Notes to Follow Thermodynamics Powerpoint File

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Oct 27, 2013 (3 years and 5 months ago)

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Thermodynamics


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Thermochemistry





The Nature of Energy and Types of Energy



A.

Definitions



1.

Energy



capacity to do work or transfer heat



2.

Work



directed energy change resulting from a process

3.

Law of Conservation of Energy



the total quantity of energy in

the
universe is assumed constant




B.

Types of Energy



1.

Kinetic



energy of motion



2.

Potential



energy of position or composition



3.

Radiant



solar energy; comes from the sun

4.

Thermal



the energy associated with the random motion of atoms
an
d molecules

5.

Chemical



stored within the structural units of chemical
substances

6.

Nuclear



energy stored within the collection of neutrons and
protons in the atom




Energy Changes in Chemical Reactions



A.

Thermal Energy versus Heat

1.

Heat



the t
ransfer of thermal energy between two bodies that are
at different temperatures



2.

Temperature



a measure of the thermal energy



3.

Temperature is not equal to thermal energy

4.

Thermal energy of a cup of coffee at 90
o
C is less than th
e thermal
energy

of a bathtub of water at 40
o
C.



B.

System versus Surroundings



1.

Thermochemistry



the study of heat change in chemical reactions

2.

System



the specific part of the universe that is of interest to

us

a.

Open System



can exchange mass and energy, u
sually in
the form of heat with its surroundings

b.

Closed System



allows the transfer of energy (heat) but
not mass

c.

Isolated System



does not allow the transfer of either mass
or energy



3.

Surroundings



the rest of the universe outside the system

Thermodynamics


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4.

Universe



system plus surroundings



C.

Exothermic versus Endother
mic




1.

Exothermic Process




a.

any process that gives off heat




b.

transfers thermal energy to the surroundings




c.





d.




2.

Endothermic Process




a.

heat has to be supplied to the system by the surroundings




b.





c.





Introduction to Thermodynamics


A.

Thermodynamics

1.

the scientific study of the interconversion

of heat and other kinds
of energy



2.

laws help explain the energy and direction of processes



3.

study the changes in the
state of a system

4
.

state of a system



the values of all relevant macroscopic
properties

(
example
:
composition, energy, tempera
ture, pressure,
volume
)

5.

state function



properties that are determined by the state of the
system, regardless of how that condition was achieved (
example
:
energy, pressure, volume, temperature)




B.

The First Law of Thermodynamics

1.

The total energy of

the universe is constant

2.

Energy can neither be created nor destroyed in the course of a
chemical reaction or physical change

3.

ΔE universe = ΔE system + ΔE surroundings = 0

4.

Units

a.

Joule (J)

1 J = 1 kg
-
m
2
/s
2

b.

Calorie (cal)

1 cal = 4.184 J

5.

State functions

a.

Depen
d only on the initial and final states of a system, not
the path by which the change occurred

b.

Most important use is to describe changes (ΔX = X final


X initial)

6.

ΔE is a state function

a.

ΔE system = E final


E initial = E products


E reactants

b.

Most common

manifestations of ΔE

Thermodynamics


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i.

Heat (q) transferred solely because of the
temperature difference between the system and
surroundings

ii.

Work (w)


energy transferred by means of a
mechanical linkage between the system and
surroundings

c.

ΔE system = q + w

i.

q > 0, system g
ains heat

ii.

q < 0, system loses heat

iii.

w > 0, work done on the system

iv.

w < 0, work done by the system

7.

Example



When gasoline burns in a care engine, the heat released
causes the products to expand, which pushes the pistons outward.
Excess heat is removed by t
he car’s cooling system. If the
expanding gases do 451 J of work on the pistons and the system
loses 325 J to the surroundings as heat, calculate the change in
energy in J, kJ, and kcal.



C.

Work and Heat

1.

Types of chemical w
ork

a
.

Electrical work



work done by moving charged particles
(electrochemistry)
; ex: a battery supplying electrons to
light the bulb of a flashlight

b
.

PV work

(mechanical work)



work done by an expanding
gas at constant pressure (the only work we are
considering
in this discussion)

2.

w =
-
PΔV


a.

P = external pressure

b.

ΔV = V final


V initial

3.

Example

-

A sample of nitrogen gas expands in volume from 1.6 L
to 5.4 L at constant temperature. What is the work done in joules if
the gas expands (a) against a vacuum and (b) against a

constant
pressure of 3.7 atm?






a.








b.











Thermodynamics


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Enthalpy of Chemical Reactions



A.

Constant Volume Conditions



1.





a.

ΔV = 0




b.

Therefore, no P
-
V work.



2.

Often inconvenient and sometimes impossible to achieve.



B.

Constant Pressure Conditions

1.





a.

P = constant




b.

P is usually atmospheric pressure



2.

Most reactions occur un
der conditions of constant pressure
.



C.

Enthalpy



1.





a.

E = internal energy of the system




b.

P = pressure of the system




c.

V = volume of the system



2
.

H has energy units just like E and PV




a.

E = J




b.

PV = (a
tm)(L)



J



1 L

·

atm = 101.3 J

c.

Therefore, H = J

3
.

E, P, and V are state functions

a.

the changes in E + PV depend only on the initial
and final
states




b
.

Therefore, H is a state function.



4.


[change in enthalpy]




a.


[change in enthalpy @ constant P]


b.


[
]



D.

Enthalpy of Reactions

1
.

Enthalpy is also a state function

a.

used to quantify the heat flow into or out of a system in a
process that occurs a
t constant pressure

b
.

ΔH (system) = H (final)


H (initial)

c
.

ΔH = H (products)


H (reactants)



2
.

ΔH can be + or






a.

ΔH > 0


endothermic




b.

ΔH < 0


exothermic



E.

Thermochemical Equations

Thermodynamics


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1.

H
2
O(s)


H
2
O(l)

Δ
H = 6.01 kJ/mole




a.

system absorbs heat from the s
urroundings




b.

endothermic




c.

ΔH > 0

d.

6.01 kJ are absorbed for every 1 mole of ice that melts at


0
o
C and 1 atm

2.

CH
4
(g) + 2O
2
(g)


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

ΔH =

890.4 kJ/mole




a.

system
gives off
heat
to

the surroundings




b.

e
xothermic




c.

ΔH
<

0

d.

890.4

kJ are
released
for every 1 mole of
methane

that

is
combusted at 25
o
C and 1 atm.

3.

Example



In each of the following cases, determine the sign of
ΔH

and
whether the reaction is exothermic or endothermic.






a.

H
2
(g) + ½ O
2
(g)


H
2
O(l) + 28
5.8 kJ






ΔH is negative and exothermic




b.

40.7 kJ + H
2
O(l)


H
2
O(g)






ΔH is positive and endothermic

4
.

Guidelines for writing and interpreting thermochemical equations:

a.

The stoichiometric coefficients always refer to the number
of moles of a s
ubstance.






H
2
O(s)


H
2
O(l)

ΔH = 6.01 kJ




b.

If you reverse a reaction, the sign of
ΔH

changes.






H
2
O(l)


H
2
O(s)

ΔH =

6.01 kJ

c.

If you multiply both sides of the equation by a factor
n
,
then
ΔH

must change by the same factor
n
.






2H
2
O(s)


2H
2
O(l)

Δ
H = 2(6.01 kJ) = 12.0 kJ

d.

The physical states of all reactants and products must be
specified in thermochemical equations.






H
2
O(s)


H
2
O(l)

Δ
H = 6.01 kJ






H
2
O(l)


H
2
O(g)

Δ
H = 44.0 kJ

5.

Stoichiometry of thermochemical equations

a.

Balanced

chemical equations with their corresponding ΔH
value

b.

Use the ΔH value in calculations as you would
stoichiometric coefficients

e.

Example



How much heat is evolved when 266 g of white
phosphorus (P
4
) burn in air?







f.

Ex
ample


When Al(s) is exposed to atmospheric oxygen,
it is oxidized to form Al
2
O
3
(s). How much heat is released
by the complete oxidation of 24.2 g Al at 25°C and 1 atm?

Thermodynamics


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


2Al
2
O
3
(s)


Δ
H =
-
3352 kJ/mole

(24.2 g Al)(1 mol Al/27.0 g Al)(1 mol

rxn/4 mol
Al)(
-
3352 kJ/mol rxn) =
-
751 kJ

751 kJ of heat is released

6.

Some important types of enthalpy change

a.

Heat of combustion (ΔH
comb
)

C
4
H
10
(l) + (13/2)O
2
(g)


4CO
2
(g) + 5H
2
O(g)

b.

Heat of formation (ΔH
f
)

K(s) + ½Br
2
(l)


KBr(s)

c.

Heat of fusion (ΔH
fus
)

NaCl(s)


NaCl(l)

d.

Heat of vaporization (ΔH
vap
)

C
6
H
6
(l)


C
6
H
6
(g)



F.

A Comparison of
ΔH and ΔE

1.

Many reactions involve little (if any) PV work (ΔH ≈ ΔE)




a.

Reactions that do not involve gases




b.

Reactions in which the moles of gas does not change

c.

Reactions in which the moles of gas does change, but q
>>> PΔV



2.

Example


ΔH =
-
367.5 kJ/mole




ΔE = ΔH


PΔV




At
25

o
C

& 1 atm,
1 mole H
2

= 2
4
.
5

L

(use PV = nRT; calculate V)








ΔE =

367.
5 kJ/mole


(8.314 J/K·mole)(1 kJ/1000 J)(298 K)(1 mole)





=

370.0 kJ/mole (assume ideal gas behavior)

3.

Example



What is ΔE for the formation of 1 mole of CO at 1 atm
and 25
o
C?




C(graphite) + ½O
2
(g)


CO(g)

Δ
H =

110.5 kJ/mole




************








ΔE =

110
.5 kJ/mole


(8.314 J/K·mole)(1 kJ/1000 J)(298 K)(
0.5

mole)





=

112.0 kJ/mole (assume ideal gas behavior)




************




ΔE = ΔH


PΔV

At 25
o
C & 1 atm, 1 mole H
2

= 24.5 L (use PV = nRT; calculate V)








ΔE =

110.5 kJ/mole


2.48 kJ/mole =

113.0 kJ/mole






Thermodynamics


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Calorimetry


A.

Calorimetry

1.

Measuring the temperature change of a known amount of
substance (usually water) of known specific heat

2.

Temperature change is a result of the

absorption or release of heat
by the process (q lost = q gained)


B.

Specific Heat and Heat Capacity

1.

Heat capacity (C)

a.

Heat needed to raise the temperature of an object 1°C

b.

C = (m)(s)

2
.

Specific heat capacity (
s
)


also called specific heat

a.

Heat

needed to raise the temperature of 1 g of substance
1°C

b.

q = (m)(
s
)(ΔT)

c.

q = (C)(ΔT)

d.

ΔT = T
final

-

T
initial

3.

Example



How much heat is given off when an 869 g iron bar
cools from 94
o
C to 5
o
C?



s of Fe = 0.444 J/g
o
C



T

= T
final

-

T
initial

= 5

o
C


94
o
C =

89
o
C

q = (m)(s)(ΔT) = (869 g)(0.444 J/g
o
C)(


89
o
C) =

34,000 J

4
.

Example



H
ow much heat in J is needed to raise the temperature
of 205 g water from 21.2°C to 91.4°C? C for water = 4.184 J/g
-
°C.


q = (m)(C)(

T) = (205 g)(4.184 J/g
-

C)(91
.4


21.2)




= 6.02 x 10
4

J

5
.

Example



H
ow much heat in calories is needed to raise the
temperature of 205 g Fe from 294.2 K to 364.4 K? C for Fe =
0.444 J/g
-
°C.





294.2 K = 21.05

C





364.4 K = 91.25

C




q = (m)(C)(

T) = (205 g)(0.444 J/g
-

C)(91
.25


21.05)




q = (6.39 x 10
3

J)(cal/4.184 J) = 1.53 x 10
3

cal



C.

Constant
-
Volume Calorimetry

1.

Uses a steel container called a “bomb” calorimeter


process
closed to the atmosphere


2.

Typical application

a.

The sample is placed in the bomb and immer
sed in a know
amount of water of known temperature



b.

The sample is ignited electrically

c.

With stirring, final temperature of water (heat given off by
the sample) is measured

Thermodynamics


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d.

q lost = q reaction



e.

q gained = q water + q calorimeter


D.

Constant
-
Pressure Calorimetry

1.

Uses a “coffee cup” calorimeter


process open to the atmosphere

2.

Typical application

a.

Solid is weighed and heated to a known temperature

b.

Add solid to a sample of water of known temperature and
mass in the calorimeter

c.

Wit
h stirring, final temperature of water (final temperature
of solid) is measured

d.

q lost = q solid

e.

q gained = q water + q calorimeter

3.

Example



3.358 kJ of heat is added to a calorimeter that contains
50.00 g water. The temperature of the water and

calorimeter,
originally at 22.34°C, increases to 36.74°C. Calculate the heat
capacity of the calorimeter.
s

for water = 4.184 J/g
-
°C.

q
lost

= q
gained

(3.358 kJ)(1000 J/kJ) = 3358 J

ΔT
w
=ΔT
c
= 36.74

C
-

22.34

C = 14.4

C

3358 J = m
w
s
w
ΔT
w

+ C
p
ΔT
c

3358 J =

(50.00 g)(4.184 J/g
-

C)(14.4
º
C) + C
p
(14.4

C)

3358 J = 3012.5 J + C
p
(14.4

C)

3358 J
-

3012.5 J = C
p
(14.4

C)

345.52 J = C
p
(14.4

C)

24.00 J/

C = C
p

4.

Example



CuSO
4
(aq) is mixed with NaOH(aq) in the calorimeter
in the previous problem. The total solu
tion mass is 102 g. The
initial temperature of the solution is 23.35°C and the final
temperature is 26.65°C. Calculate the heat evolved. Assume
s

for
the solution = 4.184 J/g
-
°C.

q
lost

= q
gained

ΔT
w
=ΔT
c
= 26.65

C
-

23.35

C = 3.3

C

q
lost

= m
s
s
s
ΔT
s

+ C
p
ΔT
c

q
lost

= (102 g)(4.184 J/g
-

C)(3.3

C) + (24.00 J/

C)(3.3

C)

q
lost

= 1408.33 J + 79.2 J = 1488 J = 1.49 kJ




Standard Enthalpy of Formation and Reaction



A.

Standard States

1.

For gases, the standard state is 1 atm with the gas behaving ideally

2.

Fo
r aqueous solutions, the standard state is 1 M concentration

3.

For pure substances, the standard state is the most stable form of
the substance at 1 atm and the temperature of interest (usually
25°C)

Thermodynamics


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B
.

Standard Enthalpy of Formation
(ΔH
f
o
)

1.

T
he heat
change that results when one mole of a compound is
formed from its elements at a pressure of 1 atm.

2.

An element in its standard state is assigned a ΔH°
f

= 0

3.

Most compounds have a negative ΔH°
f

(heat is given off when the
compound forms)



C
.

Standard
Enthalpy of Reaction
(ΔH
rxn
o
)

1
.

T
he enthalpy of a reaction carried out at 1 atm.



2
.

Hess’ Law of Summation

a.

The enthalpy change of an overall process is the sum of the
enthalpy changes of its individual steps (ΔH° rxn = ΔH°
step 1 + ΔH° step 2 + …)

b.

ΔH° rxn = ∑(n)(ΔH°
f

products)


∑(n)(ΔH°
f

reactants)

c.

See CS
2

example on slide.

d
.

Example



Two gaseous pollutants that form auto exhaust
are CO and NO. An environmental chemist is studying
ways to convert them to less harmful gases through the
follow
ing reaction

CO(g) + NO(g)


CO
2
(g) + ½N
2
(g)

Given the following information, calculate ΔH° rxn for the
above reaction.






CO(g) + ½O
2
(g)


CO
2
(g)

ΔH° =
-
283.0 kJ






N
2
(g) + O
2
(g)


2NO(g)

Δ
H° = 180.6 kJ






CO(g) + ½O
2
(g)



2
(g)


ΔH° =
-
283.0 kJ


N
O(g)


봠N
2
(g) + ½ O
2
(g)


Δ
H° = (½)(
-
180.6 kJ)

CO(g) + NO(g)


2
(g) + ½N
2
(g)

ΔH° =
-
374.3 kJ




e.

See benzene example on slide.




Heat of Solution and Dilution



A.

Heat of Solution (ΔH
soln
)

1.

The heat generated or absorbed when a certain amount of sol
ute
dissolves in a certain amount of solvent.



2.





B.

Heat of Dilution



1.

The heat change associated with the dilution process.

2.

If a certain solution process is exothermic, further dilution will also
be exothermic.

3.

If

a certain solution process is endothermic, further dilution will
also be endothermic.