Chapter 5. Thermochemistry.

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Chapter 5. Thermochemistry.
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Chapter 5. Thermochemistry.

• Thermochemistry deals with changes in energy
  that occur in chemical reactions. The study of
  energy and its transformations is known as
  thermodynamics.
• 5.1 The nature of energy.
• Different types of energy are kinetic energy and
  potential energy.

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Different forms of energy are
interconvertible
Potential Energy
Kinetic Energy
Chemical Energy
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• Kinetic energy (Ek)
• Ek ½ mv2
• Where m is the mass of the object in kg, and v
  is its speed in m/s (meters per second, also
  written as m.s-1).
• Kinetic energy is important in chemistry because
  molecules are constantly in motion and so have
  kinetic energy.

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Units of Energy

• The unit of energy is the joule (J). This is the
  energy required to move an object weighing 2 kg
  at 1 m per second.
• units of joules kg x m2
• s2
• Example. What is the energy in J of an O2
  molecule moving at 200 m/sec?
• Mass of O2 molecule 2 x 16.0 32.0 amu (1 kg
  1000 g)
• One amu 1.66 x 10-24 g 1.66 x 10-24 g x
  _1 kg_
• 1000 g
• 1.66 x 10-27 kg.

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• Mass of O2 molecule 32.0 x 1.66 x 10-27 kg
• 5.31 x 10-26 kg
• Energy in Joules is calculated as
• Ek ½ mv2
• Ek ½ x 5.31 x 10-26 kg x (200
  m/sec)2
• 1.06 x 10-21 J.

mass velocity
What is the kinetic energy of a person (50kg)
moving at a speed of 1m/s?
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Weight of person
Walking speed
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example

• What is the kinetic energy in Joules of a 45 g
  golf-ball moving at 61 m/s?
• Ek ½ mv2
• Note m has units of kg, v of m/s
• 45 g 45 g x 1 kg/1000 g 0.045 kg.
• Ek ½ x 0.045 kg x 61 m x 61 m 83.7 J.
• s s
• What happens to this energy when the ball lands
  in a sand-trap? Ans. It is converted to heat.

units check out
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Units of energy

• A joule is a very small amount of energy, and so
  one commonly uses the kJ. Energies for bonds are
  usually expressed in kJ/mol.
• The calorie Amount of energy required to raise
  temperature of 1 g of water by 1 ºC.
• Use kcal.mol-1 (kcal/mol) for bonds.
• Nutritional Calorie (note upper case) 1000 cal.

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Types of energy

• Potential energy Object has this by virtue of
  its position.
• Electrostatic energy (not covered in CHM 101).
  Potential energy due to electrostatic attraction
  or repulsion.
• Chemical Energy Due to arrangement of atoms,
  e.g. gasoline, glucose
• Thermal Energy Due to kinetic energy of
  molecules.

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Calculation of work or potential energy

• The potential energy equals the work done to
  raise the object to the height it is above the
  ground. e.g. a 5.4 kg bowling ball is raised to a
  height of 1.6 m above the ground. What is its
  potential energy? Note The force is the
  gravitational constant is g 9.8 m/s2.
• Work m x g x d
• 5.4 kg x 9.8 m x 1.6 m
• s2
• 85 kg.m2/s2 85 J
• (check J units of kg.m2/s2 so our
• calculation produces the right units).

Units Check out
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System and surroundings

• In thinking about thermodynamics, we cannot
  think about the whole universe at one time. We
  have to think about the system of interest to us,
  which for chemistry is usually the contents of
  something the size of a beaker. Thermodynamics is
  the book-keeping of energy, and so we are
  concerned with how much heat goes in or out of
  the system from the surroundings.

everything else surroundings
energy out energy in
system
an example of a system a beaker plus a solution
A system is like a bank account see below
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The System and its Surroundings
Energy can also be transferred from the
surroundings to the system
Energy can be transferred from the system to the
surroundings
OR
heat in
heat out
Hot coffee Cold soda
Heat is transferred from the hotter to the colder
object
until their temperatures are equal
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The sign of the loss or gain of energy

• We are interested in how much energy goes in or
  out of a system because this is what causes a
  chemical reaction to take place. If energy is
  lost from the system into the surroundings the
  sign of the energy change is negative, and if
  energy is gained, it is positive.

the signs of the energy changes are rather like a
bank account ve for money in, -ve for money out
energy out negative
system
energy in positive
A system is like your bank account. You only
worry about what goes in or out of it, not what
happens to the money in the surroundings, i.e.
the rest of the world.
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Work and Heat

• Energy can be transferred from one object to
  another either as work or as heat.
• Energy used to make an object move against a
  force is called work.
• w F x d ( work force x distance)
• Heat is energy transferred from a hotter to a
  colder object.

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Energy can be transferred as heat (q)
Energy can also be transferred as work (w)
When an object is moved by a force, F, over a
distance, d, energy (work) is transferred
the soccer player is doing work on the ball
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energy

• ENERGY IS THE CAPACITY TO DO WORK OR TO TRANSFER
  HEAT.
• Heat is the non-ordered transfer of energy due
  to random collisions between particles, whereas
  work is the ordered transfer of energy.

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5.2. The first Law of Thermodynamics.

• Energy is conserved
• This means that energy cannot be created or
  destroyed, but only converted from one kind of
  energy to another.

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Internal energy
molecules also have vibrational energy

• The internal energy of a system (E) is the sum
  of all kinetic and potential energies
• We dont know the internal energy of the system,
  and can generally only calculate ?E, the change
  in E that accompanies a change in the system.
• ?E Efinal - Einitial

system
kinetic energy is energy of molecules rapidly
moving about
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Relating ?E to heat and work.

• ?E q w
• Where
• q is the heat transferred to the system, and
• w is the work done on the system
• Heat transferred into the system, and work done
  on the system, are positive.

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Practice exercise

• Calculate the change in internal energy where
  the system absorbs 140 J of heat from the
  surroundings, and does 85 J of work on the
  surroundings
• q 140 J (absorbs heat ve)
• w - 85 J (does work on
• ______________ surroundings -ve)
• ?E 55 J ______________________

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Endothermic and Exothermic processes

• When a chemical process absorbs heat as it
  occurs, it is referred to as endothermic. When
  heat is given off, it is exothermic.

The burning of H2 in O2 is exothermic, because a
large amount of heat is given off 2 H2 O2 ?2
H2O heat
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Exothermic / Endothermic Processes
H2O (l)
H2O (g)
Endothermic system gains heat
Water
Water vapor
H2O (l)
H2O (s)
Exothermic system loses heat
Water ice
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State functions.

• The value of a state function depends only on
  the present conditions, not on how it got there.
  Examples of state functions are temperature and
  ?E, the change in internal energy. Q and w are
  not state functions, because one can get to a
  particular value of ?E by a variety of
  combinations of q and w. E itself is also a state
  function.

?E q w
The same value of ?E can be achieved e.g.
with large q and small w, or small q and large w
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The height difference between Denver and Chicago
is a state function.
Denver
Route A
4684 ft
Chicago
Route B
The height difference between Denver and Chicago
is a state function because it is independent of
the route taken to travel from one to the other.
The travel distance is not a state function
because one can travel by different routes.
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Work done in a chemical reaction

• The work done in a chemical reaction at constant
  pressure is given by P?V, where V is the change
  in volume during the reaction. For a reaction
  involving a gas, this can be a considerable
  contribution.

piston
increase in volume due to H2 gas given off
Zn metal
H2 formed plus air
air
Zn dissolving
Bubbles of H2
HCl
HCl
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Enthalpy (?H)

• If a reaction is carried out at constant P,
  which is true for all reactions open to the
  atmosphere, e.g. in a beaker, then the work done
  by the system is equal to -P ?V. However, the
  change in volume of a solution will generally be
  very small, and so this can be ignored. The heat
  given off or absorbed during a reaction at
  constant pressure is known as the enthalpy, and
  is given the symbol ?H.

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Enthalpy (?H)

• It can be shown that the change at constant
  pressure is given by
• ?H qp
• where the subscript p denotes constant
  pressure. When ?H is positive, the system has
  gained heat from the surroundings, and is
  endothermic. When ?H is negative, the process is
  exothermic.

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5.4. Enthalpies of reaction.

• In a thermochemical equation, the heat of
  reaction for the equation is written as
• 2 H2(g) O2(g) ? 2 H2O(g) ?H -483.6 kJ
• This is the heat given off when 2 moles of H2
  combine with one mole of O2 to give 2 moles of
  water, all in the gas phase. Note that a large
  negative value of ?H such as we have here is
  associated with a very exothermic reaction.

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1. Heat is an extensive property

• With an extensive property such as heat, the
  amount of heat given off is proportional to the
  amount of substance reacted.

small log log twice as
big burning twice as much heat
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• Enthalpy is an extensive property. If we burn
  one mol of H2 with ½ mol of O2, we will get
  483.6/2 -241.8 kJ, as shown below
• 2 H2(g) O2(g) ? 2 H2O(g) ?H -483.6
  kJ
• 2 moles 1 mole ?H
  -483.6 kJ
• 1 mole ½ mole ?H -241.8 kJ
• Factor _____moles we have______
• moles in balanced equation
• 1 mole 0.5
• 2 moles
• So multiply everything in the equation by the
  factor of 1/2 , including the enthalpy.

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2. The enthalpy of a reaction is of opposite sign
to its reverse reaction.

• If we burn H2 a large amount of heat is given
  off
• If we break H2O up into H2 and O2, an equal
  amount of heat energy has to be put into this
  reverse reaction

?H - 483.6 kJ Heat given off
2 moles H2 1 mole O2 2 moles H2O
?H 483.6 kJ Heat put back in
2 moles H2O 2 moles H2 1 mole O2
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• The enthalpy of a reaction is equal in magnitude
  but opposite in sign for the reverse reaction.
• 2 H2(g) O2(g) ? 2 H2O(g) ?H - 483.6 kJ
• but for the reverse reaction
• 2 H2O(g) ? 2 H2(g) O2(g) ?H 483.6 kJ
• For the reverse reaction one simply changes
• the sign of ?H.

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• 3. The enthalpy change for a reaction depends on
  the state of the reactants.
• 2 H2(g) O2(g) ? 2 H2O(g) ?H -483.6 kJ
• but
• 2 H2(g) O2(g) ? 2 H2O(l) ?H -659.6 kJ
• since
• H2O(l) ? H2O(g) ?H 88 kJ
• or
• H2O(g) ? H2O(l) ?H -88 kJ
• Note that -659.6 (2 x 88) -483.6 kJ
  (discussed later)

water vapor
liquid water
liquid water water vapor
water vapor liquid water
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Example

• How much heat is given off by burning 3.4 g of
  H2 in excess O2?
• 2 H2(g) O2(g) ? 2 H2O(g) ?H -483.6
  kJ
• 2 moles 1 mole
• Excess O2 means that H2 is the limiting
  reagent, and so we dont need to bother with the
  O2. So we know that 2 moles of H2 burns in O2 to
  give off -483.6 kJ, so we need to know how many
  moles of H2 we have in 3.4 g.

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Problem (contd.)

• Molecular mass H2 1.0 1.0 2.0 g/mol
• Moles H2 3.4 g x 1 mole 1.7 moles
• 2.0 g
• 2 H2(g) O2(g) ? 2 H2O(g) ?H -483.6 kJ
• 2 moles 1 mole
• 1.7 moles ?H ?
• ?H - 483.6 kJ x moles we have
• moles in balanced equation
• - 483.6 kJ x 1.7 moles
• 2 moles
• - 411.1 kJ

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Specific heat

• Specific heat is the amount of heat in joules it
  takes to raise the temperature of a substance by
  1 K. The units of specific heat are J/g.K. Some
  examples are
• Substance Specific heat (J/g.K)
• H2O(l) 4.184
• N2(g) 1.04
• Al(s) 0.90
• Fe(s) 0.45
• Hg(l) 0.14

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Calculating heat produced from rise in
temperature and a knowledge of the specific heat

• Example 5 ml of H2SO4 (at 21.2 ºC) is added to
  50 ml of water in a coffee-cup calorimeter. The
  temperature of the solution in the calorimeter
  rises from 21.2 to 27.8 ºC. How much heat was
  liberated by the dissolution of the H2SO4?
  (assume all 55 ml of solution has specific heat
  of water 4.184 J/g.K, and density of water
  1g/ml).

thermometer
Temperature 21.2 ºC
Temperature 27.8 ºC
add 5 ml H2SO4
55 ml H2O plus H2SO4
50 ml H2O
Coffee-cup calorimeter
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• Problem (contd.)
• 55 ml x 1g 55 g of solution
• 1 ml
• 4.184 heat in J (q)
• weight (g) x temperature rise (K or
  ºC)
• ________q (J)__________
• 55 g x (27.8 21.2) ºC
• q 4.184 J x 55 g x 6.6 ºC
• 1 g x 1 ºC
• -1519 J

rises in K or ºC will be the same
specific heat of water
(q is negative because heat is evolved)
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Example on calculating heat evolved per mole

• When 9.55 g of NaOH dissolves in 100.0 g of
  water in a coffee-cup calorimeter, the temp.
  rises from 23.6 to 47.4 oC. Calculate ?H for the
  process
• (Assume specific heat is as for pure water
  4.18 J/g.K.)
• NaOH (s) ? Na (aq) OH- (aq)
• We assume that when presented with the balanced
  equation we need to calculate ?H for the numbers
  of moles indicated by the coefficients, i.e. 1
  mole NaOH

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Problem (contd.)

• Wt. of solution (100.0 g 9.95 g) 109.95
  g
• change in K 47.4 - 23.6 23.8 K.
• q specifc heat x mass in g x temp. rise
  in K
• q 4.18 J x 109.95 g x 23.8 K -10938 J
• g x K
• -10.9 kJ
• F. Wt. NaOH 23 16 1 40 g/mol
• Moles NaOH 9.95 g x 1 mole
• 40 g
• 0.249 mol

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Problem (contd.)

• NaOH (s) ? Na (aq) OH- (aq)
• 1 mole 1 mole 1 mole
• 0.249 mole 0.249 mole 0.249 mol ?H
  -10.9 kJ
• ?H -10.9 kJ x 1 mol/0.249 mol
• - 43.8 kJ/mol

Note If the temperature rises in a process, then
?H will be negative.
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5.3 Calorimetry (covered in labs)
stirrer
thermometer

• The system in a coffee cup calorimeter is
  usually the solution in the calorimeter. A
  thermometer is used to monitor the temperature
  rise due to the chemical reaction being studied.
  One assumes the heat capacity of the solution is
  that of water.

two nested coffee cups to provide better
insulation
lid
the solution the system
Coffee-cup calorimeter
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5.6 Hesss Law.

• Hesss Law states If a reaction is carried out
  in a series of steps, ?H for the overall reaction
  will equal the sum of the enthalpy changes for
  the individual steps.
• Hesss Law provides a method for calculating ?H
  values that are impossible to measure directly.

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Hesss law
Increasing Enthalpy (H)
The enthalpy of going from H2O (g) (water vapor)
to H2O (s) (ice) in one step (-50 kJ) is the sum
of the two steps of going first from H2O (g) to
H2O (l) (-44kJ) and then from H2O (l) to H2O
(s) (-6 kJ)
H2O (g)
-44kJ
H2O (l)
-6 kJ
H2O (s)
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Adding enthalpies following Hess Law

• We cannot measure directly the heat of burning
  graphite to give CO. However, we can calculate
  this by combining two equations
• C(s) O2(g) ? CO2(g) ?H -395.5 kJ
• CO2(g) ? CO(g) ½O2(g) ?H 283.0 kJ
• _________________________________________________
  ___________________
• C(s) O2(g) CO2(g) ? CO2(g)
• CO(g) ½ O2(g) ?H -110.5 kJ
• C(s) ½O2(g) ? CO(g) ?H -110.5
  kJ

add the ?H values


add the equations
Cancel things that occur on both sides of equation
net equation
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The enthalpy of conversion of graphite to diamond
from Hess Law

• Graphite is the stable form of carbon
• C(graphite) ? C(diamond) ?H 1.9 kJ
• This value of ?H could not be measured directly,
  but could be obtained from the enthalpy of
  combustion of graphite and diamond using Hess
  Law
• C(graphite) O2(g) ? CO2(g) ?H
  -393.5 kJ
• C(diamond) O2(g) ? CO2(g) ?H -395.4
  kJ
• ________________________________________________
  ___________________
• C(diamond) ? C(graphite) ?H -1.9 kJ
• or C(graphite) ? C(diamond) ?H 1.9 kJ

subtract
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Using Hess Law to calculate the energy of
formation of ethylene from C (graphite) and H2
gas

• An impossible (so far) reaction to carry out
  would be
• 2 C(s) H2(g) C2H2(g) (acetylene).
• We can calculate the energy of the above by
  combining the heats of combustion of the
  components in the reaction
• C2H2(g) 2½O2(g) ? 2 CO2(g) H2O(l) ?H
  -1299.6 kJ
• C(s) O2(g) ? CO2(g) ?H
  -393.5 kJ
• H2(g) ½ O2(g) ? H2O(l)
  ?H -285.8 kJ

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• We first want to get the products on the right
  hand side, so we reverse the first equation
• C2H2(g) 2½ O2(g) ? 2 CO2(g) H2O(l) ?H
  -1299.6 kJ
• 2 CO2(g) H2O(l) ? C2H2(g) 2½O2(g) ?H
  1299.6 kJ
• Then we double the second equation because there
  are two C-atoms in the desired reaction
• C(s) O2(g) ? CO2(g) ?H -393.5
  kJ
• 2 C(s) 2 O2(g) ? 2 CO2(g) ?H -787.0 kJ

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• We now add them together in two steps (its
  easier that way)
• 2 CO2(g) H2O(l) ? C2H2(g) 2½ O2(g) ?H
  1299.6 kJ
• 2 C(s) 2 O2(g) ? 2 CO2(g) ?H -787.0 kJ
• __________________________________________________
  _____________________
• 2 C(s) H2O(l) ? C2H2(g) ½ O2(g) ?H
  512.6 kJ
• 2 C(s) H2O(l) ? C2H2(g) ½ O2(g) ?H
  512.6 kJ H2(g) ½ O2(g) ? H2O(l)
  ?H -285.8 kJ
• __________________________________________________
  _____________________
• 2 C(s) H2(g) ? C2H2(g) ?H 226.8
  kJ

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5.7. Standard enthalpies of formation

• The standard enthalpy change ?Ho is defined as
  the enthalpy change when all the reactants and
  products are in their standard states. The
  standard state is 25 oC (298 K) and 1 atm
  pressure.
• 2 H2(g) O2(g) ? 2 H2O(l) ?Hº -659.6 kJ

Superscript o indicates standard enthalpy change
Standard states for H2 and O2 Standard state
for H2O is a are gases at 25 oC and 1 atm
liquid at 25 oC and 1 atm
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Standard enthalpies of formation, ?Hof

• The standard enthalpy of formation of a compound
  ?Hof is the enthalpy of formation of one mole of
  the substance from its constituent elements, all
  being in their standard states. For elements, the
  standard state is the most stable form of the
  element at 298 K and 1 atm, e.g. C is graphite,
  not diamond. For elements in their standard state
  (e.g. C(graphite) or O2(g)), ?Hof is zero.

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• (See Table of ?Hof values on p. 192)
• ?Hof for some substances (kJ/mol)
• ______________________________________
• C2H2(g) 226.7 HCl(g) -92.3
• NH3(g) -46.19 HF(g) -268.6
• C6H6(l) 49.0 CH4(g) -74.8
• CO2(g) -393.5 AgCl(s) -127.0
• Diamond 1.88 NaCl(s) -410.9
• C2H5OH(l) -277.7 H2O(l) -285.8
• C6H12O6(s) -1273 Na2CO3(s) -1130.9
• ______________________________________

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Using Enthalpies of formation to calculate
Enthalpies of reaction

• One can show from
• Hesss Law that
• ?Horxn Sn?Hof(products) Sm?Hof(reactants)

Upper case Greek sigma means sum of
coefficients in the balanced equation
Standard sum of standard sum of standard
enthalpy heats of formation of heats of
formation of reaction all products of
all reactants
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Heat of reaction sum of heats of formation of
products minus sum of heats of formation of
reactants

• What the equation on the previous slide is
  saying is that the standard enthalpy change for a
  chemical reaction (?Horxn) is given by the sum of
  the standard heats of formation of all the
  products minus the sum of the standard heats of
  formation of all the reactants. This is best
  illustrated by some examples.

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Example the standard enthalpy of formation of
benzene
H

• Calculate the standard enthalpy change (?Horxn)
  of combustion of 1 mol of benzene from standard
  enthalpies of formation.

C
Benzene, C6H6

• write out the balanced equation
• C6H6(l) 7½ O2(g) ? 6 CO2(g) 3 H2O(l)
• 1 mole

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• Products (?Hof)
• 6 CO2(g) 6 x (-393.5) -2361.0
  kJ.mol
• 3 H2O(l) 3 x (-285.8) -857.4
  kJ/mol
• -3218.4 kJ/mol
• Reactants
• C6H6(l) (49) 49 kJ/mol
• 7 ½ O2(g) (7.5 x 0.0) 0
  kJ/mol
• 49 kJ/mol
• ?Horxn -3218.4 (49) 3267 kJ/mol
  
• (products) (reactants)