Lesson 1: Exothermic and endothermic reactions
When a chemical reaction happens, energy is transferred to or from the surroundings and often there is
a temperature change. For example, when a fire burns it transfers heat energy to the surroundings.
Objects near a fire become warmer and the temperature rise can be measured with a thermometer.
Exothermic reactions
These are reactions that transfer energy to the surroundings. The energy is usually transferred as
heat energy, causing the reaction mixture and its surroundings to get hotter. The temperature increase
can be detected using a thermometer. Some examples of exothermic reactions are:
•
•
•
Burning ( combustion)
neutralisation reactions between acids and alkalis, and
the reaction between water and calcium oxide
Endothermic reactions
These are reactions that take in energy from the surroundings. The energy is usually transferred as
heat energy, causing the reaction mixture and its surroundings to get colder. The temperature decrease
can be detected using a thermometer. Some examples of endothermic reactions are:
•
•
the reaction between barium hydroxide and ammonium chloride
the reaction between ethanoic acid and sodium carbonate
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Exothermic and endothermic reactions are used extensively in everyday life and in industry. Airbags, a
safety device in modern cars, utilize an exothermic reaction. An exothermic reaction is responsible for
the inflation of air bags in cars.
http://www.youtube.com/watch?v=dZfLOnXoVOQ&feature=related
Cold packs used to treat muscular injury involve an endothermic reaction.
Experiment 1: Exothermic and Endothermic reactions
In this experiment you will make qualitative observations of some processes that are accompanied by
release or absorption of heat energy. For each reaction, determine by feel whether the reaction mixture
becomes hotter or colder.
1. To about 10 ml of water in a test tube, add one drop of concentrated sulfuric acid. BE CAREFULCORROSIVE SUBSTANCE. DO NOT ALLOW SULFURIC ACID TO COME IN CONTACT
WITH YOUR SKIN. Record your observations. Does the solution get hotter or colder? Is the
dissolution of sulfuric acid exothermic or endothermic?
2. To about 10 ml of water in a test tube add two pellets of sodium hydroxide. Stir with a glass rod.
Record your observations. Is the dissolution of sodium hydroxide in water exothermic or endothermic?
3. To about 10ml of water in a test tube, add about 1g of solid sodiumthiodulfate-5-water,
Na2S2O3.5H2O. Stir. Record your observations. Is the dissolution of sodium thiosulfate in water
exothermic or endothermic?
4. To about 5 ml of 2 mol dm-3 sodium hydroxide solution in a test tube, add about 5 ml of 2 mol dm3 hydrochloric acid solution. Record your observations. Is the reaction that took place exothermic or
endothermic?
5. To about 5ml of 2 mol dm-3 hydrochloric acid solution in a test tube add a piece of zinc. Record
your observations. Write the equation for the reaction that occurs when zinc is put into a hydrochloric
acid solution. Is this reaction exothermic or endothermic?
6. Pour 10 ml of water into a test tube and record the temperature. Weigh out 5 g of ammonium
nitrate, pour into the test tube and stir with a glass rod. Record the coldest temperature reached. Repeat
the procedure using 5 g of calcium chloride instead of the ammonium nitrate. Approximately 325 joules
of heat are consumed per gram of ammonium nitrate in this reaction. Where does the heat come from
since there are no burners or heater involved? Explain.
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Energy level diagrams
Chemical energy
The chemical energy stored in the bonds in a substance gives us a measure of a chemical's energy level.
The higher the energy level of a substance, the more chemical energy is stored in its bonds. The
reactants and products in a chemical reaction usually have different energy levels, which are shown in
a type of graph called an energy level diagram. The vertical axis on this diagram represents the energy
level and the horizontal axis represents the progress of the reaction from reactants to products.
Energy level diagrams for exothermic reactions
In an exothermic reaction, reactants have a higher energy level than the products. The difference
between these two energy levels is the energy released to the surroundings in the reaction, and an
energy level diagram shows this as a vertical drop from a higher to a lower level:
Usually some extra energy is needed to get the reaction to start. This is called the activation energy
and is drawn in energy level diagrams as a hump. Catalysts reduce the activation energy needed for a
reaction to happen - this lower activation energy is shown by the dotted red line in the diagram here.
Energy level diagrams for endothermic reactions
In endothermic reactions the reactants have a lower energy level than the products. The difference
between these two energy levels is the energy gained from the surroundings in the reaction, represented
in an energy level diagram as a vertical jump from a lower to a higher level - the bigger the jump, the
more energy is gained.
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Exercises:
1. Classify each of the following reactions as either exothermic or endothermic.
a. 2H2O(l) + heat → 2H2(g) + O2(g)
b. Mg(s) + Cl2(g) →
MgCl2(s) + heat
1. The complete combustion of acetic acid (HC2H3O2) in oxygen gas to form water and carbon
dioxide at constant pressure releases 871.7 kJ of heat per mole of acetic acid.
a. Write a balanced chemical equation for this reaction.
b. How much heat (kJ) would be released if you burned 2.00 moles of acetic acid?
c. Draw an energy level diagram for the reaction.
3. Draw an energy level diagram for a reaction in which the total energy of the reactants is 50 kJ mol-1,
the total energy of products is 120 kJ mol-1 and the activation energy for the forward reaction is 120 kJ
mol-1. Label the diagram clearly .Is this reaction exothermic or endothermic?
4. When solid sodium hydroxide is dissolved in water, the temperature of the solution formed rapidly
increases.
a. Compare the total energy of the solid NaOH with that of the solution and state which is greater.
b. Classify this reaction as endothermic or exothermic.
5. Consider the reaction A + 2B → C
In this reaction, the total energy of the reactants is 80 kJ mol-1, the total energy of the products is -90
kj mol-1 and the activation energy for the forward reaction is 120 kj mol-1.
a. Draw a diagram of the energy profile for this reaction. Label the diagram.
B State whether the reaction is endothermic or exothermic.
c. Calculate the energy difference between the reactants and the products.
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Lesson 2: Heat of reaction
The amount of heat absorbed or released when specified amounts of substances react is
called the HEAT OF REACTION. The heat of reaction is also sometimes called the
ENTHALPY CHANGE.
Experiment 2: Heat of reaction and amount of reactants
For a particular chemical change, the heat of reaction depends upon the amounts of substances that
react. In this experiment, we will investigate how the quantity of heat released during the reaction
between an acid and a base depends on the amounts of acid and base that react.
When a solution of an acid such a hydrochloric acid is mixed with a solution of a base such as sodium
hydroxide, hydrogen ions from the acid react with hydroxide ions from the base to form water. Heat is
released, causing the temperature of the reaction mixture to rise.
H+(aq) + OH-(aq)
→ H2O + heat energy
Several reaction mixtures will be made, containing various amounts of hydrochloric acid and sodium
hydroxide. The volumes of the reaction mixtures are the same, so the magnitudes of the temperature
changes are relative measures of the quantities of heat released. The concentrations of the solutions to
be mixed are listed below. 10.0 ml of each solution is used.
Reaction
mixture
1
2
3
4
5
6
Concentration of hydrochloric acid solution
( mol dm-3)
2
1
0.5
0.1
2
0.5
Concentration of sodium hydroxide solution
( mol dm-3)
2
1
0.5
0.1
0.5
2
Draw a table in your lab notebook with the following headings:
Amount of
H+(aq) (mol)
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Amount of
OH-(aq)(mol)
Amount
reacted (mol)
Initial temp.
(oC)
Final temp.
(oC)
Temp. change
(oC)
Procedure:
1. Place 10.0 ml of hydrochloric acid solution into a test tube. Measure and
record its temperature.
2. Add 10.0 ml of the appropriate sodium hydroxide solution and stir gently
with the thermometer. Measure and record the maximum temperature
reached.
3. Calculate and record the increase in temperature of the reaction mixture.
4. Calculate and record the amount, in moles, of hydrogen ions and hydroxide
ions added.
5. Calculate and record the amount, in moles, of hydrogen and hydroxide ions
that react.
6. Repeat this procedure for each of the reaction mixtures listed.
7. Construct a graph of temperature change against amount reacted and use
the graph to draw a conclusion about the magnitude of the heat of reaction
and the amount of substance reacted.
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Lesson 3: Heat and Temperature – are they the same?
A Wrong Idea
Often the concepts of heat and temperature are thought to be the same, but they are not.
Perhaps the reason the two are incorrectly thought to be the same is because as human
beings on Earth everyday experience leads us to notice that when you heat something up,
say like putting a pot of water on the stove, then the temperature of that something goes
up. More heat, more temperature - they must be the same, right? Turns out, though, this is
not true.
Initial Definitions
Temperature is a number that is related to the average kinetic energy of the molecules of a
substance. If temperature is measured in Kelvin degrees, then this number is directly
proportional to the average kinetic energy of the molecules.
Heat is a measurement of the total energy in a substance. That total energy is made up of
not only of the kinetic energies of the molecules of the substance, but total energy is also
made up of the potential energies of the molecules.
More About Temperature (unit: Kelvin) K = oC + 273
If Temperature is measured in Kelvin, then it is directly proportional to the average kinetic
energy of the particles. Notice we did not say that temperature is the kinetic energy. We said
it is a number, if in degrees Kelvin, proportional to the average kinetic energies of the
molecules; that is, if you double the Kelvin temperature of a substance, you double the
average kinetic energy of its molecules.
More About Heat (unit: Joules)
Heat is energy. Heat is the total amount of energy possessed by the molecules in a piece of
matter. This energy is both kinetic energy and potential energy. When heat energy goes into
a substance one of two things can happen:
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1. The substance can experience a raise in temperature. That is, the heat can be used to
speed up the molecules of the substance.
2. The substance can change state.Although heat is absorbed by this change of state, the
absorbed energy is not used to speed up the molecules. The energy is used to change the
bonding between the molecules.. Heat comes in and there is an increase in the potential
energy of the molecules. Their kinetic energy remains unchanged.
So, when heat comes into a substance, energy comes into a substance. That energy can be
used to increase the kinetic energy of the molecules, which would cause an increase in
temperature. Or that heat could be used to increase the potential energy of the molecules
causing a change in state that is not accompanied by an increase in temperature.
Exercise: Using the ideas you have just learned, with a group, develop an argument to explain which
has more energy – a swimming pool of cold water or a pot of boiling water. Produce a diagram
/poster that enhances your explanation.
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Specific Heat
Substances don’t all respond the same way to heat. Some particles more readily absorb heat than
others. The specific heat of a substance is the amount of heat (per unit mass) required to raise the
temperature by one degree Celsius. The relationship between heat and temperature change is usually
expressed in the form shown below where c is the specific heat. The relationship does not apply if a
phase change is encountered, because the heat added or removed during a phase change does not
change the temperature.
The specific heat of water is 4.186 joule/gram °C which is higher than any other common substance.
As a result, water plays a very important role in temperature regulation.
Q is heat of the reaction in Joules
m is mass of the substance experiencing the temperature change in grams
C ( or Cp) is th specific heat capacity in J/g K
∆ T is change in temperature in K
Worked example:
If you drink a cold glass of water (250 g) at 0° C, how much heat is transferred to the water as it
warms to 37° C. The specific heat capacity of water is 4.187 J/g.K
q = m Cp ΔT
m = 250 g of water
Cp = 4.187 J/g•K
ΔT = 37 – 0 = 37 °C (37K)
q= (250 g) (4.187 J/g•K) (37K) = 1.5466 x 105 J = 150 kJ
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Exercises:
1. Aluminum has a specific heat of 0.902 J/g.K. How much heat is lost when a piece of aluminum
with a mass of 23.984 g cools from a temperature of 415.0 oC to a temperature of 22.0 oC?
2. The temperature of a sample of water increases by 69.5 oC when 24 500 J are applied. The
specific heat of liquid water is 4.18 J/g.K. What is the mass of the sample of water?
3. 850 joules of heat are applied to a 250 g sample of liquid water with an initial temperature of
13.0 oC. Find a) the change in temperature and b) the final temperature.
4. When 34 700 J of heat are applied to a 350 g sample of an unknown material the temperature
rises from 22.0 oC to 173.0 oC. What must be the specific heat of this material?
5. How many joules of heat are required to raise the temperature of 550 g of water from 12.0 oC to
18.0 oC?
6. How much heat is lost when a 640 g piece of copper cools from 375 oC, to 26 oC? (The specific
heat of copper is 0.38452 J/g.K)
7. The specific heat of iron is 0.4494 J/g.K. How much heat is transferred when a 24.7 kg iron
ingot is cooled from 880 oC to 13 oC?
8. How many joules of heat are necessary to raise the temperature of 350ml of water from 1.00 oC
to 5.00oC?
9. Find the mass of a piece of copper when 8000.0 J of heat are applied, causing a 45 oC increase in
temperature. (Cp of copper is 0.38452 J/g.K)
10. Find the specific heat of an unknown metal with an initial temperature of 16.0 oC, when 3500
Joules are applied to a 40.0g sample and the final temperature is 81.0 oC.
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Lesson 4: Heat content of food
Exothermic reactions are important for life. As you digest food, glucose and other molecules are
absorbed into the bloodstream and taken to cells to be metabolized. In the process of metabolism,
energy is released. We measure the energy that foods give us in Joules (or calories)
The Heat Content of Snack Foods
http://www.instructables.com/id/Energy-Of-Candy-Gummi-Bears-Vs.-MMs-Experiment/
OBJECTIVE: In this experiment, you will burn several types of snack foods in order to determine
their heat content per gram.
MATERIALS
For one team:
•
•
•
•
•
•
•
one soda can
balance
stirring rod
ring stand and iron ring
paper clip
thermometer, range to 110 degrees C
2-3g sample of each type of snack food, such as Cheetos, chips or marshmallows
HAZARDS: Be careful not to touch any of the surfaces which will be hot after heating. Let them all
cool down before handling.
PROCEDURE:
After the food burns completely, record the final temperature of the water, and determine the actual
mass of food that has burned. Repeat the procedure, using a different type of food sample.
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RESULTS:
Using your data for the mass of the water, the mass of the food that actually burned, and the initial and
final temperatures of the water, calculate the heat released per gram of nut/food burned.
DISPOSAL: Discard the ash in the waste basket. Recycle the soda cans.
CALCULATIONS: SHOW ALL WORK USED!!!
1. Calculate the change in temperature of the water.
2. Calculate the heat that was released by the food and absorbed by the water in calories and in
joules. (NOTE: 1 calorie(c) = 4.18 J)
Heat absorbed by water= (mass of water)(temperature change)(specific heat capacity of water)
3. Calculate the joules released per gram of food that burned.
4. Examine the "Nutritional Value Information" found on the package of one of the food
samples. Note that 1 Food Calorie(C) is equivalent to 4.184 kJ of heat energy. Use this
information to determine the "accepted value" for the heat content per gram of snack food.
What is the percent error for your experiment?
5. Explain how can you improve the accuracy of this experiment?
6. Compare the heat content of the various types of food tested. Use the nutritional information
on the side of their packages to determine how much fat and carbohydrates each type of food
has. Is there any correlation between these two values?
DISCUSSION
The value for kilojoules per gram of nut/food determined by this procedure is generally much lower
than the value in the literature, The literature value for the heat content of raw almonds is 28.4 kJ/g,
Brazil nuts = 30.1 kJ/g, pecans = 31.6 kJ/g, pistachios = 27.6 kJ/g, black walnuts = 28.6 kJ/g and
peanuts = 23.6 kJ/g.
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LESSON 5: Coffee Cup calorimeters
A calorimeter is a device that allows you to determine the heat involved in a chemical or physical
change. In the previous experiment, you used a metal tin as a calorimeter. The problem with a metal
cup is that it is not a very good insulator and so a lot of heat is lost to the atmosphere during the
reaction.
We can make a more efficient calorimeter using Styrofoam coffee cups - a "coffee cup" calorimeter. In
this calorimeter, the mass of the water in the calorimeter is measured. The temperature is measured
accurately once before the reaction begins and once after the reaction has taken place
Experiment 5: The heat involved in the reaction of magnesium ribbon with hydrochloric acid
Procedure:
1. Carefully measure and record the length of one of the cut pieces of polished magnesium
ribbon. Obtain the mass of a 1.00 meter length of polished magnesium ribbon from your
instructor.
2. Crumple the piece of magnesium into a small ball.
3. Pour 50-60 mL of 3 M hydrochloric acid into a 100-mL cylinder. Measure and record the
exact volume to the nearest 0.2 mL.
4. Assemble a calorimeter.
5. A calorimeter assembled from three foam cups works well. Two nested cups hold the fluid.
One fourth of the third cup is removed at the lip. A hole is made in the bottom for the
thermometer; a hole is made in the side near the upper edge through which additions can be
made. When inverted, this piece serves as a calorimeter cover.
6. Pour the measured volume of acid into the inner calorimeter cup. Cover. Insert the
thermometer. Read and record the temperature to the nearest 0.1 °C at regular intervals until
it becomes constant.
7. Add the crumpled magnesium to the acid solution. Swirl gently. Note the temperature.
8. Record the maximum temperature reached by the hydrochloric acid solution.
Assume that the specific heat of the HCl(aq) = 4.184 J/mL, calculate the energy released in Joules.
Then determine the energy released in kJ/mol
Compare your value to the accepted value of -462.0 kJ/mol at 25 °C for 1 M H+
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Lesson 5 continued: Enthalpy
Scientists use the word Enthalpy to describe the heat content of a substance. So when we measure the
heat released or absorbed in a chemical reaction, we are measuring the change in enthalpy for that
reaction. It is often times written at the end of the equation using the symbol ∆H.
∆H means the change in enthalpy for that particular reaction.
•
•
1.
For an exothermic reaction, the enthalpy of the products is less than that of the reactants so the
∆H value will be negative.
For an endothermic reaction, the enthalpy of the products is more than that of the reactants so
the ∆H value will be positive.
How much heat will be released when 6.44 grams of sulfur reacts with excess oxygen according
to the following equation? Is this reaction endothermic or exothermic?
2S + 3O2  2SO3
ΔHo = -791.4 kJ
2.
How much heat will be released when 4.72 grams of carbon reacts with excess oxygen
according to the following equation? Is this reaction endothermic or exothermic?
C + O2  CO2
ΔHo = -393.5 kJ
3.
How much heat will be absorbed when 38.2 grams of bromine reacts with excess hydrogen
according to the following equation? Is this reaction endothermic or exothermic?
H2 + Br2  2HBr
ΔHo = 72.80 kJ
4.
How much heat will be released when 1.48 grams of chlorine reacts with excess phosphorus
according to the following equation? Is this reaction endothermic or exothermic?
2P + 5Cl2  2PCl5
ΔHo = -886 kJ
5.
How much heat will be released when 4.77 grams of ethanol (C2H5OH) reacts with excess
oxygen according to the following equation? Is this reaction endothermic or exothermic?
C2H5OH + 3O2  2CO2 + 3H2O
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ΔHo = -1366.7 kJ
If you were to write the reaction in the reverse direction, then the value of ΔH would stay the same
however the sign would be reversed.
C + O2  CO2
CO2  C + O2
ΔHo = -393.5 kJ
ΔHo = +393.5 kJ
If you were to double the equation, then the value of ΔH would double
C + O2  CO2
ΔHo = -393.5 kJ
2C + 2O2  2CO2
ΔHo = -787.0 kJ
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Lesson 6: Design a lab to answer the question “Which is the Best Fuel?”
Introduction Rising energy prices, the contribution of carbon dioxide from fossil fuel burning to global
warming and concerns about carbon monoxide and nitrogen oxides as air pollutants has led to
increased interest in alternative fuels. Alcohol has been used as a fuel in vehicles since the early 1900’s
but fossil fuels (coal, gasoline, oil, natural gas) have become the dominant energy source.
The alcohols methanol, ethanol, propanol, and butanol are potential alternative fuels for vehicles
because they can be synthesized (made) biologically from the fermentation of organic material and they
have characteristics which allow them to be used in current engines. Because alcohol fuels are obtained
from biological sources, they are sometimes known as bio-alcohols (e.g. bio-ethanol), gasohol or biofuel.
Methanol (CH3OH) and ethanol (CH3CH2OH) have both been proposed as bio-alcohol fuels of the
future. Both have advantages and disadvantages over fossil fuels, such as petrol and diesel. For
instance, both alcohols don’t need additives to improve performance. Alcohols also combust more
completely to carbon dioxide and water resulting in less smog producing carbon monoxide and
nitrogen oxides being emitted.
Methanol combustion: CH3OH + 1 O2 → CO2 + 2 H2O + heat energy
Ethanol combustion: CH3CH2OH + 3 O2 → 2 CO2 + 3 H2O + heat energy
Task: Design a plan to investigate which alcohol (methanol or ethanol) would be the best fuel for the
future.
Some literature values
Alcohol
Methanol
Ethanol
Propanol
2-Propanol (iso-propyl alcohol)
Butanol
2-Butanol
2-methyl-1-propanol
Formula
CH3OH
CH3CH2OH
CH3CH2CH2OH
CH3CH(OH)CH3
CH3CH2CH2CH2OH
CH3CH2CH(OH)CH3
CH3CH(CH3)CH2OH
DHc (KJmol-1)
- 725
- 1364
- 2016
- 2003
- 2677
- 2658
- 2666
1. Show your design to the teacher as soon as you have it in draft form
2. Once it is approved, you will need to write a lab order of equipment and chemicals
needed for the next class.
3. The lab order needs to be submitted to your teacher before you leave class today.
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Lesson 7: Design experiment and lab report
1. Perform your experiment, following all necessary safety precautions
2. Record all data carefully in your lab notebook. Include uncertainties your data table.
3. When you have collected your data, clean up all apparatus thoroughly and return safety glasses and
aprons.
4. Write a lab report, using the grading rubric that follows as a guide.
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General Chemistry Lab Report Grading RUBRIC
Scoring rubric: completely met = 3 ; partially met = 2 ; not met = 1
Title
Relates the dependent and independent variables in general terms. Can be written as a question.
Introduction
Clearly stated in about two sentences. It clarifies the relationship between the factor being
investigated (independent variable) and what is being measured (dependent variable). It does
not describe the method.
A balanced chemical equation is written for the reaction under investigation (if there is one).
A list of all the important dependent, independent and controlled variables.
o Independent variable – the factor being investigated.
o Dependent variable(s) – what is measured or the type of numerical data collected about
the independent variable.
o Controlled / constant variables – what is kept constant or remains unchanged between
the different experiments into the factor being investigated.
Procedure
Written in sufficient detail that it is easy to follow, quantities are given
Clearly explains the logic of the method, student displays a sound understanding of the
principles behind the lab
Data Collection
Sufficient qualitative observations and quantitative measurements are recorded.
Uncertainties are listed for quantitative data
The quantitative data collected is presented in easily interpretable, organized and labeled
table(s). See the guidelines attached for the features of a good scientific table.
Data Processing
The processed data is presented in an easily interpretable manner so that patterns in the results
18 | P a g e
are made obvious. The average of the results is plotted not the results from individual trials.
A final uncertainty is determined ; a percent error is determined
Conclusion
A statement is made which indicates if the aim was achieved. This is stated explicitly. It is
stated whether the results supported the hypothesis.
Describe using supporting evidence the relationship found between the independent (factor
changed) and dependent (measured) variables.
Evaluation
Compare the expected (theoretical) and actual results (found in the experiment) quantitatively.
Explain the reasons for the possible difference using % error and uncertainty as evidence.
Suggest and explain improvements and modifications that could be made to minimize the
differences between the expected and the actual results described above.
References
Bibliography of all sources cited is included. This includes information sourced when planning
the lab, diagrams, graphs, literature values and expected results.
Features of good Scientific Tables

Clear column and row headings

Units of measurement included in the column heading

Borders/lines around text and numerical data

Don’t run over two pages

Columns to be compared are placed next to one another

Called tables and numbered consecutively

Concise descriptive title that relates the measured (dependent) and changed (independent)
variables on top of the table

Consistent and correct use of decimal places/significant figures for numerical data

Text and data is centered

Rows and columns are evenly distributed

11 – 12 point font size for electronic tables

Consistent font type for electronic tables

Correctly placed in data collection part of the lab report
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Lesson 8: The origins of heat of reaction
Why are some reactions exothermic and some endothermic? How can we explain why the heat
content of reactants in some reactions is more than the heat content of the products, but
sometimes the reverse is true?
We need to recall that for a chemical reaction to occur bonds in the reactants must be broken and
bonds in the products must form. We know that breaking bonds requires an input of energy and
making bonds releases energy. So as bonds break, energy is absorbed and as they form energy is
released. The heat of the reaction will depend on which of these two processes has the greatest energy
value.
Lets use the reaction between hydrogen gas and iodine gas as an example:
The bond energies involved in the reaction between hydrogen and iodine are shown in the table below.
Bond Bond energy in kJ/mole
H-H
436
I-I
151
H-I
298
1. So the energy required to break the bonds , the energy IN , is 436 + 151 kJ/mole = 587
kJ/mole
2. The energy released as bonds form, the energy OUT , is 2 x 298 kJ/mole = 596 kJ/mole
3. The energy change is therefore 587 - 596 = -9 kJ/mole
Since the energy change is negative, this is an exothermic reaction. More energy is given out as
bonds form (products) than is taken in to break the bonds.(reactants)
Table of Average Bond Dissociation Energies
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Bond
Energy (kJ/mol)
Bond
Energy (kJ/mol)
H-H
436
N-N
160
C-H
413
N=O
631
N-H
393
N triple N
941
O=O
498
N-O
201
C-C
347
C=O
805
C-O
358
O-H
464
Cl-Cl
242
H-Cl
433
C - Cl
397
O - Cl
269
C=C
607
O-O
204
Activity: Modelling bond making and breaking
In this activity, you will
•
•
Build models to investigate bond breaking and making.
Use your model making to help work out the enthalpy change in a reaction. You will need to
use the table of bond enthalpies given on the previous page.
Procedure:
1. Build models of the reactants given
2. Analyse all of the bonds in the reactants. Write down each bond and look up the energy required
to break that bond. This will determine the energy IN.
3. Build models of the products.
4. Analyse all of the bonds formed. Write down each bond and look up the energy released when
the bond forms. This will determine the energy OUT.
5. Determine the total energy of the reaction .
REACTANTS
Ethanol
Bonds broken
Oxygen
Number
broken
→
Energy IN
Energy of the reaction = energy IN - energy OUT
=
21 | P a g e
PRODUCTS
Carbon dioxide
Water
Bonds
formed
Number
formed
Energy OUT
REACTANTS
Ethene
Bonds broken
Hydrogen
Number
broken
PRODUCTS
Ethane
→
Energy IN
Bonds
formed
Number
formed
Energy OUT
Energy of the reaction = energy IN - energy OUT
=
To do this without models, follow the following example:
First write out the balanced equation for the reaction showing full structural formulae.
Now simply add together the bond enthalpies involved for the reactants to obtain the total energy in
Do the same for the products to obtain a total energy out
Finally, subtract energy out from energy in to find ∆H for the reaction.
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Bond
Number
broken
Number
formed
Average
Bond Enthalpy
kJ /mol
C-C
1
0
+347
C-H
5
0
+413
C-O
1
0
+358
O-H
1
6
+464
O=O
3
0
+498
C=O
0
4
+805
Bond breaking:
Total energy IN = (347 x 1) + (413 x 5) + (358 x 1) + (464 x 1) + (498 x 3) = 4728 kJ
Bond making:
Total energy OUT = (464 x 6) + (805 x 4) = 6004 kJ
Sum total of bond breaking and bond making=
energy IN – energy OUT = 4728kJ – 6004kJ = -1276 kJ
Exercises:
1.Use the table of bond enthalpies to determine the enthalpy of the following reactions:
a. N2(g) + 3H2(g) → 2NH3(g)
b. N2(g) + 2O2(g) → 2NO2(g)
c. ½ H2(g) + ½ Cl2(g) → HCl(g)
d. N2H4 (g) + O2 (g) → N2 (g) + 2 H2O (l)
2. Given that the enthalpy change for the reaction N2(g) + 3Cl2(g) → 2NCl3(g) is 688 kJ/mol,
calculate the bond enthalpy of the N-Cl bond.
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SUMMARY OF KEY IDEAS
Type of
Reaction
Energy
absorbed or
released
Exothermic
Endothermic
Energy is released.
Energy is a product of the reaction.
Reaction vessel becomes warmer.
Temperature inside reaction vessel increases.
Energy is absorbed.
Energy is a reactant of the reaction.
Reaction vessel becomes cooler.
Temperature inside reaction vessel decreases.
Relative
Energy of Energy of the reactants is greater than the energy Energy of the reactants is less than the energy
reactants &
of the products
of the products
products
Sign of
H
H = H(products) - H(reactants) = negative (-ve)
H= H(products) - H(reactants) = positive (+ve)
N2(g) + 3H2(g) -----> 2NH3(g) + 92.4 kJ
2NH3(g) + 92.4 kJ -----> N2(g) + 3H2(g)
Writing the
equation N2(g) + 3H2(g) ---> 2NH3(g)
H=-92.4 kJ mol 2NH3g ---> N2(g) + 3H2(g)
H=+92.4 kJ
1
-1
mol
Energy of reactants (N2 &
H2) is greater than the energy
of the products (NH3).
Energy is released.
Energy
Profile
24 | P a g e
H is negative.
H is measured from the
energy of reactants to the
energy of products on the
Energy Profile diagram.
Energy of reactants (NH3)
is less than the energy of
the products (N2 & H2).
Energy is absorbed.
H is positive.
H is measured from the
energy of reactants to the
energy of products on the
Energy Profile diagram.
Lesson 9: Hess's Law
A calorimeter is used to measure heat changes in reactions directly. For some reactions, however, it is
not possible to use a calorimeter. Another way to determine the heat of reaction is to use Hess’s Law.
Hess's Law says that the overall enthalpy change of chemical reaction is independent of the route
taken in going from reactants to products.
An example of a Hess Cycle is shown in Fig. 1.
Fig 1. The layout of a Hess Cycle
•
•
•
(1) is ∆H for forming of the reactants from its elements[reactants].
(2) is ∆H for the reaction – reactants forming products
(3) is ∆H for formation of the products from the elements
You can see from the diagram that ∆H reaction] = ∆Hproducts] − ∆Hreactants
You can re arrange the equation , for example
(2) = (3) − (1)
Worked example using Hess’s Law
Calculate ΔH for the reaction: C2H4 (g) + H2 (g) → C2H6 (g), from the following data.
C2H4 (g) + 3 O2 (g) → 2 CO2 (g) + 2 H2O (l)
ΔH = -1411. kJ
C2H6 (g) + 3½ O2 (g) → 2 CO2 (g) + 3 H2O (l)
ΔH = -1560. kJ
H2 (g) + ½ O2 (g) → H2O (l)
ΔH = -285.8 kJ
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Solution: In this example, we need to reverse the second equation so that the C2H6 is on the right
hand side, as a product. When we reverse an equation, we must change the sign of the enthalpy
value.
C2H4 (g) + 3 O2 (g) → 2 CO2 (g) + 2 H2O (l)
ΔH = -1411. kJ
2 CO2 (g) + 3 H2O (l) → C2H6 (g) + 3½ O2 (g)
ΔH = +1560. kJ
H2 (g) + 1/2 O2 (g) → H2O (l)
ΔH = -285.8 kJ
C2H4 (g) + H2 (g) → C2H6 (g)
ΔH = -137. kJ
Exercises:
1. Calculate DH for the reaction 4 NH3 (g) + 5 O2 (g) → 4 NO (g) + 6 H2O (g), from the following
data.
N2 (g) + O2 (g) →2 NO (g)
ΔΗ = -180.5 kJ
N2 (g) + 3 H2 (g) →2 NH3 (g)
ΔH = -91.8 kJ
2 H2 (g) + O2 (g) →2 H2O (g)
ΔH = -483.6 kJ
2. From the following heats of reaction
2 SO2(g) + O2(g) → 2 SO3(g)
2 S(s) + 3 O2(g) → 2 SO3 (g)
calculate the heat of reaction for
S(s) + O2(g) → SO2(g)
ΔH = – 196 kJ
ΔH = – 790 kJ
ΔH = ? kJ
3. Find ΔH for the reaction 2H2(g) + 2C(s) + O2(g)  C2H5OH(l), using the following
thermochemical data.
C2H5OH (l) + 2 O2 (g) → 2 CO2 (g) + 2 H2O (l)
ΔH = -875. kJ
C (s) + O2 (g) → CO2 (g)
ΔH = -394.51 kJ
H2 (g) + ½ O2 (g) → H2O (l)
ΔH = -285.8 kJ
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4. Calculate ΔH for the reaction CH4 (g) + NH3 (g)  HCN (g) + 3 H2 (g), given:
N2 (g) + 3 H2 (g) → 2 NH3 (g)
ΔH = -91.8 kJ
C (s) + 2 H2 (g) → CH4 (g)
ΔH = -74.9 kJ
H2 (g) + 2 C (s) + N2 (g) → 2 HCN (g)
ΔH = +270.3 kJ
Additional practice problems
Heat Problems Solve each of the following problems.
Use correct units, and show your work
1. The specific heat of ethanol is 2.46 J/goC. Find the heat energy required to raise the temperature of
193 g of ethanol from 19oC to 35oC.
2. When a 120 g sample of aluminum absorbs 9612 J of heat energy, its temperature increases from
25oC to 115oC. Find the specific heat of aluminum. Include the correct unit.
3. The specific heat of lead is 0.129 J/goC. Find the amount of heat released when 2.4 mol of lead are
cooled from 37.2oC to 22.5oC.
4. How many kJ of energy are needed to raise the temperature of 165 mol of water from 10.55oC to
47.32oC?
5. A 150 g sample of water (initially at 45.0oC) is mixed with a 200 g sample of water (initially at
84.0oC). Find the final temperature of the system.
6. A 25 g sample of iron (initially at 800.00oC) is dropped into 200 g of water (initially at 30.00oC). The
final temperature of the system is 40.22oC. Find the specific heat of iron.
7. A 440 g sample of mercury (specific heat = 0.14 J/goC, initial temperature of 22.00oC) is placed into
134 g of water (initial temperature of 35.00oC). Find the final temperature of the system.
Answers:
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1. + 7600 J
2. 0.89 J/goC
3. – 943 J
4.
5.
6.
456.5 kJ
67.3oC
0.45 J/goC
7.
33.71oC
More practice problems on Hess's Law
1. Calculate ΔH for the reaction: C2H4 (g) + H2 (g) → C2H6 (g), from the following data.
2.
C2H4 (g) + 3 O2 (g) → 2 CO2 (g) + 2 H2O (l)
ΔH = -1411. kJ
C2H6 (g) + 3½ O2 (g) → 2 CO2 (g) + 3 H2O (l)
ΔH = -1560. kJ
H2 (g) + ½ O2 (g) → H2O (l)
ΔH = -285.8 kJ
Calculate ΔH for the reaction 4 NH3 (g) + 5 O2 (g) → 4 NO (g) + 6 H2O (g), from the
following data.
N2 (g) + O2 (g) → 2 NO (g)
ΔH = -180.5 kJ
N2 (g) + 3 H2 (g) → 2 NH3 (g)
ΔH = -91.8 kJ
2 H2 (g) + O2 (g) → 2 H2O (g)
ΔH = -483.6 kJ
3. Find ΔH° for the reaction 2H2(g) + 2C(s) + O2(g) → C2H5OH(l), using the following
thermochemical data.
C2H5OH (l) + 2 O2 (g) → 2 CO2 (g) + 2 H2O (l)
ΔH = -875. kJ
C (s) + O2 (g) → CO2 (g)
ΔH = -394.51 kJ
H2 (g) + ½ O2 (g) → H2O (l)
ΔH = -285.8 kJ
4. Calculate ΔH for the reaction CH4 (g) + NH3 (g) → HCN (g) + 3 H2 (g), given:
N2 (g) + 3 H2 (g) → 2 NH3 (g)
ΔH = -91.8 kJ
C (s) + 2 H2 (g) → CH4 (g)
ΔH = -74.9 kJ
H2 (g) + 2 C (s) + N2 (g) → 2 HCN (g)
ΔH = +270.3 kJ
5. Calculate ΔΗ for the reaction 2 Al (s) + 3 Cl2 (g) → 2 AlCl3 (s) from the data.
2 Al (s) + 6 HCl (aq) → 2 AlCl3 (aq) + 3 H2 (g)
ΔH = -1049. kJ
HCl (g) → HCl (aq)
ΔH = -74.8 kJ
H2 (g) + Cl2 (g) → 2 HCl (g)
ΔH = -1845. kJ
AlCl3 (s) → AlCl3 (aq)
ΔH = -323. kJ
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Solutions
1.
2.
3.
4.
5.
Î
C2H4 (g) + 3 O2 (g) → 2 CO2 (g) + 2 H2O (l)
ΔH = -1411. kJ
RÏ
2 CO2 (g) + 3 H2O (l) → C2H6 (g) + 3½ O2 (g)
ΔH = +1560. kJ
Ð
H2 (g) + 1/2 O2 (g) → H2O (l)
ΔH = -285.8 kJ
C2H4 (g) + H2 (g) → C2H6 (g)
ΔH = -137. kJ
Î×2
2 N2 (g) + 2 O2 (g) → 4 NO (g)
ΔH = 2 (-180.5 kJ)
RÏ×2
4 NH3 (g) → 2 N2 (g) + 6 H2 (g)
ΔH = 2 (+91.8 kJ)
Ð×3
6 H2 (g) + 3 O2 (g) → 6 H2O (g)
ΔH = 3 (-483.6 kJ)
4 NH3 (g) + 5 O2 (g) → 4 NO (g) + 6 H2O (g)
ΔH = -1628. kJ
RÎ
2 CO2 (g) + 2 H2O (l) → C2H5OH (l) + 2 O2 (g)
ΔH = +875. kJ
Ï×2
2 C (s) + 2 O2 (g) → 2 CO2 (g)
ΔH = 2 (-394.51 kJ)
Ð×2
2 H2 (g) + O2 (g) → 2 H2O (l)
ΔH = 2 (-285.8 kJ)
2H2(g) + 2C(s) + O2(g) → C2H5OH(l)
ΔH° = -486. kJ
Î×½
NH3 (g) → 1/2 N2 (g) + 3/2 1H2 (g)
ΔH = ½(+91.8 kJ)
RÏ
CH4 (g) → C (s) + 2 H2 (g)
ΔH = +74.9 kJ
Ð×½
1/2 H2 (g) + C (s) + 1/2 N2 (g) → HCN (g)
ΔH = ½(+270.3 kJ)
CH4 (g) + NH3 (g) → HCN (g) + 3 H2 (g)
ΔH = +256.0 kJ
Î
2 Al (s) + 6 HCl (aq) → 2 AlCl3 (aq) + 3 H2 (g)
ΔH = -1049. kJ
Ï×6
6 HCl (g) → 6 HCl (aq)
ΔH = 6 (-74.8 kJ)
Ð×3
3 H2 (g) + 3 Cl2 (g) → 6 HCl (g)
ΔH = 3 (-1845. kJ)
RÑ×2
2 AlCl3 (aq) → 2 AlCl3 (s)
ΔH = 2 (+323. kJ)
2 Al (s) + 3 Cl2 (g) → 2 AlCl3 (s)
ΔH = -6387. kJ
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