first, system reaches equilibrium

First, system reaches equilibrium

Chemical Equilibrium Introduction

1.) Equilibria govern diverse phenomena Protein folding, acid rain action on minerals to aqueous reactions

2.) Chemical equilibrium applies to reactions that can occur in both directions:

reactants are constantly forming products and vice-versa At the beginning of the reaction, the rate that the reactants are changing into the

products is higher than the rate that the products are changing into the reactants. When the net change of the products and reactants is zero the reaction has

reached equilibrium.

At equilibrium the amount of reactants and products are constant, but not necessarily equal

Then, system continually exchanges products and reactants, while maintaining equilibrium distribution.

Reactants Product

Chemical Equilibrium Equilibrium Constant

1.) The relative concentration of products and reactants at equilibrium is a constant.

2.) Equilibrium constant (K): For a general chemical reaction

Equilibrium constant:

Where:- small superscript letters are the stoichiometry coefficients- [A] concentration chemical species A relative to standard state

ba

dc

][][

][][

BA

DCK


2.) Equilibrium constant (K): A reaction is favored when K > 1 K has no units, dimensionless

- Concentration of solutes should be expressed as moles per liter (M).- Concentrations of gases should be expressed in bars.

► express gas as Pgas, emphasize pressure instead of concentration► 1 bar = 105 Pa; 1 atm = 1.01325 bar

- Concentrations of pure solids, pure liquids and solvents are omitted► are unity► standard state is the pure liquid or solid

3.) Manipulating Equilibrium Constants

][

]][[

HA

AHK1

1'1 K/1

AH

HAK ]][[

][

Consider the following reaction:

Reversing the reaction results in a reciprocal equilibrium reaction:

K1


3.) Manipulating Equilibrium Constants

K2

K3

If two reactions are added, the new K is the product of the two individual K values:

]][[

]H][[

]][[

][C

][

]][[

CHA

CA

CH

H

HA

AHKKK 213

][

]][[

HA

AHK1

]][[

][

CH

CHK2

]][[

]H][[

CHA

CAK3


3.) Manipulating Equilibrium Constants Example:

Kw= 1.0 x 10-14

Given the reactions and equilibrium constants:

KNH3= 1.8 x 10-5

Find the equilibrium constant for the reaction:

Solution:

K1= Kw

K2=1/KNH3

K3=Kw*1/KNH3=5.6x10-10

Chemical Equilibrium Equilibrium and Thermodynamics

1.) Equilibrium constant derived from the thermodynamics of a chemical reaction.

deals with the relationships and conversions between heat and other forms of energy

2.) Enthalpy H – is the heat absorbed or released when the reaction takes place under

constant applied pressure

H = Hproducts – Hreactants

Standard enthalpy change (Ho) – all reactants and products are in their standard state. Ho – negative heat released

- Exothermic- Solution gets hot

Ho – positive heat absorbed- Endothermic- Solution gets cold


3.) Entropy Measure of a substances “disorder” Greater disorder Greater Entropy

- Relative disorder: Gas > Liquid > solid

S = Sproducts – Sreactants

So – change in entropy when all species are in standard state.- positive

product more disorder- negative

product less disorder

So = +76.4 J/(K.mol) at 25oC

More disorder for aqueous ions than solid


3.) Entropy Increase in temperature results in an increase in Entropy (S)

Increase occurs for all products and reactants

Primarily concerned with S, which is only weakly temperature dependent- generally treat S and H as temperature independent


4.) Free Energy Systems at constant temperature and pressure have a tendency toward

lower enthalpy and higher entropy

Chemical reaction is favored if:- H is negative heat given offand- S is positive more disorder

Chemical reaction is not favored if:- H is positive and S is negative

Gibbs Free Energy (G): determines if a reaction is favored or not when both H and S are positive or negative- A reaction is favored if G is negative

where T is temperature (Kelvin)

Free energy: G = H -TS


4.) Free Energy Example:

Is the following reaction favored at 25oC?

Ho = -74.85 x 103 J/molSo = -130.4 J/K.mol

Free energy: G = H –TS = (-74.85x103 J/mol) – (298.15K)(-130.4 J/K.mol)

G = -35.97 kJ/mol G negative reaction favored

Favorable influence of enthalpy is greater than unfavorable influence of entropy


5.) Free Energy and Equilibrium Relate Equilibrium constant to the energetics (H & S) of a reaction Equilibrium constant depends on G:

where R (gas constant) = 8.314472 J/(K.mol) T = temperature in kelvins

The more negative G larger equilibrium constant Example:

RTG o

eK

G = -35.97

6)K15.298)(molK/(J314472.8)(mol/J10x97.35(RTG

10x00.2eeK.3

o

Because K is very large, HCl is very soluble in water and nearly completely ionized


5.) Free Energy and Equilibrium If Go is negative or K >1 the reaction is spontaneous

- Reaction occurs by just combining the reactants

If Go is positive or K < 1, the reaction is not spontaneous- Reaction requires external energy or process to proceed

Gas flows towards a vacuum.spontaneous

A vacuum does not naturally form.nonspontaneous

Chemical Equilibrium Le Châtelier’s Principal

1.) What Happens When a System at Equilibrium is Perturbed? Change concentration, temperature, pressure or add other chemicals

Equilibrium is re-established- Reaction accommodates the change in products, reactants, temperature,

pressure, etc.- Rates of forward and reverse reactions re-equilibrate


1.) What Happens When a System at Equilibrium is Perturbed? Le Châtelier’s Principal:

- the direction in which the system proceeds back to equilibrium is such that the change is partially offset.

Consider this reaction:

At equilibrium:

Add excess CO(g):

To return to equilibrium (balance), some (not all) CO and H2 are converted to CH3OH

If all added CO was converted to CH3OH, then reaction would be unbalanced by the amount of product


2.) Example:

Consider this reaction:

C25at101CB

HCBK o11

23-3

8-272

-

]r][rO[

]][Or][r[

At one equilibrium state:

M 0.043][BrOM 1.0][Br

M 0.0030][CrM 0.10]O[CrM 5.0][H

-3

3-272


2.) Example:

What happens when:

M 0.20M 0.10]O[Cr -272 tofromincreased

According to Le Châtelier’s Principal, reaction should go back to left to off-set dichormate on right:

Use reaction quotient (Q), Same form of equilibrium equation, but not at equilibrium:

K1020030.0043.0

0.520.00.1

CB

HCBQ 11

2

8

23-3

8-272

-

]r][rO[

]][Or][r[


2.) Example:

Because Q > K, the reaction must go to the left to decrease numerator and increase denominator.

Continues until Q = K:

1. If the reaction is at equilibrium and products are added (or reactants removed), the reaction goes to the left

2. If the reaction is at equilibrium and reactants are added ( or products removed), the reaction goes to the right


3.) Affect of Temperature on Equilibrium

RS

RTH

RS

RTH

RTSTHRTG

oo

o

oo

ee

e

eeK

Combine Gibbs free energy and Equilibrium Equations:

Only Enthalpy term is temperature dependent:

RTH o

e)T(K

1. Equilibrium constant of an endothermic reaction (Ho = +) increases if the temperature is raised.

2. Equilibrium constant of an exothermic reaction (Ho = -)decreases if the temperature is raised.


3.) Affect of Temperature on Equilibrium

H = +

H = -


4.) Thermodynamics vs. Kinetics Thermodynamics predicts if a reaction will occur

- determines the state at equilibrium

Thermodynamics does not determine the rate of a reaction

- Will the reaction occur instantly, in minutes, hours, days or years?

- While reaction is spontaneous, takes millions of years to occur

G = -

spontaneous

Diamonds Graphite

Chemical Equilibrium Solubility Product

1.) Equilibrium constant for the reaction which a solid salt dissolves to give its constituent ions in solution

Solid omitted from equilibrium constant because it is in a standard state

Example:

18sp 102.1CHK 2-2

2 ]l][g[


1.) Saturated Solution – contains excess, undissolved solid

Solution contains all the solid capable of dissolving under the current conditions

Example:

Find [Cu2+] in a solution saturated with Cu4(OH)6(SO4) if [OH-] is fixed at 1.0x10-6M. Note that Cu4(OH)6(SO4) gives 1 mol of SO4

2- for 4 mol of Cu2+? 691032 .K sp


2.) If an aqueous solution is left in contact with excess solid, the solid will dissolve until the condition of Ksp is satisfied

Amount of undissolved solid remains constant Excess solid is required to guarantee ion concentration is consistent with Ksp

3.) If ions are mixed together such that the concentrations exceed Ksp, the solid will precipitate.

4.) Solubility product only describes part of the solubility of a salt Only includes dissociated ions Ignores solubility of solid salt

Common ion effect – a salt will be less soluble if one of its constituent ions is already present in the solution.

Chemical Equilibrium

Decrease in the solubility of MgF2 by the addition of NaF

PbCl2 precipitate because the ion product is greater than Ksp.

Chemical Equilibrium

Common Ion Effect

1.) Affect of Adding a Second Source of an Ion on Salt Solubility Equilibrium re-obtained following Le Châtelier’s Principal Reaction moves away from the added ion

Find [Cu2+] in a solution saturated with Cu4(OH)6(SO4) if [OH-] is fixed at 1.0x10-6M and 0.10M Na2SO4 is added to the solution.

Chemical Equilibrium Complex Formation

1.) High concentration of an ion may redissolve a solid Ion first causes precipitation Forms complex ions, consists of two or more simple ions bonded to each other

Complex forms and redissolves solid

ppt. formation


2.) Lewis Acids and Bases M+ acts as a Lewis acid accepts a pair of electrons X- acts as a Lewis base donates a pair of electrons Bond is a coordinate covalent bond

ligand adduct

Lewis acid Lewis base


3.) Affect on Solubility Formation of adducts increase solubility

Solubility equation becomes a complex mixture of reactions- don’t need to use all equations to determine the concentration of any species

All equilibrium conditions are satisfied simultaneously

Concentration of Pb2+ that satisfies any one of the equilibria must satisfy all of the equilibria

Only one concentration of Pb2+ in solution

9sp 109.7IPbK 2-2 ]][[Implies low Pb2+ solubility:

Ksp


3.) Affect on Solubility Total concentration is dependent on each individual complex species

2432

2total PbIPbI)aq(PbIPbIPbPb

Total solubility of lead depends on [I-] and the solubility of each individual complex formation.


3.) Affect on Solubility Example:

Given the following equilibria, calculate the concentration of each zinc-containing species in a solution saturated with Zn(OH)2(s) and containing [OH-] at a fixed concentration of 3.2x10-7M.

Zn(OH)2 (s) Ksp = 3.0x10-16

Zn(OH)+ 1 = 2.5 x104

Zn(OH)3- 3 = 7.2x1015

Zn(OH)42- 4 = 2.8x1015

Chemical Equilibrium Acids and Bases

1.) Protic Acids and Bases – transfer of H+ (proton) from one molecule to another

Hydronium ion (H3O+) – combination of H+ with water (H2O)

Acid – is a substance that increases the concentration of H3O+

Base – is a substance that decreases the concentration of H3O+

- base also causes an increase in the concentration of OH- in aqueous solutions

2.) Brønsted-Lowry – definition does not require the formation of H3O+

Extended to non-aqueous solutions or gas phase

Acid – proton donor

Base – proton acceptor

acid base salt

acid


3.) Salts – product of an acid-base reaction Any ionic solid Acid and base neutralize each other and form a salt Most salts with a single positive and negative charge dissociate completely

into ions in water

4.) Conjugate Acids and Bases

Products of acid-base reaction are also acids and bases

A conjugate acid and its base or a conjugate base and its acid in an aqueous system are related to each other by the gain or loss of H+


5.) Autoprotolysis – acts as both an acid and base Extent of these reactions are very small

water14

w 100.1OHHK ]][[ -

Acetic acid15105.3K

- H3O+ is the conjugate acid of water- OH- is the conjugate base of water- Kw is the equilibrium constant for the dissociation of water


6.) pH – negative logarithm of H+ concentration Ignores distinction between concentration and activities (discussed later)

A solution is acidic if [H+] > [OH-] A solution is basic if [H+] < [OH-]

An aqueous solution has a neutral pH if [H+]=[OH-]- This occurs when [H+] = [OH-] = 10-7M or pH = 7

]Hlog[pH

C25at00.14]Klog[pOHpH ow


6.) pH pH values for some common samples


6.) pH Example:

What is the pH of a solution containing 1x10-6 M H+?

What is [OH-] of a solution containing 1x10-6 M H+?


7.) Strengths of Acids and Bases Depends on whether the compound react nearly completely or partially to

produce H+ or OH-

strong acid or base completely dissociate in aqueous solution- equilibrium constants are large- everything else termed weak

Strong no undissociated HCl or KOH


7.) Strengths of Acids and Bases

weak acids react with water by donating a proton- only partially dissociated in water- equilibrium constants are called Ka – acid dissociation constant- Ka is small

weak bases react with water by removing a proton- only partially dissociated in water- equilibrium constants are called Kb – base dissociation constant- Kb is small

][

]][[

HA

AHKa

Ka

Kb

][

]][[

B

OHBHKb

Equivalent

Equivalent

Ka

Kb

ACID FORMULA Ka pKa ACID FORMULA Ka pKa

acetic acid H(C2H3O2) 1.74 E-5 4.76 hydrocyanic acid HCN 6.17 E-10 9.21

ascorbic acid (1) H2(C6H6O6) 7.94 E-5 4.10 hydrofluoric acid HF 6.31 E-4 3.20

ascorbic acid (2) (HC6H6O6)- 1.62 E-12 11.79 lactic acid H(C3H5O3) 8.32 E-4 3.08

boric acid (1) H3BO3 5.37 E-10 9.27 nitrous acid HNO2 5.62 E-4 3.25

boric acid (2) (H2BO3)- 1.8 E-13 12.7 octanoic acid H(C8H15O2) 1.29 E-4 4.89

boric acid (3) (HBO3)= 1.6 E-14 13.8 oxalic acid (1) H2(C204) 5.89 E-2 1.23

butanoic acid H(C4H7O2) 1.48 E-5 4.83 oxalic acid (2) (HC2O4)- 6.46 E-5 4.19

carbonic acid (1) H2CO3 4.47 E-7 6.35 pentanoic acid H(C5H9O2) 3.31 E-5 4.84

carbonic acid (2) (HCO3)- 4.68 E-11 10.33 phosphoric acid (1) H3PO4 6.92 E-3 2.16

chromic acid (1) H2CrO4 1.82 E-1 0.74 phosphoric acid (2) (H2PO4)- 6.17 E-8 7.21

chromic acid (2) (HCrO4)- 3.24 E-7 6.49 phosphoric acid (3) (HPO4)= 2.09 E-12 12.32

citric acid (1) H3(C6H5O7) 7.24 E-4 3.14 propanoic acid H(C3H5O2) 1.38 E-5 4.86

citric acid (2) (H2C6H5O7)- 1.70 E-5 4.77 sulfuric acid (2) (HSO4)- 1.05 E-2 1.98

citric acid (3) (HC6H5O7)= 4.07 E-7 6.39 sulfurous acid (1) H2SO3 1.41 E-2 1.85

formic acid H(CHO2) 1.78 E-4 3.75 sulfurous acid (2) (HSO3)- 6.31 E-8 7.20

heptanoic acid H(C7H13O2) 1.29 E-5 4.89 uric acid H(C5H3N4O3) 1.29 E-4 3.89

hexanoic acid H(C6H11O2) 1.41 E-5 4.84

Chemical Equilibrium Some Common Weak Acids (carboxylic acids)

Chemical Equilibrium Some Common Weak Acids (Metals cations)

Chemical Equilibrium Some Common Weak Bases (amines)

BASE FORMULA Kb pKb

alanine C3H5O2NH2 7.41 E-5 4.13

Ammonia NH3 (NH4OH) 1.78 E-5 4.75

dimethylamine (CH3)2NH 4.79 E-4 3.32

ethylamine C2H5NH2 5.01 E-4 3.30

glycine C2H3O2NH2 6.03 E-5 4.22

hydrazine N2H4 1.26 E-6 5.90

methylamine CH3NH2 4.27 E-4 3.37

trimethylamine (CH3)3N 6.31 E-5 4.20

The Ka or Kb of an acid or base may also be written in terms of “pKa” or “pKb”

As Ka or Kb increase pKa or pKb decrease

- a strong acid/base has a high Ka or Kb and a low pKa or pkb

)Klog(pK aa )Klog(pK bb


8.) Polyprotic Acids and Bases – can donate or accept more than one proton

Ka or Kb are sequentially numbered- Ka1,Ka2,Ka3 Kb1,Kb2,Kb3


8.) Relationship Between Ka and Kb

][

]][[

HA

AHKa

][

]][[

A

OHHAKb

]][[][

]][[

][

]][[

OHHA

OHHA

HA

AH

KKK baw

baw KKK


8.) Relationship Between Ka and Kb Example:

Write the Kb reaction of CN-. Given that the Ka value for HCN is 6.2x10-10, calculate Kb for CN-.

first, system reaches equilibrium

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