JF Basic Chemistry Tutorial Thermodynamics
Shane Plunkett
plunkes@tcd.ie
1. Gas Laws and the behaviour of gases
2. Enthalpy
- Exothermic and endothermic reactions
- Hess’s Law
- Calorimetry
- Bond enthalpy
2. 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/silberberg6
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
Boyle’s 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
Avogadro’s 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
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
Don’t forget to change the units!!
V = 438 L
R = 0.0821 atmLmol-1K-1
T = 21oC = 294K
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
P = nRT/V
P = (27.7mol)(0.0821 atmLmol-1K-1)(294K)
(438L)
P = 1.53 atm
Behaviour of gases
Dalton’s 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)
Dalton’s 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
Xa = na
ntotal
Where a is the one component you
are interested in
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.
Pa = Xa x Ptotal
Xa = na
ntotal
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
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
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
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.
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.
Ek  T
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.
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
Behaviour of gases
At the same temperature, lighter gases move on average faster than
heavier gases.
Behaviour of gases
Mean Free Path
The mean free path (λ) of any molecule is the average distance travelled by a
molecule between two successive collisions.
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
ΔEsys + ΔEsurr = 0
= ΔEuniv
Closed system
System
-can exchange
only energy with
surroundings
×
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
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, ΔH°rxn: change in enthalpy when reactants
in their standard states are changed to products in their standard states
Standard enthalpy of formation, ΔH°form: change in enthalpy when one
mole of product is formed from reactants in their standard states
Standard enthalpy of combustion, ΔH°comb: change in enthalpy when
one mole of reactant is burned completely in oxygen
Exothermic reactions
• release heat to their surroundings
• negative enthalpy change, ΔH < 0
Endothermic reactions
• take in heat from their surroundings
• positive enthalpy change, ΔH > 0
reactants
Heat given out
products
reactants
Heat taken in
products
Hess’s 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)
ΔH = -2816 kJ
and for the combustion of ethanol,
C2H5OH(l) + 3O2(g)
2CO2(g) + 3H2O(l)
ΔH = -1372 kJ
Calculate ΔH (in kJ) for the fermentation of glucose:
C6H12O6(s)
2C2H5OH(l) + 2CO2(g)
ΔH = ?
C6H12O6(s) + 6O2(g)
6CO2(g) + 6H2O(g)
ΔH = -2816 kJ
4CO2(g) + 6H2O(l)
2C2H5OH(l) + 6O2(g)
ΔH = +2744 kJ
C6H12O6(s)
2C2H5OH(l) + 2CO2(g)
ΔH = -72 kJ
Exothermic reaction
Example
Given that ΔH° for formation (ΔH°f) of Pb3O4 is -175.6 kJ mol-1 and
ΔH°
for the reaction 3PbO2  Pb3O4 + O2 is 22.8 kJ mol-1. What is the
ΔH°f
Step 1: Write
equation
for reactions given
-1
for PbO2 (in kJ mol )?
Enthalpy of formation:
3Pb(s) + 2O2(g)
Pb3O4(s)
ΔH°f = -175.6 kJ mol-1
3PbO2(s)
Pb3O4(s) + O2(g)
Step 2: Write down what you are looking for
Enthalpy of formation of PbO2:
Pb(s) + O2(g)
PbO2
ΔH° = 22.8 kJ mol-1
ΔH°f = ?
Step 3: Combine the equations we know to get the answer we require
3Pb(s) + 2O2(g)
Pb3O4(s)
ΔH°f = -175.6 kJ mol-1
Pb3O4(s) + O2(g) 3PbO2(s)
ΔH° = -22.8 kJ mol-1
3Pb(s) + 3O2(g)
3PbO2(s)
ΔH°f = -198.4 kJ mol-
PbO2(s)
ΔH°f = -66.13 kJ mol-
1
Pb(s) + O2(g)
1
Question
1.
Given the following information:
2C(s) + 3H2(g)  C2H6(g)
ΔH = -84.68 kJ mol-1
C(s) + O2(g)  CO2(g)
ΔH = -393.51 kJ mol-1
H2(g) + ½O2(g)  H2O(l)
ΔH = -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
2. Given the following information:
S(s) + O2(g)  SO2(g)
ΔH = -296.1 kJ mol-1
C(s) + O2(g)  CO2(g)
ΔH = -393.5 kJ mol-1
CS2(l) + 3O2(g)  CO2(g) + 2SO2(g)
ΔH = -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
By using a bomb calorimeter in the lab, we can determine the reaction
enthalpy
Mechanical stirrer
To electrical
source
Thermometer
Material
combusted
in oxygen
The equation to calculate the heat change is
q = m s ΔT
where q is the heat change
m is the mass of the sample
s is the specific heat of the sample
ΔT 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
Example
A sample of 350g of water is heated from 10.5°C to 15.0°C. The
specific
heat of water is 4.184 J g-1 °C-1. Calculate the heat change.
q = m s ΔT
m = 350g
4.5°C
s = 4.184 J g-1 °C-1 ΔT = (15.0 – 10.5)°C =
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 40°C to 78°C. 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 25°C to 1°C. 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
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)
ΔH°= +54 kJ mol-1
H2(g)  2H(g)
ΔH°= +436 kJ mol-1
I2(g)  2I(g)
ΔH°= +214 kJ mol-1
What is the bond dissociation energy for HI?
Step 1: Draw out structures for each molecule and decide what bonds
are broken
The equation we are interested in is:
1.
H
I
H
I
H
iodine
hydrogen
I
H
I
hydrogen iodide hydrogen iodide
ΔH°= +54 kJ mol-
The information we are given is:
2.
H
H
1
2H
ΔH°= +436 kJ mol-1
2I
ΔH°= +214 kJ mol-1
hydrogen
3.
I
I
iodine
Bonds broken in equation 1:
H
H
I
I
ΔH°= +436 kJ
-1
mol
ΔH°= +214 kJ
mol-1
Bonds broken, endothermic, ΔH°= +650 kJ mol-1
Step 2: Draw out structure and decide what bonds are formed
H
H
hydrogen
I
I
iodine
H
I
H
hydrogen iodide hydrogen iodide
ΔH°rxn = +54 kJ
mol-1
Bonds formed, exothermic, ΔH°= -2 (H
I)
ΔH°rxn = ΔH°bonds broken + ΔH°bonds
formed
+54 kJ mol-1 = 650 kJ mol-1 – 2 ΔH° (H
2 ΔH° (H
2 ΔH° (H
ΔH° (H
kJ mol-1
I
I) = (650 – 54) kJ mol-1
I) = 596 kJ mol-1
I) = 298
I)
Question
Given the following data:
HCl(g) + F2(g)  HF(g) + ClF(g)
ΔH°= -232 kJ mol-
1
ClF(g)  Cl(g) + F(g)
ΔH°= 256 kJ mol-1
HCl(g)  H(g) + Cl(g)
ΔH°= 427 kJ mol-1
F2(g)  2F(g)
ΔH°= 158 kJ mol-1
What is the bond enthalpyAnswer:
for HF?+561 kJ mol-1
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 25°C)
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, ΔS must be greater than zero. For
negative ΔS 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
Given the following information, calculate ΔS° for the
reaction
S° (J / mol
K) 213.7
CO
2
H2O
69.9
C3H8
269.9
O2
205.0
ΔS°rxn
= ΣS°products ΣS°reactants
= [(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
Question
Predict the sign of ΔS° and calculate its value from the following:
2NO(g) + O2(g)  2NO2(g)
S° (J / mol K)
NO
210.65
O2
205.0
NO2
239.9
Answer: should be negative
ΔS° = -146.5 J/K
Spontaneity and Gibbs Free Energy
• Gibbs Free energy is a measure of the spontaneity of a process
• ΔG is the free energy change for a reaction under standard state
conditions
• At constant temperature and pressure:
ΔG = ΔH – TΔS
– an increase in ΔS leads to a decrease in ΔG
– if ΔG < 0, the forward reaction is spontaneous
– if ΔG > 0, the forward reaction is nonspontaneous
– if ΔG = 0, the process is in equilibrium
– if ΔH and ΔS are positive, then ΔG will be negative only at high T,
i.e. forward reaction spontaneous
– if ΔH is positive and ΔS is negative, ΔG will always be positive,
i.e. forward reaction nonspontaneous
– if ΔH is negative and ΔS is positive, ΔG will always be negative,
i.e. forward reaction spontaneous
– if ΔH and ΔS are negative, ΔG is negative only at low T, i.e. forward
reaction spontaneous
Example
At 27°C, a reaction has ΔH = +10 kJ mol-1 and ΔS = +30 J K-1 mol-1.
What is the value of ΔG?
ΔG = ΔH – TΔS
T = 300 K
ΔG = (+10 kJ mol-1) – (300 K)(+30 J K-1 mol-1)
ΔG = +10 kJ mol-1 – 9000 J mol-1
ΔG = +10 kJ mol-1 – 9 kJ mol-1
ΔG = + 1.0 kJ mol-1
Question
For the reaction
4KClO3(s)  3KClO4(s) + KCl(s)
Calculate ΔG for the process at 298 K if ΔH = -144.3 kJ and ΔS = -36.8
J K-1
Answer: -133.3 kJ
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:
Rateforward
Reactants
Products
Ratebackward
At equilibrium, the rate of the forward reaction = the rate of the
backwards reaction
∆G0 = -RTlnK
∆G0 = Standard Gibb’s free energy change
R = gas contant
T = temperature
K = equilibrium constant
Le Chatelier’s Principle
When a chemical system at equilibrium is disturbed, the system shifts to
counteract the change