Transcript Chapter 6 - Colby College Wiki

Chapter 6 – Energy
Types of Systems
Energy
Energy (E)
The capacity to do work or transfer heat.
Kinetic Energy
The energy of motion.
Potential Energy
Energy due to condition, position, or composition.
Internal Energy
The sum of all kinetic & potential energy contained in a system
First Law of thermodynamics: The energy of the universe is
constant (i.e. energy is neither created nor destroyed)
Energy Units
Calorie (cal)
The quantity of heat required to change the temperature of one gram of
water by one degree Celsius.
Joule (J)
SI unit for heat
1 cal = 4.184 J (exactly)
Internal Energy Changes
ΔE = q + w
q = Heat
The transfer of energy due to temperature changes
Heat and temperature are not the same
w = Work
Force acting through a distance.
Energy can be transferred between a system and its surroundings as a result
of a temperature difference (q) and/or work being done (w).
Internal Energy Changes
The sign of q and w are from the system’s point of view
In an endothermic process, heat flows into a system - ΔE is positive
In an exothermic process, heat flows out of a system - ΔE is negative
Calculate ΔE for a system undergoing an
endothermic process in which 15.6 kJ of heat
flows and where 1.4 kJ of work is done on the
system.
Pressure – Volume Work
• The expansion/compression of the system against the
external atmosphere is pressure–volume work.
Calculating Energy Changes (ΔE)
When a reaction is run, the internal energy (kinetic, potential)
can be transferred as heat and/or pressure-volume work:
ΔE = q + w = q – PΔV
How much work is associated with the expansion
of a gas from 46L to 64L at a constant pressure
of 15 atm?
Filling (and heating) a hot air balloon takes
1.3X108 J of heat. At the same time, the
volume changes from 4.00X106 L to 4.5X106 L.
Assuming a constant pressure of 1 atm, what is
the energy change for the process?
Heat Capacities
Heat capacity, C
• Amount of heat needed to raise the temperature of
the system by one degree Kelvin
Heat Capacities
Heat capacity, C
• Amount of heat needed to raise the temperature of
the system by one degree Kelvin
Molar heat capacity, C
q = CΔT
• System = one mole of substance.
q = mcΔT
Specific heat capacity, c
• System = one gram of substance
If we add 500 J of heat to 25g of water
initially at 25 °C, what is the final
temperature of the water? (the specific heat
of water = 4.184 J/g•°C)
Heat Capacity and Conservation of Energy
If substances at two different
temperatures are mixed and allowed to
come to a constant temperature:
qa = –qb
What is the final temperature when 125 g of
iron at 92.3 °C is dropped into 50.0 g of
water at 27.7 °C? The specific heat of iron is
0.444 J/g•°C and the specific heat of water
is 4.184 J/g•°C. (Assume an isolated system)
A 150.0 gram sample of metal at 75.0 °C
is added to 150.0g of water at 15.0 °C .
The temperature of the water rises to 23.0
°C. What is the specific heat of the metal?
(Assume an isolated system)
Reaction (Bomb) Calorimetry
If we run a reaction in an isolated system (a calorimeter), we can
very accurately measure the heat transferred as a result of the
reaction. Note that this is a constant volume process.
As long as we know the
calorimeter (C):
qcal = CΔT
qrxn = –qcal = –CΔT
heat
Well insulated – considered isolated
Reactions at Constant Volume
ΔE = q – PΔV
In a system at constant volume, no pressure-volume work is done:
PΔV = P(0) = 0
Therefore, at constant volume, the internal energy change is
equal to the heat of reaction:
ΔE = q + 0 = qv
A Coffee Cup (Simple) Calorimeter
Note that this is a constant pressure process.
qrxn = -qcal
Well insulated – considered isolated
50.0 mL each of 1.0M HCl and 1.0M NaOH at
25 °C at mixed in a calorimeter. After
reaction, the temperature of the calorimeter is
31.9 °C. What is the heat generated for the
reaction? (We will estimate that the specific
heat of the solution/calorimeter is about the
same as that of water = 4.184 J/g•°C)
State Functions
Any property that has a unique value for a specified state of a
system is said to be a State Function.
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Water at 293 K and 1.00 atm is in a specified state.
In this state, the density of water is 0.99820 g/mL
This density is a unique function of the state.
It does not matter how the state was established.
Capitalized letters are used to identify State functions
Internal Energy – A State Function
ΔE has a unique value between two states
Reactions at Constant Volume
In a system at constant volume, no pressure-volume work is done:
PΔV = P(0) = 0
Therefore, at constant volume, the internal energy change is
equal to the heat of reaction:
ΔE = qv
Reactions at Constant Pressure
Normally, reactions are run at constant pressure (and changing
volume).
At constant pressure, both heat and pressure-volume work results
from energy changes:
Enthalpy Change
Because we are usually only interested in the heat of reaction
at constant pressure, we will define a new state function:
ΔE = qP - PΔV
Let
qP = ΔE + PΔV
H = E + PV
qP = ΔH = ΔE + PΔV
ΔH, the enthalpy change, is the measurement we will generally
use to describe thermal changes in a chemical system.
Exothermic and Endothermic Reactions
Negative ΔH = an exothermic reaction
Positive ΔH = an endothermic reaction
Note: Enthalpy change is an extensive property – it is directly
proportional to the amount of substances in the system
How Does a “Hand Warmer” Work?
How much energy is needed to heat the water
used in a 5 minute shower on Colby’s campus?
(Assume Colby showers are set to 2 gal/min)
Heat (Enthalpy) of Reaction
All reactions will have an accompanying enthalpy change:
ΔHrxn = Hfinal - Hinitial
ΔHrxn = ΔHproducts- ΔHreactants
For any reaction, the enthalpy change is the sum of the product
enthalpies minus the sum of the starting material enthalpies.
Why Do We Use These Energy Sources?
How much natural gas must we burn to produce
the heat (4435 kJ) needed for a single 5 minute
hot shower on Colby’s campus?
Changes in States of Matter
Why doesn’t a pot of (boiling) water at 100 °C all become
steam at once? Why do we have to continually apply heat?
Changes in States of Matter
Any change of state will cause an enthalpy change :
Unless stated otherwise, ΔH values are assumed to be per mole
Changes in States of Matter
Any change of state will cause an enthalpy change :
How much energy is required to convert 5 g of ice at 0 °C
to water at 50 °C? To steam at 100 °C?
Changes in States of Matter
Any change of state will cause an enthalpy change :
Which will cause a more damaging burn: skin exposed to 1 g of water at
100 °C or skin exposed to 1 g of steam at 100 °C?
Manipulating Reaction Enthalpies
The “reverse” of any reaction will have an equal enthalpy
change of opposite sign.
Standard States and Standard
Enthalpy Changes
• First we must define a particular state as a standard state
• ΔH° is the standard enthalpy of reaction
– The enthalpy change of a reaction in which all reactants and
products are in their standard states
• The Standard States are defined as:
– The pure element or compound at a pressure of 1 bar
(approximately 1 atm) and at the temperature “of interest”
(usually 25 °C).
Standard Enthalpies of Formation
• ΔHf° , the standard enthalpy of formation, is the
enthalpy change that occurs in the formation of one mole
of a substance in the standard state from the reference
(most common) forms of the elements in their standard
states.
• The standard enthalpy of formation of a pure element in its
standard state is 0.
Standard Enthalpies of Formation
ΔHf° for CH2O = – 108.6 kJ/mol.
ΔHf° for Al2O3 = – 1670 kJ/mol.
ΔHf° for Fe2O3 = – 822 kJ/mol.
What reactions do these heats of formation represent?
Standard Enthalpies of Formation
Standard Enthalpies of Formation
Reaction Summation – Hess’s Law
Hess’s law:
If a process occurs in stages or steps (even hypothetically),
then the enthalpy change for an overall process is the sum of
the enthalpy changes for the individual steps.
Manipulating ΔH – Hess’s Law
• The enthalpy change of a chemical transformation is
directly proportional to the amounts of substances:
• The reverse of a chemical reaction has an equal but
opposite DH:
What is the standard enthalpy of reaction for
the thermite reaction? How can we use a
table of standard heats of formation to
determine this?
ΔHf° for Al2O3 = – 1670 kJ/mol.
ΔHf° for Fe2O3 = – 822 kJ/mol.
What is the standard enthalpy of reaction for
the reaction below? How can we use a table
of standard heats of formation to determine
this?
2 NaHCO3
Na2CO3 + H2O + CO2
What is the standard enthalpy of reaction for
the reaction below? How can we use a table
of standard heats of formation to determine
this?
2 NaHCO3
Na2CO3 + H2O + CO2
ΔHrxn = ΣΔHf°products- ΣΔHf°reactants
What is the standard enthalpy of reaction for
the formation of N2O5 as shown below?
2 N2 (g) + 5O2
(g)
1/2 N2 (g) + 3/2 O2 (g) + 1/2 H2 (g)
N2O5 (g) + H2O (l)
H2 (g) + 1/2 O2 (g)
2N2O5
HNO3 (l)
(g)
DH = -174.1kJ
2 HNO3 (l) DH = -76.6kJ
H2O
(l)
DH = -285.8kJ