chapter 17a reaction rates and equilibrium. chapter 17 table of contents copyright © cengage...

Chapter 17a

Reaction Rates and Equilibrium

Chapter 17

Table of Contents

Copyright © Cengage Learning. All rights reserved 2

17.1 How Chemical Reactions Occur

17.2 Conditions That Affect Reaction Rates

17.3 The Equilibrium Condition

17.4 Chemical Equilibrium: A Dynamic Condition

Chapter 17

Table of Contents


Rates of Chemical Reactions

4 C3H5N3O9 6 N2 + 10 H2O + 12 CO2 + O2 + 5720kJNitroglycerine Exothermic

Chapter 17

Table of Contents


An Example of Reaction Rates

Fast Reaction vs. Slow Reaction

35/97 people died in 1937

Section 17.1

How Chemical Reactions Occur

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Collision Model

• Molecules must collide in order for a reaction to occur.

• Rate depends on concentrations of reactants and temperature.

Conditions That Affect Reaction Rates

Section 17.2

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• Concentration – increases rate because more molecules lead to more collisions.

• Temperature – increases rate. Why?


Section 17.2

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How to Tame Allergic Reactions

How can you slow down a histamine attack?

Histamine attacks are greater when you are hot. Cooling down affected areas can reduce allergy symptoms.


Section 17.2

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Activation Energy

• Minimum energy required for a reaction to occur.


Section 17.2

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What makes Switzerland unique?


Section 17.2

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Section 17.2

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Chemical Reactions must go over an energy hill like a mountain (Swiss Alps).


Section 17.2

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Catalyst

• A substance that speeds up a reaction without being consumed.

• Enzyme – catalyst in a biological system


Section 17.2

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Catalyst




Section 17.2

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Catalyst



Chlorofluoro Carbons (CFC’s) are acting as catalysts to decompose the ozone (O3) layer. The ozone layer is formed from cosmic radiation and protects us from UV light.


Section 17.2

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Depletion is measured by T.O.M.S.

“Total Ozone Mapping Spectrometer”

The below dark shaded are shows the amount of depletion around the Antarctica


Section 17.2

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An Amana refrigerator, one of many appliances that now use HFC-134a. This compound is replacing CFC’s, which lead to destruction of atmospheric ozone.


Section 17.2

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Use a catalytic converter to convert the polluting exhaust gases of burned lead-free gasoline into harmless gases. Platinum (Pt) is the catalysts used. Only a small amount is needed.

The Equilibrium Condition

Section 17.3

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Equilibrium

• The exact balancing of two processes, one of which is the opposite of the other.

The Equilibrium Condition

Section 17.3

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Chemical Equilibrium

• A dynamic state where the concentrations of all reactants and products remain constant.

Section 17.4

Chemical Equilibrium: A Dynamic Condition

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• On the molecular level, there is frantic activity. Equilibrium is not static, but is a highly dynamic situation.

• Macroscopically static • Microscopically dynamic

Section 17.4


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The Reaction of H2O and CO to Form CO2 and H2 as Time Passes

Equal numbers of moles of H2O and CO are mixed in a closed container.

The reaction begins to occur, and some products (H2 and CO2) are formed.

The reaction continues as time passes and more reactants are changed to products.

Although time continues to pass, the numbers of reactant and product molecules are the same as in (c). No further changes are seen as time continues to pass. The system has reached equilibrium.

Section 17.4


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Section 17.4


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• Concentrations reach levels where the rate of the forward reaction equals the rate of the reverse reaction.

Section 17.4


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Concept Check

Consider an equilibrium mixture in a closed vessel reacting according to the equation:

H2O(g) + CO(g) H2(g) + CO2(g)

You add more H2O(g) to the flask. How does the concentration of each chemical compare to its original concentration after equilibrium is reestablished? Justify your answer.

Section 17.4


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Concept Check

Consider an equilibrium mixture in a closed vessel reacting according to the equation:

H2O(g) + CO(g) H2(g) + CO2(g)

You add more H2 to the flask. How does the concentration of each chemical compare to its original concentration after equilibrium is reestablished? Justify your answer.

Section 17.4


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Reactions Rates and Equilibrium

Chapter 17b

W

Section 17.4


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17.5 The Equilibrium Constant: An Introduction

17.6 Heterogeneous Equilibria

17.7 Le Châtelier’s Principle

17.8 Applications Involving the Equilibrium Constant

17.9 Solubility Equilibria

Section 17.5

The Equilibrium Constant: An Introduction

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Consider the following reaction at equilibrium:

jA + kB lC + mD

• A, B, C, and D = chemical species.• Square brackets = concentrations of species at equilibrium.• j, k, l, and m = coefficients in the balanced equation.• K = equilibrium constant (given without units).

j

l

k

m

[B][A]

[D] [C]K =

Section 17.5


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Example

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

2

33

2 2

NH =

N HK

Section 17.5


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• K always has the same value at a given temperature regardless of the amounts of reactants or products that are present initially.

• For a reaction, at a given temperature, there are many equilibrium positions but only one value for K. Equilibrium position is a set of equilibrium

concentrations.

Section 17.5


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Concept Check

Consider the following equilibrium reaction:HC2H3O2(aq) H+(aq) + C2H3O2

–

(aq)Determine the equilibrium constant expression for the dissociation of acetic acid.

a) b)

c) d)

]OHHC[

]OHC][H[K

232

232

]OHHC[

]OHC[][HK

232

232

]OHC][H[

]OH[HC K

232

232

]H[

]OH[HC K 232

Section 17.5


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Exercise

For the reaction below, calculate the value of the equilibrium constant, given the equilibrium concentrations.N2O4(g) 2NO2(g)[N2O4] = 0. 055 M [NO2] = 0.060 M

a) K = 0.050b) K = 0.92c) K = 1.1d) K = 0.065

K = (0.060)2/0.055 = 0.065

Section 17.5


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What is the equilibrium expression for the following?

CH4(g) + 2 H2S(g) <==> CS2(g) + 4 H2(g)

Keq = ----------------[CS2][H2]4

[CH4][H2S]2

H2(g) + I2(g) <==> 2 HI(g)

Keq = ----------[HI]2

[H2][I2]

Fe3+(aq) + SCN-(aq) <==> Fe(SCN)+2(aq)

Keq = ------------------[Fe(SCN)+2]

[Fe3+][SCN-]

Section 17.6

Heterogeneous Equilibria

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Homogeneous Equilibria

• Homogeneous equilibria – involve the same phase:

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

HCN(aq) H+(aq) + CN-(aq)

Section 17.6


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• Heterogeneous equilibria – involve more than one phase:

2KClO3(s) 2KCl(s) + 3O2(g)

2H2O(l) 2H2(g) + O2(g)

Section 17.6


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• The position of a heterogeneous equilibrium does not depend on the amounts of pure solids or liquids present. The concentrations of pure liquids and solids

are constant.

2KClO3(s) 2KCl(s) + 3O2(g)

3

2 = OK

Section 17.6


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Concept Check

Determine the equilibrium expression for the reaction:

CaF2(s) Ca2+(aq) + 2F–(aq)

a) b)

c) d)

][CaF

]][F[CaK

2

-2

]CaF[

]F[][CaK

2

-2

2 -

2

[Ca ][2F ]K

[CaF ]

2-2 ]F][[CaK

Section 17.7

Le Châtelier’s Principle

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• If a change is imposed on a system at equilibrium, the position of the equilibrium will shift in a direction that tends to reduce that change.

O3 (g) + Cl (g) O2 (g) + OCl(g)

Equilibrium shifts to counter a disturbance.Hills and Valleys!

Section 17.7


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Effect of a Change in Concentration

• When a reactant or product is added the system shifts away from that added component.

• If a reactant or product is removed, the system shifts toward the removed component.

Section 17.7


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Effect of a Change in Volume

The system is initially at equilibrium.

The piston is pushed in, decreasing the volume and increasing the pressure. The system shifts in the direction that consumes CO2 molecules, lowering the pressure again.

Section 17.7


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• Decreasing the volume

The system shifts in the direction that gives the fewest number of gas molecules.

Section 17.7


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• Increasing the volume The system shifts in the direction that increases its

pressure.

Section 17.7


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Effect of a Change in Temperature

• The value of K changes with temperature. We can use this to predict the direction of this change.

• Exothermic reaction – produces heat (heat is a product) Adding energy shifts the equilibrium to the left (away

from the heat term).• Endothermic reaction – absorbs energy (heat is a

reactant) Adding energy shifts the equilibrium to the right (away

from the heat term).

Section 17.7


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Effect of Temperature on Equilibrium

NO2(g) N2O4(g)

Section 17.7


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Industrial Application-The Manufacture of Ammonia

N2(g) + 3H2(g) 2NH3(g) ΔH = -92.4 kJ mol-1

To increase production how would you manipulate the equilibrium?

2. Lower Temperature

1. Lower Volume

3. Remove Product

Section 17.7


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Concept Check

Consider the reaction: 2CO2(g) 2CO(g) + O2(g)

How many of the following changes would lead to a shift in the equilibrium position towards the reactant?

I. The removal of CO gas.

II. The addition of O2 gas.

III. The removal of CO2 gas.

IV. Increasing the pressure in the reaction by decreasing the volume of the container.

a) 1

b) 2

c) 3

d) 4

Section 17.7


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Concept Check

One method for the production of hydrogen gas can be described by the following endothermic reaction:

CH4(g) + H2O(g) CO(g) + 3H2(g)

How many of the following changes would decrease the amount of hydrogen gas (H2) produced?

I. H2O(g) is added to the reaction vessel.II. The volume of the container is doubled.III. CH4(g) is removed from the reaction vessel.IV. The temperature is increased in the reaction

vessel.

a) 1

b) 2

c) 3

d) 4

Section 17.8

Applications Involving the Equilibrium Constant

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• A value of K much larger than 1 means that at equilibrium the reaction system will consist of mostly products – the equilibrium lies to the right. Reaction goes essentially to completion.

The Extent of a Reaction

Section 17.8


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• A very small value of K means that the system at equilibrium will consist of mostly reactants – the equilibrium position is far to the left. Reaction does not occur to any

significant extent.

The Extent of a Reaction

Section 17.8


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• The value of K for a system can be calculated from a known set of equilibrium concentrations.

• Unknown equilibrium concentrations can be calculated if the value of K and the remaining equilibrium concentrations are known.

Section 17.8


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Concept Check

If the equilibrium lies to the right, the value for K is __________.

large (or >1)

If the equilibrium lies to the left, the value for K is ___________.

small (or <1)

Section 17.8


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Concept Check

At a given temperature, K = 50 for the reaction:H2(g) + I2(g) 2HI(g)

Calculate the equilibrium concentration of H2 given:[I2] = 1.5 × 10–2 M and [HI] = 5.0 × 10–1 M

a) 1.5 × 10–2 Mb) 3.0 × 10–2 Mc) 5.0 × 10–1 Md) 3.3 × 10–1 M

K = (HI)2/(H2)(I2)

50 = (5.0 × 10–1)2/(H2)(1.5 × 10–2)

(H2) = 3.3 × 10–1 M

Section 17.9

Solubility Equilibria

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• The equilibrium conditions also applies to a saturated solution containing excess solid, MX(s). Ksp = [M+][X] = solubility product constant

The value of the Ksp can be calculated from

the measured solubility of MX(s).

Section 17.9


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• Solubility product (Ksp) – equilibrium constant; has only one value for a given solid at a given temperature.

• Solubility – an equilibrium position.

Bi2S3(s) 2Bi3+(aq) + 3S2–(aq)

2 33+ 2sp = Bi S K

Section 17.9


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Concept Check

If a saturated solution of PbCl2 is prepared by dissolving some of the salt in distilled water and the concentration of Pb2+ is determined to be 1.6 × 10–2 M, what is the value of Ksp?

a) 2.6 × 10–4

b) 2.0 × 10–4

c) 3.2 × 10–2

d) 1.6 × 10–5

Ksp = [Pb2+] [Cl–]2 = (1.6 × 10–2) (3.2 × 10–2)2 = 1.6 × 10–5

Section 17.9


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Concept Check

Calculate the solubility of silver chloride in water.

Ksp = 1.6 × 10–10

a) 1.3 × 10–5 M

b) 1.6 × 10–10 M

c) 3.2 × 10–10 M

d) 8.0 × 10–11 MKsp = [Ag+][Cl–]

1.6 × 10–10 = (x)(x) = x2

x = 1.3 × 10–5 M

chapter 17a reaction rates and equilibrium. chapter 17 table of contents copyright © cengage...

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