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Thermodynamics
Energy Changes in Reactions
Thermodynamics
●
Thermodynamics is a branch of science concerned with heat & temperature
& their relation to energy and work. The behavior of these quantities is
governed by the laws of thermodynamics, irrespective of the composition or
specific properties of the material or system in question.
Energy
●
Our society is preoccupied with energy: it’s availability,
management, benefits & future sources.
●
North Americans consume ＞ ¼ of the world’s energy output.
Energy
= the ability to do work
-Primary sources: Chemical, nuclear, solar, & geothermal
-All the types of energy we observe are changes and energy transfers
Review of Changes: Chemical, Physical & Nuclear
●
Chemical Changes: When the characteristic properties of a substance are
changed. Involves making & breaking chemical bonds involving valence
electrons. Chemical energy must be initiated by an energy source.
Ex: photosynthesis, breathing, reactions with oxygen,: tarnishing rusting,
burning, oxidizing, etc.
●
Physical Changes: When the characteristic properties stay the same.
Ex: changing state of matter, crumpling, separating mixtures: distillation,
filtration, decantation, etc.
Changes in state:
●
Nuclear Change: involves a change in the structure of the nucleus of a
substance. Nuclear changes cannot be reversed.
Ex: Formation of an isotope of an element (addition or removal of neutron(s)
from the nucleus).
Ranking of energy changes:
Nuclear > Chemical > Physical
Endothermic Vs. Exothermic Changes:
●
Endothermic Change (NEEDS ENERGY): Heat gained by the system from
the surroundings.
Ex 1: Making a snowman - the snowman (the system) takes the heat from
your hand to be built.
Ex 2: Rubbing alcohol - the alcohol (the system) takes heat from your body to
evaporate & lowers your body temperature.
-Endo reactions always have energy on the reactant side:
●
Exothermic Change (RELEASES ENERGY): Heat is lost by the system to
the surroundings.
Ex: The human body releases energy to the surroundings.
-Exo reactions always have energy on the product side.
Law of Conservation of Energy:
In any chemical or physical process, energy is neither created nor destroyed. It is
converted from one form of energy to another.
Important forms of energy:
●
●
Kinetic energy: Energy of motion
Potential: Stored energy due to position.
Difference Between Heat & Temperature:
●
Heat (thermal) energy: energy which is transferred from one body to another
due to a difference in temperature between two bodies.
-Heat flows spontaneously from hotter to colder objects (spontaneous
reactions favor entropy → increased disorder).
●
Temperature (T): is a measure of the average kinetic energy of its particles.
-Measured with a thermometer in oC.
-Heat transfer continues until the temperatures are the same.
3 Types of Systems:
1.
2.
3.
Open allows exchange of matter & energy with surroundings (beaker with
reaction inside)
Closed allows exchange of energy (but not matter) with the surroundings
(balloon with air).
Isolated allows no exchange of matter & energy with surroundings (insulated
bottle with hot chocolate) for short periods of time.
Calorimeters
-are instruments used to experimentally determine the heat energy absorbed or
released during a given reaction (i.e. measure heat transfers).
-When measuring, temp change of H2O is used to calculate the heat gained by the
H2O, or the heat lost by the substance (specific heat of it).
Specific Heat Capacity (c):
It is the quantity of energy required to change the temperature of a unit mass (g) of
a substance by one degree Celsius.
Q = mcΔT
Q = Heat energy (Joules, J)
m = mass (gram, g)
(for these problems 1mL H2O = 1g)
ΔT = temperature change (degree Celsius, oC) = Tf - Ti
c = specific heat capacity (Joules per gram degrees Celsius)
Cwater = 4.19
J
g•oC
Heat required to change the temperature:
-different amounts of heat are needed to raise the temp of different substances.
-specific heat depends on how much matter (mass) is heated & how much the
temp rises.
-Specific heat varies not only from one substance to another, but from one state of
matter to another.
-high specific heat indicates that something absorbs more energy (before the temp
rises), & releases more heat if the temp falls.
-the specific heat capacity is proportional to the difficulty of raising or lowering its
temp.
-If the change in temp is negative, the substance has lost thermal energy or given
off heat.
-If the change in temp is positive, the substance has absorbed heat or gained
thermal energy
Specific Heat Capacities of Common Substances:
-Several different units are possible:
Note that c in J/goC & J/gK is the same value
since each degree is the same size in both
scales.
When doing calculations, make sure the units match.
Specific Heat Capacity of Water
It takes a great deal of energy to break the intermolecular bonds in water
molecules because they are much stronger than the bonds in most compounds.
Examples of Specific Heat Problems:
Ex 1: Calculate the quantity of thermal energy absorbed by a 5.0 kg block of
concrete to raise its temp from 17.0 oC to 35.0 oC.
m = 5.00 kg = 5000 g
ΔT = Tf - Ti = 35.5oC - 17.1oC = 18.4oC
c = 2.10 J/(g•oC)
Q = mcΔT
Ti = 17.1oC
= 5000 g • 2.10 J/g•oC•18.4oC
Tf = 35.5oC
= 193 200 J = 193.2 kJ
ΔT = ?
Q=?
The concrete block must absorb 193 kJ.
Ex 2: A 1.35 g pellet of aluminum foil is heated to 205 oC, & then removed from
the heat. After a few seconds, the pellet has released 176 J of heat. What is the
final temp?
m = 1.35 g
Q = mcΔT
c = 0.90 J/(goC)
ΔT = Q
Ti = 205oC
=
mc
-176 J
= - 144.9oC
1.35 g • 0.90 J/(goC)
T f = ? oC
ΔT = Tf - Ti
ΔT = ?
Tf = ΔT + Ti = -144.9oC + 205.0oC = 60.1oC
Q = -176 J
The final temp of the Al pellet is 60 oC.
Calculating Energy Transfer
-We assume that the heat released by the first system (Q1) is equal is = to the
heat absorbed by the second one (Q2), if they are considered to be isolated, with
no energy loss.
-when 2 systems at different temps are in contact, the thermal energy of the hotter
one (system 1) is transferred to the cooler one (system 2).
- Q1 = Q 2
OR
-Qloss = Qgain
- m1c1ΔT1 = m2c2ΔT2
Ex: How to calculate mass of H2O to cool a system:
Calculate the mass of cold water at 10oC needed to cool to 30oC a 10 g piece of
glass at 95oC.
m1 = 10 g
ΔT1 = Tf1 - Ti1 = 30oC - 95oC = - 65oC
c1 = 0.84 J/(goC)
ΔT2 = Tf2 - Ti2 = 30oC - 10oC = 20oC
Ti1 = 95oC
- m1c1ΔT1 = m2c2ΔT2 ⇨ m2 = - m1c1ΔT1
Tf1 = 30oC
m2 = ?
c2 = 4.184 J/(goC)
c2ΔT2
m2 = - (10 g • 0.84 J/(goC) • -65oC) = 6.5 g
4.184 J/(goC) • 20oC
Ex: Identifying a Metal by Measuring Specific Heat
A 59.7 g piece of metal that had been submerged in boiling water was quickly transferred
into 60.0 mL of water initially at 22.0 °C. The final temperature is 28.5 °C. Use these data to
determine the specific heat of the metal. Use this result to identify the metal.
Solution
Assuming perfect heat transfer, the heat given off by metal is the negative of the heat taken
in by water, or:
qmetal= −qwater
In expanded form, this is:
Cmetal mmetal(Tf,metal−Ti,metal) = cwater mwater (Tf,water−Ti,water)
Cmetal mmetal (Tf,metal− Ti,metal) = cwater mwater(Tf,water− Ti,water)
Noting that since the metal was submerged in boiling water, its initial temp was 100.0 °C;
and that for water, 60.0 mL = 60.0 g; we have:
(cmetal)(59.7 g)(28.5oC−100.0oC)=(4.18 J/goC)(60.0 g)(28.5oC−22.0oC)
Solving this:
cmetal= −(4.184 J/goC)(60.0 g)(6.5oC)(59.7 g)(−71.5oC) = 0.38 J/goC
Our experimental specific heat is closest to the value for copper (0.39 J/g °C), so we identify
the metal as copper.
Heat Transfers - Calculating Tf
-Energy is always transferred from the hottest substance to the coldest until
equilibrium.
-To calculate an energy transfer between 2 systems:
-Q1 = Q2
OR
-Qloss = Qgain
-To calculate the final temperature of 2 systems in contact when Ti s are known:
Tf = m2c2Ti2 + m1c1Ti1
m1c1 + m2c2
Refer to your textbook p. 139 to see how this formula is derived.
Heat Transfers
Ex: a beaker contains 400.0 mL of water at 60.0 oC. A student adds 250.0 mL of
water at 20.0 oC. Find the temperature of the final mixture. (Assume that the
density of water is 1.0 g/mL).
-Qloss
=
Qgain
-m1c1ΔT1 = m2c2ΔT2
-400.0g • 4.19 J/(goC) • (Tf - 60.0oC) = 250.0g • 4.19 J/(goC) • (Tf - 20.0oC)
Tf = m2c2Ti2 + m1c1Ti1 = 44.62oC
m1c1 + m2c2
Problem # 13 from Heat Transfer Worksheet:
Determine the final temp when 450.2 g of aluminum at 95.2oC is placed in an
insulated calorimeter with 600.0 g of water at 10.0oC.
m1 = 450.2 g
Ti1 = 95.2oC
m2 = 600 g
Ti2 = 10.0oC
Tf = m2c2Ti2 + m1c1Ti1
m1c1 + m2c2
Tf = 600.0 g • 4.19 J/(goC)•10.0oC + 0.9 J/(goC)•95.2oC
450.2 g •0.9 J/(goC) + 600.0 g • 4.19J/(goC)
= 21.8 oC
Specific Heat Lab
https://youtu.be/8gHFaL2990U?t=10
Problem # 16 from Heat Transfer Worksheet:
A piece of metal with a mass of 15.3 g has a temp of 50.0oC. When the metal is
placed in 80.2 g of water at 21.0oC, the temp rises by 4.3oC. What is the specific
heat capacity of the metal?
This problem is exactly like your specific heat capacity lab!
mmetal = 15.3 g
mH2O = 80.2 g
Timetal = 50.0oC
TiH2O = 21.0oC
Tfmetal = 25.3oC
TfH2O = 21.0oC + 4.3oC = 25.3oC
ΔTmetal = 25.3 - 50.0 = -24.7oC
ΔTH2O = 4.3oC
Contin. of problem # 16 from Heat Transfer WS:
-Qloss = Qgain
-mmetal• cmetal • ΔTmetal = mH2O• cH2O• ΔTH20
-(15.3 g) c (-24.7oC) = (80.2 g)(4.19J/goC)(4.3oC)
cmetal = (80.2 g)(4.19J/goC)(4.3oC) =
-(15.3 g)(-24.7oC)
3.82 J/goC
Enthalpy & Enthalpy Change
●
●
●
●
Enthalpy (H) is the total energy of a system, that is, the sum of all the
potential & kinetic energies of the system, at constant pressure.
Enthalpy change (ΔH) is the energy exchanged (absorbed or released)
between a system & its environment during a physical change or chemical
reaction at constant pressure. Also referred to as heat of reaction.
It is impossible to measure an individual enthalpy or the total energy of a
system (too complex), but enthalpy change can be determined from the
energy changes of the surroundings.
Enthalpy changes occur during phase changes, chemical reactions & nuclear
reactions.
Enthalpy Change
ΔH = Hp - Hr
Where
ΔH = Enthalpy change or the reaction, expressed in kJ
Hp = Product enthalpy, expressed in kJ
Hr = Reactant enthalpy, expressed in kJ
Standard Molar Enthalpy Change
●
●
●
●
Standard molar enthalpy change (ΔHo), also called molar enthalpy, is the
enthalpy change that accompanies a reaction for one mole of substance at
SATP. (The standard state of a substance is the state at which it is at SATP).
Its unit of measure is kJ/mol.
ΔHo has been determined for a large number of physical & chemical changes:
○ ΔHo of vaporization of water = +40.66 kJ/mol at 100oC
i.e. 40.66 kJ
must be supplied to one mole of water to vaporize it at normal
atmospheric pressure.
○ ΔHo of fusion of ice = 6.01 kJ/mol
A positive sign of ΔH means that for the reaction to occur, heat must be
absorbed, & a negative sign means it releases heat.
Phase Changes & Enthalpy Changes
-During a phase change, like the fusion or melting of water, the ice can be
considered the reactant & the water the product in the formula for calculating
enthalpy change.
H2O(s) ➡ H2O(l)
●
●
Phase changes cause cause forces of attraction between particles to be
broken or formed.
In general, breaking the forces of attraction between atoms or molecules
requires energy, while forming new bonds releases energy.
●
When a substance changes from a solid to a liquid phase, there is a greater
separation of its particles. Consequently, fusion (melting), sublimation &
vaporization are endothermic changes, since they require energy to break the
forces of attraction between particles.
●
Conversely, reactions in which attractions between particles become greater
are exothermic & release energy.
Phase Changes & Enthalpy Changes
●
●
●
●
A phase change is a change in state of matter without any change in chemical
composition of the system.
Phase changes always involve energy changes but they never involve
temperature changes (in fact, the constant temp at which phase changes
occur is a characteristic property).
A graph of temperature vs. time as heat is transferred to a system is called a
heating curve.
When heat is transferred from a system, the temp-time graph is called a
cooling curve.
Heating Curve
Graph of Temp Vs Time for Phase Change of Water
Phase Changes
●
●
●
●
During phase changes the temp remains constant (even though heat is being
added or removed from the system) so no change in the average kinetic
energy (Ek) of the molecules occurs.
According to the chemical bonding theory energy is required to overcome
forces or bonds that hold particles together, so the heat flowing from the
surroundings works to separate the bonded particles, which increases the
potential energy (Ep) of the separated particles.
During the opposite phase changes (freezing & condensing) the stored Ep is
released as particles rearrange to form bonds.
Enthalpy change during any phase change is explained as change in the
chemical potential energy of the system.
Comparison of Energy Changes
Energy Change
Empirical Evidence
Theoretical Explanation
Flow of heat, q
A temp change with no
change in state or
chemicals
ΔEk as a result of an
increase or decrease in the
speed of particles
Phase change, ΔH
Exo or endothermic change
forming a new state of
matter
ΔEp as a result of changes
in the intermolecular bonds
among particles
Chemical reaction, ΔH
Exo or endothermic change
forming a new chemical
substance
ΔEp as a result of changes
in the ionic or covalent
bonds among ions or atoms
Nuclear reaction, ΔH
Exo or endothermic change
forming new elements or
subatomic particles
ΔEp as a result of changes
in the nuclear bonds among
nuclear particles (nucleons)
More on Molar Enthalpy Change
●
●
●
Recall that molar enthalpy change is the enthalpy change per mole of
substance undergoing the change
It is customary to include a subscript to indicate the type of change occurring:
Ex: ΔHvap for molar enthalpy of vaporization
To calculate an enthalpy change, the molar enthalpy value is obtained from a
reference source & the amount in moles of the substance undergoing the
phase change must be determined.
ΔHvap = nHvap
Ex of Calculating ΔH:
If 500 g of freon gas CCl2F2(l) vaporized at SATP, the expected enthalpy change
ΔHvap can be calculated as follows:
ΔHvap = nHvap
↙ From a table of Hvap
CCl2F2
= 500 g x
1 mol
120.91 g
From the periodic table ↑
x
34.99 kJ
1 mol
= 145 kJ
Refer to your textbook, p. 151
Graph of Temp Vs Time for Phase Change of Water
As a pure substance is heated, the following changes can occur:
-an increase in temp of the solid (Ek = mcΔT)
-a solid to liquid phase change (Ep =nΔH), fusion point or solidification point
-an increase in the temp of the liquid (Ek = mcΔT)
-a liquid to gas phase change (Ep =nΔH)
-boiling pt. & an increase in temp of the gas
(Ek = mcΔT)
Total Energy Changes to a System
●
The total energy of a system may be calculated by adding up the the energy
associated with any temperature changes and phase changes that it
undergoes.
ΔEtotal =
q
(cooling/heating)
+
ΔH
(phase change)
+
q
(cooling/heating)
Example:
Calculate the total energy change for 1000 g of molten iron at 1700oC that
changes to solid iron at 80oC. The specific heat capacity of liquid iron is 0.82
J/(goC) & that of solid iron is 0.52 J/(goC). The molar enthalpy of solidification for
iron is 15 kJ/mol at 1535oC.
ΔEtotal = qliquid cooling+
= mcΔT
+
ΔHfreezing + qsolid cooling
nHfr
+
mcΔT
= 1000 g • 0.82 J/goC • (1700 - 1535)oC + 1000 g • 1 mol • 15 kJ
55.85 g 1 mol
+
1000 g • 0.52 J/goC • (1535 - 80)oC
= 1.16 MJ
Potential Energy Diagrams
●
●
●
A potential energy diagram is a theoretical description of an enthalpy change.
Potential energy is stored or released as the positions of particles change
during a phase change.
The enthalpy change in any change of state is theoretically described as an
increase or decrease in potential energy:
ΔEp = ΔH
molecules
●
●
system
An increase in PE of molecules describes an endothermic process in a
system.
A decrease in PE describes an exothermic process.
Energy Balance & Chemical Equations
●
●
The energy balance is the sum of the energy required to break the chemical
bonds of the reactants & the energy released at the formation of the product
bonds.
The enthalpy change for any reaction is = the energy changes associated
with the breaking & making of bonds in a reaction.
ΔH = ΔHbonds broken + ΔHbonds formed
Note: ΔH is expressed in kJ/mol
There are several ways to represent molecules &
their structures:
Propane (C3H8):
Lewis dot diagrams are helpful to visualize presence
of single, double or triple bonds:
2H2
●
●
+
O2
→
2H2O
Bond strength depends on atoms that come into play & the # of electrons
shared between them..
A double bond (where 4 electrons are shared) is stronger than a single bond.
Instructions For Drawing Lewis Dot Diagrams
Watch the following tutorial on how to draw Lewis dot structures for compounds in
5 simple steps:
https://www.google.ca/search?q=how+to+draw+lewis+structures+5+easy+steps&r
lz=1C1GGGE_enCA533CA533&oq=How+to+draw+lewis+structures+in+5&aqs=c
hrome.1.69i57j0.23220j0j8&sourceid=chrome&ie=UTF-8
It is necessary to refer to a table of bond energies:
Refer to Table 8.5 in Appendix 8 of your textbook.
Example of how to calculate enthalpy change of a
reaction involving simple bonds:
The following is a reaction between hydrogen gas (H2) gas & chlorine (Cl2) gas:
H2(g) + Cl2(g) → 2 HCl(g)
Calculate the enthalpy change of the reaction by performing an energy balance,
then indicate if the reaction is endo or exothermic.
Solution:
1.
Determine the bonds broken using Lewis dot diagrams:
1 Cl-Cl single bond
1 H-H single bond
2. Determine the bonds formed on product side using Lewis dot diagram:
2 H-Cl bonds
3. Refer to table of average bond energies:
EH-H: 436 kJ/mol
ECl-Cl: 243 kJ/mol
EH-Cl: 432 kJ/mol
4. Calculate the total bond energy of the reactants & products:
Reactants: ΔHbonds broken = EH-H + ECl-Cl = 436 kJ + 243 kJ = 679 kJ
Products: ΔHbonds formed = -(2 x EH-Cl) = -(2 x 432 kJ) = -864 kJ
5. Calculate the enthalpy change of the reaction by making an energy balance.
ΔH = ΔHbonds broken + ΔHbonds formed = 679 kJ - 864 kJ = -185 kJ
The reaction is exothermic since enthalpy change is -185 kJ for 2 mol of HCl, that
is, -92.5 kJ/mol.
Refer to your textbook p. 158 for an energy balance diagram of this reaction.
Example of how to calculate enthalpy change of a
reaction involving double bonds:
Below is the combustion of methane (CH4).
CH4(g) + 2O2(g) → CO2(g) + 2H2O(g)
Calculate the enthalpy change of this reaction by performing an energy balance &
indicate if the reaction is endo or exothermic.
1.
Determine bonds broken:
4 C-H bonds in CH4
2 O=O bonds in 2 O2
2. Determine the bonds formed:
2 C=O bonds in CO2 & 4 O=H bonds in 2 O2
3. Find the bond energies:
EC-H: 413 kJ/mol
EO=O
: 498 kJ/mol
EC=O: 745 kJ/mol
EO-H: 460 kJ/mol
4. Calculate the total bond energy of the reactants & products:
Reactants: ΔHbonds broken = 4 x EC-Hl + 2 x EO=O = 4 x 413 + 2 x 498 = 2648 kJ
kJ
Products: ΔHbonds formed = -(2 x EC=O+ 4 x EO-H) = -(2 x 745 + 4 x 460) = -3330
5. Calculate the enthalpy change of the reaction by making an energy balance.
ΔH = ΔHbonds broken + ΔHbonds formed = 2648 kJ - 3330 kJ = -682 kJ
The reaction is exothermic since enthalpy change is -682 kJ for 2 mol of methane,
CH4, that is, -682 kJ/mol.
Refer to your textbook p. 159 for an energy balance diagram of this reaction.
NOTE: You may use ΔH = ΔHbonds broken - ΔHbonds formed if you have not already assigned a ‘+’ or ‘-’ to your
ΔH values.
Calculating Enthalpy Change Using Stoichiometry:
●
●
Allows us to obtain the ΔH that accompanies a chemical reaction whose mass
of reactants & products is known.
For reactions carried out in a lab, it is important to know the energy involved
in a reaction for the quantities of reactants & products actually used.
Example of Calculating ΔH using Stoichiometry:
Consider the following reaction:
H2(g) + F2(g) ➝ 2HF(g) + 546.6 kJ
Calculate the energy released during this reaction if a mass of 5.50 g of hydrogen
(H2) is used with a sufficient amount of fluorine(F2).
Data:
Solution:
m = 5.50 g
1. Calculate the # of moles of H2:
M = 2.016 g/mol
Energy = ?
M=m
n
n=m=
M
5.50 g
2.016 g/mol
= 2.73 mol
H2(g) + F2(g) ➝ 2HF(g) + 546.6 kJ
5.50 g
2.73 mol
?
2. Calculation of the energy released for 2.73 mol of H2:
-546.6 kJ =
1 mol
? kJ
2.73 mol
? kJ = -546.6 kJ • 2.73 mol = -1492 kJ
1 mol
The reaction carried out with 5.50 g of H2 releases 1492 kJ of energy.
Another Example:
Ethane (C2H6) is one of the substances that makes up natural gas. Its combustion
occurs according to the following equation:
2C2H6(g) + 7O2(g) ➝ 4CO2(g) + 6H2O(g) + 2857 kJ
Calculate the energy needed to produce 10.0 g of water.
Data:
Solution:
m = 10.0 g
1. Calculate the # of moles of H2O:
M = 18.015 g/mol
M=m
Energy = ?
n
n=m=
M
10.0 g
18.015 g/mol
= 0.555 mol
2C2H6(g) + 7O2(g) ➝ 4CO2(g) + 6H2O(g) + 2857 kJ
10.0 g
? kJ
0.555 mol
? kJ
2. Calculation of the energy needed to produce 0.555 mol of H2O:
2857 kJ =
6 mol
? kJ
0.555 mol
? kJ = 2857 kJ • 0.555 mol = 264.3 kJ
6 mol
The reaction that produces 10.0 g of water releases 264 kJ of energy.