SCH4U
Grade 12
University Chemistry
Lesson 9 – Thermochemistry
SCH4U – Chemistry
Unit 3 - Introduction
Unit 3: Rates of Reaction
Energy transformations take place continually inside our bodies and in the world around
us. For example, you transform the food you eat into useable cellular energy or
adenosine triphosphate (ATP). Machines such as automobiles and power tools convert
fuel into mechanical energy. In this unit you will learn more about energy
transformations including the factors that affect reaction rate, and their kinetics.
Overall Expectations
•
•
•
demonstrate an understanding of the energy transformations and kinetics of
chemical changes;
determine energy changes for physical and chemical processes and rates of
reaction,
demonstrate an understanding of the dependence of chemical technologies and
processes on the energetics of chemical reactions.
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SCH4U – Chemistry
Lesson 9
Lesson 9: Thermochemistry
Heat transformations take place everyday in both living organisms and machines. Cars,
for example, convert gasoline into energy to propel the car forward. Heat is also lost in
the process. In living organisms, the process of cellular respiration converts simple
sugars into cellular energy, carbon dioxide, and water. Again, heat is lost in the
process. The study of heat transformation is called Thermochemistry, and it will be the
focus of this lesson.
What You Will Learn
After completing this lesson, you will;
•
•
•
write thermochemical equations, expressing the energy change as a H value or as a
heat term in the equation;
determine heat of reaction, and use the data obtained to calculate the enthalpy
change for a reaction
compare the energy changes resulting from physical change, chemical reactions,
and nuclear reactions (fission and fusion);
Thermochemistry
Thermochemistry is the study of energy changes involved in chemical reactions.
These changes can be physical (i.e. melting ice,) chemical (rusting iron) or nuclear
(nuclear fusion reactions in the Sun).
All energy transformations are governed by the 1st Law of Thermodynamics, which
states that the total energy of the universe is constant.
Thermodynamics Terms
There are a set of terms that are necessary to understand when discussing
thermodynamics. They are summarized following:
Thermal energy: The energy available from a substance as a result of its motion of its
molecules. For example, when you burn your foods, you break down sugars into
carbon dioxide, water, ATP and thermal energy (heat). The SI unit for measuring work
and all forms of energy is the Joule (J).
Chemical system: The set of reactants and products undergoing energy
transformations. Chemical systems can be open (energy and matter can move in or
out), closed (only energy can move in or out), or isolated (neither energy or matter can
move in or out)
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SCH4U – Chemistry
Lesson 9
Surroundings: All matter around the system that is capable of absorbing or releasing
thermal energy.
Consider the following reaction taking place in your body cells:
C6H12O6 + 6O2 Æ 6H2O +2CO2 + energy
The molecules (glucose, oxygen, water, and carbon dioxide are the chemical system,
while the surrounding are the extracellular fluid in your cells.
Heat: Amount of heat energy transferred between substances, systems or
surroundings. When a reaction occurs, heat is transferred between substances,
systems and surroundings.
Exothermic: A chemical system that releases energy from its surroundings
Endothermic: A chemical system that absorbs energy from its surroundings
Temperature: A measure of the average kinetic energy of the particles in a sample
In early work, the energy required to raise the temperature of one gram of water one
Celsius degree was defined as one calorie (from the Latin word for heat, calor). The
energy needed to raise the temperature of one kilogram of water one Celsius degree is
1000 times larger, so it was called a kilocalorie. The kilocalorie is the unit still used in
most countries to describe the energy available in food. It was commonly referred to as
one Calorie (capital C).
The calorie and the Calorie are not SI metric units. They have therefore been replaced
by the joule, which is the standard unit for any form of energy. A joule (J) is the energy
involved when a force of one Newton (1N) acts through a distance of one meter (1m).
One calorie is equal to 4.184 J, and one Calorie (kilocalorie) is equal to 4184 J or 4.184
kJ. To raise the temperature of 1.0 kg of water by 1.0oC requires 4184 J or 4.184 kJ of
energy.
Support Questions
1. Identify each of the following as physical, chemical, or nuclear change.
a)
b)
c)
d)
A gas stove cooking pasta
an ice cube melting in a glass
wax melts
ice applied to a sore back
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SCH4U – Chemistry
Lesson 9
2. Identify the system and the surroundings in each of the examples in the previous
question.
3. A cup of water at 100oC has a higher temperature than a swimming pool full of water
at 20oC, but the pool has more thermal energy. Explain.
Calorimetry
Calorimetry is an experimental technique used to measure heat energy changes in a
chemical system. A calorimeter is an instrument for measuring the heat of a reaction.
The following diagram depicts a simple calorimeter:
Figure 9.1: A simple calorimeter
Any heat energy changes that take place are measured by a change (increase or
decrease) by the temperature of the surroundings, which in most cases is water in the
cup.
However, it is important to note that we cannot just simply measure the change in
temperature. There are other factors that will depend on the amount of energy
transferred:
1. The mass of the water
2. Specific Heat Capacity
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SCH4U – Chemistry
Lesson 9
Specific Heat Capacity
The specific heat capacity is the amount of heat energy that is needed to raise the
temperature of one gram of a substance 1oC (or 1 K). Specific heat capacities vary from
substance to substance, and even for different states (solid, liquid, gas) of the same
substance.
•
•
•
Symbol is “c”, Units of J/g●oC
Larger values indicate that more energy is required to increase its temperature.
Some specific heat capacities are summarized in table 9.1 below.
Table 9.1: Specific Heat Capacities of Various Substances
Substance
Concrete
Aluminum
Wood
Specific Heat Capacity
(J/ g●oC)
0.88
0.900
1.76
Water (liquid)
4.184
Water has a very high specific heat capacity, and as such, is commonly used in
calorimetry experiments.
All of these factors, mass, specific heat capacity and temperature change must be
considered when calculating the amount of heat energy transferred in a chemical
system.
This leads us to an important equation used to calculate heat transfer (Q):
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SCH4U – Chemistry
Lesson 9
Example 1: When 500 mL of water in an electric kettle is heated from 25oC to 65oC,
how much heat energy flows into the water?
Solution 1: First calculate the mass, considering that the density of water is 1g/mL
500 mL × 1 g/mL = 500 g
Given:
m = 500g
c = 4.18J/goC
ΔT = 65oC - 25oC = 40oC
Required:
Q =?
Equation:
Q = mc ΔT
Solution:
Statement:
Q = 500 g × 4.18J / g °C × 40 °C
= 83600J or 83.6kJ
83.6kJ of heat energy flows into an electric kettle when 500 mL of water is
heated from 25EC to 65EC.
The specific heat capacities of various substances are summarized in table 9.2.
Table 9.2: Specific Heat Capacities of Various Substances
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SCH4U – Chemistry
Lesson 9
Support Questions
4. A swimming pool contains 1000 L of water. When the water is warmed by solar
energy, its temperature increases from 15.8oC to 22.1oC. How much heat does the
water absorb? (cwater(liquid)= 4.184 J/gyoC)
5. A cook heats water from 20oC to 50oC. Calculate the mass of water that could be
warmed by the addition of 8.00kJ of heat.
6. A 50% ethylene glycol solution has a specific heat capacity of 3.5J/goC. What
temperature change would be observed in a solution of 4kg of ethylene glycol if it
absorbs 250kJ of heat?
Enthalpy
•
•
•
•
the total internal energy of a substance at a constant pressure
chemists study the enthalpy change, ∆H, that accompanies a chemical process
∆H of a process is equivalent to its heat change at a constant pressure
∆H can be associated with physical changes as well as chemical reactions
Molar Enthalpies
Molar enthalpy, ΔHx is the enthalpy change associated with one mole of substance.
For example:
H2(g) + 1/2O2(g) Æ H2O(g) + 242.0kJ
The enthalpy change associated per mole of substance in the above reaction is 285kJ/mol.
Enthapy changes for exothermic reactions are given a negative(-) sign or written or
placed on the product side of the equation (with a positive (+) sign)
Enthapy changes for endothermic reactions are given a positive(+) sign or placed on
the reactant side of the equation
Stating the molar enthalpy is a method of describing the energy changes in a variety of
physical and chemical changes involving one mole of a particular reactant or product.
For example consider the vaporization of water (physical change)
H2O(l) + 40.8kJ Æ H2O(g)
ΔHvap = 40.8kJ/mol
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SCH4U – Chemistry
Lesson 9
These values are obtained empirically. Selected molar enthalpies are found in table
9.3.
Table 9.3: Molar Enthalpies of Selected Substances
Chemical Name
Formula
Molar enthalpy of
fusion (kJ/mol)
Sodium
Chlorine
Sodium chloride
Water
Ammonia
Freon-12
Methanol
Ethylene glycol
Na
Cl2
NaCl
H2O
NH3
CCl2F2
CH3OH
C2H5(OH)2
2.6
6.40
28
6.03
-
Molar enthalpy of
vaporization
(kJ/mol)
101
20.4
171
40.8
1.37
34.99
39.23
58.8
Molar Enthalpies can also be used in heat calculations, let try a sample calculation.
Example 2: Freon 12 (molar mass 120.91 g/mol) is a common refrigerant that is
vaporized in tubes inside a refrigerator, releasing heat. This results in heat being
released to the outside of the fridge. If 500g of the Freon is vaporized, what is the
expected enthalpy change, ΔH?
Solution 2:
Given:
Δ Hvap = 34.99kJ/mol (from table 9.3)
m = 500g
Required:
ΔH = ?
Equation:
η =m/M
Solution:
First calculate the moles of Freon
ΔH = η × ΔHvap
Second calculate the enthalpy change
Statement:
M = 120.91g/mol
η = 500 g / 120.91 g / mol
= 4.135 mol
ΔH = 4.135 mol × 34.99kJ / mol
= 144.7kJ
The expected enthalpy change when Freon 12 is vapourized is 144.7 kJ.
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SCH4U – Chemistry
Lesson 9
Support Questions
7. Calculate the enthalpy change ΔH for the vaporization of 150.0g of water at 100oC.
Using Calorimetry to Find Molar Enthalpies
As mentioned above, a calorimeter is a device used to measure heat energy changes
There are three assumptions used in calorimetry:
•
•
•
No heat is transferred between the calorimeter and the outside environment
Any heat absorbed or released by the calorimeter materials is neglible
A dilute aqueous solution is assumed to have a density and specific heat capacity
equal to that of pure water (1.00g/mL and 4.18J/goC)
Since this is a correspondence course, you won’t be able to perform a calorimetry
experiment, but we can practice some calculations for a typical calorimetry experiment.
Example 3: In a calorimetry experiment, 7.46 g of potassium chloride is dissolved in
100.0 mL of water at an initial temperature of 24.1°C. The final temperature of the
solution is 20oC. What is the molar enthalpy?
Solution 3:
Given:
mKCl = 7.46g
MKCl = 74.6g / mol
mwater = 100.0g (100mL = 100g )
c water = 4.18J / g °C
ΔT = 24.1°C − 20.0°C = 4.1°C
Required:
ΔHsol = ?
Equation:
η =m/M
ΔHsol =
mc ΔT
η
**
Note: Recognize the Law of Conservation of Energy (in this case, P is
constant and Work = 0, therefore, ΔH = q)
ΔH = q
ΔH (KCl dissolving) = q (calorimeter water)
Or in other words
η × ΔHsol = mc ΔT
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SCH4U – Chemistry
Solution:
Lesson 9
Start by calculating the moles of potassium chloride
ηKCl = m / M
= 7.46 g / 74.6 g / mol
= 0.100mol
Calculate the molar enthalpy.
Since, η × ΔHsol = mc ΔT , then we can rearrange to;
ΔHsol =
=
mc ΔT
η
100 g × 4.18J / g °C × 4.1°C
0.100mol
= 1.7 × 10 J / mol
4
Statement:
The final molar enthalpy of the solution is 1.7 × 10 4 J / mol .
Support Questions
8. The energy change in the process H2O(g) Æ H2O(l) could be described as a molar
enthalpy of condensation. State the type of molar enthalpy that would occur in each
of the following reactions:
a) Br2(l) Æ Br(g)
b) CH4(g) Æ CH4(s)
c) NaCl(s) Æ Na+(aq) + Cl-(aq)
d) KOH(aq) + HCl(aq) Æ KCl(aq) + H2O(l)
9. In the lab, you add 50 mL of concentrated hydrochloric acid (12 mol/L), HCl to form
250mL of dilute solution. The temperature of the solution changes from 18oC to
24oC. Calculate the molar enthalpy of dilution of hydrochloric acid.
10. What mass of lithium chloride must have dissolved if the temperature of 200g of
water increased by 6.0oC? The molar enthalpy of solution of lithium chloride is
37kJ/mol.
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SCH4U – Chemistry
Lesson 9
Representing Enthalpy Changes
Enthalpy changes can be represented in four ways
•
As part of a thermochemical equation
e.g. CH3OH(l) + 3/2 O2(g) Æ CO2(g) + H2O(g) +726 kJ
•
By writing a chemical equation and stating its enthalpy change
e.g. CH3OH(l) + 3/2 O2(g) Æ CO2(g) + H2O(g) ΔH = -726 kJ
•
By stating the molar enthalpy of a specific reaction
e.g. ΔHcomb = -726 kJ/mol CH3OH
•
By drawing a potential energy diagram
Method 1: Thermochemical Equations with Energy Terms
As previously mentioned, if the reaction is exothermic, the energy is listed with the
products, and if the reaction is endothermic, the energy is listed with the reactants (both
with a positive (+) sign).
Example 4: Write a thermochemical equation for the combustion of butane. The molar
enthalpy for the combustion of butane is -2871kJ/mol
Solution 4: Recall from unit two, that the products of a complete combustion reaction
are carbon dioxide and water. Write out and balance your equation.
2C4H10(g) + 13O2(g) Æ 8CO2(g) + 10H2O(l)
Note that the molar enthalpy is -2871kJ per one mole, and in this reaction we have two
moles, so calculate your enthalpy for two moles
ΔH = 2mol x -2871kJ/mol = -5742kJ
Since the enthalpy value, the reaction is exothermic, and the enthalpy must be listed
with the products
2C4H10(g) + 13O2(g) Æ 8CO2(g) + 10H2O(l) + 5742kJ
Method 2: Thermochemical Equations with ΔH symbols
You can also write the Thermochemical equation and then write the ΔH value beside it.
Remember that the ΔH value is negative for exothermic reactions and positive for
endothermic reactions.
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SCH4U – Chemistry
Lesson 9
Example 5: Sulphur dioxide and oxygen react to form sulphur trioxide. The molar
enthalpy for the combustion of sulphur dioxide is -98.9kJ/mol. What is the enthalpy
change for this reaction?
Solution 5:
Start off by writing the balanced thermochemical equation
2SO2 + O2 Æ 2SO3
Calculate your enthalpy for 2 moles of sulphur dioxide
ΔH = 2mol x -98.9kJ/mol
ΔH = -197.8 kJ
Now state the enthalpy value with the balanced equation
2SO2 + O2 Æ 2SO3
ΔH = -197.8 kJ
Method 3: Molar Enthalpy of Reaction
Recall that the molar enthalpy, ΔHx is the enthalpy change associated with one mole of
substance. The standard molar enthalpy, ΔHox is the energy change associated with
one mole of substance at 100kPa and usually 25oC.
Table 9.4: Molar enthalpy of Reactions
Type of molar enthalpy
Example of change
Solution ( ΔHsol )
NaBr(s) → Na + (aq) + Br −(aq)
Combustion ( ΔHcomb )
CH4(g) + 2O2(g) → CO2(g) + 2H2O(I)
Vaporization (ΔHvap )
CH3OH(I) → CH3OH(g)
Freezing
H2O(I) → H2O(s)
Neutralization ( ΔHneut ) *
2NaOH(aq) + H2SO4(aq) → 2Na 2SO4(aq) + 2H2O(I)
Neutralization ( ΔHneut ) *
1
1
NaOH(aq) + H2SO4(aq) → Na 2SO4(aq) + H2O(I)
2
2
1
Formation ( ΔHf ) * *
C(s) + 2H2(g) + O2(g) → CH3OH(I)
2
* Enthalpy of neutralization can be expressed per mole of either base or acid
consumed.
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SCH4U – Chemistry
Lesson 9
Example 6: Write an equation whose energy change in the molar enthalpy of
combustion of propanol, (C3H7OH).
Solution 6: C3H7OH(g) +
9
O2(g) Æ 3CO2(g) + 4H2O(l)
2
Method 4: Potential Energy Diagrams
The energy transferred can be communicated graphically using a potential energy
diagram.
Figure 9.1: Potential Energy Diagrams
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SCH4U – Chemistry
Lesson 9
Key Question #9
1. Solar energy can heat cold water for household hot water tanks. What quantity of
heat is obtained from solar energy if 150kg of water is heated from 15oC to 50oC?
(5 marks – remember to show your work)
2. Identify each of the following processes as exothermic or endothermic; (7 marks)
a)
b)
c)
d)
The combustion of propane in a barbeque tank.
Ag2O(s) + heat Æ 2Ag(s) + 1/2O2(g)
As solid NaNO3 dissolves in water at 25oC, the temperature changes to 22oC.
When calcium is added to water at 22oC, the temperature changes to 27oC.
1
e) Mg(s) + O2(g) Æ MgO(s) + bright light
2
f) 2H2(g) + O2(g) Æ 2H2O(l) ∆H =-571.0kJ
g) N2(g) + O2(g) Æ 2NO(g) ∆H = 180.8kJ
3. To test the properties of a fertilizer, 15.0 g of urea, NH2CONH2(s), is dissolved in
150mL of water in a simple calorimeter. A temperature change from 20.6EC to
17.8EC is measured. Calculate the molar enthalpy of solution for the fertilizer urea.
(5 marks – remember to show your work)
4. a) A laboratory technician adds 56.1mL of concentrated 12 mol/L hydrochloric acid
to water to form 500mL of dilute solution. The temperature of the solution changes
from 20.2EC to 22.3EC. Calculate the molar enthalpy of dilution of hydrochloric acid.
(5 marks – remember to show your work)
b) What effect would there be on the calculated value for the molar enthalpy of
dilution if the technician accidently used too much water so that the volume was
actually more than 500 mL. Explain. (3 marks)
5. If the molar enthalpy of combustion of ethane is -1.56 MJ/mol, how much heat is
produced in the burning of; (4 marks)
a) 5.0 mol of ethane
b) 45 g of ethane
6. In a laboratory investigation into the neutralization reaction a scientist adds
potassium hydroxide to nitric acid (balanced reaction following); (5 marks)
HNO3(aq) + KOH(s) Æ KNO3(aq) + H2O(l)
Data Collected
Mass KOH = 4.6g
Volume of nitric acid solution = 250mL
T1 = 21oC
T2 = 29.3oC
Calculate the molar enthalpy of neutralization of potassium hydroxide.
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SCH4U – Chemistry
Lesson 9
Key Question #9 (continued)
7. Draw a potential energy diagram with appropriately labelled axes to represent:
(4 marks)
a) The exothermic combustion of octane ∆Ho =-5.47MJ)
b) The endothermic formation of diborane (B2H6) from its elements ∆Ho = +36kJ)
8. Translate each of the molar enthalpies given below into a balanced thermochemical
equation, including the enthalpy change, ∆H
a) The enthalpy change for the reaction in which solid magnesium hydroxide is
formed from its elements at SATP is -925kJ/mol. (3 marks)
b) The standard molar enthalpy of combustion for pentane, C5H12, is -2018 kJ/mol
(3 marks)
9. For each of the following reactions, write a thermochemical equation including the
energy term as part of the equation. (6 marks)
a) The combustion of butane in a candle lighter is -2.86MJ/mol
b) The molar enthalpy of the transition of graphite to diamond is 2kJ.
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SCH4U
Grade 12
University Chemistry
Lesson 10 – Enthalpies of Reaction
SCH4U – Chemistry
Lesson 10
Lesson 10: Enthalpies of Reaction
In lesson 9 you learned that calorimetry is one method to determine the enthalpy of a
reaction. In this lesson, you will learn about two more methods; Hess’s Law and using
heats of formation.
What You Will Learn
After completing this lesson, you will;
•
•
•
explain Hess’s law, using examples;
apply Hess’s law to solve problems, including problems that involve data obtained
through experimentation (e.g., measure heats of reaction that can be combined to
yield the H of combustion of magnesium);
calculate heat of reaction using tabulated enthalpies of formation;
Hess’s Law of Heat Summation
Hess’s Law states that the enthalpy change of a physical or chemical process depends
only on the beginning conditions (reactants) and the end conditions (products). The
enthalpy change is independent of the pathway of the process and the number of
intermediate steps in the process.
Hess’s law allows you to determine the energy of a chemical reaction without directly
measuring it. There are two ways to accomplish this:
1. by combining chemical equations algebraically
2. by using the enthalpy of formation reactions (a special class of reactions)
Combining Chemical Equations Algebraically
Add equations for reactions with known enthalpy changes, so that their net result is the
reaction you are interested in. This may require manipulation of the known equations
by reversing or multiplying each coefficient (you need to multiply ∆Ho by the same
number)
Example 1: What is the enthalpy for the formation of one mole of diamond from one
mole of carbon?
Target sequence C (s, graphite) Æ C (s, diamond) ΔH° =? kJ
(The presence of the degree sign, (°), on the enthalpy indicates that the reaction is happening under
standard conditions.)
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SCH4U – Chemistry
Lesson 10
The following known equations, determined by calorimetry, are provided:
1. C (s, graphite) + O2(g) Æ CO2(g) ΔH° = -394 kJ
2. C (s, diamond) + O2(g) Æ CO2(g) ΔH° = -396 kJ
Solution 1: We can manipulate the above known equations, to obtain the target
sequence, and thus obtain the enthalpy for the reaction.
First off, reverse the bottom equation. This will put the C (s, diamond) on the product
side, where we need it. When you reverse an equation the sign on the enthalpy value
is also reversed. The reason behind this is that original equation is exothermic. We
know this from the negative in front of the enthalpy value. That means that the opposite,
reverse equation is endothermic. Putting in enthalpy (endothermic) is the reverse, the
opposite of exothermic (giving off enthalpy). Hence, we change the sign EVERY time
we reverse an equation.
1. C (s, graphite) + O2(g) Æ CO2(g)
ΔH° = -394 kJ
Reverse 2. CO2(g) ÆC (s, diamond) + O2(g) ΔH° = +396 kJ
If we add the two equations together, the oxygen and carbon dioxide will cancel out.
This is, of course, what we want since those two substances are not in the final, desired
equation. The enthalpy values are also added together.
CO2(g) + C (s, graphite) + O2(g) Æ CO2(g) + C (s, diamond) + O2(g) ΔH° = (-394 kJ) + (+396 kJ)
Notice the items which are the same on both sides and cancel them:
CO2(g) + C (s, graphite) + O2(g) Æ CO2(g) + C (s, diamond) + O2(g) ΔH° = (-394 kJ) + (+396 kJ)
C (s, graphite) Æ C (s, diamond) ΔH° = +2 kJ
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SCH4U – Chemistry
Lesson 10
Example 2: Calculate the enthalpy for the following reaction:
Target sequence N2(g) + 2O2(g) Æ 2NO2(g) ΔH° =? kJ
Using the following two equations:
1. N2(g) + O2(g) Æ 2NO(g)
ΔH° = +180 kJ
2. 2NO2(g) Æ 2NO(g) + O2(g) ΔH° = +112 kJ
Solution 2: Note that in the target sequence, the nitrogen dioxide, NO2, is on the right
side, so begin by reversing equation 2.
1. N2(g) + O2(g) Æ 2NO(g)
ΔH° = +180 kJ
Reverse 2. 2NO(g) + O2(g) Æ 2NO2(g) ΔH° = -112 kJ
Notice that the sign on the enthalpy has changed from positive to negative.
Next, we add the two equations together and eliminate identical items. We also add the
two enthalpies together.
N2(g) + 2O2(g) Æ 2NO2(g) ΔH° = +68 kJ
Example 3: What is the enthalpy change for the formation of one mole of butane
(C4H10) gas from its elements? The reaction is:
Target sequence: 4C(s) + 5H2(g) Æ C4H10(g) ΔH° =? kJ
The following known equations are provided:
1. C4H10(g) +
13
O2(g) Æ 4CO2(g) +5 H20 ΔH° = -2657.4 kJ
2
2. C(s) + O2(g) Æ CO2(g)
ΔH° = -393.5 kJ
3. 2H2(g) + O2(g) Æ 2H2O(g)
ΔH° = -483.6 kJ
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SCH4U – Chemistry
Lesson 10
Solution 3: Begin by reversing equation 1 and changing the enthalpy sign.
Reverse 1. 4CO2(g) +5 H2O Æ C4H10(g) +
13
O2(g) ΔH° = +2657.4 kJ
2
Since we need 4 moles of carbon in the target sequence, we will then multiply equation
2 by 4. Note that the enthalpy is also multiplied by 4.
2.x 4 4C(s) + 4O2(g) Æ 4CO2(g) ΔH° = -1574 kJ
We need 5 moles of hydrogen gas in the target sequence. In equation 3, we currently
5
have 2 moles. If we multiply equation 3 by
we will obtain 2 moles. Don’t forget to
2
5
also multiply the enthalpy by .
2
3x
5
5
5H2(g) + O2(g) Æ 5H2O(g) ΔH° = -1209kJ
2
2
Now add together all of the reactants, products, and enthalpy for your three modified
equations
4CO2(g) +5 H20 + 4C(s) + 4O2(g)+ 5H2(g) +
5
13
O2(g) Æ C4H10(g) +
O2(g) + 4CO2(g) + 5H2O(g)
2
2
ΔH° = -125.6kJ
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SCH4U – Chemistry
Lesson 10
Support Questions
11. Calculate ∆H for the reaction: C2H4(g) + H2(g) Æ C2H6(g), from the following data.
C2H4 (g) + 3O2 (g) Æ 2CO2 (g) + 2H2O (l)
∆HE = -1411. kJ
C2H6 (g) + 3½O2 (g) Æ 2CO2 (g) + 3H2O (l)
∆HE = -1560. kJ
H2 (g) + ½ O2 (g) Æ H2O (l)
∆HE = -285.8 kJ
12. Calculate ∆H for the reaction 4 NH3 (g) + 5 O2 (g) Æ 4 NO (g) + 6 H2O (g), from the
following data.
N2(g) + O2(g) Æ 2NO(g)
∆HE = -180.5 kJ
N2(g) + 3H2(g) Æ 2NH3(g)
∆HE = -91.8 kJ
2H2(g) + O2(g) Æ 2H2O(g)
∆HE = -483.6 kJ
13. Find ∆H° for the reaction 2H2(g) + 2C(s) + O2(g) Æ C2H5OH(l), using the following
thermochemical data.
C2H5OH(l) + 2O2(g) Æ 2 CO2(g) + 2H2O(l)
∆HE = -875. kJ
C(s) + O2(g) Æ CO2(g)
∆HE = -394.51 kJ
H2(g) + ½O2(g) Æ H2O(l)
∆HE = -285.8 kJ
14. Calculate ∆H for the reaction CH4(g) + NH3(g) Æ HCN(g) + 3H2(g), given:
N2(g) + 3H2(g) Æ 2NH3(g)
∆HE = -91.8 kJ
C(s) + 2H2(g) Æ CH4(g)
∆HE = -74.9 kJ
H2(g) + 2C(s) + N2(g) Æ 2HCN(g)
∆HE = +270.3 kJ
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SCH4U – Chemistry
Lesson 10
Support Questions (continued)
15. Calculate ∆H for the reaction 2Al(s) + 3Cl2(g) Æ 2AlCl3(s) from the data.
2Al(s) + 6HCl(aq) Æ 2AlCl3(aq) + 3H2(g)
∆HE = -1049. kJ
HCl(g) ÆHCl(aq)
∆HE = -74.8 kJ
H2(g) + Cl2(g) Æ 2HCl(g)
∆HE = -1845. kJ
AlCl3(s) Æ AlCl3(aq)
∆HE = -323. kJ
Standard Enthalpies of Formation
Standard molar enthalpy of formation, ∆Hof, is the quantity of heat energy that is
absorbed or released when one mole of a compound is formed directly from its
elements in their standard states (SATP = 25oC, 100 kPa). By definition, the enthalpy of
formation of an element in its standard state is zero. What does the word formation
mean?
Formation means a substance, written as the product of a chemical equation, is formed
DIRECTLY from the elements involved. The substance in question is always written
with a coefficient of one. Here are some examples:
C (s) + O2 (g) Æ CO2 (g)
C (s) + (1/2) O2 (g) Æ CO (g)
H2 (g) + O2 (g) Æ H2O2 (l)
H2 (g) + (1/2) O2 (g) Æ H2O (l)
C (s) + 2 H2 (g) + (1/2) O2 (g) Æ CH3OH (l)
* Recall that some elements (Hydrogen, Oxygen, Fluorine, Bromine, Iodine, Nitrogen,
and Chlorine or HOFBrINCl) are diatomic elements, meaning they exist in pairs. For
example, H2, O2, etc.
Writing Formation Equations
a) Write one mole of product in the given state
b) Write the reactant elements in their standard states
c) Balance your equation
Example 4: Write the formation equation for liquid ethanol, C2H5OH.
Solution 4: 2C(s) + 3H2(g) + 1/2O2(g) Æ C2H5OH(l)
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SCH4U – Chemistry
Lesson 10
Using Standard Enthalpies of Formation
The enthalpy change for the target equation equals the enthalpy of formation for the
products minus the enthalpies of formation for the reactants. This can be summarized
in an equation
ΔH = ΣηΔHof (products) –ΣηΔHof(reactants)
The standard enthalpies of formation are found in appendix A at the end of the booklet.
This table will be provided on the any exams or tests that you write.
Example 5: What is the molar enthalpy of combustion of methane fuel?
Solution 5:
CH4(g) + 2O2(g) Æ CO2(g) + 2H20(l)
ΔH = ΣηΔHof (products) –ΣηΔHof(reactants)
ΔH = [(1 mol x -393.5
kJ
kJ
kJ
kJ
) + (2 mol x -285.8
)] – [(1 mol x -74.4
) + 2 mol x 0
)]
mol
mol
mol
mol
Note that there are different enthalpy values for different states, so make sure you look
up the correct state.
ΔH = [-965.1kJ] – [-74.4kJ]
ΔH = -890.7 kJ
Sometimes enthalpy problems are multistep. They involve more than one step and the
following relationships are used:
1. Enthalpy change in the system = heat transferred to or from the surroundings
2. ΔH =ηΔHr
ΔH = ΣηΔHof (products) –ΣηΔHof(reactants)
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SCH4U – Chemistry
Lesson 10
Example 6: What mass of octane is completely burned to cause the heating of 20.0g of
aqueous ethylene glycol automobile coolant from 10oC to 90oC? The specific heat
capacity of aqueous ethylene glycol is 3.5J/(goC). Assume water is produced as a gas
and that all the heat flows into the coolant. The balanced equation is below
C8H18 (g) +
25
O2(g) Æ 8CO2(g) + 9H2O(g)
2
Solution 6: First calculate the molar enthalpy of combustion of octane
ΔH = [(8mol x -393.5kJ/mol) + (9mol x -241.8kJ/mol)] – [(1mol x -250.1kJ/mol) +
(25/2mol x 0kJ/mol)]
ΔH = -5074kJ
Now we will assume that the heat produced by this reaction will be absorbed by the
enthylene glycol (engine coolant).
Remember:
ΔHoctane = qcoolant
Therefore; ηΔHc =mcΔT
Solve for the moles of octane;
η=
ηoctane =
mc ΔT
ΔH c
20.0 g × 0.0035 kJ / g °C × 80 °C
5074.1kJ / mol
= 0.0011mol
Copyright © 2008, Durham Continuing Education
m =η / M
moctane = 1.1 mol / 114.26g / mol
= 0.126g
Page 25 of 70
SCH4U – Chemistry
Lesson 10
Support Questions
16. Write formation equations for the following:
a) benzene (C6H6)
b) potassium bromate
c) glucose
17. Use standard enthalpies of formation to calculate:
a) the molar enthalpy of combustion for pentane.
b) the enthalpy change that accompanies the reaction between solid iron (III) oxide
and carbon monoxide gas to produce solid iron metal and carbon dioxide gas.
18. For each of the following reactions, use standard enthalpies to calculate ∆H:
a) CH4(g) + H2O(g) Æ CO(g) + 3H2(g)
b) CO(g) + H2O(g) Æ CO2(g) + H2(g)
c) N2(g) + 3H2(g) Æ 2NH3(g)
19. Ammonium nitrate fertilizer is produced by the reaction of ammonia with nitric acid:
NH3(g) + HNO3(l) Æ NH4NO3(s)
Use standard enthalpies of formation to calculate the standard enthalpy change of
the reaction used to produce ammonium nitrate.
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SCH4U – Chemistry
Lesson 10
Key Question #10
1. Use Hess’s Law to predict the enthalpy change for the reaction (5 marks)
HCl(g) + NaNO2(s) Æ HNO2(g) + NaCl(s)
Using the following information:
(1) 2NaCl + H2O Æ 2HCl + Na2O
(2) NO + NO2 + Na2O Æ 2NaNO2
(3) NO + NO2 Æ 2N2O + O2
(4) 2HNO2 Æ N2O + O2 + H2O
∆Ho1 = -507kJ
∆Ho2 = -427kJ
∆Ho3 = -43kJ
∆Ho4 = 34kJ
2. Use the thermochemical equations shown below to determine the enthalpy for the
reaction: (5 marks)
C3H8(g) + 5O2(g) Æ3CO2 + 4 H2O(l)
CO2 ÆC(graphite) + O2
UHE = 221.6KJ
H2(g) + 1/2O2(g) ÆH2O(l)
UHE = -160.3KJ
3 C(graphite) + 4H2(g)ÆC3H8(g)
UHE = -58.5KJ
3. Use the thermochemical equations shown below to determine the enthalpy for the
reaction: (5 marks)
SO2(g)ÆS(s) + O2(g)
H2S(g) + 3/2O2(g)ÆH2SO3(l)
UHE = -306KJ
H2SO3(l)ÆH2O(l) +SO2(g)
UHE = 93KJ
H2S(g) + 1/2O2(g)ÆS(s) + H2O(l)
UHE = -232.5KJ
4. Write balanced equations for the formation of the following: (4 marks)
a) acetylene gas (C2H2)
b) potassium chloride (KCl)
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SCH4U – Chemistry
Lesson 10
Key Question #10 (continued)
5. Use standard enthalpies of formation to calculate the enthalpy changes in each of
the following equations: (6 marks)
a) Magnesium carbonate decomposes when strongly heated.
b) Ethene burns in air
6. A sample of acetone (C3H6O) is burned in an insulated calorimeter to produce
carbon dioxide gas and liquid water. (9 marks)
a) Use standard enthalpies of formation to calculate the theoretical value for the
molar enthalpy of combustion of acetone.
b) Suppose the experiment yielded the following results. Calculate the molar
enthalpy of combustion of acetone using the experimental data.
Quantity
Mass of water
Specific heat capacity of aluminum
Mass of aluminum can
Initial temperature of calorimeter
Final temperature of calorimeter
Mass of acetone burned
Data
120g
0.91J/goC
70g
20oC
26oC
0.087g
c) Calculate the percentage error by comparing the predicted and experimental
values.
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SCH4U
Grade 12
University Chemistry
Lesson 11 – Energy Options
SCH4U – Chemistry
Lesson 11
Lesson 11: Energy Options
Energy changes occur in our trillions of cells that make up our body. We consume food,
and then convert our food into useful energy called adenosine triphosphate or ATP. We
also rely on energy to sustain our lifestyles. In this lesson, you will learn about the
various ways energy can be harvested to meet these demands, including their
advantages and disadvantages.
What You Will Learn
After completing this lesson, you will;
•
compare conventional and alternative sources of energy with respect to efficiency
and environmental impact (e.g., burning fossil fuels, solar energy, nuclear fission);
Most of society’s energy needs are met by two sources: electricity and the combustion
of fossil fuels such as natural gas and oil. In Ontario, electricity has three sources;
hydroelectric, nuclear power, and the burning of fossil fuels. In all three sources,
potential energy is converted into kinetic energy.
All sources of energy have advantages and disadvantages. Hydroelectric power can
flood large areas and displace wildlife from its natural habitat. Burning fossil fuels can
cause smog and pollution such as acid rain. Nuclear power is very expensive and there
are concerns about safety and waste disposal. We will discuss nuclear power in more
detail below.
Nuclear Fission
In the generation of nuclear power, the most common reaction is the splitting of a
uranium atom into two small nuclei. Nuclear fission reactions provide the energy for
nuclear power generating systems.
As soon as the nucleus of the uranium atom captures the neutron, it splits into two
lighter atoms and throws off two or three new neutrons (the number of ejected neutrons
depends on how the U-235 atom splits). The process of capturing the neutron and
splitting happens very quickly, on the order of picoseconds (1x10-12 seconds).
The splitting of an atom releases an incredible amount of heat and gamma radiation, or
radiation made of high-energy photons. The two atoms that result from the fission later
release beta radiation (super fast electrons) and gamma radiation of their own as well.
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SCH4U – Chemistry
Lesson 11
There is controversy about use of nuclear power.
Advantages
Disadvantages
•
•
•
•
•
low costs
little air pollution
reduces dependence on fossil fuels
•
•
•
release of radioactive material
difficulty disposing of toxic
radioactive waste
cost of building nuclear generators
unknown health effects of low-level
radiation
thermal pollution
Key Question #11
1. There are many methods for generating electricity. These include:
Hydroelectric power
Fossil Fuels
Nuclear Power
Geothermal
Wind
Solar
a) Research the following methods of generating electricity. Include the following;
Æ advantages and disadvantages
Æ efficiency, in terms of energy per dollar spent
Æ efficiency in terms of energy output per gram of fuel used
Æ environmental impact
Æ comparison to Canadians versus other areas of the world
Your research must be summarized in a clear logical concise manner and
include references that are reputable such as government agencies. Search
engines such as Google, and Wikipedia are not reputable. Look for government
or educational agencies for gathering information.
b) Write a report expressing the direction that you feel the government should take
on energy policy. Your report must be backed up by data from your research
and should be 1-2 pages in length, in a word processed format. (25 marks – 20
for content/research, 5 for grammar/spelling)
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SCH4U
Grade 12
University Chemistry
Lesson 12 – Chemical Kinetics
SCH4U – Chemistry
Lesson 12
Lesson 12: Chemical Kinetics
Some reactions, such as the rusting of a nail, occur very slowly. Other reactions such
as the explosion of a firework occur very rapidly. Why are some reactions so fast that
they occur instantaneously, while other reactions are so slow, we don’t even know that
they are occurring?
This unit discusses the concept of reaction rates. It forms a branch of chemistry called
chemical kinetics.
What You Will Learn
After completing this lesson, you will;
•
•
•
•
describe, with the aid of a graph, the rate of reaction as a function of the change of
concentration of a reactant or product with respect to time; express the rate of
reaction as a rate law equation (first- or second-order reactions only); and explain
the concept of half-life for a reaction;
demonstrate understanding that most reactions occur as a series of elementary
steps in a reaction mechanism.
analyse simple potential energy diagrams of chemical reactions (e.g., potential
energy diagrams showing the relative energies of reactants, products, and activated
complex);
explain, using collision theory and potential energy diagrams, how factors such as
temperature, surface area, nature of reactants, catalysts, and concentration control
the rate of chemical reactions;
Expressing & Measuring Reaction Rates
Expressing Reaction Rate
Chemical Kinetics is the study of ways to make a reaction proceed faster or slower.
The pace at which the reaction occurs is termed the reaction rate. The reaction rate is
the change in amount of reactants (or products) over time. Reaction rate is usually
expressed as [concentration reactant] per unit of time (mol/Lys), and reaction rates are
always positive, by convention
Rate =
Δc
Δt
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SCH4U – Chemistry
Lesson 12
Example 1: What is the overall rate of production of nitrogen dioxide in the system
shown below if the concentration of nitrogen dioxide changes from 0.34mol/L to
0.82mol/L in 4 minutes?
N2(g) + 2O2(g) Æ 2NO2(g)
Solution 1:
Given:
Δc = 0.82mol / L − 0.34mol / L
= 0.48mol / L
Required:
Rate of production = ?
Equation:
Rate =
Solution:
ΔT = 4min
Δc
Δt
Δc
Δt
0.48mol / L
=
4min
= 0.12mol / L.min
Rate =
Average & Instantaneous Reaction Rate
•
reaction rates are not usually constant, they change with time
Average Rate - averages the change in [reactant] or [product] per unit of time
over a given time interval (secant line)
Instantaneous Rate is the rate of a reaction at a particular time (tangent line)
Considering the graph below, the average rate of formation of oxygen gas would be
calculated over a given time from (i.e. over 10 seconds), whereas the instantaneous
rate would be for a given moment (i.e. at t = 10seconds). This is determined using a
tangent.
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SCH4U – Chemistry
Lesson 12
Rate of Formation of Oxygen Gas
0.009
0.008
0.007
[O2] 0.006
0.005
(mol/L)
0.004
0.003
0.002
0.001
0
0
1000
2000
3000
4000
5000
6000
7000
Time (s)
Example: Consider the following reaction:
IO3(aq)- + 5I(aq)- + 6H+(aq)Æ 3I2(aq) + 3H2O(l)
What are the reaction rates of the various reactant and products? The reaction rate of
iodate (IO3(aq)-) ions is 1.2 x 10-2mol/L.s
In order to calculate the rate of the other reactants and products, we must consider the
moles of each reactant and product.
Thus:
I(aq)- = 1.2 x 10-2mol/L.s x 5 = 0.06 mol/L.s
H+(aq) = 1.2 x 10-2mol/L.s x 6 = 0.072 mol/L.s
I2(aq)+ = 1.2 x 10-2mol/L.s x 3 = 0.036 mol/L.s
H2O(l)+ = 1.2 x 10-2mol/L.s x 3 = 0.036 mol/L.s
Measuring Reaction Rates
There are various ways that reaction rate can be measured in the laboratory. The
method employed depends on the nature of the reaction (the reactants, how the
reaction proceeds, type of reaction). The methods used are the following:
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SCH4U – Chemistry
Lesson 12
Reactions that produce a gas
In reactions that produce a gas, the chemist can collect gas and measure the volume
and pressure changes as a reaction occurs to get an indication of rate. Generally, the
faster the reaction, the greater the change in pressure or volume in the same interval.
Reactions that Involve ions
Ions conduct electricity, so the more ions that are formed, the greater the conductivity.
Reactions that change colour
Many reactants will undergo a colour change when they form products. The intensity of
the colour can be measured using a device called a spectrophotometer, which
measures relative colour intensity.
Support Questions
20. Consider the following reaction:
MnO4-(aq) + 5Fe2+(aq) + 8H+(aq) Æ Mn2+(aq) + 5Fe3+(aq) + 4H2O(l)
Given that the rate of reaction is 3.5 x 10-2 mol/L MnO4-(aq), determine the rate of
reaction for all reactants and products in the reaction.
21. The reaction between ammonia and oxygen produces nitrogen monoxide and water
vapour.
4NH3(g) + 5O2(g) Æ 4NO(g) + 6H2O(g)
If the rate of consumption of ammonia is 1.6 x 10-2 mol/(L.s) calculate the rate of
consumption of oxygen and the formation of water vapour.
22. Rates of reactions are generally fastest at the beginning of a reaction. Explain why
this is so.
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SCH4U – Chemistry
Lesson 12
Support Questions (continued)
23. A kinetics experiment is performed in which oxygen gas is collected. The
concentration of oxygen is measured every 10 s. The table below has the data
obtained from the experiment.
Time (s)
0
10
20
30
40
50
60
70
80
90
100
Oxygen gas (mol/L)
0
0.23
0.40
0.52
0.60
0.66
0.70
0.73
0.75
0.76
0.76
a) Plot a graph of oxygen concentration vs. time
b) Calculate the average formation of oxygen gas during the first 60 seconds
c) the average rate of formation of oxygen gas between t = 20s and t = 60s
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SCH4U – Chemistry
Lesson 12
Factors Affecting Reaction Rate
There are five major factors that affect the rate of a reaction. They are;
•
•
•
•
•
Chemical nature of the reactants
Concentration of reactants
Temperature
Presence of a catalyst
Surface Area
Chemical Nature of Reactants - The rate of any reaction partially depends of the
chemical nature of the reactants. For instance, metals generally will react with acid to
produce hydrogen gas. However the rate of gas evolved varies from metal to metal.
Understanding the important properties of metals allows chemists to choose metals that
will react more slowly when trying to prevent corrosion, for example.
Concentration - Generally if the concentration of the reactants is increased, the rate of
the reaction increases. This is because there is a higher concentration of reactant
available for collision to form product.
Temperature - An increase in temperature speeds up molecule collisions and thus
speeds of reaction rates and conversely cooling the temperature will slow reaction
rates.
Presence of a Catalyst - A catalyst is a substance that speeds up a chemical reaction
without being consumed in the reaction itself. In your body, organic catalysts are called
enzymes and speed up important cellular processes such as cellular respiration. When
you consume food, your saliva releases an enzyme called amylase which speeds up
the breakdown of starch into simpler sugars. Many chemical industrial processes rely on
catalysts to speed reaction rates.
Surface Area - Increasing the surface area increases the rate of reaction. You may
have experienced this when cooking food. For example, chopping up potatoes into
smaller allows the potatoes to cook faster.
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SCH4U – Chemistry
Lesson 12
Support Questions
24. Identify the five factors that increase the rate of a reaction and provide an example
of each.
The Rate Law: Reactant Concentration and Rate
Relating Reactant Concentrations and Rate
Consider:
aA + bB → products
In general, the rate of a reaction increases when
the concentrations of the reactants increase
Rate ∝ [ A]
This relationship can be expressed in a general
equation called the
Rate = k [ A ]
m
[B ]
n
m
[B ]
n
Rate Law Equation:
•
•
the exponents m and n must be determined by experiment
they do not necessarily correspond to the stoichiometric coefficients of their
reactants (they are usually 1 or 2, but values of 0,3, or fractions can occur)
Order of the Reaction
The values of the exponents establish the order of the reaction. The sum of the
exponents (m + n) is known as the overall reaction order.
Consider the following theoretical reaction;
2X + 2Y + 3Z Æ products
Experimental evidence suggests the following rate law equation
r = k[X]1[Y]2[Z]0
The overall order of the reaction in this case is 3 (1+2+0)
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SCH4U – Chemistry
Lesson 12
Rate Law Constant, k
The magnitude of the rate constant, k, indicates the speed of a reaction
¾ small k value indicates a slow reaction
¾ large k value indicates a fast reaction
¾ k remains constant throughout a reaction under constant conditions
Determining the Rate Law Exponents
The exponents must be determined using experimental evidence. One method of
directly measuring k, m, and n is called the method of initial rates. By measuring the
initial rate (the rate near reaction time zero) for a series of reactions with varying
concentrations of one reactant at a time, we can deduce to what power the rate
depends on the concentration of each reactant. For example, let's use the method of
initial rates to determine the rate law for the following reaction:
C3H6O + Br2 Æ C2H5OBr + HBr
whose, rate law has the form:
rate = k[C3H6O]m[Br2]n
Using the following initial rates data, it is possible to calculate the order of the reaction
for both bromine and acetone:
Experiment
1
2
3
[BI2]o
0.1 M
0.2 M
0.1 M
[C3H6O]0
0.1 M
0.1 M
0.2 M
Initial Rate (M•s-1)
1.64 x 10-5
1.65 x 10-5
3.29 x 10-5
To calculate the order of the reaction for bromine, notice that experiments 1 and 2 hold
the concentration of acetone constant while doubling the concentration of bromine. The
initial rate of the reaction is unaffected by the increase in bromine concentration, so the
reaction is zero order in bromine. We can prove this mathematically by taking the ratio
of the rates from experiments 1 and 2:
rate2 k [Br2 ]2m [C3H6O ]2n (0.2M )m (0.1M )n 1.64 × 10 −5
=
=
=
=1
rate1 k [Br2 ]1m [C3H6O ]1n (0.1M )m (0.1M )n 1.64 × 10−5
As you can see in the above equations, by holding the concentrations of all but one
species constant between two experiments, you can calculate the order of the reaction
in a single reactant at a time. By similar reasoning, we can conclude that because the
rate of reaction doubled when the concentration of acetone was doubled (cf.
Copyright © 2008, Durham Continuing Education
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SCH4U – Chemistry
Lesson 12
experiments 1 and 3) the reaction must be first order in acetone. However, had the rate
quadrupled or octupled with a doubling of the acetone concentration, the reaction would
have been second or third order in acetone, respectively. In practice, you will likely
never see a reaction with an order higher than 3. If you calculate an order higher than 3
for a reaction, double check your math because that is highly unusual. If you compute a
fractional power for a reactant's order, do not be discouraged; they are quite common
(especially half-order reactions).
Therefore the rate law for this reaction is:
Rate = k [Br2]0[C3H6O]1
The overall order of the reaction is 1
Example 2: Consider the following reaction
2BrO3(aq)- + 5HSO3(aq)- Æ Br2(g) + 5SO4(aq)2- + H2O(l) + 3H+(aq)
The following experimental evidence was collected to determine the rate law
Trial
1
2
3
Initial [BrO3(aq)-]
(mmol/L)
4.0
2.0
2.0
Initial [HSO3(aq)-]
(mmol/L)
6.0
6.0
3.0
Initial rate
(mmol/L.s)
1.60
0.80
0.20
Solution 2:
Considering the experimental data, determine the rate law.
First state the rate law:
Rate = k[BrO3(aq)-]m[HSO3(aq)-]n
Begin by identifying which trials change the concentration for only one reactant.
Comparing trials one and two, note how the concentration of [BrO3(aq)-] is doubled in trial
one compared to 2, while the concentration of [HSO3(aq)-] remains the same.
According to the experimental results, if the concentration of bromate ions is doubled,
the rate of the reaction is also doubled.
This means 2m = 2 (Read as double concentration to the rate exponent m equals double
rate)
Solving m = 1 or 21 = 2. Thus, the order of the reaction for [BrO3(aq)-] ions is 1.
Copyright © 2008, Durham Continuing Education
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SCH4U – Chemistry
Lesson 12
Comparing trials two and three, note that the concentration of [BrO3(aq)-] ions remains
constant, while the concentration of [HSO3(aq)-] ions is doubled.
According to the experimental results, if the concentration of [HSO3(aq)-] is doubled, the
rate of the reaction is quadrupled.
This means 2n = 4
Solving n = 2 or 22 = 4. Thus the order of the reaction for [HSO3(aq)-] ions is 2.
Restate rate law
r = k[BrO3(aq)-]1[HSO3(aq)-]2
b) Using your rate law expression, solve for k
Pick any trial to solve for k. In this case, we will use the data for concentration and rate
from trial one:
k=
0.00160mol / L.s
0.0040mol / L × (0.0060mol / L )2
= 1.1× 104 L2 / mol 2 • s −1
Table 12.1: Summary of Reaction Rate Orders
Order of Reaction
Concentration
0
1
2
change
10 =1
11 =1
12 = 1
X1
20 =1
21 =2
22 = 4
X 2 (doubling)
0
1
3 =1
3 =3
32 = 9
X 3 (tripling)
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3
13 = 1
23 = 8
33 = 27
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SCH4U – Chemistry
Lesson 12
Support Questions
25. For a particular reaction at constant temperature,
A(g) + 2 B(g) Æ products
initial
[A]
1.00
2.00
3.00
4.00
initial
[B]
1.00
4.00
9.00
2.00
initial
rate
1.00
8.00
27.00
?
What is the value of "?" in this table?
26. What is a rate law? What is the proportionality constant called?
27. What is meant by the order of a reaction?
28. The rate law for the following reaction 2 NO + O2 Æ 2 NO2 is rate = k [NO]2[O2]. At
25oC, k = 7.1 X 109 L mol-2s-1. What is the rate of reaction when [NO] = 0.0010 mol/L
and [O2] = 0.034 mol/L?
29. The initial rate of the reaction:
BrO3-(aq) + 5 Br-(aq) + 6 H+(aq) Æ 3 Br2(l) + 3H2O(l)
Has been measured at the reactant concentrations shown (in mol/L):
Experiment
1
2
3
4
[BrO3-]
0.10
0.20
0.10
0.10
[Br-]
0.10
0.10
0.20
0.10
[H+]
0.10
0.10
0.10
0.20
Initial rate (mol/Ls)
8.0 x 10-4
1.6 x 10-3
1.6 x 10-3
3.2 x 10-3
According to these results what would be the initial rate (in mol/Ls) if all three
concentrations are:
[BrO3-] = [Br-] = [H+] = 0.20 mol/L?
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Lesson 12
Support Questions
30. The reaction of iodide ion with hypochlorite ion, OCl- (which is found in liquid
bleach), follows the equation;
OCl- + I- Æ OI- + ClIt is a rapid reaction that gives the following rate data.
Initial Concentrations
[OCl-]
[I-]
1.7 X 10-3 1.7 X 10-3
3.4 X 10-3 1.7 X 10-3
1.7 X 10-3 3.4 X 10-3
Rate of Formation
(mol L-1 s-1)
(mol/L) of Cl1.75 X 104
3.50 X 104
3.50 X 104
What is the rate law for the reaction? Determine the value of the rate constant.
Relating Reaction Rate to Time
Sometime the rate of reaction in time can be measured by a visible change, by
measuring the elapsed time before the visible change occurs. The average rate is
inversely proportional to the elapsed time:
1
rav ∝
Δt
Chemical Kinetics and Half-Life
•
•
•
•
the half-life, t1/2, of a reaction is the time that is needed for the reactant mass or
concentration to decrease by one half of its initial value
SI units for t1/2 are seconds, but it is usually expressed in whatever units of time are
appropriate to the reaction
t1/2 of any first-order reaction is always constant and it depends on k
(t1/2 is independent of initial concentration of the reactant)
t1/2 for any first order reaction can be calculated using this equation:
t1 =
2
0.693
k
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SCH4U – Chemistry
Lesson 12
Example 3: Calculating with Half-Lives
Solution 3: The radioisotope lead-212 has a half-life of 10.6h. What is the rate
constant for this isotope?
Given:
R = k ⎡⎣Pb−212 ⎤⎦
Required:
k =?
Equation:
t1 =
2
1
t 1 = 10.6h
2
0.693
k
0.693
2
k
0.693
k=
t1
t1 =
Solution:
2
0.693
10.6h
= 0.0654h −1
=
Statement:
The half-life rate constant for the radioisotope lead-212 is 0.0654h-1.
Support Questions
31. A radioisotope has a half-life of 24 s and an initial mass of 0.084g.
What mass of radioisotope will remain after;
(i) 72 s?
(ii) 192 s?
Energy Changes & Rates of Reaction
Theories of Reaction Rates
Rates of reaction can be explained with collision theory. Since molecules are held
together by chemical bonds, energy must be provided to break these bonds. The
collision theory explains why some reactions are faster than others.
Collision Theory
For a reaction to occur the reactant particles must collide. Only a certain fraction of the
total collisions cause chemical change; these are called successful collisions. The
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SCH4U – Chemistry
Lesson 12
successful collisions have sufficient energy (activation energy) at the moment of impact
to break the existing bonds and form new bonds, resulting in the products of the
reaction. Increasing the concentration of the reactants and raising the temperature bring
about more collisions and therefore more successful collisions, increasing the rate of
reaction.
Figure 12.1: A successful collision has sufficient energy and proper orientation.
Key Points about Collision Theory
•
•
•
higher concentration of reactant particles will increase the number of collisions
between the particles per second
for solid reactants, greater surface area means that more collisions can occur
collisions must be effective
9 correct orientation of reactants/collision geometry
9 sufficient collision energy/activation energy (Ea)
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Lesson 12
Activation Energy
The activation energy is the minimum
increase in potential energy of a system
required for molecules to react.
Figure 12.2: Potential energy diagram
depicting activation energy
Transition State Theory is used to explain
what happens when molecules collide in a
reaction. It examines the transition, or
change, from reactants to products. The
kinetic energy (KE) of reactants is
transferred to potential energy (PE) as the
reactants collide, due to the law of conservation of energy
•
Transition state can be represented by a potential energy diagram
9 “hill” illustrates the activation energy (Ea) barrier of the reaction
9 a slow reaction has a high Ea barrier
9 a fast reaction has a low Ea barrier
9 transition state occurs at the top of the hill; species is an activated
complex that is neither product nor reactant
Figure 12.3: Potential energy
diagram showing the transition
state in which the molecules are
in an activated state (meaning
they are neither reactant nor
products)
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Exothermic Reactions
y reactants have a higher PE than the products
Endothermic Reactions y reactants have a lower PE than products
Note: The overall change in the potential energy is the enthalpy change (∆H). There is
no way to predict the Ea of a reaction by its ∆H.
Reaction Mechanisms & Catalysts
Elementary Reactions
Most chemical reactions actually occur as a sequence of elementary steps. The overall
sequence is called the reaction mechanism.
Let’s looks at a chemical reaction involving the combustion of acetylene gas to explain
this concept:
2 C2H2 + 5 O2 Æ 4 CO2 + 2 H2O
According to the equation, two molecules of acetylene (C2H2) react with 5 molecules of
oxygen. The collision theory states these molecules must collide in order to react.
However, it is highly unlikely that 7 molecules would collide together all at once.
Instead, the reaction most likely occurs in a series of simple steps which only required
two or three molecules colliding at any one instant. Although these steps cannot always
actually be observed, chemists can often make predictions about the sequence of
events.
For example, nitrogen monoxide reacts with oxygen according to the equation
2 NO(g) + O2 → 2 NO2
This reaction does not occur in a single step, however, but rather through these two
steps:
Step 1:
2 NO Æ N2O2
Step 2:
N2O2 + O2 Æ2 NO2
Notice that if you add these two reactions together, you end up with the overall reaction:
Step 1:
2 NO Æ N2O2
Step 2:
N2O2 + O2 Æ 2 NO2
Overall:
2 NO(g) + O2 Æ 2 NO2
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Dinitrogen dioxide (N2O2) cancels out and does not appear in our overall equation.
Substances such as this are called reaction intermediates and are typically shortlived.
Given an overall reaction, it is not possible to predict what the reaction mechanism
would be. However, if you are given the steps of a reaction mechanism you will need to
be able to add together individual steps to end up with the overall reaction.
Rate Determining Step
Here is another reaction mechanism with some additional information concerning the
relative rates of each of the individual steps:
Step 1:
HBr + O2 → HOOBr
Slow
Step 2:
HOOBr + HBr → 2 HOBr
Fast
Step 3:
HOBr + HBr → H2O + Br2
Fast
Step 4:
HOBr + HBr → H2O + Br2
Fast
Overall:
4 HBr + O2 → 2 H2O + 2 Br2
We don't have values for the actual rates of these individual steps, but here is a
question for you to consider - would you consider the overall reaction to be fast or slow?
With three fast steps and only one slow step, many of you will predict that the reaction
will be fast.
But let's make up some extreme numbers and ask the question again:
Step 1:
HBr + O2 Æ HOOBr
1 year
Step 2:
HOOBr + HBr Æ 2 HOBr
0.1 s
Step 3:
HOBr + HBr Æ H2O + Br2
0.1 min
Step 4:
HOBr + HBr Æ H2O + Br2
0.1 min
Overall:
4 HBr + O2 Æ 2 H2O + 2 Br2
Now would you consider the overall reaction to be fast or slow? Clearly it is a slow
reaction, taking over a year to complete despite some fast steps.
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SCH4U – Chemistry
Lesson 12
The overall rate of any reaction depends on the rate of the slowest step. This slowest
step is called the rate determining step. If you want to speed up a reaction, this is
where you should focus your attention.
Proposing & Evaluating Mechanisms (Summary)
•
•
•
the equations for the elementary steps must combine to give the overall equation
the proposed elementary steps must be reasonable
the mechanism must support the experimentally determined rate law
9 1 elementary reaction is much slower than the rest (rate-determining step)
9 rate-determining steps can occur anywhere in a reaction mechanism
Let’s try an example:
Consider the decomposition of dinitrogen pentoxide:
2N2O5(g) Æ 2N2O4(g) + O2(g)
a) What would the rate equation be if the reaction occurred in a single step?
The rate equation would be the same as the co-efficient on the reactant. Therefore the
rate law would be
r = k[N2O5]2
b) Suppose the actual rate law is r = k[N2O5]1
What is the rate determining step?
Because the coefficient on the reactant must be the same as the exponent in the rate
equation, the rate-determining step must be
1N2O5 Æ some intermediate product
Suggest a possible mechanism for this reaction
N2O5 Æ N2O4 + O (slow)
O + N2O5 Æ N2O4 + O2 (fast)
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Support Questions
32. Consider the overall reaction involving three elements as reactants and a compound
as the product:
X + 2Y + 2Z Æ XY2Z2
When a series of reactions is performed with different initial concentrations of
reactants. The results are as follows:
Doubling the concentration of X has no effect on the overall rate
Doubling the concentration of Y multiplies the overall rate by 4
Doubling the concentration of Z doubles the overall rate
State:
a)
b)
c)
d)
The rate law for this system
The rate determining step
A possible mechanism, indicating the slow step
A possible reaction intermediate in your mechanism
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Lesson 12
Key Question #12
1. At 25oC a catalyzed solution of formic acid produces 44.6mL of carbon monoxide
gas in 30s.
a) Calculate the rate of reaction with respect to CO(g) production. (2 marks)
b) What can you state about how long you would expect the production of the same
volume to take: (4 marks)
i) at 30oC?
ii) without the catalyst
2. Chlorine dioxide and hydroxide ions react to form chlorate ions, chlorite ions, and
water.
2ClO2(aq) + 2OH-(aq) Æ ClO3(aq) + ClO2-(aq) + H2O(l)
The reaction is found to be second order with respect to chlorine dioxide and first
order with respect to hydroxide ions.
a) Write a rate equation for the reaction. (1 mark)
b) What is the overall order of reaction? (1 mark)
c) What would you expect the effect on rate to be of doubling the concentration
of chlorine dioxide? (2 marks)
3. Nitric oxide, NO(g) reacts with chlorine gas, Cl2(g), in the reaction
2NO(g) + Cl2(g) Æ 2NOCl(g)
Initial rates of reaction are determined for various combinations of initial
concentrations of reactants and recorded below;
Trial
1
2
3
a)
b)
c)
d)
Initial [NO]
(mol/L)
0.10
0.10
0.20
Initial [Cl2]
(mol/L)
0.10
0.20
0.20
Rate of production
of NOCl (mol/L.s)
1.8 x 10-2
3.6 x 10-2
1.43 x 10-1
What is the rate law equation for the reaction? (1 mark)
What is the rate-determining step? (1 mark)
Calculate a value for the rate constant, including units. (3 marks)
Calculate the expected rate of reaction if the initial concentrations of NO and Cl2
gases were 0.30 and 0.40 mol/L, respectively. (3 marks)
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Lesson 12
Key Question #12 (continued)
4. Draw a sketch, roughly to scale, of the potential energy diagram for a system in
which Ea = +80kJ and ∆H = -20kJ. Label the axes, reactants, products, the
activation energy, the activated complex, and the enthalpy change. (5 marks)
5. An investigation is performed in which the concentration of nitrogen dioxide reacting
is measured as a function of time.
[NO2](mol/L)
0.500
0.445
0.380
0.340
0.250
0.175
Time (s)
0
12
30
45
90
180
a) Plot a graph of [NO] vs. time (5 marks)
b) Determine the average rate of reaction of NO2 between 10 and 60s (2 marks)
c) As the [NO2] halved, what was the effect on the instantaneous rate of reaction?
(2 marks)
6. If the half life of a radioisotope is 3.5s, what percentage of the original isotope
remains after 14s? (5 marks)
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SCH4U
Grade 12
University Chemistry
Support Question Answers
SCH4U – Chemistry
Support Question Answers
Support Question Answers
1. Identify each of the following as physical, chemical, or nuclear change.
a) A gas stove cooking pasta – chemical
b) an ice cube melting in a glass – physical
c) wax melts – physical
d) ice applied to a sore back – physical
2. Identify the system and the surroundings in each of the examples in the previous
question.
a) gas, pasta – system, environment outside of pot – surroundings
b) ice cube – system, glass, external environment – surroundings
c) wax – system, external environment - surroundings
d) ice cube – system, back and external environment – surroundings
3. A cup of water at 100oC has a higher temperature than a swimming pool full of water
at 20oC, but the pool has more thermal energy, Explain.
There are a greater number of water molecules, thus a greater amount of potential
energy.
4. A swimming pool contains 1000 L of water. When the water is warmed by solar
energy, its temperature increases from 15.8oC to 22.1oC, how much heat does the
water absorb? (cwater(liquid)= 4.184 J/gyoC)
Given:
m = 1000L = 1000kg = 1× 106 g
(since the density of water is 1g/mL)
Required:
Q=?
Equation:
Q = mc p ΔT
c water/liquid = 4.184J / g °C
ΔT = 22.1°C − 15.8°C
= 6.3°C
Q = mc p ΔT
Solution:
= 1× 106 g × 4.184J / g °C × 6.3 °C
= 26359200J or 26359.2kJ
Statement:
The swimming pool absorbs 26359.2 kJ of heat energy when
warmed to 22.1EC.
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5. A cook heats water from 20oC to 50oC. Calculate the mass of water that could be
warmed by the addition of 8.00kJ of heat.
Given:
c water/liquid = 4.184J / g °C
Q = 8kJ or 8000J
ΔT = 50°C − 20°C
= 30°C
Required:
m=?
Equation:
Q = mc p ΔT
m=
Solution:
Statement:
or
m=
Q
c ΔT
Q
c p ΔT
8000 J
4.184 J / g °C × 30 °C
= 63.73g
=
The mass of water that could be warmed 30EC with the addition of
8kJ of heat is 63.73g.
6. A 50% ethylene glycol solution has a specific heat capacity of 3.5J/goC. What
temperature change would be observed in a solution of 4kg of ethylene glycol if it
absorbs 250kJ of heat?
Given:
cethylene glycol 50% soln = 3.5J / g °C
m = 4kg or 4000g
Q = 250kJ or 250000J
Required:
ΔT = ?
Equation:
Q = mc p ΔT
ΔT =
Solution:
=
or
ΔT =
Q
mc p
Q
mc p
250000 J
4000 g × 3.5 J / g °C
= 17.86°C
Statement:
The heat change that would result is 17.86EC.
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7. Calculate the enthalpy change ΔH for the vaporization of 150.0g of water at 100oC.
Given:
ΔHvap = 40.83kJ (from table 9.3)
M = 18.0g / mol
m = 150g
Required:
η =?
ΔH = ?
Equation:
η = m ×M
ΔH = η × ΔHvap
Solution:
First calculate the moles of water.
η = m ×M
= 150 g × 18.0 g / mol
= 8.33mol
Now calculate the enthalpy change.
ΔH = η × ΔHvap
= 8.33 mol × 40.83g / mol
= 340.11kJ
Statement:
The enthalpy change for the vapourization of 150g of water at
100EC is 340.11kJ.
8. The energy in the process H2O(g) Æ H2O(l) could be described as a molar enthalpy of
condensation. State the type of molar enthalpy that would occur in each of the
following reactions:
a) Br2(l) Æ Br(g)
b) CH4(g) Æ CH4(s)
c) NaCl(s) Æ Na+(aq) + Cl-(aq)
d) KOH(aq) + HCl(aq) Æ KCl(aq) + H2O(l)
ans: vapourization
ans: sublimation
ans: solution
ans: neutralization
9. In the lab, you add 50 mL of concentrated hydrochloric acid (12 mol/L), HCl to form
250mL of dilute solution. The temperature of the solution changes from 18oC to
24oC. Calculate the molar enthalpy of dilution of hydrochloric acid.
Start by calculating the moles of HCl (Step 1)
Given:
Required:
c = 12molL
V = 0.050L
ηHCl = ?
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Equation:
η = c ×V
Solution:
ηHCl = 12 mol / L × 0.050 mol
= 0.60mol
Recognize and apply the Law of Conservation of Energy (Step 2)
ΔH (HCl dissolving ) = q ( calorimeter water )
∴ η × ΔHsol = mc ΔT
Given:
mwater = 250.0g ( 250mL = 250g )
c water = 4.18J / g °C
ΔT = 24°C − 18°C
= 6°C
Required:
ΔHsol = ?
Equation:
η × ΔHsol = mc ΔT
η × ΔHsol = mc ΔT
ΔHsol =
Solution:
=
mc ΔT
η
250 g × 4.18J / g °C × 6 °C
0.60mol
= 10450J / mol
Statement:
The molar enthalpy of a dilution of HCl is 10450J/mol.
10. What mass of lithium chloride must have dissolved if the temperature of 200g of
water increased by 6.0oC? The molar enthalpy of solution of lithium chloride is
37kJ/mol.
Given:
mwater = 200.0g
ΔHsol = 37kJ / sol
MLiCl = 42.4g / mol
Required:
mLiCl = ?
Equation:
η × ΔHsol = mc ΔT
ΔT = 6°C
c water = 4.18J / g °C
m =η ×M
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η × ΔHsol = mc ΔT
ηLiCl =
Solution:
=
mc ΔT
ΔHsol
200 g × 4.18 J / g °C × 6 °C
37000 J / mol
= 0.1356mol
m =η ×M
mLiCl = 0.14 mol × 42.4g / mol
= 5.7g
Statement:
It would require 5.7g of LiCl to complete this reaction.
11. Calculate ∆H for the reaction: C2H4(g) + H2(g) Æ C2H6(g), from the following data.
C2H4 (g) + 3O2 (g) Æ 2CO2 (g) + 2H2O (l)
∆HE = -1411. kJ
C2H6 (g) + 3½O2 (g) Æ 2CO2 (g) + 3H2O (l)
∆HE = -1560. kJ
H2 (g) + ½ O2 (g) Æ H2O (l)
∆HE = -285.8 kJ
Solution
1
C2H4 (g) + 3 O2 (g) Æ 2 CO2 (g) + 2 H2O (l)
∆HE = -1411. kJ
Reverse 2 CO2 (g) + 3 H2O (l) Æ C2H6 (g) + 3½ O2 (g)
2
∆HE = +1560. kJ
3
∆HE = -285.8 kJ
H2 (g) + 1/2 O2 (g) Æ H2O (l)
1+R2+3 C2H4 (g) + H2 (g) Æ C2H6 (g)
∆HE = -137. kJ
12. Calculate ∆H for the reaction 4 NH3 (g) + 5 O2 (g) Æ 4 NO (g) + 6 H2O (g), from the
following data.
N2(g) + O2(g) Æ 2NO(g)
∆HE = -180.5 kJ
N2(g) + 3H2(g) Æ 2NH3(g)
∆HE = -91.8 kJ
2H2(g) + O2(g) Æ 2H2O(g)
∆HE = -483.6 kJ
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Solution
1x2
2 N2(g) + 2 O2(g) Æ 4 NO(g)
∆HE = 2 (-180.5 kJ)
Reverse 4 NH3(g) Æ2 N2 (g) + 6 H2 (g)
2
∆HE = 2 (+91.8 kJ)
3x3
6 H2(g) + 3 O2(g) Æ 6 H2O(g)
∆HE = 3 (-483.6 kJ)
4 NH3(g) + 5 O2(g) Æ 4 NO(g) + 6 H2O(g)
∆HE = -1628. kJ
13. Find ∆H° for the reaction 2H2(g) + 2C(s) + O2(g) Æ C2H5OH(l), using the following
thermochemical data.
C2H5OH(l) + 2O2(g) Æ 2 CO2(g) + 2H2O(l)
∆HE = -875. kJ
C(s) + O2(g) Æ CO2(g)
∆HE = -394.51 kJ
H2(g) + ½O2(g) Æ H2O(l)
∆HE = -285.8 kJ
Solution
Reverse 2 CO2 (g) + 2 H2O (l) Æ C2H5OH (l) + 2 O2 (g)
1
∆HE = +875. kJ
2 x2
2 C (s) + 2 O2 (g) Æ 2 CO2 (g)
∆HE = 2 (-394.51 kJ)
3x2
2 H2 (g) + O2 (g) Æ 2 H2O (l)
∆HE = 2 (-285.8 kJ)
2H2(g) + 2C(s) + O2(g) Æ C2H5OH(l)
∆H° = -486. kJ
14. Calculate ∆H for the reaction CH4(g) + NH3(g) Æ HCN(g) + 3H2(g), given:
N2(g) + 3H2(g) Æ 2NH3(g)
∆HE = -91.8 kJ
C(s) + 2H2(g) Æ CH4(g)
∆HE = -74.9 kJ
H2(g) + 2C(s) + N2(g) Æ 2HCN(g)
∆HE = +270.3 kJ
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Solution
1 x½
NH3(g) Æ 1/2 N2(g) + 3/2 1H2(g)
∆HE = ½(+91.8 kJ)
Reverse CH4(g) Æ C(s) + 2 H2(g)
2
∆HE = +74.9 kJ
3x½
1/2 H2(g) + C(s) + 1/2 N2(g) Æ HCN(g)
∆HE = ½(+270.3 kJ)
CH4(g) + NH3(g) Æ HCN(g) + 3 H2(g)
∆HE = +256.0 kJ
15. Calculate ∆H for the reaction 2Al(s) + 3Cl2(g) Æ 2AlCl3(s) from the data.
2Al(s) + 6HCl(aq) Æ 2AlCl3(aq) + 3H2(g)
∆HE = -1049. kJ
HCl(g) ÆHCl(aq)
∆HE = -74.8 kJ
H2(g) + Cl2(g) Æ 2HCl(g)
∆HE = -1845. kJ
AlCl3(s) Æ AlCl3(aq)
∆HE = -323. kJ
Solution
1
2 Al (s) + 6 HCl (aq) Æ 2 AlCl3 (aq) + 3 H2 (g)
∆HE = -1049. kJ
2x6
6 HCl (g) Æ 6 HCl (aq)
∆HE = 6 (-74.8 kJ)
3x3
3 H2 (g) + 3 Cl2 (g) Æ 6 HCl (g)
∆HE = 3 (-1845. kJ)
Reverse
4 x2
2 AlCl3 (aq) Æ 2 AlCl3 (s)
∆HE = 2 (+323. kJ)
2 Al (s) + 3 Cl2(g) Æ 2 AlCl3(s)
∆HE = -6387. kJ
16. Write formation equations for the following:
a) benzene (C6H6)
b) potassium bromate
c) glucose
a) 6C(s) + 3H2(g) Æ C6H6
b) K(s) + 1/2Br2(g) + 2O2 Æ KBrO4
c) 6C(s) + 6H2(g) + 3O2(g) Æ C6H12O6
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17. Use standard enthalpies of formation to calculate:
a) the molar enthalpy of combustion for pentane.
8
O → 5CO2( g) + 6H2O(l)
2 2(g)
ΔH = ∑ηΔH °f (products) − ∑ηΔH °f (reactants )
C5H12 +
Equation:
Solution:
kJ ⎞ ⎛
kJ ⎞ ⎤ ⎡ ⎛
kJ ⎞ ⎛ 8
kJ ⎞ ⎤
⎡⎛
ΔH = ⎢⎜ 5mol × −393.5
⎟ + ⎜ 6mol × −285.8
⎟ ⎥ − ⎢ ⎜ 1mol × −173.5
⎟ + ⎜ mol × 0
⎟
mol ⎠ ⎝
mol ⎠ ⎦ ⎣ ⎝
mol ⎠ ⎝ 2
mol ⎠ ⎥⎦
⎣⎝
⎡
⎢⎣
= −3682.30
= −3508.8
kJ ⎤ ⎡
kJ ⎤
− −173.5
⎥
⎢
mol ⎦ ⎣
mol ⎥⎦
kJ
mol
b) the enthalpy change that accompanies the reaction between solid iron (III) oxide
and carbon monoxide gas to produce solid iron metal and carbon dioxide gas.
Fe 2O3(s ) + 3CO(q) → 2Fe( s ) + 3CO2(q)
Equation:
ΔH = ∑ηΔH °f (products ) − ∑ηΔH °f (reactants )
Solution:
kJ ⎞ ⎛
kJ ⎞ ⎤ ⎡⎛
kJ ⎞ ⎛
kJ ⎞ ⎤
⎡⎛
ΔH = ⎢⎜ 3mol × −393.5
⎟ + ⎜ 2mol × 0
⎟ ⎥ − ⎢⎜ 1mol × −824.2
⎟ + ⎜ 3mol × −110.5
⎟
mol
mol
mol
mol ⎥
⎠ ⎝
⎣⎝
kJ ⎤ ⎡
kJ ⎤
⎡
− ⎢ −1155.7
= ⎢ −1180.50
⎥
mol ⎦ ⎣
mol ⎥⎦
⎣
= −24.8
⎠ ⎦ ⎣⎝
⎠ ⎝
⎠⎦
kJ
mol
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18. For each of the following reactions, use standard enthalpies to calculate ∆H.
a)
CH4(q) + H2O( q) → CO(q) + 3H2(q)
Equation:
ΔH = ∑ηΔH °f (products ) − ∑ηΔH °f (reactants )
Solution:
kJ ⎞ ⎛
kJ ⎞ ⎤ ⎡⎛
kJ ⎞ ⎛
kJ ⎞ ⎤
⎡⎛
⎟ + ⎜ 3mol × 0
⎟ ⎥ − ⎢⎜ 1mol × −74.8
⎟ + ⎜ 1mol × −241.8
⎟
mol ⎠ ⎝
mol ⎠ ⎦ ⎣⎝
mol ⎠ ⎝
mol ⎠ ⎥⎦
⎣⎝
kJ ⎤ ⎡
kJ ⎤
⎡
= ⎢ −110.5
− ⎢ −316.2
⎥
mol ⎦ ⎣
mol ⎥⎦
⎣
ΔH = ⎢⎜ 1mol × −110.5
= 206.1
kJ
mol
b)
CO(q) + H2O(q) → CO2(q) + H2(q)
Equation:
ΔH = ∑ηΔH °f (products) − ∑ηΔH °f (reactants )
Solution:
kJ ⎞ ⎛
kJ ⎞ ⎤ ⎡⎛
kJ ⎞ ⎛
kJ ⎞ ⎤
⎡⎛
ΔH = ⎢⎜ 1mol × −393.5
⎟ + ⎜ 1mol × 0
⎟ ⎥ − ⎢⎜ 1mol × −110.5
⎟ + ⎜ 1mol × −241.8
⎟
mol ⎠ ⎝
mol ⎠ ⎦ ⎣⎝
mol ⎠ ⎝
mol ⎠ ⎥⎦
⎣⎝
⎡
⎣
= ⎢ −393.5
= −41.2
kJ ⎤ ⎡
kJ ⎤
− ⎢ −352.3
⎥
mol ⎦ ⎣
mol ⎥⎦
kJ
mol
c)
N2(q) + 2H2(q) → 2NH3(q)
Equation:
ΔH = ∑ηΔH °f (products ) − ∑ηΔH °f (reactants )
Solution:
kJ ⎞ ⎤ ⎡⎛
kJ ⎞ ⎛
kJ ⎞ ⎤
⎡⎛
ΔH = ⎢⎜ 2mol × −46.1
⎟ ⎥ − ⎢⎜ 1mol × 0
⎟ + ⎜ 3mol × 0
⎟
mol
mol
mol ⎥
⎣⎝
= −92.2
⎠ ⎦ ⎣⎝
⎠ ⎝
⎠⎦
kJ
mol
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19. Ammonium nitrate fertilizer is produced by the reaction of ammonia with nitric acid:
NH3(g) + HNO3(l) Æ NH4NO3(s)
a) Use standard enthalpies of formation to calculate the standard enthalpy change
of the reaction used to produce ammonium nitrate.
ΔH = ∑ηΔH °f (products ) − ∑ηΔH °f (reactants )
Equation:
Solution:
kJ ⎞ ⎤ ⎡⎛
kJ ⎞ ⎛
kJ ⎞ ⎤
⎡⎛
ΔH = ⎢⎜ 1mol × −365.6
⎟ ⎥ − ⎢⎜ 1mol × −46.1
⎟ + ⎜ 1mol × −174.1
⎟
mol
mol
mol ⎥
⎠ ⎦ ⎣⎝
⎣⎝
kJ ⎤ ⎡
kJ ⎤
⎡
220.0
= ⎢ −365.6
−
−
mol ⎥⎦ ⎢⎣
mol ⎥⎦
⎣
= −145.4
⎠ ⎝
⎠⎦
kJ
mol
20. Consider the following reaction:
MnO4-(aq) + 5Fe2+(aq) + 8H+(aq) Æ Mn2+(aq) + 5Fe3+(aq) + 4H2O(l)
Given that the rate of reaction is 3.5 x 10-2 mol/L MnO4-(aq), determine the rate of
reaction for all reactants and products in the reaction.
MnO4Fe2+
H+
Mn2+
Fe3+
H2O
=
=
=
=
=
=
3.5 x 10-2 mol/L
0.175mol/L
0.28 mol/L
3.5 x 10-2 mol/L
0.175mol/L
0.14mol/L
21. The reaction between ammonia and oxygen produces nitrogen monoxide and water
vapour is as follows:
4NH3(g) + 5O2(g) Æ 4NO(g) + 6H2O(g)
If the rate of consumption of ammonia is 1.6 x 10-2 mol/(L.s) calculate the rate of
consumption of oxygen and the formation of water vapour.
O2
H2O
=
=
0.02mol/(L.s)
0.024mol/(L.s)
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22. Rates of reactions are generally fastest at the beginning of a reaction. Explain why
this is so.
There is a higher number of reactants, and hence a greater number of collisions,
thus speeding up reaction rate.
23. A kinetics experiment is performed in which oxygen gas is collected. The
concentration of oxygen is measured every 10 s. The table below has the data
obtained from the experiment.
Time (s)
0
10
20
30
40
50
60
70
80
90
100
Oxygen gas (mol/L)
0
0.23
0.40
0.52
0.60
0.66
0.70
0.73
0.75
0.76
0.76
a) Plot a graph of oxygen concentration vs time
b) Calculate the average formation of oxygen gas during the first 60 seconds
rate = 0.70mol/L/60s
rate = 0.117 mol/L.s
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c) the average rate of formation of oxygen gas between t=20s and t=60s
rate = 0.30mol/L/40s
rate = 7.5 x 10-3mol/L.s
24. Identify the five factors that increase the rate of a reaction and provide an example
of each.
-concentration of reactants
-temperature
-surface area
-presence of a catalyst
-chemical nature of reactants
25. For a particular reaction at constant temperature,
A(g) + 2 B(g) Æ products
initial
[A]
1.00
2.00
3.00
4.00
initial initial
[B]
rate
1.00 1.00
4.00 8.00
9.00 27.00
2.00 ?
What is the value of "?" in this table?
Value is 64. There is a trend that as [A] is increased, the rate increased to the
power of 3. There appears to be no trend with the concentration of [B]
26. What is a rate law? What is the proportionality constant called?
The rate law is the relationship among rate, the rate constant, the initial
concentration of reactants. The proportionality constant is k, it is also called the rate
constant.
27. What is meant by the order of a reaction?
The order of the reaction is the exponent value that describes the initial
concentration dependence of a particular reactant.
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28. The rate law for the following reaction 2 NO + O2 Æ 2 NO2 is rate= k [NO]2[O2]. At
25oC, k=7.1 X 109 L mol-2s-1. What is the rate of reaction when [NO] = 0.0010 mol/L
and [O2] =0.034 mol/L?
r = 7.1 X 109 L mol-2s-1 [0.0010mol/L]2[0.034mol/L]
r = 2.4 x 104L-1mols-1
29. The initial rate of the reaction:
BrO3-(aq) + 5 Br-(aq) + 8 H+(aq) Æ 3 Br2(l) + H2O(l)
Has been measured at the reactant concentrations shown (in mol/L):
Experiment [BrO3-] [Br-]
1
0.10 0.10
2
0.20 0.10
3
0.10 0.20
4
0.10 0.10
[H+] Initial rate (mol/Ls)
0.10 8.0 x 10-4
0.10 1.6 x 10-3
0.10 1.6 x 10-3
0.20 3.2 x 10-3
According to these results what would be the initial rate (in mol/Ls) if all three
concentrations are:
[BrO3-] = [Br-] = [H+] = 0.20 mol/L?
Doubling [BrO3-] doubles rate
Doubling [Br-] doubles rate
Doubling [H+] quadruples rate
r= k[BrO3-]1[Br-]1[H+]2
k = initial rate/ [BrO3-]1[Br-]1[H+]2
k = 8.0 x 10-4mol/L•s/(0.10mol/L)(0.10mol/L)(0.10mol/L)2
k = 8L3•mol-3•s-1
r = 8l3• mol-3•s-1 [0.20mol/L]1[0.20mol/L]1[0.20mol/L]2
r = 1.28 x 10-2mol/L•s
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30. The reaction of iodide ion with hypochlorite ion, OCl- (which is found in liquid
bleach), follows the equation;
OCl- + I- Æ OI- + ClIt is a rapid reaction that gives the following rate data.
Initial Concentrations
[I-]
[OCl-]
Rate of Formation
(mol L-1 s-1)
(mol/L) of Cl1.75 X 104
3.50 X 104
3.50 X 104
1.7 X 10-3 1.7 X 10-3
3.4 X 10-3 1.7 X 10-3
1.7 X 10-3 3.4 X 10-3
What is the rate law for the reaction? Determine the value of the rate constant.
Doubling [OCl-] doubles rate 21 = 2 (1st order)
Doubling [I-] doubles rate 21 =2 (1st order)
Rate = k[OCl-]1[I-]1
k=
1.75 × 10 4 mol/L • s
1
1
⎡⎣1.7 × 10 −3 mol/L ⎤⎦ ⎡⎣1.7 × 10 −3 mol/L ⎤⎦
k = 6.06 × 109 L • mol−1 • s−1
31. A radioisotope has a half-life of 24s and an initial mass of 0.084g.
a) What mass of radioisotope will remain after:
(i) 72 s?
(ii) 192 s?
i) Number of half lives in 72 s =3
m 0.084g x ½ x ½ x ½
m = 0.0105g or 10.5 mg
ii) Number of half lives in 192 s =8
m = 0.084g x ½ x ½ x ½ x ½ x ½ x ½ x ½ x ½
m = 0.33mg
32. Consider the overall reaction involving three elements as reactants and a compound
as the product:
X + 2Y + 2Z Æ XY2Z2
When a series of reactions is performed with different initial concentrations of
reactants. The results are as follows:
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Doubling the concentration of X has no effect on the overall rate
Doubling the concentration of Y multiplies the overall rate by 4
Doubling the concentration of Z doubles the overall rate
State;
a)
b)
c)
d)
The rate law for this system
The rate determining step
A possible mechanism, indicating the slow step
A possible reaction intermediate in your mechanism
Solution
a) r = k[Y]2[Z]1
b) 2Y + 1Z Æ some product
c) Many answers are possible, as long as it is consistent with above rules
X + Y Æ XZ (fast)
XZ + Y2Z Æ XY2Z2 (fast)
2Y + Z ÆY2Z (slow)
d) Two possible reaction intermediates are XZ and Y2Z
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Appendix A
Appendix A: Thermochemical Data of Selected Elements & Compounds (at 25°C and 100.000 kPa)
S°
ΔGf°
ΔHf°
S°
ΔGf°
ΔHf°
Substance (kJ/mol) (J/K‚mol) (kJ/mol)
Substance
(kJ/mol) (J/K‚mol) (kJ/mol)
Al (s)
Al2O3 (s)
Br2 (l)
HBr (g)
Ca (s)
CaCO3 (s) (calcite)
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)
Cl2 (g)
HCl g)
Cr (s)
CrCl3 (s)
Cu (s)
CuO (s)
0
-1675.7
0
-36.4
0
-1206.9
-795.8
0
1.9
-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
0
-92.3
0
-556.5
0
-157.3
28.3
50.9
151.6
198.7
41.4
92.9
104.6
5.7
2.38
216.4
309.9
201.7
186.3
200.9
219.6
229.6
269.9
172.8
126.8
160.7
159.8
197.7
213.7
283.5
237.8
223.1
186.9
23.8
123.0
33.2
42.6
0
-1582.3
0
-53.5
0
-1128.8
-748.1
0
2.90
-65.2
-60.6
-73.7
-50.7
209.2
68.2
-32.8
-23.5
124.5
-166.3
-178.8
-389.9
-137.2
-394.4
-204.6
67.1
0
-95.3
0
-486.1
0
-129.7
CuCl
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)
Pb (s)
PbCl2 (s)
Mg (s)
MgCl2 (s)
MgO (s)
Hg (l)
HgS (s)
Ne (g)
N2 (g)
-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
-359.4
0
-641.3
-601.7
0
-58.2
0
0
86.2
108.1
202.8
173.8
126.0
130.7
69.9
188.8
109.6
27.8
57.6
87.4
146.4
118.0
142.3
52.9
64.8
136.0
32.7
89.6
26.9
76.0
82.4
146.2
191.6
-119.9
-175.7
0
-273.2
0
0
-237.1
-228.6
-120.4
0
245.1
-742.2
-1015.4
-302.3
-344.0
-166.9
0
-314.1
0
-591.8
-569.4
0
-50.6
0
0
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 (g)
SF6 (g)
H2S (g)
SO2 (g)
SO3 (g)
H2SO4 (l)
Sn (s) (white)
Sn (s) (gray)
SnCl2 (s)
SnCl4 (l)
Copyright © 2008, Durham Continuing Education
-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
-428.8
0
-127.1
-124.4
0
-411.2
-425.6
-1130.7
0
192.5
121.2
94.6
151.1
210.8
240.1
219.9
304.3
155.6
161.1
205.1
238.9
164.4
91.2
310.2
311.8
345.6
228.9
110.5
64.2
82.6
143.1
78.9
42.6
96.2
140.9
51.2
72.1
64.5
135.0
31.8
-16.5
149.3
-202.9
-183.9
86.6
51.3
104.2
97.9
-80.7
231.7
0
163.2
0
-48.4
13.4
-267.8
-2247.4
-2697.7
-1119.1
0
-409.1
-296.3
-379.1
0
-109.8
-33.4
0
-384.1
-379.5
-1044.0
0
278.8
-1209.0
-20.6
-296.8
-395.7
-814.0
0
-2.1
-325.1
-155.3
167.8
291.8
205.8
248.2
256.8
156.9
51.6
44.1
122.6
258.6
238.3
-1105.3
-33.6
-300.2
-371.1
-690.0
0
0.1
-302.1
-440.1
Page 70 of 70