Thermochemistry review

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Chemistry 30

Thermochemistry

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Thermochemistry Table of Contents

AN INTRODUCTION TO TERMS...3

TWO LAWS OF THERMODYNAMICS...3

DIPLOMA EXAMINATION PRACTICE:...3

CALCULATING KINETIC ENERGY (EK) CHANGES...3

TEMPERATURE CHANGES AND HEAT CAPACITIES...3

EXAMPLES OF PROBLEMS YOU WILL BE EXPECTED TO DO FOR KINETIC ENERGY CHANGES...3

KINETIC ENERGY PRACTICE PROBLEMS:...3

DIPLOMA EXAMINATION PRACTICE...3

CALCULATING POTENTIAL ENERGY (EP) CHANGES...3

ENERGY IN CHEMICAL REACTIONS...3

EXAMPLES OF SOME PROBLEMS YOU’LL BE ASKED TO SOLVE:...3

PRACTICE PROBLEMS FOR ENERGYIN CHEMICAL REACTIONS...3

FINDING ENERGY IN CHEMICAL REACTIONS EXPERIMENTALLY: CALORIMETRY...3

1. Styrofoam calorimeter...3

2. Metal Can calorimeter...3

3. Bomb Calorimeter...3

Examples of Practice Problems for Calorimetry You’ll be Expected to Answer.3 PRACTICE PROBLEMS FOR CALORIMETRY:...3

FINDING ENERGY IN REACTIONS USING THE DATA BOOKLET...3

CATALYSTS AND ACTIVATION ENERGY...3

CATALYSTS...3

Examples Of Problems You Will Be Expected To Do For Enthalpies of Formation:...3

1. WHICH OF THE QUANTITIES IN THE ENTHALPY LEVEL DIAGRAM BELOW IS (ARE) AFFECTED BY THE USE OF A CATALYST? 3 ENERGY IN REACTIONS PRACTICE PROBLEMS...3

HESS’S LAW...3

EXAMPLES FOR HESS’ LAW:...3

PRACTICE PROBLEMS FOR HESS’S LAW USING EQUATIONS:...3

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Thermochemistry

An Introduction to Terms

Thermochemistry is the study of heat changes associated with

physical, chemical and nuclear processes.

Heat is a form of energy that flows between two samples at different temperatures (Heat always moves from hot to cold)

Temperature is the average kinetic energy of the molecules. It does

not reflect total energy, only the random movement of atoms and molecules.

Thermal Energy is a measure of the total heat energy in a substance.

Heat and temperature are related however. If something absorbs heat energy, the temperature increases (except during phase changes).

An example to show the difference between temperature and thermal energy:

CUP OF COFFEE at 80°C BATHTUB FULL OF WATER AT 50°C

Higher temperature More thermal energy (greater average kinetic energy) (more molecules of water)

TWO LAWS OF THERMODYNAMICS

1. Energy can be converted from one form to another but can’t be created or destroyed.

2. Heat flows from hot objects to colder ones until thermal equilibrium is reached.

We will apply these laws later in this unit during the calorimetry section.

Units for energy

Joule (J) (the energy in a 2.0kg body moving at a velocity of 1m/s)

For this Course Energy Comes in Two Forms:

Potential energy (Ep) is stored energy (For much of our unit, it will be the stored energy in the bonds within and between molecules – chemical energy). Potential energy is absorbed when bonds break and released when bonds form.

Kinetic energy (Ek) is the energy of motion of ions, atoms, molecules (Translational, rotational, vibrational motion). Ek is temperature dependant. When energy is absorbed, temperature of that substance increases. When energy is released, the temperature

How This Relates to the Curriculum:

30–A1.2k explain, in a general way, how stored energy in the chemical bonds of hydrocarbons originated from the sun

30–A1.4s use appropriate International System of Units (SI) notation, fundamental and derived units and significant digits

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Diploma Examination Practice:

Use the following information to answer the next two questions.

1. In the kiln, this reaction would

A. produce energy, reducing the amount of fuel required B. produce energy, increasing the amount of fuel required C. absorb energy, increasing the amount of fuel required D. absorb energy, reducing the amount of fuel required

Use the following key to answer the next question.

2. Identify the type of bond, as numbered above, that is primarily involved in each change listed below.

Dry ice sublimes __________

Hydrogen atoms fusing into helium __________ Gasoline burning in an automobile engine __________ Water vapour condensing __________ 3. Radiant energy from the Sun is stored by plants. This energy is

released when plant material undergoes a A. phase change

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Use the following information to answer the next three questions.

4. The ATAD process is A. a reduction B. exothermic C. endothermic

D. an acid–base reaction

5. Which of the following processes is always endothermic? A. Neutralization

B. Photosynthesis C. Oxidation D. Reduction

6. Which of the following molecular properties is a main component of the potential energy of matter?

A. Vibrational motion B. Intramolecular bonding

C. Movement from place to place

D. Rotation about the molecules’ centre of mass

Use the following information to answer the next question.

7. This reaction is an example of A. respiration

B. photosynthesis

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Calculating Kinetic Energy (E

k

) Changes

Temperature Changes and Heat Capacities

Remember from Science 10 (I’m not kidding…) that temperature changes involve changes in kinetic energy (Ek) for a substance.

Different substances will heat up and cool down at different rates, depending on their molecular makeup. Metals tend to heat up quickly, and will cool down quickly. Water tends to take more energy to heat up, but once it gets hot, it tends to take longer to cool down. To make this concept of heating and cooling more quantitative, we use heat

capacity. The heat capacity of a substance is a measure of the energy

required to raise that amount of substance by one degree Celsius. The units for heat capacity are J/°C. Heat capacity will change depending on the amount of substance involved. For example, it will take more energy to heat up a bathtub of water to 55 °C than a cup of water to 55 °C. We standardize our heat capacity measurements so they can be compared. Molar heat capacity (J/mol•°C) allows us to see which substances require more or less energy to heat up or cool down because they all involve the same amount. However, specific heat capacity (J/

g•°C) is used, because, in industrial applications, we want to know how

quickly something might heat up or cool down, and we can more easily measure the mass of a substance than moles.

SPECIFIC HEAT CAPACITY (Specific Heat) The amount of heat

energy needed to raise the temperature of 1 g of a substance 1°C (1K)

The specific heat is different for different substances and for different states.

H2O(l) = 4.19 J/g°C

H2O(g)= 2.02 J/g°C

H2O(s) = 2.00 J/g°C

Al(s) = 0.897 J/g°C

O2(g) = 0.917 J/g°C

From the values above, we see that water can store more energy than O2(g) or Al(s)

When 1g of water increases by 1°C it absorbs 4.19J of energy. When 1g of Al increases by 1°C, it only absorbs 0.903J of energy. i.e. It takes longer for water to heat up, but it will also take longer for it to cool down.

This makes water a good coolant (stores heat well).

As well, a dropof boiling water on your skin will not destroy skin tissue. A cup of boiling water will scald your hand.

The higher the specific heat capacity of a substance and the greater its mass, the more energy it can store (it takes longer to heat up).

How This Relates to the Curriculum: 30–A1.1k recall the application of Q =

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We can calculate the heat transferred to or from a substance if we know the heat capacity, mass and temperature change of the substance.

Ek = mc∆t

Where:

Ek = kinetic energy (J)

m = mass of substance that is heating or cooling (g) c = specific heat capacity of the substance (J/g·°C)

∆t = change in temperature (°C) where ∆t = tfinal - tinitial If you are given any 3 of the 4 values, you will be expected to find the 4th

Ek has a negative value when something loses heat (t decreases)

Ek has a positive value when something gains heat (t increases)

Remember that Ek , m , c, t values are all for the same substance

Examples of Problems You Will Be Expected To Do For

Kinetic Energy Changes

Example 1:

Find the energy needed to heat 150.0 g of water from 25.3 °C to 75.0 °C.

Ek = mc∆t

Ek = 150.0 g x 4.19 J/g°C x (75.0°C – 25.3 °C)

Ek = 3.12 x 104 J

Example 2:

5.00 x 102_{g of water was cooled from 41.5 °C to 10.0 °C. Find the}

energy released.

Ek = mc∆t

Ek = 500 g x (4.19 J/g°C) x (10.0 ˚C – 41.5 ˚C)

Ek = -6.60 x 104 J

The negative value tells us that heat is released (you can’t really have a negative energy).

Example 3:

50.0 g of water at 40.0 °C was cooled, releasing 2.5 kJ of heat. Find the new temperature.

* Remember that when heat is released, it is considered negative.

Ek = mc∆t

-2500 J = 50.0 g x 4.19 J/g°C (tf – 40.0 °C)

You may also see q = mc∆t, but we think that using Ek as a symbol

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Kinetic Energy Practice Problems:

1. 250.0 g of pure copper was heated from 21.8 °C to 55.3 °C. Find the heat energy required. (3.22 kJ)

2. 0.600 kg of tin was heated from 20.0 °C to 50.0 °C. Find the energy required. (4.09 kJ)

3. A 550 g piece of metal at 95.0 °C is cooled down to 30.0 °C and releases 1.63 x 103_{J of energy. Calculate the specific heat of}

the metal. (-4.56 * 10-2_J/g·°C)

4. Water at 78.0 °C is cooled down to 15.0 °C and is found to release 1.12 x 105_{J of energy. What is the mass of the water?}

(424 g)

5. 500.0 g of ethanol, C2H5OH(l) were heated from 25.0 °C to its

boiling point of 78.3 °C using 69.3 kJ of heat energy. Calculate the specific heat for C2H5OH(l). (2.60 J/g·°C)

6. 350 g of water at 45.0 °C release 25.0 kJ of heat. Find the new temperature. (28.0 °C)

7. Substance A has a smaller specific heat capacity than substance B. Equal mass blocks of A and B are heated on an electric hot plate which adds heat energy at a constant rate to each block. After 10 minutes which block will have the lower temperature and why?

Diploma Examination Practice

8. The human body contains about 70% water by mass. A body temperature close to 37°C is vital to survival. The property of water that allows the body to maintain an almost-constant temperature despite sudden changes in ambient temperature is its high

A. heat of fusion B. heat of vaporization C. specific heat capacity D. enthalpy of formation R esponse

9. Liquid mercury (cHg = 0.140 J/g˚C) is used in many thermometers

because it has a relatively low freezing point and a relatively high boiling point. A particular mercury thermometer contains 3.21 g of mercury. When the thermometer reading changes from 17.3°C to 101.2°C, the mercury has absorbed __________ J of energy.

10. When a 25.0 g sample of a metal is heated from 20.0 °C to 50.0 °C, 178 J of energy is absorbed from the surroundings. The specific heat capacity of the metal is

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Calculating Potential Energy (E

p

) Changes

Enthalpy (H) is a measure of the total potential energy possessed by a

substance. For this course, potential energy will be considered chemical potential energy. Enthalpy is a property of a system that reflects its capacity to exchange energy (usually heat – and this is why we use the symbol “H”, but sometimes other forms such as light or electricity) with the surroundings at constant pressure during a physical or chemical change (This course will focus on chemical changes).

Enthalpy can be different for a substance depending upon its state

(gas > liquid > solid), and it will be different at different temperatures (the substance will possess greater Ek at greater temperature).

We CAN’T measure the enthalpy of a system.

We CAN measure the CHANGE in enthalpy (∆H) of a system.

A substance will have more enthalpy if more of it takes part in a physical or chemical change (a firecracker of TNT will have less enthalpy than a large stick of dynamite); and therefore we try to standardize our enthalpy measurements so they can be compared. Molar enthalpy (kJ/mol)

allows us to see which substances release more or less energy during a change because they all involve the same amount. Specific enthalpy (kJ/g) could be used, but because we often deal with chemical reactions, the mole is a more convenient value to work with.

The enthalpy of reaction is the difference between the enthalpy of the products and the enthalpy of the reactants. Symbol is ∆Hrxn.

∆Hrxn = Hproducts - Hreactants

For exothermic reactions, the Hproducts is < Hreactants (∆Hrxn is negative)

For endothermic reactions the Hproducts is > Hreactants (∆Hrxn is positive)

Usually we calculate enthalpy indirectly by looking at temperature changes as energy is given off or absorbed.

Often ∆H will have a subscript following it to help us understand the type of change occurring. You should be prepared to recognize the different possible ∆H subscripts such as:

∆Hsol’n Enthalpy of solution ∆Hf Enthalpy of formation

∆Hcomb Enthalpy of combustion ∆Hrxn Enthalpy of reaction

∆Hdecomp Enthalpy of decomposition

So for any calculation involving potential energy, we can use the formula:

Ep = n∆H

Where Ep = Potential Energy

n = number of moles of a substance

How This Relates to the Curriculum:

30–A1.3k define enthalpy and molar enthalpy for chemical reactions

30–A1.4k write balanced equations for chemical

reactions that include energy changes

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Communicating Enthalpy:

We indicate enthalpy change (∆H) three ways:

1. With the energy term incorporated within the equation.

Exothermic reactions have the energy term on the products side. e.g. H2(g) + ½ O2(g)  H2O(l) + 285.8 kJ

Endothermic reactions have the energy term on the reactants side. e.g. H2O(l) + 285.8 kJ  H2(g) + ½ O2(g)

Enthalpy of reaction will change directly with the coefficients in a chemical reaction.

e.g. H2(g) + ½ O2(g)  H2O(l) + 285.8 kJ

2H2(g) + O2(g)  2H2O(l) + 571.6 kJ

2. With the energy term off to the right side of the equation (this communication method is called ∆H notation)

In exothermic reactions overall, energy is released:

There is more chemical potential energy in reactants for exothermic reactions.

e.g. H2(g) + ½ O2(g)  H2O(l) ∆H = -285.8 kJ

In endothermic reactions overall, energy is added:

There is more chemical potential energy in products for endothermic reactions.

e.g. H2O(l)  H2(g) + ½ O2(g) ∆H = +285.8 kJ

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Progress of Reaction

H2(g) + 1/2O2(g)

H2O(l)

∆H = -285.8 kJ

Formation of Hydrogen

Ep (kJ)

Progress of Reaction Ep (kJ)

H2(g) + ½ O2(g)

H2(g)

∆H = +285.8 kJ

Decomposition of Hydrogen

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Energy in Chemical Reactions

In chemical reactions, two things occur:

 Bonds are broken, and energy is required for this (endothermic).  Bonds are reformed, and energy is released (exothermic). Bonds break with reactants, and bonds reform with products.

Essentially all the energy required or released in a chemical reaction are the result of bonds breaking and reforming.

In endothermic reactions overall, energy is added:

33.2 kJ + ½ N2(g) + O2(g) NO2(g) ∆H is positive

There is more chemical potential energy in products for endothermic reactions.

In exothermic reactions overall, energy is released:

C(s) + O2(g)  CO2(g) + 393.5 kJ ∆H is negative

There is more chemical potential energy in reactants for exothermic reactions.

The enthalpy of reaction is the difference between the enthalpy of the products and the enthalpy of the reactants. Symbol is ∆Hrxn.

∆Hrxn = Hproducts - Hreactants

For exothermic reactions, the Hproducts is < Hreactants (∆Hrxn is negative)

For endothermic reactions the Hproducts is > Hreactants (∆Hrxn is positive)

30–A1.5k use and interpret ΔH notation to communicate and calculate energy changes in chemical reactions

30–A1.10k classify chemical reactions as endothermic or exothermic, including those for the processes of photosynthesis, cellular respiration and hydrocarbon combustion.

30–A1.3s analyze data and apply mathematical and conceptual models to develop and assess possible solutions

 compare energy changes

associated with a variety of chemical reactions through the analysis of data and energy diagrams

30–A2.3k analyze and label energy diagrams of a chemical reaction, including reactants, products, enthalpy change and activation energy

Less Ep More Ep

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Progress of Reaction Ep Reactants

Products

∆H = -ve

Exothermic Reaction

Progress of Reaction Ep

Reactants

Products

∆H = +ve

Endothermic Reaction

∆Hrxn can also be shown visually with potential energy diagrams that

show chemical potential energy change over the progress of a reaction

.

Notes to Remember:

Enthalpy of reaction will change directly with the coefficients in a chemical reaction.

H2(g) + ½ O2(g)  H2O(l) ∆H°f = -285.8 kJ

2H2(g) + O2(g)  2H2O(l) ∆H°f = -571.6 kJ

Chemical reactions are reversible, and if you reverse the reaction, then you need to reverse the sign on the ∆H as well.

H2(g) + ½ O2(g)  H2O(l) ∆H°formation = -285.8 kJ

H2O(l)  H2(g) + ½ O2(g) ∆H°decomposition = +285.8 kJ

You can be expected to communicate the enthalpy in a reaction three ways:

1 Using ∆H notation and written to the right of an equation.

2 Using ∆H notation and incorporate the energy term within the equation.

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C12H22O11(aq) + 12O2(g)

12CO2(g) + 11H2O(l)

∆H = -5640.3 kJ

Combustion of Sugar

EXAMPLES OF SOME PROBLEMS YOU’LL BE ASKED TO SOLVE:

Example 1:

Rewrite the equations expressing the enthalpy in ∆H notation for one mole of the underlined substance.

a) 2NH3(g) + 92.2 kJ  N2(g) + 3H2(g)

NH3(g)  ½ N2(g) + 3/2 H2(g)∆H = +46.1 kJ

b) C6H6(l) + 15/2 O2(g)  6CO2(g) + 3H2O(g) + 3135 kJ 1/6C6H6(l) + 15/12 O2(g)  CO2(g) + ½ H2O(g) ∆H = -522.5 kJ

Example 2:

When sucrose is burned in our bodies with excess oxygen, carbon dioxide, water and 5640.3 kJ of energy are produced according to the following equation:

C12H22O11(aq) + 12O2(g)  12CO2(g) + 11H2O(l)

Rewrite the equation by (a) using ∆H notation, (b) placing the energy term within the equation, and (c) drawing a potential energy diagram. a) C12H22O11(aq) +12O2(g)  12CO2(g) + 11H2O(l) ∆H = -5640.3 kJ

b) C12H22O11(aq) + 12O2(g)  12CO2(g) + 11H2O(l) + 5640.3 kJ

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Example 3:

For the reaction:

CO(g) + ½ O2(g)  CO2(g) ∆H = -283.0 kJ

Find the enthalpy released when 5.00 mol of O2 is consumed.

Example 4:

When 1.00 kg of pure carbon is burned in oxygen to produce carbon dioxide, 3.28 x 104_{kJ of heat energy is released. Write a balanced}

equation using the ∆H notation.

C(s) + O2(g)  CO2(g)∆H = -394 kJ

OR USING UNIT ANALYSIS:

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Practice Problems For Energy in Chemical Reactions

1. Rewrite the equations expressing the enthalpy in ∆H notation for one mole of the underlined substance.

a. 2Al(s ) + 3/2 O2(g)  Al2O3(s) + 1675.6 kJ

b. C8H18(l) + 25/2 O2(g)  8CO2(g) + 9H2O(g) + 5075 kJ

c. 2H2O(l) + 81.6 kJ  2H2O( l )

2. Given the equation for the formation of water, H2O(l):

2H2(g) + O2(g)  2H2O(l) + 571.6 kJ

a. The enthalpy given off in the above equation is: (-571.6 kJ) b. The molar enthalpy of the reaction for 1.00 mol of

H2O(l) is: (-285.8 kJ)

c. The enthalpy of reaction for the production of 0.500 moles of H2O(l) is (-143 kJ):

d. The ∆H for the production of 10.0 g H2O(l) is: (-159 kJ)

e. The ∆H for the production of 20.0 g H2O(l) is: (-317 kJ)

f. If hydrogen is formed from the decomposition of water, what is the enthalpy of reaction per mole of H2(g)

produced? (+285.8 kJ)

g. What is the enthalpy of reaction for the production of 5.00 g of H2(g)? (+707 kJ)

3. Given the equation for the formation of ethane C2H6(g)

2C(s) + 3H2(g)  C2H6(g) ∆H = -84.7 kJ

a. How much enthalpy is produced per mole of C2H6(g)

produced? (-84.7 kJ)

b. How much enthalpy is released if 0.500 mol C2H6(g) is

produced? (-42.4 kJ)

c. Find the enthalpy released if 12.37 moles of C2H6(g) are

produced. (1.05 x 103_kJ)

d. Find the enthalpy released when 3.01 g of C2H6(g) are

produced. (-8.47 kJ)

4. When 10.00 g of ethane, C2H6(g) is burned in oxygen, two

gases, CO2(g) and H2O(g) are produced and release 474.6 kJ of

heat energy.

a. Write a balanced equation for the reaction, using ∆H notation, showing the enthalpy of reaction of one mole of ethane.

b. Sketch a potential energy diagram showing the reaction for one mole of ethane.

5. When 5.00 g of nitrogen, N2(g), reacted with excess oxygen,

nitrogen dioxide, NO2(g) was produced and the reaction

required 11.82 kJ of energy.

a. Write a balanced equation using ∆H notation.

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Diploma Examination Practice

11. If this equation is rewritten to show the production of one mole of CH4(g)

and the energy is expressed as a term in the equation, then the energy will be

A. 587.7 kJ on the reactant side

B. 1763.0 kJ on the reactant side

C. 587.7 kJ on the product side

D. 1763.0 kJ on the product side

12. The student’s correct statements were

A. I and II

B. I and III

C. II and III

D. I, II, and III

13. The reason that dynamite releases a great amount of heat energy when it explodes is that the

A. products have more potential energy than the reactants in this endothermic reaction

B. reactants have more potential energy than the products in this endothermic reaction

C. products have more potential energy than the reactants in this exothermic reaction

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Use the following equations to answer the next question.

14. Which of these equations represent exothermic reactions?

A. I and II

B. I and III

C. II and IV

D. III and IV

15. In the production of 4.00 mol of NaAlO2(aq), the heat released is

_________ MJ.

Use the following equation to answer the next question.

16. The quantity of energy available when 1.00 g of sucrose reacts is __________ kJ.

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Use the following reactions to answer the next question.

18. Which of the following statements describes the reactions above?

A. Carbon is reduced in both reactions.

B. Both reactions are endothermic.

C. Reaction I could be classified as cellular respiration and reaction II could be classified as combustion.

D. The state of the water produced makes no difference when the heat of reaction is calculated.

19. Which statement is not consistent with the other three?

A. Statement I

B. Statement II

C. Statement III

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Finding Energy in Chemical Reactions Experimentally: Calorimetry

When we try to measure energy changes we make two assumptions: 1. Energy is conserved (not lost to the environment)

2. Heat flows from hot to cold until thermal equilibrium is reached. These assumptions are derived from the first two laws of thermodynamics. When an exothermic reaction takes place, energy is released and flows from the reaction to the water (or solution) in the calorimeter.

When an endothermic reaction takes place, energy is absorbed, and flows from the water (or solution) in the calorimeter to the reaction. We can measure the amount of energy released or absorbed by the reaction indirectly using calorimetry. All we need to do is find the energy change of the liquid, and assume:

Heat gained by liquid = Heat released by the reaction (and vice versa)

Water is cheap, and has a high heat capacity, so it is a great liquid to use in calorimetry. It’s high heat capacity allows us to keep water as a liquid, and then calculating energy gained or lost by water is only an Ek = mc∆t calculation.

In the lab, to help these assumptions we use a calorimeter

We are responsible to apply calorimetry equations for three types of calorimeters:

1. Styrofoam calorimeter

(AKA Thermos®).

The calorimeter is insulated, and minimizes heat movement with the outer environment. As well, because it is assumed to be perfectly insulated, the calorimeter does not need to be included in the calculations. Good for aqueous solutions, and reactions.

1. Find E absorbed (released) by the calorimeter (Ek =

mc∆t) 2. Use the Principle of Heat Transfer

Ep reaction = - Ek calorimeter

3. Find E released (absorbed) by the reaction (Ep = n∆H)

30–A1.8k use calorimetry data to determine the enthalpy changes in chemical reactions

30–A1.9k identify that liquid water and carbon dioxide gas are reactants in photosynthesis and products of cellular respiration and that gaseous water and carbon dioxide gas are the products of hydrocarbon combustion in an open system

30–A1.2s conduct investigations into relationships among observable variables and use a broad range of tools and techniques to gather and record data and information

 perform calorimetry experiments

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2. Metal Can calorimeter

Usually these calorimeters are made of copper or aluminum (good conductors).

These calorimeters are great for finding heat of a flame, or combustion reactions. You need to include mc∆T of the can as well as the water. The temperature change is the same as the water, because metal is such a good conductor that the heat transfers from the water to the can.

1. Find E absorbed (released) by the calorimeter (Ek = mc∆t)

2. Use the Principle of Heat Transfer Ep reaction = - Ek calorimeter

3. Bomb Calorimeter

Reaction occurs in oxygen gas, and the reaction centre is surrounded by liquid water. The calorimeter is sealed and can withstand great pressures resulting from gases produced in exothermic reactions. Used for explosive reactions all heat needs to be transferred to the water, and so final products are CO2(g) and

H2O(l).

1. Find E absorbed (released) by the calorimeter (Ek = C∆t)

(C is the heat capacity of the entire bomb calorimeter)

2. Use the Principle of Heat Transfer Ep reaction = - Ek calorimeter

On diploma examinations, the general principle that will be followed is that if combustion reactions are performed empirically in a bomb calorimeter, liquid water will be the product, and if the combustion occurs in an ambient (open) environment and a theoretical heat of combustion is to be determined, the product will be water vapour.

Combustion in a bomb calorimeter (or in a living system):

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As well; any solution used in a calorimeter will be considered to be very dilute, and will have all the physical properties of water (we can use the density of water as 1.0 g/mL, and specific heat capacity = 4.19 J/g˚C).

Examples of Practice Problems for Calorimetry You’ll be Expected to Answer

Example 1:

In a styrene calorimeter, the following data was collected when ammonium bromide was dissolved in water. What is the molar enthalpy of solution of ammonium bromide?

Mass of NH4Br(s) 9.80 g

tinitial (°C)(Calorimeter) 23.5

tfinal (°C)(Calorimeter) 19.4

Volume of Water in Calorimeter. 100 mL

Energy released by the calorimeter:

Ek = mc∆t

Ek = 100.0 g x 4.19 J/g°C x (19.4°C – 23.5°C) Ek = -1719.9 J

Ek = -1.72 kJ

Ereleased by the water = Egained by reaction

Energy absorbed by reaction

Ep = 1.72 kJ

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Example 2:

1.00 g of zinc powder is dissolved in 150.0 mL of 0.500 mol/L

Copper(II) sulphate (excess reagent). The initial temperature increased from 21.5 C to 43.6 C. Calculate the molar enthalpy of solution for the reaction between zinc metal and copper(II) sulphate.

Energy absorbed by the calorimeter:

Ek = mc∆t

Ek = 150.0 g x 4.19 J/g°C x (43.6°C – 21.5°C) Ek = 13889.85 J

Ek = 13.9 kJ

Egained by the water = Ereleased by reaction

Energy released by reaction

Ep = -13.9 kJ

n = 0.0153 mol Zn(s)

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Example 3:

In an aluminum calorimeter, 1.50 g methane (CH4(g)) was burned in a

camp stove and heated 500.0 mL of water at 12.8 ˚C. From the following data, find the molar enthalpy of reaction for the combustion of methane and add to the balanced equation in ∆H notation.

Data:

Mass of CH4 burned 1.50 g

Mass of inner container, Al 70.37 g Volume of water in inner container 500.0 mL Initial temperature 21.0 °C

Final temperature 55.0 °C

Reaction is:

CH4(g) + 2O2(g)  CO2(g) + 2H2O(g) ∆H = ??

Heat absorbed by the calorimeter

Note: Al container will have the same temperatures as the water.

Ek = mc∆tAl + mc∆tH2O

Ek = (70.37g x 0.897 J/gC x (34.0 °C)) + (500.0 g x 4.19 J/g°C x (34.0 °C)) Ek = 2.15 kJ + 71.2 kJ

Ek = 73.4 kJ

Eabsorbed by the water = Ereleased by reaction

Energy released by the reaction

Ep = -73.4 kJ

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Example 4:

150.0 mL of 0.200 mol/L HCl(aq) was added to 150.0 mL of 0.200 mol/L NaOH(aq) in a styrene cup. Initial temperature of both solutions was 25.0 °C. The final temperature was 27.1 °C. Write a balanced equation for the reaction and add the enthalpy of reaction in ∆H notation.

Data Given:

Mass of water = 300.0 g

*Note: total volume is 300.0 mL. The solutions are so dilute, that the volume of the actual ions are negligible, and so essentially they are dissolved in 300.0 mL of water. Mass of 300.0 mL of water is 300.0 g. HCl(aq) + NaOH(aq)  NaCl(aq) + H2O(l) ∆H = ??

We can use either NaOH or HCl for the calculations…(have enough information on both).

Energy absorbed by the water

Ek = mc∆t

Ek = 300.0 g x 4.19 J/g°C x (27.1°C – 25.0°C) Ek = 2600 J

Ek = 2.60 kJ

Ep = -2.60 kJ

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Example 5:

In a bomb calorimeter, 1.00 g of sulphur, S8(s), was burned in oxygen,

O2(g), to produce sulphur dioxide, SO2(g). Using the data given, find

the molar heat of reaction for sulphur and add to the balanced equation

Data Given:

Heat capacity of the calorimeter 419 J/˚C Initial temperature of water 21.5 °C Final temperature of water 43.6 °C S8(s) + 8O2(g)  8SO2(g) ∆H = ??

Energy Absorbed by the calorimeter:

Ek = C∆t

Ek = 419 J/°C x (43.6°C – 21.5°C) Ek = 9260 J

Ek = 9.26 kJ

Ep = -9.26 kJ

n = 0.00390 mol S8(s)

∆H = -2.37 x 103_kJ/mol

This makes 8 moles of SO2(g):

S8(s) + 8O2(g)  8SO2(g) ∆H = -2.37 x 103 kJ

So for 1 mol of SO2(g):

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Practice Problems for Calorimetry:

1. The following data set was obtained using simple Styrofoam Cup calorimetry. What is the molar enthalpy of solution of cesium fluoride? (-35.2 kJ/mol)

Mass of CsF(s) 15.19 g

tinitial (°C)(Calorimeter) 23.5

tfinal (°C)(Calorimeter) 31.9

Volume of Water in Calorimeter. 100 mL

2. In a styrene cup, some solid aluminum, Al(s), was added to an excess solution of copper(II) chloride, CuCl2(aq). Use the data below to

establish the molar enthalpy of reaction. Write a balanced equation for the reaction and include the enthalpy of reaction (∆H). (-66.7 kJ/mol)

Data

Mass of Al 1.00 g

Mass of water in calorimeter 100.0 g Initial temperature of water 21.0 °C Final temperature of water 26.9 °C

3. Propane, C3H8(g) (1.00 g) was burned, in an aluminum calorimeter (m =

13.8 g). The 200.0 mL of water in the calorimeter increased in

temperature from 21.0 ˚C to 87.6 ˚C. Find the ∆H of propane and add this to the balanced equation. (-2.50 x 103_kJ)

4. 50.0 mL of 0.800 mol/L HBr(aq) was added to 50.0 mL of KOH(aq) in a styrene cup. Initial temperature of both solutions was 23.18 °C. Final temperature was 26.38 °C. Write the balanced equation for the reaction and add the heat of reaction (∆H). (-33.5 kJ)

5. _{The molar heat of combustion of fructose (C6H12O6(s)) is -2820 kJ/mol.}

If 5.00 g of fructose is burned in a bomb calorimeter (ti= 25.00 °C, heat capacity: 29.7 kJ/°C) estimate the likely final temperature of this calorimeter. (27.6 °C)

6. _{A 2.870 g sample of naphthalene (formula C10H8(s)) undergoes}

complete combustion in a bomb calorimeter with a heat capacity of 10.2 kJ/°C. If the temperature change of the calorimeter is from 21.27 °C to 32.58 °C, what is the heat of combustion of this organic compound? (-5.15 * 103_kJ/mol)

7. _{Explain why the molar enthalpy of neutralization of LiOH(aq) by}

HNO3(aq) regardless of the concentrations of reactants; gives exactly the

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Diploma Examination Practice

20. The change that occurs in this cold pack is an

A. endothermic change, which results in an increase in temperature B. exothermic change, which results in an increase in temperature C. endothermic change, which results in a decrease in temperature D. exothermic change, which results in a decrease in temperature

21. Based on this information, a student determined that a hot pack could contain

A. calcium chloride, which undergoes an exothermic dissolving process B. calcium chloride, which undergoes an endothermic dissolving process C. ammonium nitrate, which undergoes an exothermic dissolving process D. ammonium nitrate, which undergoes an endothermic dissolving

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22. Which of the following diagrams represents the enthalpy of solution for either a cold pack or a hot pack?

23. The energy lost by the water was A. 3.26 kJ

B. 8.61 kJ C. 11.9 kJ D. 26.2 kJ

Use your recorded answer from the previous Multiple Choice to answer the next

Numerical Response.

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25. To more accurately determine the total heat released by the heat pack, the student should

A. use more water B. use a larger heat pack C. start with colder water

D. collect results for a longer period

26. If the energy change of the plastic container is not considered, the calculated energy change for the water from 0 s to 200 s is

A. 13.0 kJ B. 14.4 kJ C. 60.3 kJ D. 838 kJ

27. In a chemistry experiment, 12 g of (NH4)2SO4(s) was dissolved in 120 mL

of water in a simple calorimeter. A temperature change from 20.2°C to 17.8°C was observed. The experimental molar enthalpy of solution for ammonium sulphate was

A. –13 kJ/mol

B. –1.2 kJ/mol

C. +1.2 kJ/mol

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28. The amount of energy involved in this change is ___________ kJ.

29. The calculated energy change represents the enthalpy of A. solution

B. combustion C. neutralization D. formation

30. Two variables that could have been manipulated in this calorimetry experiment are

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31. The molar heat of solution for NaOH(s) is –44.6 kJ/mol. If 25.0 g of NaOH(s) is dissolved in water in a calorimeter, the heat released inside the calorimeter is

A. 27.9 kJ B. 71.4 kJ C. 1.12 MJ D. 1.78 MJ

32. When 24.0 g of carbon and 10.0 g of hydrogen are placed in a bomb calorimeter and reacted according to the equation 3 C(s) + 4H2(g) C3H8(g) + 103.8 kJ, the maximum amount of heat liberated by this reaction is

A. 69.1 kJ B. 128 kJ C. 257 kJ D. 619 kJ

33. The amount of energy lost by silicon in the experiment is A. 670 J

B. 1.35 kJ C. 2.25 kJ D. 3.82 kJ

34. When 0.500 g of peanut oil was burned, the temperature of 0.950 kg of water in a calorimeter increased by 4.60°C. The enthalpy of combustion of the peanut oil was ___________ kJ/g

35. When 1.65 g of ethanal (CH3CHO(l)) is burned in a calorimeter to produce H2O(l) and CO2(g), 44.7 kJ of heat energy is produced. According to this experimental data, the molar enthalpy of combustion of ethanal is

A. +1.52 x103_kJ/mol

B. –76.6 kJ/mol C. –165 kJ/mol D. –1.19 x103_kJ/mol

36. A student heated a 120.0 g sample of H2O(l) from 21.0°C to 32.5°C by adding 5.93 kJ of energy. The student then used this data to calculate the specific heat capacity of water and compared it with the standard value.

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37. The data required to determine the molar heat of combustion for ethanol, listed in numerical order, are ____ , ____ , ____ , and ___ .

38. In an experiment, a student heated 500 g of water from 25.0°C to 91.0°C using 0.133 mol of ethanol. If it is assumed that all the heat energy was absorbed by the calorimeter water, the experimental molar enthalpy of combustion for ethanol was +/–__________ MJ/ mol.

39. The balanced equation and the appropriate enthalpy change for the combustion of ethanol are

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Finding Energy In Reactions Using The Data Booklet

 We have found the ∆Hsolution, and ∆Hreaction in the laboratory.

 We can also find the enthalpy of a reaction using the Molar Enthalpy of Formation (∆H°f) of each compound.

 The reactions for the formation of many compounds from their elements are recorded already.

 We can use this information found in our data booklets (more on this later)

These reactions are balanced, and record values based on one mole of the compound.

e.g. ½ H2(g) + ½ I2(s)  HI(g) ∆H°f = +26.5 kJ

e.g. 6C(s) + 6H2(g) + 3O2(g)  C6H12O6(s) ∆H°f = -1273.1kJ

The ∆H means the energy absorbed or released (on a per mole basis) The ° means it is standard conditions (25°C and 101.3 kPa)

The “f” means formation.

It makes a difference what state the substances are in, so please record the state!!!

e.g. H2(g) + ½ O2(g)  H2O(l) ∆H°f =-285.8 kJ

versus

H2(g) + ½ O2(g)  H2O(g) ∆H°f =-241.8 kJ

 Molar enthalpies of formation are given for compounds

only, the molar enthalpy of formation for elements are assumed to be zero for this course.

Visually the standard enthalpies of formation look like this:

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½ H2(g) + ½ I2(s)

HI(g)

∆H = +26.5 kJ ½ H2(g) + ½ I2(s) HI(g) ∆H°f = +26.5 kJ

0 26.5

H2(g) + ½ O2(g)

H2O(l)

∆H =-285.8 kJ H2(g) + ½ O2(g) H2O(l) ∆H°f =-285.8 kJ

0

-285.8



Notice that numbers can be added to the y axis!

 Most values are negative for molar enthalpy of formation.

 In reality, all exothermic reactions have activation energy. This energy requirement prevents a spontaneous reaction from occurring.

Catalysts and Activation Energy

Catalysts can be used to speed up reactions that are exothermic, but react very slowly.

30–A2.1k define activation energy as the energy barrier that must be overcome for a chemical reaction to occur

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6C(s) + 6H2(g)+3O2(g)

C6H12O6(s)

∆H =-1273.1kJ Uncatalyzed Reaction (No Enzyme)

0

Activation Energy

6C(s) + 6H2(g)+3O2(g)

C6H12O6(s)

∆H =-1273.1kJ Catalyzed Reaction (Enzyme)

0 Activation Energy

H2O2(l)  H2O(l) + O2(g) ∆H°decompostion = -145.4 kJ

MnO2

H2O2(l)  H2O(l) + O2(g) ∆H°decompostion = -145.4 kJ

Catalysts simply reduce the activation energy required to move the reaction forward, and they get us to equilibrium (completion) more quickly, but do not release or absorb any more energy.

Activation energy (Ea) is the minimum required energy for a reaction

to move to completion.

Visually an uncatalyzed reaction and a catalyzed reaction look like this:

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6C(s) + 6H2(g)+3O2(g)

C6H12O6(s)

∆H =-1273.1kJ 6C(s) + 6H2(g)+3O2(g) C6H12O6(s)

0

12C(s) + 11H2(g)+ 11/2O2(g)

C12H22O11(s)

∆H =-2225.5kJ

12C(s) + 11H2(g)+ 11/2O2(g) C12H22O11(s)

0

Examples Of Problems You Will Be Expected To Do For Enthalpies of Formation:

Example 1:

Which sugar compound is more stable, glucose or sucrose? Draw potential energy graphs showing the reaction.

From the standard enthalpies of formation table: Glucose, C6H12O6(s) ∆H°f = -1273.1 kJ

Sucrose, C12H22O11(s) ∆H°f = -2225.5 kJ

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(Stability is a measure of how hard it would be to revert a compound back to its elements – the larger the negative ∆H°f, the more stable the

compound)

Example 2:

Use the following information to answer the next two questions:

1. Which of the quantities in the enthalpy level diagram below is (are) affected by the use of a catalyst?

Ea is reduced using a catalyst, therefore quantities I and II are reduced.

2. Which of the quantities is require to calculate the Activation energy for this reaction?

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Example 3:

What is the energy change when 2.00 moles of ammonia are produced in the Haber Process (shown in the following equation). Include the energy term in the equation:

3H2(g) + N2(g)  2NH3(g)

Solution:

From the Table of Standard Molar Enthalpies of Formation:

Ammonia ∆H°f =-45.9 kJ This is a per mole value (1 mol).

For two moles, we release twice the energy.

3H2(g) + N2(g)  2NH3(g) + 91.8 kJ

Example 4:

What is the energy change when 27.2 g of ammonia are produced in the Haber Process (shown in the following equation).

3H2(g) + N2(g)  2NH3(g)

Solution:

From the Table of Standard Molar Enthalpies of Formation:

Ammonia ∆H°f =-45.9 kJ This is a per mole value (1 mol).

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(kJ)

Reactants

Products

0 100 180

-400

Energy in Reactions Practice Problems

1. For the compounds listed below, write the balanced equations for the production of one mole of each compound from its elements. Include the ∆H°f in the equation.

a. Ammonia NH3(g)

b. Ethane C2H4(g)

c. Methanol CH3OH(l)

d. Potassium chlorate KClO3(s)

e. Octane C8H18(l)

2. Use the Table of Enthalpies of Formation to calculate the energy absorbed or released when 0.111 mol of calcium hydroxide is produced from its elements. (-109 kJ)

3. Find the energy exchange when 1.00 g of iron is oxidized to Fe2O3(s), rust. (-7.38 kJ/g)

4. Given the following potential energy diagram:

Give values for

a. Uncatalyzed activation energy

b. ∆H

c. Catalyzed activation energy

5. Draw and label a potential energy diagram for the formation of ethane, C2H6(g), from its elements.

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Diploma Examination Practice

40. At the treatment plant, the enzymes in the bacteria act as A. buffers

B. reducing agents C. oxidizing agents D. biological catalysts

41. Of the following metallic oxides, the one that would require the greatest energy per mole to decompose into its constituent elements is A. SnO(s)

B. PbO(s) C. Fe2O3(s)

D. MnO(s)

42. When 10.0 g of ethane gas was originally formed from its elements, the decrease in enthalpy was

A. 3.92 kJ B. 27.9 kJ C. 84.7 kJ D. 255 kJ

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44. When one mole of sodium bicarbonate is formed from its elements, 947.7 kJ of heat energy is released into the surroundings. This enthalpy change can be represented as

A. Na(s) + 1/2 H2(g) + C(s) + 3/2 O2(g) NaHCO3(s) + 947.7 kJ B. Na(s) + 1/2 H2(g) + C(s) + 3/2 O2(g) + 947.7 kJ NaHCO3(s) C. Na+_(aq)+ HCO3–_(aq)NaHCO3_(s)+ 947.7 kJ

D. Na+_(aq)+ HCO3–_(aq)+ 947.7 kJ NaHCO3_(s)

45. The thermochemical equation that represents the molar enthalpy of formation for ethanol is

A. C2H5OH(l) + 277.6 kJ 2 C(s) + 3H2(g) + 1/2 O2(g) B. C2H5OH(l) + 3O2(g) 2 CO2(g) + 3H2O(g) + 1234.8 kJ C. 2C(s) + 3H2(g) + 1/2 O2(g) C2H5OH(l) + 277.6 kJ

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Use the following diagram to answer the next question.

46. The potential energy change for this reaction is A. +170 kJ

B. +90 kJ C. –80 kJ D. –170 kJ

47. One of the byproducts of the cracking process used at Novacor is ethyne (C2H2(g)). In the presence of a palladium catalyst, the ethyne forms

ethene and ethane. This reaction is represented by the unbalanced

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Hess’s Law

There is actually a reason why we make you do all that work with those Heats of Formation Reactions and tables!

We use those Heats of Formations to apply Hess’s Law.

Hess’s Law lets us calculate the energy released or required by a reaction, simply by finding the difference in potential energy between the products and reactants (using Standard Molar Enthalpyies of Formation Values from our DataBooklet pp 4-5). Now we don’t have to perform a calorimetry experiment to find the energy absorbed or released by a reaction!

More Formally, Hess’s Law states:

If a chemical reaction can be expressed as the algebraic sum of two or more reactions, then its enthalpy of reaction (∆H rxn)is also the algebraic

sum of the separate enthalpy of reactions (∆Hf).

If you can add the equations, you can add the ∆H’s

We need to follow some basic rules to make Hess’s law work:  Write a balanced equation for the reaction for which ∆H is

needed.

 Add all the H°f values of the compounds on the product side.

 Add all of the H°f values on the reactant side.

 Remember to multiply any H°f on either the product or reactant

side by the coefficient found in the original equation!

 Subtract the sum of the reactants from the sum of the products and you have the ∆H of the reaction.

It looks like this:

These calculations are addition functions; so remember that significant digit rules for adding apply (the least accurate measure)

From the databooklet, our calculated values should have 1 decimal place.

EXAMPLES FOR HESS’ LAW:

Example 1: You could be asked to find molar enthalpy:

Find the molar enthalpy of combustion for ethane, CH (g).

30–A1.7k explain and use Hess’ law to calculate energy changes for a net reaction from a series of reactions

30–A1.3s analyze data and apply mathematical and conceptual models to develop and assess possible solutions

 compare energy changes

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= -1427.7 kJ/mol

 Remember that O2 is an element, and has a chemical

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Example 2: You could be asked to find the enthalpy of formation for a reaction, given the enthalpy of reaction:

Establish the standard enthalpy of formation for carbon disulphide, CS2(l), given the enthalpy of combustion for CS2(l) is -1075.1 kJ/mol.

Carbon disulphide burns in oxygen to produce carbon dioxide gas and sulphur dioxide gas.

Solution:

Reaction is:

CS2(l) + 3O2(g)  CO2(g) + 2SO2(g) ∆H = -1075.3 kJ

-1075.3 kJ = (1 mol(-393.5 kJ/mol) + 2 mol(-296.8 kJ/mol)) - ∆H°f CS2(l)

-88.2 kJ = -∆H°f CS2(l)

+88.2 kJ = ∆H°f CS2(l)

Example 3: You could be asked to find the enthalpy change in a reaction, given the mass of a substance (we need to use

stoichiometry here):

Natural gas consists of a mix of ethane and methane. What is the enthalpy released from burning 1.00 kg of ethane in a hot water heater?

Solution:

Reaction is:

C2H6(g) + 7/2 O2(g) _{2CO2(g) + 3H2O(g)}

∆H°rxn = (2mol(-393.5 kJ/mol) + 3mol(-241.8 kJ/mol)) - ( 1mol(-84.7 kJ/mol)) ∆H°rxn = -1427.7 kJ/mol

This means that 1427.7 kJ of heat energy are released when 1 mole of ethane burns completely to give gaseous products - as it would in a hot water heater.

We have 1.00 kg ethane

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Practice Problems for Hess’ Law:

1. Limestone is impure calcium carbonate, CaCO3(s). It is heated

to produce lime, CaO(s) and CO2(g). Establish the ∆H value for

this reaction. (+178.3 kJ/mol)

2. The heat of combustion for propanol, C3H7OH(l), is -1843.7 kJ/

mol. Find the enthalpy of formation for propanol. (-304.0 kJ/mol)

3. Hydrogen sulphide, H2S(g), burns in excess oxygen to produce

sulphur dioxide gas and liquid water. Find the heat released when 100.0 g of H2S(g) are burned. (-1649 kJ)

4. What amount of heat energy is produced when 1.00 kg of CO2(g) is produced during the combustion of methane? (1.82 * 10 4_kJ)

5. What amount of heat energy is produced when 1.00 kg of CO2(g) is produced during the combustion of octane? (14.4 MJ) 6. What amount of heat energy is produced when 1.00 kg of

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Diploma Examination Practice

49. The molar enthalpy of reaction for the fermentation of glucose is +/– _________ kJ/mol.

50. The balanced equation and the enthalpy change for the cellular respiration of glucose can be represented as

A. C6H12O6(s) + O2(g) CO2(g) + H2O(l) + 593.8 kJ B. C6H12O6(s) + 6O2 (g) + 2 802.7 kJ 6CO2(g) + 6H2O(l)

C. C6H12O6(s) + 6O2 (g) 6CO2(g) + 6H2O(l) + 2 802.5 kJ

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C. 240.0 kJ/mol D. 451.4 kJ/mol

52. The amount of energy absorbed when 0.350 mol of VCl4(l) decomposes to form VCl2(s) and Cl2(g) is _____ kJ.

53. The amount of energy released by the combustion of 100 g of C2H2(g) is __________ MJ.

54. How much heat is produced when 1.00 g of butane in a disposable lighter is completely burned to form gaseous carbon dioxide and water vapour?

A. 45.7 kJ

B. 124.7 kJ

C. 2 656.5 kJ

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Hess’s Law: Including the Equations

You will be able to use the math method for Hess’s law when you are asked to find (this works for about 85% of the problems):

 Enthalpy of reaction, when you have enthalpy of formation values.

 Enthalpy of formation for either reactants or products, BUT there can only be one compound!! Otherwise you will need to solve the problem using the equation method using heats of formation data, or enthalpy data given to us in a question. For example:

Given the following equation:

C12H22O11(aq) + 12O2(g)  12CO2(g) + 11H2O(l)

You could find the ∆Hrxn, or you could find the Hf for sugar. It

gets to be more difficult to find the Hf for either water or

carbon dioxide, because you would have two unknowns. It tends to be easier in that case to solve it using the equations and their enthalpies.

We need to follow some basic rules to make Hess’s law work using equations:

1. Write a balanced equation for the reaction for which ∆H is needed (this is the original equation).

2. Using the heats of formation table, write out the balanced equation to form one mole of each product.

3. Using the heats of formation table, write out the balanced equation to decompose one mole of each reactant (change the sign of the ∆H!).

4. Multiply any of the equations so they match up with the moles of that compound in the original equation (remember to multiply the ∆H as well!).

5. Ignore elements on either side (they will sort themselves out by the end).

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Example 1: You can find the molar enthalpy of a reaction, just like using the math method:

Find the molar enthalpy of combustion for methanol, CH3OH(l).

Original Equation

CH3OH(l) + O2(g)  CO2(g) + H2O(g) ∆H = ??

C(s) + O2(g)  CO2(g) ∆H = -393.5 kJ

2H2(g) + O2(g)  2H2O(g) ∆H = 2(-241.8 kJ)

CH3OH(l)  C(s) + 2H2(g) + 1/2O2(g) ∆H = +239.0 kJ

___________________________________________________ Net Reaction:

CH3OH(l) + O2(g)  CO2(g) + H2O(g) ∆H = -638.1 kJ

Example 2: You could be expected to find the enthalpy of formation for a reaction, given the molar enthalpy of an overall reaction.

Calculate the molar heat of formation for ammonia, NH3(g), using the

reaction given here for the combustion of ammonia.

Given:

2NH3(g) + 3/2 O2(g)  N2(g) + 3H2O(g) ∆H = -633.2kJ

We want to find:

½ N2(g) + 3/2 H2(g)  NH3(g) ∆H = ??

N2(g) + 3H2O(g)  2NH3(g) + 3/2 O2(g) ∆H = +633.2kJ

3H2(g) + 3/2 O2(g)  3H2O(g) ∆H = 3(-241.8 kJ)

___________________________________________________ Net Reaction:

N2(g) + 3 H2(g)  2NH3(g) ∆H = -92.2 kJ

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Practice Problems for Hess’s Law Using Equations:

1. Use the equations:

i. Ge(s_{) + O2(g)} GeO2(s₎ ∆H° = -534.7kJ

ii. Ge(s_{) + 1/2 O2(g)} GeO(s_{) ∆H° = -255.0 kJ}

to determine an enthalpy for the reaction

GeO(s_{) + 1/2 O2(g)}_GeO2(s₎ ∆H° = ?

(-279.7 kJ)

2. Use the following equations:

i.C(s_{) + 2H2(g)} CH4(g) ∆H° = -74.8 kJ

ii.C(s₎ C(g) ∆H° = +715.0 kJ

iii.H2(g) _{2 H(g)} ∆H° = +436.0 kJ

to determine an enthalpy for the formation of the four C- H bonds in methane:

C(g) + 4 H(g) _CH4(g) _{∆H° = ?}

(-1661.8 kJ)

3. Given the following equations:

a.N2H4(l) + 3O2(g)  2NO2(g) + 2H2O(l) ∆H = -466 kJ

b. H2O(l) + ½ O2(g)  H2O2(l) ∆H = +98 kJ

c.½ N2(g) + O2(g)  NO2(g) ∆H = +34 kJ

d.H2O(l)  H2O(g) ∆H = +40.8 kJ

What is the energy change for the following reaction: (-567 kJ)

(54)

Diploma Examination Practice

55. This diagram illustrates

A. the Law of Conservation of Mass B. an exothermic reaction

C. an endothermic reaction D. Hess’s Law

(55)