Lecture 23 - organic.cc.stonybrook.edu
Transcript of Lecture 23 - organic.cc.stonybrook.edu
Lec-25-Thermo2-Complete.ppt 10/17/2021
Thermochemistry
Chemical Reactions. Hess Law.
Calorimetry 1
1
Lecture 23Energy and Thermochemistry
Chapter 13
Office Hours:
Monday 2:30 – 4:30 PM Zoom Meeting accessed
from Blackboard.
Fernando Raineri
Chemistry Department Room 519
1
Lec-25-Thermo2-Complete.ppt 10/17/2021
Thermochemistry
Chemical Reactions. Hess Law.
Calorimetry 2
2
Enthalpy Changes: Chemical Processes
( ) ( ) ( ) 1
2
-
2 2
1H 241.8kJ mog O g H O g
2lr H =+ → −
ΔrH at any temperature and 1 bar: standard enthalpies
of reaction Δ r H0
pure reactants at pure products at , ,
A CB D
T PT P
rc Ha b d+ + → + +
2
Lec-25-Thermo2-Complete.ppt 10/17/2021
Thermochemistry
Chemical Reactions. Hess Law.
Calorimetry 3
Enthalpy Changes: Formation Reactions ΔfH
°
Standard molar enthalpies of formation:
enthalpy change ΔfH to produce a mol of substance
starting from the elements in their stable state at
standard conditions (1 bar). It is customary to
tabulate data at 25 ºC.
( ) ( ) ( ) 0 -112 2 22
H g O g H O 285.8kJ molf H+ → = −
( ) ( ) ( ) 0 -1
2 2 42C graphite 2H g C H g 52.26kJ molf H+ → =
( ) ( ) 0
2 2H g H g 0f H→ =
The enthalpy of formation of an element in its standard
state at 25°C is zero. 3
3
Lec-25-Thermo2-Complete.ppt 10/17/2021
Thermochemistry
Chemical Reactions. Hess Law.
Calorimetry 4
( ) ( )( )2C s + 2H g
( )4CH ,gf H−
4
( ) ( ) ( ) ( )2 24 2CCH g O g2O g 2H O l+ → +
( ) ( )( )22O g O g+
( )22 O ,g 0f H− =
+
Standard Enthalpy of ReactionEnthalpy is a function of state.
For any given transformation: (initial) → (final)
ΔH is independent of the path.
4
Lec-25-Thermo2-Complete.ppt 10/17/2021
Thermochemistry
Chemical Reactions. Hess Law.
Calorimetry 5
5
( ) ( ) ( ) ( )2 24 2CCH g O g2O g 2H O l+ → +
( ) ( )( )2C s + 2H g
( )4CH ,gf H−
( ) ( )( )22O g O g+
( )22 O ,g 0f H− =
+
( )22 H O,gf H ( )2CO ,gf H
Standard Enthalpy of Reaction
5
Lec-25-Thermo2-Complete.ppt 10/17/2021
Thermochemistry
Chemical Reactions. Hess Law.
Calorimetry 6
6
( ) ( ) ( ) ( )2 24 2CCH g O g2O g 2H O l+ → +
( ) ( )( )2C s + 2H g
( )4CH ,gf H−
( ) ( )( )22O g O g+
( )22 O ,g 0f H− =
+
( )22 H O,gf H ( )2CO ,gf H
Standard Enthalpy of Reaction
( ) ( )( )
( ) ( )( )
2
4 2
2CO ,g 2 H O
CH ,g 2 O
,
g
l
,f f
fc f
H
HH H
H
+
= +
−
6
Lec-25-Thermo2-Complete.ppt 10/17/2021
Thermochemistry
Chemical Reactions. Hess Law.
Calorimetry 7
7
( ) ( )0
Products a
0
R t
0
e c ants
r Rf fP HH RPH
=
−
pure reactants at p
0
, 1bar ure products at , 1bar
CA B D
T P T P
ra b Hc d
==
+ + → + +
( ) ( )( )
( ) ( )( )
0 0
sum over the p
00
sum over the reactan s
roducts
t
0 C D
A B
f
f f
r fc H d
a
H
H b
H
H
+
= + −
+
+
7
Lec-25-Thermo2-Complete.ppt 10/17/2021
Thermochemistry
Chemical Reactions. Hess Law.
Calorimetry 8
Enthalpy Changes in Chemical Reactions:
Hess’s Law
Because enthalpy is a function of state, we can
combine any number of auxiliary reactions and their
enthalpies to calculate the enthalpy of a reaction of
interest.
8
8
Lec-25-Thermo2-Complete.ppt 10/17/2021
Thermochemistry
Chemical Reactions. Hess Law.
Calorimetry 9
( ) ( ) ( ) 0
2 3 2 3 2 3CH CHCH g H g CH CH CH g 124kJ/molhH+ → = −
The standard reaction enthalpy for the conversion of
propene to propane at 25 °C is:
The standard reaction enthalpy of formation for liquid
water at 25 °C is:
( ) ( ) ( ) 0
2 2 2
1H g O g H O 286kJ/mol
2f H+ → = −
What is the standard reaction enthalpy of combustion
of propene at 25 °C?
( ) ( ) ( ) ( ) ( )0
2 3 2 2 2 2 3
9CH CHCH g O g 3CO g +3H O CH CHCH ?
2cH+ → =
9
( )1−
The standard reaction enthalpy for the combustion
propane at 25 °C is:
( ) ( ) ( ) ( ) 0
3 2 3 2 2 2CH CH CH g 5O g 3CO g 4H O 2220kJ/molcH+ → + = −
9
Lec-25-Thermo2-Complete.ppt 10/17/2021
Thermochemistry
Chemical Reactions. Hess Law.
Calorimetry 10
( ) ( ) ( ) ( ) ( )0
2 3 2 2 2 2 3
9CH CHCH g O g 3CO g +3H O CH CHCH ?
2cH+ → =
( ) ( ) ( ) 0
2 3 2 3 2 3CH CHCH g H g CH CH CH g 124kJ/molhH+ → = −
( ) ( ) ( ) 0
2 2 2
1H O H g O g 286kJ/mol
2f H→ + − =
( ) ( ) ( ) ( ) ( )0
2 3 2 2 2 2 3
9CH CHCH g O g 3CO g +3H O CH CHCH ?
2cH+ → =
( ) ( ) ( ) ( )
( )
( )
00
2 3
0
2 3
0
2
0 0
2 3 3 2 3 2
3
CH CHCH CH CH CH H O
kJ kJ kJ124 2220 286
mol m
CH CHCH
ol molCH CHCH
kJCH CHCH 2058
mol
h cc
c
c
fH H HH
H
H
− − −
−
= + −
= + −
=
10
9
2
( ) ( ) ( ) ( ) 0
3 2 3 2 2 2CH CH CH g 5O g 3CO g 4H O 2220kJ/molcH+ → + = −
10
Lec-25-Thermo2-Complete.ppt 10/17/2021
Thermochemistry
Chemical Reactions. Hess Law.
Calorimetry 11
Measuring Enthalpy Changes: Calorimetry
Constant Pressure Calorimeter
0pq =
Due to the insulating
nature of the walls
there is no heat
exchanged between
the system (the
contents of the
calorimeter) and the
outside.
The process is at
constant pressure11
11
Lec-25-Thermo2-Complete.ppt 10/17/2021
Thermochemistry
Chemical Reactions. Hess Law.
Calorimetry 12
Measuring Enthalpy Changes: Calorimetry
Constant Pressure Calorimeter
Ti
reactants
T
Extent of
reaction
Tf
products
ΔH
0pH q = =
ΔH1
ΔH2
21 0HH H= + =
( )2
final s,fi1 nal
1 H
m CH C T
H
−=
+=
−
12: heat capacity of the calorimeterC
12
Lec-25-Thermo2-Complete.ppt 10/17/2021
Thermochemistry
Chemical Reactions. Hess Law.
Calorimetry 13
Measuring Enthalpy Changes: Calorimetry: Constant V Bomb Calorimeter
bomb calorimeterq q=−
Due to the insulating
nature of the walls
there is no heat
exchanged between
the inside of the
calorimeter and the
outside.
The process in the
bomb is at constant
volume
13
: heat capacity
of the calorimeter
C
13
Lec-25-Thermo2-Complete.ppt 10/17/2021
Thermochemistry
Chemical Reactions. Hess Law.
Calorimetry 14
14
( ) ( )0 0 0
gasesr r r r rH U PV U n RT = + = +
( ) ( ) ( ) ( )7 8 2 2 2C H 9O g 7CO g 4 H O+ → +
( ) ( ) ( )2 2gasCO O 7 9 2n = − = − = −
0 0 2r rH U RT = −
As the process is at constant volume
calorimeter wat rbom er ,wateb ( )sq m C Cq TU = = =− − +
14
Lec-25-Thermo2-Complete.ppt 10/17/2021
Thermochemistry
Chemical Reactions. Hess Law.
Calorimetry 15
15
Spontaneous and Non-Spontaneous Processes
Splash!
The falling down of the
object is a spontaneous
process.
The rising up of the object
is a non-spontaneous process.But not impossible.
15
Lec-25-Thermo2-Complete.ppt 10/17/2021
Thermochemistry
Chemical Reactions. Hess Law.
Calorimetry 16
16
Spontaneous and Non-Spontaneous Processes
A spontaneous process is a process that occurs
naturally, without our help or intervention.
Some examples are:
❖ Melting of ice at 30 °C.
❖ Dissolution of small amounts of sucrose (table
sugar) in water (as in sweetening a coup of tea).
❖ Oxidation of iron in an environment with O2(g):
( ) ( )2 2 3
32 Fe O ( ) Fe O
2s g s+ →
16
Lec-25-Thermo2-Complete.ppt 10/17/2021
Thermochemistry
Chemical Reactions. Hess Law.
Calorimetry 17
17
Correspondingly, a non-spontaneous process is a
process that does not occurs naturally, unless we help
or intervene.
The reverse of the processes cited above are examples:
❖ freezing of water at 30 °C.
❖ Precipitation of dissolved sucrose from water at 60
°C (if the amount originally dissolved was small).
❖ Reduction of iron oxide to neat iron:
( ) ( )2 3 2
3Fe O 2 Fe O ( )
2s s g→ +
Notice that if we intervene (by changing the conditions),
we may be able to make a non-spontaneous process
happen. Therefore, non-spontaneous doesn’t mean
impossible.
17
Lec-25-Thermo2-Complete.ppt 10/17/2021
Thermochemistry
Chemical Reactions. Hess Law.
Calorimetry 18
Directionality of Chemical Processes
18
It would be very useful if we could predict if a chemical
reaction
or a phase transition
C+ and consta+A D+ nB+ tT Pa b c d→
will proceed spontaneously (or naturally).
Phase and constantPhase T P→
The Second Law of Thermodynamics provides us with
the tools for developing a very useful spontaneity
criterion for processes that take place at constant
temperature and pressure.
18
Lec-25-Thermo2-Complete.ppt 10/17/2021
Thermochemistry
Chemical Reactions. Hess Law.
Calorimetry 19
System and its surroundings
19
Surroundings
System
Energy can be exchanged
between system and
surroundings through the
walls that separate them.
19
Lec-25-Thermo2-Complete.ppt 10/17/2021
Thermochemistry
Chemical Reactions. Hess Law.
Calorimetry 20
Second Principle of Thermodynamics
❖ The entropy S is an extensive function of state.
❖ In a reversible process the entropy of the universe is
constant.
❖ In a spontaneous or irreversible process the entropy
of the universe increases. 20
In the following “Universe” refers to the combination
( ) ( ) ( )System Surroundings Universe+ =
universe system surroundingsS S S= +
universe 0S =
universe 0S
20