Download Potential energy (E ) is stored energy, resulting from condition

Energy
Energy is the capacity to supply
heat or do work.
∆Energy = heat + work
∆E = q + w
Chapter 6
Thermochemistry
Dr. Peter Warburton
[email protected]
http://www.chem.mun.ca/zcourses/1010.php
There are two types of energy:
kinetic or potential.
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Kinetic energy
Potential energy
Potential energy (Ep)
is stored energy,
resulting from
condition, position or
composition.
Kinetic energy (Ek) is the
energy of motion.
Ek = ½ (mass)(velocity)2
Ek = ½ mv2
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Units of energy
Units of energy
Energy can be expressed in units other
than the joule. But they all represent
energy, and can be converted from one
unit form to another.
Ek = ½ mv2 gives units of
kg m2 s-2 = joule (J)
Always pay attention to the units of the
variables you are working with! You might
need to convert!
One Tic Tac contains about
8000 J, or 8 kJ (kilojoules)
of stored (potential!)
chemical energy.
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Calories and calories
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Units of energy
Calories are associated with the energy of
food. A Calorie should actually be referred
to as a kilocalorie (kcal) because
1 Calorie (Cal) = 1000 calories (cal) = 1 kcal
By definition
1 cal = 4.184 J (exactly)
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Forms of energy
System and surroundings
The system is the part of
the universe that
undergoes a fundamental
change of interest.
The surroundings is every
other part of the universe.
Thermal energy is the kinetic
energy of randomly moving
molecules.
The average thermal energy
of the molecules is what we
measure as the temperature.
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System and surroundings
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System and surroundings
We can further divide a system into one of
three different types:
1. Open system – BOTH matter and
energy can enter (or exit) the system
from (or to) the surroundings.
2. Closed system – ONLY energy can
enter or exit the system
3. Isolated system – NEITHER matter nor
energy can enter or exit the system.
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Energy changes and conservation
Energy changes and conservation
As a person climbs (kinetic energy)
Signal Hill, they store an equivalent
amount of potential energy due to the
work they do changing their height
(distance) against the force of gravity.
If the person parachutes back down to the
harbour this potential energy becomes an
equivalent amount of kinetic energy as
the person falls.
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Internal energy of a system
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Internal energy of a system
However, measuring the total internal energy of
a system is next to impossible. It’s somewhat
like the amount of water between the bottom of
a boat and the bottom of the ocean.
The internal energy U is the
sum of the kinetic energy Ek of all
the molecular motions in the
system and the potential energy
Ep stored in any chemical bonds,
intermolecular attractions and the
attractions of electrons to nuclei
U = Ek + Ep
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Internal energy of a system
U is a state function
Usually we’re more interested in the change in
the height of the ocean (how high are the
waves), and analogously we are often more
interested in the change in internal energy ∆U
of our system.
Internal energy is a
state function just like
the height of a hill is a
state function.
No matter how you
move from the top of a
hill to the bottom (path
A or path B) your
change in height is
always the same.
∆rU =
Ufinal - Uinitial
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First Law of Thermodynamics
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Another consequence of the First Law
If the system is isolated from the
surroundings, then the internal energy of
the system can not change, regardless of
the changes that take place in the system.
∆Uisolated system = 0
First Law of Thermodynamics – The
internal energy of an isolated system
is constant.
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C (s) + O2 (g) → CO2 (g)
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Imagine the universe as a “super”
isolated system which we describe in two
pieces – a “normal” system (closed,
open or isolated) and its surroundings
(everything else). Since the internal energy
of the universe must be constant, this
implies
∆Usystem = -∆Usurroundings
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Changes in internal energy
Heat
Heat is the amount of
thermal energy transferred
between the system and
surroundings due to a
difference in temperature.
The only way the internal energy of a
system can change is if there is energy
transfer to or from the surroundings.
We’ve already seen the ONLY two types
of such energy transfer – heat and work
∆Usys = q + w
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Work
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Units of work
Work is the energy change of a force
applied over a distance
w=Fxd
So if w = F x d
w=mxaxd
since F = mass x acceleration
Work must have the same units
as energy. Is this true?
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The SI unit of force, the newton (N) is a
derived unit
1 N = 1 kg m s-2
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The unit for work then is a N m,
Nm
= (kg m s-2) m
= kg m2 s-2
= joule (J).
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Point of view of the system!
More specifically…
Net energy transfer out of the system means ∆U is
negative
Net energy transfer into the system means
∆U is positive
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Heat capacity
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Specific and molar heat capacities
The specific heat (capacity)
Cs is the amount of heat
required to raise the
temperature of 1 gram of a
specific substance by 1 K.
Heat capacity (C) is the heat (q)
required to change the temperature
(T) of a substance by a given amount
(usually 1 K or ºC – they’re the same
size!).
C = q/∆T
∆T = Tfinal - Tinitial (always!)
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q = m x Cs x ∆ T
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Specific heats
Specific and molar heat capacities
These units can also be expressed as
J g-1 K-1
OR the temperature change can be
evaluated in ºC.
This is the one time you can get
away without T conversion!
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q = n x Cm x ∆ T
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Problem
Relationship between Cs and Cm
Say we heat up a glass of water
How much heat in kilojoules is required to
raise the temperature of 237 g of cold
water from 4.0 ºC to 37.0 ºC?
q=q
n x Cm x ∆T = m x Cs x ∆T
n x Cm = m x Cs
Cm = m/n x Cs
Cm = molar mass x Cs
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The molar heat capacity Cm
is the amount of heat
required to raise the
temperature of 1 mole of a
specific substance by 1 K.
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Problem answer
Thermal energy transfer
q = 32.7 kJ
When a hot object touches a cold
object we all know what happens.
The hot object cools down and the
cold object heats up.
This thermal energy transfer (heat q)
will continue until both objects reach
thermal equilibrium – the same
temperature!
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Thermal energy transfer
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Problem
Heat lost by hot object = heat gained by cold object
-qhot = qcold
-mhot Cs,hot ∆T = mcold Cs,cold ∆T
mhot Cs,hot (Teqm – Thot) =
mcold Cs,cold (Teqm – Tcold)
A 100.0 g copper sample
at 100.0 ºC is added to
50.0 g of water at 26.5 ºC.
What is the final
temperature of the
copper/water mixture?
Ultimately we will often want to solve for Teqm!
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Problem answer
PV work
If we consider the
combustion of gasoline
with oxygen to form
carbon dioxide and
water in a car engine,
there will certainly be a
qrxn to consider for the
energy change of the
system.
Teqm = 37.9 ºC = 311.1 K
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PV work
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PV work
But for the gases to
expand against the
external pressure, it
must “push” (an action
with a force) the
cylinder out of the way.
An amount of work is
done depends on “how
far” (leading to a
volume change) the
cylinder is pushed.
However, the CO2 and
H2O gas generated will
try to expand as much
as it can against the
external barometric
pressure Pbar until the
PCO2 = PH2O = Pbar.
The gas must expand
(increase it’s volume)
until this happens.
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Amount of PV work
PV work is energy transfer
Remember from slide 34 of last chapter
(blowing up a balloon) pressure multiplied
by volume is equivalent to energy units!
Here we have a
gas in a piston
cylinder where
the piston can
move in or out.
1 Pa m3 = 1 J
1 bar L = 100 J
1 atm L = 101.325 J
‫ݓ‬௚௔௦ = −ܲ௕௔௥ ∆ܸ௚௔௦
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w = -P∆V
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Problem
Positive ∆V (expansion so V gets bigger)
means work is negative.
The system does work on the surroundings
Negative ∆V (compression so V gets smaller )
means work is positive.
The system is worked upon by the
surroundings
In compressing a gas, 355 J of work is
done on the system. At the same time 185
J of heat escapes from the system. What
is ∆U for the system?
No volume change (∆
∆V = 0).
There is no PV work done.
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Problem answer
Problem
∆U = +170 J
If the internal energy of a system
decreases by 125 J at the same time 54 J
of heat is absorbed by the system, does
the system do work or have work done on
it? How much?
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Problem answer
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Chemical energy
Chemical energy is potential
energy stored in chemical
bonds of molecules. As bonds
are broken or formed in a
chemical reaction, there will be
a heat change qrxn in the
reaction system.
w = -179 J. The system does work.
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Endothermic and exothermic
Endothermic and exothermic
Endothermic reaction - heat
moves from the surroundings
into the system.
qrxn is positive.
Exothermic reaction - system
releases heat to the
surroundings.
qrxn is negative.
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Bomb calorimeter
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Bomb calorimeter
A bomb calorimeter is an
isolated system where a
heat change of a (often
combustion) reaction
results in a temperature
increase of the calorimeter,
which is made up of water,
the bomb and any other
parts that are not part of
the reaction system
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Because the bomb is a
rigid, tightly sealed
container, the chemical
system CAN NOT change
its volume. There is no PV
work!
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Bomb calorimeter
Problem
∆rU = qrxn,V = -qcalorimeter
= -(qwater + qbomb + qother parts)
= - Ccalorimeter x ∆Tcalorimeter
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The combustion of 1.013 g of vanillin,
C8H8O3 in a bomb calorimeter with a
Ccalorimeter of 4.90 kJ K-1 results in a
temperature rise from 24.89 to 30.09 ºC.
What is the heat of combustion (∆rU) of
vanillin in kJ mol-1?
Molar mass of vanillin is 152.147 g mol-1
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Problem answer
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Internal energy and enthalpy
∆rU =qcombustion = qrxn,V
= -3.83 x 103 kJ mol-1
∆U = q + w
∆U = q - P∆
∆V
So heat transferred in a process or reaction is
q = ∆U + P∆
∆V
A process at constant volume where ∆V = 0
(like in a bomb calorimeter – slide 52) has
qV = ∆U
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Internal energy and enthalpy
Endothermic and exothermic
At a constant pressure (like an open
beaker), the volume of the system can
change, and so P∆
∆V is not always zero,
and therefore the heat is
Endothermic
reaction – the
products are higher
in enthalpy than the
reactants
∆H is positive.
Exothermic reaction
- the products are
lower in enthalpy
than the reactants
∆H is negative.
qP = ∆U + P∆
∆V = ∆H
Notice we give the heat at constant
pressure a special name and symbol.
We call it the change in enthalpy (∆
∆H)
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Enthalpy is a state function
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Internal energy and enthalpy
Enthalpy (H),
like internal energy (U), is a
state function. Since ONLY
one path can be taken “at
constant pressure”, qp behaves
like a state function!
∆H = Hfinal - Hinitial
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Enthalpy is a state function
A convenient relationship
The difference between ∆U and ∆H
is generally pretty small. We will
almost exclusively talk about ∆H
(open beakers) for reactions
unless we are dealing with a bomb
calorimeter. There any heat
change at constant volume will
give us ∆U.
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How “small” is P∆V?
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How “small” is P∆V?
The PV work done on the system is
+2.5 kJ, which happens to be RT
when we lose 1 mole of gas per
mole of rxn.
This 2.5 kJ represents a value of
work that is about 0.44% of the
enthalpy (heat) change for this
reaction!
Consider the reaction
2 CO(g) + O2 (g) 2 CO2 (g)
∆U for this reaction at 298 K and
constant V is -566.0 kJ mol-1
∆H for this reaction at 298 K and
constant P is -563.5 kJ mol-1
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In a reaction involving just gases treated
with the ideal gas law, then at a constant
temperature and constant pressure
PV = nRT
w = P∆
∆V = ∆(nRT)
w = P∆
∆V = ∆nRT
The work done at constant P and T can be
related to the change in the total number of
moles of gas!
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Problem
Problem answer
∆U = -2.008 x 103 kJ mol-1
and
∆H = -2.012 x 103 kJ mol-1
The heat of combustion of liquid
2-propanol (C3H8O) is determined to be
-33.41 kJ g-1 at 298 K in a bomb
calorimeter. For the combustion of 1 mole
of 2-propanol, what is ∆U and ∆H?
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“Coffee-cup” calorimeter
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“Coffee-cup” calorimeter
A “coffee-cup” calorimeter is
treated as an isolated
system where a heat
change of a reaction results
in a temperature change of
the calorimeter, which is
made up of water, the cup
and any other parts
that are not part of
the reaction system.
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∆rH = qrxn,P = -qcalorimeter
= -(qwater + qcoffee cup + qother parts)
We OFTEN assume the cup, other parts and any
non-water reactants and products have small heat
capacities relative to water, so
∆rH = qrxn,P ≈ -qwater
qrxn ≈ -mwater x Cs,water x ∆Twater
qrxn ≈ -(dwater x Vwater) x Cs,water x ∆Twater
where d and V are the density and
volume of the water in the cup
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Problem
Problem answer
100.0 mL of 1.00 M AgNO3 (aq) and 100.0 mL
of NaCl (aq), both at 22.4 ºC, are mixed in a
coffee-cup calorimeter. If the reaction is
Ag+ (aq) + Cl- (aq) AgCl (s)
and the temperature rises to 30.2 ºC, then what
is qrxn per mole of AgCl formed.
Assume you can ignore qcoffee cup (since
Styrofoam has a high heat capacity) and the
mixed solution has Cs and density equal to that
of pure water (1.00g mL-1)
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Enthalpy is extensive
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Enthalpy is extensive
Enthalpy is an extensive property – it
changes directly with the amount of reaction
we consider, whether stoichiometricly or in
actual amounts. If
½ N2 (g) + ½ O2 (g) NO (g)
∆rH = 90.25 kJ mol-1
then
N2 (g) + O2 (g) 2 NO (g)
∆rH = 2 x 90.25 kJ mol-1 = 180.50 kJ mol-1
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qrxn = -65 kJ mol-1
½ N2 (g) + ½ O2 (g) NO (g)
∆rH = 90.25 kJ mol-1
N2 (g) + O2 (g) 2 NO (g)
∆rH = 2 x 90.25 kJ = 180.50 kJ mol-1
Notice that regardless of how we’ve written
the balanced equation, in each case we
have an enthalpy change of
90.25 kJ mol-1 of NO formed
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Enthalpy for reversed process
State functions
If we reverse a process (switch the initial
and final states around), the sign of the
enthalpy change will change as well. If
½ N2 (g) + ½ O2 (g) NO (g)
Internal energy is a state
function.
A state function is a value
or property that depends only
on the current condition of
the system.
∆rH = 90.25 kJ mol-1
then
NO (g) ½ N2 (g) + ½ O2 (g)
∆rH = -90.25 kJ mol-1
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State functions revisited
A change in a state
function only requires
us to know the initial
state and the final state
– not the “path” that got
us from one to the
other.
∆rH = Hfinal - Hinitial
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State functions revisited
Height
above sea
level is
a state
function.
Here the enthalpy change for
A + 2B → 2D
is the same as the sum of the
enthalpy changes for the rxns
A + 2B → C
C → 2D
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Path-dependent functions
Distance
travelled is
a pathdependent
function.
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Enthalpy is a state function too!
½ N2 (g) + O2 (g) NO2 (g)
The overall change in height is -425 ft
from Signal Hill to Fort Amherst, no matter
what steps we take. It is a state function!
We’re interested in the
enthalpy change of the
overall reaction above. We
can use Hess’s Law on the
reactions:
½ N2(g) + O2(g) NO(g) + ½ O2(g)
∆rH = 90.25 kJ mol-1
NO(g) + ½ O2(g) NO2(g)
∆rH = -57.07 kJ mol-1
Hess’s Law states the overall
enthalpy change for a reaction is
equal to the sum of enthalpy
changes for all the individual steps
that lead to the same overall reaction.
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How do we use Hess’s Law?
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How do we use Hess’s Law?
- number the step reactions
- make a list of the chemicals from
the overall reaction, including
states
- in the chemical list put the
numbers of all step reactions where
the same chemical appears
We need a balanced equation for the
reaction whose enthalpy change we are
interested in. This is the overall
reaction.
We also need balanced equations with
enthalpy change data for the step
reactions.
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How do we use Hess’s Law?
How do we use Hess’s Law?
- identify the chemicals that appear in
ONLY ONE step reaction
- we want these step reaction
chemicals to match the overall reaction
both in amounts and as reactants or
products
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How do we use Hess’s Law?
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How do we use Hess’s Law?
- add the modified step reactions
- group together and cancel out
similar chemicals based on which side
of the arrow they appear
- ∆rH for this sum reaction is the
added ∆rH values of the modified step
reactions
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- to make the match we are allowed to
reverse the step reaction and multiply
the step reaction amounts
- if we reverse the step reaction we
must change the sign of ∆rH - if we
multiply the step reaction amounts we
must multiply ∆rH by the same
amount
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- if this sum reaction matches the
overall reaction, we are done and have
found ∆H
- if not, use other step reactions to
cancel out the chemicals that do not
appear in the overall reaction
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How do we use Hess’s Law?
Problem
- multiply to match amounts but have
step reaction chemicals on opposite side
of arrow (by reversing) as in the sum
reaction
- REMEMBER to change ∆rH of the step
reaction
- add sum reaction and step reactions to
give overall reaction and add ∆rH values
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Methane, the main constituent of natural gas, burns in
oxygen to yield carbon dioxide and water:
CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (l)
Use the following data to calculate ∆rH (in kJ mol-1)
CH4 (g) + O2 (g) CH2O (g) + H2O (g)
∆rH = -284 kJ mol-1
CH2O (g) + O2 (g) CO2 (g) + H2O (g)
∆rH = -518 kJ mol-1
H2O (l) H2O (g)
∆rH = 44.0 kJ mol-1
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Problem answer
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Problem
CH3CH2OH (l) + O2 (g) CH3COOH (l) + H2O (l)
∆rH = -8.90 x 102 kJ
Use the following data to find ∆rH (in kJ
mol-1)
4 C (s) + 6 H2 (g) + O2 (g) 2 CH3CH2OH (l)
∆H° = -555.2 kJ mol-1
2 C (s) + 2 H2 (g) + O2 (g) CH3COOH (l)
∆H° = -484.3 kJ mol-1
2 H2 (g) + O2 (g) 2 H2O (g)
∆H° = -483.6 kJ mol-1
H2O (l) H2O (g)
∆H° = 44.0 kJ mol-1
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Problem answer
Standard states
We must be careful when looking at
energy changes.
Any differences in the system will mean
differences in enthalpy changes as well
∆H° = -492.5 kJ
We must pay attention to differences in
amounts, chemical states, pressure,
temperature, and volume
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Why are they different?
Common experience tells us that gaseous
water has higher internal energy than
liquid water (we heat water to boil it).
1 H2O (l) 1 H2O (g)
4 H2O (l) 4 H2O (g)
Notice the one difference between these
reactions, and the differences in ∆rH
The difference between the enthalpy changes is
(-2043 kJ mol-1) – (-2219 kJ mol-1)
= 174 kJ mol-1
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∆H = 44.0 kJ mol-1
∆H = 176 kJ mol-1
Notice the change in enthalpy of 4 liquid
water to 4 gas water is essentially the
same as the difference on the last slide.
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Thermodynamic standard state
Thermodynamic standard state
- pure substances
- physical state must be explicitly mentioned
- gases at one bar of pressure
- solution concentrations of one mole per
litre
temperature is NOT PART of the standard
state definition, but must be mentioned
(usually 298.15 K)
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Standard enthalpies of formation
Property values measured at
thermodynamic standard state
conditions are emphasized by the
special symbol °.
So if we measure the enthalpy change for
a reaction at the thermodynamic
standard state conditions, then we have
the
standard enthalpy of the reaction
∆rH°
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Standard enthalpies of formation
The standard enthalpy (or heat)
of formation (∆
∆fH°) is the enthalpy
change for the formation of 1
mole of a substance in its
standard state from its
constituent elements in their
reference forms in their standard
states.
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What is ∆H°f for O2(g)?
O2 (g) O2 (g)
Reference form?
formation reaction
Obviously, no change occurs here, and
so ∆fH° for oxygen gas is zero.
In fact, for any pure element in its
reference form and standard state,
we define the standard enthalpy of
formation as zero.
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Reference form?
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∆rH° = Σνp∆fH°(prod) - Σνr∆fH°(react)
We can find the standard enthalpy
change for a chemical reaction by
∆rH° = Σνp∆fH°(prod) - Σνr∆fH°
Here ν is the stoichiometric
coefficient from the balanced
equation!
The reference form of an element is
almost always the most stable form of
the element at thermodynamic standard
state conditions.
However solid “red” phosphorus is more
stable than solid “white” phosphorus, but
we choose white phosphorous as the
reference form
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The reference form of an element is
almost always the most stable form of
the element at thermodynamic standard
state conditions.
Therefore we choose to use
C(graphite) instead of C(diamond)
and Br2(l) instead of Br2(g)
as the reference forms for these elements.
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Example
∆rH° = Σνp∆fH°(prod) - Σνr∆fH°(react)
Let’s look at the reaction
∆rH° = Σνp∆fH°(prod) - Σνr∆fH°(react)
CH3CH2OH (l) + O2 (g) CH3COOH (l) + H2O (l)
We add together νp∆fH°(prod) for ALL
products, add together νr∆fH° °(react) for
ALL reactants, and take the difference
of the sums (products minus
reactants)
For the four chemicals involved we can come up with the
formation reactions, and get ∆fH° values from the text.
∆H°f = -277.6 kJ mol-1
2 C (s) + 3 H2 (g) + ½ O2 (g) CH3CH2OH (l)
∆H°f = 0.0 kJ mol-1
O2 (g) O2 (g)
∆H°f = -484.3 kJmol-1
2 C (s) + 2 H2 (g) + O2 (g) CH3COOH (l)
∆H°f = -285.8 kJ mol-1
H2 (g) + ½ O2 (g) H2O (l)
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∆rH° = Σνp∆fH°(prod) - Σνr∆fH°(react)
106
∆rH° = Σνp∆fH°(prod) - Σνr∆fH°(react)
The result ∆H°r = -492.5 kJ looks familiar because this
is the same overall reaction we looked at in slides 92
and 93 where we applied Hess’s Law.
∆rH° = Σνp∆fH°(prod) - Σνr∆fH°(react)
is Hess’s Law!
Here the very specific set of step reactions we use
are the formation reactions for all chemicals involved
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107
The application of
the equation
represents the
hypothetical case
where we break our
reactants down into
the elements in their
standard states and
then recombine the
elements to make
the products.
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27
Problem
Problem answer
Use data from Table 6.5 to calculate the
standard enthalpy of combustion of ethanol
at 298.15 K.
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∆rH° = -1367 kJ mol-1
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Chapter 6-10 Energy Use and the
Environment
Read and study this part of the
textbook for yourself.
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