Download Thermodynamics

Thermodynamics
Honors Unit 5
Energy: Basic Principles

Thermodynamics – the study of energy
changes

Energy – the ability to do work or
produce heat

Note: Work is force acting over a
distance
Energy: Basic Principles

Kinetic Energy – energy of motion

KE =
1 2
mv
2
Potential Energy –
energy due to position
or composition

Law of Conservation of Energy

A.k.a. first Law of Thermodynamics

Energy can be converted from one form to
another but can’t be created or destroyed

This means the total energy of the
universe is CONSTANT!
Heat vs. Temperature

Temperature – measure of the random
motion of a substance


Temperature is proportional to kinetic
energy (it is a measure of the average
kinetic energy in a substance)
Heat (q) – flow of energy due to a
temperature difference
Important Aspects of Thermal
Energy & Temperature

Heat is NOT the same as temperature!

The more kinetic energy a substance has,
the greater the temperature of its atoms and
molecules.

The total thermal energy in an object is the
sum of its individual energies of all the
molecules.

For any given substance, its thermal energy
depends not only on its composition but also
on the amount of substance
System vs. Surroundings

A system is the part of the universe we
are studying.
The surroundings
are everything else
outside
of the system.

Direction of Heat Flow

Heat transfer occurs when two objects are
at two different temperatures.

Eventually the two objects reach the same
temperature

At this point, we say that the system
has reached equilibrium.
Thermal Equilibrium

Heat transfer
always occurs with
heat flowing from
the HOT object to
the COLD object.
Thermal Equilibrium
The quantity of heat lost by
the hotter object and the
quantity of heat gained by the
cooler object are EQUAL.
Exothermic vs. Endothermic
 Exothermic
process  heat is
transferred from the system to the
surroundings
 Heat is lost from the system
(temperature in system decreases)
 Endothermic
process  heat transferred
from the surroundings to the system
 Heat is added to the system
(temperature in system increases)
Exothermic Process
Endothermic Process
Units of Energy

Joule (J) is the SI unit of energy & heat
One kilojoule (kJ) = 1000 joules (J)

calorie (cal) = heat required to raise the
temperature of 1.00 g of water by 1 °C
1 calorie = 4.184 J
Units of Energy

Food is measured in Calories (also known
as kilocalories) instead of calories
1 Cal = 1 kcal = 1000 calories
Units of Energy
3800 cal = __________ Cal = _________ J
Units of Energy
The label on a cereal box indicates that 1
serving provides 250 Cal. What is the energy
in kJ?
Heat Transfer
Direction and sign of heat flow – MEMORIZE!
ENDOTHERMIC: heat is added to the
system & the temperature increases (+q)
EXOTHERMIC: heat is lost from the system
(added to the surroundings) & the
temperature in the system decreases (-q)
Heat Capacity

The quantity of heat required to raise an
object’s temperature by 1 °C (or by 1
Kelvin)

Heat capacity is an extensive property.
 Which will take more heat to raise the
temperature by 1 °C?
Specific Heat
(Specific Heat Capacity)

Specific Heat (C) - The quantity of heat
required to raise the temperature of one
gram of a substance by 1 °C
 Intensive property

Units:
 J/(g°C) or J/(g°K)
 cal/(g°C) or cal/(g°K)
Examples of Specific Heat
At the beach, which gets hotter, the sand or
the water?
Higher specific heat means the substance
takes longer to heat up & cool down!
Examples of Specific Heat

Specific heat (C)= the heat required to
raise the temperature of 1 gram of a
substance by 1 °C
Cwater = 4.184 J/(g°C)
Csand = 0.664 J/(g°C)
Calculating Changes in Thermal E
q = m x C x Δt
q = mCΔt
q = heat (cal or J)
m = mass (g)
C = specific heat capacity, J/(g°C)
Δt = change in temperature, Tfinal – Tinitial
(°C or K)
***All units must match up!!!***
Example
q = mCΔt
How much heat in J is given off by a 75.0 g
sample of pure aluminum when it cools from
84.0°C to 46.7°C? The specific heat of
aluminum is 0.899 J/(g°C).
Example
q = mCΔt
What is the specific heat of benzene if 3450 J
of heat are added to a 150.0 g sample of
benzene and its temperature increases from
22.5 °C to 35.8 °C?
Example
q = mCΔt
A 50.0 g sample of water gives off 1.025 kJ as
it is cooled. If the initial temperature of the
water was 85.0 °C, what was the final
temperature of the water? The specific heat
of water is 4.18 J/(g°C).
Calorimetry

Calorimetry: measurement of quantities of
heat
 A calorimeter is the device in which heat
is measured.
Calorimetry
 Assumptions:

Heat lost = -heat gained by the system

In a simple calorimeter, no heat is lost
to the surroundings
Coffee Cup Calorimetry
***Use styrofoam instead of a beaker to keep
heat in
Steps:
1. Add hot solid metal to cool water
2. Water will heat up (T rises) as metal cools
3. Eventually, water & metal are at same T.
qmetal + qwater = 0
qmetal = -qwater
Heat lost by metal = -heat gained by the water
Calorimetry
qmetal = -qwater
Heat lost by metal = -heat gained by water
Since q = mCΔT,
mmCmΔTm = -[mH2OCH2OΔTH2O]
Sample Problem
A 358.11 g piece of lead was heated in water
to 94.1 °C. It was removed from the water and
placed into 100. mL of water in a Styrofoam
cup. The initial temperature of the water was
18.7 °C and the final temperature of the lead
and water was 26.1 °C. What is the specific
heat of lead according to this data?
Bomb Calorimeter

Constant volume “bomb”
calorimeter


Some heat from reaction
warms water


qwater = mCH2OΔT
Some heat from reaction
warms bomb


Burn sample in O2
qbomb = CbombΔT
qrxn + qH2O + qbomb = 0
Energy & Changes of State

All changes of state involve energy
changes (more in Unit 9)

Note that fusion = melting
State Functions

A property where the change from initial
to final state does not depend on the path
taken

Ex.) The change in elevation from the top to
bottom of a ski slope is independent of the
path taken to go down from the slope
Enthalpy Changes for Chemical Rxns.

Heat of reaction
 The
heat absorbed or given off when
a chemical reaction occurs at
constant T (temp.) and P (pressure)
Enthalpy

Enthalpy (H)
 The heat content of a reaction
(chemical energy)

ΔH = change in enthalpy
 The amount of energy absorbed by or
lost from a system as heat during a
chemical process at constant P
 ΔH = ΔHfinal - ΔHinitial
Properties of Enthalpy

Enthalpy is an extensive property


Enthalpy is a state function


It does depend on quantity
Depends only on the final & initial values
Every reaction has a unique enthalpy
value since ΔH = Hproducts - Hreactants
Representation of Enthalpy as a Graph
Two Ways to Designate
Thermochemical Equations
Endothermic:
a) H2
(g) + I2 (s)  2 HI (g)
b) H2
(g) + I2 (s) + 53.0 kJ  2 HI (g)
kJ
ΔH = 53.0
Two Ways to Designate
Thermochemical Equations
Exothermic:
a)
½ CH4 (g) + O2 (g)  ½ CO2 (g) + H2O (l)
ΔH = -445.2 kJ
b)
½ CH4(g) + O2(g)  ½ CO2(g) + H2O(l) + 445 .2 kJ
Two Ways to Designate
Thermochemical Equations
Note the meaning of the sign in ΔH in the
equations above!!
Endothermic: ΔH = +
Exothermic: ΔH = -
Two Ways to Designate
Thermochemical Equations
Note the important of designating the
physical state or phase of matter. Why??
Because this will change the heat
of reaction! (ΔH)
Thermochemical Equations
What do the coefficients stand for? How can
they differ from the ones we have used
before?
Coefficients = the number of moles
(as before)
BUT
We can use fractional coefficients
now!
Thermochemical Equations
What is the standard state? How do we
designate conditions of temperature and
pressure that are not at standard state?
Standard state = 1 atm pressure & 25 °C
ΔH° = ΔH at standard state
Must show conditions over arrow if not at
standard state!
Thermochemical Equations
How can we find the enthalpy of reaction when
we reverse it?
Reverse the reaction, reverse the sign of
ΔH!
Example:
CO (g) + ½ O2 (g)  CO2 (g)
ΔH = -283 kJ
CO2 (g)  CO (g) + ½ O2 (g)
ΔH = +283 kJ
Example
Given Rxn. #1, find the ΔH for Rxns. 2 & 3
Reaction #1
2 SO2 (g) + O2 (g)  2 SO3 (g)
ΔH = +197.8 kJ
Reaction #2
SO2 (g) + ½ O2 (g)  SO3 (g)
ΔH =
Reaction #3
4 SO3 (g)  4 SO2 (g) + 2 O2(g)
ΔH =
ΔH as a Stoichiometric Quantity
Given the reaction below, how much heat is
produced when 15.0 g of NO2 are produced?
2 NO (g) + O2 (g)  2 NO2 (g)
ΔH = -114.1 kJ
ΔH as a Stoichiometric Quantity
Given:
ΔH = -283 kJ
CO (g) + ½ O2 (g)  CO2 (g)
(a) Calculate the enthalpy of the above
reaction when 3.00 g of product are formed
ΔH as a Stoichiometric Quantity
Given:
ΔH = -283 kJ
CO (g) + ½ O2 (g)  CO2 (g)
(b) If only 10.0 grams of oxygen and an
unlimited supply of CO are available to run this
reaction, how much heat will be given off?
ΔH as a Stoichiometric Quantity
Given:
ΔH = -283 kJ
CO (g) + ½ O2 (g)  CO2 (g)
(c) How many grams of carbon monoxide are
necessary (assuming oxygen is unlimited) to
produce 500 kJ of energy in this reaction?
ΔH as a Stoichiometric Quantity
Given:
ΔH = -283 kJ
CO (g) + ½ O2 (g)  CO2 (g)
(d) Calculate the heat of decomposition of two
moles of carbon dioxide.
Hess’s Law

The heat of a reaction (ΔH) is constant,
whether the reaction is carried out directly
in one step or indirectly through a number
of steps.

The heat of a reaction (ΔH) can be
determined as the sum of heats of
reaction of several steps.
Hess’s Law: Example
Consider the formation of water:
H2(g) + ½ O2(g)  H2O(g) + 241.8 kJ
(Exothermic Rxn  ΔH = -241.8 kJ)
Hess’s Law
Hess’s Law
Hess’s Law
Σ ΔH along one path =
Σ ΔH along another
path
Since ΔH is a state function!!
Hess’s Law
Given:
C(s) + O2(g)  CO2(g)
kJ
2 CO(g) + O2(g)  2 CO2(g)
kJ
ΔH = -393.5
ΔH = -577.0
Determine the heat of reaction for:
C(s) + ½ O2(g)  CO(g)
Hess’s Law
Given:
C(s) + O2(g)  CO2(g)
ΔH = -393.5 kJ
C2H4(g)+3 O2(g)  2 CO2(g)+2 H2O(l)
ΔH= -1410.9 kJ
H2(g) + ½ O2(g)  H2O(l)
ΔH = -285.8 kJ
Determine the heat of reaction for:
2 C(s) + 2 H2(g)  C2H4(g)
Standard Enthalpies of
Formation
NIST (National Institute for Standards and
Technology) gives values for
ΔHf° = standard molar heat of formation
Definition:
The heat content or enthalpy change
when one mole of a compound is formed at
1.0 atm pressure and 25 °C from its elements
under the same conditions.
Examples of Formation Equations
H2(g) + ½ O2(g)  H2O(g)
ΔHf°(H2O, g) = -241.8 kj/mol
C(s) + ½ O2(g)  CO(g)
ΔHf°(CO, g) = -111 kj/mol
***Elements/reactants  1 mol of compound
Notice units are per mole
Standard Enthalpy of Formation Values
 Can
look up values of in reference
book or textbook
 By
definition, ΔHf° = 0 for elements
in their standard states

Example: Cl2 (g)
H2 (g)
Ca (s)
Summation Equation

In general, when all enthalpies of formation
are known:
ΔHrxn° = ΣΔHf°(products) - ΣΔHf°(reactants)
Must multiply all Hf values by coefficient
from balanced equation!!!
Summation Equation Example
Use the summation equation to determine the
enthalpy of the following reaction:
4 NH3(g) + 5 O2(g)  4 NO(g) + 6 H2O(g)
ΔHreaction° = ΣΔHf°(products) - ΣΔHf°(reactants)
Unit 5: Part II Thermodynamics:
Spontaneity, Entropy and Free
Energy
Spontaneous Change
What is a spontaneous process?
A process that occurs by itself without an
outside force helping it.
[ A spark to start a process is OK
though]
Spontaneous Change
Which of the following are
spontaneous processes?
1.
Snowman melting in the sun
2.
Assembling a jigsaw puzzle
3.
Rusting of an iron object in humid air
4.
Recharging of a camera battery
Spontaneous Reactions and
Energy
Many spontaneous reactions are
exothermic, but not all!
Example:
H2O (s)  H2O (l)
is spontaneous and an ENDOTHERMIC
reaction!
(∆ H = + 6.0 kJ)
What other Factor Influences
Spontaneity?
The Randomness Factor!
Nature tends to move
spontaneously to a
more random state.
Entropy: Disorder and Spontaneity
What is entropy?
A measure of the
randomness (disorder) of
a system—a STATE
property (state function)!
Reaction of K
with water
The Second Law of Thermodynamics
The Second Law of
Thermodynamics states:
In a spontaneous process, there is a
net increase of entropy (taking into
account system and surroundings).
Spontaneous Processes result in more
random states (more disorder).
EXAMPLE:
H2O (s)  H2O (l)
Water molecules are
more disordered as a
liquid than as a solid.
Sample Problem
Predict which of the following processes
have a positive change in entropy:
(an increase in the randomness or disorder)
a. Taking dry ice from a freezer and
allowing it to warm from -80oC to room
temperature
b. dissolving blue food coloring in water
c. freezing water into ice cubes
Entropy
 Entropy
is used to quantify
randomness or disorder.
 Like
enthalpy, entropy is also a
state function.
The Third Law of Thermodynamics
The Third Law of Thermodynamics
states:
A completely ordered pure
crystalline solid has an entropy of
zero at 0 K.
Standard Molar Entropies: ΔSo
(1 mole, standard conditions):
Tells you entropy at 25oC and 1 atm
(standard state conditions)
Units: J/mol K
Note: Elements DO NOT have ΔSo = 0!
(like they did with ΔHo)
Standard Molar Entropies
For a substance, Entropy generally
increases as:
1.
Phase change occurs from
s l  g
2.
# moles of gas increase from
reactants to products
3.
T increases (KE increases)
For a reaction, entropy generally
increases as:
1. Reactants (solids or liquids)
Products (gases)
2. Total # moles of products >
Total # moles of reactants
3. Total # moles of gaseous products >
Total # moles of gaseous reactants
4. T is increasing.
Sign of
o
ΔS
for a reaction means:
+∆S  Entropy increases;
S prod > S react
-∆S  Entropy decreases;
S react > S prod
Example
Predict the sign of ΔS in each of the following
reaction, and explain your prediction.

NH3 (g) + HCl (g)  NH4Cl (s)

2 KClO3 (s)  2 KCl (s) + 3 O2 (g)

CO (g) + H2O (g)  CO2 (g) + H2 (g)
Calculating ∆S for a Reaction
∆So =  So (products) -  So (reactants)
Calculation is similar to ∆Ho
(from Part I)
Note units are JOULES not kJ as
before!
Example: Calculate ΔSo for the following
reaction using the tables in your reference
book and the summation equation.
2 H2 (g)
+
O2 (g)  2H2O (l)
Gibbs Free Energy and Free
Energy Change
The Gibbs (also known as Gibbs-Helmholtz) Equation
shows relationship between Energy, Entropy and
Spontaneity:
ΔG = ΔH - T ΔS
Change in Free Energy = Change in
Enthalpy – (Temp. x Change in Entropy)
What is free energy?
Free energy =
AVAILABLE energy
The Relationship between ΔGreaction and
Spontaneity
1.
If ΔG = positive, reaction is
NONSPONTANEOUS.
2.
If ΔG = zero, reaction is at
equilibrium (balanced).
3.
If ΔG = negative, reaction is
SPONTANEOUS.
Gibbs Free Energy, G
Spontaneous Processes
Must Have
a Negative Free Energy!
J. Willard Gibbs
1839-1903
How are these factors and
spontaneity related?
Case
#
ΔH
ΔS
ΔG
Result
1
-
+
-
spontaneous at all T
2
-
-
+
spontaneous toward low T
HOWEVER
nonspontaneous toward high T
3
+
+
+
+
+
-
nonspontaneous toward low T
HOWEVER
spontaneous toward high T
4
+
-
+
nonspontaneous at all T
EXOTHERMIC reactions with Increasing Entropy are
Always spontaneous!
Example
Predict if the reaction will be spontaneous or not. Use
ΔH as given and your estimate of the sign of ΔS.
a. C6H12O6 (s) + 6 O2 (g)  6 CO2 (g) + 6 H2O (g)
ΔH = -2540 kJ
b. Cl2 (g)  2 Cl (g)
ΔH is positive
Two methods of calculating
o
∆G
∆Go = ∆Ho - T∆So
a) Determine ∆Horxn and ∆Sorxn and use
Gibbs equation.
b) Use tabulated values of free energies
of formation, ∆Gfo.
(we will not do this calculation, since it is
similar to the ∆Ho one we did in Part I)
Standard Free Energy Change, ∆Go
o
∆G
=
o
∆H
-
o
T∆S
Note:
kJ
 The units for ΔSo generally are in J

The units for ΔHo are generally in

You must convert FIRST before
beginning the problem!

T is in K (oC + 273)
Example: Gibbs Equation
Calculate ΔGo for the reaction below, and predict whether
the reaction is spontaneous at 25oC.
C (s)+
2H2 (g) 
CH4 (g)
ΔSo = -80.8 J/mol K
ΔHo = -74.8 kJ/mol
T= 298 K
Standard Free Energy of Formation
ΔGorxn. = ΣGof(products) - ΣGof(reactants)
SAME SUMMATION EQUATION as
ΔHo !!!!!
(Use reference book for values)