This is a continuation of our discussion of Chapter 14: Equilibria

Approximations to Simplify the Math

• When the equilibrium constant is very small, the position of equilibrium favors the reactants

• For relatively large initial concentrations of reactants, the reactant concentration will not change significantly when it reaches equilibrium

[A]eq = ([A]o x) [A]o

The equilibrium concentration of reactant is assumed to be the same as the initial concentration.

Checking the Approximation and Refining as Necessary

• Compare the approximate value of x to the initial concentration

• If the approximate value of x is less than 5% of the initial concentration, the approximation is valid.

Sulfur trioxide decomposes to give sulfur dioxide and molecular oxygen. The equilibrium constant at 300˚C is 1.6 x 10-10. Calculate the equilibrium concentrations if the initial concentration of sulfur trioxide is 0.100M.

For the isomerization reaction of butane to isobutane the equilibrium constant at 25˚C is 2.50. In a 10.0 L bulb at equilibrium the concentrations of butane and isobutane are 0.140 M and 0.350 M respectively.

If 2.00 mol isobutane is added to the equilibrium mixture, what will be the new equilibrium concentrations?

Changes in the Position of

Equilibrium • Once at equilibrium, the concentrations of all the

reactants and products remain the same• If the conditions are changed, the concentrations of

all the chemicals will change until equilibrium is re-established

• The new concentrations will be different, but the equilibrium constant (K) will be the same– unless you change the temperature

Le Châtelier’s Principle:

• If a system at equilibrium is disturbed, the

position of equilibrium will shift to

minimize the disturbance

• Predicts effects changes in conditions have

on position of equilibrium.

• Look at results from the butane isomerization

The Effect of Concentration Changes on Equilibrium

• Adding a reactant: other reactants decrease and products increase until a new position of equilibrium is found

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

• K stays the same!

The Effect of Concentration Changes on Equilibrium

• Equilibrium shifts away from side with added chemicals or toward side with removed chemicals

• Important: adding more of a solid or liquid does not change its concentration – and therefore has no effect on the equilibrium

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

• Adding a gaseous reactant increases its partial pressure and concentration, causing a shift to the products

• Removing a gaseous reactant increases its partial pressure and concentration, causing a shift to the reactants

• Adding an inert gas to the mixture has no effect on the position of equilibrium

The Effect of Concentration Changes on Equilibrium

When N2O4 is added, some of itdecomposes to make more NO2

2 NO2 N2O4

Effect of Volume Change on Equilibrium

• Decreasing the size of the container increases partial pressure of all the gases P increases

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

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

• Solids, liquids, or solutions: changing the size of the container has no effect on the position of equilibrium

The Effect of Temperature Change on Equilibrium

• When T changes, K changes

• Use Le Châtelier’s Principle to predict the effect of temperature changes

The Effect of Temperature Change on Equilibrium

• Exothermic reactions release energy – Write heat as a product in an exothermic reaction– As T increases, K decreases shifting to the reactants

• Endothermic reactions absorb energy – Write heat as a reactant in an endothermic reaction– As T increases, K increases shifting to the products

Catalysts

• Catalysts provide an alternative, more efficient mechanism

• Works for both forward and reverse reactions

• Affects the rate of the forward and reverse reactions by the same factor

• Catalysts do not affect the position of equilibrium

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 re-established?

ü adding more O2 to the container

ü condensing and removing SO3

ü compressing the gasesü cooling the containerü doubling the volume of the containerü warming the mixtureü adding the inert gas helium to the containerü adding a catalyst to the mixture

Chapter 15: Acids and Bases

Properties of Acids• Sour taste• React with “active” metals

Al, Zn, Fe, but not Cu, Ag, or Au

2 Al + 6 HCl AlCl3 + 3 H2

Corrosive

• React with carbonates, producing CO2

marble, baking soda, chalk, limestone

CaCO3 + 2 HCl CaCl2 + CO2 + H2O

• Change color of vegetable dyes blue litmus turns red

• React with bases to form ionic salts (neutralization)

Common AcidsChemical Name Formula Uses Strength

Nitric Acid HNO3 explosive, fertilizer, dye, glue

Strong

Sulfuric Acid H2SO4 explosive, fertilizer, dye, glue, batteries

Strong

Hydrochloric Acid HCl metal cleaning, food prep, ore refining, stomach acid

Strong

Phosphoric Acid H3PO4 fertilizer, plastics & rubber, food preservation

Moderate

Acetic Acid HC2H3O2 plastics & rubber, food preservation, Vinegar

Weak

Hydrofluoric Acid HF metal cleaning, glass etching

Weak

Carbonic Acid H2CO3 soda water Weak

Boric Acid H3BO3 eye wash Weak

Structures of Acids

• Binary acids have acid hydrogens attached to a nonmetal atom– HCl, HF

Structure of Acids• oxy acids have acid hydrogens attached

to an oxygen atom– H2SO4, HNO3

Structure of Acids• Carboxylic acids have COOH group

– HC2H3O2, H3C6H5O7

• Only the first H in the formula is acidic– the H is on the COOH

Properties of Bases• Also called alkalis• Tastes bitter

– alkaloids = plant product that is alkaline• often poisonous

• Solutions feel slippery• Changes color of vegetable dyes

– red litmus turns blue

• React with acids to form ionic salts (neutralization)

Common BasesChemical

Name

Formula Common

Name

Uses Strength

sodium hydroxide

NaOH lye,

caustic soda

soap, plastic, petrol refining

Strong

potassium hydroxide

KOH caustic potash

soap, cotton, electroplating

Strong

calcium hydroxide

Ca(OH)2 slaked lime cement Strong

sodium

bicarbonate

NaHCO3 baking soda cooking, antacid Weak

magnesium hydroxide

Mg(OH)2 milk of magnesia

antacid Weak

ammonium hydroxide

NH4OH, {NH3(aq)}

ammonia water

detergent, fertilizer,

explosives, fibers

Weak

Structure of Bases

• Most ionic bases contain OH ions– NaOH, Ca(OH)2

• Some contain CO32- ions

– CaCO3 NaHCO3

• Molecular bases contain structures that react with H+

– NH3, other amine groups

Arrhenius Theory

• Bases dissociate in water to produce OH- ions and cations

NaOH(aq) → Na+(aq) + OH–(aq)• Acids ionize in water to produce H+ ions

and anionsHCl(aq) → H+(aq) + Cl–(aq)

HC2H3O2(aq) → H+(aq) + C2H3O2–(aq)

Arrhenius Acid-Base Reactions

• H+ from the acid combines with the OH- from the base to make a molecule of H2O

• The cation from the base combines with the anion from the acid to make a salt

acid + base → salt + water

HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)

Problems with Arrhenius Theory

• Can’t explain molecular bases like NH3 that don’t contain OH–

• Can’t explain why molecular substances, like CO2, dissolve in water to form acidic solutions – even though they do not contain H+ ions

• Does not include acid-base reactions that take place outside aqueous solution

Brønsted-Lowry Theory• In a Brønsted-Lowry Acid-Base reaction, an H+ is

transferred- Does not have to take place in aqueous solution- Broader definition than Arrhenius

• Acid is H donor, base is H acceptor- Base must have an unshared pair of electrons

• In an acid-base reaction, the acid molecule gives an H+ to the base molecule

H–A + :B :A– + H–B+

Hydronium Ion• H+ ions produced by the acid don’t exist

alone in water• They react with a water molecule(s) to

produce complex ions, mainly hydronium ion, H3O+

H+ + H2O H3O+

– there are also minor amounts of H+ with multiple water molecules, H(H2O)n

+

Brønsted-Lowry Acids• Acids are H+ donors

-Needs an H to be a Brønsted-Lowry acid

• HCl(aq) is acidic because HCl transfers an H+ to H2O, forming H3O+ ions-water acts as base, accepting H+

HCl(aq) + H2O(l) → Cl–(aq) + H3O+(aq)acid base

Brønsted-Lowry Bases• Brønsted-Lowry bases are H+ acceptors

-Needs lone pairs to act as a Brønsted-Lowry base

• NH3(aq) is basic because NH3 accepts an H+ from H2O, forming OH–(aq)-water acts as acid, donating H+

NH3(aq) + H2O(l) NH4+(aq) + OH–(aq)

base acid

Amphoteric Substances

• amphoteric substances can act as either an acid or a base– have both transferable H and atom with lone

pair

• Water acts as base, accepting H+ from HCl

HCl(aq) + H2O(l) → Cl–(aq) + H3O+(aq)• Water acts as acid, donating H+ to NH3

NH3(aq) + H2O(l) NH4+(aq) + OH–(aq)

Polyprotic Acids• polyprotic acids --Acid molecules with more than one

ionizable H – HCl, monoprotic,

– H2SO4, diprotic,

– H3PO4, triprotic

• Ionize in steps, each ionizable H removed sequentially

• Removing of the first H automatically makes removal of the second H harder – H2SO4 is a stronger acid than HSO4

Conjugate Pairs• A Brønsted-Lowry acid becomes a Brønsted-

Lowry base after it donates a proton.

• A Brønsted-Lowry base becomes a Brønsted-Lowry acid after it accepts a proton.

• Each reactant and the product it becomes is called a conjugate acid/base pair

Identify the Brønsted-Lowry Acids and Bases and Their Conjugates in the Reactions

H2SO4 + H2O HSO4– + H3O+

NH3 + H2O NH4++ OH-

Write the formula for the conjugate acid of the following

Base

H2O

NH3

CO32−

H2PO41−

Write the formula for the conjugate base of the following

Acid

H2O

NH3

CO32−

H2PO41−

Strength of Acids and Bases

•The strength of an acid is inversely related to the strength of its conjugate base

--The stronger the acid, the weaker it’s conjugate base

•The strength of an acid is inversely related to the strength of its conjugate base

--The stronger the base, the weaker it’s conjugate acid

Incr

easi

ng A

cidi

ty

Increasing Basicity

Acids Conjugate Bases HClO4 ClO4

-1 H2SO4 HSO4

-1

HI I-1

HBr Br-1

HCl Cl-1

HNO3 NO3-1

H3O+1 H2O

HSO4-1 SO4

-2

H2SO3 HSO3-1

H3PO4 H2PO4-1

HNO2 NO2-1

HF F-1

HC2H3O2 C2H3O2-1

H2CO3 HCO3-1

H2S HS-1

NH4+1 NH3

HCN CN-1

HCO3-1 CO3

-2

HS-1 S-2

H2O OH-1

CH3-C(O)-CH3 CH3-C(O)-CH2-1

NH3 NH2-1

CH4 CH3-1

OH-1 O-2

Acid Ionization Constant, Ka

• For the reaction of an acid with water

HA + H2O A-1 + H3O+1

• The equilibrium constant is called the acid ionization constant, Ka

– larger Ka = stronger acid

€

Ka =[A−1] ×[H3O

+1]

[HA]

Name Formula Ka1 Ka2 Ka3

Benzoic C6H5COOH 6.14 x 10-5

Propanoic CH3CH2COOH 1.34 x 10-5

Formic HCOOH 1.77 x 10-5

Acetic CH3COOH 1.75 x 10-5

Chloroacetic ClCH2COOH 1.36 x 10-5

Trichloroacetic Cl3C-COOH 1.29 x 10-4

Oxalic HOOC-COOH 5.90 x 10-2 6.40 x 10-5

Nitric HNO3 strong

Nitrous HNO2 4.6 x 10-4

Phosphoric H3PO4 7.52 x 10-3 6.23 x 10-8 2.2 x 10-13

Phosphorous H3PO3 1.00 x 10-2 2.6 x 10-7

Arsenic H3AsO4 6.0 x 10-3 1.05 x 10-7 3.0 x 10-12

Arsenious H3AsO3 6.0 x 10-10 3.0 x 10-14 very small

Perchloric HClO4 > 108

Chloric HClO3 5 x 102

Chlorous HClO2 1.1 x 10-2

Hypochlorous HClO 3.0 x 10-8

Boric H3BO3 5.83 x 10-10

Carbonic H2CO3 4.45 x 10-7 4.7 x 10-11

Base Equilibrium Constant, Kb

• For the reaction of a base with water

B + H2O OH- + BH+

• The equilibrium constant is called the base constant, Kb

– larger Kb = stronger base

€

Kb =[OH−] ×[BH+]

[B]

Autoionization of Water

• ~ 1 out of every 106 water molecules form ions through a process called autoionization

H2O H+ + OH–

H2O + H2O H3O+ + OH–

• All aqueous solutions contain both H3O+ and OH–

– H3O+ and OH– are equal in water

– [H3O+] = [OH–] = 10-7M @ 25°C

Ion Product of Water• The equilibrium constant for the

autoionization of water is called the ion product of water and has the symbol Kw

• Kw =[H3O+] [OH–] = 1 x 10-14 at 25°C

• As [H3O+] , [OH–] (so the product stays constant)

Acidic and Basic Solutions• Neutral solutions :

[H3O+] = [OH–] = 1 x 10-7

• Acidic solutions:

[H3O+] > 1 x 10-7 [OH–] < 1 x 10-7

• Basic solutions:

[H3O+] < 1 x 10-7 [OH–] > 1 x 10-7

Calculate the [OH] at 25°C when the [H3O+] = 1.5 x 10-9 M, and determine if the solution is acidic, basic, or neutral

pH• The acidity/basicity of a solution is often

expressed as pH

• pH = -log[H3O+] [H3O+] = 10-pH

• pH < 7 is acidic; pH > 7 is basic, pH = 7 is neutral

Sig. Figs. & Logs• When you take the log of a number written in scientific

notation, the digit(s) before the decimal point come from the exponent on 10, and the digits after the decimal point come from the decimal part of the number

log(2.0 x 106) = log(106) + log(2.0)= 6 + 0.30303… = 6.30303...

• The part of the scientific notation number that determines the significant figures is the decimal part, the sig figs are the digits after the decimal point in the log

log(2.0 x 106) = 6.30