Transcript Chemistry: The Study of Change
Chapter 6: Thermochemistry
Ch.6 HW: 2, 8, 11, 15, 23, 24 a-b , 25, 34, 38, 39, 51, 53 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
Energy
is the capacity to do
work
or produce
heat
.
Energy can exist in
many forms
: (not an exhaustive list) • Kinetic •
Potential
• Mechanical •
Chemical
• Magnetic •
Radiant
(Electromagnetic) •
Nuclear
• Gravitational •
Thermal
• Electrical • Acoustic • Electrostatic
First law of thermodynamics
– energy can be converted, but can never be created or destroyed.
Methane combustion in a Bunsen burner is an exothermic chemical reaction (generates thermal energy) Chemical energy
lost
by combustion
=
Thermal energy
gained
by the surroundings
Mechanical energy
is sum of
kinetic
(energy of motion) &
potential
(energy of position) of
macroscopic objects
.
Trebuchet
converting potential energy to kinetic energy to perform work (moving matter).
Thermal energy
is the sum
kinetic
energy associated with the random motion of
microscopic atoms and molecules
.
Less thermal energy More thermal energy
Electromagnetic (radiant) energy
is transmitted through electric/magnetic waves travelling at the speed of light. • Varies in energy content: Gamma, X-ray, Visible, IR, Radio, etc.
• Characterized by a specific wavelength ( l ) and frequency ( n ) • Earth’s fundamental energy source (from the sun).
Chemical energy
is the energy
stored within the bonds
of chemical substances.
• Bonds are a type of potential energy • Breaking absorbs/Creating bonds emits energy • Vary in energy content depending on type of bond (covalent, ionic, intermolecular)
Nuclear energy
is the energy stored within the collection of neutrons and protons in the atom.
• Required to hold nucleons together • Converts measurable amount of matter to energy (mass defect) • Incredibly exothermic
Thermochemistry
is the study of heat change in chemical processes.
Heat
is the transfer of
thermal energy.
• Occurs between bodies at different temperatures • Transfers from high to low temperature
Temperature
is a
relative measurement
of the thermal energy.
(°F, °C, K) Temperature = Thermal Energy Temperature Thermal Energy As Temp. ↑, Thermal Energy ↑ Crash Course: Energy & Chemistry www.youtube.com/watch?v=GqtUWyDR1fg
The
system
is the specific part of the universe that is of interest in the study.
The
surroundings
is every other part of the universe that
could exchange mass or energy
with the system
System Type: Exchange:
open mass & energy closed energy isolated nothing
An exothermic process
gives off thermal energy and transfers it to the surroundings. (Feels hot to the touch) Combustion: 2H 2 (
g
) + O 2 (
g
) → 2H 2 O (
l
) + energy Freezing of water: H 2 O (
l
) → H 2 O (
s
) + energy
An endothermic process
takes in thermal energy from the surroundings. (Feels cool to the touch) Dissolution of NH 4 NO 3 : energy + NH 4 NO 3 (
s
) H O → NH 4 +1 (aq) + NO 3 -1 (aq) Melting of ice: energy + H 2 O (
s
) → H 2 O (
l
)
Schematic of Exothermic and Endothermic Processes Combustion
Thermodynamics
is the scientific study of the interconversion of heat and other kinds of energy.
State functions
are properties that are determined by the state of the system, regardless of how that condition was achieved.
energy , pressure, volume, temperature
U: Internal energy
D
U
=
U final
-
U initial
Potential energy of hiker 1 and hiker 2 is the same even though they took different paths.
1 st Law of Thermodynamics
D
U system
=
-
D
U surroundings
D
P
=
P final
-
P initial
D
V
=
V final
-
V initial
D
T
=
T final
-
T initial
Another form of the
first law
for D
U
system D
U
=
q
+
w
D
U
is the change in
internal energy
of a system q is the
heat exchange
between the system and the surroundings
w
is the
work performed
on (or by) the system Both forms of energy transfer combine cumulatively to show
net energy transfer
to/from system
Work Done by/on the System The expansion/compression of gases performs mechanical work
Work
=
Force × distance w
=
F × d
=
P
D
V F d 2 × d 3 = F x d = w Equivalency: w
= 1 L • atm = 101.3 J *Note: No work is done if not expanding against pressure (
w
= 0 if
P
= 0 atm)
w
= -
P
D
V
initial final Negative because acting on surroundings (losing energy)
Example
6.1
A certain gas expands in volume from 2.0 L to 6.0 L at constant temperature.
Calculate the work done
by the gas if it expands (a) against a vacuum
(a) w
= −
P
D
V
= −(0)(6.0 − 2.0) L = 0 J (no work is performed) (b) against a pressure of 1.2 atm
(b) w
= −
P
D
V
= −(1.2 atm) (6.0 − 2.0) L = −4.8 L · atm
Example
6.2
The work done when a gas is compressed in a cylinder is 462 J. During this process, there is a heat transfer of 128 J from the gas to the surroundings.
Calculate the energy change
( D U) for this process.
→
System is compressed
(worked upon) so work is
positive
(+462 J)
Heat is released
by the gas,
q
is
negative
(-128 J) D
U
=
q
+
w
= −128 J + 462 J = 334 J There is a
net gain of energy
the system in this process for
Crash Course: Energy and Chemistry
www.youtube.com/watch?v=GqtUWyDR1fg
Enthalpy (H):
a quantification of heat flow as a state function at constant pressure.
Work (
w
) and heat (
q
) are
not
state functions ( D
w
≠
w
final –
w
initial ) Fortunately, most reactions occur under conditions of
constant pressure
D
U
= (
q
+
w) = (q – P
D
V) q p
= D
U
+
P
D
V (constant pressure)
Since heat is written in terms of U/P/V, which are all state functions, so is q p q p is now referred to as Enthalpy (H) D
H
=
H
(products) –
H
(reactants) D
H
= D
U
+
P
D
V
D
H
= heat given off or absorbed during a reaction
at constant pressure
Enthalpy (H)
is used to quantify the heat flow into or out of a system in a process that occurs at constant pressure.
H
products <
H
reactants D
H
< 0
H
products >
H
reactants D
H
> 0 Crash Course: Enthalpy www.youtube.com/watch?v=SV7U4yAXL5I
Thermochemical Equations
: show enthalpy changes.
Is D
H
negative or positive?
System absorbs heat Endothermic; D
H
> 0 6.01 kJ are absorbed for every 1 mole of ice that melts at 0 0 C and 1 atm.
H 2 O (
s
) H 2 O (
l
) D
H
= 6.01 kJ/mol 17
• • • • Thermochemical Equations The stoichiometric coefficients always refer to the number of moles of a substance 1 H 2 O (
s
) H 2 O (
l
) D
H
= 6.01 kJ/ 1 mol If you reverse a reaction, the sign of D
H
changes H 2 O (
l
) H 2 O (
s
) D
H
= 6.01
kJ/mol If you multiply both sides of the equation by a factor
n
, then D
H
must change by the same factor
n
.
2H 2 O (
s
) 2H 2 O (
l
) D
H
= 2 x 6.01 = 12.0 kJ The physical states must be specified in thermochemical equations.
H 2 O (
s
) H 2 O (
l
) D
H
= 6.01 kJ/mol H 2 O (
l
) H 2 O (
g
) D
H
= 44.0 kJ/mol
Thermochemical Equations Is D
H
negative or positive?
System gives off heat Exothermic; D
H
< 0 CH 4 890.4 kJ are released for every 1 mole of methane that is combusted at 25 0 C and 1 atm.
(
g
) + 2O 2 (
g)
CO 2 (
g) +
2H 2 O (
l
) D
H
= -890.4 kJ/mol 19
Example
6.3
Given the thermochemical equation:
2SO 2(g)
D H = -198.2 kJ/mol
+ O 2(g) → 2SO 3(g)
Calculate the heat
evolved when 87.9 g of SO 2 is converted to SO 3
Solution
We first calculate moles of SO 2 in 87.9 g of the compound and then find the number of kJ produced from the exothermic reaction.
Example
In the melting process water has the following thermochemical process: H 2 O (
s
) H 2 O (
l
) D
H
= 6.01 kJ/mol
Calculate the heat
of 3 moles of liquid water freezing to a solid.
A Comparison of D
H
and D
U
2Na(
s
) + 2H 2 O(
l
) 2NaOH(
aq
) + H 2 (
g
) D
H
= -367.5 kJ/mol D
U
= D
H
-
P
D
V
At 25 o C, 1 mole H 2 = 24.5 L at 1 atm
P
D
V
= 1 atm x 24.5 L = 2.5 kJ D
U
= -367.5 kJ/mol – 2.5 kJ/mol = -370.0 kJ/mol D
U
≈ D
H Work
done is
often insignificant
compared to
Heat
flow 22
Example
6.4
Calculate the change in internal energy when 2 moles of CO are converted to 2 moles of CO 2 at 1 atm and 25°C: D U = D H - P D V From the ideal gas law we know: P D V = D nRT D U = D H - RT D n The reaction uses 3 moles of gas to produce 2 moles of gas D n = 2 – 3 = -1
Carbon monoxide burns in air to form carbon dioxide.
*Note: Again, work
done is insignificant compared to
heat
(enthalpy)
The
specific heat (s)
is the amount of heat (
q
) required to raise the temperature of
one gram
of the substance by
1
°C.
• Metals have
low specific heats
which is why they heat up and cool down
rapidly
.
• Water takes longer to heat up due to a
larger specific heat
. Takes
more thermal energy
to raise its
temperature.
Experimentally determined
The
heat capacity
(
C
) is the amount of heat (
q
) required to raise the temperature of
a given quantity
(
mass
) by
1
° C .
C
=
m × s
Heat (
q
) absorbed or released:
q
=
m × s × q
=
C ×
D
t
D
t
D
t
=
t
final -
t
initial
Example
6.5
A 466-g sample of water is heated from 8.50°C to 74.60°C. Calculate the amount of heat absorbed (in kJ) by water.
Heat capacity (s) of water: 4.184 J/g•
°C (from reference table) D
t
=
74.60
–
8.50 = 66.1
°C
q
=
m × s ×
D
t Positive value
indicates energy was
absorbed
via heat
Calorimetry: measurement of heat changes in physical or chemical processes (using a calorimeter).
Bomb calorimeter
: constant-volume If
volume is constant
D
U
= (
q
then
zero work
+
w) = q sys
can be done
(only heat) q
sys =
q
water +
q
bomb +
q
rxn
q
sys = 0
q
rxn
q
= - ( water
q
water = +
q
bomb )
m ×
s
×
D
t q
bomb =
C bomb ×
D
t Standards
Reaction at Constant
V
D
H
=
q
rxn D
H
~
q
rxn No heat enters/leaves the system 26
Example
6.6
Naphthalene (C 10 H 8 ) is the pungent-smelling substance used in moth repellents. A quantity of 1.435 g was burned in a constant-volume bomb calorimeter.
Consequently, the temperature of the water rose from 20.28°C to 25.95°C. If the heat capacity (C cal ) of the bomb plus water was 10.17 kJ/°C,
What is the molar heat of combustion of naphthalene?
Solution
The heat absorbed by the bomb and water is equal to the product of the heat capacity and the temperature change.
q
cal = −
q
rxn = -57.66 kJ
Chemistry in Action:
Energy content
of food is measured with bomb calorimeters C 6 H 12 O 6 (
s
) + 6O 2 (
g
) 6CO 2 (
g
) + 6H 2 O (
l
) D
H
= -2801 kJ/mol 2801 kJ of energy are released from glucose whether burned or digested 1 Cal = 1,000 cal = 4184 J Substance Apple Beef Bread Cheese Butter D
H combustion
(kJ/g) -2 -8 -11 -18 -34 *Note: Shown per unit gram (not mole)
No heat enters/leaves CrashCourse Calorimetry: www.youtube.com/watch?
v=JuWtBR-rDQk Constant-Pressure Calorimetry
q
sys =
q
water +
q
cal +
q
rxn
q
sys = 0
q
rxn
q
water = - ( =
q
water
m ×
s +
q
cal )
×
D
t q
cal =
C cal ×
D
t Standards
Reaction at Constant
P
D
H
=
q
rxn
Example
6.7
A lead (Pb) pellet having a mass of 26.47 g at 89.98°C was placed in a constant-pressure calorimeter of negligible heat capacity containing 100.0 g of water. The water temperature rose from 22.50°C to 23.17°C.
What is the specific heat of the lead pellet
?
Example
6.7 Solution
Solution
The heat gained by the water is given by
m
= 100.0 g of water
s H2O = 4.184 J/g•
°C (from reference table) D
t
=
23.17
−
22.50 = 0.67
°C The heat
gained by water
was
lost by the mass of lead: q
Pb = −280.3 J
Example
6.8
100. mL of 0.50
M
HCl was mixed with 100. mL of 0.50
M
NaOH in a constant-pressure calorimeter of negligible heat capacity. Upon mixing the
temperature rose
from 22.50°C to 25.86°C. Calculate the
molar heat change
for the neutralization reaction: Assume that the densities and specific heats of the solutions are the same as for water (1.00 g/mL and 4.184 J/g · °C, respectively).
q
rxn = −
q
soln , where
q
soln is the heat absorbed by the combined solution.
Example
6.8 Solution
Because
q
rxn = −
q
soln ,
q
rxn = −2.81 kJ.
From the molarities given, the number of moles in solution is Therefore, the heat of neutralization when 1.00 mole of HCl reacts with 1.00 mole of NaOH is
The
Heat of Solution
( D
H
soln
) is the heat generated or absorbed when a solute dissolves in a solvent.
Energy given off
once ions are stabilized by hydration
Required energy intake
to break electrostatic forces * * Large intake of energy lowers temp.
Useful for instant icepacks D
H
soln = Step 1 + Step 2 = 788 – 784 = 4 kJ/mol
Standard state
:
designated conditions
used as a reference point to calculate various properties under different conditions.
(1 atm,
usually
25°C)
Standard enthalpy of formation
when
one mole
( D
H
f 0 ) is the heat change that results of a compound is
formed from its elements
at a pressure of 1 atm.
D The
standard enthalpy of formation
of any
element
in its
most stable form is zero
. (neutral, non-compound)
H
D
H
0 0 f f (O (O 2 3 ) = 0 ) = 142 kJ D
H
D
H
0 0 f f (C (C graphite diamond ) = 0 ) = 1.90 kJ D
H
0 f (Na) = 0 D
H
0 f (Na + ) = -239.66 kJ Using recorded D
H
f 0 data eliminates the need to experimentally measure enthalpy change for every reaction of interest.
Much like altitude is set to 0 ft at sea level (can be + or -) Most stable element form is set to 0, increases for less stable allotropes.
Will decrease (-) when forming compounds indicating more stable. 36
The
standard enthalpy of reaction
( D
H
0 rxn ) is the enthalpy of a reaction carried out at 1 atm.
a
A +
b
B
c
C +
d
D D
H
0 rxn = [
c
D
H
0 f ( C ) +
d
D
H
0 f ( D ) ] [
a
D
H
0 f ( A ) +
b
D
H
0 f ( B ) ] D
H
0 rxn = S
n
D
H
0 f (products) S
m
D
H
0 f (reactants)
Hess’s Law:
When reactants are converted to products, the change in enthalpy is the same whether the reaction takes place in
one step or in a series of steps.
(Enthalpy is a state function. It doesn’t matter how you get there, only where you start and end.) 37
Example
6.10 Heat of Reaction
The thermite reaction is shown below: 2Al (s) + Fe 2 O 3(s) → Al 2 O 3(s) + 2Fe (l) This reaction is highly exothermic and the liquid iron formed is used to weld metals.
Calculate the molar Heat of Reaction
The D H f ° for Fe (
l
) is 12.40 kJ/mol.
*Note: Values of
D
H f ° can be found in Appendix 3
The molten iron formed is run down into a mold between the ends of two railroad rails to weld them together.
Determining Standard Heat of Reaction ( D
H
0 rxn ) Direct Method C (graphite) D H f ° : 0 + O 2(
g
) → CO 2(
g
) 0 -393.5 kJ D
H
0 rxn = D
H
0 (CO2) D
H
0 (C) D
H
0 (O2) D
H
0 rxn = -393.5
kJ - 0 kJ - 0 kJ D
H
0 rxn = -393.5
kJ Indirect Method C (graphite) + 1/2O 2 (
g
) → CO (
g
) CO (
g
) + 1/2O 2 (
g
) → CO 2 (
g
) D H° rxn -110.5 kJ -283.0 kJ C (graphite) + O 2 (
g
) → CO 2 (
g
) -393.5 kJ
Example
6.9 Indirect Method
Calculate the standard enthalpy of formation of acetylene (C 2 H 2 : The equations for each step and corresponding enthalpy changes are:
Solution
Looking at the synthesis of C 2 H 2 , we need 2 moles of graphite as reactant. So we multiply Equation (a) by 2 to get Next, we need 1 mole of H 2 as a reactant and this is provided by (b).
Example
6.9 Solution
Last, we need 1 mole of C 2 H 2 as a product. If we flip the reaction direction then we also must change the sign Adding Equations (d), (b), and (e) together, we get 1 mole of C 2 H 2 synthesized from C and H 2 the surroundings. (Endothermic process) absorbs 226.6 kJ of heat from
Ch. 18 Entropy and Free Energy
Spontaneous
Physical and Chemical Processes • A waterfall runs downhill • At 1 atm, water freezes below 0 o C and ice melts above 0 o C • A lump of sugar dissolves in a cup of hot coffee • Heat flows from a hotter object to a colder object • A gas expands in an evacuated bulb • Iron exposed to O 2 & H 2 O rust spontaneous non-spontaneous 42
Review: Enthalpy (H)
is used to quantify the
heat flow
into or out of a system in a process that occurs at constant pressure.
D
H
= heat given off or absorbed during a reaction Combustion Melting D
H
< 0 Exothermic D
H
> 0 Endothermic • •
Bond disruption
is associated with a
positive
ΔH (endothermic).
Bond formation
is associated with a
negative
ΔH (exothermic).
• Chemical reactions involve
both
bond breaking and formation.
Processes
usually
lead to a lower energy state.
Does a decrease in enthalpy
always
mean a reaction proceeds spontaneously?
*Nope -D
H
0 Spontaneous processes CH 4 (
g
) + 2O 2 (
g
) CO 2 (
g
) + 2H 2 O (
l
) D
H
0 = -890.4 kJ/mol H + (
aq
) + OH N 2 (g) + 3H 2 (
g
) (
aq
) H 2 O (
l
) D
H
0 = -56.2 kJ/mol 2NH 3 (
g
) D
H
0 = -92.6 kJ/mol Each Exothermic H 2 O (
s
) NH 4 NO 3 (
s
) H 2 H 2 O O (
l
) D
H
0 = 6.01 kJ/mol NH 4 + (
aq
) + NO 3 (
aq
) D
H
0 Both Endothermic = 25 kJ/mol
* A second thermodynamic quantity, known as Entropy, is also needed to determine reaction spontaneity.
Entropy (S)
is a measure of the
randomness or disorder
of a system among the
possible ways
a system can contain energy/matter.
disorder
S
For processes we consider the change in entropy: D
S
=
S
f -
S
i If the change from initial to final results in an increase in randomness
S
f >
S
i D
S
> 0 For any substance, the solid state is more ordered than the liquid state and the liquid state is more ordered than gas state.
H 2 O (
s
)
S
solid <
S
liquid <<
S
gas H 2 O (
l
) D
S
> 0
Macroscopic Entropy Analogies
Disorder naturally occurs spontaneously It would take an input of energy to increase the order of the system
First Law of Thermodynamics (Ch. 6 Review)
Energy can be converted from one form to another but energy cannot be created or destroyed.
Second Law of Thermodynamics
(Minute Physics: Arrow of Time)
www.youtube.com/watch?v=GdTMuivYF30 The entropy of the
universe
increases in a spontaneous process and remains unchanged in an equilibrium process.
Spontaneous process: D
S
univ = D
S
sys + D
S
surr > 0 *Note: Neither
must
be greater than 0
Third Law of Thermodynamics
The entropy of a perfect crystalline substance is zero at 0 Kelvin. • As temperature (energy) increases, entropy (disorder) increases.
Processes that lead to an increase in entropy ( D
S
> 0) Melting Chemical reactions that increase the moles of gas 2NH 3(g) → 3H 2(g) + N 2(g) Vaporizing Diffusion
+
Heating
The Entropy Factor Increases with Temperature (
T
D
S)
If I only have 1 quarter there are only 2 ways to be disordered As the number of quarters increase there are a greater number of ways to be disordered Likewise, at low temperatures, there are less ways for energy to be dispersed (low entropy).
As temperature increases, there are more available states for energy and matter to exist leading to larger possible entropy. TedEd: What Triggers a chemical reaction?
www.youtube.com/watch?v=8m6RtOpqvtU
The Hydrophobic effect
• Non-polar solutes are difficult to dissolve in water • It can not form favorable interactions such as H bonds • Instead, water forms a strongly
ordered
cage about the non-polar molecule called a
clathrate
.
•
Clathrate formation
results in a large
decrease in entropy
• This is unfavorable, but can occur for small molecules TedEd: Why don't oil and water mix?
www.youtube.com/watch?v=h5yIJXdItgo
Atheistic view on Entropy
• Despite the
increasing order
of biological processes brought about by evolution, living organisms generate and put off heat (entropy). D
S
univ = D
S
sys + D
S
surr > 0 • As long as D
S
sys < D
S
surr • Our existence is the result of our metabolism’s capacity to transfer heat into the universe; therefore increasing entropy elsewhere. • We exist today only to increase universal entropy.
Contrasting view
• I believe Entropy supports creationism. Saying that as time goes on things get more chaotic. This implies that at one time everything was in perfect order.
• All science has shown the universe is constantly expanding, spreading out from a common origin (increasing entropy).
• The Big bang theory describes at one “time” all energy/mass was found in a ultra-dense point.
• There has never been a consistent theory as to why an infinitely dense mass existed
or
what “kick-started” the universe.
I believe the answer to both is God
Gibbs Free Energy
Entropy
and
Enthalpy
factors
combine to determine if a reaction is spontaneous
.
For a constant temperature and pressure process:
Gibbs free energy (G)
D
G
= D
H
sys -
T
D
S
sys Free-energy change ( D G) must be
negative
for a reaction to be spontaneous.
D
G
< 0 The reaction is spontaneous in the forward direction.
D
G
> 0 The reaction is non-spontaneous as written. The reaction is spontaneous in the reverse direction.
D
G
= 0 The reaction is at equilibrium.
D
G
=
D
H
-
T
D
S Exothermic reactions (-
D
H) and increasing Entropy (+
D
S) drive reactions, but also dependent on temperature
• In biological systems, T is assumed to be constant • ΔG is the amount of usable energy that may be extracted from a process • The value of ΔG is
not
related to the rate of a process (*Know this chart) 54
The Energetics of Boiling Water
H 2 O (l) → H 2 O (g) • • It is an
endothermic
process; D H = + Takes energy to reach a higher kinetic state This typically disfavors a process from occurring • • The process also increases entropy; D S = + This typically favors a reaction from occurring The entropy factor is temperature-dependent ( D S T) At
low temperatures
the
entropy factor
is smaller than the D H resulting in + D G D H > D S T Not Boiling At
temperatures above 100 °C
the
entropy factor
is greater than the D H resulting in D G D H < D S T Boiling
The Energy of dissolving Table Salt
•
Gibbs Free Energy
At equilibrium ΔG = 0.
Note that this means that entropy is at a maximum.
• Equilibrium is the state that all processes tend to approach. • Your body will not reach equilibrium until you are dead.
“Life is (among other things), an ultimately futile attempt to avoid equilibrium.” - Mark Brandt *Sad take on life in my opinion Crash Course: Entropy and Free energy www.youtube.com/watch?v=ZsY4WcQOrfk
Coupled Reactions Could the metal
weight A spontaneously rise
?
It would be a
non-spontaneous process
.
A If a heavier
weight B
is coupled,
weight A
can move upward spontaneously by coupling it with the falling of a larger weight.
B • Many biological reactions are
energetically unfavorable.
• They rely on coupling with other molecules to release a surplus of free energy.
ADP ATP D
G
0 = -31 kJ
Coupled Reactions Example: Alanine + Glycine Alanylglycine D
G
0 = +29 kJ
non-spontaneous
ATP + H 2 O + Alanine + Glycine ADP + H 3 PO 4 + Alanylglycine D
G
0 = -2 kJ
spontaneous
The Hydrophobic effect
• Non-polar solutes are difficult to dissolve in water • It can not form favorable interactions such as H-bonds • Instead, water forms a strongly
ordered
cage about the non-polar molecule called a
clathrate
.
•
Clathrate formation
results in a large
decrease in entropy
• This is unfavorable, but can occur for small molecules