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Thermochemistry

Slide 3 / 118 Table of Contents

· The Nature of Energy

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· State Functions** · Enthalpy · Enthalpies of Formation · Energy in Foods and Fuels · Hess's Law · Measuring Enthalpy Changes: Calorimetry · Energy Associated with Changes of State · Enthalpies of Reaction

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The Nature of Energy

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Slide 5 / 118

Thermochemistry

Pr es en at We know chemical and physical processes release and absorb

• energy. We use these thermochemical principles to design air

conditioners and refrigerators as well as foot warmers that allow us to stay comfortable as we "go big" on the hill!

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Potential Energy is the energy that objects have energy due to their position. Gravitational Potential Energy GPE = mgh Elastic Potential Energy EPE = 1/2 kx2 Electric Potential Energy UE = kQ1Q2 r2

A Review of Energy from Physics

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Kinetic Energy is the energy that an object has by virtue of its motion: KE = 1/2 mv2 Work is defined by the formula W = Fdparallel

A Review of Energy from Physics Slide 8 / 118

Algebraically, these two statements combine to become: E0 + W = Ef Since Ef - Eo = ∆E, this can also be written as ∆E = W

A Review of Energy from Physics

work The total energy of an isolated system is constant. An outside force can change the energy of a system by doing work on it.

Slide 9 / 118

Another unit of energy is the calorie (cal).

1 cal = 4.184 J

The energy of food is measured in Calories (C). [note the capital "C"]

1 Calorie = 1000 calories = 4184 Joules

Units of Energy

The SI unit of energy is the Joule (J).

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1 A reaction produces 3.8 cal of energy. How many joules

• f energy is produced?

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2 A reaction uses 235 J of energy. How many calories have been burned?

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3 A 20 ounce coke contains 240 Calories. How many kilojoules of energy are present in a 20 ounce Coke?

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From last year, we know that ∆E = W. This year, we extend that by adding another way to change the energy of a system; by the flow of Heat (q). When two objects of different temperature are in contact, heat flow results in an increase of the energy of the cooler object and an identical decrease of the energy of the hotter object.

∆E = w + q

*Note, we use a lower case "w" in chemistry.

Energy & Heat A B

heat flow T = 20℃ T = 10℃

Slide 14 / 118 The First Law of Thermodynamics

∆E = w + q

Energy is neither created nor destroyed. In other words, the total energy of the universe is a constant; if the system loses energy, it must be gained by the surroundings, and vice versa.

Initial state Final state E of system decreases Internal energy, E E < E0

∆E < 0 (-)

E E0 Energy lost to surroundings Initial state Final state E of system increases Internal energy, E Energy gained from surroundings E > E0

∆E > 0 (+)

E E0

Slide 15 / 118 System and Surroundings

The system includes the reactants and products (here, the hydrogen, oxygen and water molecules). The surroundings are everything else (here, the cylinder and piston). Surroundings system When considering energy changes, we need to focus on a well- defined, limited part of the universe. The portion we focus on is called the system and everything else is called the surroundings. Consider the following reaction occurring within a metal cylinder.

2H2(g) + O2(g) --> 2H2O(g)

Slide 16 / 118 Changes in Internal Energy

If ∆E > 0, Efinal > Einitial The system absorbed energy from the surroundings. If ∆E < 0, Efinal < Einitial The system released energy from the surroundings. H2 (g) + O2 (g)

Internal energy, E

∆E < 0

(negative)

∆E >0

(positive)

H2 O(l)

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4 Ten grams of table salt in dissolved in water in a 250 mL

• beaker. Which of the following is a component of the

system? A NaCl B water C Na+ D beaker E A and B F A, B, and C G A, B, and D

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5 When a strong acid is added to a flask containing water the flask becomes warm to the touch. This is because... A the reaction performed work on the flask B the system absorbed heat from the surroundings C the system released heat to the surroundings D the surroundings released heat to the system

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6 When a strong acid is added to a flask containing water the flask becomes warm to the touch. Which correctly describes the change in energy? A ∆Esys is positive and ∆Esur is negative B ∆Esys is positive and ∆Esur is positive C ∆Esys is negative and ∆Esur is positive D ∆Esys is negative and ∆Esur is negative

Slide 20 / 118 Changes in Internal Energy

System ∆E>0 Heat q > 0 Work w > 0 Surroundings When energy is exchanged between the system and the surroundings, it is either exchanged as either heat (q)

• r work (w).

∆E = q + w

Slide 21 / 118 q, w, ∆E, and Their Signs

q

+

system gains heat

• system loses heat

w

+

work done on system

• work done by system

∆E

+

net gain of energy by system

• net loss of energy by system

Sign Conventions for q, w and ∆E

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7 The ∆E of a system that gains 50 kJ of heat and performs 24 kJ of work on the surroundings is ________ kJ. A

• 74

B

• 26

C D +26 E +74

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8 The ∆E of a system that releases 120 J of heat and does 40 J of work on the surroundings is ________ J. A

• 80

B

• 160

C D +80 E +160

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9 The ∆E of a system that absorbs 120 J of heat and does 120 J of work on the surroundings is ________ J. A

• 240

B

• 120

C D

+120

E +240

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10 The ∆E of a system that absorbs 12,000 J of heat and the surrounding does 12,000 J of work on the system is _______ J. A

• 24000

B

• 12000

C D +12000 E +24000

Slide 26 / 118 Exchange of Heat between System and Surroundings

Recall, when heat is absorbed by the system from the surroundings, the process is endothermic . System Surroundings Heat

• q

System Surroundings Heat +q When heat is released by the system into the surroundings, the process is exothermic.

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11 The reaction that occurs inside the foot warmer packet is endothermic? True False

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12 What will happen when a hot rock is put into cold water? A the water and rock will both gain energy B the water and rock will both lose energy C the rock will gain energy and the water will lose energy D the rock will lose energy and the water will gain energy

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13 If you put a hot rock in cold water, and your system is the rock, the process is _______. A exothermic B endothermic C neither, there is no net change of energy D it depends on the exact temperatures

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14 If you put a hot rock in cold water, and your system is the water, the process is _______. A exothermic B endothermic C neither, there is no net change of energy D it depends on the exact temperatures

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15 If you put an ice cube in water, and your system is the ice, the process is _______. A exothermic B endothermic C neither, there is no net change of energy D it depends on the exact temperatures

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16 If you put an ice cube in water, and your system is the water, the process is _______. A exothermic B endothermic C neither, there is no net change of energy D it depends on the exact temperatures

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17 When NaOH dissolves in water, the temperature of solution increases. This reaction is________. A exothermic B endothermic

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18 When CaCl 2 dissolves in water the temperature of water drops. This reaction is _____. A endothermic B exothermic

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19 Water droplets evaporating from the skin surface will make you feel cold. This process is _____. A exothermic for water B endothermic for skin C exothermic for skin D endothermic for water E A and C F C and D

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State Functions**

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Slide 37 / 118 State Functions

The internal energy of a system is independent of the path by which the system achieved that state. Below, the water could have reached room temperature from either direction...it makes no difference to the final energy of the system if it reached its final temperature by heating up or cooling down. 50g 100C 50g 25C 50g 0C

Cooling Heating

# #

#

Slide 38 / 118 State Functions

Internal energy is a state function. It depends only on the present state of the system, not

• n the path by which the system arrived at that state.

D∆E depends only on Einitial and Efinal # #

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These multiple paths explains how engines, air conditioners, batteries, heaters, etc. operate.

State Functions

They move between energy states while performing some task. # #

Slide 40 / 118 State Functions

Understanding thermodynamics enables us to harness energy flow for useful purposes. # #

Charged battery Heat Work

Energy lost by battery

Discharged battery

#E

Heat

A B

For instance, whether the battery is shorted out or is discharged by running the fan, its #E is the same. But q and w are different in the two cases. If the battery shorts

• ut, all of the energy is lost as

heat, whereas if it is used to run the fan, some energy is used to do work. Q and w are NOT state functions.

A B

Slide 41 / 118 Work Done by a Gas

When a gas expands it does work on its surroundings: W = Fd. In this case: F = PA and d = ∆h. W = Fd W = (PA)∆h W = P∆V A = cross sectional area P= F/A P= F/A

∆h ∆V

h Initial state Final state # #

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Since the gas expands, and does work on its surroundings, the energy

• f the gas decreases, this is

considered negative work. W = - P∆V

Work Done By a Gas

Initial state Final state A = cross sectional area P= F/A P= F/A

∆h

h

∆V

# # (-) work is done by system

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If the surroundings compress the gas, decreasing its volume, this increases the energy of the gas, it is considered positive work. W = + P∆V

Work Done By a Gas

P= F/A P= F/A h

∆V

Initial state Final state A = cross sectional area

∆h

# # (+) work is done on system

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We can measure the work done by the gas if the reaction is done in a vessel fitted with a piston.

Work Done by a Gas

# # HCl Solution HCl Solution + Zinc

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Enthalpy

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Slide 46 / 118 Enthalpy

The word "enthalpy" is derived from the Greek noun "enthalpos" which means heating. The enthalpy of a system (H) is a combination of the internal energy of a system (E) plus the work that needs to be done to create the space for the substance to occupy. It is impossible to measure enthalpy, H, directly. Only the change (∆H) can be measured.

H = E + W ∆H = ∆E + W

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At constant pressure, the change in enthalpy is the heat gained or lost by the system.

Enthalpy

Note: This is only true at constant pressure. See ** for more information (W = P∆V)

∆H = (q+W) - W ∆H = q

Slide 48 / 118 Enthalpy

The enthalpy of a system (H) is a combination of the internal energy of a system (E) plus the work that needs to be done to create the space for the substance (PV) to occupy. It is impossible to measure enthalpy, H, directly. Only the change (∆H) can be measured.

H = E + PV ∆H = ∆E + PV

**

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If a process takes place at constant pressure (as most processes we study do), we can account for heat flow during the process by measuring the enthalpy of the system. At constant pressure, the change in enthalpy is the heat gained or lost by the system.

Enthalpy

∆H = ∆E + ∆(PV) ∆H = ∆E + P∆V ∆H = (q+w) - w ∆H = qp (at constant pressure)

** Slide 50 / 118

Enthalpy is an extensive property; its value depends on the quantity of the substance present. Combustion of 16 grams of CH4 ∆H = -891 kJ Combustion of 32 grams of CH4 ∆H = -1782 kJ

∆H for a reaction in the forward direction is equal in size,

but opposite in sign, to ∆H for the reverse reaction. CH4(g) + 2O2(g) --> CO2(g) + 2H2O(g) ∆H = -891 kJ 2H2O(g) + CO2(g) --> CH4(g) + 2O2(g) ∆H = +891 kJ

∆H for a reaction depends on the state of the products and

the state of the reactants. CH4(g) + 2O2(g) --> CO2(g) + 2H2O(g) ∆H = -891 kJ CH4(g) + 2O2(g) --> CO2(g) + 2H2O(l) ∆H = -979 kJ

Enthalpy Slide 51 / 118

20 When 114 grams of gasoline combust, the ∆H is equal to

• 5,330 kJ. What is the ∆H for combustion of 57 grams of

gasoline?

Slide 52 / 118 Endothermic and Exothermic Processes

A process is endothermic when ∆H is positive. ∆H>0 System Endothermic Surroundings Heat +q ∆H<0 System Exothermic Surroundings Heat

• q

A process is exothermic when ∆H is negative.

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21 The reaction A + B --> C is endothermic. The DH for this reaction is +50 J. What is the DH for the reaction C --> A + B? A Cannot be determined. B 0.02 J C

• 50 J

D 100 J

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22 The reaction A + B --> C is exothermic. The ∆H for this reaction is -150 J. What is the ∆H for the reaction C --> A + B? A +150 B zero C

• 150

D this reaction will not happen

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23 Dissolving NaOH in water will increase the temperature of the solution. This reaction is _____

A

exothermic

B

endothermic

C

adiabatic

D

isothermal

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24 NH3NO3 + H2O --> NH4+ + NO3- + OH- ∆H = +25.69 kJ/mol. This reaction is exothermic.

Yes No

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Measuring Enthalpy Changes: Calorimetry

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Since we cannot know the exact enthalpy of the reactants and products, we measure ∆H through calorimetry, the measurement of heat flow by making use of the expression: ∆H = qp

Measuring Enthalpy Changes: Calorimetry

The subscript p on q means the process is occurring at constant pressure. No work is being done only heat is being

• exchanged. You will not always see the subscript but it is

implied when we are speaking of Enthalpy.

Slide 59 / 118 Heat Capacity and Specific Heat

The amount of energy required to raise the temperature of a substance by 1 K (1°C) is its heat capacity. The amount of energy required to raise the temperature of one gram of a substance by 1 K (1°C) is its specific heat (c).

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25 Heat capacity is an example of an A Intensive property B Extensive property

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26 Specific heat is an example of an A Intensive property B Extensive property

Slide 62 / 118 Heat Capacity and Specific Heat

Specific heat = heat transferred mass x temperature change

m∆T c = q

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27 The specific heat of marble is 0.858 J/g-K. How much heat (in J) is required to raise the temperature of 20g of marble from 22 °C to 45 °C?

q = mc∆T

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28 An 26 g sample of wood (c = 1.674 J/(g-K)) absorbs 200 J of heat, upon which the temperature of the sample increases from 20.0 °C to ______°C.

q = mc∆T

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29 A sample of silver (c = 0.236 J/g-K) absorbs 800 J of heat, upon which the temperature of the sample increases from 50.0 °C to 80°C. What is the mass of the sample?

q = mc∆T

Slide 66 / 118

sparknotes.com

thermometer styrofoam cup stirrer insulated cover

Constant Pressure Calorimetry

Many chemical reactions happen in aqueous solutions. The apparatus to the left is a calorimeter. How could you use it to measure the heat change for a chemical reaction in an aqueous solution?

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Because the specific heat for water is well known (4.184 J/g-K), we can measure ∆H for the reaction with this equation: D∆H = q = mc∆T (at constant pressure)

Constant Pressure Calorimetry

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thermometer styrofoam cup stirrer insulated cover

Slide 68 / 118 Constant Pressure Calorimetry

Example: A student wishes to determine the enthalpy change when ammonium chloride (NH4Cl) dissolves in water. The student masses

• ut 20.00 grams of ammonium chloride and adds it to 500 grams of

water in a styrofoam cup at a temperature of 16.1 Celsius. The student

• bserves the temperature to decrease to 13.2 Celsius. What is the

enthalpy change for the dissolution of ammonium chloride?

• 1. Find enthalpy change of solution using q = mc∆T

= 520 g x -2.9 C x 4.2 J = -6,334 J

• 2. Since heat released by surroundings (solution) is equal

to the heat gained by system (ammonium chloride). ∆H for dissolving of NH4Cl = +6334 J g℃

Slide 69 / 118 Bomb Calorimetry

Reactions can be carried

• ut in a sealed “bomb”

such as this one. The heat absorbed (or released) by the water is a very good approximation of the enthalpy change for the reaction. Thermometer Ignition wires Oxygen atmosphere Sample Water

Slide 70 / 118

Because the volume in the bomb calorimeter is constant, what is measured is really the change in internal energy, ∆E, not ∆H. For most reactions, the difference is very small.

Bomb Calorimetry (Constant Volume)

Thermometer Ignition wires Oxygen atmosphere Sample Water

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30 The reaction below takes place in a bomb calorimeter. If the student found that the temperature of the water was 12.2 Celsius to start and the heat capacity of the calorimeter was 34 J/C, what must be the enthalpy change of the reaction if the temperature of the water increased to 15.6 Celsius? 4Fe(s) + 3O2(g) --> 2Fe2O3(s)

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31 A mysterious meteorite is discovered in your backyard. To determine its identity, you determine its specific heat. The 164.6 gram sample of metal is heated to 90 C and then dumped in 300 grams of water in a styrofoam cup at an initial temperature of 10 C. After the metal is added, the temperature rises to 11.3 C. Identify the metal. A B C Metal Specific Heat (J/gC) Cu 0.386 Au 0.126 Al 0.900

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Energy Changes Associated with Changes of State

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Slide 74 / 118 Energy Changes Associated with Changes of State

Chemical and physical changes are usually accompanied by changes in energy. Recall the following terms: When energy is put into the system, the process is called _____________. When energy is released by the system, the process is called _____________.

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Endothermic processes Exothermic processes

Energy is taken into the system from the surroundings (∆H > 0) Energy is released from the system to surroundings (∆H < 0)

Energy Changes Associated with Changes of State

Answer

melting a solid boiling or evaporating a liquid sublimation of a solid freezing a liquid condensing a gas deposition of a gas Fill in the blanks

Slide 76 / 118

Solid Gas Liquid Vaporization Condensation Melting Sublimation Freezing Deposition Energy of system

Phase Changes Slide 77 / 118

32 Which of the following is/are exothermic?

• I. boiling
• II. melting
• III. freezing

A I only B I and II only C III only D I, II and III

Slide 78 / 118 Energy Changes Associated with Changes of State

The heat of fusion (∆Hfus) is the energy required to change a solid at its melting point to a liquid. The heat of vaporization (∆Hvap) is defined as the energy required to change a liquid at its boiling point to a gas. Heat of fusion(H2O) = 6.0 kJ/mol Heat of vaporization(H2O) = 41.0 kJ/mol Class Question: Why is the heat of vaporization much higher than the heat of fusion for a substance? In order to change phase from a solid to liquid, the particle attractions need only be strained somewhat. When a material changes from a liquid to a gas, the particle attractions must be essentially broken. move for answer

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10 40 80 Heat of vaporization Heat of fusion 24 5 29 41 58 23 6 7

Energy Changes Associated with Changes of State

Note that these quantities are usually per 1.00 mole, whereas q = mc∆T involves mass in grams. Butane Diethyl ether Water Mercury Heat of fusion and vaporization kJ/mol

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33 The heat of vaporization for butane is 24 kJ/mol. How much energy is required to vaporize 2 mol of butane? A 2kJ B 12 kJ C 22 kJ D 48 kJ

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34 The heat of vaporization for butane is 24 kJ/

• mol. How much energy is required to vaporize

0.33 mol of butane? A 8kJ B 12 kJ C 16 kJ D 72 kJ

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35 How much energy is required to melt 0.5 mol of water? A 2kJ B 3 kJ C 6 kJ D 12 kJ Heat of fusion for H2O (s) Heat of vaporization for H2O (l) 6 kJ/mol 41 kJ/mol

Answer

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36 How much energy is released when 3.0 mol of water freezes? A 2kJ B 3 kJ C 18 kJ D 123 kJ Heat of fusion for H2O (s) Heat of vaporization for H2O (l) 6 kJ/mol 41 kJ/mol

Answer

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37 How much energy is needed to melt 10.0 mol solid Hg? A 2.3 kJ B 5.8 kJ C 23 kJ D 230 kJ Heat of fusion for Hg (s) Heat of vaporization for Hg (l) 23 kJ/mol 58 kJ/mol E 580 kJ

Answer

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38 How much energy is needed to vaporize 2.0 mol Hg (l)? Heat of fusion for Hg (s) Heat of vaporization for Hg (l) 23 kJ/mol 58 kJ/mol

Answer

Slide 86 / 118 Energy Changes Associated with Changes of State

This graph is called a heating curve. It illustrates how temperature changes over time as constant heat is applied to a pure solid substance. Temperature (

0C)

Heat added (each division corresponds to 4 kJ) Liquid water Ice and liquid water (melting) Liquid water and vapor (vaporization) Water vapor

• 25

25 75 125 50 100

Ice A B C D E F

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From A to B, ice is heating up from

• 25oC to 0oC.

Energy Changes Associated with Changes of State

Since Q = mc∆T ∆T = Q So the slope is (mc) 1 (mc) Temperature (

0C)

Heat added (each division corresponds to 4 kJ) Liquid water Ice and liquid water (melting) Liquid water and vapor (vaporization) Water vapor

• 25

25 75 125 50 100

Ice A B C D E F

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From B to C, ice is melting. The added heat is breaking the hydrogen bonds of the solid, so the temperature is constant. That's why the slope is zero.

Energy Changes Associated with Changes of State

Temperature (

0C)

Heat added (each division corresponds to 4 kJ) Liquid water Ice and liquid water (melting) Liquid water and vapor (vaporization) Water vapor

• 25

25 75 125 50 100

Ice A B C D E F

Slide 89 / 118

From C to D, liquid water is heating up from 0 C to 100 C. Once again, the slope is 1/(mc). But "c" is different for all the phases

• f a substance, so

the slope is different for solid, liquid and gaseous H2 O.

Energy Changes Associated with Changes of State

Temperature (

0C)

Heat added (each division corresponds to 4 kJ) Liquid water Ice and liquid water (melting) Liquid water and vapor (vaporization) Water vapor

• 25

25 75 125 50 100

Ice A B C D E F

Slide 90 / 118

From D to E, liquid water is boiling into vapor. The added heat is breaking the IM forces

• f the liquid, so the

temperature is constant. That's why the slope is zero.

Energy Changes Associated with Changes of State

Temperature (

0C)

Heat added (each division corresponds to 4 kJ) Liquid water Ice and liqui d water (melting) Liquid water and vapor (vaporization) Water vapor

• 25

25 75 125 50 100

Ice A B C D E F

Slide 91 / 118

From E to F water vapor, steam, is heating up from 100 C to 125 C. Once again, the slope is 1/(mc). But "c" is different for all the phases

• f a substance,

so the slope is different for for solid, liquid and gaseous H2 O.

Energy Changes Associated with Changes of State

Temperature (

0C)

Heat added (each division corresponds to 4 kJ) Liquid water Ice and liquid water (melting) Liquid water and vapor (vaporization) Water vapor

• 25

25 75 125 50 100

Ice A B C D E F

Slide 92 / 118

Recall that slope = where m = mass and C = specific heat of the substance

Energy Changes Associated with Changes of State

1 mC (q) Energy Added (Joules) (DT) Temperature (C) A B C D This graph shows heat transfer versus change in temperature for 1 gram of 4 different substances.

Slide 93 / 118

39 Which substance has the lowest specific heat? A A B B C C D D

(DT) Temperature (C) A B C D (q) Energy Added (Joules)

Slide 94 / 118

40 Which substance requires the highest amount of heat added to raise the temperature? A A B B C C D D

(DT) Temperature (C) A B C D (q) Energy Added (Joules)

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41 Which segment(s) contain solid sodium chloride? A AB only B AB and BC C AB, BC and CD D BC, CD and DE

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42 Which segment(s) contain liquid sodium chloride? A AB only B AB and BC C AB, BC and CD D BC, CD and DE

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43 What is the melting point (in oC) of sodium chloride?

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44 What is the freezing point (in oC) of sodium chloride?

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45 Which is greater: the specific heat of solid NaCl,

• r the specific heat of molten (liquid) NaCl?

A solid B liquid C Cannot be determined. D They are equal.

Slide 100 / 118

Segments AB, CD, EF slope = nonzero T increasing KE increasing PE constant apply q = mc∆T Segments BC & DE slope = 0 ∆T = 0 KE constant PE increasing apply ∆Hfus or ∆Hvap

Features of a Heating Curve

Temperature (

0C)

Heat added (each division corresponds to 4 kJ) Liquid water Ice and liquid water (melting) Liquid water and vapor (vaporization) Water vapor

• 25

25 75 125 50 100 Ice A B C D E F

Slide 101 / 118

Recall that any given substance has a different value for specific heat as a solid, as a liquid and as a gas. Additionally, melting one mole of a substance (∆Hfus) and vaporizing one mole of the same substance (∆Hvap) require different amounts of energy.

Energy Changes Associated with Changes of State Slide 102 / 118

Specific heat of ice 2.09 J/g-℃ Specific heat of water 4.18 J/g-℃ Specific heat of steam 1.84 J/g-℃ Heat of fusion (∆Hfus) of water at 0℃ 6.01 kJ/mol or 330 J/g Heat of vaporization (∆Hvap) of water at 100℃ 40.7 kJ/mol or 2600 J/g

Specifics about Water

Slide 103 / 118 Calculating Energy Changes from a Heating Curve

Sample Problem: Calculate the enthalpy change (in kJ) for converting 10.0 g of ice at -15.0°C to water at 25.0 °C.

It is a good idea to sketch out the segments first on a graph.

Slide 104 / 118 Calculating Energy Changes from a Heating Curve

Sample Problem: Calculate the enthalpy change (in kJ) for converting 10.0 g of ice at -15.0°C to water at 25.0 °C. Segment 1 Warming the ice from -15.0°C up to the substance's MP which is 0°C.

• 14
• 4

10 20 6 12 24

Temperature (C) Time (s)

Slide 105 / 118 Calculating Energy Changes from a Heating Curve

Sample Problem: Calculate the enthalpy change (in kJ) for converting 10.0 g of ice at -15.0°C to water at 25.0 °C. Segment 2 Melting the ice at 0°C.

• 14
• 4

10 20 6 12 24

Temperature (C) Time (s)

Slide 106 / 118 Calculating Energy Changes from a Heating Curve

Sample Problem: Calculate the enthalpy change (in kJ) for converting 10.0 g of ice at -15.0°C to water at 25.0 °C. Segment 3 Warming the liquid from 0°C to 25°C.

• 14
• 4

10 20 6 12 24

Temperature (C) Time (s)

Slide 107 / 118

q1 = mc∆T q1 = (10.0g) (2.09 J/g-oC) (15.0oC) q1 = 313.5 J q1 = 0.314 kJ or 313.5J

We are now ready to apply either the calorimetry equation or ∆Hfus or ∆Hvap. Sample Problem: Calculate the enthalpy change (in kJ) for converting 10.0 g of ice at -15.0°C to water at 25.0 °C.

Calculating Energy Changes from a Heating Curve

Segment 1 Warming the ice from -15.0°C up to the substance's MP which is 0°C.

Slide 108 / 118

Sample Problem: Calculate the enthalpy change (in kJ) for converting 10.0 g of ice at -15.0°C to water at 25.0 °C.

Calculating Energy Changes from a Heating Curve

q2 = (∆Hfus ) (# moles) q2 = (6.01 kJ/mol) [(10.0 g)/ 18.0 g/mol)] q2 = 3.34 kJ

• r

6.01/18 J/g x 10g = 3.34 kJ = 3340J

Segment 2 Melting the ice at 0 °C.

Slide 109 / 118

Sample Problem: Calculate the enthalpy change (in kJ) for converting 10.0 g of ice at -15.0°C to water at 25.0 °C.

Calculating Energy Changes from a Heating Curve

q3 = mc∆T q3 = (10.0g) (4.18 J/g-oC) (25.0oC) q3 = 1045 J q3 = 1.05 kJ

Segment 3 Warming the water from 0 °C to 25.0 °C.

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Sample Problem: Calculate the enthalpy change (in kJ) for converting 10.0 g of ice at -15.0°C to water at 25.0 °C.

Calculating Energy Changes from a Heating Curve

Now, we add the heat changes for all 3 segments. First, make sure that all quantities are in kilojoules.

Total ∆H = (q1 + q2 + q3) kJ ∆H = (0.314 + 3.34 + 1.05) kJ ∆H = 4.6 kJ or 4698.5J

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46 Calculate the enthalpy change in J of converting 75 g of ice at -11 ℃ to liquid water at 22℃. A 98,654 B 33,371 J C 8,621 D 26,474 E 35,096 J Specific heat of ice

2.09 J/g-

℃

Specific heat of water 4.18 J/g-

℃

Specific heat of steam 1.84 J/g-

℃

Heat of fusion (∆Hfus)

• f water at 0℃

6.01 kJ/mol or 330 J/g Heat of vaporization (∆Hvap) of water at 100℃ 40.7 kJ/mol or 2600 J/g

Answer

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47 Calculate the enthalpy change of converting 25 g of water vapor at 110 ℃ to liquid water at 50℃. A -69,765 B 69,765 C -59,315 D -70,685 J E -13,935 Specific heat of ice

2.09 J/g-

℃

Specific heat of water 4.18 J/g-

℃

Specific heat of steam 1.84 J/g-

℃

Heat of fusion (∆Hfus)

• f water at 0℃

6.01 kJ/mol or 330 J/g Heat of vaporization (∆Hvap) of water at 100℃ 40.7 kJ/mol or 2600 J/g

Answer

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48 Calculate the enthalpy change in kJ of converting 31.8 g

• f ice at 0℃ to water vapor at 140℃.

A 109 kJ B 96 kJ C 104 kJ D - 88 kJ E -109 kJ Specific heat of ice

2.09 J/g-

℃

Specific heat of water 4.18 J/g-

℃

Specific heat of steam 1.84 J/g-

℃

Heat of fusion (∆Hfus)

• f water at 0℃

6.01 kJ/mol or 330 J/g Heat of vaporization (∆Hvap) of water at 100℃ 40.7 kJ/mol or 2600 J/g

Answer

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Enthalpies of Reaction

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Slide 115 / 118 Enthalpy of Reaction

The change in enthalpy, ∆H, is the enthalpy of the products minus the enthalpy of the reactants:

D∆H = Hproducts − Hreactants

CH4(g) + 2 O2(g) CO2(g) + 2H2O(l) ∆H1 =

• 890 kJ

Enthalpy ∆H2 = 890 kJ

Slide 116 / 118

This quantity, ∆H, is called the enthalpy of reaction or the heat of

• reaction. The combustion of hydrogen in the

balloon below is an exothermic reaction and the energy released is equal to the heat of reaction.

Enthalpy of Reaction

2H2(g) + O2(g) 2H2O(g) ∆H<0 exothermic Enthalpy

Slide 117 / 118

How much energy is released when 4.0 mol of HBr is formed in this reaction? H2(g) + Br2(g) --> 2HBr(g) ∆H = -72 kJ

Enthalpy of Reaction

Example Problem #1 X 4.0 mol HBr(g)

• 72kJ

2.0 mol HBr(g) X = -72kJ (4/2) X = -144kJ 144kJ of energy are released =

Slide 118 / 118

How much energy is released when 4.0 mol of HBr is formed in this reaction? H2(g) + Br2(g) --> 2HBr(g) ∆H = -72 kJ

Enthalpy of Reaction

Example Problem #1 4.0 mol HBr x -72 kJ 2 mol HBr = -144kJ or 144 kJ released