chemical equilibrium - wordpress.com · 2018. 12. 18. · at equilibrium the amount of reactants...

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


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




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

+−

=




➢ 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


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➢ DH – is the heat absorbed or released when the reaction takes place under

constant applied pressure

DH = Hproducts – Hreactants

➢ Standard enthalpy change (DHo) –

all reactants and products are

in their standard state.

➢ DHo – negative → heat released

- Exothermic

- Solution gets hot

➢ DHo – positive → heat absorbed

- Endothermic

- Solution gets cold



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

➢ Greater disorder → Greater Entropy

- Relative disorder: Gas > Liquid > solid

DS = Sproducts – Sreactants

➢ DSo – change in entropy when all species are in standard state.

- positive→

product more disorder

- negative →

product less disorder

DSo = +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 DS, which is only weakly temperature dependent

- generally treat DS and DH 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:- DH is negative → heat given off

and

-DS is positive → more disorder

➢ Chemical reaction is not favored if:

- DH is positive and DS is negative

➢ Gibbs Free Energy (DG): determines if a reaction is favored or not when both

DH and DS are positive or negative

- A reaction is favored if DG is negative

where T is temperature (Kelvin)

Free energy: DG = DH -TDS



4.) Free Energy➢ Example:

Is the following reaction favored at 25oC?

DHo = -74.85 x 103 J/mol

DSo = -130.4 J/K.mol

Free energy: DG = DH –TDS = (-74.85x103 J/mol) – (298.15K)(-130.4 J/K.mol)

DG = -35.97 kJ/mol → DG 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 (DH & DS) of a reaction

➢ Equilibrium constant depends on DG:

where

R (gas constant) = 8.314472 J/(K.mol)

T = temperature in kelvins

➢ The more negative DG → larger equilibrium constant

➢ Example:

RTG o

eKD−

=

DG = -35.97

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

10x00.2eeK.3

o

=== −−−D

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



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

- Reaction occurs by just combining the reactants

➢ If DGo is positive or K < 1, the reaction is not spontaneous

- Reaction requires external energy or process to proceed

Gas flows towards a vacuum.

spontaneousA vacuum does not naturally form.

nonspontaneous


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:

( )( )( )

( )( )K102

0030.0043.0

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

RTSTH

RTG

oo

o

oo

ee

e

eeK

DD

DD

DDD

=

=

==

−

+−

−−−

Combine Gibbs free energy and Equilibrium Equations:

Only Enthalpy term is temperature dependent:

RTH o

e)T(KD−

1. Equilibrium constant of an endothermic reaction (DHo = +) increases if

the temperature is raised.

2. Equilibrium constant of an exothermic reaction (DHo = -)decreases if the

temperature is raised.



3.) Affect of Temperature on Equilibrium

DH = +

DH = -

D

D



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

DG = -

spontaneous

Diamonds Graphite


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

−+ == 2-22 ]l][g[


Solubility Product

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 SO42- for

4 mol of Cu2+?

691032 −= .K sp


Solubility Product

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.


Decrease in the solubility of MgF2 by

the addition of NaFPbCl2 precipitate because the

ion product is greater than Ksp.


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.


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


Complex 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


Complex Formation

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


Complex Formation

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.


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)+ b1 = 2.5 x104

Zn(OH)3- b3 = 7.2x1015

Zn(OH)42- b4 = 2.8x1015


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


Acids and Bases

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+


Acids and Bases

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

water14

w 100.1OHHK−+ == ]][[ -

Acetic acid15

105.3K−=

- H3O+ is the conjugate acid of water

- OH- is the conjugate base of water

- Kw is the equilibrium constant for the dissociation of water


Acids and Bases

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[pOHpHo

w =−=+


Acids and Bases

6.) pH➢ pH values for some common samples


Acids and Bases

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+?


Acids and Bases

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


Acids and Bases

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


Some Common Weak Acids (carboxylic acids)


Some Common Weak Acids (Metals cations)


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 −=


Acids and Bases

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


Acids and Bases

8.) Relationship Between Ka and Kb

][

]][[

HA

AHKa

−+

=

][

]][[

−

−

=A

OHHAKb

]][[][

]][[

][

]][[ −+

−

−−+

==

=

OHHA

OHHA

HA

AH

KKK baw

baw KKK =


Acids and Bases

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-.

chemical equilibrium - wordpress.com · 2018. 12. 18. · at equilibrium the amount of reactants...

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