Chem 17 Concepts and Equations

2nd LE Topics

The Equilibrium Constant

• Similarly, for the general reaction:

we can define a constant

( ) ( )( ) ( )

tscoefficien

tscoefficien

products ofactivity

reactants ofactivity b

B

a

A

d

D

c

CeqK

←

←=

aa

aa

(aq)(aq)k

k

(aq)(aq) dD cC bB aA

r

f

+←→

+

NOTE: Each species is raised to their respective coefficients in the

balanced chemical equation when expressing the Keq value.

[ ]CγCC =a

[ ]DγDD =a

[ ]AγAA =a

[ ]BγBB =a

Activity of species i -

[ ]ia speciesγ ii = i

ia

i

i

species oft coefficienactivity γ

species ofactivity

−

−

At low concentrations

(dilute, ideal solutions):

( ) ( )( ) ( )bB

a

A

d

D

c

CeqK

aa

aa=

[ ]

1 γ i

iai →∴

→

[ ] [ ][ ] [ ]ba

dc

cBA

DCKK ==

• Kc - Equilibrium constant in terms of concentrations. It is the

product of the equilibrium concentrations (in M) of the

products, each raised to a power equal to its stoichiometric

coefficient in the balanced equation, divided by the product of

the equilibrium concentrations (in M) of the reactants, each

raised to a power equal to its stoichiometric coefficient in the

balanced equation.

Keq becomes Kc � ONLY at ideal dilute conditions

• Kc values are dimensionless

Relationship of K to the Balanced Chemical

Equation

• Reverse an equation, invert the value of K.

• Multiply the coefficients in a balanced equation by a common factor, equilibrium constant is raised to the corresponding power

• Divide the coefficients in a balanced equation by a common factor, take the corresponding root of the equilibrium constant (square root, cube root, )

N2(g) + 3H2(g) � 2NH3(g) K = 5.8 × 105

NH3(g)� 1/2N2(g) + 3/2H2(g) Knew = 1.313 × 10-3

Combining Equilibrium Constant

Expressions• When individual equations are combined (that is,

added), their equilibrium constants are multiplied to obtain the equilibrium constant for the overall reaction.

N2O(g) + 1/2O2(g) � 2NO(g) Koverall = 8.54 × 10-13

2N2(g) + O2(g) � 2N2O(g) K = 2.9 × 10-37 (inverse, take square root)

N2(g) + O2(g) � 2NO(g) K = 4.6 × 10-31(retain)

Solving Equilibrium Problems: How to use

ICE table

Example: At a given temperature 0.80 mole of N2 and 0.90 mole of H2 were placed in an evacuated 1.00-liter container. At equilibrium 0.20 mole of NH3 was present. Calculate Kc for the reaction.

[ ][ ][ ]

( )( )( )

26.060.070.0

20.0

HN

NHK

M 0.20 M 0.60 M 0.70 mEquilibriu

M 0.20+ M 0.30- M 0.10- Change

0 M 0.90 M 0.80 Initial

2NH 3H N

3

2

3

22

2

3c

3(g)2(g)2(g)

===

+ →

←

The Reaction

Quotient, Q

• The major difference between Q and Kc is that the concentrations used in Q are not equilibrium values.

• Q helps predict how the equilibrium will respond to an applied stress - compare Q with Kc.

When

Q = Kc : the system is in equilibrium

Q > Kc : the system goes to the left (), towards reactants

Q < Kc : the system goes to the right (����), towards products

[ ] [ ]

[ ] [ ]bnonequila

nonequil

d

nonequil

c

nonequil

BA

DCQ

)conditions mequilibriu-non(at

dD+cC bB+aA

=

→

←

Le Chatelier’s Principle

� If a change of conditions (stress) is applied to a system in equilibrium, the system responds in the way that best tends to reduce the stress in reaching a new state of equilibrium.

� Some possible stresses to a system at equilibrium are:

1. Changes in concentration of reactants or products.

2. Changes in pressure or volume (for gaseous reactions)

3. Changes in temperature (effect depends on sign of ∆H)

reactants) gaseous of moles of (#-products) gaseous of moles of (#=n∆

Relationship Between Kp and Kc

( ) ( ) n

pc

n

cp RTKKor RTKK∆−∆

==

For gas phase reactions the equilibrium constants can be expressed in partial pressures rather than concentrations.

Gibbs Free Energy and Equilibrium

General equation: THERMODYNAMIC

FORM

All gaseous reactants and products

All solutions of reactants and products

Used when both gaseous and solution

forms appear in the chemical equation

∆Gorxn K Spontaneity condition

< 0 > 1Forward reaction spontaneous,

More products than reactants at equilibrium

= 0 = 1 IDEAL system, very RARE

> 0 < 1Reverse reaction spontaneous,

More reactants than products at equilibrium

Gibbs Free Energy and Equilibrium

EVALUATION OF EQUILIBRIUM CONSTANTS AT

DIFFERENT TEMPERATURES

• From the value of ∆Ho and K1 at one temperature, T1, we can use the van’t Hoff equation to estimate the value of K2 at another temperature, T2.

OR

Gibbs Free Energy: Standard and Non-

standard forms

quotientreaction =Q

Kelvinin re temperatuabsolute = T

K-mol

J 8.314constant gas universal =R

Q log RT 303.2G=G

or lnQ RTG=G

o

o

+∆∆

+∆∆

� ∆Go - standard free energy change.

� ∆G - free energy change at nonstandard conditions

eq

o

eq

o

K log RT 2.303 -=G

or Kln RT -=G

∆

∆ � If concentrations and partial pressures of

species are in equilibrium, then ∆G = 0, and the

equations at left follows

ACID-BASE EQUILIBRIA

[HA]

]][AH[K

-

a

+

=

[B]

]][BHOH[Kb

+−

=

HA ⇄ H⇄ H⇄ H⇄ H++++ + A+ A+ A+ A----

B + H2O ⇄ OH⇄ OH⇄ OH⇄ OH---- + BH+ BH+ BH+ BH++++

• Relating Ka to Kb

HA + H2O ⇄ A- + H3O+ Ka (HA is acid)

A- + H2O ⇄ HA + OH- Kb (A- is base)

H2O + H2O ⇄ H3O+ + OH- Kw

baw K KK ×=

a

wb

K

KK =

b

wa

K

KK =

The stronger the

acid/base, the

weaker is its

conjugate

11-

a2(aq)3

-2

3(aq)(l)2

-

3(aq)

-7

a1(aq)3

-

3(aq)(l)23(aq)2

10 x 4.7 K OH COOH HCO

10 x 4.4K OH + HCOOHCOH

=+↔+

=↔+

+

+

8-

b2(aq)3(aq)2(l)2

-

3(aq)

-4

b1(aq)

-

3(aq)(l)2

-2

3(aq)

10 x 2.3 K HO COHOH HCO

10 x 1.2K HO + HCOOHCO

=+↔+

=↔+

−

−

b2a2

w

b1a1

KK

K

KK

12-

b3(aq)4(aq)3(l)2

-

4(aq)2

7-

b2(aq)

-

4(aq)2(l)2

-2

4(aq)

2

b1(aq)

-2

4(aq)(l)2

-3

4(aq)

101.33 K HO POHOH POH

101.63 K HO POHOH HPO

1078.2K HO + HPOOHPO

×=+→+

×=+→+

×=→+

−

−

−−

b3a3

b2wa2

b1a1

KK

KKK

KK

13-

a3(aq)3

-3

4(aq)(l)2

-2

4(aq)

8-

a2(aq)3

-2

4(aq)(l)2

-

4(aq)2

-3

a1(aq)3

-

4(aq)2(l)24(aq)3

10 3.60 K OH POOH HPO

10 6.20 K OH HPOOH POH

10 50.7K OH + POHOHPOH

×=+→+

×=+→+

×=→+

+

+

+

Strengths of Acids

• BINARY Acids HX - acid strength increases with

decreasing H-X bond strength.

• Down a group: ����size, ����energy to break H- bond (����electronegativity),

����acidity

• Across a period: ����electronegativity, ����acidity

• TERNARY ACIDS HXOn - hydroxides of nonmetals

that produce H3O+ in water.

• For acids with same central element, increase acidity with increase

oxidation state of central element, or increase in attached O atoms

• For acids with different central element but with same number of

O atoms, increase acidity with increase elctronegativity of central

element

Strength of Amine Bases

� The electronic

properties of the

substituents (alkyl

groups enhance the

basicity, aryl groups

diminish it).

� Steric hindrance

offered by the groups

on nitrogen.

Buffers and Henderson-Hasselbalch Equation

HHE in acidic form

HA + H2O ⇄ A⇄ A⇄ A⇄ A---- + H+ H+ H+ H3333OOOO++++

B + H2O ⇄ BH⇄ BH⇄ BH⇄ BH++++ + HO+ HO+ HO+ HO----

HHE in basic form

Preparation of Buffers

• Buffer components must be chosen (acid and conjugate-base, or base and conjugate-acid) based on how near the pKa (of pKb) is near the required buffer pH

• 2 equations needed when solving buffer systems

base conjugate of mole acid of mole componentsbuffer moles total +=

acid moles

base conjugate moleslog pK pH a +=

Other Acid-Base Equations

[ ]( ) HAlog pK2

1 pH initiala -=

pH of weak acid solution,

assumptions valid

[ ][ ]

HA

AlogpK pH

-

a += pH of weak acid solution

added with base (limiting

reactant); forms a buffer

2

pK pK pH a2a1 +=

pH of an amphiprotic weak

acid, amphiprotic form

between the 1st and 2nd Kas

(e.g. HCO32-)

Acid-Base Titration curve• Weak Acid (analyte) titrated with Strong base (titrant)

SOLUBILITY EQUILIBRIA

Example:

• Note: When comparing the solubility of

different salts, look at molar solubility

derived from the Ksp, instead of just the

Ksp value.

• Which of the following salts is more soluble in

water?

Ag3AsO4 27(molar solubility s)4 = Ksp (1.00 × 10-22)

FeAsO4 (molar solubility s)2 = Ksp (5.70 × 10-21)

Cu3(AsO4)2 108(molar solubility s)5 = Ksp (7.60 × 10-36)

The Common Ion Effect in Solubility Calculations

• Calculate the molar solubility of barium sulfate, BaSO4, in 0.010 M sodium sulfate, Na2SO4, solution at 25oC.

The Reaction Quotient in Precipitation

Reactions

• The reaction quotient, Qsp, and the Ksp of a compound are used to analyze whether or not a precipitate will form upon mixing of two ionic species

�Set-up the Ksp reaction

�Calculate the Qsp

�If Qsp > Ksp, then PRECIPITATION happens (← reaction toward ppt formation)

�If Qsp < Ksp, then DISSOLUTION happens (→ reaction toward dissociation into ions)

Fractional Precipitation

• The method of precipitating some ions from solution

while leaving others in solution is called fractional

precipitation.

If a solution contains 0.010 M (each) Cu+, Ag+, and Au+ at

100.0 mL, and a solution of NaCl is added (0.010 M), each ion

can be precipitated as chlorides.

[ ][ ][ ][ ][ ][ ] 13

sp

-

(aq)(aq)(s)

10

sp

-

(aq)(aq)(s)

7

sp

-

(aq)(aq)(s)

100.2Cl AuK Cl Au AuCl

108.1Cl AgK ClAgAgCl

109.1Cl CuK ClCu CuCl

−−++

−−++

−−++

×==+↔

×==+↔

×==+↔

• To determine which species will precipitate first upon addition of

Cl-, calculate the molar solubility

– ↓[molar solubility], will precipitate first (since it is least soluble)

– ↑[molar solubility], will precipitate last (since it is more soluble, it will

stay more in solution)

• For CuCl ⇌ Cu(�)� + Cl(�)

:

[molar solubility of CuCl] = x = [Cu+]eq = [Cl-]eq

∴ Ksp = 1.97 × 10-7 = x ∙ x

∴ x = Ksp = 4.44 × 10-4

• AgCl ⇌ Ag(�)� + Cl(�)

:

[molar solubility of AgCl] = x = [Ag+]eq = [Cl-]eq

∴ Ksp = 1.80 × 10-10 = x ∙ x

∴ x = Ksp = 1.34 × 10-5

• AuCl ⇌ Au(�)� + Cl(�)

:

[molar solubility of AuCl] = x = [Au+]eq = [Cl-]eq∴ Ksp = 2.00 × 10-13 = x ∙ x

∴ x = Ksp = 4.47 × 10-7

Determining [1st ion to be precipitated]

remaining when the 2nd ion starts to

precipitate

• For two ions in solution in which the Solubility Product expressions can be written as 1:1 cation:anion dissolution process

1stion(tobeprecipitated)remaininginsolution

=K!"#!$%&'

K!"(')%&'× 2ndionabouttobeprecipitated

NOTE: This expression changes when solubility

expression is not 1:1 cation:anion dissolution