Copyright 2000, PRENTICE HALL Chapter 51 These are the remaining slides from class, but beginning...

Copyright 2000, PRENTICE HALL

Chapter 5 1

These are the remaining slides from class, but beginning with slide 22, there is another presentation from another instructor.


Chapter 5 2

For each of the following, define a system and its surroundings and give the direction of energy transfer.

a) Methane is burning in a Bunsen burner in a laboratory.

b) Water drops, sitting on your skin after swimming, evaporate.

c) Two chemicals mixing in a beaker give off heat.

React 4


Chapter 5 3

• Hydrogen gas and oxygen gas react violently to form water.

• Which is lower in energy: a mixture of hydrogen and oxygen gases, or water?

• Sketch an energy level diagram for this reaction.

React 5


Chapter 5 4

• Gas A2 reacts with gas B2 react to form gas AB. The bond energy of AB is much greater than the bond energy of A2 or B2.

• Is the formation of AB endothermic or exothermic?

• Which is lower in energy, the reactants or the products?

React 6


Chapter 5 5

Which of the following performs more work?

• A gas expanding against a pressure of 2 atm from 1.0 L to 4.0 L

• A gas expanding against a pressure of 3 atm from 1.0 L to 3.0 L

React 7


Chapter 5 6

Determine the sign of E for each of the following with the listed conditions:

• An endothermic process that performs work.– *work* > *heat*

– *work* < *heat*

• Work is done on a gas and the process is exothermic.– *work* > *heat*

– *work* < *heat*

React 8


Chapter 5 7

Enthalpy and Calorimetry


Chapter 5 8

A Coffee Cup Calorimeter Made of Two Styrofoam Cups


Chapter 5 9

A 100.0 g sample of water at 90°C is added to a 100.0 g sample of water at 10°C.

The final temperature of the water is:a) Between 50°C and 90°C.

b) 50°C

c) Between 10°C and 50°C.

Calculate the final temperature of the water.

React 9


Chapter 5 10

A 100.0 g sample of water at 90°C is added to a 500.0 g sample of water at 10°C.


b) 50°C



React 10


Chapter 5 11

You have a Styrofoam cup with 50.0 g of water at 10C. You add a 50.0 g iron ball at 90C to the water.


b) 50°C



React 11


Chapter 5 12

Change in Enthalpy

• Can be calculated from enthalpies of formation of reactants and products.

Hrxn° = npHf(products) - nrHf(reactants)


Chapter 5 13

Hess’s Law


Chapter 5 14

The Principle of Hess’s Law


Chapter 5 15

Standard Enthalpies of Formation


Chapter 5 16

Standard States

• Compound– For a gas, pressure is exactly 1 atmosphere.– For a solution, concentration is exactly 1 molar.– Pure substance (liquid or solid)

• Element– The form [N2(g), K(s)] in which it exists at 1

atm and 25°C.


Chapter 5 17

A Schematic Diagram of the Energy Changes for the Reaction

CH4(g) + 2O2(g) CO2(g) + 2H2O(l)


Chapter 5 18

Using the following data, calculate the heat of formation of the compound ICl (g) at 25°C, and show your work:

H° (kJ/mol)

Cl2 (g) 2Cl (g) 242.3

I2 (g) I (g) 151.0

ICl (g) I (g) + Cl (g) 211.3

I2 (s) I2 (g) 62.8

React 12


Chapter 5 19

Present Sources of Energy


Chapter 5 20

Energy Sources Used in the United States


Chapter 5 21

The Earth’s Atmosphere


Chapter 5 22

Thermochemistry

Chapter 6Chapter 6


Chapter 5 23

The Nature of Energy

• Thermodynamics is the study of energy and its transformations.

• Thermochemistry is the study of the relationships between chemical reactions and energy changes.

• Kinetic Energy and Potential Energy– Kinetic energy is the energy of

motion:

– Potential energy is the energy an object possesses by virtue of its position.


Chapter 5 24

Units of Energy

• SI unit is the joule, J.

• Traditionally, we use the calorie as a unit of energy.

• 1 cal = 4.184 J (exactly)

• The nutritional Calorie, Cal = 1000 cal.


Chapter 5 25

Exercise

• Suppose you consume a six-pack of diet coke (1 nutritional Calorie per can). Convert the energy content of the six-pack to kJ.


Chapter 5 26

System and Surroundings• A system is the part of the universe we are

interested in studying.• Surroundings are the rest of the universe (i.e., the

surroundings are the portions of the universe not involved in the system).

Example: If we are interested in the interaction between hydrogen and oxygen in a cylinder, then the H2 and O2 in the cylinder form a system.


Chapter 5 27

Transferring Energy: Work and Heat

• Force is a push or pull on an object.• Work is the amount of force applied to an object

over a distance:

w = F d• Heat is the energy transferred from a hotter object

to a colder one. • Energy is the capacity to do work or to transfer

heat.


Chapter 5 28

• The first law of thermodynamics states that energy cannot be created or destroyed.

• The first law of thermodynamics is the law of conservation of energy.– That is, the energy of (system + surroundings) is

constant.– Thus any energy transferred from a system must

be transferred to the surroundings (and vice versa).

The First Law of Thermodynamics


Chapter 5 29

Internal Energy• The total energy of a system is called the internal

energy.– It is the sum of all the kinetic and potential energies

of all components of the system. • Absolute internal energy cannot be measured, only

changes in internal energy.

• Change in internal energy; E = Efinal - Einitial.


Chapter 5 30

Example

• A mixture of H2(g) and O2(g) has a higher internal energy than H2O(g).

• Going from H2(g) and O2(g) to H2O(g) results in a negative change in internal energy:

H2(g) + O2(g) 2H2O(g) E < 0

• Going from H2O(g) to H2(g) and O2(g) results in a positive change in internal energy:

2H2O H2(g) + O2(g) E > 0


Chapter 5 31

Heat and Work


Chapter 5 32

Relating E to Heat and Work• From the first law of thermodynamics:

– When a system undergoes a physical or chemical change, the change in internal energy is given by the heat added to or absorbed by the system plus the work done on or by the system:

E = q + w

• Heat flowing from the surroundings to the system is positive (i.e., the system feels cold to the touch because it is absorbing heat from your hand), q > 0.

• Work done by the surroundings on the system is positive, w > 0.


Chapter 5 33

Endothermic and Exothermic Processes

• An endothermic process is one that absorbs heat from the surroundings.

• An endothermic reaction feels cold.

• An exothermic process is on that transfers heat to the surroundings.

• An exothermic reaction feels hot.


Chapter 5 34

Exercise 5.16

For the following process, calculate the change in the internal energy of the system and determine if the process is exothermic or endothermic. (a). A balloon is heated by adding 670 J of heat. It expands, doing 332 J of work on the atmosphere.


Chapter 5 35

State Functions

• A state function depends only on the initial and final states of a system.

• Example: The altitude difference between Denver and Chicago does not depend on whether you fly or drive, only on the elevation of the two cities above sea level.

• Similarly, the internal energy of 50 g of H2O(l) at 25C does not depend on whether we cool 50 g of H2O(l) at 100C or heat 50 g of H2O(l) at 0C.


Chapter 5 36

State Functions• A state function does not

depend on how the internal energy is used.

• Example: A battery in a flashlight can be discharged by producing heat and light. The same battery in a toy car is used to produce heat and work. The change in internal energy of the battery is the same in both cases.


Chapter 5 37

Enthalpy• The heat transferred between

the system and the surroundings during a chemical reaction carried out under constant pressure is called enthalpy, H.

• Again, we can only measure the change in enthalpy, H.

• Mathematically,

H = Hfinal - Hinitial = qp


Chapter 5 38

• Heat transferred from the surroundings to the system has a positive enthalpy (i.e., H > 0 for an endothermic reaction).

• Heat transferred from the system to the surroundings has a negative enthalpy (i.e., H < 0 for an exothermic reaction).

• Enthalpy is a state function.

Enthalpy


Chapter 5 39

Energy, Enthalpy, and P-V Work

Consider:

A cylinder of cross-sectional area A,

A piston exerting a pressure, P = F/A, on a gas inside the cylinder,


Chapter 5 40

• The volume of gas expanding through V while the piston moves a height

h = hf – hi ,

• The magnitude of work done = F x h = P x A x h = P x V.

• Since work is being done by the system on the surroundings,

w = - PV

• From the first law of thermodynamics,

E = q - PV

• If the reaction is carried out under constant volume,

V = 0 and E = qv.

• If the reaction is carried out under constant pressure,

E = qp - PV, or

qp = H = E + PV

and E =H - PV

Energy, Enthalpy, and P-V Work


Chapter 5 41

Enthalpies of Reaction• The enthalpy change that accompanies a reaction is called the

enthalpy of reaction.

• For a reaction, Hrxn = H(products) - H(reactants).

• Consider the thermochemical equation for the production of water:

2H2(g) + O2(g) 2H2O(g) H = -483.6 kJ

• The equation tells us that 483.6 kJ of energy is released to the surroundings when water is formed.

• Enthalpy is an extensive property.

• Therefore, the magnitude of enthalpy is directly proportional to the amount of reactant consumed.


Chapter 5 42

• Example: If 1 mol of CH4 is burned in oxygen to produce CO2 and water, 890 kJ of heat is released to the surroundings. If 2 mol of CH4 is burned, then 1780 kJ of heat is released.

• The sign of H depends on the direction of the reaction.

• The enthalpy changes for a reaction and its reverse reaction are equal in magnitude but opposite in sign.

Enthalpies of Reaction


Chapter 5 43

CH4(g) + 2O2(g) CO2(g) + 2H2O(l)

H = -890 kJ,

but CO2(g) + 2H2O(l) CH4(g) + 2O2(g)

H = +890 kJ.

Enthalpy change depends on state.

2H2O(g) 2H2O(l) H = -88 kJ

Enthalpies of Reaction


Chapter 5 44

Exercise 5.28

Consider the following reaction:CH3OH(g) CO(g) + 2H2(g) H = + 90.7 kJ(a). Is the reaction exothermic or endothermic?(b). Calculate the amount of heat transferred when 45.0 g of methanol decomposes by this reaction at constant pressure. (c). If the enthalpy change is 16.5 kJ, how many grams of hydrogen gas are produced?


Chapter 5 45

Heat Capacity and Specific Heat

• Heat capacity is the amount of energy required to raise the temperature of an object by 1°C.

• Molar heat capacity is the heat capacity of 1 mol of a substance.

• Specific heat, or specific heat capacity is the amount of energy required to raise the temperature of 1 g of a substance by 1°C.

Heat, q = (specific heat) x (grams of substance) x T.• Be careful of the sign of q.


Chapter 5 46

Calorimetry• Calorimetry is a measurement of heat flow.

• A calorimeter is an apparatus that measures heat flow.

Constant-Pressure Calorimetry

• Most common technique: use atmospheric pressure as the constant pressure.

• Recall that H = qp.

• Easiest method: use a coffee-cup calorimeter.

qsoln = (specific heat of solution) x (grams of solution) x T = -qrxn


Chapter 5 47

Bomb Calorimetry (Constant-Volume Calorimetry)

Reactions can be carried out under conditions of constant volume instead of constant pressure.

Constant-volume calorimetry is carried out in a bomb calorimeter.

Calorimetry


Chapter 5 48

• The most common type of reaction studied under these conditions is combustion.

• If we know the heat capacity of the calorimeter, Ccal, then the heat of reaction,

qrxn = -Ccal x T.

• Since the reaction is carried out under constant volume, q corresponds to E rather than H.

• For most reactions the difference between E and H is small.

Calorimetry


Chapter 5 49

Exercise 5.40

The specific heat of toluene, C7H8, is 1.13J/g-K. How many joules of heat are needed to raise the temperature of 40.0 g of toluene from 10.4oC to 28.0oC?


Chapter 5 50

Exercise 5.43

A 2.200 g sample of quinone (C6H4O2) is burned in a bomb calorimeter with a total heat capacity of 7.854kJ/oC. The temperature of the calorimeter increases from 23.44oC to 30.57oC.

What is the heat of combustion per gram of quinone?

What is the heat of combustion per mole of quinone?


Chapter 5 51

Hess’s Law• If a reaction is carried out

in a series of steps, H for the reaction is the sum of the H’s for each of the steps.

• The total change in enthalpy is independent of the number of steps.

• Total H is also independent of the nature of the path.


Chapter 5 52

Hess’s LawExample:

CH4(g) + 2O2(g) CO2(g) + 2H2O(g) H = -802 kJ

2H2O(g) 2H2O(l) H = -88 kJ______________________________________________________________

CH4(g) + 2O2(g) CO2(g) + 2H2O(l) H = -890 kJ

Therefore, for the reaction

CH4(g) + 2O2(g) CO2(g) + 2H2O(l), H= -890 kJ.

Note that H is sensitive to the states of the reactants and products.

Hess’s law allows us to calculate enthalpy data for reactions that are difficult to carry out directly: C(s) + O2(g) produces a mixture of CO(g) and CO2(g).


Chapter 5 53

Exercise 5.52

From the following heats of reaction

2H2(g) + O2(g) 2H2O(g) H= -483.6 kJ3O2(g) 2O3(g) H= +284.6 kJ

calculate the enthalpy for the reaction

3H2(g) + O3(g) 3H2O(g)


Chapter 5 54

Enthalpies of Formation

Consider the formation of CO2(g) and 2H2O(l) from CH4(g) and 2O2(g).

If the reaction proceeds in one step:

CH4(g) + 2O2(g) CO2(g) + 2H2O(l),

then H1 = -890 kJ.

However, if the reaction proceeds through a CO intermediate:

CH4(g) + 2O2(g) CO(g) + 2H2O(l) + ½O2(g) H2 = -607 kJ

CO(g) + 2H2O(l) + ½O2(g) CO2(g) + 2H2O(l)H3 = -283 kJ,

Then H for the overall reaction is:

H2 + H3 = -607 kJ - 283 kJ = -890 kJ = H1


Chapter 5 55

• If a compound is formed from its constituent elements, then the enthalpy change for the reaction is called the enthalpy of formation, Hf.

• Standard conditions (standard state) refer to the substance at in its pure form at 1 atm at the temperature of interest. We usually work with 25°C (298 K).

• Standard enthalpy, H°, is the enthalpy measured when everything is in its standard state.



Chapter 5 56


Chapter 5 57

• Standard enthalpy of formation of a compound, H°f, is the enthalpy change for the formation of 1 mol of compound with all substances in their standard states.

• If there is more than one state for a substance under standard conditions, the most stable state is used.

• Example: When dealing with carbon we use graphite because graphite is more stable than diamond or C60.

• The standard enthalpy of formation of the most stable form of an element is zero.



Chapter 5 58

Using Enthalpies of Formation to Calculate Enthalpies of

ReactionExample: Calculate H for C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(l)


Chapter 5 59

We start with the reactants, decompose them into elements, then rearrange the elements to form products. The overall enthalpy change is the sum of the enthalpy changes for each step (Hess’s Law).

Decomposing into elements (note O2 is already elemental, so we concern ourselves with C3H8); we have:

C3H8(g) 3C(s) + 4H2(g) H1 = -H°f[C3H8(g)]

Next we form CO2 and H2O from their elements:

3C(s) + 3O2(g) 3CO2(g) H2 = 3H°f[CO2(g)]

4H2(g) + 2O2(g) 4H2O(l) H3 = 4H°f[H2O(l)]

Using Enthalpies of Formation to Calculate Enthalpies of Reaction


Chapter 5 60

We look up the values and add:

H°rxn = -1(-103.85 kJ) + 3(-393.5 kJ) + 4(-285.8 kJ) = -2220 kJ

Notice that CO2 and H2O are products and propane is the reaction !

In general:

H°rxn = nH°f (products) - mH°f (reactants)

Where n and m are the stoichiometric coefficients

Using Enthalpies of Formation to Calculate Enthalpies of Reaction


Chapter 5 61

Exercise 5.60

Many cigarette lighters contain liquid butane, C4H10(l). Using enthalpies of formation, calculate the quantity of heat produced when 1.0 g of butane is completely combusted in air.


Chapter 5 62

Exercise 5.67FG05_025.JPG

One of the cleanest burning octanes is sketched above. The complete combustion of 1 mol of this compound to H2O(g) and CO2(g) leads to Ho = -5069 kJ.

(a). Write a balanced equation for the complete combustion of this compound.

(b). By using the information in this problem and the data in Table 5.3, calculate the standard enthalpy of formation of this compound at 298 K.


Chapter 5 63

Foods and Fuels• Fuel value is the energy released when 1 g of substance

is burned.

Foods• Fuel value is usually measured in Calories (1 nutritional

Calorie, 1 Cal = 1000 cal).• Most energy in our bodies comes from the oxidation of

carbohydrates and fats.• In the intestines carbohydrates are converted into

glucose, C6H12O6, or blood sugar.


Chapter 5 64

• In the cells glucose reacts with O2 in a series of steps which ultimately produce CO2, H2O, and energy.

C6H12O6(s) + 6O2(g) 6CO2(g) + 6H2O(l) H° = -2803 kJ

• Fats, for example, tristearin, react with O2 as follows:

2C57H110O6(s) + 163O2(g) 114CO2(g) + 110H2O(l) H° = -75,250 kJ

• Fats contain more energy than carbohydrates. Fats are not water soluble.

• Therefore, fats are good for energy storage.

Foods and Fuels


Chapter 5 65

Fuels• In the United States we use

about 1.0 x 106 kJ of energy per person per day.

• Most of this energy comes from petroleum and natural gas.

• The remainder of the energy comes from coal, nuclear, and hydroelectric sources.

• Coal, petroleum, and natural gas are fossil fuels. They are not renewable.


Chapter 5 66

Natural Gas• Natural gas consists largely of carbon and hydrogen.

Compounds such as CH4, C2H6, C3H8, and C4H10 are typical constituents.

• Petroleum is a liquid consisting of hundreds of compounds. Impurities include S, N, and O compounds.

• Coal contains high molecular weight compounds of C and H. In addition compounds containing S, O, and N are present as impurities.

• Common fuels have typical fuel values around 30 kJ/g.

• Hydrogen has great potential as a fuel with a fuel value of 142 kJ/g.


Chapter 5 67

Other Energy Sources• Nuclear energy: energy released in splitting or fusion of nuclei of

atoms.

• Fossil fuels and nuclear energy are nonrenewable sources of energy.

• Renewable energy sources include:– Solar energy

– Wind energy

– Geothermal energy

– Hydroelectric energy

– Biomass energy

• These are virtually inexhaustible and will become increasingly important as fossil fuels are depleted.

Copyright 2000, PRENTICE HALL Chapter 51 These are the remaining slides from class, but beginning...

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