating Equilibrium Concentration using K – Perfect Squares

I

E C

nitial

hange

quilibirum

0.100M 0.100M

-x -x

0.100M - x 0.100M - x

H2 + I2 ⇌ 2HI

The reaction between H2 and I2 produces HI. Assume 0.100mols of H2 and 0.100 moles of I2 are placed into a 1.00L container. The gas react and are allowed to reach

equilibrium. Calculate the equilibrium concentration of all the gases. Kc for the reaction at 443°C is 50.5.



•Since we don’t know the change, we, can substitute with a variable x.

•Equilibrium is just the initial plus the change


0M

+2x

0M + 2x

50.5]][I[H

[HI]

12

2

c KSince…

50.5x)-x)(0.100-(0.100

(2x)2

Perfect Square

50.5x-0.100

2x2

50.5

x-0.100

2x2

or

Calculating Equilibrium Concentration using K – Perfect Squares

I

E C

nitial

hange

quilibirum

0.100M 0.100M

-x -x

0.100M - x 0.100M - x

H2 + I2 ⇌ 2HI

0M

+2x

0M + 2x

50.5x-0.100

2x2

7.11x)-(0.100

2x

7.11x-0.7112x

9.11

0.7119.11x

x = 0.0780

Since [H2]eq and [I2]eq = 0.100-x [HI]eq = 2x

[H2]eq and [I2]eq = 0.022M [HI]eq = 0.156M Double-check by working backward

50.3022)(0.022)(0.

(0.156)2

c K

5/27/2014

15

Calculating Equilibrium Concentration using K – Quadratic Formula

I

E C

nitial

hange

quilibirum

We can calculate equilibrium concentrations if we know initial concentrations and the equilibrium constant, Kc.

•Write the balanced equation N2O4 ⇌ 2NO2


If we start with an initial concentration of N2O4 of 5.0M before decomposition , how

much NO2 will there be at equilibrium? Kc for the reaction is 4.6x10-3.

5.0M 0M •Since we don’t know the change, we, can

substitute with a variable x. -x +2x

5.0M - x 2x •Equilibrium is just the initial plus

the change

3

42

22

c 4.6x10]O[N

][NO K


Since… 32

c 4.6x10x)-(5.0

(2x) K Solve for x…


I

E C

nitial

hange

quilibirum

N2O4 ⇌ 2NO2

5.0M 0M

-x +2x

5.0M - x 2x

32

c 4.6x10x)-(5.0

(2x) K

(2x)2 = (5.0-x)(0.0046)

4x2 = -0.0046x + 0.023

4x2 + 0.0046x - 0.023 = 0

2a

4acbb 2

Quadratic Formula

This is in the form of a quadratic

ax2 + bx + c = 0

General form of a Quadratic

4x2 + 0.0046x - 0.023 = 0

2(4)

3)4(4)(-0.02(0.0046)0.0046- 2

5/27/2014

16

2(4)

3)4(4)(-0.02(0.0046)0.0046 2

x = 0.075 and x = -0.076

[NO2]eq = 0.15M 9150.004

4.9

(0.15)2

c K


This value for x does not make “chemical sense” because you cannot

have negative concentrations

[NO2]eq = 2x [N2O4]eq = 5.0M-x

[N2O4]eq = 4.9M


7

2

2

1.67x102x)-(0.0250

(x)(2x)

Calculating Equilibrium Concentration using K – Simplifying Assumptions

I

E C

nitial

hange

quilibirum

2H2S ⇌ 2H2 + S2

0.0250M 0M

-2x +2x

0.0250M - 2x 2x

At this point, you had better know which steps to take!!!

0M

+x

+x

7

22

22

2c 1.67x10

S][H

][S][H K

Consider the decomposition of H2S, to give H2 and S2. If Kc = 1.67x10-7 at 800°C, determine the equilibrium concentrations for all gases if the initial amount of H2S is

0.0250M.

7

2

2

1.67x102x)-(0.0250

(x)(2x)

This looks like it is going to be a cubic function; 3rd degree polynomial!!! That is: ax3 + bx2 + cx +d = 0. I don’t feel like solving this!!! Although you can do this via a graphing calculator (which you can’t use

on an exam, of course…..) or using successive approximations.

Let’s make some simplifying assumptions. For example, since K is so small, I’m going to assume that the change in x will be small also. So…..

…Let’s get rid of this..since it should

small

7

2

2

1.67x10(0.0250)

(x)(2x) HHMMM….Better……

5/27/2014

17

Calculating Equilibrium Concentration using K – Simplifying Assumptions

I

E C

nitial

hange

quilibirum

2H2S ⇌ 2H2 + S2

0.0250M 0M

-2x +2x

0.0250M - 2x 2x

0M

+x

+x

7

2

2

1.67x10(0.0250)

(x)(2x)

7

4-

3

1.67x106.25x10

4x

4

))(6.25x10(1.67x10x

473

42.97x10x

3 113 3 2.61x10x

[H2S]eq = 0.0250M – 2x

[H2]eq = 2x

[S2]eq = x

[H2S]eq = 0.0244M

[H2]eq = 5.94x10-4M

[S2]eq = 2.97x10-4M 7-

2

-42-4

c 1.76x10 (0.0244)

)(2.97x10)(5.94x10K


Slightly off because of

approximation. But O.K.

Is our approximation appropriate? Calculate what percentage x is of your

original []’s. If it is 5% or smaller, it is OK.

1.19%x1000.0250

2.97x10-4

Our approximation is

appropriate

chapter 15 equilibrium - welcome to web.gccaz.eduweb.gccaz.edu/~jaszi38221/2014/summer/chm 152...

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