Transcript Thermochemistry I

Chem 1215
Requirements for All Six Units
Writing the chemical equation for the reaction
or process of interest.
Writing math equations correctly, and then
manipulating them correctly. This involves
showing the steps you take.
Use complete dimensional analysis whenever possible.
UNIT 1
Thermochemistry
Thermochemistry
Thermochemistry is the study of the
relationships between chemical reactions
and energy changes involving heat.
Energy comes in two forms:
potential energy (P.E.) – stored energy
chemical, electrostatic, positional
kinetic energy (K.E.) – energy of motion
Temperature is a measure of the kinetic energy of atoms and molecules.
System and Surroundings
Thermodynamics is the
study of energy and its
transformations.
To apply thermodynamics to our
study of thermochemistry, we
have to be very careful about
what we call the system and
what we call the surroundings.
What we single out for study is
the system. Everything else is
the surroundings.
The universe is the
sum of the system and
the surroundings.
Closed Systems vs. Open Systems
Closed systems are the most
straightforward to study. They can
exchange energy but not matter
with their surroundings.
open system
closed system
Open systems can exchange
both energy and matter with
their surroundings.
Internal Energy
Energy is the capacity to do work or to transfer
heat.
The internal energy (E) is the sum of all the
kinetic and potential energies of all the
components of a system.
E can be difficult to measure. ΔE, the change
in internal energy, can be measured.
ΔE = Efinal – Einitial
FYI: internal energy = translational + vibrational + rotational +
electronic + chemical potential energies
ΔE: Work and Heat
A closed system exchanges energy (ΔE) with its
surroundings through work and heat.
1. Work is energy used to move an object against a
force.
work = force x distance
d
after expansion
w = Fd
Closed system
performing work
on the
surroundings:
expanding gas
moves a piston.
ΔE: Work and Heat
1. Work is energy used to move an object against a
force.
work = force x distance
w = Fd
There are several types of work.
• Pressure-volume: w = -PΔV
• Electrical: w = -nFE
• Elongation: w = fl
• Surface expansion: γσ
Work and Heat
Closed systems exchange energy with their
surroundings in the form of work and heat.
1. Work is energy used to move an object against a
force.
2. Heat is the energy transferred from a hotter
object to a colder one as the result of a
temperature difference.
Work and heat are what bring about
changes in the internal energy
(ΔE).
First Law of Thermodynamics
The energy of the universe does not change.
ΔEuniv = 0
This is the 1st Law as it
applies to the universe.
Note the “univ” subscript.
It says that energy changes
in the system and the
surroundings must balance.
First Law of Thermodynamics
Another way to state the first law is in
terms of the system we are studying.
The system and the
surroundings can exchange
heat q and mechanical
energy (work) w.
w>0
q>0
• A positive value for heat
means heat goes into the
system.
• A positive value for work
means work is done on the
system.
First Law of Thermodynamics as it
Applies to a System
If the energy of the universe does not change, that
means the change in the energy of the system, which
is the change in internal energy ΔE, must be equal to
the heat that goes into the system and the work done
on the system.
ΔE = q + w
Note the lack of subscripts. No
subscripts means we are referring
to the system.
This is the 1st Law as it
applies to a system.
The 1st Law of Thermodynamics:
ΔE = q + w
The signs of
thermodynamic variables
are as important as their
values:
ΔE > 0 means the
internal energy
increased.
q > 0 means heat was
transferred into the
system.
w > 0 means work was
done on the system.
w>0
q>0
ΔE
Calculating ΔE
ΔE = q + w
EXAMPLE: H2(g) and O2(g) are placed in a cylinder/piston
arrangement and ignited. The gases in the system expand against
the piston and lose 1150 J of heat to the surroundings.
The joule (J) is the SI unit
for energy. In terms of SI
base units,
1J=𝟏
𝒌𝒈 𝒎𝟐
𝒔𝟐
An older unit of energy is
the calorie: 1 cal = 4.184 J
q = -1150 J
Calculating ΔE
ΔE = q + w
EXAMPLE: H2(g) and O2(g) are placed in a cylinder/piston
arrangement and ignited. The gases in the system expand against
the piston and lose 1150 J of heat to the surroundings.
The expanding gases from the
reaction cause the piston to rise,
doing 480 J of work against the
atmosphere.
w = - 480 J
ΔE = q + w = -1150 - 480
= -1630 J
q = -1150 J
The internal energy
decreased,
was evolved.
and heat
Exothermic Processes
A process in which the system evolves heat is
exothermic.
The reaction between H2(g) and O2(g) to form
H2O(l) is exothermic.
2H2(g) + O2(g) 2H2O(l) + heat
Einitial = internal energy of the reactants, H2(g) and O2(g)
Efinal = internal energy of the product, H2O(l)
Internal Energy and Energy
Diagrams
What does ΔE<0
mean?
That the internal
energy of the
products is lower
than the internal
energy of the
reactants.
Internal Energy and Energy
Diagrams
What does ΔE<0 mean?
ΔE < 0, and
ΔE = Eprod – Ereac,*
so Eprod – Ereac <0, or
Eprod <Ereac
It means that the internal energy of the
products is lower than the internal energy
of the reactants.
* Is ΔE = Eprod – Ereac the 1st Law?
Calculating ΔE
ΔE = q + w
EXAMPLE: Calculate the change in the internal energy ΔE for a
process in which the system absorbs 141 J of heat from the
surroundings and does 85 J of work on the surroundings.
ΔE = q + w = 141 - 85
ΔE = 56 J
What does ΔE>0 mean?
1. Efinal > Einitial , or
2. The internal energy E
increased.
In this example, heat was
absorbed by the system.
w = - 85 J
q = 141 J
Endothermic Processes
A process in which the system absorbs heat is
called endothermic.
The physical process that occurs when a
cold pack is activated is endothermic:
NH4NO3(s) + H2O(l) + heat  NH4NO3(aq)
The elephant and the piston
A large cylinder/piston arrangement filled with air at 1
bar and 25°C was just sitting there, minding its own
business, when an elephant stepped on it,
compressing the air inside to 36% of its original
volume and raising the pressure to 9.1 bar, which
equated to 14365 J of work. It happened so fast that
no heat could enter or leave the cylinder, resulting in a
T rise to 715°C.
What is the value of the change in internal energy of
the system?
Was the process endothermic or exothermic?
Can you use Kinetic-Molecular Theory to explain
what’s happening?
The elephant and the piston
A large cylinder/piston arrangement filled with air at 1
bar and 25°C was just sitting there, minding its own
business, when an elephant stepped on it,
compressing the air inside to 36% of its original
volume and raising the pressure to 9.1 bar, which
equated to 14365 J of work. It happened so fast that
no heat could enter or leave the cylinder, resulting in a
T rise to 715°C.
What is the value of the change in internal energy of
the system?
∆E = q + w = 0 + 14365 J = 14365 J
The elephant and the piston
A large cylinder/piston arrangement filled with air at 1
bar and 25°C was just sitting there, minding its own
business, when an elephant stepped on it,
compressing the air inside to 36% of its original
volume and raising the pressure to 9.1 bar, which
equated to 14365 J of work. It happened so fast that
no heat could enter or leave the cylinder, resulting in a
T rise to 715°C.
Was the process endothermic or exothermic?
The gas got hotter, right? So…exothermic, right?
NOT! q = 0 The process was neither.
This type of process is called adiabatic.
The elephant and the piston
Can you use Kinetic-Molecular Theory to explain
what’s happening?
The pressure increase comes from the decrease in
volume and the consequent increase in collisions with
the wall.
The temperature rise is due to the fact that work was
done on the system, causing the internal energy to
increase. For an ideal gas, KE = 3/2 RT. If KE is
increased (from the work done on the system) and
there is no loss from heat, then the temperature must
rise.