Transcript Chapter 5: Thermochemistry

Chapter 5:
Thermochemistry
Problems: 5.1-5.95, 5.97-98, 5.100, 5.106,
5.108, 5.118-5.119, 5.121, 5.126
Energy: Basic Concepts and
Definitions
energy: capacity to do work or to produce heat
thermodynamics: study of energy and its transformations
from one form to another
thermochemistry: study of heat flow accompanying
chemical reactions
heat: energy that is transferred from a body at a higher
temperature to one at a lower temperature  heat always
transfers from the hotter to the cooler object!
– "heat flow" means heat transfer
Heat Transfer and
Temperature
One becomes hotter by gaining heat.
One becomes colder by losing heat—i.e., when you “feel
cold”, you are actually losing heat!
Example:
a. You burn your hand on a hot frying pan.
________________ loses heat, and ______________ gains the heat.
b. Your tongue feels cold when you eat ice cream.
_______________ loses heat, and ______________ gains the heat.
Heat Transfer and
Temperature
Example: A student puts a few drops of rubbing alcohol on her
palm then spreads it out with her finger and notices that the
area with the alcohol felt cooler. A minute later her palm is dry.
Explain what physical change has occurred and why her palm
felt cooler.
Types of Energy
work (w): w = F  d
Work is done whenever a force (F) moves an object a
specified distance (d)
kinetic energy (KE): energy associated with an object’s
motion
Example: a car moving at 75 mph has much greater KE
than the same car moving at 15 mph
 Greater damage if the car crashes at 75 mph than at
15 mph
KE = ½ mv2
m=mass, v=velocity
Types of Energy
potential energy (PE): energy due to position or its
composition (chemical bonds)
A 10-lb bowling ball has higher PE when it is 10 feet off the
ground compared to 10 inches off the ground
 Greater damage on your foot after falling 10 feet
compared to falling only 10 inches
positional P.E. = mass  force of gravity  height
In terms of chemical bonds, the stronger the bond,
 more energy is required to break the bond,
 the higher the potential energy of the bond
Energy is a State Function
state function: a property that is based only on the physical
state and chemical composition of a substance, so it is
independent of the path followed to achieve that state or
composition.
For example:
Potential energy is a state function because the potential
energy of two skiers at the top of a given hill would be the
same whether they climbed that hill or took a ski lift to the
top.
Units of Energy
1 kg m2
joule (J): 1 J =
s2
• SI (Système Internationale or standard) unit of energy
– To recognize the size of a joule, note that 1 watt = 1 J/s
– So a 100-watt light bulb uses 100 J every second.
• Heat is also often reported in kilojoules (kJ), where 1 kJ =
1000 J
• 1 J = 0.239 calories; 1000 calories = 1 kcal = 1 food calorie
• How many Joules in a banana?
– A medium banana has about 105 kcal = 439,330 J
Conservation of Energy
Energy is neither created nor destroyed but converted from
one form to another.
For example, the kinetic energy of a car can cause
considerable damage if the car is stopped suddenly in a
crash.
• Example: Calculate the kinetic energy (in joules) for a
3.63103 kg Hummer H2 traveling at 25 miles per hour.
(Use 1 mph = 0.4469 m/s and 1 J = kg m2/s2).
Conservation of Energy
Example (cont’d): At what speed (in miles per hour)
must a 7.3102 kg Smart Car move to have the same
kinetic energy as the Hummer H2?
At the Molecular Level
• Temperature governs the motion of particles at the
molecular level.
– At higher temperature, particles move faster and have
higher kinetic energy.
• Temperature is a measure of the average kinetic energy for
a substance.
• Thermal energy is the kinetic energy associated with the
motion of particles.
– Proportional to the temperature for any given substance
– Increases with the size of a sample
• The particles in a cup of boiling water at 100°C and those in
a pot of boiling water (also at 100°C) have the same average
kinetic energy, but the pot of boiling water has more
thermal energy than the cup of boiling water because it
contains more water molecules.
At the Molecular Level
• Electrostatic potential energy is due to the electrostatic
interactions due to the charges or dipoles in atoms, ions,
and molecules.
Q1 Q2
– Coulomb's law (E ∝ r ) is used to determine the
strength of ionic bonds.
– Thus, at the molecular level, potential energy is
determined in terms of the strength of the bonds holding
atoms, ions, and molecules together in various
substances.
Systems, Surroundings and
Energy Transfer
system: that part of the universe being studied
surroundings: the rest of the universe outside the system
Systems can be isolated, closed or open.
Isolated, Closed and Open
Systems
• isolated system: exchanges neither energy nor matter
with the surroundings
– Hot soup in a perfectly insulated thermos that does
not allow any heat to escape.
• closed system: exchanges energy but no matter with
the surroundings
– Hot soup in a cup with a lid allows heat to escape
to the surroundings but no matter.
• open system: exchanges energy and matter with the
surroundings
– Hot soup in an open cup allows both heat and
water vapor (steam) to escape to the surroundings
Direction and Sign of Heat Flow
Let q = heat flow,
• q is + when heat flows into the system from the
surroundings
• q is – when heat flows out of the system into the
surroundings
endothermic change: a physical or chemical change that
requires energy or heat to occur
Boiling water requires energy:
H2O(l) + heat  H2O(g)
Electrolysis of water requires energy:
2 H2O(l) + electrical energy  2 H2(g) + O2(g)
Direction and Sign of Heat Flow
exothermic change: a physical or chemical change that
releases energy or heat
Water condensing releases energy:
H2O(g)
 H2O(l) + heat
Hydrogen burning releases energy:
2 H2(g) + O2(g)  2 H2O(g) + heat
For physical changes, consider whether the reactants or
products have more kinetic energy.
– If the reactants have greater kinetic energy than the
products  exothermic process.
– If the products have greater kinetic energy than the
reactants  endothermic process.
Heat of Reaction
What causes the Heat of Reaction?
Bond Energy
• energy required to break a particular bond in 1 mol of
gaseous molecules
– always positive since breaking a bond always requires
energy
– a quantitative measure of the strength of a bond (i.e.
stability of compound)
Breaking and Forming Bonds
• Energy is absorbed by reactants when their bonds are
broken, and energy is released by products when their
bonds are formed.
heat of reaction (qreaction): heat associated with a chemical
reaction
Heat of Reaction
If
 energy required 


to break

 is greater
 reactants’ bonds 


 energy released 


 w henproducts’ 
 bonds are formed


 Endothermic Reaction
If
 energy required 


to break

 is less than
 reactants’ bonds 


 energy released 


 w henproducts’ 
 bonds are formed


 Exothermic Reaction
Heat of Reaction
For chemical changes, observe if the surroundings (including
you) feel hotter or colder after the reaction has occurred.
– If the surroundings are hotter, the reaction released heat
 exothermic reaction.
– If the surroundings are colder, the reaction absorbed
heat  endothermic reaction.
Examples
Which of the following are endothermic changes:
freezing
vaporizing
sublimation
deposition
melting
condensation
When a student dissolves ammonium chloride in a large test
tube, he notices the test tube feels colder. Explain what is
releasing heat and what is gaining heat.
1st Law of Thermodynamics
The energy of the universe is constant. The energy gained or
lost by a system must equal the energy lost or gained by the
surroundings.
Essentially, the Law of Conservation of Energy: Energy can
neither be created or destroyed but converted from one form
to another.
A system’s Internal Energy (E) = kinetic energy (KE) +
potential energy (PE) of all the particles in the system.
• While the values of KE and PE at a given instant are difficult
to determine, changes in KE and/or PE can be determined
by measuring any temperature changes for the system.
Heat and Work
A system’s internal energy (E) can be changed using heat (q),
work (w), or both:
• The total increase in the energy of a system is the sum of
heat flowing into it and the work done to it.
∆E = q + w
In this course, we will focus on work that involves the
expansion or compression of gases.
• For example, consider the work done during the combustion
of fuel in an engine…
Consider the following sign conventions for the change in
volume, ∆V, for the system:
• If a gas is created, the system’s volume expands.  ∆V is
positive
• The system does work to expand into the surroundings. 
w is negative
• If a gas is compressed, the system’s volume is compressed.
 ∆V is negative
 Work is done to the system.  w is positive
Atmospheric pressure, P, is positive
 work is defined as w = –P∆V
Thus, for the expansion or compression of a gas, the change in
internal energy (∆E) can be shown as:
∆E = q + w
or
∆E = q – P∆V
Examples
Consider the combustion of octane (C8H18), a primary
component of gasoline.
2 C8H18(l) + 25 O2(g)  18 H2O(g) + 16 CO2(g)
a. Calculate the change in volume (in L) due to the total
number of moles of gases produced when 1.000 gal (~3.784 L)
of octane undergoes combustion at 1.00 atm and 25.00°C. The
density of octane at 25.00°C is 0.703 g/mL.
Examples
b. What is the volume of the gases at 475 K (the temperature
of a car engine)?
c. Use the equation, w = – P∆V, to calculate the work (in kJ)
done by the system (the reaction) during the combustion of
1.000 gal of octane. (Use 1 Latm = 101.325 J).
Examples
d. If 1 kJ 1 kWs (kilowatt·second), then calculate the energy (in
kJ) in a kilowatt·hour (kWh).
e. Consider a new kind of vehicle that could be powered by an
electrical current similar to that used in our homes. Calculate
the cost of electricity needed to produce the same amount of
energy (determined in part c) as the combustion of 1.00 gal of
gasoline if Seattle City light charges 9.55 cents/kWh.
Examples
f. The combustion of 1.000 gal of octane produces about
1.2105 kJ of heat. Compare the change in internal energy due
to work calculated in part c with the heat of the reaction. How
much of the internal energy change is due to the work done by
the system due to gas expansion? (Hint: Compare the absolute
value for heat versus work.)
For the remainder of the chapter, we will focus mainly on the heat of
reaction and assume the change in energy due to gas expansion work is
negligible for the reactions considered.
Enthalpy and Enthalpy Changes
Enthalpy (H): the sum of a system’s internal energy and the
product of its pressure and volume
H = E + PV
Thus, the enthalpy change (∆H) is: ∆H = ∆E + ∆(PV)
or at constant atmospheric pressure is: ∆H = ∆E + P∆V
Since the previous definition of internal energy, ∆E = q – P∆V,
the equation can be rewritten as
∆H = q – P∆V + P∆V = qp
(the subscript p means “at constant pressure”)
Enthalpy and Enthalpy Changes
Thus, the enthalpy change (∆H) refers to heat flow into and
out of a system under constant pressure (usually the case
since reactions occur under atmospheric pressure),
qreaction = ∆H = Hproducts – Hreactants
For an endothermic reaction:
∆H = positive
For an exothermic reaction:
∆H = negative
Heating Curves and Heating
Capacity
Consider the changes that H2O undergoes when a block of ice
is taken from a freezer and heated in a pan until it is
completed converted into steam.
– A heating-cooling curve shows the changes in physical
state with temperature and heat added to or removed
from any system.
100 °C
Temperature
0 °C
Heat Added
Specific Heat
specific heat (cs): amount of heat necessary to raise the
temperature of 1 gram of any substance by 1°C; has units of
J/g°C
• Water has a relatively high specific heat (4.184 J/g·°C)
compared to the specific heats of rocks and other solids (1.3
J/g·°C for dry Earth, 0.9 J/g·°C for concrete, 0.46 J/g·°C for
iron)
• Because water covers most of the Earth, water can absorb a
lot more energy before its temperature starts to rise.
– Water helps to regulate temperatures on Earth within a
comfortable range for humans.
– Why coastal regions have less extreme temperatures
compared to desert regions.
Heat Capacity
molar heat capacity (cp): heat capacity per mole of a
substance (in J/mol·°C)
heat capacity: amount of heat necessary to raise the
temperature of a given amount of any substance by 1°C; in
units of J/°C
Use the following equations to solve calorimetry problems:
q = n cp ∆T
or
q = cs m ∆T
where ∆T=change in temperature, n=# of moles, and m=mass.
Examples
a. If 279.9 J is required to raise the temperature from 23.0°C to
99.5°C for a 15.5-g sample of silver, what is the specific heat of
silver?
b. A beaker with 100.0 g of water is heated from 25.0°C to its
boiling point. If the specific heat of water is 4.184 J/g·°C, how
much heat is required to heat the water?
Examples
c. Determine the final temperature for a 100.0 g sample of iron
at 25.0°C heated with the amount of heat calculated in a. given
the molar heat capacity of iron is 25.1 J/mol·°C.
Units of Energy
calorie (cal): unit of energy used most often in the US. Is
equal to the amount of energy required to raise the
temperature of 1 g of water by 1˚C
1 cal  4.184 J
(Note: This is EXACT!)
But a nutritional calorie (abbreviated Cal) is 1000 cal:
1 Cal = 1 kcal = 4.184 kJ
Example
When consuming an ice-cold drink, one must raise the
temperature of the beverage to 37.0°C (normal body
temperature). Can one lose weight by drinking ice-cold
beverages if the body uses up about 1 calorie per gram of
water per degree Celsius (i.e. the specific heat of water = 1.00
cal/g·°C) to consume the drink?
a. Calculate the energy expended (in Cal) to consume a 12-oz
beer (about 355 mL) if the beer is initially at 4.0°C. Assume the
drink is mostly water and its density is 1.01 g/mL.
Example
b. If the label indicates 103 Cal, what is the net calorie gain or
loss when a person consumes this beer? Is this a viable weight
loss alternative?
c. Calculate the amount of heat (in kJ) required to heat 1.00 kg
(~1 L) of water at 25°C to its boiling point.
Units of Energy
Another energy unit is the British thermal unit (abbreviated
Btu). A Btu is the energy required to raise the temperature of 1
pound of water by 1°F when water is most dense (at 39°C).
The heating power of many gas cooktops is often given in
Btu’s. Calculate the time (in minutes) required to heat 1.00 kg
of water at 25°C to boiling using a 12,000 Btu per hour burner.
(Assume complete energy transfer from the burner to the
water.) Use 1 kWh = 3412 Btu and 1 kJ = 1 kWs.