In This Lesson:
Thermochemistry
(Lesson 2 of 3)
Today is Tuesday,
th
December 8 , 2015
Pre-Class:
What are the three
temperature scales
we know?
Please turn in your
homework if you did
it on paper.
Stuff You Need:
Calculator
Today’s Agenda
• Thermochemistry!
– Especially five-step diagrams
– (We’ll start the whole lesson by drawing a picture)
• Where is this in my book?
– P. 505 and following…
By the end of this lesson…
• You should be able to analyze and calculate
energy (heat) flow through reactions and
physical changes within a system.
Video Introduction
• TED: George Zaidan and Charles Morton – All
the Energy in the Universe
Demo Introduction
• Self-Lighting Candle
– Sulfuric acid decomposing sugar into carbon and
oxygen (plus HEAT).
Keep this in mind…
• I want you to think of heat like money.
• If I find $10 on the ground, it means someone
else lost $10.
• The same goes for heat transfer. When
something loses, something gains.
– More to come.
Energy
• What is energy?
– Technically, it’s the capacity to do work.
• It can take many forms:
– Electrical Energy
– Potential Energy
– Mechanical Energy
– Thermal Energy
• We’re going to focus on the last one, which is
also known as heat.
Heat Energy?
• Heat is not the same as temperature.
• Heat is a measure of thermal energy; how
much energy something has.
• Temperature is a measure of the average
kinetic energy of molecules.
– Similar, but not the same.
First things first…
• In your notes, leave a wide-open space
(probably the front of a whole piece of paper)
under your heading for Thermochemistry.
• You’ll be putting equations in this space along
with a large graph/diagram.
• This page will be your best friend.
– Looks like you’ll need to find it on Facebook.
Temperature 
H2O Heating/Cooling Curve
Condensing
E
Vaporizing
Boiling Pt.
100°C
Melting
0°C
D
Freezing
Freezing Pt.
A
B
C
Vaporization
Liquid
Fusion
Solid
Heat 
Gas
Thermochemistry Diagram
• Stuff to note:
– At freezing and boiling points, the temperature
plateaus.
• Thus, when an ice cube is actually in the process of
melting, it’s not getting any warmer. Weird.
– Melting and freezing points are the same.
– Boiling and condensation points are the same.
Second things second…
• There is no such thing as “cold.”
– There, I said it.
• In chemistry, there is only heat and the
absence of heat. Cold does not exist.
– If something feels cold, it simply means it has less
heat (less energy) in its particles.
– So when you think about it, your freezer is just a
low-energy zone.
Third things third…
• Imagine you’ve put a skillet on the stove to
make an omelet. Obviously, it’s hot.
• In a moment of unparalleled brilliance, you
decide to touch it just to make sure the stove’s
on.
• Which way does the heat flow? From what to
what?
Heat
• In chemistry, heat is represented with two
different symbols:
– q (that’s a lowercase Q)
– H (often written as ΔH, meaning “change in heat”)
• So, ΔH is written for heat change, q for just plain ol’ heat.
• So, when you touch a hot skillet, heat is
transferred from the skillet to your finger.
– The skillet lost heat; you gained heat.
• Important: The heat transfer continues until
both reach the same temperature.
Enthalpy, System, and Surroundings
• In the skillet example, let’s consider your hand
“the system.”
– The system is what we focus on.
• That means the skillet is part of the
surroundings.
– The surroundings are everything else.
• Both the system and the surroundings
experienced the change in enthalpy.
– Enthalpy is the amount of heat in a system.
Endothermic and Exothermic
Summary of the following info coming soon…
• In this case, the system (your hand) gained heat.
– When the system gains heat, the reaction is
endothermic.
• The amount of heat gained/lost by the system
must equal that gained/lost by the surroundings
according to the Law of Conservation of Energy.
– Just like the Law of Conservation of Mass/Matter
• Do you see what I did there?
Let’s try another…
Summary of the following info coming soon…
• You fill an ice cube tray with water and put it
into the freezer.
• When the cubes (the system) freeze, they lose
heat.
– This is an exothermic reaction.
• Endothermic:
– When the system gains heat (ΔH = positive)
• Exothermic:
– When the system loses heat (ΔH = negative).
Endothermic or Exothermic?
• A system gives off heat over time.
– Exothermic.
• The surroundings get warmer.
– Exothermic.
• The change in heat is positive.
– Endothermic.
• There is a –ΔH.
– Exothermic.
• The test tube feels warm to the touch.
– Exothermic.
• The energy of the products is greater than the energy
of the reactants.
– Endothermic.
One last thing…
• Reactions have a certain activation energy –
the energy needed to get the reaction going.
– Remember this from biology – enzymes?
– Enzymes lower activation energy to make
reactions more favorable.
• Activation energy is the difference between
the energy of the reactants and the point at
which the reaction begins (see graph).
Activation Energy
• Think of it this way: Suppose you and a friend are
going to hang out tonight.
• Suddenly, someone you don’t know invites you to
a professional shuffleboard tournament. Are you
going to go?
– Probably not.
• But if that same person also coaxes you with
$500, maybe you’d consider it.
• This is like activation energy – the energy needed
to coax some molecules into bonding in other
ways.
High and Low Energy
• High Activation Energy
– Like needing a LOT of money to want to change
your plans (a.k.a. bonds).
• Low Activation Energy
– Like needing little to no money to want to change
your bonds.
– This is sort of like if your favorite musician offered
you a chance to be backstage for free. You’d
probably take the deal instantly.
Endothermic Reactions
• Energy is absorbed during the reaction and
the surroundings get cooler.
Exothermic Reactions
• Energy is released during the reaction and the
surroundings get warmer.
IMPORTANT: ON THE FINAL
• This is an old concept, but do you remember
what kind of “thing” lowers activation energy?
• In biology, it’s called an enzyme.
• In chemistry, it’s called a catalyst.
Demo – Exothermic Reaction
• Glowing Test Tube (a.k.a. instant omelette)
– Heat prompting calcium metal and sulfur flour to
combine (and release more HEAT).
Endothermic and Exothermic
Summary
• In an endothermic reaction:
– Energy (heat) is absorbed.
– Heat enters the system.
– Heat is a reactant (not a product).
• Reactant + E  Product
– Change in energy is positive.
• In an exothermic reaction:
– Energy is emitted.
– Heat exits the system.
– Heat is a product.
• Reactants  Products + E
– Change in energy is negative.
Temperature 
H2O Heating/Cooling Curve
Condensing
E
Vaporizing
Boiling Pt.
100°C
Melting
0°C
D
Freezing
Freezing Pt.
A
B
C
Vaporization
Liquid
Fusion
Solid
Heat 
Gas
Great, so heat moves. So?
• Remember, heat energy is not the same as
temperature.
• Chemists measure heat in either joules (J) or
calories (cal) – NOT in temperature scales.
– Joules are named for the English physicist/brewer
James Joule.
• I know you’ve heard of calories before – here’s
an explanation of what it means…
calories and Calories
• A calorie (small c) is the amount of energy
needed to raise the temperature of 1 g (1 mL)
of water by 1 °C.
• A Calorie (big C) is 1000 calories.
– Calories are also called kilocalories (kcal).
– The amount of energy needed to raise 1 kg by 1 °C.
• So, you literally are burning calories.
http://www.dailyfork.com/Outback%20Bloomin%20Onion.jpg
Aside: High Calorie Foods
• Recommended daily intake of Calories is 2000 for
an adult woman, 2500 for an adult male
(average).
– White Castle: Chocolate Shake [Large]
• 1680 Calories.
– Outback Steakhouse: Bloomin’ Onion
• 2210 Calories.
– Uno Chicago Grill: Classic Deep Dish Pizza
• 2310 Calories
– and 162g Fat, 123g Carbs, 4470mg Sodium.
– Cheesecake Factory: Bistro Shrimp Pasta
• 3120 Calories
Calorimetry – How to Measure Heat
• Chemists can measure the enthalpy change of
a reaction using a calorimeter.
How Does a Calorimeter Work?
• A substance is burned, dissolved, or immersed in
either water or an inner container.
– The “bomb” part of a bomb calorimeter.
• The burning, dissolving, or immersion process heats
or cools a surrounding known quantity of water.
• By measuring the temperature change of the water, a
chemist can figure out how much heat the water
gained/lost.
• The amount of heat the water gained/lost exactly
equals the (opposite) amount of heat the substance
lost/gained.
Joule and calorie Conversions
•
•
•
•
IMPORTANT:
1 cal = 4.184 J
1 kJ (kilojoule) = 1000 J
1 kcal (kilocalorie) = 1000 cal = 1 Cal
– Only food chemists use Calories.
– Chemists in general use kilocalories to avoid
confusion.
Joule and calorie Conversions
• How many calories are in 100.0 J?
• 1 cal = 4.184 J
– So: 100.0 J
1
cal
= 23.90 cal
4.184 J
• How many joules are in 522 calories?
• 1 cal = 4.184 J
– So:
522 cal
4.184 J
=
1 cal
2180 J
Heat Capacity
• Okay, let’s imagine something else. You still
have that same skillet, only now you also have
a smaller one.
– Same material, just smaller.
• You decide to heat them to the same
temperature. Which one takes longer to get
there?
– Of course, the biggin’. It has a higher capacity for
heat. A higher heat capacity.
Heat Capacity
• Heat capacity is the amount of heat needed to
raise a substance’s temperature by one degree
(Celsius).
– Water’s heat capacity is 1 calorie or 4.184 J,
remember?
• As you can predict, though, different size
objects and/or different materials can affect
heat capacity, which makes it difficult to use.
Specific Heat
• Since objects all have their own heat
capacities and size will affect them, a better
measurement is specific heat.
• Specific heat is the amount of heat needed to
raise 1 gram of a substance by one degree.
– Since it eliminates the “bigger objects take longer
to heat” concept, we can now compare different
substances of any size.
– Water’s specific heat is still 4.184 J, by the way.

Specific Heat
• Specific heat is given by the symbol Cp
(specific heat at a constant pressure).
– Sometimes it’s shown as just C.
• Specific heat’s units are usually:
J
g oC
cal
g oC
J
gK
cal
gK


Heat Capacity
• Objects with low heat capacities gain AND lose heat
faster than those with high heat capacities, by the
way.
– Water has a relatively high heat capacity.
– That’s why it takes so darn long to boil water.
• Except in Denver.
• Another way to think about it:
– At the beach, the ocean and sand are exposed to the same
source of heat – the sun. However, which feels hotter on
your feet?
– The sand, because it has a lower heat capacity than the
water. Its temperature rises (and falls) more easily than
the water.
Specific Heat Chart
Substance
Specific Heat (J/g·K)
Water (liquid)
4.184
Ethanol (liquid)
2.44
Water (solid)
2.06
Water (steam)
2.02
Aluminum (solid)
0.897
Carbon (graphite - solid)
0.709
Iron (solid)
0.449
Copper (solid)
0.385
Mercury (liquid)
0.140
Lead (solid)
0.129
Gold (solid)
0.129
Finally…
• After all those slides of introduction, we get to
the Two Big Equations of Thermochemistry.
• Naturally, there are more than these two later
on, but for now these are the two you’re going
to want to come to know.
Temperature 
H2O Heating/Cooling Curve
Condensing
E
Vaporizing
Boiling Pt.
100°C
Melting
0°C
D
Freezing
Freezing Pt.
A
B
C
Vaporization
Liquid
Fusion
Solid
For A, C, E:
ΔH = mCΔT
Heat 
Gas
Big Equation No. 1
• On your five-step thermochemistry diagram, look at
Steps A, C, and E.
– At these steps, temperature changes but phase doesn’t.
• Here, to calculate change in heat (q or ΔH), use:
• ΔH = m  Cp  ΔT
•
•
•
•
ΔH – change in enthalpy/heat (sometimes listed as q)
m – mass (in grams)
Cp – specific heat
ΔT – change in temperature (Celsius or Kelvin)
• ΔT = (Final Temp – Starting Temp)
• Typically, this equation provides answers in J.
Common Student Error
• The most common student error when using
big equation #1 is using it with numbers from
different substances.
• MAKE SURE ALL VALUES ARE FOR THE SAME
CHEMICAL!
– For example, if you want to find the heat change
for water:
• m is the mass of water.
• C is the specific heat of water.
• ΔT is the change in temperature of the water.
Calculating ΔT
• If an object goes from 20°C to 50°C, what is
ΔT?
• 50°C – 20°C = 30°C
• If an object goes from 152°C to 13°C, what is
ΔT?
• 13°C – 152°C = -139°C
Finding Specific Heat
• What is the specific heat of copper if 95.4
grams absorbs 849 J while rising from 25°C to
48°C?
• ΔH = m  Cp  ΔT
• 849 = 95.4  Cp  (48-25)
• 849 = 2194.2  Cp
• Copper’s specific heat is 0.387 J/g  °C.
Big Equation No. 1
• To practice Big Equation Number One (more
appropriately called change in heat with no
phase change), let’s try numbers 1 and 2 in Part A
on your Energy Calculations worksheets.
– There’s a reference table at the bottom of the page.
• How do I find units?
– Check your specific heat (C) value for a hint.
– Remember, heat is given in J or cal (or related units)!
Energy Calculations W.S. A1
• ΔH = m  C  ΔT
• ΔH = 68  (0.222 J/g°)  (80°C-25°C)
• ΔH = 830.28 J
• Note that the units are joules because the table
lists specific heats in units of J/g°.
68 g
0.222 J
g  °C
55°C
= 830.28 J
Energy Calculations W.S. A2
•
•
•
•
5 L H2O = 5000 mL H2O = 5000 g H2O.
ΔH = m  C  ΔT
ΔH = 5000  4.184 J/g°  (0°C - 100°C)
ΔH = 5000  4.184 J/g°  (-100°C)
– Note that ΔH is negative because the substance
cooled.
– You can make ΔT negative in the equation or
change it later.
• ΔH = -2,092,000 J or -2092 kJ.
Big Equation No. 2
• Big Equation No. 2 comes in for steps B and D.
• The important thing about them is that
temperature doesn’t change, so chemists needed
another way to calculate enthalpy change that
doesn’t partially depend on temperature.
• FYI, heat involved in these phase changes is called
latent heat, whereas change in heat accompanied
without phase changes is sensible heat.
Big Equation No. 2
• To introduce Big Equation No. 2, first you need
to know some statistics used for latent heat:
• Molar Heat of Fusion (ΔHfus) is the energy that
must be absorbed to convert one mole of solid
to liquid at melting point.
• Molar Heat of Solidification (ΔHsolid) is the
energy that must be removed to convert one
mole of liquid to solid at freezing point.
• ΔHfus = -ΔHsolid
Big Equation No. 2
• Molar Heat of Vaporization (ΔHvap) is the
energy that must be absorbed to convert one
mole of liquid to gas at boiling point.
• Molar Heat of Condensation (ΔHcond) is the
energy that must be removed to convert one
mole of gas to liquid at condensation point.
• ΔHvap = -ΔHcond
Temperature 
H2O Heating/Cooling Curve
Condensing
100°C
Freezing
ΔHfus
0°C
ΔHvap
Boiling Pt.
Freezing Pt.
A
C
Melting
B
E
Vaporization
Liquid
Fusion
Solid
For A, C, E:
ΔH = mCΔT
Gas
Vaporizing
D
ΔHsolid
ΔHcond
ΔHfus = -ΔHsolid
ΔHvap = -ΔHcond
For B, D:
ΔH = molΔHphase change
Heat 
Big Equation No. 2
• To calculate q or ΔH during phase changes:
• ΔH = mol  (ΔHphase change)
• ΔH – change in energy
• mol – moles of substance
• ΔHphase change
• Can use ΔHfus or ΔHsolid
• Can use ΔHvap or ΔHcond
• Typically, this equation provides answers in kJ.
Latent Heat Practice Problem
• The molar heat of fusion of water is 6.009 kJ/mol.
How much energy is needed to convert 60.00
grams of ice at 0°C to liquid water at 0°C?
• We need to find moles of water first:
• 60.00 g H2O = 3.331 mol H2O
• Now multiply:
• ΔH = mol  ΔHfus
• 3.331 mol * 6.009 kJ/mol = 20.02 kJ
Big Equation No. 2
• To practice Big Equation Number Two (more
appropriately called change in heat with
phase change), let’s try numbers 1 and 2 in
Part B on your worksheet.
Energy Calculations W.S. B1
• 20 g H2O = 1.11 mol H2O
• ΔH = mol * ΔHphase change
• ΔH = 1.11 mol * (6.02 kJ/mole)
• ΔH = 6.6822 kJ.
• Note that ΔH is calculated here in kJ.
• 1 kJ = 1000 J.
Energy Calculations W.S. B2
• 60 g H2O = 3.33 mol H2O
• ΔH = mol * (ΔHfus or ΔHvap)
• ΔH = 3.33 mol * (-40.7 kJ/mole)
• ΔH = -135.531 kJ.
• Note that ΔHvap is negative here because the problem
states that condensation is happening, not
vaporization.
• If the same amount of water had been vaporized in the
problem, the answer would have been 135.531 kJ.
Formula Summary
• Enthalpy Change (no phase change)
• ΔH = m  C  ΔT
• Enthalpy Change (phase change)
• ΔH = mol  ΔHphase change
• Let’s practice – try #1-2 in “More Problems” at
the bottom of the Energy Calculations worksheet.
Calculating Heat Across Multiple Steps
• When a problem asks you to calculate heat
evolved over multiple steps (phase changes
and temperature changes), use both
equations as you would normally.
– However, you have to go in steps (solid heating,
solid melting, liquid heating, et cetera).
• Then, at the end, make sure all equations are
the same units and add together.
• Let’s try Part C to see how this works.
Temperature 
H2O Heating/Cooling Curve
Condensing
100°C
Freezing
ΔHfus
0°C
ΔHvap
Boiling Pt.
Freezing Pt.
A
C
Melting
B
E
Vaporization
Liquid
Fusion
Solid
For A, C, E:
ΔH = mCΔT
Gas
Vaporizing
D
ΔHsolid
ΔHcond
ΔHfus = -ΔHsolid
ΔHvap = -ΔHcond
For B, D:
ΔH = molΔHphase change
Heat 
Energy Calculations W.S. C
• Part A (heating the solid from -5°C to 0°C)
•
•
ΔH = m  C  ΔT
ΔH = (6 g) (2.06 J/g°C) (5°C) = 61.8 J
• Part B (melting)
•
•
ΔH = mol  ΔHphase change
ΔH = (0.333 mol) (6.02 kJ/mol) = 2.0047 kJ = 2004.7 J
• Part C (heating the liquid from 0°C to 100°C)
•
•
ΔH = m  C  ΔT
ΔH = (6 g) (4.184 J/g°C) (100°C) = 2510.4 J
• Part D (vaporizing)
•
•
ΔH = mol  ΔHphase change
ΔH = (0.333 mol) (40.7 kJ/mol) = 13.5531 kJ = 13553.1 J
• Part E (heating the gas from 100°C to 110°C)
•
•
ΔH = m  C  ΔT
ΔH = (6 g) (2.02 J/g°C) (10°C) = 121.2 J
• Part F (add the total) (1000 J = 1 kJ)
• 61.8 J + 2004.7 J + 2510.4 J + 13553.1 J + 121.2 J = 18251.2 J OR 18.25 kJ
Energy Calculations W.S. #3
• 6.81 kg of steel cools from
2720°C to 625°C. The freezing
point is 1300°C. How much
heat is removed?
• We need to solve this in three
and a half steps:
A. The liquid steel cooling.
B. The liquid steel freezing into a
solid.
C. The solid steel cooling.
D. Adding it all up.
• Also, you need to use steel in
grams:
•
6.81 kg = 6810 g
2720°C
1300°C
625°C
Energy Calculations W.S. #3
• Part A (cooling the liquid from 2720°C to 1300°C)
•
•
ΔH = m  C  ΔT
ΔH = (6810 g) (0.719 J/g°C) (-1420°C) = -6,952,874 J
= -6952.87 kJ
• Part B (freezing)
•
•
ΔH = mol  ΔHphase change
ΔH = (121.93 mol) (-15.4 kJ/mol) = -1877.72 kJ
• Part C (cooling the solid from 1300°C to 625°C)
•
•
ΔH = m  C  ΔT
ΔH = (6810 g) (0.450 J/g°C) (-675°C) = -2,068,537 J =
-2068.54 kJ
• Part D (add the total)
•
-6952.87 kJ + -1877.72 kJ + -2068.54 kJ = -10,899.13
kJ OR -10,899,130 J
More Practice
• Now that we’ve gotten the basics down, let’s
practice on our own.
• Remember, the key is to identify which
variables are “knowns” and which are
“unknowns.”
– Specific Heat Worksheet #1-2
• Now a little conceptual (as a class):
– Specific Heat Worksheet #6 and 8
Calorimetry Problems
• Sometimes you’ll be asked to solve problems
involving calorimetry. How to spot them?
– It’s nearly always an object being placed in water to cool.
• How to solve them? [write this]
– The final temperature of the water is the final
temperature of the object.
– Figure out how much energy the water gained/lost using
ΔHwater = mwaterCwaterΔTwater
– That value, made opposite, becomes the new ΔH value
for the object: ΔHobject = mobjectCobjectΔTobject
– Solve what you need to solve.
• It’s like a see-saw – one end (water) goes up, the
other must have gone down, and vice versa.
Calorimetry Example
• A 20 g metal is heated to 100°C and then is
dropped into 50 g of H2O. The water, in response,
goes from 23°C to 25°C. How much heat did the
metal lose?
• ΔHwater = m  C  ΔT
• ΔHwater = (50 g) (4.184 J/g°C) (2°C)
• ΔHwater = 418.4 J
• The water gained 418.4 J, which means the metal
lost 418.4 J.
• ΔHmetal = -418.4 J
Calorimetry Example
• In the previous example, the water went from
23°C to 25°C when the 100°C metal was put in it.
• If the 20 g metal from the previous example lost
418.4 J, what is the specific heat of the metal?
• Since the water rose to 25°C, that’s the temperature to
which the metal cooled. It started at 100°C and ended
at 25°C.
• ΔHmetal = m  C  ΔT
• -418.4 J = (20 g) (C) (-75°C)
• Cmetal = 0.2789 J/g°C
More Practice
• Now some calorimetry problems (individual):
– Specific Heat Worksheet #7, 9
• Now a little more involved (individual):
– Specific Heat Worksheet #3-5
Calorimetry Boss
• Copper’s specific heat is 0.092 cal/g•°C. If a 50 g Cu
sample is heated to 130°C and placed in 100 g of 24°C
H2O, what is the final temperature?
• We know the final temperature and heat transfer must
be the same, so we can use two mCΔT equations to our
advantage:
•
•
•
•
•
•
•
ΔHwater = -ΔHCu
mwaterCwaterΔTwater = -(mCuCCuΔTCu)
(100 g)(1)(Tfinal – 24°C) = -((50 g)(0.092)(Tfinal – 130°C))
100T – 2400 = -(4.6T – 598)
100T – 2400 = -4.6T + 598
104.6T = 2998
T = 28.66°C
Calorimetry Lab
• And now for your first calorimetry lab!
– Lab – Specific Heat of a Metal
Thermochemical Equations
• Because you’ve now seen that many
equations or interactions involve changes of
heat, chemists need a way to include these
details in your reactions.
• They do this with thermochemical equations.
• A thermochemical equation is a balanced
chemical equation that includes the enthalpy
change.
Thermochemical Equations
• A good example is the combustion of 1 mole
of methane (CH4):
– CH4 (g) + 2O2 (g)  CO2 (g) + 2H2O (l)
•
•
•
•
•
This reaction releases 890 kJ of heat:
CH4 (g) + 2O2 (g)  CO2 (g) + 2H2O (l)
ΔH = -890 kJ
So it’s written like this:
CH4 (g) + 2O2 (g)  CO2 (g) + 2H2O (l) + 890 kJ
Thermochemical Equations
• Remember, in exothermic reactions, heat is
given off (-ΔH) and is thus listed on the
product side.
• In endothermic reactions, heat is absorbed
(+ΔH) and is thus listed on the reactant side.
Thermochemical Equation Practice
• Sodium bicarbonate decomposes when heated
according to the following equation:
• 2NaHCO3 (s) + 129 kJ  Na2CO3 (s) + H2O (g) + CO2 (g)
• How much heat (in kJ) is needed to decompose 2.24
moles of sodium bicarbonate?
• Well, the equation only tells us the heat necessary
for decomposing 2 moles of NaHCO3 (129 kJ).
– We need to calculate for 2.24 moles, so we’ll need to
make an adjustment.
2NaHCO3 (s) + 129 kJ 
Na2CO3 (s) + H2O (g) + CO2 (g)
• Use a mole ratio to find the heat absorbed by
2.24 moles of NaHCO3:
∆H =
2.24 mol
• ΔH = 144.48 kJ
129 kJ
2 mol
Let’s try another…
• When carbon disulfide is formed from its elements,
89.3 kJ heat is absorbed.
– Write the reaction.
– Is the reaction endothermic or exothermic?
– Calculate the amount of heat in kJ that is absorbed when
5.66 g of carbon disulfide is formed.
• Let’s start by writing the reaction.
• C + 2S + 89.3 kJ  CS2
• Since heat is absorbed, it must be endothermic.
C + 2S  CS2
• Calculate the amount of heat in kJ that is
absorbed when 5.66 g of carbon disulfide is
formed (89.3 kJ of heat absorbed per mole).
• We’ll need moles for this, so let’s convert:
• 5.66 g CS2 = 0.074 mol CS2
• If 89.3 kJ are absorbed per mole, let’s use a
mole ratio.
0.074 mol 89.3 kJ
∆H =
1 mol
• ΔH = 6.608 kJ
Hess’s Law
• In reality, while overall reactions may
look like the ones on the previous
slides (reactant directly to product),
they may not always proceed that way.
• Other times, the reactions may occur
so slowly that measurements are
difficult.
• Enter Hess’s Law, an indirect way to
measure heats of reaction.
– Named for Germain Henri Hess, by the
way.
• Who was not German, by the way. SwissRussian, actually.
Germain Hess
Hess’s Law Example
• Imagine we are measuring heat transfer in a
reaction that forms strontium carbonate from its
elements:
– Sr + C + O2  SrCO3
– Balanced: 2Sr + 2C + 3O2  2SrCO3
• How much heat is absorbed or released from this
reaction when 1.00 mol SrCO3 forms, given the
following intermediate equations:
– 2Sr + O2  2SrO
– SrO + CO2  SrCO3
– CO2  C + O2
∆H = -1184 kJ
∆H = -234 kJ
∆H = 394 kJ
2Sr + 2C + 3O2  2SrCO3
•
•
•
•
2Sr + O2  2SrO
∆H = -1184 kJ
SrO + CO2  SrCO3
∆H = -234 kJ
CO2  C + O2
∆H = 394 kJ
So, let’s do some color-coding.
2Sr + 2C + 3O2  2SrCO3
•
•
•
•
•
2Sr + O2  2SrO
∆H = -1184 kJ
SrO + CO2  SrCO3
∆H = -234 kJ
CO2  C + O2
∆H = 394 kJ
Nice.
So now you can see that our desired equation
features:
– Sr, C, and O2 on the left.
– SrCO3 on the right.
• But our intermediate equations aren’t that way.
2Sr + 2C + 3O2  2SrCO3
•
•
•
•
2Sr + O2  2SrO
∆H = -1184 kJ
SrO + CO2  SrCO3
∆H = -234 kJ
CO2  C + O2
∆H = 394 kJ
The underlined one is the problem. Let’s flip it.
2Sr + 2C + 3O2  2SrCO3
•
•
•
•
2Sr + O2  2SrO
∆H = -1184 kJ
SrO + CO2  SrCO3
∆H = -234 kJ
C + O2  CO2
∆H = -394 kJ
Notice that now that I’ve flipped it, I had to
change the sign next to the heat quantity.
• Next task – make sure we satisfy the Law of
Conservation of Mass/Matter, let’s count the
moles of each of our substances.
2Sr + 2C + 3O2  2SrCO3
•
•
•
•
2Sr + O2  2SrO
∆H = -1184 kJ
SrO + CO2  SrCO3
∆H = -234 kJ
C + O2  CO2
∆H = -394 kJ
DESIRED
• INTERMEDIATE
– Sr = 2 moles
– C = 2 moles
– O2 = 3 moles
– SrCO3 = 2 moles
–
–
–
–
Sr = 2 moles
C = 1 mole
O2 = 2 moles
SrCO3 = 1 mole
2Sr + 2C + 3O2  2SrCO3
•
•
•
•
2Sr + O2  2SrO
∆H = -1184 kJ
SrO + CO2  SrCO3
∆H = -234 kJ
C + O2  CO2
∆H = -394 kJ
Let’s try multiplying the entire middle and
bottom equations by two…
2Sr + 2C + 3O2  2SrCO3
•
•
•
•
2Sr + O2  2SrO
∆H = -1184 kJ
2SrO + 2CO2  2SrCO3 ∆H = -468 kJ
2C + 2O2  2CO2
∆H = -788 kJ
DESIRED
• INTERMEDIATE
– Sr = 2 moles
– C = 2 moles
– O2 = 3 moles
– SrCO3 = 2 moles
–
–
–
–
Sr = 2 moles
C = 2 mole
O2 = 3 moles
SrCO3 = 2 moles
2Sr + 2C + 3O2  2SrCO3
•
•
•
•
•
2Sr + O2  2SrO
∆H = -1184 kJ
2SrO + 2CO2  2SrCO3 ∆H = -468 kJ
2C + 2O2  2CO2
∆H = -788 kJ
Great!
Now let’s just check the other stuff:
– SrO is balanced.
– CO2 is balanced.
• And so we go to the last step…
Quick Note
• Just a quick heads-up…
– Did you notice how we were checking to see whether
SrO and CO2 were balanced even though they weren’t
part of the overall equation?
– Yes?
– We did that to abide by the law of conservation of
mass.
– However, what that also means is that occasionally
you’ll see intermediate equations that seemingly don’t
have much to do with the overall equation.
– Just an FYI…
2Sr + 2C + 3O2  2SrCO3
•
•
•
•
2Sr + O2  2SrO
∆H = -1184 kJ
2SrO + 2CO2  2SrCO3 ∆H = -468 kJ
2C + 2O2  2CO2
∆H = -788 kJ
So now we just need to add the terms on the
right.
• -1184 kJ + -468 kJ + -788 kJ = -2440 kJ
• That means -2440 kJ evolved when 2 mol SrCO3
formed.
– So for 1.00 mol SrCO3, -1220 kJ evolved.
Practice
• Hess’s Law Worksheet
– All (NOTE: #5 is a bit challenging, and #4 has some
weird stuff going on. You’ll be fine.)
• #2 will be solved on the next slides.
Hess’s Law W.S. #2
• 4 NH3 (g) + 5 O2 (g)  4 NO (g) + 6 H2O (g)
• N2 (g) + O2 (g)  2 NO (g)
• N2 (g) + 3 H2 (g)  2 NH3 (g)
• 2 H2 (g) + O2 (g)  2 H2O (g)
∆H = -180.5 kJ
∆H = -91.8 kJ
∆H = -483.6 kJ
• Flip intermediate equations:
• N2 (g) + O2 (g)  2 NO (g)
• 2 NH3 (g)  N2 (g) + 3 H2 (g)
• 2 H2 (g) + O2 (g)  2 H2O (g)
∆H = -180.5 kJ
∆H = +91.8 kJ
∆H = -483.6 kJ
4 NH3 (g) + 5 O2 (g)  4 NO (g) + 6 H2O (g)
• Flip intermediate equations:
• N2 (g) + O2 (g)  2 NO (g)
• 2 NH3 (g)  N2 (g) + 3 H2 (g)
• 2 H2 (g) + O2 (g)  2 H2O (g)
∆H = -180.5 kJ
∆H = +91.8 kJ
∆H = -483.6 kJ
• Balance intermediate equations:
• 2 N2 (g) + 2 O2 (g)  4 NO (g)
• 4 NH3 (g)  2 N2 (g) + 6 H2 (g)
• 6 H2 (g) + 3 O2 (g)  6 H2O (g)
• Add ∆H values:
∆H = -361 kJ
∆H = +183.6 kJ
+ ∆H = -1450.8 kJ
∆H = -1628.2 kJ
One last thing…
• Sometimes you’ll be asked to calculate heat of
reaction, which is how much heat a reaction gives off
or takes in, usually as a rate of kJ/mol.
• To find this, divide kJ evolved by mol reactant.
• HUGELY IMPORTANT:
– This is the number that ultimately goes in the
thermochemical equation.
• Example 1: (easy as π)
– If, during the course of a reaction, 0.22 mol of Ca react and
produce 0.019 kJ of heat, what is the heat of reaction in
kJ/mol?
• 0.019 kJ/0.22 mol = 0.0863 kJ/mol
• Ca + 2H2O  Ca(OH)2 + H2 + 0.0863 kJ
Heat of Reaction Example
This is on the core!
• 1.16 g of calcium react with water to form calcium
hydroxide and hydrogen gas.
• If the reaction gives off 4.3 kJ in the process, write the
thermochemical equation.
• Ca + 2H2O  Ca(OH)2 + H2
• We can’t just add in the 4.3 kJ because our reaction
shows what happens when 1 mol of Ca reacts. We
need to find kJ/mol with our data.
• 1.16 g Ca = 0.029 mol Ca
• -4.3 kJ / 0.029 mol Ca = -148.27 kJ/mol
• Ca + 2H2O  Ca(OH)2 + H2 + 148.27 kJ
Heat of Reaction Example Deux
This is on the core!
• 4.5 L of oxygen react with ethanol to form carbon
dioxide and water, giving off 18 kJ.
• 2C2H5OH + 5O2  4CO2 + 6H2O
• Write the thermochemical equation and determine
what mass of ethanol is required to produce 100 kJ of
heat.
• 4.5 L O2 = 0.2 mol O2
• -18 kJ / 0.2 mol = -90 kJ/mol
• 2C2H5OH + 5O2  4CO2 + 6H2O + 90 kJ
• Note that we don’t need to adjust the kJ amount even
though there’s a “5” in front of O2.
– That “5” will be accounted for in stoichiometric calculations.
Heat of Reaction Example Deux
This is on the core!
• 2C2H5OH + 5O2  4CO2 + 6H2O + 90 kJ
• What mass of ethanol will make 100 kJ of heat?
100 kJ
2 mol C2H5OH
90
2.22 mol
kJ
46.1864 g
1
mol
= 2.22 mol C2H5OH
= 102.54 g C2H5OH
Heat of Formation
• Example 2: (more like Hess’s Law)
• Br2 (g)  Br2 (l)
• What is the heat of formation (ΔHf0)?
• To answer this question, note the following:
• ΔHf0 is the symbol for heat of formation/reaction.
– For free elements (e.g. C, Fe, etc.) or diatomic elements
(BrINClHOF) have a ΔHf0 = 0.*
• * as long as they’re in their natural states.
– Others need to be looked up.
• ΔHf0products - ΔHf0reactants = ΔHf0reaction
Heat of Formation
• Anyway, back to the example:
• Br2 (g)  Br2 (l)
• What is the heat of formation (ΔHf0)?
• First, remember that Br is a liquid at room
temperature, so ΔHf0 = 0 for Br2 (l).
• Next, looking up ΔHf0 for Br2 (g) shows 30.91
kJ/mol.
• ΔHf0products - ΔHf0reactants = ΔHf0reaction
• 0 – 30.91 kJ = -30.91 kJ
Heats of Formation/Solution
•
•
•
•
•
Here’s another example:
C2H2 + O2  CO2 + H2O
ΔHf0 CO2 = -393.5 kJ/mol
ΔHf0 H2O = -241.8 kJ/mol
ΔHf0 C2H2 = 227 kJ/mol
• First step? Balance it!
2C2H2 + 5O2  4CO2 + 2H2O
• ΔHf0 CO2 = -393.5 kJ/mol
• ΔHf0 H2O = -241.8 kJ/mol
• ΔHf0 C2H2 = 227 kJ/mol
• Okay, now let’s think logically for a moment. All
these values are in kJ/mol.
• According to the equation, do we have one mole
of everything?
• No, so we need to adjust them slightly.
2C2H2 + 5O2  4CO2 + 2H2O
• ΔHf0 CO2 = -393.5 kJ/mol = -1574 kJ (for 4 mol)
• ΔHf0 H2O = -241.8 kJ/mol = -483.6 kJ (for 2 mol)
• ΔHf0 C2H2 = 227 kJ/mol = 454 kJ (for 2 mol)
• ΔHf0 O2 = 0 kJ/mol (remember, it’s a free element)
• Now let’s add up the products and subtract the
reactants.
454 kJ
0 kJ
-1574 kJ
-483.6 kJ
2C2H2 + 5O2  4CO2 + 2H2O
•
•
•
•
ΔHf0 CO2 = -393.5 kJ/mol = -1574 kJ (for 4 mol)
ΔHf0 H2O = -241.8 kJ/mol = -483.6 kJ (for 2 mol)
ΔHf0 C2H2 = 227 kJ/mol = 454 kJ (for 2 mol)
ΔHf0 O2 = 0 kJ/mol (remember, it’s a free element)
• (-1574 kJ + -483.6 kJ) – (454 kJ + 0 kJ) = -2511.6 kJ
Quick FYI
• Remember those intermediate equations from
Hess’s Law?
–
•
•
•
You know, stuff like…
N2 (g) + O2 (g)  2 NO (g)
N2 (g) + 3 H2 (g)  2 NH3 (g)
2 H2 (g) + O2 (g)  2 H2O (g)
∆H = -180.5 kJ
∆H = -91.8 kJ
∆H = -483.6 kJ
• Did you notice that one side of the equation
always uses terms whose heats of formation
(ΔHf0) are 0?
• Thus, each intermediate equation is perhaps
better known as the standard enthalpy of
formation equation for the ending compound.
Heats of Formation/Solution
• One more example:
• Al (s) + Fe3O4 (s)  Al2O3 (s) + Fe (s)
• ΔHf0 Fe3O4 = -1120.9 kJ/mol
• ΔHf0 Al2O3 = -1669.8 kJ/mol
•
•
•
•
Balanced: 8Al (s) + 3Fe3O4 (s)  4Al2O3 (s) + 9Fe (s)
Reactants: 3 * -1120.9 kJ/mol = -3362.7 kJ
Products: 4 * -1669.8 kJ/mol = -6679.2 kJ
(-6679.2 kJ) – (-3362.7 kJ) = -3316.5 kJ
Heat of Solution
• Sometimes chemicals give off or absorb heat
while dissolving.
– This is called heat of solution.
Substance
NaOH
NH4NO3
Heat of Solution (kJ/mol)
-44.51
+25.69
KNO3
+34.89
HCl
-74.84
Phase Change Diagram
• Remember that phase change diagrams
represent phases as a function of temperature
and pressure.
• Critical Temperature is the temperature above
which vapor cannot be liquefied (condensed).
• Critical Pressure is the pressure required to
liquefy (condense) at the critical temperature.
• Critical Point is the critical temperature and
pressure.
– For water, Tc is 374 °C and 218 atm.
Phase Change Diagram
Closure Part Deux
• Thermochemistry Round Robin Review
– #5-6
– #1, 2, 4, 10, 15
Temperature 
H2O Heating/Cooling Curve
Condensing
100°C
Freezing
ΔHfus
0°C
ΔHvap
Boiling Pt.
Freezing Pt.
A
C
Melting
B
E
Vaporization
Liquid
Fusion
Solid
For A, C, E:
ΔH = mCΔT
Gas
Vaporizing
D
ΔHsolid
ΔHcond
ΔHfus = -ΔHsolid
ΔHvap = -ΔHcond
For B, D:
ΔH = molΔHphase change
Heat 