Chapter 17: Thermodynamics
Thermochemistry
Chapter 17: Thermodynamics
Energy is the capacity to do work.
•
Radiant energy comes from the sun and is
earth’s primary energy source
•
Thermal energy is the energy associated with
the random motion of atoms and molecules
•
Chemical energy is the energy stored within the
bonds of chemical substances
•
Nuclear energy is the energy stored within the
collection of neutrons and protons in the atom
•
Potential energy is the energy available by virtue
of an object’s position
•
Kinetic energy is the energy of motion
Chapter 17: Thermodynamics
Energy Changes in Chemical Reactions
Heat is the transfer of thermal energy between
two bodies that are at different temperatures.
Temperature is a measure of the
thermal energy.
Temperature = Thermal Energy
Chapter 17: Thermodynamics
Thermochemistry is the study of heat change in chemical
reactions.
The system is the specific part of the universe that is of
interest in the study.
Surrounding:
the rest of
the universe.
open
Exchange: mass & energy
closed
energy
isolated
nothing
Chapter 17: Thermodynamics
Collision Theory
• To explain factors affecting rxn rate
1. For two molecules to react, they must
come in contact or collide.
2. When molecules collide,
1. Ineffective collision occurs: molecules may
not orient in right position. No Rxn.
2. Effective collision occurs: molecules
possess enough kinetic energy to position.
Chapter 17: Thermodynamics
What Happen Just Before Rxn
• Molecules approaches each other.
• Repulsive force increases as they
approach each other.
• There is a minimum energy to push the
molecules to collide
 Activation Energy (Ea)
• Supplying enough kinetic energy will “kickstart” the reaction.
Chapter 17: Thermodynamics
Graphing Rxn: Spontaneous
Chapter 17: Thermodynamics
Gibbs Free Energy (ΔG)
• If ΔG is positive, or ΔG > 0
Indication: Rxn  Nonspontaneous
• If ΔG is negative, or ΔG < 0
Indication: System  Spontaneous
• The previous example is spontaneous.
Chapter 17: Thermodynamics
Rate Influencing Factor
1.
2.
3.
4.
Surface Area
Temperature
Concentration
Presence of Catalysts
Chapter 17: Thermodynamics
Factor 1: Surface Area
• Similar to solubility
• More surface area enhances collision
frequency
• Homogeneous Rxn
– The reactants are in the same physical state.
For example, gas reacting with gas
• Heterogeneous Rxn
– The reactants are in DIFFERENT physical state
For example: Solid reacting with liquid/aq. soln
Chapter 17: Thermodynamics
Factor 2: Temperature
• Increase in temperature provide more
kinetic energy for reaction to proceed
• Fast-moving molecules have better
chance of “effective collision.”
• It also enhances frequency of collision.
• Refrigerator is designed to SLOW reaction
rate.
Chapter 17: Thermodynamics
Factor 3: Concentration
• Higher chemical concentration generally
promotes reaction rate
• Concentration is directly proportional to
the frequency of collision, in general.
• Just like medicine, we need higher dosage
for severely-ill patient.
Chapter 17: Thermodynamics
Factor 4: Catalyst
• In human body, we have enzymes acting
as catalyst.
• It does not undergo permanent change.
• It provides an alternative reaction pathway
of Lower Activation Energy.
• Inhibitor: reduces a reaction rate by
preventing the reaction occurring in the
usual way.
– Example: Food preservatives
Chapter 17: Thermodynamics
Effect of Catalyst on Rxn
Chapter 17: Thermodynamics
The First Law of Thermodynamics
Relating DE to Heat(q) and Work(w)
• Energy cannot be created or destroyed.
• Energy of (system + surroundings) is constant.
• Any energy transferred from a system must be
transferred to the surroundings (and vice versa).
• From the first law of thermodynamics:
DE  q  w
Chapter 17: Thermodynamics
First Law of Thermodynamics
Energy can be converted from one form to another but
energy cannot be created or destroyed.
Second Law of Thermodynamics
The entropy of the universe increases in a spontaneous
process and remains unchanged in an equilibrium process.
Spontaneous process:
DSuniv = DSsys + DSsurr > 0
Equilibrium process:
DSuniv = DSsys + DSsurr = 0
Chapter 17: Thermodynamics
Enthalpy Changes
• The difference between the potential
energy of the reactants and the products
during a physical or chemical change is
the Enthalpy change or ∆H.
• AKA: Heat of Reaction at constant
pressure
Chapter 17: Thermodynamics
Thermochemical Equations
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
Chapter 17: Thermodynamics
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
Chapter 17: Thermodynamics
Graphing Rxn: Endothermic
The temperature goes down
Chapter 17: Thermodynamics
Graphing Rxn: Exothermic
The temperature goes up
Chapter 17: Thermodynamics
How does ΔH and ΔS affect spontaneity?
-ΔH
+ΔH
+ΔS
Always
spontaneous
Spontaneity
depends
on temp
-ΔS
Spontaneity
depends on
temp
never
spontaneous
Chapter 17: Thermodynamics
Cold and hot packs
• How do instant hot and cold packs work?
Chapter 17: Thermodynamics
Hot pack
• Pressing the bottom , the diaphragm breaks.
• Calcium chloride dissolves in water and warms it.
• The beverage gets warm.
Chapter 17: Thermodynamics
Exothermic process
• Heat flows into the surroundings from the
system in an exothermic process.
Surroundings
Energy
Hot pack
Temperature rises
Chapter 17: Thermodynamics
Cold pack
• Water and ammonium nitrate are kept in separate
compartments.
• Pressing the wrapper, the ammonium nitrate dissolves in
water and absorbs heat.
• The pack becomes cold.
It is used to treat
sports injuries.
Chapter 17: Thermodynamics
Endothermic process
• Heat flows into the system from the
surroundings in an endothermic process.
Surroundings
Cold
pack
Energy
Temperature falls
Chapter 17: Thermodynamics
Explosions
This reaction is exothermic!
Chapter 17: Thermodynamics
Photosyntesis
This reaction is endothermic!
Chapter 17: Thermodynamics
Combustions
These reactions are exothermic!
Chapter 17: Thermodynamics
Enthalpy Changes in Reactions
• All chemical reactions require bond
breaking in reactants followed by
bond making to form products
• Bond breaking requires energy
(endothermic) while bond formation
releases energy (exothermic)
Chapter 17: Thermodynamics
Chapter 17: Thermodynamics
Enthalpy Changes in Reactions
endothermic reaction - the energy
required to break bonds is greater than
the energy released when bonds form.
exothermic reaction - the energy
required to break bonds is less than the
energy released when bonds form.
Chapter 17: Thermodynamics
Enthalpy Changes in Reactions
∆H can represent the enthalpy change for
a number of processes
1. Chemical reactions
∆Hrxn – enthalpy of reaction
∆Hcomb – enthalpy of combustion
Chapter 17: Thermodynamics
2. Formation of compounds from elements
∆Hof – standard enthalpy of formation
The standard molar enthalpy of formation
is the energy released or absorbed when
one mole of a compound is formed
directly from the elements in their
standard states.
Chapter 17: Thermodynamics
3. Phase Changes
∆Hvap – enthalpy of vaporization
∆Hfus – enthalpy of melting
∆Hcond – enthalpy of condensation
∆Hfre – enthalpy of freezing
4. Solution Formation
∆Hsoln – enthalpy of solution
Chapter 17: Thermodynamics
q = mc∆T
140
100
q = n∆H
q = mc∆T
Temp.
(°C )
q = n∆H
0
q = mc∆T
-40
Time
Chapter 17: Thermodynamics
Data:
cice = 2.01 J/g.°C
cwater = 4.184 J/g.°C
csteam = 2.01 J/g.°C
ΔHfus = +6.02 kJ/mol
ΔHvap = +40.7 kJ/mol
Chapter 17: Thermodynamics
warming ice:
q = mc∆T
= (40.0)(2.01)(0 - -40)
= 3216 J
warming steam:
q = mc∆T
= (40.0)(2.01)(140 -100)
= 3216 J
warming water:
q = mc∆T
= (40.0)(4.184)(100 – 0)
= 16736 J
Chapter 17: Thermodynamics
moles of water:
n = 40.0 g
18.02 g/mol
= 2.22 mol
melting ice:
q = n∆H
= (2.22 mol)(6.02 kJ/mol)
= 13.364 kJ
boiling water:
q = n∆H
= (2.22 mol)(40.7 kJ/mol)
= 90.354 kJ
Chapter 17: Thermodynamics
Total Energy
90.354 kJ
13.364 kJ
3216 J
3216 J
16736 J
127 kJ
Chapter 17: Thermodynamics
Sample Problem
• How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg C?
Step 1: Heat the ice
Q=mcΔT
Q = 36 g x 2.06 J/g deg C x 8 deg C = 593.28 J = 0.59 kJ
Step 2: Convert the solid to liquid
ΔH fusion
Q = 2.0 mol x 6.01 kJ/mol = 12 kJ
Step 3: Heat the liquid
Q=mcΔT
Q = 36g x 4.184 J/g deg C x 100 deg C = 15063 J = 15 kJ
Chapter 17: Thermodynamics
Sample Problem
• How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg C?
Step 4: Convert the liquid to gas
Q = 2.0 mol x 44.01 kJ/mol =
Step 5: Heat the gas
ΔH vaporization
88 kJ
Q=mcΔT
Q = 36 g x 2.02 J/g deg C x 20 deg C = 1454.4 J = 1.5 kJ
Now, add all the steps together
0.59 kJ + 12 kJ + 15 kJ + 88 kJ + 1.5 kJ
= 118 kJ
Chapter 17: Thermodynamics
Heat/Enthalpy Calculations
specific heat capacity - the amount of
energy , in Joules (J,SI unit), needed to
change the temperature of one gram (g)
of a substance by one degree Celsius
(°C).
• c: does not depend on mass, so intensive
property.
• The symbol for specific heat capacity is a
lowercase c
• The unit is J/ g°C or kJ/ g°C
Chapter 17: Thermodynamics
• A substance with a large value of c can
absorb or release more energy than a
substance with a small value of c.
• with the larger c will undergo a smaller
temperature change with the same
amount of heat applied.
Chapter 17: Thermodynamics
FORMULA
q = mc∆T
q = heat (J)
m = mass (g)
c = specific heat capacity
∆T = temperature change
= T2 – T1
= Tf – Ti
Chapter 17: Thermodynamics
eg. How much heat is needed to raise the
temperature of 500.0 g of water from
20.0 °C to 45.0 °C?
Solve
q = m c ∆T
for c, m, ∆T, T2 & T1
Chapter 17: Thermodynamics
heat capacity - the quantity of energy , in
Joules (J, SI unit), needed to change
the temperature of a substance by one
degree Celsius (°C)
• The symbol for heat capacity is
uppercase C
• C: depends on mass, so extensive
property.
• The unit is J/ °C or kJ/ °C
Chapter 17: Thermodynamics
FORMULA
C = mc
q = C ∆T
C = heat capacity
c = specific heat
capacity
m = mass
∆T = T2 – T1
Chapter 17: Thermodynamics
Calorimetry
•
•
•
•
•
•
Heat Capacity and Specific Heat
Calorimetry = measurement of heat flow (follow Law of
conservation of energy).
Heat lost by one substance = Heat gained by other
substance
Calorimeter = apparatus that measures heat flow.
Heat capacity = the amount of energy required to raise the
temperature of an object (by one degree).
Molar heat capacity = heat capacity of 1 mol of a
substance.
Specific heat = specific heat capacity = heat capacity of 1
g of a substance.
q  specific heat  grams of substance  DT
Chapter 17: Thermodynamics
Calorimetry
Constant Pressure Calorimetry
qrxn  S  ( total mass of solution)  DT
DH rxn
qrxn

mol *
* moles of species of interest
The heat capacity (C) of a substance
C = ms
Heat (q) absorbed or released:
q = msDt
q = CDt
Dt = tfinal - tinitial
Chapter 17: Thermodynamics
How much heat is given off when an 869 g iron bar cools
from 940C to 50C?
s of Fe = 0.444 J/g • 0C
Dt = tfinal – tinitial = 50C – 940C = -890C
q = msDt = 869 g x 0.444 J/g • 0C x –890C = -34,000 J
6.5
Chapter 17: Thermodynamics
More practice
1. 5 g of copper was heated from 20C to 80C.
How much energy was used to heat the Cu?
q=cmDT = 0.38 J/(gC) x 5 g x 60 C = 114 J
2. If a 3.1 g ring is heated using 10.0 J, it’s temp.
rises by 17.9C. Calculate the specific heat
capacity of the ring. Is the ring pure gold?
q=cmDT
10.0 J
c = q/mDT= 3.1 g x 17.9C = 0.18 J/(g°C)
The ring is not pure. Gold is 0.13 J/(gC)
Chapter 17: Thermodynamics
Thermochemical Equations
•
The stoichiometric coefficients always refer to the number
of moles of a substance
DH = 6.01 kJ/mol ΔH = 6.01 kJ
H2O (s)
H2O (l)
• If you reverse a reaction, the sign of DH changes
H2O (l)
•
H2O (s)
DH = -6.01 kJ
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 mol x 6.01 kJ/mol = 12.0 kJ
Chapter 17: Thermodynamics
Standard enthalpy (heat) of formation (DH0) is the
f
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.
DH0f (O2) = 0
DH0f (C, graphite) = 0
DH0f (O3) = 142 kJ/mol
DH0f (C, diamond) = 1.90 kJ/mol
Chapter 17: Thermodynamics
Chapter 17: Thermodynamics
0 ) (heat of reaction)
The standard enthalpy of reaction (DHrxn
is the enthalpy of a reaction carried out at 1 atm.
aA + bB
cC + dD
DH0rxn = [ cDH0f (C) + dDH0f (D) ] - [ aDH0f (A) + bDH0f (B) ]
0 (reactants)
DH
S
DH0rxn = S DH0 (products)
f
f
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.
ΔH for the overall rxn will equal to the sum of the
enthalpy changes for the individual steps.
Chapter 17: Thermodynamics
17.4 Section Quiz.
1. Calculate DH for
NH3 (g) + HCl (g)  NH4Cl (s).
Standard heats of formation:
NH3(g) = - 45.9 kJ/mol,
HCl(g) = - 92.3 kJ/mol,
NH4Cl(s) = - 314.4 kJ/mol
a.
b.
c.
d.
176.2 kJ
360.8 kJ
176.2 kJ
268 kJ .
Chapter 17: Thermodynamics
Benzene (C6H6) burns in air to produce carbon dioxide and liquid water. How
much heat is released per mole of benzene combusted? The standard
enthalpy of formation of benzene is 49.04 kJ/mol.
2C6H6 (l) + 15O2 (g)
12CO2 (g) + 6H2O (l)
DH0rxn = S DH0 f(products) - S DH0 f(reactants)
DH0rxn = [ 12DH0f (CO2) + 6DH0f (H2O)] - [ 2DH0f (C6H6)]
DH0rxn = [ 12 × -393.5 + 6 × -285.8 ] – [ 2 × 49.04 ] = -6535 kJ
-6535 kJ
= - 3267 kJ/mol C6H6
2 mol
Chapter 17: Thermodynamics
17.4 Section Quiz.
2. The heat of formation of Cl2(g) at 25°C is
a.
b.
c.
d.
the same as that of H2O at 25°C.
larger than that of Fe(s) at 25°C.
undefined.
zero.
Chapter 17: Thermodynamics
Hess’s Law
Example:
CH4(g) + 2O2(g)  CO2(g) + 2H2O(g)
2H2O(g)  2H2O(l)
CH4(g) + 2O2(g)  CO2(g) + 2H2O(l)
DH = -802 kJ
DH= - 88 kJ
DH = -890 kJ
Given:
C(s) + ½ O2(g)  CO(g)
DH = -110.5 kJ
CO2(g)  CO(g) + ½ O2(g)
DH = 283.0 kJ
Calculate DH for:
C(s) + O2(g)  CO2(g)
Chapter 17: Thermodynamics
Calculate the standard enthalpy of formation of CS2 (l) given that:
C(graphite) + O2 (g)
CO2 (g) DH0rxn = -393.5 kJ
S(rhombic) + O2 (g)
CS2(l) + 3O2 (g)
SO2 (g)
DH0rxn = -296.1 kJ
CO2 (g) + 2SO2 (g)
0 = -1072 kJ
DHrxn
1. Write the enthalpy of formation reaction for CS2
C(graphite) + 2S(rhombic)
CS2 (l)
2. Add the given rxns so that the result is the desired rxn.
C(graphite) + O2 (g)
2S(rhombic) + 2O2 (g)
+ CO2(g) + 2SO2 (g)
CO2 (g) DH0rxn = -393.5 kJ
2SO2 (g) DH0rxn = -296.1x2 kJ
CS2 (l) + 3O2 (g)
0 = +1072 kJ
DHrxn
C(graphite) + 2S(rhombic)
CS2 (l)
DH0rxn= -393.5 + (2x-296.1) + 1072 = 86.3 kJ
Chapter 17: Thermodynamics
17.4 Section Quiz.
1. According to Hess’s law, it is possible to
calculate an unknown heat of reaction by
using
a. heats of fusion for each of the compounds in the
reaction.
b. two other reactions with known heats of reaction.
c. specific heat capacities for each compound in the
reaction.
d. density for each compound in the reaction.
Chapter 17: Thermodynamics
Chapter 17: Thermodynamics
The critical temperature (Tc) is the temperature above which
the gas cannot be made to liquefy, no matter how great the
applied pressure.
The critical pressure
(Pc) is the minimum
pressure that must be
applied to bring about
liquefaction at the
critical temperature.
Chapter 17: Thermodynamics
Where’s Waldo?
Can you find…
The Triple Point?
Critical pressure?
Critical
temperature?
Where fusion
occurs?
Where vaporization
occurs?
Melting point
(at 1 atm)?
Carbon Dioxide
Boiling point
(at 6 atm)?
Chapter 17: Thermodynamics
The melting point of a solid
or the freezing point of a
liquid is the temperature at
which the solid and liquid
phases coexist in equilibrium
Freezing
H2O (l)
Melting
H2O (s)
Chapter 17: Thermodynamics
Molar heat of sublimation
(DHsub) is the energy required
to sublime 1 mole of a solid.
DHsub = DHfus + DHvap
( Hess’s Law)
Deposition
H2O (g)
Sublimation
H2O (s)
Chapter 17: Thermodynamics