Chemistry 30
Unit 2:
Energy Changes in
Chemical Reactions
Teacher
1
Energy:
In science, energy is often defined as “the ability to do work” or “the capacity to produce change”.

Potential energy is energy possessed by a body because of its position

Kinetic energy is the energy of motion
o Temperature measures the average kinetic energy of the substance, not the total energy,
since potential energy is not measured.
o Temperature is measured in Celsius or Kelvin. To change Celsius to Kelvin add 273 and to
change Kelvin to Celsius, subtract 273.
Ex. Determine the temperature change in Celsius and in Kelvin if a container of water cools
from 98oC to 23oC.
∆t = tf – ti = 23oC – 98oC= -75oC
∆t = tf – ti =296K – 371K = -75K
Notice that the change in temperature is the same value if we use Celsius OR Kelvin.

Thermal energy is the sum of the kinetic energy and potential energy an object has.
o If you have a cup of water at 80oC and a bathtub of water at 30oC, which has more KE and
which has more PE? Which would have more thermal energy?

Cup has more kinetic energy

Bathtub has more potential energy

Bathtub has more thermal energy
o Heat energy is the transfer of thermal energy from one object to another. Heat always flows
from the warmer object to the colder object. The transfer of energy can be detected by
measuring the resulting temperature change.
o When energy is transferred to molecules, the molecules move faster, hit each other more
often, and transfer more energy to one another.

Radiant energy is energy transmitted through space as electromagnetic waves. The most familiar
form is light.

Chemical energy is the energy necessary to keep atoms joined by chemical bonds. During a
chemical reaction, chemical energy may be stored, released as heat, converted to other forms of
energy, or converted to work.
Whatever changes take place, energy is always conserved. It is a fundamental law of nature (the law of
conservation of energy) that energy can neither be created nor destroyed but may be converted from one
form to another. Energy in the universe is constant.
The unit used to measure energy is the Joule.
1J =
1𝑘𝑔∙𝑚2
𝑠2
1 J = 2.78 x 10-7 kW∙h
1 J = 2.39 x 10-4 kcal
2
Heat:



Heat is thermal energy in transit.
If you touch something hot, energy is transferred from the hot object to your hand. If you touch
something cold, heat energy passes from your hand to the cold object.
We should only use the term heat to describe energy that is flowing from one substance to another
because of a temperature difference.
Specific Heat:


The specific heat of a substance is the amount of heat energy required to raise the temperature of 1
gram of a substance by 1 degree Celsius or 1 kelvin.
Units used to measure specific heat = J•g−1•K−1 or J•g−1•°C−1
or
𝐽
𝑔∙𝐾

or
𝐽
𝑔∙℃
Different substances have different specific heats, this is because substances vary in their ability to
store energy, and they also gain or lose heat at different rates.
Ex. Comparing water to aluminum. It takes 4.18 J to raise the temperature of liquid water 1°C, whereas
for aluminum only 0.903 J are required. For the same increase in temperature, liquid water requires
over 4 times as much energy as the same mass of aluminum.
Why do you think water would be used as a coolant instead of another liquid (like mercury)?

Water can absorb a large quantity of heat with only a slight rise in temperature.
Water has the greatest heat capacity of all liquids.
Substance Specific heat (J/g·°C)
Al(s)
0.903
Fe(s)
0.449
Hg(l)
0.139
H2O(l)
4.184
O2(g)
0.917
He(g)
5.19
CO2(g)
0.843
Substance Specific heat (J/g·°C)
Au(s)
0.129
Na(s)
1.24
Cu(s)
0.385
H2(g)
14.4
H2O(g)
1.86
Ag(s)
0.235
H2O(s)
2.087
Heat Required to Change Temperature:

The formula to calculate heat required to change the temperature of a substance is:
Q = (m) (c) (∆t)
Where Q= heat amount (Joules)
m = mass (g)
∆t = tfinal – tinitial
*Losing heat would give you a negative Q*
c = specific heat capacity (J/g∙oC)
∆t = temperature change (oC or K)
3
Ex. Determine the heat required to raise the temperature of 100.0 mL of water from 298.0 K to 373.0 K.
Q = (m) (c) (∆t)
Q = (m) (c) (∆t)
Q=?
m = 100g (100mL = 100g of water)
c = 4.18J/g∙K
∆t = tf – ti = 373.0 K − 298.0 K = 75.0 K
= (4.18 J/g•K)(100 g)(75.0 K)
= 31.4 kJ
Heat capacity:

Heat capacity is the heat energy required to raise the temperature of a given quantity of a
substance by one degree Celsius or one kelvin.
*Heat capacity is not the same as SPECIFIC heat capacity*
Heat capacity (J/°C) = specific heat × mass
∴ Heat Capacity = Q/∆t
Ex. Determine the heat capacity of 1 cup of water (250mL).
*Remember the density of water is ~1g/mL*
Heat capacity = specific heat x mass
= (4.18J/g∙K)(250g)
= 1045J/K or 1045J/oC
Ex. Calculate the heat capacity of a piece of iron that releases 3500J of heat into a container of water, and
the temperature of the iron drops from 100oC to 24oC.
Heat Capacity = Q/∆t
= (-3500J/[24oC – 100oC])
Heat Capacity = 46J/oC
Molar Heat Capacity:

Molar heat capacity of a substance is the amount of energy required to raise the temperature of one
mole of a substance by one Celsius degree or one kelvin.
Molar heat capacity (J/mol•°C) = specific heat × molar mass
Ex. Calculate the molar heat capacity of water, given that the specific heat of water is 4.18 J/g•°C.
The molar mass of water is 18.02 g/mol
Molar heat capacity = specific heat x molar mass
= 4.18 J/ g•°C × 18.02 g/mol
= 75.3 J/mol•°C
Heat Capacity Assign
4
Molecular Motion:
There are three types of motion that molecules undergo:

Translational Motion: Molecules are free to move along linear pathways from one place to
another

Rotational Motion: When a molecule rotates about an axis through its center of mass.

Vibrational Motion: molecules oscillate along the direction of a bond.
Comparing types of motions among solids, liquids and gases:
Phase
Translational
Rotational
Vibrational
Gas
Free
Free
Free
Liquid
Restricted
Restricted
Free
Solid
Absent
Very Restricted
Free
Heat Capacities of gas molecules:

Monatomic gases such as helium and neon have the lowest molar heat capacity
o

Diatomic gases such as hydrogen and nitrogen have higher molar heat capacities
o

Have only translational motion since there is no bond about which to rotate or on which to
vibrate, therefore adding heat makes the atoms move faster, which increases the
temperature.
Have vibrational and rotational motion as well as translational motion, thus requiring more
energy. Diatomic gases have similar heat capacities.
Triatomic gases such as water vapour and carbon dioxide have even higher heat capacities
o These molecules can vibrate and rotate in more ways than diatomic gases, thus requiring
even more energy.
5
Enthalpy:

Enthalpy is defined as the heat content in a chemical reaction or physical process. It is symbolized
by the letter H.

The heat content if a substance cannot be measured directly. Instead, we measure enthalpy
difference between the initial and final states. A chemical reaction undergoes a change in enthalpy
when the reaction either releases energy to the surroundings or absorbs energy from the
surroundings. This heat change is called the enthalpy of reaction and is symbolized by ∆H.

A thermochemical equation is written with the value of ΔH included:
 CuCl2(s)
Cu(s) + Cl2(g) 
ΔH = −220.1 kJ
or
 CuCl2(s) + 220.1 kJ
Cu(s) + Cl2(g) 
Ex. The exothermic reaction of gaseous hydrogen and oxygen at constant pressure releases 241.8 kJ of
heat energy for every mole of water vapour formed. Write the thermochemical equation for the
production of one mole of water vapour.
 H2O(g)
H2(g) + ½ O2(g) 
ΔH = −241.8 kJ/mol
Exothermic and Endothermic Reactions:

In a chemical reaction, the chemical
components involved in the process are
known as the system. Everything outside
the
system
then
becomes
the
surroundings.

Since energy is always conserved, energy
lost by the system must be absorbed by
the surroundings. Similarly, energy
absorbed by the system must come from
the surroundings.

A reaction in which the system releases heat to the surroundings is known as an exothermic
reaction. If the system absorbs heat from the surroundings, it is an endothermic reaction.
Exothermic Reactions
Endothermic Reactions
Enthalpy of products is less than enthalpy of
reactants
Hproducts < Hreactants
Energy is released
∆H is negative
Less energy is needed to break bonds than the
energy that is released when bonds form
Enthalpy of reactants is less than enthalpy of
products
Hreactants < Hproducts
Energy is absorbed
∆H is positive
More energy is needed to break bonds than the
energy that is released when bonds form
6
Research into the Thermochemistry of Hot and Cold Packs
*Adapted from Exp 2 of Chen 115.3 Laboratory Manual 2005-2006*
Purpose:
To develop a prototype hot or cold pack using a dissolved salt that produces an exothermic or
endothermic process.
Background Information:
In a chemical reaction where there is an energy change, heat can either be released or absorbed by the
reaction. If energy (in the form of heat) is released into the surroundings, the surroundings get warmer.
This is known as an exothermic reaction. If heat is absorbed from the surroundings, the surroundings get
cooler. This is known as an endothermic reaction. The amount of heat transferred between a system and
its surroundings (at constant pressure) is represented by the symbol ∆H and called enthalpy. Exothermic
processes have a negative ∆H and endothermic have a positive ∆H.
In this lab you will be working with a research team to develop either a hot pack or a cold pack. If you
choose to develop a hot pack, you must create one that will reach a maximum temperature of 45oC. If you
choose a cold pack the minimum temperature it must reach is 5oC.
The way you will develop your hot or cold pack is by dissolving either CaCl2 or NH4NO3 in 100mL of water.
Since dissolving CaCl2 in water is an exothermic process, it would increase the temperature of the water.
Dissolving NH4NO3 in water is an endothermic process, so it would cool the water by absorbing the heat
from the water. Before you can start developing your hot/cold pack, you must calculate the appropriate
amount of salt to dissolve in the water, so you know where to begin.
1.00 mole of CaCl2 releases 82.9KJ of heat energy when dissolved in water (∆Hosoln= -82.9KJ/mole) and
1.00 mole of NH4NO3 absorbs 25.7KJ of heat energy when dissolved in water (∆Hosoln= +25.7KJ/mole).
In this lab, in order to determine the amount of salt needed to raise/lower the temperature you will need
to try several different amounts of salts. Since you will be changing the amount of salts, this is the variable
in the experiment and should be the only thing you change (in order to accurately evaluate the data).
Everything else in the experiment should remain constant.
Procedure:
1. Complete the beginning of your formal lab report (up to the data/observations section- don’t forget
your hypothesis!).
2. Gather a team of 3 or 4 people to perform this experiment with. Determine whether you will create a
hot or cold pack.
3. Calculate the amount of heat energy that the water needs to release/absorb to lower/raise the
temperature of 100mL of water to the appropriate temperature. Remember the specific heat of water
is 4.184J/goC. This means that is takes 4.184J of energy to raise the temperature of 1g of water by 1oC.
The density of water at 20oC is 0.998g/mL. The initial temperature we will use for these calculations is
20oC.
4. Calculate the amount of salt you will need to use to raise/lower the temperature of the water to the
desired value. To do this, first convert the molar enthalpy of solution to enthalpy in terms of heat
7
energy released or absorbed by 1.00g of each salt (divide the molar enthalpy value by the molar mass
of the salt). This value is the amount of heat absorbed/released by 1.00g of salt dissolving in water. Use
this value to determine the amount of salt needed to release/absorb the amount of energy you
calculated in step three.
5. Depending on the size of your group, choose 6-8 different amounts of solid to use for this experiment
(the amount you calculate, 2 or 3 smaller amounts and 3 or 4 larger amounts using 5g increments). The
largest amount used should be less than 60g. Prepare your observations section of your lab report as
follows:
State whether your group will be creating a hot or cold pack.
Salt used:
Appearance:
Mass of _____________
(g)
Initial
Temperature of
H2O
Ti (oC)
Max/Min
Temperature
of Solution
Tm (oC)
Person who made the
measurements
Change in
temperature
∆T (oC)
Prototype Data:
Initial Temperature of Water:
Temperature Change Required:
Mass of ___________ Required:
Final Temperature of Water:
6. Each member of your group will choose 2 of the 6-8 selected amounts of solid and complete the
following steps. Complete the observations section of your lab report using the data your team collects.
a. In a beaker, weigh the solid to the nearest 0.1g. Remember to try to keep as many variables
constant as possible.
b. Using a graduated cylinder, measure out 100mL of distilled water. Record the initial
temperature of the water (remember to always estimate one digit).
c. Add the water to the solid and stir (DO NOT USE THE THEMOMETER TO STIR THE SOLUTION).
Observe the temperature increase/decrease and record the maximum/minimum temperature,
Tm (Note: this will note be the final temperature, as the temperature of the room will warm/cool
the solution after it reaches the max or min).
7. On graph paper, plot ∆T (y-axis) vs. mass of solid (x-axis) in each of the 6-8 trials. Each team member
should make their own graph. The plot should give a straight line.
8. Create a prototype pack: Measure the initial temperature of the water to be used for the prototype
pack. Determine the ∆T required to change the temperature to 45/5oC. From the graph you created,
determine the weight of solid you need to make the hot/cold pack.
9. Place the experimentally determined amount of solid (from the graph) into a Ziploc bag. Pour the water
into a sandwich bag (new bag), close the bag with a small rubber band and cut off the excess bag. Place
8
the sealed packet of water into the Ziploc bag, remove as much air as possible and seal the Ziploc bag.
Be sure the bag is sealed completely.
10. Activate the hot/cold pack.
WARNING: The bag may break open. If the concentrated solution gets on your skin, be sure to
wash it off immediately with lots of water.
11. Determine the final temperature of the hot/cold pack by wrapping it around the end of the
thermometer.
Analysis:
Write a paragraph that explains why dissolving a salt in water changes the temperature of the water.
1. Describe how well the prototype worked. Did it perform as planned? If not explain why. If you think
further research could be completed, explain what that might be.
2. Explain why there was a difference observed between the calculated theoretical amount of chemical
required, and the amount determined experimentally.
3. Explain why there was a difference observed (if you saw a difference) between the expected final
temperature and the actual final temperature of the hot/cold pack.
4. What would you expect to observe if you used a finely powdered solid rather than pelletized solid?
Would the mass of solid required change?
5. What do you expect would happen if you used the same amount of solid but used 50mL of water in
your hot/cold pack instead of 100mL? Explain.
Conclusion:
Summarize your results from this lab.
9
Types of Enthalpy Equations:





Enthalpy of Formation Equations
o Each equation shows the forming of one mole of product from its elements and also contains
the loss or gain of heat energy when the compound is formed.
Enthalpy of Combustion Equations
o Each equation shows the burning of one mole of substance in O2 and includes a heat term.
The enthalpy of combustion equation is always has a loss of heat so ∆H is negative
Enthalpy of Solution Equations
o Each equation shows the dissolving on one mole of compound in water and includes a heat
term.
Enthalpy of Vaporization Equations
o Each equation shows the energy involved in changing one mole of liquid to a gas at its
boiling point.
Enthalpy of Melting Equations.
o Each equation shows the energy involved in changing one mole of solid to a liquid at its
melting point.
Bond Energy & Enthalpy of Formation:

Atoms in molecules or formula units are held together by the electrons being attracted to the
nuclei. These bonds can be easily broken by adding energy. This amount of energy is known as
bond energy.
Ex. H-H + 436.4kJ  2H
When a bond between two hydrogen atoms break, it requires 436.4kJ of energy to occur.
Alternatively, if 2 hydrogen atoms where to bond, 436.4kJ of energy would be released.
2H  H-H + 436.4kJ
Bond Energies at 298K


Bond energies can be used to determine
whether heat will be released or absorbed
and how much when a mole of a
compound forms from its elements.
Bond energies are given in units kJ and
this represents the amount of energy
released/absorbed per mole of substance.
Bond
H-N
H-O
H-S
H-P
H-H
H-F
H-Cl
H-Br
H-I
C-H
C-C
C=C
C≡C
C-N
Bond
Energy
(KJ∙mol-1)
393
460
368
326
436.4
568.2
431.9
366.1
298.3
414
347
620
812
276
Bond
C=N
C≡N
C-O
C=O
C-P
C-S
C=S
N-N
N=N
N≡N
N-O
N-P
O-O
Bond
Energy
(KJ∙mol-1)
615
891
327
804
263
255
477
393
418
941.4
176
209
142
Bond
O=O
O-P
O=S
P-P
P=P
S-S
S=S
F-F
Cl-Cl
Br-Br
I-I
C-Cl
N-F
Bond
Energy
(KJ∙mol-1)
498.7
502
469
197
489
268
352
150.4
242.7
192.5
151.0
326
275
10
Ex. Assume the following reaction takes place in a series of steps and energy is absorbed when bonds
break and released when bonds form. Determine whether the reaction is endothermic or exothermic, and
the amount of energy released or absorbed.
H2 + Cl2  2HCl
Step 1: H2  2H (bond breaks ∴ absorbs 436kJ of energy):
H2 + 436kJ  2H
Step 2: Cl2  2Cl (bond breaks ∴ absorbs 242kJ of energy):
Cl2 + 242kJ  2Cl
Step 3: H + Cl  HCl (bond forms ∴ releases 431kJ of energy. This occurs twice so 862kJ is released)
2H + 2Cl  2HCl + 826kJ
To determine final reaction we add these three steps together and cancel anything that occurs in reactants
and products:
H2 + 436kJ  2H
Cl2 + 242kJ  2Cl
2H + 2Cl  2HCl + 862kJ
H2 + Cl2 + 678kJ-678kJ  2HCl + 862kJ-678kJ
Final Reaction: H2 + Cl2  2HCl + 184kJ ∴ exothermic and ∆H = - 184kJ for the formation of one mole of
hydrochloric acid.
We have now used bond energies to determine the amount of energy released when 1 mole of a
compound is formed from its elements. This is known as the enthalpy of formation.
See Bond Energies Assignment
Enthalpy of Formation and Hess’s Law:



The standard enthalpy of formation of a substance is the loss or gain in heat energy when one mole
of the substance is formed from its elements in their standard states, ΔH°f.
o The superscript ° indicates that the value of ΔH is measured when all substances are in their
standard states.
o The subscript f indicates that the energy is associated with the formation of one mole of a
compound from its elements in their standard states.
o The enthalpy of formation for different compounds is found using the bond energy
breaking/formation, just like we did in the last assignment.
The more negative ΔH°f, the more stable the compounds, because these compounds require a lot of
energy to break the bonds.
Many chemical reactions occur in steps. For example, when carbon is burned in the presence of
excess oxygen, carbon dioxide is produced. In this reaction 393.5 kJ of heat energy is liberated for
every mole of carbon used. This means that the enthalpy change for the reaction is −393.5 kJ/mol.
 CO2(g)
C(s) + O2(g) 
ΔH = −393.5 kJ
OR
(1)
(2)
 CO(g)
C(s) + ½ O2(g) 
 CO2(g)
CO(g) + ½ O2(g) 
 CO2(g)
C(s) + O2(g) 
ΔH = −110.5 kJ
ΔH = −283.0 kJ
ΔH = −393.5 kJ
The end result is the same whether it is formed in a one-step reaction or in a multi-step process. This is an
illustration of Hess’s Law of Constant Heat Summation: The change in enthalpy for a chemical reaction is
constant, whether the reaction occurs in one step or several.
11
*When elements are formed, there is no change in enthalpy so ∆H is 0 kJ/mol*
Substance
ΔHf (kJ/mol)
Substance
ΔHf (kJ/mol)
Al(s)
Al2O3(s)
Br2(l)
HBr(g)
Ca(s)
CaCO3(s)
CaCl2(s)
C(s) (graphite)
C(s) (diamond)
CCl4(l)
CCl4(g)
CHCl3(l)
CH4(g)
C2H2(g)
C2H4(g)
C2H6(g)
C3H8(g)
C6H6(l)
CH3OH(l)
C2H5OH(l)
CH3CO2H(l)
CO(g)
CO2(g)
COCl2(g)
CS2(g)
CS2(l)
Cl2(g)
HCl(g)
Cr(s)
CrCl3(s)
Cu(s)
CuO(s)
CuCl(s)
CuCl2(s)
F2(g)
HF(g)
He(g)
H2(g)
H2O(l)
H2O(g)
H2O2(l)
Fe(s)
FeO(s)
Fe2O3(s)
Fe3O4(s)
FeCl2(s)
FeCl3(s)
FeS2(s) (pyrite)
Pb(s)
0
−1675.7
0
−36.4
0
−1206.9
−795.8
0
+1.90
−135.4
−96.0
−134.5
−74.8
+226.7
+52.3
−84.7
−103.8
+49.0
−238.7
−277.7
−484.5
−110.5
−393.5
−218.8
+117.4
+89.70 kJ
0
−92.3
0
−556.5
0
−157.3
−137.2
−220.1
0
−271.1
0
0
−285.8
−241.8
−187.8
0
−272.0
−824.2
−1118.4
−341.8
−399.5
−178.2
0
PbCl2(s)
Mg(s)
MgCl2(s)
MgO(s)
Hg(l)
HgS(s)
Ne(g)
N2(g)
NH3(g)
N2H4(l)
NH4Cl(s)
NH4NO3(s)
NO(g)
NO2(g)
N2O(g)
N2O4(g)
HNO3(l)
O(g)
O2(g)
O3(g)
P4(s) (white)
P4(s) (red)
PH3(g)
PCl3(g)
P4O6(s)
P4O10(s)
H3PO4(s)
K(s)
KCl(s)
KClO3(s)
KOH(s)
Ag(s)
AgCl(s)
AgNO3(s)
Na(s)
NaCl(s)
NaOH(s)
Na2CO3(s)
S(s) (rhombic)
S(s)
SF6(g)
H2S(g)
SO2(g)
SO3(g)
H2SO4(l)
Sn(s) (white)
Sn(s) (grey)
SnCl2(s)
SnCl4(l)
−359.4
0
−641.3
−601.7
0
−58.2
0
0
−46.1
+50.6
−314.4
−365.6
+90.3
+33.2
+82.1
+9.2
−174.1
+249.2
0
+142.7
0
−70.4
+5.4
−287.0
−2144.3
−2984.0
−1279.0
0
−436.7
−397.7
−424.8
0
−127.1
−124.4
0
−411.2
−425.6
−1130.7
0
+278.8
−1209.0
−20.6
−296.8
−395.7
−814.0
0
−2.1
−325.1
−511.3
12
Finding ΔH for an Equation (Hess’s Law cont.):

If two or more thermochemical equations are added to give a final equation, then the enthalpies can
be added to give the enthalpy for the final equation.
Ex. Use the given intermediate steps (1 and 2) for the production of one mole of tetraphosphorus
decaoxide, to determine the ΔH for the overall reaction from its elements.
Intermediate steps:
(1)
 P4O6(s)
4 P(s) + 3 O2(g) 
ΔH = −1640 kJ
(2)
P4O6(s) + 2 O2(g) 
 P4O10(s)
ΔH = −1344 kJ
The overall reaction is the sum of (1) and (2):
(1)
(2)
 P4O6(s)
4 P(s) + 3 O2(g) 
 P4O10(s)
P4O6(s) + 2 O2(g) 
 P4O6(s) + P4O10(s)
4 P(s) + 3 O2(g) + P4O6(s) + 2 O2(g) 
Cancelling out species that appear on both sides of the reaction, we are left with the equation for the
 P4O10(s)
overall reaction:
4 P(s) + 5 O2(g) 
The enthalpy of the overall reaction is the sum of the enthalpies of the intermediate steps.
ΔH = (−1640 kJ) + (−1344 kJ) = −2984 kJ
Since ΔH° < 0, the reaction is exothermic.

Sometimes you may be required to reverse an intermediate step in order for the reactants and
products you want to be on the appropriate side. If you reverse an intermediate step, don’t
forget to reverse the sign of the ΔH.

You may also need to multiply or divide the intermediate equations so that the quantities of
reactants/products match the quantities you want in your final reactions. If you multiply of
divide an intermediate step, remember to multiply or divide your heat value as well.
Ex. The enthalpy changes for the following reactions are
(1)
 2 CO2(g) + H2O(l)
C2H2(g) + 5/2 O2(g) 
ΔH = −1299 kJ
(2)
 6 CO2(g) + 3 H2O(l)
C6H6(l) + 15/2 O2(g) 
ΔH = −3267 kJ
Find ΔH° for the following reaction:
3 C2H2(g) 
 C6H6(l)
Is the reaction endothermic or exothermic?
(1)
 6 CO2(g) + 3 H2O(l)
3 C2H2(g) + 15/2 O2(g) 
ΔH = 3(−1299 kJ)
(2)
 C6H6(l) + 15/2 O2(g)
6 CO2(g) + 3 H2O(l) 
ΔH = +3267 kJ
3C2H2(g) 
 C6H6(l)
ΔH = −630 kJ
ΔH is negative so it is exothermic
See Hess’s Law Assign
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Enthalpy Change in a Reaction
From Hess’s law, we now know the enthalpy change for a reaction can be determined by subtracting the
sum of the heat of formation of the reactants from the sum of heat of formations of the products.
ΔH°reaction =
 H

f products
−
 H

f reactants
Ex. Using the ΔH°f values from your tables, determine the enthalpy change for the combustion reaction of
benzene, C6H6(l).
 12 CO2(g) + 6 H2O(l)
2 C6H6(l) + 15 O2(g) 
ΔH° =
 H

f products
−
 H

f reactants
ΔH° = [12(−393.5) + 6(−285.8)] − [2(+49.0) + 15(0)] = −3267 kJ
See Enthalpy Change in Rxn Assign
Enthalpy and Phase Change:

In all solids the particles vibrate back and forth weakly. They have low kinetic energy. Heating a
solid results in greater vibration of particles so their kinetic energy increases. The solid particles
move further away from one another, which causes their potential energy to increase.

These energy changes can be observed by an increase in temperature and an increase in volume.
o Note that the physical appearance of the solid will not change.

The heat energy (Q) involved in a solid changing temperature can be calculated using mc∆t.

A phase change occurs when the physical state of a substance changes. Melting, freezing,
evaporation and condensation involve phase changes.

Both melting and vaporization are endothermic processes: heat is absorbed from the
surroundings.

During the reverse processes, in
which steam condenses or water
freezes, heat energy is released to
the surroundings. These are
exothermic processes. These
enthalpy values are negative.

During a phase change, the
temperature of the substance does
not change even though you are
adding/removing energy. This is
because the energy being added
or removed is working to change
the state of the substance, not
change the temperature.
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
The heat energy (Q) involved in changing the state of one mole of a substance at its boiling/melting
point is known as the molar enthalpy of phase change.
Molar enthalpies of phase change (kJ•mol−1)
Substance
ΔHvap ΔHcond ΔHmelt ΔHfre
Water
+40.7
-40.7
+6.02
-6.02
Methane
+10.4
-10.4
+0.94
-0.94
Mercury
+59.3
-59.3
+2.3
-2.3
Sodium chloride
+207
-207
+27.2
-27.2
*Note: values for melting/freezing or vaporization/condensation are the same, but have opposite signs *

When you have molar enthalpies of phase change you can use these values to calculate the heat
energy. The formula is as follows:
Q = ∆Hvap, cond, melt, fre ∙ n
Where Q = heat energy
∆Hvap, cond, melt, fre = molar enthalpy of phase change
n = # moles
Ex. Calculate the enthalpy change involved when 17.2 g of liquid mercury is vaporized.
First calculate moles of Hg:
1 𝑚𝑜𝑙
17.2𝑔 𝐻𝑔 × 200.6𝑔 = 0.0857 mol Hg
ΔH = 0.0857 mol Hg(59.3 kJ/mol) = 5.08 kJ
See Energy Needed to Melt Ice Lab &
Enthalpy and Phase Changed Assignment
Calorimetry

The branch of thermochemistry
concerned with measurement of
such heat changes is known as
calorimetry. Any device used for
measuring a heat change is called a
calorimeter.

When we measure heat changes
with this calorimeter, we assume
that all the heat is absorbed by the
water, and that the polystyrene cup
does not absorb heat and that no
heat is lost to the surroundings. If the quantity of heat emitted during the reaction is not large, such a
“coffee cup” calorimeter will give reasonably accurate results. If large quantities of heat are involved,
then heat losses to the surroundings will lead to inaccuracies.
15
Ex. A mass of 100 g of water is placed in a coffee cup calorimeter. The solution temperature is measured to
be 14.4°C. A mass of 0.412 g of calcium metal is placed in the calorimeter. When the reaction is complete
the temperature is recorded as 24.6°C. Calculate the standard molar enthalpy change for this reaction:
Start by finding the heat amount that the water changed by (Q), then determine the amount of
heat release by the reaction. Next we will determine molar enthalpy by dividing the heat that
the reaction released by the the mole s of calcium used in the reaction.
 Ca(OH)2(s) + H2(g)
Ca(s) + 2 H2O(l) 
Q = m c ∆t
Heat absorbedwater
= (4.18 J/g•C)(100 g)(10.2°C)
= 4.26 × 103 J or 4.26 kJ
Therefore, heat released by calcium= −4.26 kJ
# of mol Ca= 0.412 g Ca(1 mol/40.08 g) = 0.0103 mol Ca
Molar enthalpy =
4.26 kJ
0.0103mol
= −414 kJ/mol
The Flame Calorimeter:
Some calorimeters may not be fully insulated. This means that the water in the calorimeter and the
calorimeter itself both absorb heat.
16
Ex. A flame calorimeter composed of steel and with a mass of 322 g is filled with 225 g of water. The
original temperature of the water and calorimeter is 10.6°C. When 1.02 g of ethanol, C2H5OH, is burned,
the temperature rises to 38.4°C. Calculate the molar enthalpy of combustion of ethanol.
Specific heat of steel = 0.44 J/g•C
The heat released by the combustion process will be absorbed (mainly) by the water and the calorimeter
as the temperature rises by 27.8°C (the change in temperature).
Heat absorbed by weater and calorimeter: Q = mc∆t for water + mc∆t for steel
Q = (225g)(4.18 J/goC)(27.8oC) + (322g)(0.44 J/goC)(27.8oC)
= (+26100 J) + (+3900 J)
= 30000 J
Since 30,000 J is absorbed by the water and calorimeter, this means that 30,000J must have
been released by the combustion of ethanol. Therefor Qcomb = - 30.0kJ
# of moles of C2 H 5OH  1.02 g C2 H 5OH 
1 mol C2 H 5OH
46.07 g C2 H 5OH
 0.0221 mol C2 H 5OH
molar enthalpy of combustion =
30000 J
J
kJ
 1400000
or  1400
0.0221mol
mol
mol
See Calorimetry Assignment .
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