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TRANSCRIPT
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Table of Contents
· The Nature of EnergyClick on the topic to go to that section· State Functions**
· Enthalpy
· Enthalpies of Formation
· Energy in Foods and Fuels
· Hess's Law
· Measuring Enthalpy Changes: Calorimetry
· Energy Associated with Changes of State
· Enthalpies of Reaction
Slide 5 / 118
Thermochemistry Presenat
We know chemical and physical processes release and absorb energy. We use these thermochemical principles to design air
conditioners and refrigerators as well as foot warmers that allow us to stay comfortable as we "go big" on the hill!
Slide 6 / 118
Potential Energy is the energy that objects have energy due to their position.
Gravitational Potential Energy
GPE = mgh
Elastic Potential Energy
EPE = 1/2 kx2
Electric Potential Energy
UE = kQ1Q2
r2
A Review of Energy from Physics
Slide 7 / 118
Kinetic Energy is the energy that an object has by virtue of its motion:
KE = 1/2 mv2
Work is defined by the formula
W = Fdparallel
A Review of Energy from Physics
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Algebraically, these two statements combine to become:
E0 + W = Ef
Since Ef - Eo = ∆E, this can also be written as
∆E = W
A Review of Energy from Physics
work
The total energy of an isolated system is
constant.
An outside force can change the energy of a system by
doing work on it.
Slide 9 / 118
Another unit of energy is the calorie (cal).
1 cal = 4.184 J
The energy of food is measured in Calories (C).[note the capital "C"]
1 Calorie = 1000 calories = 4184 Joules
Units of Energy
The SI unit of energy is the Joule (J).
Slide 10 / 118
1 A reaction produces 3.8 cal of energy. How many joules of energy is produced?
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2 A reaction uses 235 J of energy. How many calories have been burned?
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3 A 20 ounce coke contains 240 Calories. How many kilojoules of energy are present in a 20 ounce Coke?
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From last year, we know that ∆E = W.
This year, we extend that by adding another way to change the energy of a system; by the flow of Heat (q).
When two objects of different temperature are in contact, heat flow results in an increase of the energy of the cooler object and an identical decrease of the energy of the hotter object.
∆E = w + q
*Note, we use a lower case "w" in chemistry.
Energy & Heat
A B
heat flow
T = 20℃ T = 10℃
Slide 14 / 118
The First Law of Thermodynamics∆E = w + q
Energy is neither created nor destroyed.
In other words, the total energy of the universe is a constant; if the system loses energy, it must be gained by the surroundings,
and vice versa.
Initialstate
Finalstate
E of system decreases
Inte
rnal
ene
rgy,
E
E < E0
∆E < 0 (-)
E
E0
Energy lost tosurroundings
Initialstate
Finalstate
E of system increases
Inte
rnal
ene
rgy,
EEnergy gained fromsurroundings
E > E0
∆E > 0 (+)
E
E0
Slide 15 / 118
System and Surroundings
The system includes the reactants and products (here, the hydrogen, oxygen and water molecules).
The surroundings are everything else (here, the cylinder and piston).
Surroundings
system
When considering energy changes, we need to focus on a well-defined, limited part of the universe. The portion we focus on is
called the system and everything else is called the surroundings.
Consider the following reaction occurring within a metal cylinder.
2H2(g) + O2(g) --> 2H2O(g)
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Changes in Internal Energy
If ∆E > 0, Efinal > Einitial
The system absorbed energy from the surroundings.
If ∆E < 0, Efinal < Einitial
The system released energy from the surroundings.
H2 (g) + O2 (g)
Inte
rnal
ene
rgy,
E ∆E < 0(negative)
∆E >0(positive)
H2 O(l)
Slide 17 / 118
4 Ten grams of table salt in dissolved in water in a 250 mL beaker. Which of the following is a component of the system?
A NaCl
B water
C Na+
D beaker
E A and B
F A, B, and C
G A, B, and D
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5 When a strong acid is added to a flask containing water the flask becomes warm to the touch. This is because...
A the reaction performed work on the flask
B the system absorbed heat from the surroundings
C the system released heat to the surroundings
D the surroundings released heat to the system
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6 When a strong acid is added to a flask containing water the flask becomes warm to the touch. Which correctly describes the change in energy?
A ∆Esys is positive and ∆Esur is negative
B ∆Esys is positive and ∆Esur is positive
C ∆Esys is negative and ∆Esur is positive
D ∆Esys is negative and ∆Esur is negative
Slide 20 / 118
Changes in Internal Energy
System
∆E>0
Heat q > 0
Work w > 0
Surroundings
When energy is exchanged between the system and the surroundings, it is either exchanged as either heat (q) or work (w).
∆E = q + w
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q, w, ∆E, and Their Signs
q + system gains heat - system loses heat
w + work done on system - work done by system
∆E + net gain of energy by system - net loss of energy by system
Sign Conventions for q, w and ∆E
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7 The ∆E of a system that gains 50 kJ of heat and performs 24 kJ of work on the surroundings is ________ kJ.
A -74
B -26
C 0
D +26
E +74
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8 The ∆E of a system that releases 120 J of heat and does 40 J of work on the surroundings is ________ J.
A -80
B -160
C 0
D +80
E +160
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9 The ∆E of a system that absorbs 120 J of heat and does 120 J of work on the surroundings is ________ J.
A -240
B -120
C 0
D +120
E +240
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10 The ∆E of a system that absorbs 12,000 J of heat and the surrounding does 12,000 J of work on the system is _______ J.
A -24000
B -12000
C 0
D +12000
E +24000
Slide 26 / 118
Exchange of Heat between System and Surroundings
Recall, when heat is absorbed by the system from the surroundings, the process is endothermic .
System
Surroundings
Heat -qSystem
Surroundings
Heat+q When heat is released by the system into the surroundings, the
process is exothermic.
Slide 27 / 118
11 The reaction that occurs inside the foot warmer packet is endothermic?
True
False
Slide 28 / 118
12 What will happen when a hot rock is put into cold water?
A the water and rock will both gain energy
B the water and rock will both lose energy
C the rock will gain energy and the water will lose energy
D the rock will lose energy and the water will gain energy
Slide 29 / 118
13 If you put a hot rock in cold water, and your system is the rock, the process is _______.
A exothermic
B endothermic
C neither, there is no net change of energy
D it depends on the exact temperatures
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14 If you put a hot rock in cold water, and your system is the water, the process is _______.
A exothermicB endothermicC neither, there is no net change of energyD it depends on the exact temperatures
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15 If you put an ice cube in water, and your system is the ice, the process is _______.
A exothermicB endothermicC neither, there is no net change of energyD it depends on the exact
temperatures
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16 If you put an ice cube in water, and your system is the water, the process is _______.
A exothermicB endothermicC neither, there is no net change of energyD it depends on the exact
temperatures
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17 When NaOH dissolves in water, the temperature of solution increases. This reaction is________.
A exothermic B endothermic
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18 When CaCl 2 dissolves in water the temperature of water drops. This reaction is _____.
A endothermicB exothermic
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19 Water droplets evaporating from the skin surface will make you feel cold. This process is _____.
A exothermic for waterB endothermic for skin
C exothermic for skinD endothermic for water E A and CF C and D
Slide 37 / 118
State Functions The internal energy of a system is independent of the path by which the system achieved that state.
Below, the water could have reached room temperature from either direction...it makes no difference to the final energy of the system if it reached its final temperature by heating up or cooling down.
50g100C
50g25C
50g0C
Cooling Heating
##
#
Slide 38 / 118
State Functions
Internal energy is a state function.
It depends only on the present state of the system, not on the path by which the system arrived at that state.
D∆E depends only on Einitial and Efinal
##
Slide 39 / 118
These multiple paths explains how engines, air conditioners, batteries, heaters, etc. operate.
State Functions
They move between energy states while performing some task.
##
Slide 40 / 118
State Functions Understanding thermodynamics enables us to harness
energy flow for useful purposes.
##
Charged battery
Heat
Work
Energy lost bybattery
Discharged battery
#EHeat
A B
For instance, whether the battery is shorted out or is discharged by running the fan, its #E is the same.
But q and w are different in the two cases. If the battery shorts out, all of the energy is lost as heat, whereas if it is used to run the fan, some energy is used to do work. Q and w are NOT state functions.
A B
Slide 41 / 118
Work Done by a Gas When a gas expands it does work on its surroundings:
W = Fd.
In this case: F = PA
and d = ∆h.
W = Fd
W = (PA)∆h
W = P∆V
A = cross sectional area
P= F/A P= F/A
∆h∆V
h
Initial state Final state
##
Slide 42 / 118
Since the gas expands, and does work on its surroundings, the energyof the gas decreases, this is considered negative work.
W = - P∆V
Work Done By a Gas
Initial state Final state
A = cross sectional area
P= F/A P= F/A
∆h
h
∆V
##
(-) work is done by system
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If the surroundings compress the gas, decreasing its volume, this increases the energy of the gas, it is considered positive work.
W = + P∆V
Work Done By a Gas
P= F/AP= F/A
h
∆V
Initial state Final state
A = cross sectional area
∆h
##
(+) work is done on system
Slide 44 / 118
We can measure the work done by the gas if the reaction is done in a vessel fitted with a piston.
Work Done by a Gas ##
HCl Solution HCl Solution + Zinc
Slide 46 / 118
EnthalpyThe word "enthalpy" is derived from the Greek noun "enthalpos" which means heating.
The enthalpy of a system (H) is a combination of the internal energy of a system (E) plus the work that needs to be done to create the space for the substance to occupy.
It is impossible to measure enthalpy, H, directly. Only the change (∆H) can be measured.
H = E + W
∆H = ∆E + W
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At constant pressure, the change in enthalpy is the heat gained or lost by the system.
Enthalpy
Note: This is only true at constant pressure. See ** for more information (W = P∆V)
∆H = (q+W) - W
∆H = q
Slide 48 / 118
Enthalpy
The enthalpy of a system (H) is a combination of the internal energy of a system (E) plus the work that needs to be done to create the space for the substance (PV) to occupy.
It is impossible to measure enthalpy, H, directly. Only the change (∆H) can be measured.
H = E + PV
∆H = ∆E + PV
**
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If a process takes place at constant pressure (as most processes we study do), we can account for heat flow during the process by measuring the enthalpy of the system.
At constant pressure, the change in enthalpy is the heat gained or lost by the system.
Enthalpy
∆H = ∆E + ∆(PV)∆H = ∆E + P∆V∆H = (q+w) - w∆H = qp (at constant pressure)
**
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Enthalpy is an extensive property; its value depends on the quantity of the substance present.
Combustion of 16 grams of CH4 ∆H = -891 kJ Combustion of 32 grams of CH4 ∆H = -1782 kJ
∆H for a reaction in the forward direction is equal in size, but opposite in sign, to ∆H for the reverse reaction.
CH4(g) + 2O2(g) --> CO2(g) + 2H2O(g) ∆H = -891 kJ 2H2O(g) + CO2(g) --> CH4(g) + 2O2(g) ∆H = +891 kJ
∆H for a reaction depends on the state of the products and the state of the reactants.
CH4(g) + 2O2(g) --> CO2(g) + 2H2O(g) ∆H = -891 kJ CH4(g) + 2O2(g) --> CO2(g) + 2H2O(l) ∆H = -979 kJ
Enthalpy
Slide 51 / 118
20 When 114 grams of gasoline combust, the ∆H is equal to -5,330 kJ. What is the ∆H for combustion of 57 grams of gasoline?
Slide 52 / 118
Endothermic and Exothermic Processes
A process is endothermic when ∆H is positive.
∆H>0
System
Endothermic
Surroundings
Heat+q
∆H<0
System
Exothermic
Surroundings
Heat -q
A process is exothermic when ∆H is negative.
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21 The reaction A + B --> C is endothermic. The DH for this reaction is +50 J. What is the DH for the reaction C --> A + B?
A Cannot be determined.B 0.02 JC - 50 JD 100 J
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22 The reaction A + B --> C is exothermic. The ∆H for this reaction is -150 J. What is the ∆H for the reaction C --> A + B?
A +150B zeroC -150D this reaction will not happen
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23 Dissolving NaOH in water will increase the temperature of the solution. This reaction is _____
A exothermic
B endothermic
C adiabatic
D isothermal
Slide 56 / 118
24 NH3NO3 + H2O --> NH4+ + NO3
- + OH- ∆H = +25.69 kJ/mol. This reaction is exothermic.
Yes
No
Slide 58 / 118
Since we cannot know the exact enthalpy of the reactants and products, we measure ∆H through calorimetry, the measurement of heat flow by making use of the expression:
∆H = qp
Measuring Enthalpy Changes: Calorimetry
The subscript p on q means the process is occurring at constant pressure. No work is being done only heat is being exchanged. You will not always see the subscript but it is implied when we are speaking of Enthalpy.
Slide 59 / 118
Heat Capacity and Specific HeatThe amount of energy required to raise the temperature of a substance by 1 K (1°C) is its heat capacity.
The amount of energy required to raise the temperature of one gram of a substance by 1 K (1°C) is its specific heat (c).
Slide 60 / 118
25 Heat capacity is an example of an
A Intensive property
B Extensive property
Slide 61 / 118
26 Specific heat is an example of an
A Intensive property
B Extensive property
Slide 62 / 118
Heat Capacity and Specific Heat
Specific heat =heat transferred
mass x temperature change
m∆Tc = q
Slide 63 / 118
27 The specific heat of marble is 0.858 J/g-K. How much heat (in J) is required to raise the temperature of 20g of marble from 22 °C to 45 °C?
q = mc∆T
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28 An 26 g sample of wood (c = 1.674 J/(g-K)) absorbs 200 J of heat, upon which the temperature of the sample increases from 20.0 °C to ______°C.
q = mc∆T
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29 A sample of silver (c = 0.236 J/g-K) absorbs 800 J of heat, upon which the temperature of the sample increases from 50.0 °C to 80°C. What is the mass of the sample?
q = mc∆T
Slide 66 / 118
sparknotes.com
thermometer
styrofoam cup
stirrer
insulated cover
Constant Pressure Calorimetry
Many chemical reactions happen in aqueous solutions. The apparatus to the left is a calorimeter.
How could you use it to measure the heat change for a chemical reaction in an aqueous solution?
Slide 67 / 118
Because the specific heat for water is well known (4.184 J/g-K), we can measure ∆H for the reaction with this equation:
D∆H = q = mc∆T
(at constant pressure)
Constant Pressure Calorimetry
sparknotes.com
thermometer
styrofoam cup
stirrer
insulated cover
Slide 68 / 118
Constant Pressure CalorimetryExample: A student wishes to determine the enthalpy change when ammonium chloride (NH4Cl) dissolves in water. The student masses out 20.00 grams of ammonium chloride and adds it to 500 grams of water in a styrofoam cup at a temperature of 16.1 Celsius. The student observes the temperature to decrease to 13.2 Celsius. What is the enthalpy change for the dissolution of ammonium chloride?
1. Find enthalpy change of solution using q = mc∆T
= 520 g x -2.9 C x 4.2 J = -6,334 J
2. Since heat released by surroundings (solution) is equal to the heat gained by system (ammonium chloride).
∆H for dissolving of NH4Cl = +6334 J
g℃
Slide 69 / 118
Bomb Calorimetry
Reactions can be carried out in a sealed “bomb” such as this one.
The heat absorbed (or released) by the water is a very good approximation of the enthalpy change for the reaction.
ThermometerIgnition wires
Oxygen atmosphere
Sample
Water
Slide 70 / 118
Because the volume in the bomb calorimeter is constant, what is measured is really the change in internal energy, ∆E, not ∆H.
For most reactions, the difference is very small.
Bomb Calorimetry (Constant Volume)
ThermometerIgnition wires
Oxygen atmosphere
Sample
Water
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30 The reaction below takes place in a bomb calorimeter. If the student found that the temperature of the water was 12.2 Celsius to start and the heat capacity of the calorimeter was 34 J/C, what must be the enthalpy change of the reaction if the temperature of the water increased to 15.6 Celsius?
4Fe(s) + 3O2(g) --> 2Fe2O3(s)
Slide 72 / 118
31 A mysterious meteorite is discovered in your backyard. To determine its identity, you determine its specific heat. The 164.6 gram sample of metal is heated to 90 C and then dumped in 300 grams of water in a styrofoam cup at an initial temperature of 10 C. After the metal is added, the temperature rises to 11.3 C. Identify the metal.
A
B
C
Metal Specific Heat (J/gC)
Cu 0.386
Au 0.126
Al 0.900
Slide 74 / 118
Energy Changes Associated with Changes of State
Chemical and physical changes are usually accompanied by changes in energy. Recall the following terms:
When energy is put into the system, the process is called _____________.
When energy is released by the system, the process is called _____________.
Slide 75 / 118
Endothermic processes Exothermic processes
Energy is taken into the system from the surroundings (∆H > 0)
Energy is released from the system to surroundings (∆H < 0)
Energy Changes Associated with Changes of State
Ans
wer
melting a solidboiling or evaporating a liquid
sublimation of a solid
freezing a liquidcondensing a gasdeposition of a gas
Fill in the blanks
Slide 76 / 118
Solid
Gas
Liquid
Vaporization Condensation
Melting
Sublimation
Freezing
Deposition
Ene
rgy
of s
yste
mPhase Changes
Slide 77 / 118
32 Which of the following is/are exothermic? I. boiling II. melting III. freezing
A I only
B I and II only
C III only
D I, II and III
Slide 78 / 118
Energy Changes Associated with Changes of State
The heat of fusion (∆Hfus ) is the energy required to change a solid at its melting point to a liquid.
The heat of vaporization (∆Hvap ) is defined as the energy required to change a liquid at its boiling point to a gas.
Heat of fusion(H2O) = 6.0 kJ/mol
Heat of vaporization(H2O) = 41.0 kJ/molClass Question: Why is the heat of vaporization much
higher than the heat of fusion for a substance?
In order to change phase from a solid to liquid, the particle attractions need only be strained somewhat. When a material changes from a liquid to a gas, the particle
attractions must be essentially broken. move for answer
Slide 79 / 118
10
40
80Heat of vaporizationHeat of fusion
24
5
29 41
58
2367
Energy Changes Associated with Changes of State
Note that these quantities are usually per 1.00 mole, whereas q = mc∆T involves mass in grams.
But
ane
Die
thyl
et
her
Wat
er
Mer
curyH
eat o
f fus
ion
and
vap
oriz
atio
n k
J/m
ol
Slide 80 / 118
33 The heat of vaporization for butane is 24 kJ/mol. How much energy is required to vaporize 2 mol of butane?
A 2kJ
B 12 kJ
C 22 kJ
D 48 kJ
Slide 81 / 118
34 The heat of vaporization for butane is 24 kJ/mol. How much energy is required to vaporize 0.33 mol of butane?
A 8kJ
B 12 kJ
C 16 kJ
D 72 kJ
Slide 82 / 118
35 How much energy is required to melt 0.5 mol of water?
A 2kJB 3 kJC 6 kJD 12 kJ
Heat of fusion
for H2O (s)
Heat of vaporization
for H2O (l)
6 kJ/mol 41 kJ/mol
Ans
wer
Slide 83 / 118
36How much energy is released when 3.0 mol of water freezes?
A 2kJB 3 kJC 18 kJD 123 kJ
Heat of fusion for H2O (s)
Heat of vaporization
for H2O (l)
6 kJ/mol 41 kJ/mol
Ans
wer
Slide 84 / 118
37 How much energy is needed to melt 10.0 mol solid Hg?
A 2.3 kJB 5.8 kJC 23 kJD 230 kJ
Heat of fusion for Hg (s)
Heat of vaporization
for Hg (l)
23 kJ/mol 58 kJ/mol E 580 kJ
Ans
wer
Slide 85 / 118
38How much energy is needed to vaporize 2.0 mol Hg (l)?
Heat of fusion for Hg (s)
Heat of vaporization
for Hg (l)
23 kJ/mol 58 kJ/mol
Ans
wer
Slide 86 / 118
Energy Changes Associated with Changes of State
This graph is called a heating curve.
It illustrates how temperature changes over time as constant heat is applied to a pure solid substance. Te
mpe
ratu
re (0 C
)
Heat added (each division corresponds to 4 kJ)
Liquid water
Ice and liquid water (melting)
Liquid water and vapor (vaporization)
Water vapor
-25
25
75
125
0
50
100
Ice
A
B C
DE
F
Slide 87 / 118
From A to B, ice is heating up from -25oC to 0oC.
Energy Changes Associated with Changes of State
Since
Q = mc∆T∆T = Q
So the slope is
(mc)
1 (mc)
Tem
pera
ture
(0 C)
Heat added (each division corresponds to 4 kJ)
Liquid water
Ice and liquid water (melting)
Liquid water and vapor (vaporization)
Water vapor
-25
25
75
125
0
50
100
Ice
A
B C
DE
F
Slide 88 / 118
From B to C, ice is melting.
The added heat is breaking the hydrogen bonds of the solid, so the temperature is constant.
That's why the slope is zero.
Energy Changes Associated with Changes of State
Tem
pera
ture
(0 C)
Heat added (each division corresponds to 4 kJ)
Liquid water
Ice and liquid water (melting)
Liquid water and vapor (vaporization)
Water vapor
-25
25
75
125
0
50
100
Ice
A
B C
DE
F
Slide 89 / 118
From C to D, liquid water is heating up from 0 C to 100 C.
Once again, the slope is 1/(mc).
But "c" is different for all the phases of a substance, so the slope is different for solid, liquid and gaseous H2 O.
Energy Changes Associated with Changes of State
Tem
pera
ture
(0 C)
Heat added (each division corresponds to 4 kJ)
Liquid water
Ice and liquid water (melting)
Liquid water and vapor (vaporization)
Water vapor
-25
25
75
125
0
50
100
Ice
A
B C
DE
F
Slide 90 / 118
From D to E, liquid water is boiling into vapor.
The added heat is breaking the IM forces of the liquid, so the temperature is constant.
That's why the slope is zero.
Energy Changes Associated with Changes of State
Tem
pera
ture
(0 C)
Heat added (each division corresponds to 4 kJ)
Liquid water
Ice and liqui d water (melting)
Liquid water and vapor (vaporization)
Water vapor
-25
25
75
125
0
50
100
Ice
A
B C
DE
F
Slide 91 / 118
From E to F water vapor, steam, is heating up from 100 C to 125 C.
Once again, the slope is 1/(mc). But "c" is different for all the phases of a substance, so the slope is different for for solid, liquid and gaseous H2 O.
Energy Changes Associated with Changes of State
Tem
pera
ture
(0 C)
Heat added (each division corresponds to 4 kJ)
Liquid water
Ice and liquid water (melting)
Liquid water and vapor (vaporization)
Water vapor
-25
25
75
125
0
50
100
Ice
A
B C
DE
F
Slide 92 / 118
Recall that slope =
where m = mass and C = specific heat of the substance
Energy Changes Associated with Changes of State
1mC
(q) Energy Added (Joules)
(DT)
Tem
pera
ture
(C)
ABCD
This graph shows heat transfer versus change in temperature for 1 gram of 4 different substances.
Slide 93 / 118
39 Which substance has the lowest specific heat?
A AB BC CD D
(DT)
Tem
pera
ture
(C)
ABCD
(q) Energy Added (Joules)
Slide 94 / 118
40 Which substance requires the highest amount of heat added to raise the temperature?
A A
B BC CD D
(DT)
Tem
pera
ture
(C)
ABCD
(q) Energy Added (Joules)
Slide 95 / 118
41 Which segment(s) contain solid sodium chloride?A AB only B AB and BC
C AB, BC and CDD BC, CD and DE
Slide 96 / 118
42 Which segment(s) contain liquid sodium chloride?
A AB only B AB and BC
C AB, BC and CDD BC, CD and DE
Slide 97 / 118
43 What is the melting point (in oC) of sodium chloride?
Slide 98 / 118
44 What is the freezing point (in oC) of sodium chloride?
Slide 99 / 118
45 Which is greater: the specific heat of solid NaCl, or the specific heat of molten (liquid) NaCl?
A solid B liquid
C Cannot be determined. D They are equal.
Slide 100 / 118
Segments AB, CD, EF
slope = nonzeroT increasingKE increasingPE constant
apply q = mc∆T
Segments BC & DE
slope = 0∆T = 0KE constantPE increasing
apply∆Hfus or∆Hvap
Features of a Heating Curve
Tem
pera
ture
(0 C)
Heat added (each division corresponds to 4 kJ)
Liquid water
Ice and liquid water (melting)
Liquid water and vapor (vaporization)
Water vapor
-25
25
75
125
0
50
100
Ice
A
B C
DE
F
Slide 101 / 118
Recall that any given substance has a different value for specific heat as a solid, as a liquid and as a gas.
Additionally, melting one mole of a substance (∆Hfus) and vaporizing one mole of the same substance (∆Hvap) require different amounts of energy.
Energy Changes Associated with Changes of State
Slide 102 / 118
Specific heat of ice 2.09 J/g-℃
Specific heat of water 4.18 J/g-℃
Specific heat of steam 1.84 J/g-℃
Heat of fusion (∆Hfus) of water at 0℃ 6.01 kJ/mol or 330 J/g
Heat of vaporization (∆Hvap) of water at 100℃ 40.7 kJ/mol or 2600 J/g
Specifics about Water
Slide 103 / 118
Calculating Energy Changes from a Heating Curve
Sample Problem: Calculate the enthalpy change (in kJ) for converting 10.0 g of ice at -15.0°C to water at 25.0 °C.
It is a good idea to sketch out the segments first on a graph.
Slide 104 / 118
Calculating Energy Changes from a Heating Curve
Sample Problem: Calculate the enthalpy change (in kJ) for converting 10.0 g of ice at -15.0°C to water at 25.0 °C.
Segment 1Warming the ice from -15.0°C upto the substance's MP which is 0°C.
-14
-40
10
20
6 12 24Tem
pera
ture
(C)
Time (s)
Slide 105 / 118
Calculating Energy Changes from a Heating Curve
Sample Problem: Calculate the enthalpy change (in kJ) for converting 10.0 g of ice at -15.0°C to water at 25.0 °C.
Segment 2Melting the ice at 0°C.
-14
-40
10
20
6 12 24Tem
pera
ture
(C)
Time (s)
Slide 106 / 118
Calculating Energy Changes from a Heating Curve
Sample Problem: Calculate the enthalpy change (in kJ) for converting 10.0 g of ice at -15.0°C to water at 25.0 °C.
Segment 3Warming the liquid from 0°C to 25°C.
-14
-40
10
20
6 12 24Tem
pera
ture
(C)
Time (s)
Slide 107 / 118
q1 = mc∆Tq1 = (10.0g) (2.09 J/g-oC) (15.0oC)q1 = 313.5 Jq1 = 0.314 kJ or 313.5J
We are now ready to apply either the calorimetry equation or ∆Hfus or ∆Hvap.
Sample Problem: Calculate the enthalpy change (in kJ) for converting 10.0 g of ice at -15.0°C to water at 25.0 °C.
Calculating Energy Changes from a Heating Curve
Segment 1Warming the ice from -15.0°C up to the substance's MP which is 0°C.
Slide 108 / 118
Sample Problem: Calculate the enthalpy change (in kJ) for converting 10.0 g of ice at -15.0°C to water at 25.0 °C.
Calculating Energy Changes from a Heating Curve
q2 = (∆Hfus ) (# moles)q2 = (6.01 kJ/mol) [(10.0 g)/ 18.0 g/mol)]q2 = 3.34 kJ
or
6.01/18 J/g x 10g = 3.34 kJ = 3340J
Segment 2Melting the ice at 0 °C.
Slide 109 / 118
Sample Problem: Calculate the enthalpy change (in kJ) for converting 10.0 g of ice at -15.0°C to water at 25.0 °C.
Calculating Energy Changes from a Heating Curve
q3 = mc∆Tq3 = (10.0g) (4.18 J/g-oC) (25.0oC)q3 = 1045 Jq3 = 1.05 kJ
Segment 3Warming the water from 0 °C to 25.0 °C.
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Sample Problem: Calculate the enthalpy change (in kJ) for converting 10.0 g of ice at -15.0°C to water at 25.0 °C.
Calculating Energy Changes from a Heating Curve
Now, we add the heat changes for all 3 segments. First, make sure that all quantities are in kilojoules.
Total ∆H = (q1 + q2 + q3) kJ ∆H = (0.314 + 3.34 + 1.05) kJ ∆H = 4.6 kJ or 4698.5J
Slide 111 / 118
46 Calculate the enthalpy change in J of converting 75 g of ice at -11 ℃ to liquid water at 22℃.
A 98,654
B 33,371 J
C 8,621
D 26,474
E 35,096 J
Specific heat of ice 2.09 J /g-℃
Specific heat of water 4.18 J /g-℃Specific heat of steam 1.84 J /g-℃
Heat of fusion (∆Hfus) of water at 0℃
6.01 kJ /mol or 330 J /g
Heat of vaporization (∆Hvap) of water at 100℃
40.7 kJ /mol or 2600 J /g
Ans
wer
Slide 112 / 118
47 Calculate the enthalpy change of converting 25 g of water vapor at 110 ℃ to liquid water at 50℃.
A -69,765
B 69,765
C -59,315
D -70,685 J
E -13,935
Specific heat of ice 2.09 J /g-℃
Specific heat of water 4.18 J /g-℃Specific heat of steam 1.84 J /g-℃
Heat of fusion (∆Hfus) of water at 0℃
6.01 kJ /mol or 330 J /g
Heat of vaporization (∆Hvap) of water at 100℃
40.7 kJ /mol or 2600 J /g
Ans
wer
Slide 113 / 118
48 Calculate the enthalpy change in kJ of converting 31.8 g of ice at 0℃ to water vapor at 140℃.
A 109 kJ
B 96 kJ
C 104 kJ
D - 88 kJ
E -109 kJ
Specific heat of ice 2.09 J /g-℃
Specific heat of water 4.18 J /g-℃Specific heat of steam 1.84 J /g-℃
Heat of fusion (∆Hfus) of water at 0℃
6.01 kJ /mol or 330 J /g
Heat of vaporization (∆Hvap) of water at 100℃
40.7 kJ /mol or 2600 J /g
Ans
wer
Slide 115 / 118
Enthalpy of Reaction
The change in enthalpy, ∆H, is the enthalpy of the products minus the enthalpy of the reactants:
D∆H = Hproducts − Hreactants
CH4(g) + 2 O2(g)
CO2(g) + 2H2O(l)
∆H1 =-890 kJ
Ent
halp
y ∆H2 =890 kJ
Slide 116 / 118
This quantity, ∆H, is called the enthalpy of reaction or the heat of reaction. The combustion of hydrogen in the balloon below is an exothermic reaction and the energy released is equal to the heat of reaction.
Enthalpy of Reaction
2H2(g) + O2(g)
2H2O(g)
∆H<0exothermic
Ent
halp
y
Slide 117 / 118
How much energy is released when 4.0 mol of HBr is formed in this reaction?
H2 (g) + Br2 (g) --> 2HBr(g) ∆H = -72 kJ
Enthalpy of ReactionExample Problem #1
X 4.0 mol HBr(g) -72kJ 2.0 mol HBr(g)
X = -72kJ (4/2)
X = -144kJ
144kJ of energy are released
=
Slide 118 / 118
How much energy is released when 4.0 mol of HBr is formed in this reaction?
H2 (g) + Br2 (g) --> 2HBr(g) ∆H = -72 kJ
Enthalpy of ReactionExample Problem #1
4.0 mol HBr x -72 kJ
2 mol HBr = -144kJ or 144 kJ released