1 | P a g e

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

2 | P a g e

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 CAREFUL- CORROSIVE 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 dm-3 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.

3 | P a g e

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.

4 | P a g e

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.

5 | P a g e

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

Concentration of hydrochloric acid solution ( mol dm-3)

Concentration of sodium hydroxide solution ( mol dm-3)

1 2 2 2 1 1 3 0.5 0.5 4 0.1 0.1 5 2 0.5 6 0.5 2

Draw a table in your lab notebook with the following headings:

Amount of H+(aq) (mol)

Amount of OH-(aq)(mol)

Amount reacted (mol)

Initial temp. (oC)

Final temp. (oC)

Temp. change (oC)

6 | P a g e

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.

7 | P a g e

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 i s 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:

8 | P a g e

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.

9 | P a g e

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

10 | P a g e

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.

11 | P a g e

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.

12 | P a g e

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.

13 | P a g e

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+

14 | P a g e

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.

• 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.

1. 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 ΔHo = -1366.7 kJ

15 | P a g e

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 ΔHo = -393.5 kJ CO2 C + O2 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

16 | P a g e

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 bio-fuel.

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 Formula DHc (KJmol-1) Methanol CH3OH - 725 Ethanol CH3CH2OH - 1364 Propanol CH3CH2CH2OH - 2016 2-Propanol (iso-propyl alcohol) CH3CH(OH)CH3 - 2003 Butanol CH3CH2CH2CH2OH - 2677 2-Butanol CH3CH2CH(OH)CH3 - 2658 2-methyl-1-propanol CH3CH(CH3)CH2OH - 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.

17 | P a g e

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.

18 | P a g e

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

19 | 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

20 | P a g e

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

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

21 | P a g e

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 PRODUCTS Ethanol Oxygen → Carbon dioxide Water

Bonds broken Number broken

Energy IN Bonds formed

Number formed

Energy OUT

Energy of the reaction = energy IN - energy OUT

=

22 | P a g e

REACTANTS PRODUCTS Ethene Hydrogen → Ethane

Bonds broken Number broken

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.

23 | P a g e

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.

24 | P a g e

SUMMARY OF KEY IDEAS

Type of Reaction

Exothermic

Endothermic

Energy absorbed or

released

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

reactants & products

Energy of the reactants is greater than the energy of the products

Energy of the reactants is less than the energy of the products

Sign of H H = H(products) - H(reactants) = negative (-ve)

H= H(products) - H(reactants) = positive (+ve)

Writing the equation

N2(g) + 3H2(g) -----> 2NH3(g) + 92.4 kJ

N2(g) + 3H2(g) ---> 2NH3(g) H=-92.4 kJ mol-1

2NH3(g) + 92.4 kJ -----> N2(g) + 3H2(g)

2NH3g ---> N2(g) + 3H2(g) H=+92.4 kJ mol-1

Energy Profile

Energy of reactants (N2 & H2) is greater than the energy

of the products (NH3). Energy is released.

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.

25 | P a g e

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

26 | P a g e

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) ΔH = – 196 kJ 2 S(s) + 3 O2(g) → 2 SO3 (g) ΔH = – 790 kJ

calculate the heat of reaction for

S(s) + O2(g) → SO2(g) 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

27 | P a g e

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: 1. + 7600 J 4. 456.5 kJ 7. 33.71oC 2. 0.89 J/goC 5. 67.3oC 3. – 943 J 6. 0.45 J/goC

28 | P a g e

More practice problems on Hess's Law

1. 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

2. 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

29 | P a g e

Solutions

1. 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

2. 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

3. 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

4. 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

5. 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