Chapter 15 Chemical Equilibrium - myCCSDmy.ccsd.net/userdocs/documents/NIrIc4uc0Gynqbpr.pdf ·...

Chemical Equilibrium

Mr. Matthew Totaro Legacy High School

AP Chemistry © 2012 Pearson Education, Inc.

Equilibrium

The Concept of Equilibrium

Chemical equilibrium occurs when a reaction and its reverse reaction proceed at the same rate.

© 2012 Pearson Education, Inc.

Equilibrium

The Concept of Equilibrium • As a system approaches

equilibrium, both the forward and reverse reactions are occurring.

• At equilibrium, the forward and reverse reactions are proceeding at the same rate.


Equilibrium

A System at Equilibrium

Once equilibrium is achieved, the amount of each reactant and

product remains constant.


Equilibrium

Dynamic Equilibrium • As the forward reaction slows and the reverse

reaction accelerates, eventually they reach the same rate

• Dynamic equilibrium is the condition wherein the rates of the forward and reverse reactions are equal

• Once the reaction reaches equilibrium, the concentrations of all the chemicals remain constant – because the chemicals are being consumed and made at

the same rate

5

Equilibrium

Depicting Equilibrium

Since, in a system at equilibrium, both the forward and reverse reactions are being carried out, we write its equation with a double arrow:

N2O4(g) 2NO2(g)


Equilibrium

When Country A citizens feel overcrowded, some will emigrate to Country B

An Analogy: Population Changes

7

However, after a time, emigration will occur in both directions at the same rate, leading to populations in Country A and Country B that are constant, though not necessarily equal

Equilibrium

The Equilibrium Constant


Equilibrium


• Forward reaction: N2O4(g) → 2NO2(g)

• Rate law: Rate = kf[N2O4]


Equilibrium


• Reverse reaction: 2 NO2(g) → N2O4(g)

• Rate law: Rate = kr[NO2]2


Equilibrium


• Therefore, at equilibrium Ratef = Rater

kf[N2O4] = kr[NO2]2 • Rewriting this, it becomes

kf kr

[NO2]2 [N2O4]

=


Equilibrium


The ratio of the rate constants is a constant at that temperature, and the expression becomes

Keq = kf kr

[NO2]2 [N2O4]

=


Equilibrium


• Consider the generalized reaction

• The equilibrium expression for this reaction would be

Kc = [C]c[D]d [A]a[B]b

aA + bB cC + dD


Equilibrium

Equilibrium Constant • Even though the concentrations of reactants and

products are not equal at equilibrium, there is a relationship between them

• The relationship between the chemical equation and the concentrations of reactants and products is called the Law of Mass Action

• For the general equation aA + bB → cC + dD, the Law of Mass Action gives the relationship below – the lowercase letters represent the coefficients of the balanced

chemical equation – always products over reactants

• K is called the equilibrium constant – unitless

14

Tro: Chemistry: A Molecular Approach, 2/e

Equilibrium

Writing Equilibrium Constant Expressions

• For the reaction aA(aq) + bB(aq) ⇔ cC(aq) + dD(aq) the equilibrium constant expression is

• So for the reaction 2 N2O5(g) ⇔ 4 NO2(g) + O2(g) the equilibrium constant expression is

15


Equilibrium

What Does the Value of Keq Imply? • When the value of Keq >> 1, we know that when the

reaction reaches equilibrium there will be many more product molecules present than reactant molecules – the position of equilibrium favors products

• When the value of Keq << 1, we know that when the reaction reaches equilibrium there will be many more reactant molecules present than product molecules – the position of equilibrium favors reactants

16

Equilibrium

A Large Equilibrium Constant

17

Equilibrium

A Small Equilibrium Constant

18

Equilibrium

Practice – Write the equilibrium constant expressions, K, and predict the position of

equilibrium for the following

2 SO2(g) + O2(g) ⇔ 2 SO3(g) K = 8 x 1025

N2(g) + 2 O2(g) ⇔ 2 NO2(g) K = 3 x 10−17

favors products

favors reactants

19

Equilibrium

Relationships between K and Chemical Equations

• When the reaction is written backward, the equilibrium constant is inverted

For the reaction aA + bB ⇔ cC + dD the equilibrium constant expression is

For the reaction cC + dD ⇔ aA + bB the equilibrium constant expression is

20

Equilibrium


• When the coefficients of an equation are multiplied by a factor, the equilibrium constant is raised to that factor

For the reaction aA + bB ⇔ cC the equilibrium constant expression is

For the reaction 2aA + 2bB ⇔ 2cC the equilibrium constant expression is

21

Equilibrium


• When you add equations to get a new equation, the equilibrium constant of the new equation is the product of the equilibrium constants of the old equations

For the reactions (1) aA ⇔ bB and (2) bB ⇔ cC the equilibrium constant expressions are

For the reaction aA ⇔ cC the equilibrium constant expression is

22

Equilibrium

Example: Compute the equilibrium constant at 25 °C for the reaction NH3(g) ⇔ 0.5 N2(g) + 1.5 H2(g) for

N2(g) + 3 H2(g) → 2 NH3(g), K = 3.7 x 108 at 25 °C

Equilibrium

Kbackward = 1/Kforward, Knew = Koldn

Example: Compute the equilibrium constant at 25 °C for the reaction NH3(g) ⇔ 0.5 N2(g) + 1.5 H2(g)

24

for N2(g) + 3 H2(g) ⇔ 2 NH3(g), K = 3.7 x 108 at 25 °C K for NH3(g) ⇔ 0.5N2(g) + 1.5H2(g), at 25 °C

Solution:

Conceptual Plan:

Relationships:

Given:

Find:

N2(g) + 3 H2(g) ⇔ 2 NH3(g) K1 = 3.7 x 108

K’ K

2 NH3(g) ⇔ N2(g) + 3 H2(g)

NH3(g) ⇔ 0.5 N2(g) + 1.5 H2(g)

Equilibrium

Calculate Kc for the reaction SnO2(s) + 2 CO(g) → Sn(s) + 2 CO2(g)

given the following information:

SnO2(s) + 2 H2(g) → Sn(s) + 2 H2O(g) Kc = 8.12 H2(g) + CO2(g) → H2O(g) + CO(g) Kc = 0.771

Equilibrium


Since pressure is proportional to concentration for gases in a closed system, the equilibrium expression can also be written

Kp = (PCc) (PD

d) (PA

a) (PBb)


Equilibrium

Relationship Between Kc and Kp

• From the ideal-gas law we know that

• Rearranging it, we get PV = nRT

P = RT n V


Equilibrium

Relationship Between Kc and Kp

Plugging this into the expression for Kp,,for each substance, the relationship between Kc and Kp becomes

where

Kp = Kc (RT)∆n

∆n = (moles of gaseous product) − (moles of gaseous reactant)


Equilibrium

Example: Find Kc for the reaction 2 NO(g) + O2(g) ⇔ 2 NO2(g), given Kp = 2.2 x 1012 @ 25 °C

29

K is a unitless number because there are more moles of reactant than product, Kc

should be larger than Kp, and it is

Kp = 2.2 x 1012

Kc

Check:

Solution:

Conceptual Plan:

Relationships:

Given:

Find:

Kp Kc

2 NO(g) + O2(g) ⇔ 2 NO2(g) ∆n = 2 − 3 = −1


Equilibrium

Practice – Calculate the value of Kp or Kc for each of the following at 27 °C

2 SO2(g) + O2(g) ⇔ 2 SO3(g) Kc = 8 x 1025

N2(g) + 2 O2(g) ⇔ 2 NO2(g) Kp = 3 x 10−17

30

Equilibrium

Equilibrium Can Be Reached from Either Direction

As you can see, the ratio of [NO2]2 to [N2O4] remains constant at this temperature no matter what the initial

concentrations of NO2 and N2O4 are.


Equilibrium


These are the data from the last two trials from the table on the

previous slide.


Equilibrium


It doesn’t matter whether we start with N2 and H2 or whether we start with NH3, we will have

the same proportions of all three substances at equilibrium.


Equilibrium

Heterogeneous Equilibrium


Equilibrium

The Concentrations of Solids and Liquids Are Essentially Constant

Both the concentrations of solids and liquids can be obtained by multiplying the

density of the substance by its molar mass—and both of these are constants at

constant temperature.


Equilibrium

The Concentrations of Solids and Liquids Are Essentially Constant

Therefore, the concentrations of solids and liquids do not appear in the

equilibrium expression.

Kc = [Pb2+] [Cl−]2

PbCl2(s) Pb2+(aq) + 2Cl−(aq)


Equilibrium

As long as some CaCO3 or CaO

remain in the system, the amount of CO2 above the solid will

remain the same.

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


Equilibrium

HNO2(aq) + H2O(l) ⇔ H3O+(aq) + NO2

−(aq) K = 4.6 x 10−4

Ca(NO3)2(aq) + H2SO4(aq) ⇔ CaSO4(s) + 2 HNO3(aq) K = 1 x 104

Practice – Write the equilibrium constant expressions, K, and predict the position of

equilibrium for the following

38

Equilibrium

Equilibrium Calculations


Equilibrium

An Equilibrium Problem A closed system initially containing 1.000 x 10−3 M H2 and 2.000 x 10−3 M

I2 at 448 °C is allowed to reach equilibrium. Analysis of the equilibrium mixture shows

that the concentration of HI is 1.87 x 10−3 M. Calculate Kc at

448 °C for the reaction taking place, which is

H2(g) + I2(s) 2HI(g)


Equilibrium

What Do We Know?

[H2], M [I2], M [HI], M

Initially 1.000 x 10−3 2.000 x 10−3 0

Change

At equilibrium 1.87 x 10−3


Equilibrium

[HI] Increases by 1.87 x 10−3 M

[H2], M [I2], M [HI], M

Initially 1.000 x 10−3 2.000 x 10−3 0

Change +1.87 x 10−3



Equilibrium

Stoichiometry tells us [H2] and [I2] decrease by half as much.

[H2], M [I2], M [HI], M

Initially 1.000 x 10−3 2.000 x 10−3 0

Change −9.35 x 10−4 −9.35 x 10−4 +1.87 x 10−3



Equilibrium

We can now calculate the equilibrium concentrations of all three compounds

[H2], M [I2], M [HI], M

Initially 1.000 x 10−3 2.000 x 10−3 0

Change −9.35 x 10−4 −9.35 x 10−4 +1.87 x 10−3

At equilibrium 6.5 x 10−5 1.065 x 10−3 1.87 x 10−3


Equilibrium

and, therefore, the equilibrium constant:

Kc = [HI]2 [H2] [I2]

= 51

= (1.87 x 10−3)2 (6.5 x 10−5)(1.065 x 10−3)


Equilibrium

Example: Find the value of Kc for the reaction 2 CH4(g) ⇔ C2H2(g) + 3 H2(g) at 1700 °C if the initial [CH4]

= 0.115 M and the equilibrium [C2H2] = 0.035 M

46

+0.035

Equilibrium

Example: Find the value of Kc for the reaction 2 CH4(g) ⇔ C2H2(g) + 3 H2(g) at 1700 °C if the initial [CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M

47

+0.035 −2(0.035) +3(0.035)

0.045 0.105

Equilibrium

Practice –The following data were collected for the reaction 2 NO2(g) ⇔ N2O4(g) at 100 °C. Complete

the table and determine values of Kp and Kc.

48

Equilibrium

Practice – A second experiment was done and the following data were collected for the reaction 2 NO2(g) ⇔ N2O4(g) at 100 °C. Complete the table and determine values of Kp and

Kc. Compare them to the first experiment.

49


Equilibrium

The Reaction Quotient (Q)

• Q gives the same ratio the equilibrium expression gives, but for a system that is not at equilibrium.

• To calculate Q, one substitutes the initial concentrations on reactants and products into the equilibrium expression.


Equilibrium

The Reaction Quotient • If a reaction mixture, containing

both reactants and products, is not at equilibrium; how can we determine in which direction it will proceed?

• The answer is to compare the current concentration ratios to the equilibrium constant

• The concentration ratio of the products (raised to the power of their coefficients) to the reactants (raised to the power of their coefficients) is called the reaction quotient, Q

for the gas phase reaction aA + bB ⇔ cC

+ dD the reaction quotient is:

51

Equilibrium

The Reaction Quotient: Predicting the Direction of Change

• If Q > K, the reaction will proceed fastest in the reverse direction

– the [products] will decrease and [reactants] will increase • If Q < K, the reaction will proceed fastest in the forward

direction – the [products] will increase and [reactants] will decrease

• If Q = K, the reaction is at equilibrium – the [products] and [reactants] will not change

• If a reaction mixture contains just reactants, Q = 0, and the reaction will proceed in the forward direction

• If a reaction mixture contains just products, Q = ∞, and the reaction will proceed in the reverse direction

52

Equilibrium

If Q = K, the system is at equilibrium.


Equilibrium

If Q > K, there is too much product, and the equilibrium shifts to the left.


Equilibrium

If Q < K, there is too much reactant, and the equilibrium shifts to the right.


Equilibrium

Q, K, and the Direction of Reaction

56

Equilibrium

If Q = K, equilibrium; If Q < K, forward; If Q > K, reverse

Example: For the reaction below, which direction will it proceed if PI2 = 0.114 atm, PCl2 = 0.102 atm. & PICl = 0.355 atm

57

for I2(g) + Cl2(g) ⇔ 2 ICl(g), Kp = 81.9 direction reaction will proceed

Solution:

Conceptual Plan:

Relationships:

Given:

Find:

I2(g) + Cl2(g) ⇔ 2 ICl(g) Kp = 81.9

Q PI2, PCl2, PICl

because Q (10.8) < K (81.9), the reaction will proceed to the right

Equilibrium

Practice – For the reaction CH4(g) + 2 H2S(g) ⇔ CS2(g) + 4 H2(g) Kc = 3.59 at 900 °C. For each of the following measured concentrations

determine whether the reaction is at equilibrium. If not at equilibrium, in which direction will the reaction proceed to get to equilibrium?

58

Equilibrium

Practice – A sample of PCl5(g) is placed in a 0.500 L container and heated to 160 °C. The PCl5 is decomposed into PCl3(g) and Cl2(g). At equilibrium, 0.203 moles of PCl3 and Cl2 are formed. Determine the equilibrium concentration of PCl5 if Kc = 0.0635

PCl5 ⇔ PCl3 + Cl2

59

0.203 mol0.500 L

0.203 mol0.500 L

Equilibrium

Le Châtelier’s Principle


Equilibrium

Le Châtelier’s Principle “If a system at

equilibrium is disturbed by a change in

temperature, pressure, or the concentration of

one of the components, the

system will shift its equilibrium position so

as to counteract the effect of the disturbance.”


Equilibrium

Le Châtelier’s Principle

• Le Châtelier's Principle guides us in predicting the effect various changes in conditions have on the position of equilibrium

• It says that if a system at equilibrium is disturbed, the position of equilibrium will shift to minimize the disturbance – disturbances all involve making the system open

62

Equilibrium

The result will be people moving from Country B into Country A faster than

people moving from Country A into Country B. This will continue until a

new equilibrium between the populations is established; the new populations will have different numbers of people than

the old ones.

An Analogy: Population Changes

63

When an influx of population enters Country B from somewhere outside Country A, it disturbs the

equilibrium established between Country A and Country B.

When the populations of Country A and Country B are in equilibrium, the emigration rates between the two countries are equal so the populations stay constant.

Equilibrium

Disturbing Equilibrium: Adding or Removing Reactants

• After equilibrium is established, a reactant is added – as long as the added reactant is included in the equilibrium

constant expression • i.e., not a solid or liquid

• How will this affect the rate of the forward reaction? • How will it affect the rate of the reverse reaction? • How will it affect the value of K?

64

Equilibrium

Disturbing Equilibrium: Adding Reactants • Adding a reactant initially increases the rate of the forward

reaction, but has no initial effect on the rate of the reverse reaction

• The reaction proceeds to the right until equilibrium is restored

• At the new equilibrium position, you will have more of the products than before, less of the non-added reactants than before, and less of the added reactant

– but not as little of the added reactant as you had before the addition • At the new equilibrium position, the concentrations of

reactants and products will be such that the value of the equilibrium constant is the same

65

Equilibrium

Disturbing Equilibrium: Adding or Removing Reactants

• After equilibrium is established, a reactant is removed – as long as the added reactant is included in the equilibrium

constant expression • i.e., not a solid or liquid

• How will this affect the rate of the forward reaction? • How will it affect the rate of the reverse reaction? • How will it affect the value of K?

66


Equilibrium

Disturbing Equilibrium: Removing Reactants • Removing a reactant initially decreases the rate of the forward

reaction, but has no initial effect on the rate of the reverse reaction.

– so the reaction is going faster in reverse • The reaction proceeds to the left until equilibrium is restored • At the new equilibrium position, you will have less of the

products than before, more of the non-removed reactants than before, and more of the removed reactant

– but not as much of the removed reactant as you had before the removal • At the new equilibrium position, the concentrations of

reactants and products will be such that the value of the equilibrium constant is the same

67

Equilibrium

The Effect of Concentration Changes on Equilibrium

• Adding a reactant will decrease the amounts of the other reactants and increase the amount of the products until a new position of equilibrium is found – that has the same K

• Removing a product will increase the amounts of the other products and decrease the amounts of the reactants. – you can use this to drive a reaction to completion!

• Equilibrium shifts away from side with added chemicals or toward side with removed chemicals – Remember, adding more of a solid or liquid does not change its

concentration – and therefore has no effect on the equilibrium 68

Equilibrium


When NO2 is added, some of it combines to make more N2O4

69

Equilibrium


When N2O4 is added, some of it decomposes to make more NO2 70

Equilibrium

The Haber Process The transformation of nitrogen and hydrogen into

ammonia (NH3) is of tremendous significance in agriculture, where ammonia-based fertilizers are of utmost

importance.


Equilibrium

The Haber Process

If H2 is added to the system,

N2 will be consumed and

the two reagents will form more

NH3.


Equilibrium

The Haber Process This apparatus

helps push the equilibrium to the right by

removing the ammonia

(NH3) from the system as a

liquid.


Equilibrium

The Effect of Adding a Gas to a Gas Phase Reaction at Equilibrium

• Adding a gaseous reactant increases its partial pressure, causing the equilibrium to shift to the right – increasing its partial pressure increases its

concentration – does not increase the partial pressure of the other

gases in the mixture • Adding an inert gas to the mixture has no effect

on the position of equilibrium – does not affect the partial pressures of the gases in

the reaction 74

Equilibrium

Effect of Volume Change on Equilibrium • Decreasing the volume of the container increases the

concentration of all the gases in the container – increases their partial pressures – It does not change the concentrations of solutions!!

• If their partial pressures increase, then the total pressure in the container will increase

• According to Le Châtelier’s Principle, the equilibrium should shift to remove that pressure

• The way the system reduces the pressure is to reduce the number of gas molecules in the container

• When the volume decreases, the equilibrium shifts to the side with fewer gas molecules 75

Equilibrium

Disturbing Equilibrium: Changing the Volume

• After equilibrium is established, the container volume is decreased

• How will it affect the concentration of solids, liquid, solutions, and gases?

• How will this affect the total pressure of solids, liquid, and gases?

• How will it affect the value of K?

76

Equilibrium

Disturbing Equilibrium: Reducing the Volume • Decreasing the container volume increases the concentration of

all gases, but not solids, liquid, or solutions – same number of moles, but different number of liters, resulting in a

different molarity • Decreasing the container volume will increase the total pressure

– Boyle’s Law – if the total pressure increases, the partial pressures of all the gases will

increase – Dalton’s Law of Partial Pressures • Because the total pressure increases, the position of equilibrium

will shift to decrease the pressure by removing gas molecules – shift toward the side with fewer gas molecules

• At the new equilibrium position, the partial pressures of gaseous reactants and products will be such that the value of the equilibrium constant is the same

77


Equilibrium

The Effect of Volume Changes on Equilibrium

78

Because there are more gas molecules on the reactants

side of the reaction, when the pressure is increased the

position of equilibrium shifts toward the side with fewer molecules to decrease the

pressure

When the pressure is decreased by increasing the

volume, the position of equilibrium shifts toward the side with the greater number of molecules – the reactant

side

Equilibrium

The Effect of Temperature Changes on Equilibrium Position

• Exothermic reactions release energy and endothermic reactions absorb energy

• If we write Heat as a product in an exothermic reaction or as a reactant in an endothermic reaction, it will help us use Le Châtelier’s Principle to predict the effect of temperature changes – even though heat is not matter and not written

in a proper equation 79

Equilibrium

The Effect of Temperature Changes on Equilibrium for Exothermic Reactions

• For an exothermic reaction, heat is a product • Increasing the temperature is like adding heat • According to Le Châtelier’s Principle, the

equilibrium will shift away from the added heat • Adding heat to an exothermic reaction will decrease

the concentrations of products and increase the concentrations of reactants

• Adding heat to an exothermic reaction will decrease the value of K

• How will decreasing the temperature affect the system?

aA + bB ⇔ cC + dD + Heat 80

Kc =C

c× D

d

A a× B

b

Equilibrium

The Effect of Temperature Changes on Equilibrium for Endothermic Reactions

• For an endothermic reaction, heat is a reactant • Increasing the temperature is like adding heat • According to Le Châtelier’s Principle, the

equilibrium will shift away from the added heat • Adding heat to an endothermic reaction will

decrease the concentrations of reactants and increase the concentrations of products

• Adding heat to an endothermic reaction will increase the value of K

• How will decreasing the temperature affect the system?

Heat + aA + bB ⇔ cC + dD 81

Kc =C

c× D

d

A a× B

b

Equilibrium

The Effect of Temperature Changes on Equilibrium

82

Equilibrium

Catalysts


Equilibrium

Catalysts Catalysts increase the rate of both the forward and reverse reactions.


Equilibrium

Catalysts When one uses a catalyst, equilibrium is achieved faster,

but the equilibrium composition remains unaltered.


Equilibrium

Not Changing the Position of Equilibrium: The Effect of Catalysts

• Catalysts provide an alternative, more efficient mechanism

• Catalysts work for both forward and reverse reactions

• Catalysts affect the rate of the forward and reverse reactions by the same factor

• Therefore, catalysts do not affect the position of equilibrium

86

Equilibrium

Practice – Le Châtelier’s Principle • The reaction 2 SO2(g) + O2(g) ⇔ 2 SO3(g) with

∆H° = −198 kJ is at equilibrium. How will each of the following changes affect the equilibrium concentrations of each gas once equilibrium is restored? – adding more O2 to the container – condensing and removing SO3 – compressing the gases – cooling the container – doubling the volume of the container – warming the mixture – adding the inert gas helium to the container – adding a catalyst to the mixture

87

Equilibrium

Practice – Le Châtelier’s Principle • The reaction 2 SO2(g) + O2(g) ⇔ 2 SO3(g) with

∆H° = −198 kJ is at equilibrium. How will each of the following changes affect the equilibrium concentrations of each gas once equilibrium is restored? – adding more O2 to the container shift to SO3 – condensing and removing SO3 shift to SO3 – compressing the gases shift to SO3 – cooling the container shift to SO3 – doubling the volume of the container shift to SO2 – warming the mixture shift to SO2 – adding helium to the container no effect – adding a catalyst to the mixture no effect

88

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