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Chapter 6

Thermochemistry

Section 6.1

The Nature of Energy

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(6.1) The nature of energy
(6.2) Enthalpy and calorimetry
(6.3) Hess’s law
(6.4) Standard enthalpies of formation
(6.5) Present sources of energy
(6.6) New energy sources

Chapter 6

Table of Contents

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Energy

Capacity to do work or to produce heat
Law of conservation of energy: Energy can be converted from one form to another but can be neither created nor destroyed

Energy of the universe is constant

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The Nature of Energy

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Classification of Energy

Potential energy (PE): Energy due to position or composition

Can result from attractive and repulsive forces

Kinetic energy (KE): Energy due to the motion of an object

Depends on the mass of the object (m) and its velocity (v)

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Conversion of Energy

Consider the following image:

Because of its higher initial position, ball A has more PE than ball B

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The Nature of Energy

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Conversion of Energy (continued 1)

PE of A is changed to KE as the ball rolls down the hill
Part of this KE is then transferred to B, causing it to be raised to a higher final position

Thus, PE of B has been increased

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The Nature of Energy

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Conversion of Energy (continued 2)

Since the final position of B is lower than the original position of A, some of the energy is still unaccounted for
Part of the original energy stored as PE in A has been transferred through work to B, thereby increasing B’s PE

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The Nature of Energy

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Methods of Transferring Energy

Heat: Transfer of energy between two objects due to a temperature difference

Temperature is a property that reflects random motion of particles in a substance

Work: Force acting over a distance

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The Nature of Energy

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Pathway

Specific conditions that define the path by which energy is transferred
Energy change is independent of the pathway
Work and heat are dependent on the pathway

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The Nature of Energy

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Energy as a State Function

State function (state property): Property that does not depend in any way on the system’s past or future

Value depends on the characteristics of the present state
While transitioning from one state to another, the change in state property is independent of the pathway taken between the two states

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The Nature of Energy

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The Process of Heat

Energy is always transferred from the hot body to the cold body

Temperature is a measure of the average KE of particles in solids, liquids, and gases
Energy is transferred as the initially faster particles decrease in motion and the initially slower molecules increase in motion

Particles will have the same average KE and thus the same temperature

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The Nature of Energy

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Figure 6.2 - Particles at Different Temperatures in Adjoining Chambers

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The Nature of Energy

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Parts of the Universe

System: Part of the universe on which one wishes to focus his/her attention

Example - System can be defined as the reactants and products of a reaction

Surroundings: Include everything else in the universe

Example - Surroundings consist of anything else other than the reactants and products

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The Nature of Energy

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Types of Reactions

Exothermic: Reaction that results in the evolution of heat

Energy flows out of the system
Example - Combustion of methane

Endothermic: Reaction that results in the absorption of energy from the surroundings

Heat flows into a system
Example - Formation of nitric oxide

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The Nature of Energy

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Reaction Mechanism

Energy gained by the surroundings must be equal to the energy lost by the system
Exothermic reactions

PE stored in chemical bonds is converted to thermal energy via heat

Bonds in the products are stronger than those of the reactants

Net result - Quantity of energy Δ(PE) is transferred to the surroundings through heat

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The Nature of Energy

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Figure 6.3 - Energy Diagram for the Combustion of Methane, an Exothermic Process

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The Nature of Energy

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Reaction Mechanism (continued)

Endothermic reactions

Energy that flows into the system as heat is used to increase the PE of the system
Products have higher PE than reactants

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The Nature of Energy

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Figure 6.4 - Energy Diagram for the Formation of Nitric Oxide, an Endothermic Process

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The Nature of Energy

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Thermodynamics

Study of energy and its interconversions
First law of thermodynamics: Energy of the universe is constant

Known as the law of conservation of energy

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The Nature of Energy

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Internal Energy (E)

Sum of kinetic and potential energies of all particles in the system
E of a system can be changed by flow of work, heat, or both

ΔE - Change in the system’s internal energy
q - Heat
w - Work

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The Nature of Energy

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Parts of Thermodynamic Quantities

Number - Gives the magnitude of change
Sign - Indicates the direction of flow and reflects the system’s point of view

In an endothermic system, q = +x

Positive sign indicates that the system’s energy is increasing

In an exothermic system, q = –x

Negative sign indicates that the system’s energy is decreasing

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Parts of Thermodynamic Quantities (continued)

Conventions that apply to the flow of work

When a system does work on the surroundings, w is negative
When the surroundings do work on the system, w is positive

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The Nature of Energy

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Interactive Example 6.1 - Internal Energy

Calculate ΔE for a system undergoing an endothermic process in which 15.6 kJ of heat flows and where 1.4 kJ of work is done on the system

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The Nature of Energy

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Interactive Example 6.1 - Solution

We use the following equation:

q = + 15.6 kJ, since the process is endothermic
w = + 1.4 kJ, since work is done on the system

Thus, the system has gained 17.0 kJ of energy

Section 6.1

The Nature of Energy

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Work

Types of work associated with a chemical process

Work done by a gas through expansion
Work done to a gas through compression

Example

In an automobile engine, heat from the combustion of gasoline expands the gases in the cylinder to push back the piston

This motion is then translated to the motion of the car

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The Nature of Energy

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Deriving the Equation for Work

Consider a gas confined to a cylindrical container with a movable piston

F is the force acting on the piston of area A
Pressure is defined as force per unit area

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The Nature of Energy

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Deriving the Equation for Work (continued 1)

Work is defined as force applied over a distance

If the piston moves a distance of Δh, work done is

Since P = F/A or F = P × A,

Volume of the cylinder equals the area of the piston times the height of the cylinder

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The Nature of Energy

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Deriving the Equation for Work (continued 2)

Change in volume ΔV resulting from the piston moving a distance Δh is

Substituting ΔV = A × Δh into the expression for work gives the magnitude of the work required to expand a gas ΔV against a pressure P

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The Nature of Energy

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Deriving the Equation for Work (continued 3)

Since the system is doing work on the surroundings, the sign of work should be negative

For an expanding gas, ΔV is a positive quantity because the V is increasing

When a gas is compressed, ΔV is a negative quantity because V decreases, which makes w a positive quantity

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The Nature of Energy

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Critical Thinking

You are calculating ΔE in a chemistry problem

What if you confuse the system and the surroundings?

How would this affect the magnitude of the answer you calculate? The sign?

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The Nature of Energy

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Interactive Example 6.2 - PV Work

Calculate the work associated with the expansion of a gas from 46 L to 64 L at a constant external pressure of 15 atm

Section 6.1

The Nature of Energy

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Interactive Example 6.2 - Solution

For a gas at constant pressure,

In this case P = 15 atm and ΔV = 64 – 46 = 18 L

Note that since the gas expands, it does work on its surroundings

Reality check

Energy flows out of the gas, so w is a negative quantity

Section 6.1

The Nature of Energy

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Interactive Example 6.3 - Internal Energy, Heat, and Work

A balloon is being inflated to its full extent by heating the air inside it

In the final stages of this process, the volume of the balloon changes from 4.00×10⁶ L to 4.50×10⁶ L by the addition of 1.3×10⁸ J of energy as heat
Assuming that the balloon expands against a constant pressure of 1.0 atm, calculate ΔE for the process

To convert between L · atm and J, use 1 L · atm = 101.3 J

Section 6.1

The Nature of Energy

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Interactive Example 6.3 - Solution

Where are we going?

To calculate ΔE
What do we know?

V₁ = 4.00×10⁶ L
q = +1.3×10⁸ J
P = 1.0 atm
1 L · atm = 101.3 J
V₂ = 4.50×10⁶ L

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The Nature of Energy

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Interactive Example 6.3 - Solution (continued 1)

What do we need?

How do we get there?

What is the work done on the gas?

What is ΔV?

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The Nature of Energy

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Interactive Example 6.3 - Solution (continued 2)

What is the work?

The negative sign makes sense because the gas is expanding and doing work on the surroundings

To calculate ΔE, we must sum q and w

However, since q is given in units of J and w is given in units of L · atm, we must change the work to units of joules

Section 6.1

The Nature of Energy

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Interactive Example 6.3 - Solution (continued 3)

Then,

Reality check

Since more energy is added through heating than the gas expends doing work, there is a net increase in the internal energy of the gas in the balloon

Hence ΔE is positive

Section 6.1

The Nature of Energy

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Exercise

A balloon filled with 39.1 moles of helium has a volume of 876 L at 0.0°C and 1.00 atm pressure

The temperature of the balloon is increased to 38.0°C as it expands to a volume of 998 L, the pressure remaining constant
Calculate q, w, and ΔE for the helium in the balloon

The molar heat capacity for helium gas is 20.8 J/°C · mol

q = 30.9 kJ, w = –12.4 kJ, and ΔE = 18.5 kJ

Section 6.1

The Nature of Energy

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Join In (1)

Calculate the work for the expansion of an ideal gas from 2.0 L to 5.0 L against a pressure of 2.0 atm at constant temperature

6.0 L h ⋅ atm
–6.0 L h ⋅ atm
0
3.0 L h ⋅ atm
–3.0 L h ⋅ atm

Section 6.1

The Nature of Energy

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Join In (2)

Label the following process as exothermic or endothermic: Water freezes

Exothermic
Endothermic

Section 6.1

The Nature of Energy

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Join In (3)

Label the following process as exothermic or endothermic: Your hand gets cold when you touch ice

Exothermic
Endothermic

Section 6.1

The Nature of Energy

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Join In (4)

Label the following process as exothermic or endothermic: Ice increases in temperature when you touch it

Exothermic
Endothermic

Section 6.1

The Nature of Energy

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Join In (5)

Label the following process as exothermic or endothermic: Water vapor condenses on a cold pipe

Exothermic
Endothermic

Section 6.1

The Nature of Energy

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Join In (6)

Gas A₂ reacts with gas B₂ to form gas AB

The bond energy of AB is much greater than the bond energy of either A₂ or B₂
Is the reaction for the formation of AB exothermic or endothermic?

Exothermic
Endothermic

Section 6.1

The Nature of Energy

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Join In (7)

For a particular process, q = –10 kJ and w = 25 kJ

Which of the following statements is true?

Heat flows from the surroundings to the system
The surroundings do work on the system
ΔE = –35 kJ

Section 6.1

The Nature of Energy

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Join In (8)

Which of the following statements is correct?

Internal energy of a system increases when more work is done by the system than when heat was flowing into the system
Internal energy of a system decreases when work is done on the system and heat is flowing into the system
System does work on the surroundings when an ideal gas expands against a constant external pressure

Section 6.1

The Nature of Energy

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Enthalpy (H)

A state function that is defined as:

E - Internal energy of the system
P - Pressure of the system
V - Volume of the system

Section 6.2

Enthalpy and Calorimetry

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Change in Enthalpy (ΔH)

For a process carried out at constant pressure and where the only work allowed is that from a volume change:

q_P is the heat at constant pressure
At constant pressure, the change in enthalpy ΔH of the system is equal to the energy flow as heat

Section 6.2

Enthalpy and Calorimetry

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Enthalpy and PV Work

For a chemical reaction, the enthalpy change is given by the following equation:

When H_products > H_reactants, ΔH is positive

Heat is absorbed by the system, and the reaction is endothermic

When H_products < H_reactants, ΔH is negative

Overall decrease in enthalpy is achieved by the generation of heat, and the reaction is exothermic

Section 6.2

Enthalpy and Calorimetry

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Interactive Example 6.4 - Enthalpy

When 1 mole of methane (CH₄) is burned at constant pressure, 890 kJ of energy is released as heat

Calculate ΔH for a process in which a 5.8-g sample of methane is burned at constant pressure

Section 6.2

Enthalpy and Calorimetry

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Interactive Example 6.4 - Solution

Where are we going?

To calculate ΔH
What do we know?

q_P = ΔH = –890 kJ/mol CH₄
Mass = 5.8 g CH₄
Molar mass CH₄ = 16.04 g

Section 6.2

Enthalpy and Calorimetry

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Interactive Example 6.4 - Solution (continued 1)

How do we get there?

What are the moles of CH₄ burned?

What is ΔH?

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Enthalpy and Calorimetry

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Interactive Example 6.4 - Solution (continued 2)

Thus, when a 5.8-g sample of CH₄ is burned at constant pressure,

Reality check

In this case, a 5.8-g sample of CH₄ is burned

Since this amount is smaller than 1 mole, less than 890 kJ will be released as heat

Section 6.2

Enthalpy and Calorimetry

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Exercise

The overall reaction in a commercial heat pack can be represented as

How much heat is released when 4.00 moles of iron are reacted with excess O₂?

1650 kJ is released

Section 6.2

Enthalpy and Calorimetry

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Exercise (continued)

How much heat is released when 1.00 mole of Fe₂O₃ is produced?

How much heat is released when 1.00 g iron is reacted with excess O₂?

How much heat is released when 10.0 g Fe and 2.00 g O₂ are reacted?

826 kJ released

7.39 kJ released

34.4 kJ released

Section 6.2

Enthalpy and Calorimetry

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Join In (9)

Which of the following statements is true?

Change in enthalpy is a state function
In exothermic reactions, the reactants are lower in potential energy than the products
A chemist takes the surroundings point of view when determining the sign for work or heat
Heat of reaction and change in enthalpy can always be used interchangeably

Section 6.2

Enthalpy and Calorimetry

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Calorimetry

Science of measuring heat

Based on observations of temperature change when a body absorbs or discharges energy as heat

Calorimeter: Device used experimentally to determine the heat associated with a chemical reaction

Section 6.2

Enthalpy and Calorimetry

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Heat Capacity (C)

Specific heat capacity: Energy required to raise the temperature of one gram of a substance by one degree Celsius

Units - J/°C · g or J/K · g

Section 6.2

Enthalpy and Calorimetry

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Table 6.1 - The Specific Heat Capacities of Some Common Substances

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Enthalpy and Calorimetry

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Heat Capacity (C) (continued)

Molar heat capacity: Energy required to raise the temperature of one mole of a substance by one degree Celsius

Units - J/°C · mol or J/K · mol

Heat capacities of metals are different from that of water

It takes less energy to change the temperature of a gram of a metal by 1°C than for a gram of water

Section 6.2

Enthalpy and Calorimetry

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Coffee-Cup Calorimeter

Contains two nested Styrofoam cups with a cover through which a stirrer and thermometer can be inserted

Outer cup is used to provide extra insulation
Inner cup holds the solution in which the reaction occurs

Section 6.2

Enthalpy and Calorimetry

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Constant-Pressure Calorimetry

Atmospheric pressure remains constant during the process
Used to determine changes in enthalpy for reactions that occur in solution

ΔH = q_P

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Enthalpy and Calorimetry

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Constant-Pressure Calorimetry (continued 1)

If two reactants at the same temperature are mixed and the resulting solution gets warmer, this means the reaction taking place is exothermic

Endothermic reaction cools the solution

Section 6.2

Enthalpy and Calorimetry

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Constant-Pressure Calorimetry: Determining Change in Enthalpy for a Neutralization Reaction

When twice the amount of solution has been mixed, twice as much heat would be produced

Heat of a reaction is an extensive property

Depends directly on the amount of substance

Section 6.2

Enthalpy and Calorimetry

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Interactive Example 6.5 - Constant-Pressure Calorimetry

When 1.00 L of 1.00 M Ba(NO₃)₂ solution at 25.0°C is mixed with 1.00 L of 1.00 M Na₂SO₄ solution at 25.0°C in a calorimeter, the white solid BaSO₄ forms, and the temperature of the mixture increases to 28.1°C

Section 6.2

Enthalpy and Calorimetry

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Interactive Example 6.5 - Constant-Pressure Calorimetry (continued)

Assume that:

The calorimeter absorbs only a negligible quantity of heat
The specific heat capacity of the solution is 4.18 J/°C · g
The density of the final solution is 1.0 g/mL

Calculate the enthalpy change per mole of BaSO₄ formed

Section 6.2

Enthalpy and Calorimetry

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Interactive Example 6.5 - Solution

Where are we going?

To calculate ΔH per mole of BaSO₄ formed
What do we know?

1.00 L of 1.00 M Ba(NO₃)₂
1.00 L of 1.00 M Na₂SO₄
T_initial = 25.0°C and T_final = 28.1°C
Heat capacity of solution = 4.18 J/ °C · g
Density of final solution = 1.0 g/mL

Section 6.2

Enthalpy and Calorimetry

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Interactive Example 6.5 - Solution (continued 1)

What do we need?

Net ionic equation for the reaction

The ions present before any reaction occurs are Ba²⁺, NO₃^–, Na⁺, and SO₄^2–
The Na⁺ and NO₃^– ions are spectator ions, since NaNO₃ is very soluble in water and will not precipitate under these conditions
The net ionic equation for the reaction is:

Section 6.2

Enthalpy and Calorimetry

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Interactive Example 6.5 - Solution (continued 2)

How do we get there?

What is ΔH?

Since the temperature increases, formation of solid BaSO₄ must be exothermic; ΔH is negative

Section 6.2

Enthalpy and Calorimetry

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Interactive Example 6.5 - Solution (continued 3)

What is the mass of the final solution?

What is the temperature increase?

How much heat is evolved by the reaction?

Section 6.2

Enthalpy and Calorimetry

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Interactive Example 6.5 - Solution (continued 4)

Thus,

What is ΔH per mole of BaSO₄ formed?

Since 1.0 L of 1.0 M Ba(NO₃)₂ contains 1 mole of Ba²⁺ ions and 1.0 L of 1.0 M Na₂SO₄ contains 1.0 mole of SO₄^2– ions, 1.0 mole of solid BaSO₄ is formed in this experiment
Thus the enthalpy change per mole of BaSO₄ formed is

Section 6.2

Enthalpy and Calorimetry

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Constant-Volume Calorimetry

Used to study the energy changes in reactions under conditions of constant volume

No work is done because V must change for pressure–volume work to be performed

Device used - Bomb calorimeter

Weighed reactants are placed within a rigid steel container and ignited
Energy change is determined by the increase in temperature of the water and other calorimeter parts

Section 6.2

Enthalpy and Calorimetry

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Figure 6.7 - A Bomb Calorimeter

Section 6.2

Enthalpy and Calorimetry

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Constant-Volume Calorimetry (continued 1)

For a constant-volume process, ΔV = 0

Therefore, w = –PΔV = 0

Energy released by the reaction

= temperature increase × energy required to change the temperature by 1°C� = ΔT × heat capacity of calorimeter

Section 6.2

Enthalpy and Calorimetry

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Constant-Volume Calorimetry (continued 2)

Since no work is done in this case, ΔE is equal to the heat

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Enthalpy and Calorimetry

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Example 6.6 - Constant-Volume Calorimetry

It has been suggested that hydrogen gas obtained by the decomposition of water might be a substitute for natural gas (principally methane)

To compare the energies of combustion of these fuels, the following experiment was carried out using a bomb calorimeter with a heat capacity of 11.3 kJ/°C

Section 6.2

Enthalpy and Calorimetry

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Example 6.6 - Constant-Volume Calorimetry (continued)

When a 1.50-g sample of methane gas was burned with excess oxygen in the calorimeter, the temperature increased by 7.3°C
When a 1.15-g sample of hydrogen gas was burned with excess oxygen, the temperature increase was 14.3°C

Compare the energies of combustion (per gram) for hydrogen and methane

Section 6.2

Enthalpy and Calorimetry

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Example 6.6 - Solution

Where are we going?

To calculate ΔH of combustion per gram for H₂ and CH₄
What do we know?

1.50 g CH₄ ⇒ ΔT = 7.3°C
1.15 g H₂ ⇒ ΔT = 14.3°C
Heat capacity of calorimeter = 11.3 kJ/°C

What do we need?

ΔE = ΔT × heat capacity of calorimeter

Section 6.2

Enthalpy and Calorimetry

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Example 6.6 - Solution (continued 1)

How do we get there?

What is the energy released for each combustion?

For CH₄, we calculate the energy of combustion for methane using the heat capacity of the calorimeter (11.3 kJ/°C) and the observed temperature increase of 7.3°C

Section 6.2

Enthalpy and Calorimetry

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Example 6.6 - Solution (continued 2)

For H₂,

Section 6.2

Enthalpy and Calorimetry

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Example 6.6 - Solution (continued 3)

How do the energies of combustion compare?

The energy released in the combustion of 1 g hydrogen is approximately 2.5 times that for 1 g methane, indicating that hydrogen gas is a potentially useful fuel

Section 6.2

Enthalpy and Calorimetry

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Join In (10)

Two different metals of equal mass with different heat capacities are subjected to the same amount of heat

Which undergoes the smallest change in temperature?

The metal with the higher heat capacity shows the smallest change in temperature
The metal with the lower heat capacity shows the smallest change in temperature
Because they have equal mass, both metals undergo the same change in temperature. distilled mixture

Section 6.2

Enthalpy and Calorimetry

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Join In (11)

If 5.0 kJ of energy is added to a 15.5-g sample of water at 10.°C, the water is:

boiling
completely vaporized
frozen solid
still a liquid

Section 6.2

Enthalpy and Calorimetry

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Join In (12)

A 50.0-g sample of water at 80°C is added to a 50.0-g sample of water at 20°C

The final temperature of the water should be:

between 20°C and 50°C
50°C
between 50°C and 80°C

Section 6.2

Enthalpy and Calorimetry

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Join In (13)

A 50.0-g sample of water at 80°C is added to a 100.0-g sample of water at 20°C

The final temperature of the water should be:

between 20°C and 50°C
50°C
between 50°C and 80°C

Section 6.2

Enthalpy and Calorimetry

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Join In (14)

You have a 50.0-g sample of water at 20°C

To it, you add a 50.0-g sample of iron at 80°C
The final temperature of the water should be:

between 20°C and 50°C
50°C
between 50°C and 80°C

Section 6.2

Enthalpy and Calorimetry

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Hess’s Law

In going from a particular set of reactants to a particular set of products, the change in enthalpy is the same whether the reaction takes place in one step or in a series of steps

Section 6.3

Hess’s Law

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Characteristics of Enthalpy Changes

If a reaction is reversed, the sign of ΔH is also reversed
Magnitude of ΔH is directly proportional to the quantities of reactants and products in a reaction

If the coefficients in a balanced reaction are multiplied by an integer, the value of ΔH is multiplied by the same integer

Section 6.3

Hess’s Law

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Critical Thinking

What if Hess’s law were not true?

What are some possible repercussions this would have?

Section 6.3

Hess’s Law

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Problem-Solving Strategy - Hess’s Law

Work backward from the required reaction

Use the reactants and products to decide how to manipulate the other given reactions at your disposal

Reverse any reactions as needed to give the required reactants and products
Multiply reactions to give the correct numbers of reactants and products

Section 6.3

Hess’s Law

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Interactive Example 6.8 - Hess’s Law II

Diborane (B₂H₆) is a highly reactive boron hydride that was once considered as a possible rocket fuel for the U.S. space program

Calculate ΔH for the synthesis of diborane from its elements, according to the following equation:

Section 6.3

Hess’s Law

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Interactive Example 6.8 - Hess’s Law II (continued)

Use the following data:

Reaction ΔH

–1273 kJ

–2035 kJ

–286 kJ

44 kJ

Section 6.3

Hess’s Law

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Interactive Example 6.8 - Solution

To obtain ΔH for the required reaction, we must somehow combine equations (a), (b), (c), and (d) to produce that reaction and add the corresponding ΔH values

This can best be done by focusing on the reactants and products of the required reaction

The reactants are B(s) and H₂(g), and the product is B₂H₆(g)

Section 6.3

Hess’s Law

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Interactive Example 6.8 - Solution (continued 1)

How can we obtain the correct equation?

Reaction (a) has B(s) as a reactant, as needed in the required equation

Thus reaction (a) will be used as it is

Reaction (b) has B₂H₆(g) as a reactant, but this substance is needed as a product

Thus reaction (b) must be reversed, and the sign of ΔH must be changed accordingly

Section 6.3

Hess’s Law

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Interactive Example 6.8 - Solution (continued 2)

Up to this point we have:

Deleting the species that occur on both sides gives

ΔH = 762 kJ

Section 6.3

Hess’s Law

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Interactive Example 6.8 - Solution (continued 3)

We are closer to the required reaction, but we still need to remove H₂O(g) and O₂(g) and introduce H₂(g) as a reactant

We can do this using reactions (c) and (d)

Multiply reaction (c) and its ΔH value by 3 and add the result to the preceding equation

Section 6.3

Hess’s Law

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Interactive Example 6.8 - Solution (continued 4)

ΔH = –96 kJ

Section 6.3

Hess’s Law

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Interactive Example 6.8 - Solution (continued 5)

We can cancel the 3/2 O₂(g) on both sides, but we cannot cancel the H₂O because it is gaseous on one side and liquid on the other

This can be solved by adding reaction (d), multiplied by 3:

ΔH = +36 kJ

Section 6.3

Hess’s Law

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Interactive Example 6.8 - Solution (continued 6)

This gives the reaction required by the problem

Thus ΔH for the synthesis of 1 mole of diborane from the elements is +36 kJ

Section 6.3

Hess’s Law

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Join In (15)

Given the following equation, which of the following statement(s) is (are) true?

The reaction is exothermic
When 0.500 mol sulfur is reacted, 148 kJ of energy is released
When 32.0 g of sulfur is burned, 2.96 × 10⁵ J of energy is released

Section 6.3

Hess’s Law

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Join In (15) (continued)

All are true
None is true
I and II are true
I and III are true
Only II is true

Section 6.3

Hess’s Law

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Standard Enthalpy of Formation (ΔH_f°)

Change in enthalpy that accompanies the formation of 1 mole of a compound from its elements with all substances in their standard states

Degree symbol on a thermodynamic function indicates that the corresponding process has been carried out under standard conditions
Standard state: Precisely defined reference state

Section 6.4

Standard Enthalpies of Formation

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Conventional Definitions of Standard States for a Compound

Standard state for a gaseous substance is a pressure of exactly 1 atm
For a pure substance in a condensed state (liquid or solid), the standard state is the pure liquid or solid
For a substance in solution, the standard state is a concentration of exactly 1 M

Section 6.4

Standard Enthalpies of Formation

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Conventional Definitions of Standard States for an Element

Standard state of an element is the form in which the element exists under conditions of 1 atm and 25°C

Examples

Standard state for oxygen is O₂(g) at a pressure of 1 atm
Standard state for sodium is Na(s)
Standard state for mercury is Hg(l)

Section 6.4

Standard Enthalpies of Formation

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Table 6.2 - Standard Enthalpies of Formation for Several Compounds at 25°C

Section 6.4

Standard Enthalpies of Formation

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Problem-Solving Strategy - Enthalpy Calculations

When a reaction is reversed, the magnitude of ΔH remains the same, but its sign changes
When the balanced equation for a reaction is multiplied by an integer, the value of ΔH for that reaction must be multiplied by the same integer

Section 6.4

Standard Enthalpies of Formation

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Problem-Solving Strategy - Enthalpy Calculations (continued)

Change in enthalpy for a given reaction can be calculated from the enthalpies of formation of the reactants and products:

Elements in their standard states are not included in the ΔH_reaction calculations

ΔH_f° for an element in its standard state is zero

Section 6.4

Standard Enthalpies of Formation

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Interactive Example 6.10 - Enthalpies from Standard Enthalpies of Formation II

Using enthalpies of formation, calculate the standard change in enthalpy for the thermite reaction:

This reaction occurs when a mixture of powdered aluminum and iron(III) oxide is ignited with a magnesium fuse

Section 6.4

Standard Enthalpies of Formation

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Interactive Example 6.10 - Solution

Where are we going?

To calculate ΔH for the reaction
What do we know?

ΔH_f° for Fe₂O₃(s) = – 826 kJ/mol
ΔH_f° for Al₂O₃(s) = – 1676 kJ/mol
ΔH_f° for Al(s) = ΔH_f°for Fe(s) = 0

What do we need?

Section 6.4

Standard Enthalpies of Formation

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Interactive Example 6.10 - Solution (continued)

How do we get there?

This reaction is so highly exothermic that the iron produced is initially molten

This process is often used as a lecture demonstration and also has been used in welding massive steel objects such as ships’ propeller

Section 6.4

Standard Enthalpies of Formation

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Critical Thinking

For ΔH_reaction calculations, we define ΔH_f° for an element in its standard state as zero

What if we define ΔH_f° for an element in its standard state as 10 kJ/mol?

How would this affect your determination of ΔH_reaction?
Provide support for your answer with a sample calculation

Section 6.4

Standard Enthalpies of Formation

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Example 6.11 - Enthalpies from Standard Enthalpies of Formation III

Until recently, methanol (CH₃OH) was used as a fuel in high-performance engines in race cars

Using data from the table containing standard enthalpies of formation for several compounds at 25°C, compare the standard enthalpy of combustion per gram of methanol with that per gram of gasoline
Gasoline is actually a mixture of compounds, but assume for this problem that gasoline is pure liquid octane (C₈H₁₈)

Section 6.4

Standard Enthalpies of Formation

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Example 6.11 - Solution

Where are we going?

To compare ΔH of combustion for methanol and octane
What do we know?

Standard enthalpies of formation from the table containing standard enthalpies of formation for several compounds at 25°C

Section 6.4

Standard Enthalpies of Formation

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Example 6.11 - Solution (continued 1)

How do we get there? (for methanol)

What is the combustion reaction?

What is the ΔH°_reaction?

Use the standard enthalpies of formation from the table containing standard enthalpies of formation for several compounds at 25°C and the following equation:

Section 6.4

Standard Enthalpies of Formation

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Example 6.11 - Solution (continued 2)

Section 6.4

Standard Enthalpies of Formation

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Example 6.11 - Solution (continued 3)

What is the enthalpy of combustion per gram?

Thus 1.45×10³ kJ of heat is evolved when 2 moles of methanol burn
The molar mass of methanol is 32.04 g/mol, which means that 1.45×10³ kJ of energy is produced when 64.08 g methanol burns
The enthalpy of combustion per gram of methanol is:

Section 6.4

Standard Enthalpies of Formation

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Example 6.11 - Solution (continued 4)

How do we get there? (for octane)

What is the combustion reaction?

What is the ΔH°_reaction?

Use the standard enthalpies of formation from the table containing standard enthalpies of formation for several compounds at 25°C and the following equation:

Section 6.4

Standard Enthalpies of Formation

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Example 6.11 - Solution (continued 5)

Section 6.4

Standard Enthalpies of Formation

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Example 6.11 - Solution (continued 6)

What is the enthalpy of combustion per gram?

This is the amount of heat evolved when 2 moles of octane burn
Since the molar mass of octane is 114.22 g/mol, the enthalpy of combustion per gram of octane is

Section 6.4

Standard Enthalpies of Formation

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Example 6.11 - Solution (continued 7)

The enthalpy of combustion per gram of octane is approximately twice that per gram of methanol

On this basis, gasoline appears to be superior to methanol for use in a racing car, where weight considerations are usually very important
Methanol was used in racing cars because it burns much more smoothly than gasoline in high-performance engines, and this advantage more than compensates for its weight disadvantage

Section 6.4

Standard Enthalpies of Formation

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Join In (16)

Given the following equation, which of the following statement(s) is (are) true?

The reaction is endothermic
The reverse reaction will have a ΔH_f = +75 kJ/mol
The enthalpy for H₂ is zero

Section 6.4

Standard Enthalpies of Formation

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Join In (16) (continued)

All are true
None is true
I and II are true
II and III are true
Only II is true

Section 6.4

Standard Enthalpies of Formation

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Figure 6.12 - Energy Sources Used in the United States

Section 6.5

Present Sources of Energy

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Petroleum and Natural Gas

Remains of marine organisms that lived approximately 500 million years ago

Petroleum: Thick, dark liquid that is mainly composed of hydrocarbons
Natural gas: Consists mostly of methane but it also contains significant amounts of ethane, propane, and butane

Section 6.5

Present Sources of Energy

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Table 6.3 - Names and Formulas for Some Common Hydrocarbons

Section 6.5

Present Sources of Energy

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Natural Gas Reserves

Exist in shale deposits

Shale is impermeable, and the gas does not flow out on its own
Accessed via hydraulic fracturing or fracking

Involves the injection of a slurry of water, sand, and chemical additives under pressure through a well bore drilled into the shale
Allows the gas to flow out into wells
Poses environmental concerns

Section 6.5

Present Sources of Energy

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Petroleum

Mainly consists of hydrocarbons that have chains containing 5 to more than 25 carbons
For efficient use, petroleum must be separated into fractions by boiling

Lighter molecules will boil away, and the heavier molecules will be left behind

Section 6.5

Present Sources of Energy

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Table 6.4 - Uses of the Various Petroleum Fractions

Section 6.5

Present Sources of Energy

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The Petroleum Era

Began during the Industrial Revolution

Demand for lamp oil outstripped the traditional sources
Edwin Drake drilled the first oil well in 1859 at Titusville, Pennsylvania

Used the petroleum to produce kerosene, which served as an excellent lamp oil

Section 6.5

Present Sources of Energy

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The Petroleum Era - Gasoline Age

Began with:

Development of electric light, which decreased the need for kerosene
Advent of the horseless carriage with its gasoline-powered engine

William Burton

Invented pyrolytic (high-temperature) cracking to increase the yield of gasoline obtained from each barrel of petroleum

Section 6.5

Present Sources of Energy

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The Petroleum Era - Gasoline Age (continued)

Tetraethyl lead (C₂H₅)₄Pb

Effective antiknock agent that was added to gasoline to promote smoother burning
Use of lead has been phased out owing to environmental concerns

Section 6.5

Present Sources of Energy

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Coal

Formed from the remains of plants that were buried and subjected to high pressure and heat over a long period of time

Plants contain cellulose (CH₂O), a complex molecule with high molar mass (500,000 g/mol)

After the plants die, chemical changes decrease the oxygen and hydrogen content of cellulose molecules

Section 6.5

Present Sources of Energy

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Coal (continued)

Stages of coal maturation

Lignite, subbituminous, bituminous, and anthracite

Each stage has a higher carbon-to-oxygen and carbon-to-hydrogen ratio
Elemental compositions vary depending on the age and location

Energy available from the combustion of a given mass of coal increases as the carbon content increases

Section 6.5

Present Sources of Energy

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Table 6.5 - Elemental Composition of Various Types of Coal

Section 6.5

Present Sources of Energy

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Coal as a Fuel - Disadvantages

Expensive
Dangerous to mine underground
Burning high-sulfur coal yields air pollutants, which, in turn, can lead to acid rain
Carbon dioxide is produced when coal is burned

Significantly affects the earth’s climate

Section 6.5

Present Sources of Energy

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Energy Received by the Earth

Earth receives a tremendous quantity of radiant energy from the sun, which is reflected back into space by the earth’s atmosphere

Remaining energy passes through the atmosphere to the earth’s surface, and most of it is absorbed by soil, rocks, and water

Increases the earth’s surface temperature

Energy is in turn radiated from the heated surface mainly as infrared radiation, often called heat radiation

Section 6.5

Present Sources of Energy

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The Greenhouse Effect

Earth’s atmosphere is transparent to visible light but absorbs infrared radiation

Molecules such as H₂O and CO₂ strongly absorb this radiation and radiate it back toward the earth
Net amount of thermal energy is retained by the earth’s atmosphere, causing the earth to become warmer

Section 6.5

Present Sources of Energy

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Figure 6.13 - Schematic of the Greenhouse Effect

Section 6.5

Present Sources of Energy

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The Earth’s Surface

Atmosphere’s water content is controlled by the water cycle

Increase in CO₂ levels have contributed to warming of the earth’s atmosphere
CO₂, methane, and other greenhouse gases block infrared radiation from leaving the earth’s atmosphere

Causes dramatic climate changes and affects growth of food crops

Section 6.5

Present Sources of Energy

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Figure 6.14 - Atmospheric CO₂ Concentration and the Average Global Temperature over the Last 250 Years

Note the significant increase in CO₂ concentration in the last 50 years

Section 6.5

Present Sources of Energy

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Potential Energy Sources

Sun

Nuclear processes

Biomass

Synthetic fuels

Section 6.6

New Energy Sources

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Figure 6.15 - Coal Gasification

Section 6.6

New Energy Sources

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Reactions in Coal Gasification

Exothermic reactions

Endothermic reaction

Section 6.6

New Energy Sources

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Coal Gasification

Desired product is a mixture of syngas and methane (CH₄)

Syngas (synthetic gas): Mixture of CO and H

Useful fuel since all the components of this product can react with oxygen to release heat in a combustion reaction

Section 6.6

New Energy Sources

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Uses of Syngas

Can be used directly as fuel
Essential raw material in the production of other fuels

Syngas can be directly converted to methanol, which is used in the production of synthetic fibers and plastics and is also used as a fuel

Syngas can also be converted directly to gasoline

Section 6.6

New Energy Sources

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Other Uses of Coal

Formation of coal slurries

Slurry - Suspension of fine particles in a liquid
Coal slurries contain a mixture of pulverized coal and water, and the slurry can be handled, stored, and burned in a manner similar to that used for residual oil

Section 6.6

New Energy Sources

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Hydrogen as a Fuel

Combustion reaction that demonstrates hydrogen’s potential as a fuel:

Has a real advantage over fossil fuels

Only product of hydrogen combustion is water

Section 6.6

New Energy Sources

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Hydrogen as a Fuel - Problems

Cost of production

Main source of hydrogen gas is from the treatment of natural gas with steam, which is highly endothermic

Section 6.6

New Energy Sources

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Alternate Sources of Hydrogen

Electrolysis of water

Involves passing electric current through water
Not economically feasible owing to the current cost of electricity

Corn

Starch from corn is fermented to produce alcohol, which is then decomposed in a reactor at 140°C with a rhodium and cerium oxide catalyst to give hydrogen

Section 6.6

New Energy Sources

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Alternate Sources of Hydrogen (continued)

Thermal decomposition

Involves heating water to several thousand degrees, where it decomposes into hydrogen and oxygen
Expensive to attain high temperatures required for this process

Thermochemical decomposition

Chemical reactions, as well as heat, are used to split water into its components

Section 6.6

New Energy Sources

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Hydrogen as a Fuel - Problems (continued)

Storage and transportation

H₂ molecules decompose to atoms on metal surfaces

Atoms may migrate into the metal and cause structural changes that make it brittle

Relatively small amount of energy is available per unit volume of hydrogen gas
Accidental leakage of hydrogen gas into the atmosphere could pose a threat

Section 6.6

New Energy Sources

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Example 6.12 - Enthalpies of Combustion

Compare the energy available from the combustion of a given volume of methane and the same volume of hydrogen at the same temperature and pressure

Section 6.6

New Energy Sources

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Example 6.12 - Solution

We know that the heat released for the combustion of methane is 55 kJ/g CH₄ and of hydrogen is 141 kJ/g H₂
We also know from our study of gases that 1 mole of H₂(g) has the same volume as 1 mole of CH₄(g) at the same temperature and pressure (assuming ideal behavior)

Section 6.6

New Energy Sources

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Example 6.12 - Solution (continued 1)

For molar volumes of both gases under the same conditions of temperature and pressure,

Section 6.6

New Energy Sources

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Example 6.12 - Solution (continued 2)

Thus about three times the volume of hydrogen gas is needed to furnish the same energy as a given volume of methane

Section 6.6

New Energy Sources

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Interactive Example 6.13 - Comparing Enthalpies of Combustion

Assuming that the combustion of hydrogen gas provides three times as much energy per gram as gasoline, calculate the volume of liquid H₂ (density = 0.0710 g/mL) required to furnish the energy contained in 80.0 L (about 20 gal) of gasoline (density = 0.740 g/mL)

Calculate also the volume that this hydrogen would occupy as a gas at 1.00 atm and 25°C

Section 6.6

New Energy Sources

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Interactive Example 6.13 - Solution

Where are we going?

To calculate the volume of H₂(l) required and its volume as a gas at the given conditions
What do we know?

Density for H₂(l) = 0.0710 g/mL
80.0 L gasoline
Density for gasoline = 0.740 g/mL
H₂(g) ⇒ P = 1.00 atm, T = 25°C = 298 K

Section 6.6

New Energy Sources

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Interactive Example 6.13 - Solution (continued 1)

How do we get there?

What is the mass of gasoline?

How much H₂(l) is needed?

Since H₂ furnishes three times as much energy per gram as gasoline, only a third as much liquid hydrogen is needed to furnish the same energy

Section 6.6

New Energy Sources

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Interactive Example 6.13 - Solution (continued 2)

Since density = mass/volume, then volume = mass/density, and the volume of H₂(l) needed is

Thus, 277 L of liquid H₂ is needed to furnish the same energy of combustion as 80.0 L of gasoline

Section 6.6

New Energy Sources

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Interactive Example 6.13 - Solution (continued 3)

What is the volume of the H₂(g)?

To calculate the volume that this hydrogen would occupy as a gas at 1.00 atm and 25°C, we use the ideal gas law:

In this case:

P = 1.00 atm
T = 273 + 25°C = 298 K
R = 0.08206 L · atm/K · mol

Section 6.6

New Energy Sources

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Interactive Example 6.13 - Solution (continued 4)

What are the moles of H₂(g)?

Thus,

At 1 atm and 25°C, the hydrogen gas needed to replace 20 gal of gasoline occupies a volume of 239,000 L

Section 6.6

New Energy Sources

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Storing Hydrogen as a Fuel

Use of metals that absorb hydrogen to form solid metal hydrides

Hydrogen gas would be pumped into a tank containing the solid metal in powdered form

In the tank, hydrogen would be absorbed to form the hydride, whose volume would be little more than that of the metal alone

Section 6.6

New Energy Sources

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Storing Hydrogen as a Fuel (continued)

Hydrogen would then be available for combustion in the engine by release of H₂(g) from the hydride as needed

Section 6.6

New Energy Sources

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Other Energy Alternatives

Oil shale deposits

Consist of a complex carbon-based material called kerogen
Disadvantage - Trapped fuel is difficult to extract

Ethanol (C₂H₅OH)

Produced via fermentation
Disadvantage - Difficult to develop an efficient process for conversion of cellulose to ethanol

Section 6.6

New Energy Sources

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Other Energy Alternatives (continued)

Methanol (CH₃OH)

Used successfully for many years in race cars

Seed oil

Oil seeds are processed to produce an oil that is mainly composed of carbon and hydrogen

This oil would react with oxygen to produce carbon dioxide, water, and heat

Advantage - Renewability

Section 6.6

New Energy Sources