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Energy
Chem Honors Chapter 10
7 Forms of Energy
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Sound- from vibration of sound waves
Chemical- fuel, gas, wood, battery
Radiant (light)- electromagnetic energy
Electrical energy- electrons moving among
atoms- as in the conductive wire of an electrical
cord
• Atomic- nuclear (from nucleus of an atom)
• Mechanical- walk, run
• Thermal- heat
What is energy?
• Energy is the ability to do work or
produce heat
• Two types:
–Potential energy
–Kinetic energy
• Potential energy- energy due
to position or composition
• Ex:
–water behind a dam
–Attractive and repulsive
forces
• Kinetic energy – is energy due to
the motion of the object and
depends on the mass of the
object and depends on the mass
(m) and velocity (v)
• KE = ½
2
mv
Law of Conservation of Energy
• States that energy can be
converted from one form to
another but can never be created
or destroyed
Energy form going in- electrical
Energy form going out- heat and light
Chemical energy stored in bonds of gasoline
molecules
converted to mechanical and heat energy
through combustion
Electrical energy converted to
mechanical energy
Hoover Dam- hydropower plant converts
mechanical energy (flowing water) to
electromagnetic, which is transported to homes,
and then converted back into mechanical energy
in a blender
• Work- force acting over a distance
State Function
• Property of a system that does not depend on
the pathway to its present state.
– Ex: Displacement is a state function: I send 2
students to the cafeteria…the cafeteria is a
specific distance from here….even though the 2
students take different routes to get to the
cafeteria, they bith end up the same distance
away from this room.
• Energy is a state function :
• Work and heat are not state functions:
Temperature and Heat
• What is the difference between warm water
and cold water?
• Thermal energy: random motions of
components of an object
• Temperature: measures the random motions
of the components of a substance (thermal
energy)
• Heat: flow of energy due to a temperature
difference
System and Surroundings
• System- part of the universe that
we are focusing our attention on
• Surroundings- everything else in
the universe
• Exothermic process – a process
that results in the evolution of
heat- energy flows out of the
system
• Endothermic process- a process
that absorbs energy from the
surroundings- energy flows into
the system
Exothermic or endothermic?
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1. Your hand gets cold when you touch ice
2. ice melts when you touch it
3. Ice cream melts
4. Propane is burning in a propane torch.
5. Water drops on your skin evaporate after
swimming
• 6. Two chemicals mixing in a beaker give off
heat
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Exothermic
Endothermic
Endothermic
Exothermic
Endthermic
exothermic
• Energy from an exothermic
reaction comes from the
difference in potential energy
between the products and
reactants
• Which has lower energy, the
reactants or products?
Burning match
• -Total energy is conserved
• -exothermic so energy
flows from system to
surroundings
• - So the energy gained by
the surroundings must be
equal to the energy lost
by the system
Burning match cont.
• The potential energy lost from the
burned match came from the stored
energy in the bonds of the reactants
• In any exothermic reaction some of the
potential energy stored in the chemical
bonds is converted to thermal energywhich is random kinetic energy- as
heat
Thermodynamics
• Thermodynamics- study of energy
• Law of Conservation of energy is also known
as
• THE FIRST LAW OF THERMODYNAMICS!!
–The energy of the universe is constant.
1st Law- A Quantitative application
• We can use the first law to analyze energy
changes in chemical systems
• Internal energy (E) of a system is the sum of all
of the PE (potential energy) and KE (kinetic
energy) of a system
E = PE + KE
ΔE > 0
ΔE < 0
Internal
Energy, E
H2 (g) , O2 (g)
H2O (l)
• The internal energy of a system can be
changed by a flow of work or heat or
both so mathematically we have
ΔE = q + w
Δ = change in
q = heat added or liberated by
the system
w = work done on the system
Thermodynamic quantities have
–Number – indicates the
magnitude of the change
–Sign (+ or -) – indicates the
direction of the flow from the
SYSTEM’s point of view
q and w
• When heat is added to a system, or work
is done on a system, the internal energy
increases so
– When heat is transferred from the
surroundings to the system q is positive
– When work is done on a system by the
surroundings w is positive
q and w
• Work and heat created by the system
transferred to the surroundings lowers
the internal energy of the system
– When work is done by the system on the
surroundings, w is negative
– When heat flows from the system to the
surroundings, q is negative
• In an exothermic process, what is
the sign of q?
• In an endothermic process, what is
the sign of q?
Measuring Energy Changes
• Common units of energy change
–Calorie – amount of energy (heat)
required to raise the temperature of
one gram of water one degree Celsius
–Joule – in terms of a calorie
• 1 calorie = 4.184 J
Measuring Energy Changes cont.
• Express 60.1 cal of energy as Joules
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•
60.1 cal x 4.184 J = 251 J
1 cal
Measuring Energy Changes Cont.
• The energy (heat) required to change the
temperature of a substance depends on
– The amount of substance being heated (grams)
– The temperature change
– Identity of the substance
Measuring Energy changes cont
• Identity of the substance- definition
of a calorie is based on water, what
about other substances?
Measuring Energy Changes cont.
• Specific heat – the amount of energy required
to change the temperature of one gram of any
substance by one degree Celsius
– Water
– Iron (s)
– Silver (s)
4.184 J/g °C
0.45 J/g °C
0.24 J/g °C
Measuring Energy Changes cont.
• q = mcΔT
q = energy in the form of heat
m = mass
c = specific heat
ΔT = change in temperature in °C
Measuring Energy Changes cont.
• Determine the amount of energy (heat) in
Joules required to raise the temperature of
7.40 g of water from 29°C to 46°C
Measuring Energy Changes cont.
• Ex: What quantity of energy (in Joules) is
required to heat a piece of iron weighing 1.3 g
from 25°C to 46°C ?
Calorimetry
• A calorimeter is used to determine the heat
associated with a chemical reaction
Calorimeter Constant
• All parts of a calorimeter heat up or cool down
as heat is released or absorbed in the
chemical process
• Heat capacity of the calorimeter, C, also
known as the calorimeter constant
– Defined as the sum of the products of the specific
heat and the mass of all components of the
calorimeter
Calorimeter Constant cont.
• Formula for the heat energy produced in the
calorimeter is
q = Ccalorimeter ΔT
Calorimeter Constant Ex
• A calorimeter has a heat capacity of 1265 J/°C
• A reaction causes the temperature of the
calorimeter to change from 22.34 C to 25.12
C. How many joules of heat were released in
this process?
• 3517 J
Calorimetry
• Can determine the temperature change of the
reaction, and use the heat capacity of the
calorimeter to determine the ΔH of the
reaction
• What is H?
Enthalpy (H)
• For most chemical reactions chemists are
interested in the heat generated at constant
pressure, which is denoted as qp = H
• H is the heat content of a compound
• ΔH= which is the difference in heat between
the reactants and products
• so ΔH= Hproducts - Hreactants
Thermochemistry
• Because chemical reactions occur at constant
pressure, ΔH = qp for chemical reaction
• ΔH is a state function- independent of the
path
Energy Diagram
Energy Diagram with Catalyst
Energy Diagram
Hess’s Law
• The enthalpy of a reaction is the sum of the
enthalpies of the combined reactions- known
as Hess’s Law
• Hess’s Law states that, whatever mathematical
operations are performed on a chemical
reaction, the same mathematical operations
are applied also to the heat of reaction :
Hess’s Law
• Enthalpies of reaction have been measured
and tabulated- which allows us to calculate ΔH
of a reaction without making calorimetric
measurements for all reactions
• Because enthalpy is a state function, the
change in enthalpy is NOT dependent on the
path…..doesn’t matter if the reaction takes
place in one step, or multiple steps, change in
enthalpy is the same
Hess’s Law continued
• Standard heat of reaction, ΔH°rxn ( ° indicates
standard conditions of 25°C and 1 atm) is the
heat produced when the specified number of
moles in the balanced equation reacts
• Ex:
• C3H8(g) + 5O2(g)  3CO2(g) + 4H2O(g)
• ΔH° of rxn = -2044 kJ when 1 mol of propane
reacts with 5 mol of oxygen
Hess’s Law cont.
• Note that the physical states of the reactants
and products are indicated (s,l,g, etc)
• H2(g) + ½ O2(g)  H2O(g) ΔH = - 241.8 kJ
• H2(g) + ½ O2(g)  H2O(l) ΔH = -285.8 kJ
*Hess’s Law*
• 1. If the coefficients of a chemical reaction are
all multiplied by a constant, ΔH° rxn is
multiplied by the same constant- once this is
done, the reaction is no longer standard, drop
the °
• Recall combustion of propane….
• C3H8(g) + 5O2(g)  3CO2(g) + 4H2O(g)
• ΔH°rxn = -2044 kJ when 1 mol of propane
reacts with 5 mol of oxygen
• If we multiply the reaction by 2,
2C3H8(g) + 10 O2(g)  6CO2(g) + 8H2O(g)
ΔHrxn = -4088 kJ
• If we multiply the reaction by ½,
½ C3H8(g) + 5/2 O2(g)  3/2CO2(g) + 2H2O(g)
ΔHrxn = -4088 kJ
• 2. If two or more reactions are added
together to obtain an overall reaction, the
heats of these reactions are also added to give
the heat of the overall reaction
Hess’s Law
• Ex:
N2(g) + O2(g)
2NO2(g) ΔH = 68 kJ
We could also break down the rxn
into 2 steps….
• N2(g) + O2(g)  2NO(g)
ΔH2 = 180 kJ
2NO(g) + O2(g)  2NO2(g)
ΔH3 = -112 kJ
ΔH2 + ΔH3 = 68 kJ
*Hess’s Law*
• 3. ΔH°forward rxn = - ΔH°reverse rxn
• Ex: C3H8(g) + 5O2(g)  3CO2(g) + 4H2O(g)
ΔH°rxn = -2044 kJ
• when 1 mol of propane reacts with 5 mol of oxygen In
this ex, burning propane produced a large amount of
heat. It would be almost impossible to experimentally
calculate the reverse reaction. But the Hess’s Law
states hat if we reverse the reaction, we can reverse
the sign of ΔH
• So the energy needed to produce the reverse reaction
would be + 2044 kJ
Hess’s Law Example
• Ex: pg 286
• Using the enthalpies of combustion for
graphite and diamond, calculate the ΔH°rxn for
the conversion of graphite to diamond
• Cgraphite(s)  Cdiamond(s)
Hess’s Law example cont.
• The combustion reactions are:
• Cgraphite(s) + O2(g)  CO2(g) ΔH° = -394 kJ
• Cdiamond(s) + O2(g)  CO2(g) ΔH° = -396 kJ
Hess’s Law Example continued
• If we reverse the second reaction and add
them together the compounds that are on
both sides of the reaction cancel out
• Cgraphite(s) + O2(g)  CO2(g)
ΔH = -394kJ
• +
CO2(g)  Cdiamond(s) + O2(g) ΔH = -(-396kJ)
Cgraphite(s)  Cdiamond(s)
ΔHrxn = +2kJ
Hess’s Law Another Example
• Given the following data:
• 4CuO(s)  2Cu2O(s) + O2(g) ΔH° = 288 kJ
• Cu2O(s)  Cu(s) + CuO(s) ΔH° = 11 kJ
• Calculate the ΔHrxn° for the following reaction
• 2Cu(s) + O2(g)  2CuO(s)
Specific ΔH
• ΔH°f = heat energy absorbed or released
during synthesis of one mole of a compound
from its elements at standard conditions
• ΔH°sol = heat energy absorbed or released
when a substance dissolves in a solvent
• ΔH°comb = heat energy released when a
substance reacts with oxygen to form CO2 and
H2O
Second Law of Thermodynamics
• States that the entropy of the universe
is always increasing
–Entropy (S) is a measure of disorder
or randomness
–Spontaneous process- occurs in
nature without outside intervention