Download Chapter 15 – Energy and Chemical Change

1/6/2015
Chapter 15 – Energy and
Chemical Change
Section 1:
Section 2:
Section 3:
Section 4:
Section 5:
Chemical reactions usually absorb or release energy.
Energy
Heat
Thermochemical Equations
Calculating Enthalpy Change
Reaction Spontaneity
Essential Questions
• Section 1: Energy can change form and flow, but it is always
conserved
• Section 2: The enthalpy change for a reaction is the
enthalpy of the products minus the enthalpy of the reactants.
• Section 3: Thermochemical equations express the amount
of heat released or absorbed by chemical reactions.
• Section 4: The enthalpy change for a reaction can be
calculated using Hess’s law.
• Section 5: Changes in enthalpy and entropy determine
whether a process is spontaneous.
• What is energy?
• How do potential and kinetic energy differ?
• How can chemical potential energy be related to the heat lost or
gained in chemical reactions?
• How is a calorimeter used to measure energy that is absorbed or
released?
• What do enthalpy and enthalpy change mean in terms of chemical
reactions and processes?
• How are thermochemical equations for chemical reactions and
processes written?
• How is energy lost or gained during changes of state?
• How is the heat that is absorbed or released in a chemical reaction
calculated?
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Essential Questions continued
• How is Hess’s law applied to calculate the enthalpy change in a
reaction?
• What is the basis for the table of standard enthalpies of formation?
• What is the enthalpy change for a reaction using standard
enthalpies of formation data?
• What is the difference between spontaneous and nonspontaneous
processes?
• How do changes in entropy and free energy determine the
spontaneity of chemical reactions and other processes?
Vocabulary
New continued
•
•
•
•
•
thermochemical equation
enthalpy (heat) of combustion
molar enthalpy (heat) of
vaporization
molar enthalpy(heat) of fusion
Hess’s law
Vocabulary
Review
•
•
•
•
•
temperature
pressure
combustion reaction
allotrope
vaporization
New
• energy
• law of conservation of
energy
• chemical potential
energy
• heat
New continued
•
•
•
•
•
•
•
•
•
•
calorie
joule
specific heat
calorimeter
thermochemistry
system
surroundings
universe
enthalpy
enthalpy (heat) of reaction
Section 1: Energy
New continued
•
•
•
•
•
standard enthalpy (heat) of
formation
spontaneous process
entropy
second law of thermodynamics
free energy
Energy can change form and flow, but it is always conserved
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Essential Questions
Vocabulary
• What is energy?
• How do potential and kinetic energy differ?
• How can chemical potential energy be related to the heat lost
or gained in chemical reactions?
Review
New
•
•
•
•
•
•
•
•
temperature
energy
law of conservation of energy
chemical potential energy
heat
calorie
joule
specific heat
The Nature of Energy
Energy is the abiltiy to do work or
produce heat.
• Energy exists in two basic forms:
potential energy and kinetic
energy.
• Potential energy is energy due to
composition or position.
• Kinetic energy is energy of
motion.
The Nature of Energy
The law of conservation of energy states that in any chemical reaction
or physical process, energy can be converted from one form to another,
but it is neither created nor destroyed. Also known as the first law of
thermodynamics.
Chemical potential energy is energy stored in a substance because of
its composition.
•
Chemical potential energy is important in chemical reactions.
Heat is energy that is in the process of flowing from a warmer object to a
cooler object.
•
q is used to symbolize heat.
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Convert Calories to calories
•
A calorie is defined as the amount of energy required to raise the
temperature of one gram of water one degree Celsius.
Use with Example Problem 1.
Problem
Relationships among Energy Units
• The energy content of
food is measured in
Calories, or 1000
calories. Also known as
a kilocalorie.
• A joule is the SI unit of
heat and energy,
equivalent to 0.2390
calories.
SOLVE FOR THE UNKNOWN
CONVERT ENERGY UNITS
Measuring Heat
Relationship
Conversion Factor
1 J = 0.2390 cal
1 J/0.2390 cal OR
0.2390 cal/1 J
1 cal = 4.184 J
1 cal/4.184 J OR
4.184 J/1 cal
1 Cal = 1 kcal
1 Cal/1000 cal OR
1000 cal/1 Cal
1 Cal = 4.184 kJ
1 Cal/4.184 kJ OR
4.184 kJ/1 Cal
Apply the relationship 1 Calorie = 1000 cal
1000
230
= 2.310 1
A breakfast of cereal, orange juice, and
milk might contain 230 Calories. Convert
this energy to joules.
Convert calories to joules
Response
EVALUATE THE ANSWER
ANALYZE THE PROBLEM
The minimum number of significant figures in
the conversion is two, and the answer correctly
has two digits. A value on the order of 105 or 106
is expected because the given number of
kilocalories is of the order of 102 and it must be
multiplied by 103 to convert it to calories. Then,
the calories must be multiplies by a factor of 4.
•
You are given an amount of energy in nutritional
Calories. You must convert Calories to calories,
and then convert calories to joules.
KNOWN
Amount of enegy =
230 Calories
Apply the relationship 1 calorie = 4.184 joules
4.184
2.310 = . 1
Therefore, the answer is reasonable.
UNKNOWN
Amount of energy =
Specific Heat
The specific heat of any substance is
the amount of heat required to raise one
gram of that substance one degree
Celcius.
• Some objects required more heat
than others to raise their
temperature.
Specific Heats at 298 K (25⁰C)
Substance
c in J/gK
Aluminum
0.897
Bismuth
0.123
Copper
0.386
Brass
0.380
Gold
0.126
Lead
0.128
Silver
0.233
Tungsten
0.134
Zinc
0.387
Mercury
0.140
Alcohol(ethyl)
2.44
Water
4.186
Ice (-10 C)
2.05
Granite
0.803
Concrete
0.840
?J
Specific Heat
Calculating heat absorbed and released
Equation for calculating heat
q=m x c x ΔT
q represents the heat absorbed or released
m represents the mass of the sample in grams
c represents the specific heat of the substance
ΔT
T is the change in temperature in Celsius or Kelvin
The quantity of heat absorbed or released by a substance is equal to the product of
its specific heat, the mass of the substance, and the change in its temperature.
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KNOWN
CALCULATE SPECIFIC HEAT
Energy released = 114 J
Ti = 50.4 ⁰C
Mass of iron = 10.0 g Fe
Tf = 25.0 ⁰C
Use with Example Problem 2.
UNKNOWN
Problem
Specific heat of iron,
In the construction of bridges and skyscrapers,
gaps must be left between adjoining steel
beams to allow for the expansion and
contraction of the metal due to heating and
cooling. The temperature of a sample of iron
with a mass of 10.0 g changed from 50.4 ⁰C to
25.0 ⁰C with the release of 114 J. What is the
specific heat of iron?
Response
Calculate ΔT.
ΔT = 50.4 − 25.0 = 25.4 •
•
ANALYZE THE PROBLEM
You are give the mass of the sample, the initial
and final temperatures, and the quantity of heat
released. You can calculate the specific heat of
iron by rearranging the equation that relates
these variables to solve for c.
Essential Questions
c = ? J/g⁰C
SOLVE FOR THE UNKNOWN
•
•
Review
State the equation for calculating heat
= Δ
solve for c
=
Δ
Substitute q=114 J, m=10.0 g and ∆T = 25.4⁰C
114
=
10.0"25.4 solve for c
= . ## $ %
• What is energy?
• How do potential and kinetic energy differ?
• How can chemical potential energy be related to the heat lost
or gained in chemical reactions?
Vocabulary
•energy
•law of conservation of
energy
•chemical potential
energy
•heat
•calorie
•joule
•specific heat
Section 2: Heat
Essential Questions
The enthalpy change for a reaction is the enthalpy of the products minus
the enthalpy of the reactants.
• How is a calorimeter used to measure energy that is absorbed
or released?
• What do enthalpy and enthalpy change mean in terms of
chemical reactions and processes?
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Calorimetry
Vocabulary
Review
New
•
•
•
•
•
•
•
•
pressure
calorimeter
thermochemistry
system
surroundings
universe
enthalpy
enthalpy (heat) of reaction
A calorimeter is an insulated device used for measuring the amount of heat
absorbed or released in a chemical reaction or physical process.
KNOWN
CALCULATE SPECIFIC HEAT
Energy absorbed = 256 J
∆T = 182 ⁰C
Chemical Energy and the Universe
Mass of metal = 4.68 g
Use with Example Problem 3.
UNKNOWN
Problem
Specific heat
A piece of metal with a mass of 4.68 g absorbs
256 J of heat when its temperature increases by
182 ⁰C. What is the specific heat of the
metal?
ANALYZE THE PROBLEM
You are give the mass of the sample, the
change in temperature, and the quantity of heat
absorbed. You can calculate the specific heat of
the metal by rearranging the equation that
relates these variables to solve for c.
c = ? J/g⁰C
SOLVE FOR THE UNKNOWN
•
•
•
Response
Thermochemistry is the study of heat changes that accompany chemical
reactions and phase changes.
State the equation for calculating heat
= Δ
solve for c
=
Δ
Substitute q=256 J, m=4.68 g and ∆T = 182⁰C
256
=
= . ' 4.68"182 $ %
The system is the specific part of the universe that contains the reaction or
process you wish to study.
•
•
The surroundings are everything else other than the system in the universe
The universe is defined as the system plus the surroundings
EVALUATE THE ANSWER
The three quantities used in the calculation
have three significant figures, and the answer is
correctly stated with three digits. The
calculations are correct and yield the expected
unit.
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Chemical Energy and the Universe
Chemical Energy and the Universe
Chemists are interested in changes in energy during reactions.
Enthalpy is the heat content of a system at constant pressure.
• Enthalpy changes for exothermic reactions are always negative.
• Enthalpy changes for endothermic reactions are always positive.
Enthalpy (heat) of reaction is the change in enthalpy during a reaction.
Symbolized as ∆Hrxn.
∆Hrxn = Hfinal – Hinitial
∆Hrxn = Hproducts – Hreactants
Chemical Energy and the Universe
Review
Essential Questions
•
•
How is a calorimeter used to measure energy that is absorbed or
released?
What do enthalpy and enthalpy change mean in terms of chemical
reactions and processes?
Vocabulary
•calorimeter
•thermochemistry
•system
•surroundings
•universe
•enthalpy
•enthalpy (heat) of
reaction
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Section 3: Thermochemical Equations
Thermochemical equations express the amount of heat released or
absorbed by chemical reactions.
• How are thermochemical equations for chemical reactions
and processes written?
• How is energy lost or gained during changes of state?
• How is the heat that is absorbed or released in a chemical
reaction calculated?
Writing Thermochemical Equations
Vocabulary
Review
New
•
•
•
combustion reaction
Essential Questions
•
•
thermochemical equation
enthalpy (heat) of
combustion
molar enthalpy (heat) of
vaporization
molar enthalpy(heat) of
fusion
A thermochemical equation is a balanced chemical equation that includes the
physical states of all reactants and products, and energy change
The enthalpy (heat) of combustion of a substance is the enthalpy change for
the complete burning of one mole of the substance.
Standard Enthalpies of Combustion
Substance
Sucrose (table sugar)
Octane (in gasoline)
Formula
⁰ΔHcomb (kJ/mol)
C12H22O11 (s)
-5644
C8H18 (l)
-5471
C6H12O6 (s)
-2808
Propane
C3H8 (g)
-2219
Methane (found in natural gas)
CH4 (g)
-891
Glucose (simple sugar)
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Changes of State
Changes of State
Molar enthalpy (heat) of vaporization refers to the heat required to vaporize
(or boil) one mole of a liquid substance
Molar enthalpy (heat) of fusion refers to the heat required to melt one mole of
a solid substance
Standard Enthalpies of Vaporization and Fusion
Substance
Formula
⁰ΔHvap (kJ/mol)
⁰ΔHfus (kJ/mol)
Water
H2O
40.7
6.01
Ethanol
C2H5OH
38.6
4.94
Methanol
CH3OH
35.2
3.22
Acetic Acid
CH3COOH
23.4
11.7
Ammonia
NH3
23.3
5.66
CALCULATING ENTHALPY
Response
Use with Example Problem 4.
Problem
A bomb calorimeter is useful for
measuring the energy released in
combustion reactions. The reaction is
carried out in a constant-volume bomb
with a high pressure of oxygen. How
much heat is released with 54.0 g of
glucose (C6H12O6) is burned according
to this equation?
C6H12O6 + 6O2 6CO2 + 6H2O
∆Hcomb = -2808 kJ
KNOWN
UNKNOWN
Mass of glucose =
54.0 g C6H12O6
q = ? kJ
∆Hcomb = -2808 kJ
CALCULATING ENTHALPY
ANALYZE THE PROBLEM
You are given a mass of glucose, the
equation for the combustion of glucose, and
∆Hcomb . You must convert grams of glucose
to moles of glucose. Because the molar mass
of glucose is more than three times the mass
of the glucose burned, you can predict that
the energy released will be less than one
third of ∆Hcomb .
EVALUATE THE ANSWER
SOLVE FOR THE UNKNOWN
Multiply moles of C6H12O6 by the enthalpy of
combustion, ∆Hcomb.
•
Multiply moles of glucose by
0.300)"(
)*+
,-,./
.
01234356278
28089
= :#;<
1)"(
)*+
All values in the calculation have three
significant figures, so the answer is
correctly stated with three digits. As
predicted, the released energy is less
than one third of ∆Hcomb.
SOLVE FOR THE UNKNOWN
Convert grams of C6H12O6 to moles of C6H12O6
•
Divide by the molar mass.
1)"(
)*+
54.0""(
)*+
180.18""(
)*+
= 0.300)"(
)*+
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Combustion Reactions
Review
Combustion is the reaction of a fuel with oxygen.
Food is the fuel in combustion reactions in biological systems.
Essential Questions
•
•
•
How are thermochemical equations for chemical reactions and
processes written?
How is energy lost or gained during changes of state?
How is the heat that is absorbed or released in a chemical reaction
calculated?
Vocabulary
•thermochemical
equation
•enthalpy (heat) of
combustion
Section 4: Calculating Enthalpy Change
The enthalpy change for a reaction can be calculated using
Hess’s law.
•molar enthalpy (heat) •molar enthalpy(heat) of
fusion
of vaporization
Essential Questions
• How is Hess’s law applied to calculate the enthalpy
change in a reaction?
• What is the basis for the table of standard enthalpies of
formation?
• What is the enthalpy change for a reaction using standard
enthalpies of formation data?
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Hess’s Law
Vocabulary
Review
New
•
•
•
allotrope
Hess’s law
standard enthalpy (heat) of
formation
KNOWN
CALCULATE SPECIFIC HEAT
a
2H2 + O2 2 H2O
∆H = -572 kJ
b
H2 + O2 H2O2
∆H = -188 kJ
Use with Example Problem 3.
UNKNOWN
Problem
∆H = ? kJ
Use thermochemical equations a and b below to
determine ∆H for the decomposition of hydrogen
peroxide (H2O2), a compound that has many
uses from bleaching hair to powering rocket
engines.
2H2O2 2H2O + O2
a
2H2 + O2 2 H2O
∆H = -572 kJ
b
H2 + O2 H2O2
∆H = -188 kJ
Response
ANALYZE THE PROBLEM
SOLVE FOR THE UNKNOWN
H2O2 is a reactant
•
Reverse equation b and change the sign of ∆H
H2O2 H2 + O2
∆H = + 188 kJ
Two moles of H2O2 are needed
•
Hess’s law states that if you can add two or more thermochemical
equations to produce a final equation for a reaction, then the sum of the
enthalpy changes for the individual reactions is the enthalpy change for the
final reaction.
CALCULATE SPECIFIC HEAT
Add equations a and new b, canceling any terms
that appear on opposite sides
• Write equation a
2H2 + O2 2 H2O
∆H = −572 kJ
•
Write new equation b
2 H2O2 2H2 + 2O2
•
Add the equations and cancel.
2 H2O2 2H2 + 2O2
∆H = + 376 kJ
2H2 + O2 2 H2O
∆H = −572 kJ
•
Add the ∆H.
2H2O2 2H2O + O2
Multiply Reversed equation b and ∆H by 2
2 H2O2 2H2 + 2O2
∆H = + 376 kJ
EVALUATE THE ANSWER
The two equations produce the desired
equation. All values are accurate to the ones
place, so ∆H is correctly stated.
∆H = + 376 kJ
∆H = −196 kJ
You have been given two chemical equations
and their enthalpy changes. These two
equations contain all the substances found in
the desired equation.
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Standard Enthalpy (Heat) of Formation
The standard enthalpy (heat)
of formation is defined as the
change in enthalpy that
accompanies the formation of
one mole of the compound in its
standard state from its elements
in their standard states.
• Elements in their standard
states have a ∆Hf⁰ of 0.0 kJ.
• The formation of compounds
are placed above or below
elements in their standard
states.
Standard Enthalpy (Heat) of Formation
Standard enthalpies of formation can be used to calculate the enthalpies for
many reactions under standard conditions by using Hess’s law.
• The summation equation: ∆Hrxn⁰ = Σ∆Hf⁰(products) - Σ∆Hf⁰(reactants)
Select Standard Enthalpies of Formation
Substance
Formation Equation
⁰ΔHf⁰ (kJ/mol)
H2S (g)
H2(g) + S(s) H2S(g)
-21
-273
HF(g)
0.5H2(g) + 0.5F2(g) HF(g)
SO3(g)
S(s) + 1.5O2 SO3(g)
-396
SF6 (g)
S(s) + 3F2 SF6 (g)
-1220
ENTHALPY CHANGES
Use with Example Problem 6.
Problem
Use Standard Enthalpies of Formation to
calculate ∆Hrxn⁰ for the combustion of
methane.
CH4(g) + 2 O2(g) CO2(g) + 2 H2O(l)
Response
ANALYZE THE PROBLEM
You are given an equation and asked to
calculate the change in enthalpy. The formula
∆Hrxn⁰ = Σ∆Hf⁰(products) - Σ∆Hf⁰(reactants)
can be used with data from table R-11, which
is on the next slide.
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ENTHALPY CHANGES
KNOWN
UNKNOWN
∆Hf⁰ CO2 = -394 kJ
∆Hrxn⁰ = ? kJ
Use with Example Problem 6.
∆Hf⁰ H2O(l) = -286 kJ
Problem
∆Hf⁰ CH4 = -75 kJ
Use Standard Enthalpies of Formation to
calculate ∆Hrxn⁰ for the combustion of
methane.
SOLVE FOR THE UNKNOWN
Use the formula ∆Hrxn⁰ = Σ∆Hf⁰(products) Σ∆Hf⁰(reactants). Expand the formula to include a
term
for each reactant and product. Multiply each
ANALYZE THE PROBLEM
term
by the coefficient of the substance in the
You are given an equation and asked to
calculate the change in enthalpy. The formula balanced chemical equation.
Response
∆Hrxn⁰ = Σ∆Hf⁰(products) - Σ∆Hf⁰(reactants)
can be used with data from table R-11, which
is on the next slide.
SOLVE FOR THE UNKNOWN
Substitute ∆Hf⁰ CO2 = -394 kJ, ∆Hf⁰ H2O(l) = -286
kJ, ∆Hf⁰ CH4 = -75 kJ, and ∆Hf⁰ O2 = 0.0 kJ into the
equation
∆Hf⁰ O2 = 0.0 kJ
CH4(g) + 2 O2(g) CO2(g) + 2 H2O(l)
ENTHALPY CHANGES
∆Hrxn⁰ = (-394 kJ + (2 x -286 kJ)) – (-75 kJ + (2 x 0.0 kJ))
EVALUATE THE ANSWER
All values are accurate to the ones place.
Therefore, the answer is correct as
stated. The calculated value is the same
as the Enthalpy of Combustion value
given on the next slide.
∆Hrxn⁰ = (-966 kJ) – (-75 kJ) = -966 kJ + 75 kJ = -891 kJ
The combustion of 1 mole CH4 releases 891 kJ.
Substitute CO2 and H2O for the products, CH4 and
O2 for the reactants. Multiply H2O and O2 by 2.
∆Hrxn⁰ = (∆Hf⁰ CO2 + (2 x ∆Hf⁰ H2O)) - (∆Hf⁰ CH4 + (2 x ∆Hf⁰ O2))
ENTHALPY CHANGES
Review
Essential Questions
Standard Enthalpies of Combustion
Substance
Formula
⁰ΔHcomb (kJ/mol)
•
•
•
How is Hess’s law applied to calculate the enthalpy change in a reaction?
What is the basis for the table of standard enthalpies of formation?
What is the enthalpy change for a reaction using standard enthalpies of
formation data?
Sucrose (table sugar)
C12H22O11 (s)
-5644
Octane (in gasoline)
C8H18 (l)
-5471
Vocabulary
C6H12O6 (s)
-2808
Propane
C3H8 (g)
-2219
•Hess’s law
•standard enthalpy
(heat) of formation
Methane (found in natural gas)
CH4 (g)
-891
Glucose (simple sugar)
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Section 5: Reaction Spontaneity
Essential Questions
Changes in enthalpy and entropy determine whether a
process is spontaneous.
• What is the difference between spontaneous and
nonspontaneous processes?
• How do changes in entropy and free energy determine the
spontaneity of chemical reactions and other processes?
Vocabulary
Spontaneous Process
Review
New
•
•
•
•
vaporization
•
spontaneous process
entropy
second law of
thermodynamics
free energy
A spontaneous process is a physical or chemical change that once begun,
occurs with no outside intervention.
• Many spontaneous processes requires some energy from the
surroundings to start the process.
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Spontaneous Process
Spontaneous Process
Entropy is a measure of the possible ways that the energy of a system
can be distributed, and this is related to the freedom of the system’s
particles to move and the number of ways they can be arranged.
The second law of thermodynamics states that spontaneous processes
always proceed in such a way that the entropy of the universe increases.
• Entropy is sometimes considered a measure of disorder or randomness
of the particles in a system
• The more spread out the particles are, the more disorder
• Entropy changes associated with changes in state can be predicted.
• Entropy increases as a substance changes from a solid to a liquid to a
gas.
• Dissolving a gas in a solvent always results in a decrease in entropy.
Spontaneous Process
Entropy, the Universe, and Free Energy
Assuming no change in physical state occurs, the entropy of a system
usually increases when the number of gaseous product particles is greater
than the number of gaseous reactant particles.
• With some exceptions, entropy increases when a solid or liquid is
dissolved in a solvent
• The random motion of particles of a substance increases as its
temperature increases.
In nature, the change in entropy tends to be positive when:
• The reaction or process is exothermic, which raises the temperature of
the surroundings
• The entropy of the system increases.
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Entropy, the Universe, and Free Energy
Entropy, the Universe, and Free Energy
Free energy is energy that is available to do work.
• If the sign of the free energy change, ∆G⁰, is positive, the reaction is non
spontaneous.
• If the sign of the free energy change is negative, the reaction is spontaneous.
Equation for calculating heat
∆Gsystem = ∆Hsystem - T∆Ssystem
∆Gsystem represents the free energy change
∆Hsystem represents the change in enthalpy
∆T is the temperature in Kelvin
∆Ssystem represents the change in entropy
ΔH
ΔS
Positive
Negative
Spontaneous?
NEVER
Negative
Positive
ALWAYS
Negative
Negative
Only at low temperatures
Postive
Positive
Only at high temperatures
The free energy released or absorbed in a chemical reaction is equal to the difference
between the enthalpy change and the product of the change in entropy and the
temperature.
DETERMINE REACTION
SPONTANEITY
SOLVE FOR THE UNKNOWN
•
Convert ∆Ssystem to kJ/K
19
322 = 0.3229/>
> 1000
Use with Example Problem 7.
Solve the free energy equation
Problem
For a process, ∆Hsystem = 145 kJ and ∆Ssystem =
322 J/K. Is the process spontaneous at 382 K?
Response
ANALYZE THE PROBLEM
•
Essential Questions
•
State the Gibbs Free Energy Equation
∆Gsystem = ∆Hsystem - T∆Ssystem
•
Substitute in T = 382 K, ∆Hsystem = 145 kJ, and
∆Ssystem =0.322 kJ/K
ΔGsystem = 145 kJ – (382 K x 0.322 kJ/K)
•
Multiply and subtract
You must calculate ∆Gsystem to determine
spontaneity.
Review
KNOWN
UNKNOWN
∆Gsystem = 145 kJ – 123 kJ
T = 382 K
Sign of ∆Gsystem = ?
∆Gsystem = 22 kJ
•
What is the difference between spontaneous and nonspontaneous
processes?
How do changes in entropy and free energy determine the spontaneity of
chemical reactions and other processes?
Vocabulary
•spontaneous process
•entropy
•second law of thermodynamics
•free energy
∆Hsystem = 145 kJ
∆Ssystem = 322 J/K
EVALUATE THE ANSWER
Because ∆Gsystem is positive, the reaction is
nonspontaneous at that temperature
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