chapter 14 powerpoint Cahng

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Chemical EquilibriumAcid and bases

equilibriaequilibria

Chapter 5

Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

Chapter 5

Equilibrium is a state in which there are no observable changes as time goes by.

Chemical equilibrium is achieved when:

the rates of the forward and reverse reactions are equal and• the rates of the forward and reverse reactions are equal and

• the concentrations of the reactants and products remain constant

Physical equilibrium

H O (l) H O (g)H2O (l)

Chemical equilibrium

N2O4 (g)

14.1

H2O (g)

2NO2 (g)

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N2O4 (g) 2NO2 (g)

equilibrium

equilibrium

equilibriumequilibrium

Start with NO2 Start with N2O4 Start with NO2 & N2O4

14.1

constant

14.1

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N2O4 (g) 2NO2 (g)

= 4.63 x 10-3K = [NO2]2

[N2O4]

aA + bB cC + dD

K = [C]c[D]d

[A]a[B]bLaw of Mass Action

K >> 1

K << 1

Lie to the right Favor products

Lie to the left Favor reactants

Equilibrium Will

14.1

Homogenous equilibrium applies to reactions in which all reacting species are in the same phase.

N2O4 (g) 2NO2 (g)

[NO ]2 NOP2

Kc = [NO2]2

[N2O4]Kp =

NO2P

N2O4P

In most cases

Kc ≠ Kp

aA (g) + bB (g) cC (g) + dD (g)(g) (g) (g) (g)

14.2

Kp = Kc(RT)∆n

∆n = moles of gaseous products – moles of gaseous reactants= (c + d) – (a + b)

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Homogeneous Equilibrium

CH3COOH (aq) + H2O (l) CH3COO- (aq) + H3O+ (aq)

[CH COO ][H O+]Kc =‘ [CH3COO-][H3O+]

[CH3COOH][H2O][H2O] = constant

Kc = [CH3COO-][H3O+]

[CH3COOH] = Kc [H2O]‘

General practice not to include units for the equilibrium constant.

14.2

The equilibrium concentrations for the reaction between carbon monoxide and molecular chlorine to form COCl2 (g) at 740C are [CO] = 0.012 M, [Cl2] = 0.054 M, and [COCl2] =0.14 M. Calculate the equilibrium constants Kc and Kp.

CO (g) + Cl2 (g) COCl2 (g)CO (g) + Cl2 (g) COCl2 (g)

Kc = [COCl2]

[CO][Cl2]=

0.140.012 x 0.054

= 220

Kp = Kc(RT)∆n

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

Kp = 220 x (0.0821 x 347)-1 = 7.7

14.2

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The equilibrium constant Kp for the reaction

is 158 at 1000K. What is the equilibrium pressure of O2 if the PNO = 0.400 atm and PNO = 0.270 atm?2

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

PNO PO2

Kp = 2PNO PO

PNO2

2

PO2 = KpPNO

22

PNO2

14.2

PO2 = 158 x (0.400)2/(0.270)2 = 347 atm

Heterogenous equilibrium applies to reactions in which reactants and products are in different phases.

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

Kc =‘ [CaO][CO2][CaCO3]

[CaCO3] = constant[CaO] = constant

Kc = [CO2] = Kc x‘ [CaCO3][CaO] Kp = PCO2

The concentration of solids and pure liquids are not included in the expression for the equilibrium constant.

14.2

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

PCO2 = Kp

PCO2 does not depend on the amount of CaCO3 or CaO

14.2

Consider the following equilibrium at 295 K:

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

NH4HS (s) NH3 (g) + H2S (g)

Kp = PNH3 H2SP = 0.265 x 0.265 = 0.0702

Kp = Kc(RT)∆n

Kc = Kp(RT)-∆n

∆n = 2 – 0 = 2 T = 295 K

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

14.2

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A + B C + D

C + D E + F

A + B E + F

Kc =‘ [C][D][A][B] Kc =‘‘ [E][F]

[C][D]

[E][F][A][B]

Kc =

Kc‘Kc‘‘Kc

Kc = Kc‘‘Kc‘ x

If a reaction can be expressed as the sum of two or more reactions, the equilibrium constant for the o erall reaction is gi en bconstant for the overall reaction is given by the product of the equilibrium constants of the individual reactions.

14.2

N2O4 (g) 2NO2 (g)

= 4.63 x 10-3K = [NO2]2

[N2O4]

2NO2 (g) N2O4 (g)

K = [N2O4][NO2]2

‘ =1K = 216

When the equation for a reversible reaction is written in the opposite direction, the equilibrium constant becomes the reciprocal of the original equilibrium constant.

14.2

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Writing Equilibrium Constant Expressions

• The concentrations of the reacting species in the condensed phase are expressed in M. In the gaseous phase, the concentrations can be expressed in M or in atm.

Th t ti f lid li id d l t• The concentrations of pure solids, pure liquids and solvents do not appear in the equilibrium constant expressions.

• The equilibrium constant is a dimensionless quantity.

• In quoting a value for the equilibrium constant, you must specify the balanced equation and the temperature.

• If a reaction can be expressed as a sum of two or more reactions, the equilibrium constant for the overall reaction is given by the product of the equilibrium constants of the individual reactions.

14.2

Chemical Kinetics and Chemical Equilibrium

A + 2B AB2

kf

kr

ratef = kf [A][B]2

rater = kr [AB2]

Equilibriumratef = rater

kf [A][B]2 = kr [AB2]

kf [AB2]= K =

14.3

kr [A][B]2= Kc =

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The reaction quotient (Qc) is calculated by substituting the initial concentrations of the reactants and products into the equilibrium constant (Kc) expression.

IF• Qc > Kc system proceeds from right to left to reach equilibriumc c y g

• Qc = Kc the system is at equilibrium

• Qc < Kc system proceeds from left to right to reach equilibrium

14.4

Calculating Equilibrium Concentrations

1. Express the equilibrium concentrations of all species in terms of the initial concentrations and a single unknown x, e s o e a co ce a o s a d a s g e u o ,which represents the change in concentration.

2. Write the equilibrium constant expression in terms of the equilibrium concentrations. Knowing the value of the equilibrium constant, solve for x.

3. Having solved for x, calculate the equilibrium concentrations of all species.

14.4

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At 12800C the equilibrium constant (Kc) for the reaction

Is 1.1 x 10-3. If the initial concentrations are [Br2] = 0.063 M and [Br] = 0.012 M, calculate the concentrations of these species at equilibrium.

Br2 (g) 2Br (g)

Br2 (g) 2Br (g)

Let x be the change in concentration of Br2

Initial (M)

Change (M)

0.063 0.012

-x +2xEquilibrium (M) 0.063 - x 0.012 + 2x

[Br]2[Br2]

Kc = Kc = (0.012 + 2x)2

0.063 - x = 1.1 x 10-3 Solve for x

14.4

Kc = (0.012 + 2x)2

0.063 - x = 1.1 x 10-3

4x2 + 0.048x + 0.000144 = 0.0000693 – 0.0011x4x2 + 0.0491x + 0.0000747 = 0

ax2 + bx + c =0 -b ± b2 – 4ac√2x =ax + bx + c 0 2ax

Br2 (g) 2Br (g)

Initial (M)

Change (M)

0.063 0.012

-x +2x

x = -0.00178x = -0.0105

Change (M)

Equilibrium (M)

x +2x0.063 - x 0.012 + 2x

At equilibrium, [Br] = 0.012 + 2x = -0.009 M or 0.00844 MAt equilibrium, [Br2] = 0.062 – x = 0.0648 M

14.4

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If an external stress is applied to a system at equilibrium, the system adjusts in such a way that the stress is partially offset as the system reaches a new equilibrium position.

Le Châtelier’s Principle

Ch i C t ti• Changes in Concentration

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

AddEquilibrium shifts left to NH3shifts left to offset stress

14.5


• Changes in Concentration continuedAddAddRemove Remove

Change Shifts the Equilibrium

aA + bB cC + dD

Increase concentration of product(s) leftDecrease concentration of product(s) right

Decrease concentration of reactant(s)Increase concentration of reactant(s) right

left14.5

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• Changes in Volume and Pressure

A (g) + B (g) C (g)

Change Shifts the Equilibrium

Increase pressure Side with fewest moles of gasDecrease pressure Side with most moles of gasI l Sid i h l fDecrease volumeIncrease volume Side with most moles of gas

Side with fewest moles of gas

14.5


• Changes in Temperature

Change Exothermic Rx Endothermic Rx

Increase temperature K decreasesDecrease temperature K increases

K increasesK decreases

14.5colder hotter

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• Adding a Catalyst• does not change K• does not shift the position of an equilibrium system• system will reach equilibrium sooner


y q

uncatalyzed catalyzed

14.5

Catalyst lowers Ea for both forward and reverse reactions.

Catalyst does not change equilibrium constant or shift equilibrium.

Chemistry In Action

Life at High Altitudes and Hemoglobin Production

Kc = [HbO2]

[Hb][O2]

Hb (aq) + O2 (aq) HbO2 (aq)

[ ][ 2]

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Chemistry In Action: The Haber Process

N2 (g) + 3H2 (g) 2NH3 (g) ∆H0 = -92.6 kJ/mol


Change Shift EquilibriumChange Equilibrium

Constant

Concentration yes noConcentration yes no

Pressure yes no

Volume yes no

Temperature yes yes

Catalyst no noCatalyst no no

14.5

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Acids and BasesAcids and Bases

Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

Acids

Have a sour taste. Vinegar owes its taste to acetic acid. Citrusfruits contain citric acidfruits contain citric acid.

React with certain metals to produce hydrogen gas.

React with carbonates and bicarbonates to produce carbon dioxide gas

Bases

Have a bitter taste.

Feel slippery. Many soaps contain bases.

Bases

4.3

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Arrhenius acid is a substance that produces H+ (H3O+) in water

Arrhenius base is a substance that produces OH- in water

4.3

A Brønsted acid is a proton donorA Brønsted base is a proton acceptor

acidbase acid base

15.1

acid conjugatebasebase conjugate

acid

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Acid-Base Properties of Water

H2O (l) H+ (aq) + OH- (aq)

autoionization of water

O

H

H + O

H

H O

H

H H OH-+[ ] +

base conjugate

H2O + H2O H3O+ + OH-

acid conjugatebase

base j gacid

15.2

H2O (l) H+ (aq) + OH- (aq)

The Ion Product of Water

Kc =[H+][OH-]

[H2O] [H2O] = constant

K [H2O] = K = [H+][OH-]Kc[H2O] Kw [H ][OH ]

The ion-product constant (Kw) is the product of the molar concentrations of H+ and OH- ions at a particular temperature.

[H+] [OH ]Solution Is

t lAt 250C

Kw = [H+][OH-] = 1.0 x 10-14

[H+] = [OH-][H+] > [OH-][H+] < [OH-]

neutralacidicbasic

15.2

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What is the concentration of OH- ions in a HCl solution whose hydrogen ion concentration is 1.3 M?

Kw = [H+][OH-] = 1.0 x 10-14

[H+] = 1.3 M

[OH-] =Kw

[H+]1 x 10-14

1.3= = 7.7 x 10-15 M

15.2

pH – A Measure of Acidity

pH = -log [H+]

[H+] = [OH-][H+] > [OH-][H+] < [OH-]

Solution Isneutralacidicbasic

[H+] = 1 x 10-7

[H+] > 1 x 10-7

[H+] < 1 x 10-7

pH = 7pH < 7pH > 7

At 250C

pH [H+]

15.3

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OH l [OH ]pOH = -log [OH-]

[H+][OH-] = Kw = 1.0 x 10-14

-log [H+] – log [OH-] = 14.00

pH + pOH = 14 00

15.3

pH + pOH 14.00

The pH of rainwater collected in a certain region of the northeastern United States on a particular day was 4.82. What is the H+ ion concentration of the rainwater?

pH = -log [H+]

[H+] = 10-pH = 10-4.82 = 1.5 x 10-5 M

The OH- ion concentration of a blood sample is 2.5 x 10-7 M. What is the pH of the blood?

pH + pOH = 14.00

pOH = -log [OH-] = -log (2.5 x 10-7) = 6.60

pH = 14.00 – pOH = 14.00 – 6.60 = 7.40

15.3

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Strong Electrolyte – 100% dissociation

NaCl (s) Na+ (aq) + Cl- (aq)H2O

Weak Electrolyte – not completely dissociatedWeak Electrolyte not completely dissociated

CH3COOH CH3COO- (aq) + H+ (aq)

Strong Acids are strong electrolytes

HCl (aq) + H O (l) H O+ (aq) + Cl- (aq)HCl (aq) + H2O (l) H3O (aq) + Cl (aq)

HNO3 (aq) + H2O (l) H3O+ (aq) + NO3- (aq)

HClO4 (aq) + H2O (l) H3O+ (aq) + ClO4- (aq)

H2SO4 (aq) + H2O (l) H3O+ (aq) + HSO4- (aq)

15.4

HF (aq) + H2O (l) H3O+ (aq) + F- (aq)

Weak Acids are weak electrolytes


HSO H O H O SO 2HSO4- (aq) + H2O (l) H3O+ (aq) + SO4

2- (aq)

H2O (l) + H2O (l) H3O+ (aq) + OH- (aq)

Strong Bases are strong electrolytes

NaOH (s) Na+ (aq) + OH- (aq)H2O( ) ( q) ( q)

KOH (s) K+ (aq) + OH- (aq)H2O

Ba(OH)2 (s) Ba2+ (aq) + 2OH- (aq)H2O

15.4

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F- (aq) + H2O (l) OH- (aq) + HF (aq)

Weak Bases are weak electrolytes

NO2- (aq) + H2O (l) OH- (aq) + HNO2 (aq)

Conjugate acid-base pairs:

• The conjugate base of a strong acid has no measurable strength.

• H3O+ is the strongest acid that can exist in aqueous solution.

• The OH- ion is the strongest base that can exist in aqeous solution.

15.4

15.4

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Strong Acid Weak Acid

15.4

What is the pH of a 2 x 10-3 M HNO3 solution?

HNO3 is a strong acid – 100% dissociation.


Start 0.002 M 0.0 M 0.0 M3 ( q) 2 ( ) 3 ( q) 3 ( q)

pH = -log [H+] = -log [H3O+] = -log(0.002) = 2.7

End 0.002 M 0.002 M0.0 M

What is the pH of a 1.8 x 10-2 M Ba(OH)2 solution?

Ba(OH)2 is a strong base – 100% dissociation.( )2 g

Ba(OH)2 (s) Ba2+ (aq) + 2OH- (aq)Start

End

0.018 M

0.018 M 0.036 M0.0 M

0.0 M 0.0 M

pH = 14.00 – pOH = 14.00 + log(0.036) = 12.5615.4

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HA (aq) + H2O (l) H3O+ (aq) + A- (aq)

Weak Acids (HA) and Acid Ionization Constants

HA (aq) H+ (aq) + A- (aq)

Ka =[H+][A-]

[HA]

Ka is the acid ionization constant

Kaweak acidstrength

15.5

15.5

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What is the pH of a 0.5 M HF solution (at 250C)?

HF (aq) H+ (aq) + F- (aq) Ka =[H+][F-]

[HF] = 7.1 x 10-4

HF (aq) H+ (aq) + F- (aq)

Initial (M)

Change (M)

Equilibrium (M)

0.50 0.00

-x +x0.50 - x

0.00

+xx x

Ka =x2

0 50 - x = 7.1 x 10-4 0.50 – x ≈ 0.50Ka << 10.50 x

Ka ≈x2

0.50 = 7.1 x 10-4 x2 = 3.55 x 10-4 x = 0.019 M

[H+] = [F-] = 0.019 M pH = -log [H+] = 1.72[HF] = 0.50 – x = 0.48 M 15.5

When can I use the approximation?

0.50 – x ≈ 0.50Ka << 1

When x is less than 5% of the value from which it is subtracted.L th 5%

x = 0.019 0.019 M0.50 M x 100% = 3.8%

Less than 5%Approximation ok.

What is the pH of a 0.05 M HF solution (at 250C)?

Ka ≈x2

0 05 = 7.1 x 10-4 x = 0.006 M0.05

0.006 M0.05 M x 100% = 12%

More than 5%Approximation not ok.

Must solve for x exactly using quadratic equation or method of successive approximation. 15.5

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Solving weak acid ionization problems:

1. Identify the major species that can affect the pH.

• In most cases, you can ignore the autoionization of water.

Ignore [OH-] because it is determined by [H+]• Ignore [OH-] because it is determined by [H+].

2. Use ICE to express the equilibrium concentrations in terms of single unknown x.

3. Write Ka in terms of equilibrium concentrations. Solve for xby the approximation method. If approximation is not valid, solve for x exactlysolve for x exactly.

4. Calculate concentrations of all species and/or pH of the solution.

15.5

What is the pH of a 0.122 M monoprotic acid whose Ka is 5.7 x 10-4?


Initial (M) 0.122 0.00 0.00

Change (M)

Equilibrium (M)

-x +x0.122 - x

+xx x

Ka =x2

0.122 - x= 5.7 x 10-4

x2

0.122 – x ≈ 0.122Ka << 1

Ka ≈x

0.122 = 5.7 x 10-4 x2 = 6.95 x 10-5 x = 0.0083 M

0.0083 M0.122 M x 100% = 6.8%

More than 5%Approximation not ok.

15.5

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Ka =x2

0.122 - x= 5.7 x 10-4 x2 + 0.00057x – 6.95 x 10-5 = 0

ax2 + bx + c =0 -b ± b2 – 4ac√2ax =

x = 0 0081 x = 0 0081x = 0.0081 x = - 0.0081


Initial (M)

Change (M)

Equilibrium (M)

0.122 0.00

-x +x0 122 x

0.00

+xx xEquilibrium (M) 0.122 - x x x

[H+] = x = 0.0081 M pH = -log[H+] = 2.09

15.5

percent ionization = Ionized acid concentration at equilibriumInitial concentration of acid

x 100%

For a monoprotic acid HA

Percent ionization = [H+]

[HA]0x 100% [HA]0 = initial concentration

15.5

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NH3 (aq) + H2O (l) NH4+ (aq) + OH- (aq)

Weak Bases and Base Ionization Constants

Kb =[NH4

+][OH-][NH ]Kb [NH3]

Kb is the base ionization constant

Kbweak base

strength

15.6

Solve weak base problems like weak acids except solve for [OH-] instead of [H+].

15.6

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Ionization Constants of Conjugate Acid-Base Pairs


A- (aq) + H2O (l) OH- (aq) + HA (aq)

Ka

Kb

H2O (l) H+ (aq) + OH- (aq) Kw

KaKb = Kw

Weak Acid and Its Conjugate Base

15.7

Weak Acid and Its Conjugate Base

Ka = KwKb

Kb = KwKa

15.8

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Molecular Structure and Acid Strength

H X H+ + X-

The stronger

The weaker g

the bond the acid

HF << HCl < HBr < HI

15.9


Z O H Z O- + H+δ- δ+

The O-H bond will be more polar and easier to break if:

• Z is very electronegative or

• Z is in a high oxidation state

15.9

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1. Oxoacids having different central atoms (Z) that are from the same group and that have the same oxidation number.

Acid strength increases with increasing electronegativity of ZAcid strength increases with increasing electronegativity of Z

H O Cl O

O••

•••••• ••••

••

••••

H O Br O

O••

•••••• ••••

••

••••Cl is more electronegative than Br

HClO3 > HBrO3

15.9


2. Oxoacids having the same central atom (Z) but different numbers of attached groups.

Acid strength increases as the oxidation number of Z increasesAcid strength increases as the oxidation number of Z increases.

HClO4 > HClO3 > HClO2 > HClO4 3 2

15.9

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Acid-Base Properties of SaltsNeutral Solutions:

Salts containing an alkali metal or alkaline earth metal ion (except Be2+) and the conjugate base of a strongacid (e.g. Cl-, Br-, and NO3

-).3

NaCl (s) Na+ (aq) + Cl- (aq)H2O

Basic Solutions:

Salts derived from a strong base and a weak acid.

H ONaCH3COOH (s) Na+ (aq) + CH3COO- (aq)H2O

CH3COO- (aq) + H2O (l) CH3COOH (aq) + OH- (aq)

15.10

Acid-Base Properties of SaltsAcid Solutions:

Salts derived from a strong acid and a weak base.

NH Cl ( ) NH + ( ) + Cl ( )H2ONH4Cl (s) NH4+ (aq) + Cl- (aq)2

NH4+ (aq) NH3 (aq) + H+ (aq)

Salts with small, highly charged metal cations (e.g. Al3+, Cr3+, and Be2+) and the conjugate base of a strong acid.

Al(H2O)6 (aq) Al(OH)(H2O)5 (aq) + H+ (aq)3+ 2+

15.10

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Acid Hydrolysis of Al3+

15.10

Acid-Base Properties of SaltsSolutions in which both the cation and the anion hydrolyze:

• Kb for the anion > Ka for the cation, solution will be basic

• Kb for the anion < Ka for the cation, solution will be acidic

• Kb for the anion ≈ Ka for the cation, solution will be neutral

15.10

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Oxides of the Representative ElementsIn Their Highest Oxidation States

15.11

CO2 (g) + H2O (l) H2CO3 (aq)

N2O5 (g) + H2O (l) 2HNO3 (aq)

Arrhenius acid is a substance that produces H+ (H3O+) in water

A Brønsted acid is a proton donor

A L i id i b t th t t i f l t

Definition of An Acid

A Lewis acid is a substance that can accept a pair of electrons

A Lewis base is a substance that can donate a pair of electrons

H+ H O H••••

+ OH-••••••acid base

N H••

H

H

H+ +

acid base 15.12

N H

H

H

H+

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Lewis Acids and Bases

N H••

H

F B

F

+ F B

F

N H

H

N H

H

acid base

F F

N H

H

No protons donated or accepted!No protons donated or accepted!

15.12

Chemistry In Action: Antacids and the Stomach pH Balance

NaHCO3 (aq) + HCl (aq)NaCl (aq) + H2O (l) + CO2 (g)

Mg(OH)2 (s) + 2HCl (aq)MgCl2 (aq) + 2H2O (l)

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