Download Chemistry: The Study of Change

Chapter 6:
Thermochemistry
Ch.6 HW: 2, 8, 11, 15, 23,
24a-b, 25, 34, 38, 39, 51, 53
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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:
open
closed
isolated
Exchange:
mass & energy
energy
nothing
An exothermic process gives off thermal energy and
transfers it to the surroundings. (Feels hot to the touch)
Combustion: 2H2 (g) + O2 (g) → 2H2O (l) + energy
Freezing of water: H2O (l) → H2O (s) + energy
An endothermic process takes in thermal energy from
the surroundings. (Feels cool to the touch)
Dissolution of NH4NO3:
H2O
energy + NH4NO3 (s) → NH4+1 (aq) + NO3-1 (aq)
Melting of ice: energy + H2O (s) → H2O (l)
Schematic of Exothermic and Endothermic Processes
Combustion
Melting
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
DU = Ufinal - Uinitial
1st Law of Thermodynamics
DUsystem = -DUsurroundings
DP = Pfinal - Pinitial
DV = Vfinal - Vinitial
Potential energy of hiker 1 and hiker 2 is the
same even though they took different paths.
DT = Tfinal - Tinitial
Another form of the first law for DUsystem
DU = q + w
DU 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 DV
F
Because: P × V = d2 × d3 = 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 DV
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 = −PDV
= −(0)(6.0 − 2.0) L
= 0 J (no work is performed)
(b) against a pressure of 1.2 atm
(b) w = −PDV
= −(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 (DU) 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)
DU = q + w
= −128 J + 462 J
= 334 J
There is a net gain of energy for
the system in this process
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 (Dw ≠ wfinal – winitial)
Fortunately, most reactions occur under conditions of constant pressure
DU = (q + w) = (q – PDV)
qp = DU + PDV (constant pressure)
Since heat is written in terms of U/P/V, which are all state functions, so is qp
qp is now referred to as Enthalpy (H)
DH = DU + PDV
DH = H (products) – H (reactants)
DH = 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.
Hproducts < Hreactants
DH < 0
Hproducts > Hreactants
DH > 0
Crash Course: Enthalpy
www.youtube.com/watch?v=SV7U4yAXL5I
Thermochemical Equations: show enthalpy changes.
Is DH negative or positive?
System absorbs heat
Endothermic; DH > 0
6.01 kJ are absorbed for every 1 mole of ice that melts at
00C and 1 atm.
H2O (s)
H2O (l)
DH = 6.01 kJ/mol
17
Thermochemical Equations
•
The stoichiometric coefficients always refer to the number of
moles of a substance
DH = 6.01 kJ/ 1 mol
1 H2O (s)
H2O (l)
•
If you reverse a reaction, the sign of DH changes
H2O (l)
H2O (s)
DH = -6.01 kJ/mol
•
If you multiply both sides of the equation by a factor n, then DH
must change by the same factor n.
2H2O (s)
2H2O (l) DH = 2 x 6.01 = 12.0 kJ
•
The physical states must be specified in thermochemical equations.
H2O (s)
H2O (l)
DH = 6.01 kJ/mol
H2O (l)
H2O (g)
DH = 44.0 kJ/mol
Thermochemical Equations
Is DH negative or positive?
System gives off heat
Exothermic; DH < 0
890.4 kJ are released for every 1 mole of methane that is
combusted at 250C and 1 atm.
CH4 (g) + 2O2 (g)
CO2 (g) + 2H2O (l) DH = -890.4 kJ/mol
19
Example 6.3
Given the thermochemical equation: 2SO2(g) + O2(g) → 2SO3(g)
DH = -198.2 kJ/mol
Calculate the heat evolved when 87.9 g of SO2 is converted to SO3
Solution We first calculate moles of SO2 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:
H2O (s)
H2O (l)
DH = 6.01 kJ/mol
Calculate the heat of 3 moles of liquid water freezing to a solid.
A Comparison of DH and DU
2Na(s) + 2H2O(l)
DU = DH - PDV
2NaOH(aq) + H2(g) DH = -367.5 kJ/mol
At 25 oC, 1 mole H2 = 24.5 L at 1 atm
PDV = 1 atm x 24.5 L = 2.5 kJ
DU = -367.5 kJ/mol – 2.5 kJ/mol = -370.0 kJ/mol
DU ≈ DH
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 CO2 at 1 atm and 25°C:
DU = DH - PDV
From the ideal gas law we know: PDV = DnRT
DU = DH - RTDn
The reaction uses 3 moles of gas to produce 2 moles of gas
Dn = 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.
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:
Experimentally
determined
q = m × s × Dt
q = C × Dt
Dt = tfinal - tinitial
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)
Dt = 74.60 – 8.50 = 66.1°C
q = m × s × Dt
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 then zero work can be done
DU = (q + w) = qsys (only heat)
qsys = qwater + qbomb + qrxn
qsys = 0
qrxn = - (qwater + qbomb)
qwater = m × s × Dt
qbomb = Cbomb × Dt
Standards
Reaction at Constant V
DH = qrxn
DH ~ qrxn
No heat enters/leaves the system
26
Example 6.6
Naphthalene (C10H8) 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 (Ccal) 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.
qcal = −qrxn = -57.66 kJ
Chemistry in Action:
Energy content of food is measured with bomb calorimeters
C6H12O6 (s) + 6O2 (g)
6CO2 (g) + 6H2O (l) DH = -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
DHcombustion (kJ/g)
-2
-8
-11
-18
-34
*Note: Shown per unit gram (not mole)
Constant-Pressure Calorimetry
qsys = qwater + qcal + qrxn
qsys = 0
qrxn = - (qwater + qcal)
qwater = m × s × Dt
qcal = Ccal × Dt
Standards
Reaction at Constant P
DH = qrxn
No heat
enters/leaves
CrashCourse Calorimetry:
www.youtube.com/watch?
v=JuWtBR-rDQk
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
sH2O = 4.184 J/g•°C (from reference table)
Dt = 23.17 − 22.50 = 0.67°C
The heat gained by water was lost by the mass of lead: qPb = −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).
qrxn = −qsoln, where qsoln is the heat absorbed by the combined solution.
Example 6.8 Solution
Because qrxn = −qsoln, qrxn = −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 (DHsoln) is the heat generated or
absorbed when a solute dissolves in a solvent.
Required energy intake
to break electrostatic
forces
Energy given off once
ions are stabilized by
hydration
*
* Large intake of energy
lowers temp.
Useful for instant icepacks
DHsoln = 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 (DHf0) is the heat change that results
when one mole of a compound is formed from its elements at a
pressure of 1 atm.
The standard enthalpy of formation of any element in its most stable
form is zero. (neutral, non-compound)
DH0
f (O2)
=0
DH0f (O3) = 142 kJ
DH0f (Cgraphite) = 0
DH0f (Na) = 0
DH0f (Cdiamond) = 1.90 kJ
DH0f (Na+) = -239.66 kJ
Using recorded DHf0 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 (DH0rxn) is the enthalpy
of a reaction carried out at 1 atm.
aA + bB
cC + dD
DH0rxn = [ cDH0f (C) + dDH0f (D) ] - [ aDH0f (A) + bDH0f (B) ]
DH0rxn = S nDH0f (products) - S mDH0f (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) + Fe2O3(s) → Al2O3(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 DHf° for Fe(l) is 12.40 kJ/mol.
*Note: Values of DHf° 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 (DH0rxn)
Direct Method
C(graphite) + O2(g) → CO2(g)
0
-393.5 kJ
DHf° : 0
DH0rxn = DH0(CO2) - DH0(C) - DH0(O2)
DH0rxn = -393.5 kJ - 0 kJ - 0 kJ
DH0rxn = -393.5 kJ
DH°rxn
Indirect Method
C(graphite) + 1/2O2(g) → CO(g) -110.5 kJ
CO(g) + 1/2O2(g) → CO2(g) -283.0 kJ
C(graphite) + O2(g) → CO2(g)
-393.5 kJ
Example 6.9 Indirect Method
Calculate the standard enthalpy of formation of acetylene (C2H2:
The equations for each step and corresponding enthalpy changes are:
Solution Looking at the synthesis of C2H2, we need 2 moles of graphite
as reactant. So we multiply Equation (a) by 2 to get
Next, we need 1 mole of H2 as a reactant and this is provided by (b).
Example 6.9 Solution
Last, we need 1 mole of C2H2 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 C2H2 synthesized from C and H2 absorbs 226.6 kJ of heat from
the surroundings. (Endothermic process)
Ch. 18 Entropy and Free Energy
Spontaneous Physical and Chemical Processes
• A waterfall runs downhill
• At 1 atm, water freezes below 0 oC and ice melts above 0 oC
• 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 O2 & H2O 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.
DH = heat given off or absorbed during a reaction
Combustion
Melting
DH < 0
Exothermic
DH > 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
-DH0
Spontaneous processes
CH4 (g) + 2O2 (g)
CO2 (g) + 2H2O (l) DH0 = -890.4 kJ/mol
H+ (aq) + OH- (aq)
H2O (l) DH0 = -56.2 kJ/mol
N2(g) + 3H2(g)
2NH3 (g) DH0 = -92.6 kJ/mol
H2O (s)
DH0
NH4NO3 (s)
H2O (l)
H 2O
= 6.01 kJ/mol
Each
Exothermic
Both
Endothermic
NH4+(aq) + NO3- (aq) DH0 = 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: DS = Sf - Si
If the change from initial to final results in an increase in randomness
DS > 0
Sf > S i
For any substance, the solid state is more ordered than the liquid
state and the liquid state is more ordered than gas state.
Ssolid < Sliquid << Sgas
H2O (s)
H2O (l)
DS > 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:
DSuniv = DSsys + DSsurr > 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 (DS > 0)
Melting
Chemical reactions that
increase the moles of gas
2NH3(g) → 3H2(g) + N2(g)
Vaporizing
Diffusion
Heating
+
The Entropy Factor Increases with Temperature (TDS)
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 Hbonds
• 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).
DSuniv = DSsys + DSsurr > 0
• As long as DSsys < DSsurr
• 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)
DG = DHsys - TDSsys
Free-energy change (DG) must be negative for a reaction to be spontaneous.
DG < 0 The reaction is spontaneous in the forward direction.
DG > 0 The reaction is non-spontaneous as written. The
reaction is spontaneous in the reverse direction.
DG = 0 The reaction is at equilibrium.
DG = DH - TDS
Exothermic reactions (-DH) and increasing Entropy (+DS)
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
H2O(l) → H2O(g)
It is an endothermic process; DH = +
• Takes energy to reach a higher kinetic state
• This typically disfavors a process from occurring
The process also increases entropy; DS = +
• This typically favors a reaction from occurring
• The entropy factor is temperature-dependent (DST)
At low temperatures the
entropy factor is smaller than
the DH resulting in +DG
DH > DST
Not Boiling
At temperatures above 100 °C
the entropy factor is greater
than the DH resulting in -DG
DH < DS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
B
If a heavier weight B is coupled, weight A can
move upward spontaneously by coupling it
with the falling of a larger weight.
• Many biological reactions are energetically unfavorable.
• They rely on coupling with other molecules to release a
surplus of free energy.
ADP
ATP
DG0 = -31 kJ
Coupled Reactions
Example:
Alanine + Glycine
DG0 = +29 kJ
ATP + H2O + Alanine + Glycine
DG0 = -2 kJ
Alanylglycine
non-spontaneous
ADP + H3PO4 + Alanylglycine
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