27 Oct 97Chemical Equilibrium1 Chemical Bonding and Molecular Structure (Chapter 9) Ionic vs....

27 Oct 97 Chemical Equilibrium 1

Chemical Bonding and Molecular Structure

(Chapter 9)

• Ionic vs. covalent bonding• Molecular orbitals and the covalent bond (Ch. 10)• Valence electron Lewis dot structures

octet vs. non-octetresonance structuresformal charges

• VSEPR - predicting shapes of molecules• Bond properties

bond order, bond strengthpolarity, electronegativity

Bond Polarity

HCl is POLAR because it has a positive end and a negative end (partly ionic).

Polarity arises because Cl has a greater share of the bonding electrons than H.

•••H••

••

Calculated charge by CAChe:

H (red) is +ve (+0.20 e-)

Cl (yellow) is -ve (-0.20 e-).

(See PARTCHRG folder in MODELS.)

• Due to the bond polarity, the H—Cl bond energy is GREATER than expected for a “pure” covalent bond.

•••H••

••

Bond Polarity (2)

BOND ENERGY

“pure” bond 339 kJ/mol calculated

real bond 432 kJ/mol measured

ELECTRONEGATIVITY, .

Difference 92 kJ/mol.

This difference is the contribution of IONIC bondingIt is proportional to the difference in

Electronegativity,

is a measure of the ability of an atom in a molecule to attract electrons to itself.

Concept proposed byLinus Pauling (1901-94)Nobel prizes:Chemistry (54), Peace (63)

See p. 425; 008vd3.mov (CD)

• F has maximum .

• Atom with lowest is the center atom in most molecules.

• Relative values of determines BOND POLARITY (and point of attack on a molecule).

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 180

Electronegativity,

Figure 9.7

Bond Polarity

(A) - (B) 3.5 - 2.1

+ -+-O—FO—H

H 2.1O F3.5 4.0

Also note that polarity is “reversed.”

Which bond is more polar ? (has larger bond DIPOLE)

O—H O—F

3.5 - 4.0 0.5

(O-H) > (O-F) Therefore OH is more polar than OF

Molecular Polarity

• Molecules such as HCl and H2O are POLAR• They have a DIPOLE MOMENT. • Polar molecules turn to align their dipole with an

electric field.POSITIVE

NEGATIVE

H—Cl

POSITIVE

NEGATIVE

H—Cl • A molecule will be polar

ONLY if

a) it contains polar bonds ANDb) the molecule is NOT “symmetric”

Symmetric molecules

••

+polar

Molecular Polarity: H2O

Water is polar because:

a) O-H bond is polarb) water is non-symmetric

The dipole associated with polar H2O is the basis for absorption of microwaves used in cooking with a microwave oven

B—F bonds are polarmolecule is NOT symmetric

B—F bonds are polarmolecule is symmetric

Molecular Polarity in NON-symmetric molecules

B +ve F -ve

Atom Chg. B +ve 2.0H +ve 2.1F -ve 4.0

BF3 is NOT polar HBF2 is polar

Fluorine-substituted Ethylene: C2H2F2

CIS isomer

• both C—F bonds on same side

molecule is POLAR.

C—F bonds are MUCH more polar than C—H bonds.

TRANS isomer

• both C—F bonds on opposite side

molecule is NOT POLAR.

(C-F) = 1.5, (C-H) = 0.4

CHEMICAL EQUILIBRIUMChapter 16

• equilibrium vs. completed reactions• equilibrium constant expressions• Reaction quotient• computing positions of equilibria: examples• Le Chatelier’s principle - effect on equilibria of:

• addition of reactant or product• pressure• temperature

YOU ARE NOT RESPONSIBLE for section 16.7 (relation to kinetics)

16_CoCl2.mov(16z01vd1.mov)

Properties of an Equilibrium

Equilibrium systems are• DYNAMIC (in constant motion)• REVERSIBLE • can be approached from either

direction

Pink to blueCo(H2O)6Cl2 ---> Co(H2O)4Cl2 + 2 H2O

Blue to pinkCo(H2O)4Cl2 + 2 H2O ---> Co(H2O)6Cl2

Co(H2O)6Cl2 (aq) Co(H2O)6Cl2 (aq) + 2 H2O

Chemical Equilibrium

• After a period of time, the concentrations of reactants and products are constant.

• The forward and reverse reactions continue after equilibrium is attained.

Fe3+ + SCN- FeSCN2+

16_FeSCN.mov16m03an1.mov

FeCl3 (aq) NaSCN(aq)

FeSCN (aq)

CaCO3(s) + H2O(l) + CO2(g)

Chemical Equilibria

At a given T and pressure of CO2,

[Ca2+] and [HCO3-] can be found from the

EQUILIBRIUM CONSTANT.

Ca2+(aq) + 2 HCO3-(aq)

For any type of chemical equilibrium of the type

THE EQUILIBRIUM CONSTANTTHE EQUILIBRIUM CONSTANT

K =[C]c [D]d

[A]a [B]b

conc. of products

conc. of reactantsequilibrium constant

If K is known, then we can predict concentrations of products or reactants.

a A + b B c C + d D

the following is a CONSTANT (at a given T) :

Determining K

Place 2.00 mol of NOCl is a 1.00 L flask. At equilibrium you find 0.66 mol/L of NO. Calculate K.

Solution

1. Set up a table of concentrations:

[NOCl] [NO] [Cl2]

2 NOCl(g) 2 NO(g) + Cl2(g)

Before 2.00 0 0Change -0.66 +0.66 +0.33Equilibrium 1.34 0.66 0.33

K [NO]2[Cl2 ]

[NOCl]2

Calculate K from equil. [ ]

2 NOCl(g) 2 NO(g) + Cl2(g)

[NOCl] [NO] [Cl2]

Before 2.00 0 0

Change -0.66 +0.66 +0.33

Equilibrium 1.34 0.66 0.33

K = (0.66)2(0.33)

(1.34)2 = 0.080

Writing and ManipulatingEquilibrium Expressions

Solids and liquids NEVER appear in equilibrium expressions.

K [SO2 ][O2 ]

S(s) + O2(g) SO2(g)

K [NH4

+][OH- ][NH3 ]

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

Ktot = K1 x K2

S(s) + 3/2 O2(g) SO3(g)

Adding equations for reactions

S(s) + O2(g) SO2(g)

Manipulating K: adding reactions

NET EQUATION

SO2(g) + 1/2 O2(g) SO3(g) [SO3]

[SO2][O2]1/2K2 =

K1 = [SO2] / [O2]

[O2]3/2Ktot =

ADD REACTIONS MULTIPLY K

Changing directionK

[SO2 ]

Knew [O2 ][SO2 ]

S(s) + O2(g) SO2(g)

SO2(g) S(s) + O2(g)

Manipulating K: Reverse reactions

Chemistry of SulfurElemental S : stable form is S8 (s)

Oxides of S : SO2 (g) and SO3 (g) - significant in atmospheric pollution

Industrially: Oxides generated as needed; ‘stored’ as the hydrate

SO3 (g) + H2O (l) H2SO4 (aq)

Sulfuric acid is HIGHEST VOLUME chemical (fertilizers, refining, manufacturing)

sources: desulfurizing natural gas

roasting metal sulfides

Concentration Units

We have been writing K in terms of mol/L. These are designated by Kc

But with gases, P = (n/V)•RT = conc • RT

P is proportional to concentration, so we can write K in terms of PARTIAL PRESSURES.

These constants are called Kp. Kc and Kp have DIFFERENT VALUES

(unless same number of species on both sides of equation)

Manipulating K : Kp for gas rxns

Concentration of products is much greater than that of reactants at equilibrium.

The Meaning of K1. Can tell if a reaction is

product-favored or reactant-favored.

K =p P(H2O)2

P(H2)2 P(O2) = 1.5 x 1080

2 H2(g) + O2(g) 2 H2O (g)

The reaction is strongly product-favored.

K >> 1

What about the reverse reaction ?

This reaction is strongly reactant-favored.

Conc. of products is much less

than that of reactants at equilibrium.

Meaning of K: AgCl rxn

Kc = K << 1

Ag+(aq) + Cl-(aq) AgCl(s)

Krev = Kc-1 = 5.6x104. It is strongly product-favored.

AgCl(s) Ag+(aq) + Cl-(aq)

[Ag+] [Cl-] = 1.8 x 10-5

CH3 —C —C —CH3

n-butane

HCH3—C—CH3

iso-butane

2. Can tell if a reaction is at equilibrium. If not, which way it moves to approach equilibrium.

Meaning of K : butane isomerization

K = [iso]

[n] = 2.5

If [iso] = 0.35 M and [n] = 0.25 M, is the system at equilibrium?

If not, which way does the rxn “shift” to approach equilibrium?

All reacting chemical systems can be characterized by their REACTION QUOTIENT, Q.

Q = =0.350.25

= 1.40

Q - the reaction quotient

If Q = K, then system is at equilibrium.

Reaction is NOT at equilibrium.

[Iso] must INCREASE and [n] must DECREASE.

Q = 1.4 which is LESS THAN K =2.5

To reach EQUILIBRIUM

[iso][n]

Q has the same form as K, . . . but uses existing concentrations

For n-Butane iso-Butane

Typical EQUILIBRIUM Calculations2 general types: a. Given set of concentrations, is system at equilibrium ?

Calculate Q compare to K

IF:Q > K or Q/K > 1 REACTANTS

Q < K or Q/K < 1 PRODUCTS

Q=K at EQUILIBRIUM

Examples of equilibrium questions

Place 1.00 mol each of H2 and I2 in a 1.00 L flask. Calculate equilibrium concentrations.

H2(g) + I2(g) 2 HI(g)

b. From an initial non-equilibrium condition, what are the concentrations at equilibrium?

H2(g) + I2(g) 2 HI(g) Kc = 55.3

Initial 1.00 1.00 0

DEFINE x = [H2] consumed to get to equilibrium.

Change -x -x +2x

At equilibrium 1.00-x 1.00-x 2x

Kc = [HI]2

[H2 ][I2 ] = 55.3

Step 1. Set up table to define EQUILIBRIUM concentrationsin terms of initial concentrations and a change variable

[H2] [I2] [HI]

Step 2 Put equilibrium concentrations into Kc expression.

Kc = [2x]2

[1.00 - x][1.00 - x] = 55.3

Step 1 Define equilibrium condition in terms of initial condition and a change variable

[H2] [I2] [HI]

At equilibrium 1.00-x 1.00-x 2x

H2(g) + I2(g) 2 HI(g) Kc = 55.3

[H2] = [I2] = 1.00 - x = 0.21 M

[HI] = 2x = 1.58 M

Step 3. Solve for x. 55.3 = (2x)2/(1-x)2

In this case, take square root of both sides.

7.44=2x

1.00 x-

H2(g) + I2(g) 2 HI(g) Kc = 55.3

Solution gives: x = 0.79Therefore, at equilibrium

“...if a system at equilibrium is disturbed, the

system tends to shift its equilibrium position

to counter the effect of the disturbance.”

EQUILIBRIUM AND EXTERNAL EFFECTS

• The position of equilibrium is changed when there is a change in:

– pressure– changes in concentration– temperature

• The outcome is governed by

LE CHATELIER’S PRINCIPLEHenri Le Chatelier

1850-1936

- Studied mining engineering

- specialized in glass and ceramics.

• If concentration of one species changes, concentrations of other species CHANGESto keep the value of K the same (at constant T)

• no change in K - only position of equilibrium changes.

Shifts in EQUILIBRIUM : Concentration

ADDING REACTANTS- equilibrium shifts to PRODUCTS

ADDING PRODUCTS- equilibrium shifts to REACTANTS

REMOVING PRODUCTS - often used to DRIVE REACTION TO COMPLETION

- GAS-FORMING; PRECIPITATION

Solution

A. Calculate Q with extra 1.50 M n-butane.

INITIALLY: [n] = 0.50 M [iso] = 1.25 M

CHANGE: ADD +1.50 M n-butane What happens ?What happens ?

Q < K . Therefore, reaction shifts to PRODUCT

Q = [iso] / [n] = 1.25 / (0.50 + 1.50) = 0.63

n-Butane Isobutane

Effect of changed [ ] on an equilibrium

16_butane.mov(16m13an1.mov)

K = [iso]

[n] = 2.5

Solution

B. Solve for NEW EQUILIBRIUM - set up concentration table [n-butane] [isobutane]Initial 0.50 + 1.50 1.25Change - x + xEquilibrium 2.00 - x 1.25 + x

Butane/Isobutane

K = 2.50 = [isobutane]

[butane]

1.25 + x2.00 - x

x = 1.07 M. At new equilibrium position, [n-butane] = 0.93 M [isobutane] = 2.32 M.

Equilibrium has shifted toward isobutane.

Effect of Pressure (gas equilibrium)

Increase P in the system by reducing the volume.

Kc = [NO2 ]2

[N2O4 ] = 0.0059 at 298 K

NN22OO44(g) (g) 2 NO 2 NO22(g)(g)

16_NO2.mov(16m14an1.mov)

Increasing P shifts equilibrium to side with fewer molecules (to try to reduce P). Here, reaction shifts LEFT PN2O4 increases

See Ass#2 - question #6

PNO2 decreases

• Temperature change change in K• Consider the fizz in a soft drinkCO2(g) + H2O(liq) CO2(aq) + heat

• Increase TEquilibrium shifts left: [CO2(g)] [CO2 (aq)] K decreases as T goes up.

•Decrease T[CO2 (aq)] increases and [CO2(g)] decreases.K increases as T goes down

Kc = [CO2(aq)]/[CO2(g)]HIGHER T

LOWER T

• Change T: New equilib. position? New value of K?

Temperature Effects on Chemical Equilibrium

Kc = 0.00077 at 273 K

Kc = 0.00590 at 298 KKc

[NO2 ]2

[N2O4 ]

N2O4 + heat 2 NO2 (colorless) (brown)

Horxn = + 57.2 kJ

Increasing T changes K so as to shift equilibrium in ENDOTHERMIC direction

16_NO2RX.mov(16m14an1.mov)

• Add catalyst ---> no change in K• A catalyst only affects the RATE of approach

to equilibrium.

Catalytic exhaust system

CHEMICAL EQUILIBRIUMChapter 16

• equilibrium vs. completed reactions• equilibrium constant expressions• Reaction quotient• computing positions of equilibria: examples• Le Chatelier’s principle - effect on equilibria of:

• addition of reactant or product• pressure• temperature

YOU ARE NOT RESPONSIBLE for section 16.7 (relation to kinetics)

27 Oct 97Chemical Equilibrium1 Chemical Bonding and Molecular Structure (Chapter 9) Ionic vs....

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