# thermodynamics

Mechanics

Oct 28, 2013 (4 years and 8 months ago)

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270

Thermodynamics

271

Test Content
from AP Chemistry Course Description

III. Reactions (35
-
40%)

E.

Thermodynamics

1.

State functions

2.

First law: change in enthalpy; heat of formation; heat of reaction; Hess's law;
heats of
vaporization and fusion; calorimetry

3.

Second law: entropy; free energy of formation; free energy of reaction; dependence of
change in free energy on enthalpy and entropy changes

4.

Relationship of change in free energy to equilibrium constants

and electrode
potentials

Types of Calculations found on Exam

from AP Chemistry Course Description

Summary of types of problems either explicitly or implicitly included in the preceding
material.

8. Equilibrium constants and their
applications, including their use for simultaneous equilibria

10. Thermodynamic and thermochemical calculations

THERMODYNAMICS

-

The study of energy in matter

-

Thermodynamics allows us to predict whether a chemical reaction occurs or not.

-

Thermodynamic
s tells us nothing about how fast a reaction occurs.

-

i. e., thermodynamics can’t explain kinetics and vice versa

STATE FUNCTIONS

A
State Function

is a thermodynamic quantity whose value depends only on the state at the
moment, i. e., the temperature, p
ressure, volume, etc…

The value of a state function is independent of the history of the system.

The fact that internal energy is a state function is extremely useful because it we can measure
the energy change in the system by knowing the initial energy

and the final energy.

272

TYPES OF ENERGY AND ENERGY CHANGES

Two types of energy changes

1.
Heat

q

-

chaotic change in molecular motion

-

related to temperature

-

heating increases (or decreases) molecular motion in all directions

-

not a state function (must know history)

-

sign convention

+q = heat gained by system

-

q = heat lost by system

2.
Work

w

-

concerted change in molecular motion

-

Work = Force

distance

w = F

d

-

movement against force is work

-

work increases (or decreases) molecular motion in a specific direction

-

in gases, w =
-

p

V

-

not a state function (must know history)

-

sign convention

+w = work done on system

(compression)

-

w = work done by system

(expansion)

Three typ
es of energy

1.

Internal Energy

E

-

Internal Energy

is the sum of kinetic and potential energy in a thermodynamic system.

-

state function

2.

Enthalpy

H

-

modified form of internal energy

-

H = E + pV

-

state function

-

value is
very close to value of internal energy for most chemical systems.

-

change in enthalpy for constant pressure process is equivalent to heat

-

H = H
f

H
i

= q
p

2.

Gibbs Free Energy

G

-

modified form of enthalpy

-

G = H

TS

(S

entropy)

-

1.

Predict whether reaction is spontaneous

2.

Give amount of useful work

3.

Relate how completely a reaction will proceed

-

much more later in the chapter!

273

REVIEW OF FIRST LAW OF THERMODYNAMICS

-

The change in
the energy of a system is due to heat and/or work.

E = q + w

-

The first law is a conservation of energy statement.

-

We will see that although energy is conserved, not all of it will be useful.

-

i. e., some energy in a process will always be wasted.

-

Be sure to review first law, especially sign convention of heat and work.

DEFINITIONS OF PROCESSES

Spontaneous Processes

-

A process that occurs without outside help.

Reversible Processes

-

A reversible process is a
process that is always in equilibrium.

Examples

1.) Ice melting at O

C

H
2
O (s)

H
2
O (l)

-

-

Water can go from solid to liquid to solid to liquid etc…

2.) Haber process in a closed
container

N
2

(g) + 3 H
2

(g)

2 NH
3

(g)

-

Production of ammonia can increase or decrease by adjusting the external pressure. (also
temperature)

-

Aside: Haber process is extremely important for support of human population. Natural
fertilizer is insuf
ficient. Thank goodness for synthetic fertilizer.

Spontaneous

Pants fall to floor

Raw egg becomes hard boiled in water

Dry ice su
blimates at room temp.

CH
4

+ 2 O
2

CO
2

+ 2 H
2
O

Nonspontaneous

Pants hang themselves in closet

Hard
-
boiled egg becomes raw egg

CO
2

gas deposits as solid at room temp.

CO
2

+ 2 H
2
O

CH
4

+ 2 O
2

274

Irreversible Processes

-

A process not in equilibrium

Examples:

1.) Ice melting at 25

C

2.) Precipitation of AgCl

Ag
+

(aq) + Cl
-

(aq)

AgCl (s)

-

As a solution of Ag
+

and a solution of Cl
-

are mixed, the system is out of equilibrium.

-

Thus, precipitation occurs

-

The solid AgCl will never redissociate into ions.

-

However, once precipitate forms, system is in equilibrium. Now the formation of solid
can be consider
ed reversible (e. g. addition or subtraction ammonia causes amount of
solid to change.

AgCl(s)

Ag
+

(aq) + Cl
-

(aq)

AgCl(s) + 2 NH
3

(aq)

[Ag(NH
3
)
2
]
+

(aq) + Cl
-

(aq)

ENTROPY

-

Review notes from solutions chapter about entropy.

-

Entropy is measure of
how energy can be dispersed or spread out.

-

Entropy makes systems more disordered.

-

Increasing the number of ways that a particle in a system can distribute energy, increases the
entropy of the system.

Example: Which has more entropy: 1 mole of gas in
10 L or 1 mole of gas in 20 L?

-

Larger volume means molecules can be in more places, the gas in the 20 L has more
entropy.

-

When heat exchange is not involved, spontaneous processes always increase entropy.

Consider diffusion of gas in two connected g
as bulbs.

-

Gas spontaneous expands, increasing entropy.

Thermodynamic definition of entropy

-

Entropy is defined via a change in heat (i. e., a change in choatic
motion)

-

q is often difficult to measure for irreversible processes, but is usually easy to measure for
reversible processes.

open stopcock

1 atm

0 atm

½ atm

½ atm

q
rev

heat in a reversible process

275

Example: Calculate the change of entropy when 36.02 g of ice mel
ts at 0.00

C. The molar
enthalpy of melting for ice is 6.008 kJ/mol.

Using the thermodynamic definition of entropy, we need to find q
rev

and T.

T = 0.00

C = 273.15 K

Melting of ice is
a reversible process; thus

Therefore the increase of the entropy of the water is

SECOND LAW OF THERMODYNAMICS

In any spontaneous processes, the total entropy of the universe increases for irreversible
processes and is zero for reversible processes.

For a thermodynamic process, the universe can be broken into two parts.

1.) System

contains what we’re interested in

2.) Surroundings

everything else

Example: For a
beaker of melting ice

System

beaker and ice

Surroundings

laboratory and everything else

Example: Measuring the enthalpy of neutralization in a coffee cup calorimeter

System

reaction e.g. NaOH + HCl

Surroundings

water and calorimeter

or

System

reaction, water and calorimeter

Surroundings

laboratory

*Defining system and surroundings can be a matter of perspective.*

Since universe can be broken into two parts, 2nd Law can be rewritten as

Note that

S
sys

can be negative (system becomes more ordered), if

S
surr

is more positive.

Note: Always use Kelvin for
thermodynamic calculations.

example of an
isolated system

276

Example: Water freezing at

10

C.

System

water

Surroundings

refrigerator

Water spontaneously decreases entropy, but 2nd law says

S
univ

>
0

Therefore,

S
surr

must be much greater than zero

-

i. e.,

S
surr

>> 0

Water transfers heat (choatic motion) to air in refrigerator.

Heating of air increases entropy of “surroundings”.

Molecular motion and entropy

-

As the motion of molecule
increases, its entropy increases.

-

According to kinetic theory of matter, motion is proportional to temperature.

-

Therefore, as the temperature of a system increases, its entropy increases.

Types of molecular motion

1. Translational

-

Particle moves i
n a straight line, i. e., particle is translated.

-

Accounts for most entropy in gases.

-

Accounts for substantial amount of entropy in liquids.

-

Has no contribution to the entropy of solids.

2. Vibrational

-

Atoms in a bond vibrate as if on a sprin
g.

-

Accounts for substantial part of entropy of liquid.

-

vibrations occur within molecules

-

vibrations occur between molecules (librations)

-

Accounts for all entropy in solid.

Vibrations of water

3. Rotational

-

Molecule rotates on axis

-

Minor contribution to entropy except at low and very high temperatures for
gases and
liquids.

O

H

H

O

H

H

symmetric stretch

O

H

H

H

antisymmetric stretch

O

H

H

bending

O

H

H

O

H

H

277

Molecular motion and degrees of freedom

Degrees of freedom

number of independent options for movement

In general for N atoms in a molecule

Translational DOF

3

Rotational DOF

Linear molecule

2

Nonlinear molecule

3

Vibrational DOF

Linear molecule

3N

5

Nonlinear molecule

3N

6

Total DOF

3N

Miscellaneous notes on entropy and molecular motion

1.)

Entropy increases during phase changes

s

l

l

g

s

g

2.)

Entropy increases when
number of particles increases.

BF
3

(g) + NH
3

(g)

BF
3
NH
3

(s)

[negative

S]

N
2
O
4

(g)

2 NO
2

(g) [positive

S]

3.)

Entropy increases when temperature increases.

-

Number of modes of motion increases

4.)

Entropy increases when a gas is produced i
n a chemical reaction.

HNO
3

(aq) + Rb
2
CO
3

(s)

2 RbNO
3

(aq) + CO
2

(g) + H
2
O (l)

2 HBr (aq) + Cd (s)

CdBr (aq) + H
2

(g)

5.)

Entropy increases when the volume of a gas increased.

Example: Which has more entropy in a 1 liter volume: 1 atm of gas or

2 atm of gas?

The volume with 2 atm of gas has more entropy since it has more molecules and these
molecules have many more states in which they can reside.

MC Question:

Of the following reactions, which involves the largest decrease in entropy?

(A)

CaCO
3
(s)

CaO(s) + CO
2
(g)

(B)

2 CO(g) + O
2
(g)

2 CO
2
(g)

(C)

Pb(NO
3
)
2
(aq) + 2KI(aq)

PbI
2
(s) 2 2KNO
3
(aq)

(D)

C
3
H
8
(g) + 5 O
2
(g)

3 CO
2
(g) + 4 H
2
O(l)

(E)

4 La(s) + 3 O
2
(g)

2 La
2
O
3
(s)

278

Entropy for a phase transition

Since a phase change at the
transition temperature is a reversible process, calculation of the
entropy change of phase transition is a straight forward application of the thermodynamics
definition of entropy.

Example:

R
-
134a is a refrigerant used automotive air conditioning and ha
s the formula,
CH
2
FCF
3
. A typical air conditioning unit holds 28 oz (800 g) of refrigerant. If the
molar enthalpy of vaporization of R
-
134a is 22.0 kJ/mol at
-
26.6

C,

a)

Calculate the entropy of the phase change.

Using the thermodynamic
definition of entropy, we need to find q
rev

and T.

T =
-
26.6

C = 246.55 K

The evaporation of a liquid at its boiling point is a reversible process; thus

Therefore
the increase of the entropy of the R
-
134a is

b)

Calculate the molar entropy of the phase change.

To calculate the molar entropy, we need the molar heat, which in this case is the
enthalpy of vaporization.

Note: Always use Kelvin for
thermodynamic calculations.

279

THE THIRD LAW OF THERMODYNAMICS

In other words, if we have a perfect crystal, we must be at absolute zero.

Consequences of the Third Law

-

Absolute zero is
unattainable.

-

Entropy of all substances at absolute zero is zero.

-

At temperature above zero, crystal will not be perfect.

-

Vibrational motion introduces imperfections

-

To remove imperfection takes some sort of motion

-

But introducing motion keep
s crystal imperfect

-

It’s a no
-
win situation!

CALCULATION OF ENTROPY CHANGES

The entropy change of chemical reaction can be calculated from a table of standard entropies.

-

Standard entropies are measured at 1 atm or 1 M and 298 K

-

Note:
S

is not zero

for elements in standard state. (Different than

H

and

G

)

For the general reaction a A + b B

c C + d D

S

rxn

= c S

(C) + d S

(D)
-

a S

(A)
-

b S

(B)

GIBBS FREE ENERGY

Reconsider 2nd Law of Thermo.

Using the
thermodynamic definition of entropy, the entropy change of the surroundings can be
related to the heat of the system.

At constant pressure:

Therefore the second law can be rewritten as

Define Gibbs Free Energy as G = H

TS

At constant temperature:

G =

H

T

S

The entropy of a perfect crystal is zero at absolute zero (0 K).

280

Second Law in terms of Gibbs Free Energy:

G
sys

< 0

G is useful to decide if a reaction occurs

G < 0

rxn is spontaneous

G = 0

system is at equilibrium

Recall

S
univ

= 0 for reversible process

G > 0

rxn is nonspontaneous; i.e., rxn does not happen

Negative sign of

G is consistent with idea that nature always chooses lowest energy.

-

System going from high energy t
o low energy must release energy.

Magnitude of

G is also important.

G = w
max

-

As

G gets more negative, more work can be done by the system.

Note:

When
-
10 kJ/mol <

G < 10 kJ/mol, reactions do not go to completion but may
favor left side (
-

G)
or right side (+

G) of the chemical equation.

Consider combustion of fuels

CH
4

(g) + 2 O
2

(g)

CO
2

(g) + 2 H
2
O (g)

G

=
-

801.14 kJ/mol =
-

49.92 kJ/g

C
8
H
18

(l) + 25/2 O
2

(g)

8 CO
2

(g) + 9 H
2
O (g)

G

=
-

5214.1 kJ/mol =
-

45.65 kJ/g

1 mol of
octane is able to do more work than 1 mol of methane.

1 g of methane is able to do more work than 1 g of octane.

281

CALCULATING STANDARD GIBBS FREE ENERGY CHANGES

For the general reaction a A + b B

c C + d D

G

rxn

= c

G
f

(C) + d

G
f

(D)
-

a

G
f

(A)
-

b

G
f

(B)

G
f

= 0; for elements in their standard state by definition.

Temperature dependence of Gibbs free energy

Recall for spontaneous processes,
usually

H is negative and

S is positive.

However, the only reliable measure of spontaneity is

G.

Putting negative

H and positive

S into equation for

G yields negative

G.

Gibbs
-
Helmholtz Equation

G =

H

T

S

Consider possibility when

S is negative such as

4 Fe (s) + 3 O
2

(g)

2 Fe
2
O
3

(s)

S

=
-

0.5437 kJ/mol

K

H

=
-

822.16 kJ/mol

G

=
-

822.16 kJ/mol

(298.0 K)(
-

0.5437 kJ/mol

K) =
-

660.1 kJ/mol

In this reaction, heat evolved (increasing disorder of universe) compensates for system
becoming more ordered.

In increasing the
temperature, the entropy has a greater effect on the system.

We can increase temperature until the entropy effects become more important than enthalpy
effects.

From the preceding, we see that the spontaneity of a reaction is affected by temperature.

Ex
ample: At what temperature is the rusting of iron nonspontaneous?

We need to find T where entropy effects are balanced with enthalpy effects

G =

H

T

S = 0

As long as we keep iron above 1512 K, it won’t rust.

282

Example:

For the reaction CaCO
3

(s)

CaO (s) + CO
2

(g), The enthalpy of reaction is 178.3
kJ/mol and the entropy of reaction is 160.5 J/mol

K.

a) Calculate the Gibbs free energy at 450 K using the Gibbs

Helmholtz equation.

G =

H

T

S =

178.3 kJ/mol

450 K(0.1605 kJ/mol

K) = 106.1 kJ/mol

b) Calculate the Gibbs free energy at 2110 K using the Gibbs

Helmholtz equation.

G =

H

T

S = 178.3 kJ/mol

2110 K(0.1605 kJ/mol

K) =

160.4 kJ/mol

c) Calculate the temperature where the

reaction becomes spontaneous.

G = 0

Temperature dependence of free energy can be summarized in table.

H

S

G

-

+

-

s灯pane潵s

-

-

-

s灯pane潵s 扥l潷s灥cific em瀮

+

+

-

s灯pane潵s a扯bes灥cific em瀮

+

-

+

never s灯pane潵s

Qesi潮: 

As摥n 灥rf潲ms anex灥rime湴 漠摥ermine hem潬ar enhal灹 潦s潬i潮

2
NCONH
2
. The student places 91.95 g of water at 25°C into a coffee cup calorimeter
and immerses a thermometer in the water. After

50 s, the student adds 5.13 g of solid urea, also
at 25°C, to the water and measures the temperature of the solution as the urea dissolves. A plot
of the temperature data is shown in the graph below.

283

a)

Determine the change in temperature of the solution that results from the dissolution of the
urea.

Δ
T
= 21.8 − 25.0 =
−3.2

C

b)

According to the data, is the dissolution of urea in water an endothermic process or an

The process is endothermic. The
decrease in temperature indicates that the process for
the dissolution of urea in water requi
res energy
.

c)

Assume that the specific heat capacity of the calorimeter is negligible and that the specific
heat capacity of the solution of urea and water is 4.2 J g
−1

°C
−1

throughout the experiment.

(i) Calculate the heat of dissolution of the urea in joules.

Assuming that no heat energy is lost from the calorimeter and given that the calorimeter
has a negligible heat capacity, the sum of the heat of dissolution,
q
soln

and the change in
heat energy of the urea
-
water mixture must equal zero.

q
soln

+
mc
Δ
T
= 0

q
soln

= −
mc
Δ
T m
soln

= 5.13 g + 91.95 g = 97.08 g

q
soln
= −(97.08 g)(4.2 J g
−1
°C
−1
)(−3.2°C) =
1.3 × 10
3

J

(ii) Calculate the molar enthalpy of solution,
, of urea in kJ mol
−1
.

molar mass of urea = 4(1.0) + 2(14.0) + 12.0 + 16.0 = 60.0 g mol
−1

d)

Using the information in the table below, calculate the value of the molar
entropy of
solution,

Accepted Value

of urea

14.0 kJ mol
-
1

of urea

-
6.9 kJ mol
-
1

284

e)

The student repeats the experiment and the time obtains a result for

of urea that is
11 percent below the accepted value. Calculate the value of

that the student
obtain
ed from the second trial.

Error = (0.11)(14.0 kJ mol
-
1
) = 1.54 kJ mol
-
1

14.0 kJ mol
-
1

1.54 kJ mol
-
1

=
12.5 kJ mol
-
1

f)

The student performs a third trial of the experiment but this time adds urea that has been
taken directly from a refrigerator at 5

C. What effect, if any, would using the cold urea

C have on the experimentally obtained value of
? Justify your

There would be
an increase in the obtained value for

bec
ause the colder urea
would have caused a larger negative temperature change
.

Concentration dependence of Gibbs Free Energy

Standard concentrations are

p

= 1 atm

c

= 1 M

The Gibbs free energy for the partial pressure of a gaseous substance is

The Gibbs free energy for the molar concentration of an aqueous substance is

Note: The higher the concentration, the
higher the free energy

285

For a general reaction: a A + b B

c C + d D

-

Where

G and

G
0

are stoichiometric sums and Q is reaction quotient

-

As reaction goes to right, Q increases; thus

G
increases

-

As reaction goes to left, Q decreases; thus

G decreases

Example:

For the reaction N
2
O
4

(g)

2 NO
2

(g) at 298 K the standard free energy of reaction
is 5.40 kJ/mol.

a) calculate the free energy of the reaction when p(N
2
O
4
) = 1 atm and
p(NO
2
) = 1
atm

b) calculate the free energy of the reaction when p(N
2
O
4
) = 0.905 atm and p(NO
2
) = 0.115
atm

286

c)
calculate the free energy of the reaction when p(N
2
O
4
) = 0.444 atm and p(NO
2
)
= 0.224 atm

System is at equilibrium!

**At equilibrium,

G=‰⨪

0 =

G

+ RT ln K

G

-
RT⁬nK

Thus the free energy is related to the equilibrium constant

Ah ha! Free energy becomes even more useful.

N
2
(
g
) + 3 F
2
(
g
)

2 NF
3
(
g
)

;

FR

Question:

The following questions relate to
the synthesis reaction represented by the
chemical equation in the box above.

a)

Calculate the value of the standard free energy change,
, for the reaction.

b)

Determine the temperature at which the
equilibrium constant, K
eq
, for the reaction is equal
to 1.00. (Assume that
, and

are independent of temperature.)

When K
eq

= 1, then

c)

Calculate
the standard enthalpy change,
, that occurs when a 0.256 mol sample of
NF
3
(
g
) is formed from N
2
(
g
) and F
2
(
g
) at 1.00 atm and 298 K.

One point is earned for multiplying

H
0

by the number of moles
of NF
3
.

One point is earned for recognizing that 2 mol
NF
3

are produced
for the reaction as it is written.

One point is earned for the answer (including kJ or J).

287

The enthalpy change in a chemical reaction is the differen
ce between energy absorbed in
breaking bonds in the reactants and energy released by bond formation in the products.

d)

How many bonds are formed when two molecules of NF
3

are produced according to the
equation in the box above?

There are six N

F bonds

formed.

e)

Use both the information in the box above and the table of average bond enthalpies below
to calculate the average enthalpy of the F

F bond.

Bond

Average Bond Enthalpy

(kJ mol
-
1
)

N

N

㤴9

=
c
=
㈷2
=

=
c
=
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=
=
=
=
=
=
bxam灬eW= =
=
Calculate= the=e煵ilibri畭= c潮stant= f潲=the=reacti潮=
=
CaCl
3

(s)

CaO (s) + CO
2

(g) at 25

C, given a table of Gibbs free energies

G
f

(CaCO
3
(s)) =

-
1128.76 kJ/mol

G
f

(CaO(s)) =

-
604.17 kJ/mol

G
f

(CO
2
(g)) =

-
394.4 kJ/mol

G

= (

604.17 kJ/mol) + (

394.4 kJ/mol)

(

1128.76 kJ/mol) = 130.2 kJ/mol

K
p

= p
CO
2

= 1.5

10
-
23

atm

Example: Calculate

G

for the equilibrium at 25

C

AgCl (s)

Ag
+

(aq) + Cl
-

(aq)

K
sp

= 1.8

10
-
10