Chemistry 30
Energy Changes in Chemical Reactions
Reference:
Theory:
Chapter 17, Thermodynamics
1.
Differentiate between heat and temperature.
2.
What is the assumption made when a thermometer is used ?
3.
Define the term specific heat.
4.
Use heat equations to calculate :
a)
b)
c)
changes in heat content of a substance as a result of heating and cooling, and
changes of state.
mass of substances given heat and temperature change data.
specific heat of a substance given heat, temperature change and mass data.
5.
Discuss applications of the heat of vaporization and the heat of fusion.
6.
Discuss molecular explanation for heat changes.
7.
Discuss the concept of heat of reaction in terms of reorganization of bonds.
8.
Discuss Hess's Law.
9.
Use Hess's Law and standard heat of formation to calculate energy changes resulting from
chemical reactions.
10.
Be able to draw a potential energy diagram.
11.
Use the concepts of enthalpy and entropy to discuss Gibb's free energy and the
spontaneity of chemical reactions.
Vocabulary:
thermodynamics
specific heat
condensation
heat of fusion
kilocalorie
Standard heat of formation
temperature
melting
heat of vaporization
calorie
Joule
entropy
heat
fusion
vaporization
endothermic
exothermic
enthalpy
Formulas:Changes in heat content as a result of temperature change:
Amount of heat = (mass)(Specific heat )(temperature change)
Q =m c T
2
Changes in heat content as a result of change of state:
For melting and fusion:
Amount of heat = heat of fusion x mass
Q = Hf x m
For condensation and vaporization:
Amount of heat = heat of vaporization x mass
Q = Hv x m
Changes of heat content as a result of chemical change:
See text, p 481 to 484, as well as additional notes.
Significant Charts:
Specific heat of various substances - handout
Standard heat of formation for several reactions - handout
3
CHEMISTRY 30
THERMODYNAMICS
Themochemistry: is the study of the transfers of energy as heat that
accompany chemical reactions and physical reactions.
Laws of Heat Movement
a) 1st law of Thermodynamics: Law of conservation of energy.
Energy is never lost or gained , it only changes form. This law states
that energy can change form but is never really lost in any closed system.
When an exothermic reaction occurs some of the molecular enthalpy
(energy in the molecules ) of the system is converted into heat or light.
This energy is then released by the system to the surroundings. The total
energy of the system decreases, but the energy of the surroundings
increases by the same amount.
Energy ( system) = - Energy (surroundings)
Energy is the capacity to do work.
Energy comes in many forms:
a)
kinetic energy is the energy of motion.
b)
potential energy is the energy stored in stressed objects, like a
bent bow or compressed spring, or water at the top of a
waterfall.
c)
chemical energy is the energy stored in chemical bonds,
which is released when a chemical reaction, like burning,
occurs.
d)
nuclear energy is the energy stored within the nucleus of an
atom, which can be released under certain conditions when the
4
nucleus is made to break apart (fission), or when two nuclei are
forced together (fusion).
Heat can be thought of as the energy transferred between samples of
matter because of a difference in their temperature.
Heat is the product of motion. All matter has heat in it since it is
composed of particles which move. When particles stop moving they do
not contain any heat. This only occurs at a special temperature called
absolute zero.
Absolute zero is the lowest temperature that can be reached:
- it is equal to - 273 degrees Celsius
- it is also equal to 0 Kelvins (K)
b) 2nd law of Thermodynamics:
Heat Energy always travels spontaneously from a warmer body (body
with higher temperature) to a colder body.
-droping a hot rock in water
-frostbite
Different masses or volumes of a substance require different
amounts of heat to produce a similar rise in temperature.
Different substances of the same mass require different amounts of
heat to produce a similar rise in temperature. This is called the
SPECIFIC HEAT CAPACITY: is the amount of heat required to raise
the temp of one gram of a substance by one degree Celsius or Kelvin.
Every substance has a different specific heat capacity and is measured in
Joules per gram degree Celsius (J/g·°C) or (J/g·K). P 513 table 17-1
Heat and temperature are related, but are not the same things:
5
Temperature is a measure of the average kinetic energy of a substance; it
looks at how fast the particles of a substance are moving. The faster the
particles are moving, the higher is the temperature.
To determine the amount of heat of a substance, the formula used:
Q = m c ΔT
Q = mc (Tf - Ti)
where
Q = the amount of heat in joules (J)
c = the heat capacity of the substance in joules
per kilogram degree Celsius (J/g·°C) or
Kelvin
m = the mass of the body in grams (g)
Tf = the final temperature
Ti = the initial temperature
Note that we are concerned with the temperature change, not the
temperature at which the substance happens to be.
Ex: Determine the heat required to raise the temperature of 100.g of
water from 298.0 K to 373.0 K .
Q = m c ΔT
Q = 100.g (4.184 J/g K) ( 373.0 K – 298.0 K)
Q = 418.4 J/K (75 K)
Q = 31350 J
Q = 31.4 kJ
Heat Capacity: is the heat energy reqired to raise the temperature of a
given quantity of a substance by one degree.
Heat capacity = (specific heat )( mass)
Ex: Heat capacity for 12.0 g of Aluminum
Heat capacity = (0.903 J/g °C)( 12.0g)
6
= 10.836 J/°C
= 10.8 J/°C
Ex: When 1.5 x 103 J of heat energy is absorbed by a beaker of water,
its temperature rises by 3.1°C . What is the heat capacity ( mc) of the
beaker of water?
Q = (m c) ΔT
Heat capacity= Q
ΔT
Heat capacity= 1.5 x 103J
3.1°C
Heat capacity = 480 J/°C
Molar Heat Capacity: ( Cp0 ) of a substance is the amount of energy
required to raise the temperature of one mole of a substance by one
Celsius degree.
Molar heat capacity = (specific heat) (molar mass)
Ex: Calculate the molar heat capacity of water, given that the specific
heat water is 4.184 J/g°C.
Cp0 = (specific heat)(molar mass)
= (4.184 J/g°C)(18.02g/mol)
= 75.40 J/mol °C
Heat released or absorbed during reaction = Q
Number of moles = n
Molar enthalpy (enthalpy change per mole) for the substance in the
reaction = H substance
The molar enthalpy for many substances in certain reactions have been
determined by chemists and recorded in tables like the one below. (Such
tables can be found in many handbooks or textbooks)
7
For example:
1) Vaporization (phase change from liquid to gas) of the substance or
2) Fusion (phase change from solid to liquid) of the substance.
Molar Enthalpy's of Vaporization and Fusion
(under standard conditions)
H vap (kJ /
H fus (kJ/
Substance Formula
mol )
mol )
Ammonia
NH3
+ 23.3
+ 5.66
Ethanol
C2H5OH
+ 38.6
+ 4.94
Methanol
CH3OH
+ 35.2
+ 3.22
Water
H2O
+ 40.7
+ 6.01
3) Combustion (burning of substance in oxygen to produce carbon
dioxide and water) of the substance.
Molar enthalpies of combustion
for selected substances
H
Substance Formula
combustion
Carbon
C
- 394
(graphite)
Ethanol
C2H5OH
- 1367
Glucose
C6H12O6
- 2800
Hydrogen
H2
- 286
Methane
CH4
- 891
Methanol
CH3OH
- 726
Propane
C3H8
- 2219
4) For other substances in reactions where molar enthalpies are not
recorded we can use the following procedure to calculate them.
8
To determine the H substance (kJ /mol) when given an
equation we need only divide the
H given by the equation by the balance of the substance.
Question:
What is the molar enthalpy of CO2 (g) in the reaction
for the burning of butane below?
2 C4H10 + 13 O2
8 CO2 + 10 H2O
H = -5315 kJ
Answer:
Molar enthalpy is the enthalpy change in equation
divided by the balance of CO2
Molar enthalpy,
H
substance
= 5315 kJ ÷ 8 mol
= 664 kJ / mol.
To solve heat calculations of enthalpy
using the equation method we need to do
the following:
1) Determine the information given.
Important 2) Calculate the # of moles substance and
procedure the molar enthalpy from the equation, or
take it from a table of values, if needed.
3) Plug values into the equation and
calculate missing value.
Question:
How much heat will be released if 65
grams of butane is burned in a lighter
according the equation in the example
above.
Answer:
1) Given
65 grams of butane
9
H reaction = 5315 kJ
2) a) Moles of butane
mass ÷ gram molecular weight ( butane )
65 grams ÷ 58.14 g/mol = 1.12 moles
b) Molar enthalpy
5315 kJ ÷ 2 mol C4H10 = 2657.5 kJ/mol
3) Plug values into equation.
Q = n * H substance
Q = 1.12 mol * 2657.5 kJ/mol = 2976.4 kJ
Considering significant figures = 3.0 MJ
Calculate the heat released
when 120 grams of Iron (III)
oxide is formed by the
following equation
2 Fe2O3(s)
4 Fe(s) + 3 O2(g)
H = 1625 kJ
1) Given information
mass of Iron(III) oxide = 120 grams
H reaction = 1625 kJ
2) Required information
Heat released (Q)
3) Calculations
a) Calculate molar enthalpy: H substance
H substance = 1625 kJ ÷ 2 mol of Fe2O3 = 812.5 kJ/mol
10
b) Calculate moles of Iron(III) oxide
Moles = mass ÷ molar mass (gram molecular wt)
Moles = 120 g ÷ 159.70 g/mol = 0.7514 mol.
4) Plug values into formula and solve
Q = n H substance
Q = (0.7514 mol )( 812.5 kJ/mol) = 610.5 kJ
Considering Significant figures = 610 kJ
Calculate the answers to the following questions using the
equation method.
1) How much heat is produced when 95 grams of methane is burned in
oxygen by the following equation?
CH4(g) + 2O2(g)
CO2(g) + 2H2O(g)
H = - 891 kJ
Answer: 5300 kJ
2) What mass of carbon dioxide must form to create 1200 kJ of heat
when the following reaction occurs?
C6H12O6(s) + 6O2(g)
6CO2(g) + 6H2O(l)
H= - 2808kJ
Answer: 110 grams
3) What mass of oxygen is needed to completely react and release 550 kJ
of heat in the following reaction?
4Fe(s) + 3O2(g)
Answer: 32 grams
2 Fe2O3 (s)
H
=
- 1625 kJ
11
Thermochemistry Heat Problems:
Q = mcΔT
1. If 10.5 g of iron, at 25°C, absorbs 128 J of heat, what will be the final temperature of the
metal? Specific heat of Iron is 0.449 J/g°C. Ans: 52°C
2. Calculate the molar heat capacity of ethanol, C2H5OH(l). The specific heat of ethanol is
2.46 J/g°C. Ans: 113 J/ mol °C.
3. Calculate the molar heat capacity of mercury. The specific heat of mercury is 0.139
J/g°C. Ans: 27.9 J/ mol °C.
4. Calculate the heat necessary to raise 27.0 g of water from 10.0°C to 90.0°C. Cp of H2O is
4.184 J/g°C. Ans: 9.04 kJ.
5. How many kJ are required to heat 45.0 g of water at 25 °C and then boil it all away?
Specific heat of H2O is 4.184 J/g°C. Ans: 14 kJ.
6. When 15.0 g of steam drops in temperature from 275.0 °C to 250.0 °C, how much heat
energy is released. Specific heat of steam is 1.87 J/g°C.
Ans: 758 J.
7. A certain mass of water was heated with 41840 J, raising its temperature form 22.0 °C to
28.5 °C. Find the mass of the water. Ans: 1540 g H2O.
8. If it takes 41.72 J to heat a piece of gold weighing 18.69 g from 10.0 °C to 27.0°C, what
is the specific heat of gold? Ans 0.131 J/g°C.
12
Heat Changes in Chemical Reactions
The difference between physical change and chemical change is
that chemical change involves bond breakage, while physical change
does not. As mentioned before, chemical bonds involve energy, and
energy is the main reason why chemical reactions occur.
This energy relationship in chemical reactions is summarized by
the term enthalpy. Enthalpy is the energy difference between reactants
and products in a chemical reaction; that is, the difference in bond
energies between reactants and products. Nature's aim is to achieve the
lowest possible energy state, (lowest enthalpy). In chemical reactions
this means that compounds with the lowest possible energy in the bonds
are favoured.
In a chemical reaction if the products have a lower energy in their
bonds (enthalpy) than the reactants, energy is released to the
environment as the reaction progresses. This reaction is exothermic.
If the products have a higher enthalpy than the reactants, energy
from the environment flows into the system as the reaction progresses.
The reaction is endothermic.
Enthalpy is represented by the symbol H.
(A change in enthalpy is H.)
 In an exothermic reaction energy is released into the environment;
energy is lost from the system, therefore, H is negative.
 In an endothermic reaction energy is absorbed from the
environment; energy is gained by the system. H is positive.
It is impossible to calculate the absolute enthalpy (H) of any
molecule. It is possible only to determine enthalpy change (H). This
can be done in two ways, by calorimetry or through the use of tables.
Calorimetry involves actually mixing the chemicals and measuring the
energy released or absorbed.
13
1) Using a balanced equation and energy as a term.
There is a direct relationship between the amount of substances that
reacts and forms in an equation. The factor that determines the exact
relationship is the equation factor or mole ratio.
2H2 (g) + 1 O2 (g) ------> 2 H2O (l)
The heat absorbed and produced in a chemical reaction also varies
directly as the amount of substance that reacts and the exact amount is
determined by the heat change for the reaction ( H) .
2H2 (g) + 1 O2 (g) -----> 2H2O (l)
If one mole of oxygen when reacted releases 572 kJ of energy ,
then 2 moles will release 1046 kJ
 If the reaction is exothermic the energy is written the products.
 If the reaction is endothermic the energy is written in the reactants.
Example:
a) Exothermic: written in the products.
2Al (s) + 3Cl2 (g) ------> 2 AlCl3 (s) + 1408 kJ
If we double the amount of the reactants and products, the heat doubles
4Al (s) + 6Cl2 (g) ------> 4 AlCl3 (s) + 2816 kJ
b) Endothermic: written in the reactants.
H2O + 286 kJ ------> H2 + 1/2 O2 (g)
If we double the amount of the reactants and products we double the
heat.
2 H2O + 572 kJ ------> 2 H2 + 1 O2 (g)
2) " H" notation
-used by chemists to represent enthalpy change is called " H" notation.
In this case the enthalpy is written separately from the reaction.
14
If the reaction is exothermic the H is negative.
If the reaction is endothermic the H is positive.
a) Exothermic: (- H value)
2Al (s) + 3Cl2 (g) ------> 2 AlCl3 (s)
H = - 1408 kJ
b) Endothermic: (+ H value)
2 H2O ------> 2 H2 + 1 O2 (g)
H = + 572 kJ
What would happen if we cut the # of moles of reactant and product
in half in each of the above reactions?
Assignment; Communicating Enthalpy
1) For each of the following rewrite the equation in " H " notation,
for one mole of the underlined substance.
Fe2O3
+ 3CO(g)
(s)
3CO2(g) + 2Fe(s) + 25 KJ
Answer:
1/3
Fe2O3 + CO(g)
(s)
CO2(g) +
2/3
Fe(s)
H = - 8.3
KJ
4 NH3
5 O2
4 NO
6 H2O
+
+
+ 1170 KJ
(g)
(g)
(g)
(l)
2 HCl
96
Cl2
2.
+
H2(g) +
(g)
KJ
(g)
N2
3 H2
2
92
3.
+
+
(g)
(g)
NH3(g)
KJ
2 CO2
2 CO
O2
4.
+ 566 KJ
+
(g)
(g)
(g)
5. 4 Al + 3 O2
2 Al2O3 + 3360
1.
15
(s)
(g)
(s)
KJ
For each of the following reactions written in " H" notation
rewrite the equation using whole number balances and energy as a
term.
3
FeCl3
(s)
3
+
FeCl2(s)
3/2
Cl2
(g)
H
=
+ 173
KJ
Answer;
6 FeCl3
+
(s)
346
KJ
6
+
FeCl2(s)
3 Cl2
(g)
Questions;
1/2 H2
1/2 Br2
+
(g)
(g)
SO2
2. H2O(g) +
(g)
1/2
3/4 C
3.
+
Fe2O3
(s)
(s)
1.
1/2 C2H5OH 3/2 O2
4.
+
(l)
(g)
1/2 N2H4
5.
(g)
1/2 O2
+
(g)
HBr
- 36
H
=
(g)
KJ
H2S
+ 3/2
+ 520
H =
(g) O2 (g)
KJ
Fe
3/4 CO2
+
(s)
(g)
CO2
+
(g)
3/2
H2O
(l)
1/2 N2
HO
+ 2
(g)
(l)
H + 117
=
KJ
H
=
684
KJ
H
1530
=
KJ
16
Heats of Formation of Several Chemical Reactions (kJ/mol)
Elements
Product Name
Heat of reaction
(kJ/mol of product)
H2(g) + 1/2 O2(g)  H2O(g)
water vapor
- 242
H2(g) + 1/2 O2(g)  H2O(l)
water
- 286
sulfur dioxide
- 297
H2(g) + S(s) + 2 O2(g)  H2SO4(l)
sulfuric acid
- 812
S(s) + 3/2 O2(g)  SO3(g)
sulfur trioxide
- 395
1/2 N2(g) + 1/2 O2(g)  NO(g)
nitric oxide
+ 90.4
1/2 N2(g) + O2(g)  NO2(g)
nitrogen dioxide
+ 33.1
1/2 N2(g) + 3/2 H2(g)  NH3(g)
ammonia
- 46.0
C(s) + 1/2 O2(g)  CO(g)
carbon monoxide
- 110
S(s) + O2(g)
 SO2(g)
C(s) + O2(g)
 CO2(g)
carbon dioxide
- 394
C(s) + 2 H2(g)
 CH4(g)
methane
- 74.5
2 C(s) + 3 H2(g)  C2H6(g)
ethane
- 84
3 C(s) + 4 H2(g)  C3H8(g)
propane
- 104
1/2 H2(g) + 1/2 I2(g)  HI(g)
hydrogen iodide
+ 25.9
4 C(s) + 4 H2(g) + O2(g)  C3H7COOH(l) propionic acid
- 522
2 C(s) + 2 H2(g) + O2(g)  CH3COOH(l) acetic acid
- 487
2Al(s) + 3/2 O2(g)  Al2O3(s)
aluminum oxide
-1676
Cu(s) + 1/2 O2(g)  CuO(s)
copper (II) oxide
- 155
Mg(s) + 1/2 O2(g)  MgO(s)
magnesium oxide
- 602
Na(s) + 1/2 Cl2(g)  NaCl(s)
sodium chloride
- 411
This table lists the heats of formation of several compounds from
their elements. Some reactions are exothermic, others are endothermic.
17
This table has uses, however, far beyond simply looking at the H of the
listed reactions. To deal with this we must first discuss Hess's law of
constant heat summation, or the law of additivity of heats of reaction.
Hess suggested that the sum of the enthalpies of
Hess's the steps of a reaction will equal the enthalpy of
Law the overall reaction.
For instance, carbon and oxygen can react to form carbon monoxide:
(1)
C(s) + 1/2 O2(g)  CO(g)
H =
- 110 kJ
Carbon monoxide can react further with oxygen to form carbon dioxide:
(2)
CO(g) + 1/2 O2(g)  CO2(g)
H = - 294 kJ
A third possible reaction involves carbon and oxygen reacting to form
carbon dioxide directly:
(3)
C(s) + O2(g)
 CO2(g)
H = - 394 kJ
You will note that the heat of reaction from (1) plus the heat of reaction
from (2) exactly equals the heat of reaction from (3). This makes sense
when you consider what is going on; chemically, reactions (1) and (2)
perform in two steps what reaction (3) does in one step. Note what
happens when reactions (1) and (2) are added up:
C(s) + 1/2 O2(g)  CO(g)
H =
- 110 kJ
CO(g) + 1/2 O2(g)  CO2(g) H = - 294 kJ
C(s) + O2(g)
 CO2(g)
This result is stated by Hess's Law:
H = - 394 kJ
18
The enthalpy change for any reaction depends only on the
products and reactants and is independent of the pathway or
the number of steps between the reactant and product.
This law allows us to take several reactions from the heats of formation
table, add them up and get H values for reactions that do not appear
directly on the table.
Calculating Heats of Reaction using Hess's Law
1) Write the overall equation for the reaction if not given.
2) Manipulate the given equations for the steps of the reaction so they
add up to the overall equation.
3) Add up the equations canceling common substances in reactant and
product.
4) Add up the heats of the steps = heat of overall reaction.
Calculate the H of the following reactions using Hess's Law
and the steps indicated
1) Calculate the heat released by the burning of sulfur in oxygen
given the following steps
2S(s) +
3O2
(g)
Given Steps
Step
S(s) +
1
Step
2
2SO3(g)
O2
(g)
SO2
(g)
2SO2
+ O2 (g)
(g)
2SO3
(g)
Answer:
1) The overall reaction is written above.
H = ?
-297
kJ
H= 198
kJ
H=
19
Note that sulfur (S) is a reactant and sulfur trioxide (SO3) is the
product. Therefore neither reaction in steps 1 or 2 needs to be
reversed.
Note that the overall equation has 2 moles of S and 2 moles of
SO3. Therefore the first reaction must be multiplied by two , but
the second can be left alone.
2) Manipulate equations
2SO2
Step
2
2O2
H = - 594
+
1
S(s)
(g)
kJ
(g)
2 SO3
Step
2
H = - 198
+ O2 (g)
2
SO2(g)
kJ
(g)
Substances in bold above will be products
3) Add equations and heats
2S
3O2
H = 2SO2 + 2SO3
Addition:
+
+ 2SO2
(s)
(g)
(g) 792 kJ
(g)
(g)
Substances in bold above are common and are therefore canceled
2S
3O2
2 SO3
H = Net
+
(s)
(g)
(g)
792 kJ
2) Calculate the H of the following reaction, which describes the
production of coal gas from carbon, given the steps below.
H2O (g) + C (s)
CO (g) + H2 (g)
Given Steps
H2
(g)
+
2CO
(g)
1/2
O2
(g)
2 C
(s)
+
H2O (g)
H =
- 242.0
kJ
O2 (g)
H =
+ 221.0
kJ
Answer:
1) The overall equation is written above.
Note that the H2O (g) is a reactant and in step #1 it is a product.
Thus step one needs to be reversed and the sign of the heat
20
changed.
Note that 1 mole of CO (g) is a product in the overall equation, but
2 moles of CO (g) is a reactant in step 2 . Thus step two needs to
be reversed, and divided by two.
2) Manipulated equations
1/2 O2
H2O
H2 (g) +
H = 242.0
(g)
(g)
kJ
+
1/2 O2
CO (g)
C (s) +
H = 110.5
(g)
kJ
Substances in bold above will be products
3) Add equations and heats
1/2
1/2
H2O
C
H
CO
H = -131.5
+
+ O2
O2
+ 2
+
(g)
(s)
(g)
(g) kJ
(g)
(g)
Substances in bold above are common and therefore are canceled
Net Equation:
H2O
C
H2
CO
H = +
+
(g)
(s)
(g)
(g)
131.5 kJ
Top of Page
3) Calculate the heat of reaction for the following equation
C3H8 (g) + 5 O2 (g) -----> 3 CO2 (g) + 4 H2O(g)
given the following steps in the reaction mechanism.
Step #1: 3C (s)+ 4 H2 (g) -------> C3H8 (g)
Step #2: 2 H2(g) + O2 (g) -------> 2H2O (g)
Step #3: C (s) + O2 (g) --------> CO2 (g)
Answer
21
1) Overall balanced equation is written above. Note that C3H8 is a
reactant and there are 3 moles of CO2 and 4 moles of H2O as
products.
2) We can manipulate the equations by:
a) Reversing equation #1
b) Multiplying equation #2 by 2
c) Multiplying equation #3 by 3
manipulated equations we get
Manipulated equations
4 H2
C3H8 (g)
+ 3C (s)
(g)
4 H2 O
2 O2
4 H2 (g) +
(g)
(g)
3 C (s)
+
3 O2
(g)
3 CO2
(g)
H
+103.8
kJ
-968
kJ
1180.5
kJ
the substances in bold print are products
3) Adding the equations we get
H = 2044.7 kJ
C3H8
(g)
+
4 H2
3
5O2
+
+
(g)
(g)
C(s)
3 C
4 H2
4 H2O
+
+
+ 3 CO2 (g)
(g)
(s)
(g)
the substances in bold print are common substances and are
canceled to produce the net equation
C3H8
5O2
4 H2O
3 CO2
H = +
+
(g)
(g)
(g)
(g)
2044.7 kJ
22
Enthalpy and Entropy
Enthalpy (H) has been discussed. It is the heat content of a
compound, found in the bonds. Nature tends towards a minimum
energy.
-so a reaction with a negative H will be spontaneous.
Entropy (S) is the amount of disorder in a system. Nature does
not like to be organized. Thus a reaction which involves substances
going from a less disordered state to one which is more disordered will
be favoured.
-Note we are discussing disorder; an increase in entropy involves an
increase in the disorder of a system.
-A reaction with a positive entropy change (S) will be more likely
to go forward spontaneously.
a)
b)
c)
a gas has more disorder than a liquid which has
more disorder than a solid.
the side with more particles is more disordered.
a dissolved substance is more disordered than a
solid.
Gibb's Free Energy (G). This is a measure of the maximum
possible work that can be obtained by a reaction.
G = H - TS
T stands for temperature in Kelvins.
 G < 0 (NEGATIVE): reaction is spontaneous
 G = 0: reaction is at equilibrium
 G > 0 (POSITIVE): reaction is not spontaneous
There are four combinations of enthalpy and entropy possible, each
with a distinct effect on Gibb's free energy:
23
H
S
positive
positive
G will be negative only if TS is very
large; spontaneous only at high temperature.
positive
negative
G cannot be negative under any conditions.
This reaction will not proceed
spontaneously.
negative
positive
G will be negative under all conditions.
This reaction will proceed spontaneously.
negative
negative
G will be negative only if TS is very
small; this reaction will be spontaneous only
at low temperature.
Effect on the Reaction
Calculate using Gibb's Free energy if the following reaction
spontaneous at 25°C if H = - 40.8 KJ /mol and S = 118.9 J/Kmol ?
H2O (l) <----> H2O (g)
Always change the temperature to Kelvin
25 C = 298.14 K
G = H - TS
G = (-40800 J/mol) - (298.15 K )( 118.9 J/K mol
= - 76250 J
= - 76.3 KJ
If the value is negative the reaction is spontaneous, even at 25 C.
24
Bond Energies (for chemical reactions)
When chemical change occurs bonds are rearranged. The bonds in the
reactants are broken and new bonds are formed to produce the products.
The energy required to break these bonds is known as the bond
dissociation. The heat of the reaction (∆H) is the difference between the
energy needed to break the bonds in the reactants and the energy
released in the formation of the bonds in the products.
∆H is always positive when the bonds are broken because energy is
absorbed during the process. The stronger the bond the more energy
needed to break the bond.
Ex:
a) The dissociation of hydrogen (hydrogen gas becoming 2 separate
hydrogen atoms).
H2 (g)
2 H (g) H = + 436.4 kJ
The bond dissociation for a H __ H bond is 436.4 kJ.
b) The bond energy of a C__ H bond in methane
CH4(g)
C(s) + 4 H (g)
H = + 1650 kJ
Since four C __ H bonds have to be broken the average energy for each
bond is + 412.5 kJ.
When the same bond is formed the bond energy is equal and opposite.
Ex: a C __ H bond forms, the energy released is - 412.5 kJ.
Hess's Law formula using bond energies
H
=
∑
Energy of
bonds
broken
__
∑
Energy of
bonds made
25
Calculating H using bond energies.
Step 1) Determine the number (in moles) and type of bonds broken and
formed from the balanced equation.
Step 2) Multiply the number of bonds by the average bond energy given
in the table to determine the energy change.
Step 3) Plug values into the above formula to calculate the H of the
overall reaction.
Example : Calculate the energy of the reaction
for the burning of methane in oxygen to form
carbon dioxide gas and water gas, using heats of
formation. The balanced equation is given below.
Use the following list of bond energies
C3H8(g) + 5 O2(g) ----> 3 CO2 (g) + 4 H2O(g)
Steps 1 and 2
1) Determine the number and types of bonds
broken and formed.
2) Determine
Bonds Broken
Type #
C - C 2
C - H 8
O = O 5
the energy change.
(reactants)
Bond Energy
347 kJ/mol
413 kJ/mol
498 kJ/mol
Total
Bonds formed (products)
O - H 8
464
C = O 6
805
Total
Energy
+ 694
+3,320
+2,490
+6,488
kJ
kJ
kJ
kJ
3,712 kJ
4,830 kJ
8,542 kJ
26
Step 3)
H = ∑Energy of Bonds Broken - ∑Energy of Bonds
formed
H = 6,488 kJ - 8,542 kJ
H = - 2,054 kJ
Ex
1/2H2(g) + 1/2 Cl2(g) ---->
HCl (g)
∆H = ∑Energy of Bonds -∑Energy of Bonds
Broken
formed
∆H
∆H
= {1/2(H - H) + 1/2(Cl - Cl)}-{(H - Cl)}
= {1/2(435kJ) + 1/2(243kJ)}-{(431kJ)}
= 339 kJ – 431 kJ
= -92 kJ
Try:
Use the average bond energies to calculate the value of ∆H for the
reaction:
2 NH3 (g) +
3 Cl2 (g)
 N2 (g)
+ 6 HCl (g)
∆H = ∑Energy of Bonds -∑Energy of Bonds
Broken
formed
∆H
∆H
= {6(N - H)+3(Cl - Cl)}-{(N  N)+(6(H - Cl)}
= {6(389kJ) + 3(243kJ)}-{(941kJ)+ 6(431kJ)}
= {2334 kJ + 729 kJ} – {941 kJ + 2586 kJ}
=
-464 kJ
27
Combustion
CH4
CH3OH
C8H18
H2 (g)
Heat of
Bond Energies
Formation
- 803.1 kJ/mol - 818 kJ/mol.
- 1276.8 kJ/mol
-2040kJ/mol
- 5075.8 kJ/mol - 5144 kJ/mol.
- 242 kJ/mol - 243 kJ/mol.
28
Chemistry 30
MEASURING QUANTITIES OF HEAT
Equation 1:
The formula for determining the amount of heat a body holds
Q = m c (Tf - Ti)
where Q =
C=
M=
Tf =
Ti =
the amount of heat in kilojoules (kJ)
the heat capacity of the substance in kilojoules per kilogram
degree Celsius (kJ/kg·°C)
the mass of the body in kilograms (kg)
the final temperature in degrees Celsius (°C)
the initial temperature in degrees Celsius (°C)
Given that the specific capacity of water is 4.184 kJ/kg·°C and that 1 g of water has a
volume of 1 mL, calculate the amount of heat gained or lost in each of the following situations:
1.
15 g of water is heated from 12°C to 32°C
2.
the temperature of 24 g of water rises by 40°C
3.
150 g of water is cooled from 80°C to 60°C
4.
the temperature of 48 mL of water drops by 30°C
29
Determine the temperature change in each of the following situations:
5.
10 g of water gains 1890 J
6.
25 mL of water gains 1050 J
7.
80 mL of water loses 1680 J
8.
100 g of water loses 6720 J
Try to use the following information to determine the specific heat capacities of the substances
involved:
9.
3000 J of heat added to 50 g of the substance causes its temperature to rise by 20°C
10.
80 g of the substance loses 2880 J of heat, causing its temperature to drop 15°C
30
The heat of fusion and heat of vapourization are also important. For water these are:
Heat of fusion
334 kJkg-1
Heat of vapourization 2260 kJkg-1
11.
Calculate the heat required to melt 25 kg of water.
12.
Calculate the heat required to condense 200 g of water.
13.
Calculate the heat required to freeze 50 mol of water.
It is also important to be able to combine both calculations:
14.
Calculate the amount of heat required to convert 2.5 kg of ice at -20C to water at 50C.
15.
Calculate the amount of heat required to covert 300 g of steam at 300C to water at 90C.
31
16.
Calculate the amount of heat required to convert 1.5 kg of ice at -35C to steam at 250C.
32
Chemistry 30
HEATING OF DIFFERENT MATERIALS
Following are a list of the specific heats of several materials
(in kJ/kg·°C):
Aluminum
Copper
Ethylene glycol
Gold
Ice
Iron
Lead
Magnesium
1.
0.900
0.390
2.200
0.130
2.100
0.450
0.130
0.980
Methanol
Oxygen
Paraffin oil
Sand
Silver
Water
Water vapour
Zinc
2.500
0.920
2.100
0.800
0.240
4.200
2.000
0.390
Find the quantity of heat in kiloJoules needed to raise the temperature of 2.5 kg of the
following substances from 60°C to 80°C:
a) sand
b) silver
c) gold
d) copper
e) methanol
f) water
g) paraffin oil
33
2.
Find the quantity of heat in kiloJoules needed to raise the temperature of 20 kg of lead
from 120°C to 160°C.
3.
Find the quantity of heat needed to raise the temperature of 200 g of ice from -120°C to
0°C:
4.
Given equal masses of each material and an equal input of heat, which of the materials in
the table on the first page will experience the greatest temperature rise? Which will
experience the lowest temperature rise? Give reasons for your answers.
5.
If equal masses of antifreeze (ethylene glycol) and methanol at 80°C are added to separate
beakers containing 500 g of water at room temperature, which mixture will have the
greatest final temperature? Justify your answer, either by calculation or in words.
6.
Explain why the climate near a large body of water is more moderate than the climate
inland. (HINT: What are the heat capacities of water versus sand?)
34
Chemistry 30
Enthalpy Change Calculations
Find the heat of reaction (H) for each of the following chemical
reactions and note whether each reaction is exothermic or endothermic:
1.
H2O(l)  H2O(g)
2.
SO2(g) + 1/2 O2(g)  SO3(g)
3.
SO3(g) + H2O(l)  H2SO4(l)
4.
2 HI(g)  H2(g) + I2(g)
5.
4 NH3(g) + 5 O2(g)  4 NO(g) + 6 H2O(l)
6.
CH4(g) + 2 O2(g)  CO2(g) + 2 H2O(g)
7.
CH4(g) + 2 O2(g)  CO2(g) + 2 H2O(l)
8.
C3H8(g) + 5 O2(g)  3CO2(g) + 4 H2O(g)
9.
2 Al(s) + Fe2O3(s)  Al2O3(s) + 2 Fe(s)
where Hf for Fe2O3(s) = -823 kJ/mol
10. SiO2(s) + C(s)  CO2(g) + Si(s)
where Hf for SiO2(s) = -861 kJ/mol
11. P4O10(s) + 6 H2O(l)  4 H3PO4(s)
where Hf for P4O10(s) = -3024 kJ/mol
and Hf for H3PO4(s) = -1273 kJ/mol
35
Suggested activities and ideas for research projects
1) Heat of reaction
Add a few pellets NaOH(s) to about 50 mL dilute H2SO4(aq) or HCl(aq). Stir, and note the
temperature change.
2) Hydration of copper sulfate
Spread about 1 g anhydrous CuSO4(s) in a thin layer on a piece of paper. Anhydrous CuSO4(s)
can be formed by grinding bluestone crystals (CuSO4•5H2O) and drying in a 100oC oven or
under a heat lamp. Put a small drop of water on part of the layer of chemical. Observe the effect
of the water. Describe the differences between the anhydrous CuSO4(s) and the hydrated form.
3) Endothermic Reaction
Heat of solutionMix Ba(OH)2·8H2O(s) and NH4SCN(s) in a 2:1 mass ratio. Stir the mixture,
noting the temperature change.
4) Study of Heats
Use the case study "History of Modern Ideas About Heat" from Science: Process and Discovery
(Field, 1985). 22 questions and problems for investigation accompany the study.
5) Uses of Heat effects
Brainstorm to produce a list of uses made of the heat effects of chemical reactions. Some are:
a) space heating by combustion of fuels
b) "instant cold" compresses
c) oxidation of glucose in the body to maintain constant temperature
6) Thermal Pollution
Identify locations where thermal pollution exists. Analyze sources, effects and reasons why such
pollution occurs.
Consider the local and global environmental implications of burning fossil fuels.
7) Pick a chemical reaction. Design a procedure to determine whether that reaction produces or
absorb heat.
8) Heat of Reactions
When equal volumes of 0.10M HCl(aq) and 0.10M NaOH(aq) are mixed, heat is produced. Will
twice as much heat be produced if equal volumes of 0.20M HCl(aq) and 0.10M NaOH(aq) are
mixed? How about if equal volumes of 0.20M HCl(aq) and 0.20M NaOH(aq) are mixed?
9) Physical and nuclear reactions
Identify ways of producing heat other than by chemical reaction. For what purposes is such heat
currently used? List places where heat produced by chemical reaction is now used. Could the
other sources of heat identified be substituted for heat produced by chemical reaction? Could
heat produced by chemical reaction substitute for any of the other sources?
10) Calorimetry
Using an insulated cup calorimeter, dissolve 3.00 g KNO3(s) in 50 mL of water. Record the
36
temperature change. Rinse the calorimeter and repeat, using 3.00 g NH4Cl(s) in 50 mL water.
Record the temperature change. Compare the two temperature changes. How many grams of
NH4Cl(s) must be mixed with 3.00 g KNO3(s) so that when the mixture is added to water, no
temperature change is noticed?
11) Calorimetry 2
Determine the molar heat of combustion of paraffin by heating a beaker of water on a stand with
a paraffin candle. Use a metal can, open at both ends, with a few vent holes on the side as a
chimney to reduce heat loss from the burning candle.
Based on the mass lost by the burning candle, and the temperature change of the known volume
of water, calculate the molar heat of combustion of paraffin.
12) Calorimetry 3
Burn a beeswax candle, using the same appartus and procedure, to get data to compare the heat
of combustion of paraffin and beeswax.
13) Hess's Law
Here is a set of reactions which can be used to illustrate Hess's Law. Measure the temperature
change of each reaction and use that data to calculate the H for each. Use NaOH pellets which
have not been exposed to the air. To know the exact mass of the NaOH(S) is important, but it is
not important that it be exactly 2.00 grams. The volumes of solutions and water have been
adjusted to give a constant volume of 100 mL.
a) 2 g NaOH(s) + 100 mL H2O(l)
b) 50 mL 1.0M NaOH(aq) + 50 mL 1.0M HCl(aq)
c) 2 g NaOH(s) + 100 mL 0.5M HCl(aq)
14) Study of Fossil Fuels
Burning fossil fuels produces most of the heat that we use in North America. What are other
sources of heat used? What percentage does each supply? What is the most common fossil fuel?
For solid and liquid fossil fuels, compare their efficiency of heat production on the basis of kJ per
mole and kJ per gram burned. For gaseous fossil fuels, compare their efficiency of heat
production on the basis of kJ per mole and kJ per litre of gas at SATP. How much energy does it
cost to extract, refine, transport and distribute each fossil fuel? (Express the energy cost in $ per
million kJ.)
Sample ideas for evaluation and for encouraging thinking
1) 3C(s) + 2Fe2O3(s) + 463.1 kJ 4Fe(s) + 3CO2(g)
Rewrite this equation using H notation for one mole of carbon dioxide product.
2) How can energy be released when a bond is formed? If energy is released when bonds form,
why aren't all reactions exothermic?
3) What would happen if energy was released when bonds were broken and absorbed when they
are formed? What would happen if energy was released when bonds were broken and formed?
37
4) Why is the law of conservation of energy considered to be valid?
5) A plateau in the heating curves of a liquid can indicate the boiling point of the liquid. What
becomes of the heat being added to the system during the time period of the plateau, where the
temperature doesn't rise?
6) The following are two examples of chemical reactions
a) 2Fe(s) + 1½O2(g) Fe2O3(s)
H= -824 kJ/mol Fe2O3
b) C(s) + O2(g) CO2(g)
H= -393.5 kJ/mol CO2
Comment on the relative merits of burning Fe(s) and C(s) as fuels.
7) Compare the following two reactions.
a) C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(l)
H= -2220 kJ/mol C3H8
b) CH4(g) + 2O2(g) CO2(g) + 2H2O(l)
H= -890 kJ/mol CH4
Since propane (C3H8(g)) gives off 2½ times as much heat per mol of fuel, why is natural gas
(CH4(g)) a more popular fuel?
8) To measure the heat of reaction between Zn(s) and HCl(aq) , it would be better to use 6.54
grams of zinc and 250 mL of 1.0 M HCl(aq) than to use 6.54 grams Zn(s) and 25 mL of 10.0 M
HCl(aq). Why?
9) Use charts of thermodynamic data to calculate what percentage of potential heat loss there is
when natural gas burns with insufficient oxygen to form H2O(l) , and equal mols of C(s) and
CO(g) , compared to when it burns to form CO2(g) and H2O(l).
To make this question easier, the equations
2CH4(g) + 2½O2(g) C(s) + CO(g) + 4H2O(l) and
CH4(g) + 2O2(g) CO2(g) + 2H2O(l) could be given.
10) Energy is absorbed during an exothermic reaction. Where does the energy absorbed come
from? Where does it go? Can it ever be recovered?
11) How does sweating keep us cool during hot weather or after strenuous exertion? The heat
which makes hot weather originates in a nuclear reaction. The heat responsible for heating us
during strenuous exercise (and other times too) originates in a chemical reaction. What reactions
are involved in each of these cases?
12) Why do chemical reactions produce or consume heat as they occur? Where does the heat
energy come from? Where does it go when it gets into the air?
38
Chemistry 30 - Thermodynamics Lab
Your challenge: To determine the amount of heat released per mole of reactant, for three
chemical reactions.
Materials:
-
foam cups - 1 per group
1.0 mol/L HCl - 50 mL per group (reaction 2)
1.0 mol/L NaOH - 50 mL per group (reaction 2)
0.5 mol/L HCl - 100 mL per group (reaction 3)
NaOH pellets - 2 g per reaction (reactions 1 and 3)
100 mL graduated cylinder
thermometers
safety glasses
weighing paper
balance
deionized water
Reactions:
1.
NaOH(s)  NaOH(aq)
2.
HCl(aq) + NaOH(aq) 
NaCl(aq) + H2O(l)
3.
HCl(aq) + NaOH(s)

NaCl(aq) + H2O(l)
Procedure:
Write up a procedure to perform the three reactions and measure the amount of heat
released in each one.
Hints:
a)
Remember that you are heating water. Use the formula Q = C x M x T
b)
Watch the thermometer during the reaction (you can use it as a stirring rod, since the
reaction will take place in a styrofoam cup);
- the initial temperature is the temperature before the reaction.
- the final temperature is the highest temperature reached during the reaction.
c)
Calculate the heat per mole of NaOH in each reaction.
d)
Report the heat released as a H number.
39
Questions:
1.
How are the three reactions related?
2.
How would you expect the H numbers to compare in the three reactions?
3.
What chemical law are you testing in this lab?
40
Chemistry 30
Specific Heat of Materials
Your Challenge:
To determine the specific heat of three pieces of metal.
Materials:
-
styrofoam cup
pieces of metal (iron, aluminum, brass, copper)
kettle
water
thermometer
Procedure:
Use the principles of calorimetry to write up a procedure.
Hints:
1)
Use the formula Q = C x M x T to determine the heat released by each piece of metal.
Remember that you are heating water.
2)
Make sure you know the volume and initial temperature of the water you use.
3)
Remember that the units of specific heat are kJkg-1C-1
Calculations:
1)
Determine specific heats for each of the 3 metals tested.
2)
Determine the % difference from the actual values given you in class.
Questions:
1)
If you measured the specific heat of two pieces of the same metal, one having a mass of
10 g and the other 150 g, how should their specific heats compare?
2)
Explain any differences in your measured specific heat values and the actual values you
were given, in terms of experimental error.
41
Lab 17B - Heat of Fusion
- need: - foam cups - 1 per group, 15 per class
- ice cubes - 2 per group, 30 per class
- additional:
-
600 mL beaker
thermometer
wire gauze
ring stand
burner
stirring rod
safety glasses
balance
water bottles
Calculations
Determine the heat of reaction for the following chemical reactions:
a)
C3H7COOH(l) + 5 O2(g)  4 CO2(g) + 4 H2O(l)
b)
3 CH3COOH(l) + 11/2 O2(g)  5 CO2(g) + CO(g) + 6 H2O(l)