Download Chapter 5. Thermochemistry.

Chapter 5. Thermochemistry.
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.
Different forms of energy are
interconvertible:
Potential Energy
Kinetic Energy
Chemical Energy
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.
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.
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 =
Ek
½ mv2
mass
velocity
= ½ x 5.31 x 10-26 kg x (200 m/sec)2
=
1.06 x 10-21 J.
What is the kinetic energy of a person (50kg) moving
at a speed of 1m/s?:
1
2
E k  mv
2
Weight of person
Walking speed
1
 m
 (50 kg )  1 
2
 s
kg m
 25
2
s
2
2
 25 J
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
units
check
out
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 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.
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.
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
Units
Check
out
=
=
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).
System and surroundings
everything else
= ‘surroundings’
system
energy
out
energy
in
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 bookkeeping of energy, and so we
are concerned with how much
heat goes in or out of the
system from the surroundings.
an example of a
system – a beaker
plus a solution
A system is like a bank account –see below
The System and its Surroundings:
Energy can be transferred
from the system to the
OR
surroundings
Energy can also be transferred
from the surroundings to the
system
heat
in
heat
out
Hot coffee
Cold soda
Heat is transferred from the hotter to the colder object
… until their temperatures are equal
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
energy out
from the system into the
= negative
system
surroundings the sign of
the energy change is
energy in
negative, and if energy is
= positive
gained, it is 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.
the signs of the energy
changes are rather like a
bank account: +ve for money
in, -ve for money out
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.
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
w  F d
the soccer player
is doing work
on the ball
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.
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.
molecules
also have
vibrational
energy
Internal energy
The internal energy of a system
(E) is the sum of all kinetic and
potential energies
system
kinetic energy is
energy of molecules
rapidly moving about
We don’t know the internal
energy of the system, and can
generally only calculate ΔE, the
change in E that accompanies a
change in the system.
ΔE
= Efinal - Einitial
Relating ΔE to heat and work.
ΔE
=
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.
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)
ΔE =
+ 55 J
______________________
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
Exothermic / Endothermic Processes
H2O (l)
H2O (g)
Endothermic:
system gains heat
Water
Water vapor
H2O (l)
H2O (s)
Exothermic:
system loses heat
Water
ice
State functions.
ΔE
=
q+w
The same value
of ΔE can be
achieved e.g. with
large q and small
w, or small q and
large w
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 ΔE, the change in
internal energy. Q and w are
not state functions, because
one can get to a particular
value of ΔE by a variety of
combinations of q and w. E
itself is also a state function.
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.
Work done in a chemical reaction:
The work done in a chemical reaction at constant
pressure is given by PΔV, where V is the change in
volume during the reaction. For a reaction involving
a gas, this can be a considerable contribution.
piston
}
Zn
metal
H2 formed
plus air
air
increase in
volume due
to H2 gas
given off
Zn
dissolving
HCl
HCl
Bubbles of H2
Enthalpy (ΔH):
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 ΔV.
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 ΔH.
Enthalpy (ΔH):
It can be shown that the change at constant
pressure is given by
ΔH =
qp
where the subscript ‘p’ denotes constant
pressure. When ΔH is positive, the system
has gained heat from the surroundings, and
is endothermic. When ΔH is negative, the
process is exothermic.
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)
ΔH = -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 ΔH such as we have
here is associated with a very exothermic
reaction.
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
burning
log twice as big
= twice as much heat
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)
2 moles
1 mole
1 mole
½ mole
Factor
=
=
ΔH = -483.6 kJ
ΔH = -483.6 kJ
ΔH = -241.8 kJ
_____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.
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:
ΔH = - 483.6 kJ
Heat given off
2 moles H2 1 mole O2
2 moles H2O
If we break H2O up into H2 and O2, an equal
amount of heat energy has to be put into this
reverse reaction:
ΔH = + 483.6 kJ
Heat put back in
2 moles H2O
2 moles H2 1 mole O2
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)
ΔH = - 483.6 kJ
but for the reverse reaction:
2 H2O(g) → 2 H2(g) + O2(g)
ΔH = + 483.6 kJ
For the reverse reaction one simply changes
the sign of ΔH.
3. The enthalpy change for a reaction depends
on the state of the reactants.
2 H2(g) + O2(g) →2 H2O(g)
ΔH = -483.6 kJ
water vapor
but
2 H2(g) + O2(g) →2 H2O(l)
since
H2O(l)
liquid water
or
H2O(g)
water vapor
ΔH = -659.6 kJ
liquid water
→
H2O(g)
ΔH = +88 kJ
H2O(l)
ΔH = -88 kJ
water vapor
→
liquid water
Note that -659.6 + (2 x 88) = -483.6 kJ (discussed later)
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) ΔH = -483.6 kJ
2 moles
1 mole
‘Excess O2’ means that H2 is the limiting
reagent, and so we don’t 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.
Problem (contd.)
Molecular mass H2 = 1.0 + 1.0 = 2.0 g/mol
Moles H2
=
3.4 g x
1 mole
2.0 g
2 H2(g) + O2(g) → 2 H2O(g)
2 moles 1 mole
1.7 moles
ΔH
=
- 483.6 kJ x
=
- 483.6 kJ
=
- 411.1 kJ
= 1.7 moles
ΔH = -483.6 kJ
ΔH = ?
moles we have
moles in balanced equation
x
1.7 moles
2 moles
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
H2O(l)
N2(g)
Al(s)
Fe(s)
Hg(l)
Specific heat (J/g.K)
4.184
1.04
0.90
0.45
0.14
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).
Temperature
= 21.2 ºC
50 ml H2O
thermometer
add 5 ml
H2SO4
Coffee-cup
calorimeter
Temperature
= 27.8 ºC
55 ml H2O
plus H2SO4
Problem (contd.)
55 ml
4.184
specific heat
of water
x
1g
1 ml
=
=
=
55 g of solution
heat in J (q)
weight (g) x temperature rise (K or ºC)
________q (J)__________
55 g
q
=
=
rises in K or
ºC will be the
same
x
4.184 J
1 g x 1 ºC
-1519 J
(27.8 – 21.2) ºC
x 55 g
x
6.6 ºC
(q is negative because
heat is evolved)
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 ΔH 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 ΔH for the numbers of
moles indicated by the coefficients, i.e. 1 mole NaOH
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
gxK
= -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
Problem (contd.)
NaOH (s) → Na+ (aq)
+
1 mole
1 mole
0.249 mole 0.249 mole
OH- (aq)
1 mole
0.249 mol
ΔH = -10.9 kJ
ΔH = -10.9 kJ x 1 mol/0.249 mol
= - 43.8 kJ/mol
Note: If the temperature rises in a process, then ΔH
will be negative.
5.3 Calorimetry (covered in labs):
thermometer
stirrer
two nested
coffee cups
to provide
better
insulation
lid
the solution =
the ‘system’
Coffee-cup
calorimeter
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.
5.6 Hess’s Law.
Hess’s Law states: If a reaction is
carried out in a series of steps, ΔH for
the overall reaction will equal the sum
of the enthalpy changes for the
individual steps.
Hess’s Law provides a method for
calculating ΔH values that are
impossible to measure directly.
Hess’s law
Increasing
Enthalpy (H)
H2O (g)
-44kJ
-50kJ
H2O (l)
-6 kJ
H2O (s)
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)
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
add the
equations:
ΔH values
{
O2(g) → CO2(g)
ΔH
→ CO(g) + ½O2(g) ΔH
= -395.5 kJ
= 283.0 kJ
C(s) + O2(g) + CO2(g) → CO2(g) +
CO(g) + ½ O2(g)
ΔH
= -110.5 kJ
C(s) +
CO2(g)
____________________________________________________________________
add the
equations
net
equation
Cancel things that occur on both sides of equation
C(s) + ½O2(g) → CO(g)
ΔH
= -110.5 kJ
}
The enthalpy of conversion of graphite to
diamond from Hess’ Law
Graphite is the stable form of carbon:
C(graphite) → C(diamond) ΔH
= +1.9 kJ
This value of ΔH could not be measured directly, but
could be obtained from the enthalpy of combustion of
graphite and diamond using Hess’ Law:
subtract
C(graphite) + O2(g) → CO2(g)
C(diamond) + O2(g) → CO2(g)
ΔH = -393.5 kJ
ΔH = -395.4 kJ
C(diamond) → C(graphite)
C(graphite) → C(diamond)
ΔH =
ΔH =
___________________________________________________________________
or
-1.9 kJ
+1.9 kJ
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) ΔH = -1299.6 kJ
C(s)
+ O2(g)
→ CO2(g)
ΔH = -393.5 kJ
H2(g) + ½ O2(g) → H2O(l)
ΔH = -285.8 kJ
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) ΔH = -1299.6 kJ
2 CO2(g) + H2O(l) → C2H2(g) + 2½O2(g) ΔH = +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)
2 C(s) + 2 O2(g) → 2 CO2(g)
ΔH = -393.5 kJ
ΔH = -787.0 kJ
We now add them together in two steps (it’s easier that
way):
2 CO2(g) + H2O(l) → C2H2(g) + 2½ O2(g) ΔH = +1299.6 kJ
2 C(s) + 2 O2(g) → 2 CO2(g)
ΔH = -787.0 kJ
_______________________________________________________________________
2 C(s) + H2O(l) → C2H2(g) + ½ O2(g) ΔH = +512.6 kJ
2 C(s) + H2O(l) → C2H2(g) + ½ O2(g) ΔH = +512.6 kJ
H2(g) + ½ O2(g) → H2O(l)
ΔH = -285.8 kJ
_______________________________________________________________________
2 C(s) + H2(g)
→ C2H2(g)
ΔH = +226.8 kJ
5.7. Standard enthalpies of formation
The standard enthalpy change ΔHo 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.
Superscript ‘o’ indicates standard enthalpy change
2 H2(g) + O2(g) → 2 H2O(l) ΔHº = -659.6 kJ
Standard states for H2 and O2
are gases at 25 oC and 1 atm
Standard state for H2O is a
liquid at 25 oC and 1 atm
Standard enthalpies of formation,
ΔHof
The standard enthalpy of formation of a
compound ΔHof 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)), ΔHof is zero.
(See Table of ΔHof values on p. 192)
ΔHof 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
______________________________________
Using Enthalpies of formation to
calculate Enthalpies of reaction:
One can show from
Hess’s Law that:
Upper case Greek ‘sigma’
means ‘sum of ’
coefficients in the
balanced equation
ΔHorxn = ΣnΔHof(products) – ΣmΔHof(reactants)
Standard
enthalpy
of reaction
sum of standard
heats of formation of
all products
sum of standard
heats of formation
of all reactants
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 (ΔHorxn)
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.
Example – the standard enthalpy of
formation of benzene:
H
Calculate the standard
enthalpy change (ΔHorxn) of
combustion of 1 mol of
benzene from standard
enthalpies of formation.
C
Benzene, C6H6
1)write out the balanced equation:
C6H6(l) + 7½ O2(g)
1 mole
→
6 CO2(g) + 3 H2O(l)
Products:
(ΔHof)
6 CO2(g) = 6 x (-393.5)
3 H2O(l) = 3 x (-285.8)
=
=
-2361.0 kJ.mol
-857.4 kJ/mol
-3218.4 kJ/mol
=
=
+49 kJ/mol
0 kJ/mol
+ 49 kJ/mol
Reactants:
C6H6(l)
7 ½ O2(g)
= (+49)
= (7.5 x 0.0)
ΔHorxn = -3218.4 – (+49) =
(products) (reactants)
3267 kJ/mol