JF Basic Chemistry Tutorial Thermodynamics

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JF Basic Chemistry Tutorial Thermodynamics
Shane Plunkett plunkes_at_tcd.ie

• Gas Laws and the behaviour of gases
• Enthalpy
• - Exothermic and endothermic reactions
• - Hesss Law
• - Calorimetry
• - Bond enthalpy
• Entropy
• - Second and Third Laws of Thermodynamics
• 3. Spontaneity and Gibbs Free Energy

• Recommended reading
• M.S. Silberberg, Chemistry, The Molecular Nature
  of Matter and Change, 3rd Ed., Chapter 6
• P. Atkins L. Jones, Molecules, Matter and
  Change, 3rd Ed., Chapter 2
• Multiple choice tests http//www.mhhe.com/silberb
  erg6

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Laws of Thermodynamics - relationships
Charles Law
At constant pressure, the volume occupied by a
fixed amount of gas is directly proportional to
the temperature
V ? T
P and n fixed
Boyles Law
At constant temperature, the volume occupied by a
fixed amount of gas is inversely proportional to
the applied (external) pressure
V ? 1/P
T and n fixed
Avogadros Law
At a fixed temperature and pressure, equal
volumes of any gas contain equal numbers of
particles (or moles)
V ? n
T and P fixed
Ideal Gas Law
Combination of these laws
PV nRT
R is the universal gas constant
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Gas Law Problem
A steel tank has a volume of 438 L and is filled
with 0.885 kg of O2. Calculate the pressure of
O2 at 21oC.
PV nRT
Dont forget to change the units!!
V 438 L
R 0.0821 atmLmol-1K-1
n number of moles Given mass 0.885 kg n
mass in g/molar mass of O2 n 885g/32gmol-1
27.7 mol of O2
T 21oC 294K
P nRT/V P (27.7mol)(0.0821 atmLmol-1K-1)(294K)
(438L) P 1.53 atm
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Behaviour of gases
Daltons Law of Partial Pressure

• Gases mix homogeneously in any proportions
• Each gas in a mixture behaves as if it were the
  only gas present (as long as there is no chemical
  reaction)

Daltons Law In a mixture of unreacting gases,
the total pressure is the sum of the partial
pressures of the individual gases
Ptotal P1 P2 P3 .
Each component in a mixture contributes a
fraction of the total number of moles in the
mixture.the mole fraction (X) of that component
Pa Xa x Ptotal
Where a is the one component you are interested in
Xa na ntotal
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Question
A mixture of noble gases consisting of 5.5 g of
He, 15.0 g of Ne and 35.0 g of Kr is place in an
tank at STP. Calculate the partial pressure of
each gas.
Xa na ntotal
Pa Xa x Ptotal
The gases are noble gases, therefore they are
stable (inert) and will not react with each
other. Need to work out mole fraction (X) of
each gas.
Molar mass of He 4.003 gmol-1 Mass of He 5.5
g nHe 1.37 mol
XHe nHe / ntotal 1.37 / 2.53 0.54
PHe XHe x Ptotal 0.54 x 1 atm 0.54
atm XNe nNe / ntotal 0.74 / 2.53
0.29 PNe XHe x Ptotal 0.29 x 1 atm
0.29 atm XKr nKr / ntotal 0.42 / 2.53
0.17 PKr XKr x Ptotal 0.17 x 1 atm
0.17 atm
Molar mass of Ne 20.18 gmol-1 Mass of Ne 15.0
g nNe 0.74 mol
Molar mass of Kr 83.8 gmol-1 Mass of Kr 35.0
g nKr 0.42 mol
ntotal nHe nNe nKr 1.37 mol
0.74 mol 0.42 mol 2.53 mol Ptotal 1
atm
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Behaviour of gases
Kinetic-Molecular Theory
An observation of the behaviour of gases at the
molecular level.looking at their motion and
their speed also helps us to make sense of the
gas laws. The theory is based on 3 assumptions.

1. Particle volume. Any gas consists of a large
   amount of individual particles. The volume of an
   individual particle is extremely small compared
   with the volume of the container therefore the
   theory says that the gas particles have mass but
   no volume.
2. Particle motion. Gas particles are in contant,
   random, straight line motion, except when they
   collide with the container walls or each other.
3. Particle collisions. Collisions are elastic.
   This means that the colliding molecules exchange
   energy (but do not lose energy through friction).
   Therefore their total kinetic energy (Ek)is
   contant. In between collisions, the molecules do
   not influence each other at all.

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Behaviour of gases
The average kinetic evergy of the molecules is
proportional to the absolute temperature. At a
given temperature, the molecules of all gases
have the same average kinetic energy. Therefore,
if two different gases are at the same
temperature, their molecules have the same
average kinetic energy. If the temperature of a
gas is doubled, the average kinetic energy of its
molecules is doubled.
Ek ? T
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Behaviour of gases
Although the molecules in a sample of gas have an
average kinetic energy and therefore an average
speed, the individual molecules move at various
speeds, i.e. they exhibit a distribution of
speeds some move fast, others relatively
slowly. Collisions can change individual
molecular speeds but the distribution of speeds
remains the same.
Ek ? T
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Behaviour of gases
At the same temperature, lighter gases move on
average faster than heavier gases.
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Behaviour of gases
Mean Free Path
The mean free path (?) of any molecule is the
average distance travelled by a molecule between
two successive collisions.
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Every time we carry out a reaction in the lab, we
get a change in energy taking place. We study
thermodynamics in order to keep track of these
energy changes.

• Our first task is to identify the system in the
  universe we are interested
• in observing

Open system -can exchange mass or energy with
surroundings
Surroundings
Closed system -can exchange only energy with
surroundings
System
?Esys ?Esurr 0 ?Euniv

Isolated system - No transfer of either energy or
mass
Law of conservation of energy states that energy
cannot be made or destroyed, but can be coverted
from one form to another
Conclusion The total energy of the universe
remains constant
The First Law of Thermodynamics
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Enthalpy method for measuring energy changes
during chemical reactions

• At constant pressure, the change in enthalpy of a
  system is the amount
• of energy released or absorbed

Standard reaction enthalpy, ?Hrxn change in
enthalpy when reactants in their standard states
are changed to products in their standard states
Standard enthalpy of formation, ?Hform change
in enthalpy when one mole of product is formed
from reactants in their standard states
Standard enthalpy of combustion, ?Hcomb change
in enthalpy when one mole of reactant is burned
completely in oxygen
reactants

• Exothermic reactions
• release heat to their surroundings
• negative enthalpy change, ?H lt 0

Heat given out
products

• Endothermic reactions
• take in heat from their surroundings
• positive enthalpy change, ?H gt 0

reactants
Heat taken in
products
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Hesss Law

• The overall reaction enthalpy is equal to the sum
  of the individual enthalpies for the reactions
  which make it up

Example
Given that for the combustion of
glucose C6H12O6(s) 6O2(g) 6CO2(g)
6H2O(g) ?H -2816 kJ and for the combustion of
ethanol, C2H5OH(l) 3O2(g) 2CO2(g)
3H2O(l) ?H -1372 kJ Calculate ?H (in kJ) for
the fermentation of glucose C6H12O6(s)
2C2H5OH(l) 2CO2(g) ?H ?
C6H12O6(s) 6O2(g) 6CO2(g) 6H2O(g) ?H
-2816 kJ 4CO2(g) 6H2O(l)
2C2H5OH(l) 6O2(g) ?H 2744 kJ
C6H12O6(s) 2C2H5OH(l) 2CO2(g) ?H -72 kJ
Exothermic reaction
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Example

• Given that ?H for formation (?Hf) of Pb3O4 is
  -175.6 kJ mol-1 and ?H
• for the reaction 3PbO2 ? Pb3O4 O2 is 22.8 kJ
  mol-1. What is the ?Hf
• for PbO2 (in kJ mol-1)?

Step 1 Write equation for reactions given
Enthalpy of formation 3Pb(s) 2O2(g)
Pb3O4(s) ?Hf -175.6 kJ mol-1
3PbO2(s) Pb3O4(s) O2(g) ?H 22.8 kJ mol-1
Step 2 Write down what you are looking for
Enthalpy of formation of PbO2
Pb(s) O2(g) PbO2 ?Hf ?
Step 3 Combine the equations we know to get the
answer we require
3Pb(s) 2O2(g) Pb3O4(s) ?Hf -175.6 kJ mol-1
Pb3O4(s) O2(g) 3PbO2(s) ?H -22.8 kJ mol-1
3Pb(s) 3O2(g) 3PbO2(s) ?Hf -198.4 kJ mol-1
Pb(s) O2(g) PbO2(s) ?Hf -66.13 kJ mol-1
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Question

• Given the following information
• 2C(s) 3H2(g) ? C2H6(g) ?H -84.68 kJ mol-1
• C(s) O2(g) ? CO2(g) ?H -393.51 kJ mol-1
• H2(g) ½O2(g) ? H2O(l) ?H -285.83 kJ mol-1
• Calculate the standard enthalpy of combustion of
  ethane
• C2H6(g) 3½O2 ? 2CO2(g) 3H2O(l)

Answer -1559.8 kJ mol-1

• Given the following information
• S(s) O2(g) ? SO2(g) ?H -296.1 kJ mol-1
• C(s) O2(g) ? CO2(g) ?H -393.5 kJ mol-1
• CS2(l) 3O2(g) ? CO2(g) 2SO2(g) ?H -1072
  kJ mol-1
• Calculate the enthalpy of formation of carbon
  disulfide, CS2
• C(s) 2S(s) ? CS2(l)

Answer 86.3 kJ mol-1
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• By using a bomb calorimeter in the lab, we can
  determine the reaction
• enthalpy

Mechanical stirrer
Thermometer
To electrical source
Material combusted in oxygen
The equation to calculate the heat change is
q m s ?T
where q is the heat change m is the mass of the
sample s is the specific heat of the sample ?T
is the temperature change during reaction
The specific heat of a substance is the amount of
heat required to raise the temperature of one
gram of that substance by one degree Celsius
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Example

• A sample of 350g of water is heated from 10.5C
  to 15.0C. The specific
• heat of water is 4.184 J g-1 C-1. Calculate the
  heat change.

q m s ?T
m 350g s 4.184 J g-1 C-1 ?T (15.0
10.5)C 4.5C
q (350 g)(4.184 J g-1 C-1)(4.5 C)
6589.8 J 6.59 kJ
Question
A 560g sample of mercury is heated from 40C to
78C. The specific heat of mercury is 0.139 J
g-1 C-1. What is the heat change for the
reaction?
Answer 2.96 kJ
A 782g sample of water is cooled from 25C to
1C. The specific heat of water is 4.184 J g-1
C-1. What is the heat change for the reaction?
Answer -78.5 kJ
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Bond Enthalpy

• Measure of stability of molecule
• Enthalpy change required to break a given bond in
  1 mole of gaseous molecules
• Bond formation exothermic process, negative sign
  enthalpy
• Bond breakage endothermic process, positive sign
  enthalpy
• Average values since energy of given bond varies
  from molecule to molecule, e.g. due to
  electronegative atoms

Example
Given the following data H2(g) I2(s) ?
2HI(g) ?H 54 kJ mol-1 H2(g) ? 2H(g) ?H
436 kJ mol-1 I2(g) ? 2I(g) ?H 214 kJ
mol-1 What is the bond dissociation energy for
HI?
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Step 1 Draw out structures for each molecule and
decide what bonds are broken

• The equation we are interested in is

1.
?H 54 kJ mol-1
The information we are given is
2.
2H
?H 436 kJ mol-1
?H 214 kJ mol-1
3.
2I
Bonds broken in equation 1
H H
?H 436 kJ mol-1
I I
?H 214 kJ mol-1
Bonds broken, endothermic, ?H 650 kJ mol-1
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Step 2 Draw out structure and decide what bonds
are formed
?Hrxn 54 kJ mol-1
Bonds formed, exothermic, ?H -2 (H I)
?Hrxn ?Hbonds broken ?Hbonds formed
54 kJ mol-1 650 kJ mol-1 2 ?H (H I)
2 ?H (H I) (650 54) kJ mol-1
2 ?H (H I) 596 kJ mol-1
?H (H I) 298 kJ mol-1
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Question

• Given the following data
• HCl(g) F2(g) ? HF(g) ClF(g) ?H -232 kJ
  mol-1
• ClF(g) ? Cl(g) F(g) ?H 256 kJ mol-1
• HCl(g) ? H(g) Cl(g) ?H 427 kJ mol-1
• F2(g) ? 2F(g) ?H 158 kJ mol-1
• What is the bond enthalpy for HF?

Answer 561 kJ mol-1
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Entropy

• Measure of disorder of a system
• May be increased by increasing number of ways of
  arranging components. Explained by Boltzmann
  equation
• S k lnW
• where S entropy of system
• k Boltzmann constant
• W number of possible arrangements
• Has relationship with spontaneous change. Second
  Law of Thermodynamics spontaneous processes
  (those which occur naturally without any external
  influence) are accompanied by an increase in
  entropy of the universe
• Absolute entropies may be determined from Third
  Law of Thermodynamics At zero degrees Kelvin,
  the entropy of a perfect crystal is zero
• Because this starting point exists, can measure
  standard molar entropies entropy change for 1
  mol of a pure substance at 1 atm pressure
  (usually 25C)

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Predicting Entropy Changes

• An increase in temperature leads to greater
  kinetic energy of moving particles, more motion
  and hence greater S
• Going from solid to liquid to gas (i.e. to less
  ordered systems) leads to an increase in S
• For spontaneous change, ?S must be greater than
  zero. For negative ?S values, the process is
  spontaneous in the reverse direction

Example
Predict whether the entropy change for the
following reaction will be positive or
negative C3H8(g) 5O2(g) ? 3CO2(g) 4H2O(l)
6 gas molecules ? 3 gas molecules 4 liquid
molecules
A decrease of the more disordered gas system
indicates the entropy change for the reaction
should be negative
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Given the following information, calculate ?S
for the reaction
S (J / mol K)
CO2
213.7
H2O
69.9
C3H8
269.9
O2
205.0
?Srxn SSproducts - SSreactants
(3 mol CO2)(S of CO2) (4 mol H2O)(S of
H2O) (1 mol C3H8)(S of C3H8) (5 mol
O2)(S of O2)

• (3 mol)(213.7 J/molK) (4 mol)(69.9
  J/molK)
• (1 mol)(269.9 J/molK) (5 mol)(205.0
  J/molK)

641.1 J/K 279.6 J/K (269.9 J/K 1025 J/K)
920.7 J/K 1294.9 J/K
-374.2 J/K
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Question

• Predict the sign of ?S and calculate its value
  from the following
• 2NO(g) O2(g) ? 2NO2(g)

S (J / mol K)
210.65
NO
O2
205.0
NO2
239.9
Answer should be negative ?S -146.5 J/K
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Spontaneity and Gibbs Free Energy

• Gibbs Free energy is a measure of the spontaneity
  of a process
• ?G is the free energy change for a reaction under
  standard state conditions
• At constant temperature and pressure
• ?G ?H T?S
• an increase in ?S leads to a decrease in ?G
• if ?G lt 0, the forward reaction is spontaneous
• if ?G gt 0, the forward reaction is
  nonspontaneous
• if ?G 0, the process is in equilibrium
• if ?H and ?S are positive, then ?G will be
  negative only at high T, i.e. forward reaction
  spontaneous
• if ?H is positive and ?S is negative, ?G will
  always be positive,
• i.e. forward reaction nonspontaneous
• if ?H is negative and ?S is positive, ?G will
  always be negative,
• i.e. forward reaction spontaneous
• if ?H and ?S are negative, ?G is negative only
  at low T, i.e. forward reaction spontaneous

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Example

• At 27C, a reaction has ?H 10 kJ mol-1 and ?S
  30 J K-1 mol-1.
• What is the value of ?G?

?G ?H T?S T 300 K
?G (10 kJ mol-1) (300 K)(30 J K-1 mol-1)
?G 10 kJ mol-1 9000 J mol-1
?G 10 kJ mol-1 9 kJ mol-1
?G 1.0 kJ mol-1
Question
For the reaction 4KClO3(s) ? 3KClO4(s)
KCl(s) Calculate ?G for the process at 298 K if
?H -144.3 kJ and ?S -36.8 J K-1
Answer -133.3 kJ
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Kinetics
Refers to reaction rates the speed of the
reaction, i.e. the change in concentration of
reactants (or products) with respect to time
Equilibrium
Refers to the extent of the reation when no
further change occurs, i.e. the (final)
concentration of the product given unlimited time
An equilibrium reaction Reactants Products
Rateforward
Ratebackward
At equilibrium, the rate of the forward reaction
the rate of the backwards reaction
?G0 Standard Gibbs free energy change R gas
contant T temperature K equilibrium constant
?G0 -RTlnK
Le Chateliers Principle
When a chemical system at equilibrium is
disturbed, the system shifts to counteract the
change