Unit 10 - Thermochemistry

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Thermodynamics &
Thermochemistry
Unit 10
Overview
• Energy
–
–
–
–
–
–
Units
Heat /temperature
System/surrounding
Calorimeters
1st Law Thermodynamics
Heat Capacity
• Enthalpy
–
–
–
–
–
–
∆H and ∆H°
Endothermic/exothermic
Heat of formation
Stoichiometry
Bond energy
Hess’s Law
• Entropy
–
–
–
–
∆S and ∆S°
Spontaneous reactions
Entropy for phase changes
2nd Law of Thermodynamics
• Gibbs Free Energy
– ∆G and ∆G°
– ∆G = ∆H - T∆S
– Spontaneous reactions
• State functions
What is Thermodynamics?
The study of the energy and its transformations
What is Thermochemistry?
The study of the transfers of energy as heat that
accompany chemical reactions and physical
changes
Energy
• The capacity to do work or to transfer heat
– Work: the energy used to cause an object with mass
to move against a force
– Heat: the energy used to cause the temperature
change in an object
• Units of Energy
– Joule (J) = 1 kg∙m2/s2
– Calorie (cal) = 4.184 J
• In nutrition 1 calorie (Cal) = 1000 cal
• (notice difference is that nutrition calorie is capitalized)
• In chemistry 1000 cal = 1 kcal
What is the difference between
heat and temperature?
What’s the difference?
HEAT
• Energy transferred between
samples of matter because
of a difference in their
temperatures
• Cannot be measured
directly (determined by
temperature change)
• Joules (J) = 1 kg∙m2/s2
• Moves from high to low
temperature
TEMPERATURE
• Measure of the average
kinetic energy of the
particles in a sample of
matter
• Directly measureable
(affected by heat transfer)
• Celsius (°C) or Kelvin (K)
• Higher kinetic energy =
higher temperature
How do we measure energy?
Calorimeter
Measures energy absorbed or released as heat in
a chemical/physical change
Constant Pressure
Example “coffee-cup”
calorimeter uses unsealed
calorimeter where reaction
occurs under atmospheric
pressure
How do we measure energy?
Constant Volume
Example “bomb”
calorimeter uses
pressurized sealed vessel
called “bomb” submerged
in water
System and Surroundings
• System: portion we single out for studying
– Open system: matter and energy exchanged with
surroundings
– Closed system: energy exchanged within system
• Surroundings: everything else that is not
included in the system
Work
E = q + w
E = change in internal energy of a system
q = heat flowing into or out of the system
-q if energy is leaving the system
+q if energy is entering the system
w = work done by, or on, the system
-w if work is done by the system on the surroundings
+w if work is done on the system by the surroundings
Pressure-Volume Work
• Most common type of work.
– Example: A sample of gas is held by a pressure of
2.4 atm. The pressure is then changed to 1.3 atm.
The gas expands. How much work is done?
– Note that because the gas is expanding, the
system is doing work on the surroundings so work
must be negative
Pressure-Volume Work
When the volume (V) of
a system increases by
an amount of ΔV
against an external
pressure (p), the system
pushes back, and thus
does pV work on the
surroundings
w = -PΔV
Pressure-Volume Work
w   PV
Expansion
Compression
+V (increase)
-V (decrease)
-w results
E
system
decreases
Work has been done by the system
on the surroundings
+w results
E
system
increases
Work has been done on the system
by the surroundings
FIRST LAW OF THERMODYNAMICS
Energy is neither created
nor destroyed
(conservation of energy)
FIRST LAW OF THERMODYNAMICS
• Energy is transferred but not lost
• Energy lost by a system must be gained
by surroundings and vice versa
What determines how much
heat is transferred during a
temperature change?
Amount depends on…
1. Nature of material changing temperature.
2. Mass of material changing temperature.
3. Size of the temperature change.
Example 1
Example 2
One gram of each of two metals (iron and
silver) are heated to 100°C and cooled to 50°C
in a beaker of water.
Iron transfers 22.5 J of energy.
Silver transfers 11.8 J of energy.
What makes the difference?
Heat Capacity
• Heat Capacity (Cp) – temperature change
experienced by an object when it absorbs a
certain amount of heat
– The greater the heat capacity, the greater the heat
required to produce given increase in temperature
Heat Capacity
Cp =
∆H
∆T
Cp = heat capacity
∆H = heat added (J or cal)
∆T = temperature change (°C or K)
Heat Capacity (Pure Substances)
• Molar Heat Capacity (Cm): Amount of energy required
to raise the temperature of 1 mole of a substance by 1
degrees – Celsius or Kelvin
• Specific Heat (Cs): Amount of energy required to raise
the temperature of 1 gram of a substance by 1 degrees
– Celsius or Kelvin
• Higher specific heat means…
– Higher heat capacity
– Heats up slower
– Transfers more heat
Specific Heat
Explain in terms of specific heat
Calculations
q = mc∆T
q = heat added (J or cal)
m = mass of substance (g or kg)
c = specific heat
∆T = temperature change (°C or K)
Specific Heat (Units)
J/g°C
or
J/gK
Example 1
A 4.0 g sample of glass was heated from 274
K to 314 K and was found to have absorbed
32 J of heat.
A. What is the specific heat of this type of glass?
B. How much energy will the same glass sample
gain when it is heated from 314 K to 344 K?
Example 2
A piece of copper alloy with a mass of 85.0 g
is heated from 30.°C to 45°C. In the process,
it absorbs 523 J of energy as heat.
A. What is the specific heat of this copper alloy?
B. How much energy will the same sample lost if it
is cooled from 45°C to 25°C?
Enthalpy
What is Enthalpy?
The amount of energy absorbed by a system
as heat during a process at constant pressure.
How does it apply to reactions?
Enthalpy Change
Difference between enthalpies of products
and reactants
H = Hproducts - Hreactants
Enthalpy of Reaction
Quantity of energy transferred as heat during
chemical reaction (“heat of reaction”)
What’s the difference?
Endothermic
Exothermic
• Heat absorbed
• Heat released
• Energy gained
• Energy lost
When you gamble and lose your money, do you
end up with less money or more?
….do you have a “positive or negative”
amount of money in your pocket after?
When you gamble and win money, do you end
up with less money or more?
….do you have a “positive or negative”
amount of money in your pocket after?
What’s the difference?
Endothermic
Exothermic
• Heat absorbed
• Heat released
• Energy gained
• Energy lost
• H = positive (+)
• H = negative (-)
• H2O + 483.6 kJ  H2 + O2
• H2O  H2 + O2 H = +483.6 kJ
• H2 + O2  H2O + 483.6 kJ
• H2 + O2  H2O H = -483.6 kJ
Energy Diagrams
• Activation Energy –
energy needed for
reactants to transition to
products (become
activated complex)
• At activated complex, all
bonds have been broken
but no bonds formed
(highest energy/lowest
stability)
• Exothermic diagram  energy of products are lower
than energy of reactants (∆H is negative)
Energy Diagrams
• Activation Energy –
energy needed for
reactants to transition to
products (become
activated complex)
• At activated complex, all
bonds have been broken
but no bonds formed
(highest energy/lowest
stability)
• Endothermic diagram  energy of products are
higher than energy of reactants (∆H is positive)
Most reactions in nature
are exothermic…
Hf = negative
When Hf positive or slightly positive,
the compound is unstable…
…very positive
Hf can even be
explosive
For example: Mercury fulminate is used as a detonator
in explosives.
HgC2N2O2 Hf = + 270 kJ/mol
Standard State Conditions
•
•
•
•
•
All gases are at 1 atm
All liquids are pure
All solids are pure
All solutions are at 1 M concentration
Temperature is room temperature (25°C or 298K)
unless told otherwise
• Energy of formation of element in its normal state = 0
• Standard state conditions indicated by superscript 0
– Example: ∆H° means standard state conditions
Formation of Compounds
Enthalpy of Formation (Hf) – Heat of Formation
Enthalpy change that occurs when 1 mole of a
compound is formed from its elements
(sometimes called molar enthalpy of formation)
To represent standard state use
Hf0
Elements in their standard state have Hf0 = 0
Heat of Formation
• To form compounds from elements or compounds in
their standard states break apart compounds into
individual components
– Balance by using fractions on standard state
elements/compounds rather than whole numbers
Example 1: Equation for formation of NaCl(s)
Na(s) + ½ Cl2(g)  NaCl(s)
Example 2: Equation for formation of CaCO3(s)
Ca(s) + C(s) + 3/2 O2(g)  CaCO3(s)
Heat of Formation
Hf0 = ∑ Hf0 products - ∑ Hf0 reactants
Example:
2CH3OH(g) + 3O2(g)  2CO2(g) + 4H2O(g)
Hf0 = [2(-394) + 4(-242)] – [2(-201) + 3(0)]
Hf0 = [-1756] – [-402]
Compound
Hf0 = -1354 kJ
CH3OH(g)
∆H (kJ/mol)
-201
O2(g)
0
CO2(g)
-394
H2O(g)
-242
Enthalpy Stoichiometry
How much heat is released when 5.40 g of methane
gas is burned in a constant-pressure system?
CH4(g) + 2O2(g)  CO2(g) + 2H2O(l)
∆H = -890 kJ
4.50 g CH4 × 1 mol CH4 × -890 kJ = - 250 kJ
16.0 g CH4 1 mol CH4
Bond Energy
• Energy required to break a bond.
– Breaking a bond requires energy (endothermic)
• Bond energy is positive
– Forming a bond releases energy (exothermic)
• Bond energy is negative
H0 = ∑ Bond Energies of bonds broken – ∑Bond Energies of bonds formed
H0 = ∑ Bond Energyreactants – ∑ Bond Energyproducts
Bond Energy
H0 = ∑ Bond Energies of bonds broken – ∑Bond Energies of bonds formed
Example:
2H2(g) + O2(g)  2H2O(g)
H0 = [2(H-H) + 1(O-O)] – [4(O-H)]
H0 = [2(436) + 1(499)] – [4(463)]
H0 = -481 kJ/mol
Bond
Bond Energy (kJ/mol)
H-H
436
O=O
499
O-H
463
Sometimes finding the enthalpy change
in reactions is difficult or even
impossible due to the reaction.
How can we find
Hf for these
types of reactions?
Hess’s Law
The overall enthalpy change in a reaction
equals the sum of the enthalpy changes for
individual steps in the process.
Can be used to find the
Hf for reactions that are
difficult or impossible to
measure.
Example
C(s) + 2H2(g)  CH4(g)
Hf0 = ?
We can use the following to help…
C(s) + O2(g)  CO2(g)
H2(g) + ½O2(g)  H2O(l)
CH4(g) + 2O2(g)  CO2(g) + 2H2O(l)
Hf0 = -393.5 kJ
Hf0 = -285.8 kJ
Hf0 = -890.8 kJ
Example
Rules to remember:
1. When a reaction is reversed, the sign of Hf must
be reversed (+/-).
2. When coefficients are multiplied by a number,
Hf must be multiplied by the same number.
Example
C(s) + O2(g)  CO2(g)
H2(g) + ½O2(g)  H2O(l)
CH4(g) + 2O2(g)  CO2(g) + 2H2O(l)
C(s) + 2H2(g)  CH4(g)
Hf0 = -393.5 kJ
Hf0 = -285.8 kJ
Hf0 = -890.8 kJ
Hf0 = ?
Example
C(s) + O2(g)  CO2(g)
2H2(g) + O2(g)  2H2O(l)
CH4(g) + 2O2(g)  CO2(g) + 2H2O(l)
C(s) + 2H2(g)  CH4(g)
Hf0 = -393.5 kJ
Hf0 = 2(-285.8 kJ)
Hf0 = -890.8 kJ
Hf0 = ?
Multiply reaction and Hf0 by 2.
Example
C(s) + O2(g)  CO2(g)
2H2(g) + O2(g)  2H2O(l)
CO2(g) + 2H2O(l)  CH4(g) + 2O2(g)
C(s) + 2H2(g)  CH4(g)
Hf0 = -393.5 kJ
Hf0 = 2(-285.8 kJ)
Hf0 = +890.8 kJ
Hf0 = ?
Reverse reaction and change the sign of Hf0
Example
C(s) + O2(g)  CO2(g)
2H2(g) + O2(g)  2H2O(l)
CO2(g) + 2H2O(l)  CH4(g) + 2O2(g)
C(s) + 2H2(g)  CH4(g)
Hf0 = -393.5 kJ
Hf0 = 2(-285.8 kJ)
Hf0 = +890.8 kJ
Hf0 = -74.3 kJ
Add the Hf0 of each reaction in the
process to determine the final Hf0 .
Elements that appear on both
sides of the reaction cancel out.
Practice
2N2(g) + 5O2(g)  2N2O5(g)
Hf0 = ?
Determine the Hf0 for the above reaction
using the following process…
H2(g) + ½O2(g)  H2O (l)
N2O5 (g) + H2O (l)  2HNO3(l)
½N2(g) + O2(g) + ½H2(g)  HNO3(l)
Hf0 = -285.8 kJ
Hf0 = -76.6 kJ
Hf0 = -174.1 kJ
What predicts whether a reaction
will occur spontaneously?
1. Change in energy (H)
Most reactions in
nature are exothermic…
The goal is to end up in
a lower energy state
(products having lower
energy than reactants)
Can spontaneous reactions be endothermic?
Think about ice…
Energy is transferred
from the warm air to
the ice
The well ordered ice
cube crystal melts to the
less ordered liquid phase
having a higher energy
What is Entropy?
Measure of the degree of randomness of
particles in a system (S)
Increase in entropy usually
means increase in…
• Temperature
• Volume
• Number of independently
moving particles
The more random the system,
the higher the entropy.
Low Entropy
Higher Entropy
Highest Entropy
•Particles in solution have higher entropy values than particles in solids
•Two moles of a substance have a higher entropy value than 1 mole
Entropy (S) for phase changes
∆S =
∆H
T
• ∆S = entropy of fusion or vaporization (J/K)
• ∆H = heat of vaporization or fusion (J/mol)
• T = Kelvin temperature
(Multiply ∆H by moles to get units of J)
Entropy (S) for phase changes
Example: The normal freezing temperature for liquid mercury is
-38.9°C and its molar enthalpy of fusion, ∆Hfusion, is
2.29kJ/mol. What is the entropy change of the system when
50.0 g of liquid mercury freezes at the normal freezing point?
50.0 g Hg = 0.25 moles Hg
∆H
∆S =
T
-2.29 kJ/mol = -2290 J/mol
(0.25 mole)(-2290 J/mol)
=
= - 2.44 J/K
234.3 K
Change in Entropy (S)
The difference between the entropy of the
products and reactants
S0 = ∑ S0 products - ∑ S0 reactants
Increase in entropy  S is positive (+)
Decrease in entropy  S is negative (-)
Units: kJ/molK
Standard Molar Entropy (S0)
• S0 of elements in their normal state at 298K are
not zero (unlike H0)
• Gases > liquids > solids
• Increase with increasing molar mass
• Increase with increasing number of atoms in the
formula of a substance
SECOND LAW OF THERMODYNAMICS
The entropy of the universe
always increases during
spontaneous reactions
(If a reaction is spontaneous in one direction it
cannot be spontaneous in the reverse direction)
SECOND LAW OF THERMODYNAMICS
∆Suniverse = ∆Ssystem + ∆Ssurroundings
• Any irreversible process results in an
increase in entropy (∆S > 0)
• Any reversible process results in no overall
change in entropy (∆S = 0)
What predicts whether a reaction
will occur spontaneously?
1. Change in energy (H)
2. Randomness of particles (S)
Which is preferred?
LEAST Enthalpy (H)
_______
MOST Entropy (S)
_______
The interaction between
the two determines the
change in a system
(spontaneous or not)
Gibbs Free Energy (G)
The combined enthalpy-entropy function
in a system
Natural processes
have a lower free
energy (negative)
Gibbs Free Energy
G0 = ∑ Gf0 products - ∑ Gf0 reactants
∆ G, ∆ H, and ∆S
Difference between change in enthalpy (H)
and the product of the Kelvin temperature
and the entropy change (TS)
G = H - TS
T = Kelvin temperature
What makes a reaction spontaneous?
If
G < 0 the reaction is spontaneous
H
S
G
–
–
+
+
–
+
–
– (low temp)
– (high temp)
+
–
+
If G = 0 then the reaction is at equilibrium.
Example
For the following reaction
NH4Cl(s)  NH3(g) + HCl(g)
At 298.15 K, H0 = 176 kJ/mol and S0 = 0.285
kJ/molK. Calculate G0 and tell whether the reaction
is spontaneous.
State Functions
• Property of a system determined by the system’s
condition (state) in terms of pressure, temperature,
etc.
– Depends only on present state of system, not the path it
took to get there
– Enthalpy, entropy, Gibbs Free Energy are state functions
• Example: When traveling to Denver from Chicago,
Denver is 5280 ft above sea level and Chicago is 596 ft
above sea level. No matter what route you take, the
altitude change will be 4684 ft. Altitude is a state
function. (distance is not)

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