Introduction to Thermochemistry That’s Me! Okay…not really …Or is it?

Introduction to Thermochemistry
That’s Me!
Okay…not really …Or is it?
Why study thermochemistry?
• We depend on energy for nearly every aspect
of our lives
• Think about how you use energy every day…
Electricity
Cooking
Automobiles
Heating/Cooling
Batteries
• Cars, cell phones, iPods, watches
Human Body
• Temperature regulation
• Conversion of food into usable cellular energy
Seriously, why are we studying this?
• Energy crisis in the world!
• Over-use/dependence on fossil fuels (petroleum,
gasoline, diesel, methane, propane)
• Research/Development of alternative energy
resources
•
•
•
•
•
Nuclear
Fuel Cells (H2, Methanol, Solid oxide)
Hydro
Wind
Solar
NEWS FLASH: This is going to be OUR problem!
Major Sources of Energy
over the last 150 years
An interesting quote…
“So called “global warming” is just a secret
ploy by wacko tree-huggers to make America
energy independent, clean our air and water,
improve the fuel efficiency of our vehicles,
kick-start 21st century industries, and make
our cities safer and more livable. Don’t let
them get away with it!”
- Chip Giller
Founder of Grist.com, where environmentally
minded people gather online.
Organizing the Universe
Any process can be defined by the following…
• Universe = everything, the big enchilada, the
whole kit and caboodle, all that and a bag of
chips, the
whole
all that we
jazz,
tout,
• System
= the
partshabang,
of the universe
are
studying
todos,
alles,
everything-ay,
Chuck Norris,
• i.e. a sarra,
particular
chemical
reaction, etc.
• etc.
Surroundings = everything else in the universe that
is NOT being studied
• i.e. the environment, the lab, the students, your mom, etc.
Universe
System
Surroundings
Internal Energy
• Total Energy = Kinetic Energy + Potential Energy
due to motion
stored energy
• Endothermic process: energy enters the system
(from the surroundings)
System
Energy
Surroundings
• Exothermic process: energy leaves the system
(to the surroundings)
System
Energy
Surroundings
Kinetic Energy in Atoms & Molecules
is proportional to

• Temperature  average KE of particles in a sample
• Components of Kinetic Energy:
Translation
Rotation
• KEtot = all components added together
Vibration
Potential Energy (PE)
• Chemical Change: intramolecular potential energy
•
Breaking and forming chemical bonds (PEnucleus-electron)
Example: 2H2 + O2  2H2O
H
H
H
O
H
O
H O
O
H
H
H
• Physical Change: intermolecular potential energy
•
Disrupting or forming intermolecular attractions (PEIMF)
Example: H2O(ℓ)  H2O(g)
H
H
O
H
H
H
H
H
O
O
H
O
O
H
H
O
H
H
The Law of Conservation of Energy
is also known as
The First Law of Thermodynamics
• Energy cannot be created or destroyed
ΔEuniverse = 0
(reminder: Δ = “change in”)
ΔEuniverse = ΔEsystem + ΔEsurroundings = 0
ΔEsystem = - ΔEsurroundings
Energy can be transferred as: heat (q)
work (w)
ΔEsys = q + w
The First Law of Thermodynamics
Endothermic process:
Universe
System
Energy
Surroundings
Exothermic process:
Universe
System
Energy
Surroundings
In either case, the energy of the universe remains constant
Conservation of Energy
in a Chemical Reaction
In this example, the energy
Endothermic
of the reactants
Reaction
and products increases,
while the energy of the surroundings decreases.
Reactant + Energy
Product
In every case, however, the total energy does not change.
Surroundings
Energy
Surroundings
System
System
Myers, Oldham, Tocci, Chemistry, 2004, page 41
Before
reaction
After
reaction
Conservation of Energy
in a Chemical Reaction
In this example, the energy
Exothermic
of the reactants
Reaction
and products decreases,
while the energy of the surroundings increases.
Reactant
Product + Energy
In every case, however, the total energy does not change.
Energy
Surroundings
Myers, Oldham, Tocci, Chemistry, 2004, page 41
System
Before
reaction
Surroundings
System
After
reaction
Heat (q)
• Heat: the transfer of energy between objects due to a
temperature difference
•
Flows from higher-temperature object to lower-temperature
object
System
(T1)
System
(T1)
Heat
Heat
Surroundings
(T2)
Surroundings
(T2)
If T1 > T2
q system = q surroundings = +
exothermic
If T1 < T2
q system = +
q surroundings = endothermic
Pressure/Volume Work (PV work)
work = force  distance
force = pressure  area = P  m2
distance = m
work = P  m2  m = P  V
Assuming external pressure (Pext) is
constant, the volume can change (ΔV)
wsystem = -PextΔV (in L atm)
Where ΔV = Vf – Vi (in L)
Sign Conventions
In chemistry, always take the system’s
perspective (physics takes the surr’s view)
• Heat (q)
q>0
q<0
heat is added to the sys by surr (q is +)
heat is leaving sys to surr (q is -)
• Work (w)
w>0
w<0
work done on sys by surr (work added to
sys by surr, so w is +)
work done by sys on surr (work is
leaving surr, so w is -)
Sign Conventions: Heat and Work
Surroundings
-w
+w
Work
Work
Heat
System
Heat
+ q (endo)
- q (exo)
Practice with Work and Heat
An ideal gas expands from 5.0 L to 12.0 L
against a constant external pressure of 2.0 atm.
How much work is done?
w = -PextΔV
Pext = 2.0 atm
Pext = 2.0 atm
w = -(Pext)(Vfinal – Vinitial)
w = -(2.0 atm)(12.0 L – 5.0 L)
w = -14 L atm
1 L atm = 101.325 J
w = -14 L atm  (101.325 J/1 L atm)
w = -1400 J
If ΔE = 0 for this process, what is q?
ΔE = q + w = q + (-1.42  103 J) = 0
q = 1400 J
V1 = 5.0 L
V2 = 12.0 L
Practice with Work and Heat
Calculate the change in energy for a system
that loses 15 kJ of heat and expands from a
volume of 10. L to 200. L under a constant
Remember:
external pressure of 2.0 atm.
ΔE = q + w
101.325 J = 1 L atm
w = -Pext ΔV
1000 J = 1 kJ
w = -(2.0 atm)(200L -10L) = -380 L atm
w = -380 L atm (101.325 J/1 L atm) (1 kJ/1000 J)
= -38.5035 kJ
ΔE = -15 kJ + -38.5035 kJ
= -53.5035 kJ -54 kJ
Calorimetry (Calorie Measuring)
• Calorimetry: the science of measuring heat
• C = heat capacity
• the heat required to raise temp by 1°C (or 1 K)
• Units → Joules/oC (J/oC or J/K)
Because the temperature increase
depends on the amount of “stuff”…
• Specific heat capacity (Cp)
• heat capacity per gram = J/goC or J/gK
•
Molar heat capacity (Cm)
• heat capacity per mole = J/moloC or J/mol K
Various Heat Capacities
Substance
Specific
heat capacity
(J/K g)
Molar mass
(g/mol)
Molar
heat capacity
(J/K mol)
Gold
0.129
197.0
25.4
Silver
0.235
107.9
25.4
Copper
0.385
63.55
24.5
Iron
0.449
55.85
25.1
Aluminum
0.897
26.98
24.2
H2O(l)
4.184
18.02
75.3
H2O(s)
2.03
18.02
36.6
H2O(g)
1.998
18.02
36.0
Using heat capacities…
q = m  C  ΔT
q (J) = mass (g)  C (J/goC)  ΔT (oC)
q = joules (J)
q (J) = moles (mol)  C (J/K mol)  ΔT (K)
q = joules (J)
Mnemonic device: q = m “CAT”
Heating Curve of water
Gas
Temperature (oC)
140
120
l↔g
100
80
Boiling Point
60
40
20
0
-20
Liquid
s↔l
Melting Point
-40
-60
-80
Solid
-100
Energy Added
Heating Curves
• Temperature Change within phase
• change in KE (molecular motion)
• depends on heat capacity of phase
C H2O (l) = 4.184 J/goC (requires the most heat)
C H2O (s) = 2.03 J/goC
C H2O (g) = 1.998 J/goC (requires the least heat)
• Phase Changes (s ↔ l ↔ g)
• change in PE (molecular arrangement)
• temperature remains constant
• overcoming intermolecular forces
Calculating Energy Changes
Temperature (oC)
140
120
q= ?
q= ?
100
80
q = mass x Cgas x DT
60
40
20
0
q = mass x Cliquid x DT
-20
-40
-60
-80
q = mass x Csolid x DT
-100
Energy Added
Same Formulas
Different C Values
Phase Transitions
• During phase transitions, added heat is
used to overcome intermolecular forces
rather than to increase temperature
ΔHfusion = energy needed to convert solid to liquid
• For water, ΔHfusion = 6.02 kJ/mol
• For liquid to solid, ΔH = - 6.02 kJ/mol
ΔHvaporization = energy needed to convert liquid to gas
• For water, ΔHvap = 40.7 kJ/mol
• For gas to liquid, ΔH = - 40.7 kJ/mol
Depending on the process occurring, all q’s
are calculated individually and then added
together
Note: steps 2 and 4 → kJ
Unit
Conversions!
steps 1, 3, 5 → J
4. q = mol x DHvap
Temperature (oC)
140
120
2. q = mol x DHfus
100
80
5. q = mass x Cgas x DT
60
40
20
0
3. q = mass x Cliquid x DT
-20
-40
-60
-80
1. q = mass x Csolid x DT
-100
Energy Added
Heating Curve of Water
From Ice to Steam in Five Easy Steps
4
5
q1: Heat the ice to 0°C
q = m Cice ΔTice
3
2
q2: Melt the ice into a liquid at 0°C
q = mol ΔHfus
q3: Heat the water from 0°C to 100°C
q = m Cwater ΔTwater
1
Heat
q4: Boil the liquid into a gas at 100°C
Heat q = mol ΔHvap
q5: Heat the gas above 100°C
q = m Csteam ΔTsteam
Heating Curve Practice
How much energy (J) is required to heat 12.5
g of ice at –10 oC to water at 0.0 oC?
4
5
3
2
1
q1: Heat the ice from -10 to 0°C
q = 12.5 g (2.03 J/g oC)(0.0 - -10.0 oC) = 253.75 J
q2: Melt the ice at 0°C to liquid at 0 oC
q = 12.5 g ice 1 mol
6.02 kJ
= 4.177 kJ
18.016 g 1 mol
qtot = q1 + q2 = 253.75 J + 4,177 J =
4,430 J
Heating Curve Practice
How much energy (J) is required to heat 25.0
g of ice at –25.0 oC to water at 95.0 oC?
4
3
2
1
5
Notice that your q values are
positive because heat is added…
q1: Heat the ice from -25 to 0°C
q = 25.0 g (2.03 J/g oC)(0.0 - -25.0 oC) = 1268.75 J
q2: Melt the ice at 0°C to liquid at 0 oC
q = 25.0 g ice 1 mol
6.02 kJ
= 8.352 kJ
18.016 g 1 mol
q3: Heat the water from 0°C to 95 °C
q = 25.0 g (4.184 J/g oC)(95.0 – 0.0oC) = 9937 J
qtot = q1 + q2 + q3 = 1268.75 J + 8,352 J + 9937 J = 19,560 J
Heating Curve Practice
How much energy (J) is removed to cool 50.0 g of steam at
115.0 oC to ice at -5.0 oC?
4
3
2
1
5
Notice that your q values are
negative because heat is removed…
q5: Cool the steam from 115.0 to 100°C
q = 50.0 g (1.998 J/g oC)(100.0 - 115.0 oC) = -1498.5 J
q4: Condense the steam into liquid at 100°C
q = 50.0 g H2O
- 40.7 kJ
1 mol
= -112.96 kJ
18.016 g
1 mol
q3: Cool the water from 100°C to 0 °C
q = 50.0 g (4.184 J/g oC)(0.0 – 100.0oC) = -20920 J
q2: Freeze the water into ice at 0 °C
q = 50.0 g H2O 1 mol
- 6.02 kJ
= -16.71 kJ
18.016 g 1 mol
q1: Cool the ice from 0°C to – 5.0 °C
q = 50.0 g (2.03 J/g oC)(- 5.0 – 0.0oC) = -507.5 J
qtot = q1 + q2 + q3+ q4 + q5 = -1498.5 J + -112,960 J + -20920 J + -16,710 J + -507.5 J =
-153,000 J
Heating Curve Challenge Problems
1. A sample of ice at
is
placed into 75 g of water
initally at 85oC. If the final
temperature of the mixture
is 15oC, what was the mass
of the ice?
52.8 g ice
Temperature (oC)
-25oC
140
120
100
80
60
40
20
0
-20
-40
-60
-80
-100
DH = mol x DHvap
DH = mol x DHfus
Heat = mass x Dt x Cp, gas
Heat = mass x Dt x Cp, liquid
Heat = mass x Dt x Cp, solid
Time
2. A 38 g sample of ice at -5oC is placed into 250 g of water
at 65oC. Find the final temperature of the mixture
assuming that the ice sample completely melts.
45.6 oC
3. A 35 g sample of steam at 116oC are bubbled into 300 g
water at 10oC. Find the final temperature of the system,
assuming that the steam condenses into liquid water.
76.6 oC
What will happen over time?
Zumdahl, Zumdahl, DeCoste, World of Chemistry 2002, page 291
Let’s take a closer look…
Zumdahl, Zumdahl, DeCoste, World of Chemistry 2002, page 291
Eventually, the temperatures will equalize
Zumdahl, Zumdahl, DeCoste, World of Chemistry 2002, page 291
Calorimetry
• Allows us to measure the flow of heat
• Think back to the ΔHfus expt..
q system = - q surroundings
• In other words, whatever energy is lost by
one is gained by the other (direct E transfer)
• Major Assumption: No E is lost to other
parts of the surroundings
Typical Apparatus
A Coffee Cup
Calorimeter
Used for calorimetry under
constant pressure
Thermometer
Styrofoam
cover
Styrofoam
cups
i.e. normal lab conditions
Stirrer
Zumdahl, Zumdahl, DeCoste, World of Chemistry 2002, page 302
Constant-Pressure Calorimetry
Example:
5.8 g of NaCl dissolves in 50.0 mL H2O
NaCl(s)  Na+(aq) + Cl-(aq)
Ti = 22.2 °C
Tf = 20.5 °C
What is qrxn?
Coffee cup calorimeter
In this case, of the
surroundings…
We can’t directly
measure q sys
qrxn = qsystem = -qsurroundings
qsurr = m  C  ΔT
qsurr = (55.8 g)(4.184 J/°C g)(20.5 °C – 22.2°C)
qsurr = -397 J
qsystem = -qsurr = 397 J
qrxn?
Bomb Calorimeter
• Constant V calorimetry (as opposed to P)
• “Bomb” has own heat capacity (kJ/0C)
• Multiply heat capacity by the ΔT to find the
total heat transferred
• Factor heat into # of
g or mol burned
(kJ/g or kJ/mol)
Bomb Calorimeter Example
A 1.800 g sample of sugar, C12H22O11, was burned in a
bomb calorimeter whose total heat capacity is 15.34 kJ/oC.
Once burned, the temperature of the calorimeter plus
contents increased from 21.36oC to 28.78oC. What is the
heat of combustion per gram of sugar? per mole of sugar?
q = Cbomb x ΔT
More sugar = more energy!
q = 15.34 kJ/oC (28.78 oC – 21.36 oC)
q = 113.8 kJ
113.8 kJ released per 1.800 g sugar
113.8 kJ / 1.800 g = 63.23 kJ/g
63.23 kJ 342.3 g
g
mol
= 21640 kJ/mol
Food and Energy
Caloric Values
Food
joules/grams
calories/gram
Protein
17,000
4,000
4
Fat
38,000
9,000
9
Carbohydrates 17,000
4,000
4
1 calorie = 4.184 joules
“Calories”/gram
1000 calories = 1 “Calorie”
"science"
"food"
or… 1 Kcal = 1 “Calorie”
Smoot, Smith, Price, Chemistry A Modern Course, 1990, page 51
Does water have negative calories…?
How many Calories (nutritional) will you burn by
drinking 1.0 L of water, initially at 36.5 oF (standard
refrigeration temperature)? Assume that the body
must expend energy to heat the water to body
temperature at 98.6 oF. 37 oC
1 L = 1000 mL 2.5 oC
5
C  F  32 
1 mL = 1 g
9
1 calorie = 4.184 joules
q  mCDT 1000 calories = 1 “Calorie”
q = 1.0 x 103 g (4.184 J/g oC)(37 oC - 2.5 oC) = 144,348 J
144348 J
1 cal
1 “Cal”
4.184 J 1000 cal
= 35 Cal
Heat Transfer Experiments
Cu
What is the final
temperature, Tf, of
the mixture?
20.0 g
150. mL
250.0 °C
20.0 °C
C = 0.385 J/°C g
- qCu = qwater
q = m x C x ΔT for both cases, although specific values differ
Plug in known information for each side
Density of water = 1 g/mL(150 mL H2O) = 150 g H2O
-mCuCCuΔT = mwaterCwater ΔT
-20 g (0.385 J/goC)(Tf – 250 oC) = 150 g (4.184 J/goC)(Tf – 20oC)
Solve for Tf ...
Tf = 22.8 °C
240. g of water (initially at 20.0oC) are mixed with an unknown
mass of iron initially at 500.0oC (CFe = 0.4495 J/goC). When
thermal equilibrium is reached, the mixture has a temperature of
42.0oC. Find the mass of the iron.
Fe
T = 500oC
mass = ? grams
T = 20oC
mass = 240 g
- LOSE heat = GAIN heat
-q1 = q2
- [ (mass) (CFe ) (DT)] = (mass) (CH2O) (DT)
- [ (X g) (0.4495 J/goC) (42oC - 500oC)] = (240 g) (4.184 J/goC) (42oC - 20oC)]
- [ (X) (0.4495) (-458)] = (240 g) (4.184) (22)
205.9 X = 22091
X = 107 g Fe
A 97.0 g sample of gold at 785oC is dropped into 323 g of water,
which has an initial temperature of 15.0oC. If gold has a specific
heat of 0.129 J/goC, what is the final temperature of the mixture?
Assume that the gold experiences no change in state of matter.
Au
T = 785oC
mass = 97.0 g
T = 15.0 oC
mass = 323 g
- LOSE heat = GAIN heat
- [(C Au) (mass) (DT)] = (C H2O) (mass) (DT)
- [(0.129 J/goC) (97 g) (Tf - 785oC)] = (4.184 J/goC) (323 g) (Tf - 15oC)]
- [(12.5) (Tf - 785oC)] = (1.35) (Tf - 15oC)]
-12.5 Tf + 9.82 x 103 = 1.35 x 103 Tf - 2.02 x 104
3 x 104 = 1.36 x 103 Tf
Tf = 22.1oC
If 59.0 g of water at 13.0 oC are mixed with 87.0 g of
water at 72.0 oC, find the final temperature of the system.
T = 72.0 oC
mass = 87.0 g
T = 13.0 oC
mass = 59.0 g
- LOSE heat = GAIN heat
- [ (mass) (C H2O) (DT)] = (mass) (C H2O) (DT)
- [ (59 g) (4.184 J/goC) (Tf - 13oC)] = (87 g) (4.184 J/goC) (Tf - 72oC)]
- [(246.8) (Tf - 13oC)] = (364.0) (Tf - 72oC)]
-246.8 Tf + 3208 = 364 Tf - 26208
29416 = 610.8 Tf
Tf = 48.2oC