Bài giảng hoá đại cương Equilibrium

What is Equilibrium?• The state in which the rate of the forward reaction equals the rate of the reverse reaction • The state in which the concentrations of all the reactants and produc

Chapter 15

Thermondyanamics, Kinetics,

and now Equilibrium

• Thermodynamics

✓ Is a reaction spontaneous? Why?

✓ What are the driving forces that push reactions?

• Kinetics

✓ What is the speed or rate of a reaction?

✓ What factors affect reaction rate?

• Equilibrium

✓ What is the extent of a reaction?

✓ What are the concentrations of reactant and product

once the quantites are no longer changing.

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What is Equilibrium?

• The state in which the rate of the forward reaction equals

the rate of the reverse reaction

• The state in which the concentrations of all the reactants

and products remain constant with time.

• However, don’t be fooled, although it may appear as if

everything has stopped, at the molecular level there is

frantic activity It is a dynamic equilibrium.

• Very few chemical reactions proceed only in one direction,

in fact most reactions are reversible.

• In a closed vessel most reactions will reach equilibrium,

however some reactions so favor the products that the

reaction appears to have gone to completion - far to the

right.

• A system that appears to not even started may actually be

at equilibrium far to the left.

So What’s Equal when

“Equalibrium is reached?”

• Usually the quantities of reactants and

products are not equal It is possible that they could be equal, but far more likely

that they will not be equal.

• The rate of the forward and reverse

reactions are equal.

• Any pressure, concentrations, or

quantities remain constant (equal).

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The Equilibrium K onstant

• The equilibrium constant is a number equal to a particular

ratio of equilbrium concentrations of the products to

reactants at a particular temperature.

• The magnitude of K is an indication of how far a reaction

proceeds toward product at a given temperature

• Some reactions so favor the products that the reaction

appears to have “gone to completion” − Equilibrium lies far

to the right.

✓ K will be large, perhaps 1 x 10 25

• A system that appears that there is “no reaction” may

actually be at equilibrium far to the left.

✓ K will be small, perhaps as small as 5 x 10−30

• When significant quantities of both reactant and product are

present, K will have an intermediate value

Chemical and Physical

• Many different types of chemical reactions can

reach equilibrium

✓ weak acid equilbrium: HC 2 H 3 O 2(aq) H + + C 2 H 3 O 2−

✓ gas phase equilbrium: 2NO 2(g) N 2 O 4(g)

• Demo

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2NO 2 N 2 O 4

Equilibrium

Name the brown gas, NO2

and name the colorless gas, N2O4

• NO2 =

• N2O4 =

8 8

Name the brown gas, NO2

and name the colorless gas, N2O4

From the observations of the tubes

containing the equilibrium, select the true statement(s).

1 NO2 N2O4 is exothermic.

2 N2O4 NO2 ∆H is negative.

3 N2O4 + Energy NO2

4 NO2 N2O4 removing energy favors

the forward reaction.

both the forward and reverse reactions.

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From the observations of the tubes containing the

equilibrium, select the true statement(s).

1 NO2 N2O4 is exothermic.

2 N2O4 NO2 ∆H is negative.

3 N2O4 + Energy NO2

4 NO2 N2O4 removing energy favors

the forward reaction.

5 adding energy increases the speed of

both the forward and reverse reactions.

What speeds up when heat is added?

• N 2 O 4 + Energy 2NO 2

• When plunged into hot water, both the forward and

reverse reactions will speed up because heat makes molecules move faster.

• Upon moving from cold to hot, initially, the forward

reaction will speed up more than the reverse because

the system is moving in a direction to use up the the external energy being applied, LeChatelier’s Principle.

• As more NO 2 builds up, the reverse reaction speeds

up more while the loss of N 2 O 4 causes the forward

reaction to slow down until the forward and reverse

reactions are at the same rate (albeit a faster rate than the previous equilibrium state) and a new equilibrium position is established with a different K value. 12

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Chemical and Physical

• Many chemical reactions can reach

equilibrium

✓ weak acid equilbrium: HC 2 H 3 O 2(aq) H + + C 2 H 3 O 2−

‣ which will be studied intensively in unit G

✓ gas phase equilbrium: NO 2(g) N 2 O 4(g)

• Some physical processes can also reach

equilibrium

✓ vapor equilbrium: H 2 O (L) H 2 O (g)

Equilibrium Situations

that you already know and love

• Vapor above a liquid in a

closed container will reach

equilibrium.

• The rate of the condensation

and rate of evaporation will

Chemical and Physical

• Many chemical reactions can reach

equilibrium

✓ weak acid equilbrium: HC 2 H 3 O 2(aq) H + + C 2 H 3 O 2−

‣ which will be studied intensively in unit G

✓ gas phase equilbrium: NO 2(g) N 2 O 4(g)

• Some physical processes can also reach

equilibrium

✓ vapor equilbrium: H 2 O (L) H 2 O (g)

✓ solubility equibrium: HgBr 2(s) Hg 2+ + Br −

Equilibrium Situations

that you already know and love

• Saturated solution

✓ A solution that contains as

much dissolved solid as possible (at a given

temperature) with some undissolved solid on the bottom.

• The rate of the dissolution (red

arrow) and rate of

crystallization (yellow arrow)

will be equal.

• Solubility is the maximum

quantity of solid that can be

dissolved and exist in

equilibrium at a particular temp.

Take a brief look at the Phet Demo

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2HI ⇄ H 2 + I 2

Consider three separate trials with different starting quantities

1 Exp 1: 1 M of HI, none of H 2 and I 2

2 Exp 2: 1 M each of H 2 and I 2 , none of HI

3 Exp 3: 1 M of all three, H 2 , I 2 and HI

After Reaching Equilibrium

• All three trials will have different quantities when the

system reaches equilibrium.

• On first inspection, there appears to be no relationship

between these equilibrium concentrations, however

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Mass Action Expression, Q

• For any reaction:

✓ aA + bB  cC + dD

• A mass action expression can be written:

✓ The reaction quotient, Q =

• If the reaction is at equilibrium, Q is equal to K

✓ and K ( aka K eq ) = the equilibrium expression

• Unlike rate laws, the equilibrium expression

depends only on the stoichiometry not on the

Calculating the Equilibrium Constant

• Apply the Law of Mass Action

K eq =

• and an Equilibrium Constant

will emerge 2HI ⇄ H2 + I2

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Just How Constant is the

Equilibrium Constant?

• The constant does vary with temperature.

• It does not depend on the initial amounts of

reactants or products.

• It is not affected by the presence of any other

materials (as long as they don’t react with any

reactants or products in the equilibrium expression).

• The equilibrium condition can be reached from

either the forward or reverse directions.

Distinguish Equilibrium Position

from Equilibrium Constant

• For a particular temperature, there is only one

equilibrium constant

• But there are an infinite number of sets of

various equilibrium concentrations, each one called an equilibrium position

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Just what is the concentration of

a pure liquid or solid?

• Because pure solids and liquids are condensed

phases (constant density), the amount of moles will respond proportionately to the volume, yielding a

consistent concentration.

• Because the concentration of a pure substance, stays constant, you should NOT include them in the

equilibrium expression.

• Equilibrium that involve more than one phase is

called heterogeneous equilibrium.

• Remember that if a pure solid or a pure liquid is in

an equilibrium reaction, they do not show up in the equilibrium expression.

✓ the AP exam will try to trick you into using them.

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What determines the position

of the Equilibrium Position?

• The equilibrium “position” can be “left, right, or

somewhere in between.”

• The equilibrium position is determined by

✓ the initial concentrations

✓ relative bond energies of reactants and products

✓ relative degree of “organization” of the reactants and

products

• Energy and organization (entropy) are important

because molecules try to achieve minimum energy and maximum disorder.

• Recall from Unit C that there is a relationship

between thermodynamics and equilibrium.

K eq and Direction of Reaction

• The equilibrium expression for a reaction

is the reciprocal for that reaction in

K eq and Stoichiometry

• When the balanced equation for a reaction is

multiplied by some factor n, the equilibrium

constant for the new reaction is the original K eq

raised to the nth power

• So when you change the coefficients by some

factor, raise the K eq to that same factor.

K eq and Hess’ Law

• Suppose you knew K for these two reactions:

• Note that when using Hess’ Law, that while the

reactions are added, the K eq values are multiplied.

• This will be shown in more detail on the next slide >>>

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Verify K eq and Hess’ Law

For the reaction A(g) + B(g) C(g) + D(g)

K was determined to be 0.5 at 500ºC

In an equilibrium mixture of these four gases at 500ºC, which one of the

following is true?

1 [D] will always be greater than [B]

2 [D] will always be less than [B]

3 [D] will always be equal to [B]

4 [D] will never be equal to [B]

5 none of the above are true

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For the reaction A(g) + B(g) C(g) + D(g)

K was determined to be 0.5 at 500ºC

In an equilibrium mixture of these four gases at 500ºC, which one of the

following is true?

1 [D] will always be greater than [B]

2 [D] will always be less than [B]

3 [D] will always be equal to [B]

4 [D] will never be equal to [B]

5 none of the above are true

Units on K eq ?

• Different texts handle this issue in different ways.

✓ Some texts indicate that units exist and vary from K eq to

K eq based on what cancels and what does not.

• Your text never puts units on their Keq

✓ They rationalize that all measurements: pressure or

molarities are divided by a standard reference value (of 1 atm or 1 M) which will cancel out the units.

✓ They also say that the molarity reference used for pure

liquids or solids is the molarity of the pure liquid or solid itself thereby reducing the concentration of any pure

(solid or liquid) substance to 1 M.

• The bottom line? The AP exam will consider K values to be

dimensionless numbers and will never ask you about units.

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✓ K p as calculated in atmospheres (atm)

✓ K c as calculated for units in molarity [M]

• However, in our Chapter 15, Brown and LeMay do not

differentiate for K p and use K eq almost exclusively to

mean K p You will know by the context of the problem

which K (K p or K c) you are calculating.

K eq , they mean K p

• Next in this unit we will study K sp

• In later chapters we will learn about K a , K b , K w

That’s All for Now

Let’s Look at Problem Types

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Write the equlibrium expression for the following equilibria

CH4(g) + H2O(g) ⇄ CO(g) + 3H2(g)

for this reaction, Kc = 0.26 at 1200 K

At equilibrium, a 0.32 L flask contains 0.26 mol

CO, 0.091 mol H2, and 0.041 mol of CH4.

What is the [H2O]?

36 36

CH4(g) + H2O(g) ⇄ CO(g) + 3H2(g)

for this reaction, Kc = 0.26 at 1200 K

At equilibrium, a 0.32 L flask contains 0.26 mol CO, 0.091 mol H2, and 0.041 mol of CH4.

[H2O] = 0.53 M

4 moles of NO2 are placed in a 1.0 L

container and maintained at a constant

temperature After equilibrium was

established, it was found that 50% of the NO2

had dissociated into NO and O2 The

Solving Keq Problems

• Write a balanced equation (if not given one).

• Write an equilibrium expression.

• Use the RICE box as necessary to solve for

information that is not known in the problems.

4 moles of NO2 are placed in a 1.0 L

container and maintained at a constant

temperature After equilibrium was

established, it was found that 50% of the NO2

had dissociated into NO and O2 The

→

K eq = [NO]2[O2 ]

[NO2 ]2

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4 moles of NO2 are placed in a 1.0 L

container and maintained at a constant

temperature After equilibrium was

established, it was found that 50% of the NO2

had dissociated into NO and O2 The

CO(g) + H2O(g) ⇄ CO2(g) + H2(g)

placed in a 125 ml flask at 900 K, what is the composition of the equilibrium mixture?

42 42

CO(g) + H2O(g) ⇄ CO2(g) + H2(g)

for this reaction, Kc = 1.56 at 900 K

If 0.250 mol each of CO and H2O

gases were placed in a 125 ml flask at

900 K, what is the composition of the

equilibrium mixture?

[CO] = [H2O] = 0.89 M (0.111 mol)

[CO2] = [H2] = 1.11 M (0.139 moles)

CO(g) + H2O(g) ⇄ CO2(g) + H2(g)

placed in a 125 ml flask at 900 K, what is the composition of the equilibrium mixture?

44 44

CO(g) + H2O(g) ⇄ CO2(g) + H2(g)

for this reaction, Kc = 1.56 at 900 K

If 2.0 M CO and 1.0 M H2O gases were placed in

a 125 ml flask at 900 K, what is the composition

of the equilibrium mixture?

x = 7.6 M and 0.73 M, but only one makes sense

[CO] = 1.27 M

[H2O] = 0.27 M

[CO2] = [H2] = 0.73 M

N2O4(g) ⇄ 2NO2(g) Kc = 0.21 at 100ºC

At a point during the reaction,

[N2O4] = 0.12 M and [NO2] = 0.55 M

Is the reaction at equilibrium?

If not, in which direction is it

progressing?

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Reaction Quotient, Q

• When given values that may not be at

equilibrium, you can substitute them into the

equilibrium expression to solve for Q

• If Q = Keq then you know those values are

actually at equilibrium

• If Q > Keq the right side is too large and the

system will shift to the left to reach

equilibrium

• If Q < Keq the right side is too small and the

system will shift to the right to reach

equilibrium

N2O4(g) ⇄ 2NO2(g) Kc = 0.21 at 100ºC

At a point during the reaction,

[N2O4] = 0.12 M and [NO2] = 0.55 M

Is the reaction at equilibrium? No

If not, in which direction is it

progressing? Q > K, thus the reaction will proceeed to the left.

48 48

• Consider the hypothetical reaction:

• A (g) + B (g) 2C (g) for which K p = 4

• Suppose 2 atm of each substance was placed in a 1 L

container, is the system at equilibrium?

• If not, what is the pressure of each substance at equilibrium

• Consider the hypothetical reaction:

• A (g) + B (g) 2C (g) for which K p = 4

• Suppose 2 atm of each substance was placed in a 1 L

container, is the system at equilibrium?

• What is the pressure of each substance at equilibrium

• Write the equilibrium expression.

• Put in initial values and solve for Q, compare Q to K eq to determine the direction of the shift.

• Use x to name the changes

• Put in equilibrium values using x

• Solve.

R

I C E

A + B 2 C

50

• Consider the hypothetical reaction:

• A (g) + B (g) 2C (g) for which K p = 4

• Suppose 2 atm of each substance was placed in a 1 L

container, is the system at equilibrium?

• What is the pressure of each substance at equilibrium

• Write the equilibrium expression.

• Put in initial values and solve for Q, compare Q to K eq to determine the direction of the shift

• Use x to name the changes

• Put in equilibrium values using x

NH3 is placed in a flask with a partial

pressure of 3 atm, and is allowed to

decompose into its elemental gases At

equilibrium, the partial pressure of nitrogen

First, write the balanced equation

• 2 NH3(g)  N2(g) + 3 H2(g)

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