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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

Total energy content of the universe is constant

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Section 6.1

The Nature of Energy

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Classification of Energy

Potential energy: Energy due to position or composition

Can result from attractive and repulsive forces

Kinetic energy: Energy due to the motion of an object

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

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Section 6.1

The Nature of Energy

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Conversion of Energy

Consider the following image:

Due to its higher initial position, ball A has more potential energy than ball B

Section 6.1

The Nature of Energy

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Conversion of Energy (Continued)

After A has rolled down the hill, the potential energy lost by A has been converted to:

Random motions of the components of the hill (frictional heating)
The increase in the potential energy of B

Section 6.1

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 reflects random motion of particles in a substance

Work: Force acting over a distance

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Section 6.1

The Nature of Energy

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Pathway

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

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Section 6.1

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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Section 6.1

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 - The reactants and products of a reaction

Surroundings: Include everything else in the universe

Example - Things other than the reactants and products

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Section 6.1

The Nature of Energy

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Types of Reactions

Exothermic reaction: Results in the evolution of heat

Energy flows out of the system
Example - Combustion of methane

Endothermic reaction: Results in the absorption of energy from the surroundings

Heat flows into a system
Example - Formation of nitric oxide

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Section 6.1

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

Endothermic reactions result from a lowered potential energy of the reaction system
In exothermic reactions, potential energy stored in chemical bonds is converted to thermal energy via heat

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The Nature of Energy

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Δ(PE)

Change in potential energy stored in the bonds of products as compared with the bonds of reactants

For an exothermic process, more energy is released while forming new bonds than is consumed while breaking the bonds in the reactants
In an endothermic reaction, energy that flows into the system as heat is used to increase the potential energy of the system

Section 6.1

The Nature of Energy

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Figure 6.2 - Energy Diagram for the Combustion of Methane

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The Nature of Energy

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Figure 6.3 - Energy Diagram for the Formation of Nitric Oxide

Section 6.1

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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Section 6.1

The Nature of Energy

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Internal Energy (E)

Sum of kinetic and potential energies of all particles in 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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Section 6.1

The Nature of Energy

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Parts of Thermodynamic Quantities

Number - Gives the magnitude of change
Sign - Indicates the direction of flow

Reflects the system’s point of view

In an endothermic system, q is equal to +x
In an exothermic system, q is equal to –x
When a system does work on surroundings, w is negative
When the surroundings do work on the system, w is positive

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Section 6.1

The Nature of Energy

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Endothermic and Exothermic Systems and Energy

Section 6.1

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

Section 6.1

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 - Motion of a car

In an automobile engine, heat from the combustion of gasoline expands the gases in the cylinder to push back the piston, and this results in the motion of the car

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Section 6.1

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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Section 6.1

The Nature of Energy

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Deriving the Equation for Work (Continued 1)

Consider that the piston moves a distance of Δh

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

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

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Section 6.1

The Nature of Energy

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Deriving the Equation for Work (Continued 2)

Substitute the expression derived for ΔV into the expression for work

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

ΔV is a positive quantity since volume is increasing
Therefore,

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Section 6.1

The Nature of Energy

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Deriving the Equation for Work (Continued 3)

For a gas expanding against an external pressure P, w is a negative quantity as required

Work flows out of the system

When a gas is compressed, ΔV is a negative quantity (the volume decreases)

This makes w a positive quantity (work flows into the system)

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Section 6.1

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?

Section 6.1

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

Section 6.1

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?

Section 6.1

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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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

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Section 6.2

Enthalpy and Calorimetry

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Enthalpy and PV Work

At constant pressure, ΔH = q_P

q_P - Heat at constant pressure

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
Reaction is endothermic

Section 6.2

Enthalpy and Calorimetry

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Enthalpy and PV Work (Continued)

When H_products < H_reactants, ΔH is negative

Overall decreased in enthalpy is achieved by the generation of heat
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?

Section 6.2

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?

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1650 kJ heat 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?

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826 kJ of heat is released

7.39 kJ of heat is released

34.4 kJ of heat is released

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 in the form of heat

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

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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

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Section 6.2

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

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

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Section 6.2

Enthalpy and Calorimetry

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Figure 6.5 - A Coffee-Cup Calorimeter

Section 6.2

Enthalpy and Calorimetry

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Table 6.1 - The Specific Heat Capacities of Some Common Substances

Section 6.2

Enthalpy and Calorimetry

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Constant-Pressure Calorimetry

Example - Coffee-cup calorimeter

Atmospheric pressure remains constant during the process

Used to determine enthalpy changes in reactions that occur in a solution

In a solution, ΔH = q_P

Section 6.2

Enthalpy and Calorimetry

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Constant-Pressure Calorimetry (Continued 1)

When two reactants at the same temperature are mixed:

An exothermic reaction warms the solution
An endothermic reaction cools the solution

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Section 6.2

Enthalpy and Calorimetry

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Constant-Pressure Calorimetry (Continued 2)

Calculation of heat for a neutralization reaction

Energy (as heat) released by the reaction

= Energy (as heat) absorbed by the solution

= Specific heat capacity×mass of solution ×change in temperature

=

Heat of a reaction is an extensive property

Depends entirely on the amount of substance

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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
Heat evolved by the reaction

= heat absorbed by the solution

= specific heat capacity×mass of solution×increase in temperature

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 in conditions when experiments are to be performed under constant volume

No work is done since V must change for PV work to be performed

Bomb calorimeter

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

Section 6.2

Enthalpy and Calorimetry

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Figure 6.6 - A Bomb Calorimeter

Section 6.2

Enthalpy and Calorimetry

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Constant-Volume Calorimetry (Continued)

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 the calorimeter

Section 6.2

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

Section 6.2

Enthalpy and Calorimetry

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Example 6.6 - Solution (Continued 1)

What do we need?

ΔE = ΔT×heat capacity of calorimeter

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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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

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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

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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

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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

Reaction (a) will be used as it is

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

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

Conclusion

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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Standard Enthalpy of Formation (ΔH_f°)

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

Standard state: Precisely defined reference state
A degree symbol on a thermodynamic function indicates that the corresponding process that is carried out under standard conditions

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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

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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 that element exists under conditions of 1 atm and 25°C

Examples

Standard state for oxygen is O₂(g)
Standard state of sodium is Na(s)
Standard state of mercury is Hg(l)

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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

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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

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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

Section 6.4

Standard Enthalpies of Formation

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Interactive Example 6.10 - Solution (Continued 1)

What do we need?

We use the following equation:

How do we get there?

Section 6.4

Standard Enthalpies of Formation

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Interactive Example 6.10 - Solution (Continued 2)

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

Used as a lecture demonstration
Used in welding massive steel objects such as ships’ propellers

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 Table 6.2

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 ?

Use the standard enthalpies of formation from Table 6.2 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?

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 ?

Use the standard enthalpies of formation from Table 6.2 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

Gasoline appears to be superior to methanol for use in a racing car, where weight considerations are usually very important
Methanol is used in racing cars since it burns much more smoothly than gasoline in high-performance engines

This advantage compensates for its weight disadvantage

Section 6.4

Standard Enthalpies of Formation

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Figure 6.11 - Energy Sources Used in the United States

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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

Mostly consists of methane but 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

Hydraulic fracturing or fracking is used to access shale deposits

Involves the injection of a slurry of water, sand, and chemical additives under pressure through a well bore
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

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

Section 6.5

Present Sources of Energy

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The Petroleum Era (Continued)

Gasoline age - Marked by the decreased need for kerosene due to the invention of electric lights and the advent of cars

William Burton invented pyrolytic cracking

Pyrolytic (high-temperature) cracking - Process used to separate gasoline from petroleum

Tetraethyl lead was added to gasoline to promote smoother burning

Use of lead has been phased out due to environmental concerns

Section 6.5

Present Sources of Energy

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Coal

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

Plants contain cellulose (empirical formula 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

Composition varies depending on the age and location
Available energy from combustion increases as the carbon content of a given mass of coal 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 leads to air pollution

Contributes 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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The Greenhouse Effect

The earth’s atmosphere is transparent to visible light from the sun

Does not permit all infrared radiation to pass back into space

Molecules such as H₂O and CO₂ strongly absorb this radiation and radiate it back to the earth
Earth becomes warmer due to the amount of thermal energy retained by the atmosphere

Section 6.5

Present Sources of Energy

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Figure 6.12 - Schematic of the Greenhouse Effect

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Section 6.5

Present Sources of Energy

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The Earth’s Surface

Temperature is controlled by water content of the atmosphere and greenhouse gases

The atmosphere’s water content is controlled by the water cycle

Extensive usage of fossil fuels has increased carbon dioxide concentration in the atmosphere

Increases the earth’s average temperature

Causes dramatic climate changes and affects agriculture

Section 6.5

Present Sources of Energy

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Potential Energy Sources

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Sun

Nuclear processes

Biomass

Synthetic fuels

Section 6.6

New Energy Sources

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Figure 6.14 - 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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Syngas (Synthesis Gas)

Mixture of carbon monoxide and hydrogen
Uses

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

Syngas can be directly converted to methanol

Section 6.6

New Energy Sources

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Formation of Coal Slurry

Suspension of fine particles in a liquid
Contains a mixture of pulverized coal and water
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:

Since the reaction’s product is water, hydrogen has an advantage over fossil fuels

Section 6.6

New Energy Sources

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Hydrogen as a Fuel - Major Problem

Cost of production

Main source of hydrogen gas is from the treatment of natural gas with steam

The reaction is highly endothermic

Most methods for producing hydrogen require tremendous amounts of energy, which is not economically feasible

Section 6.6

New Energy Sources

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Alternate Sources of Hydrogen

Electrolysis of water

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

Corn

Starch is fermented to produce alcohol that is 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 and heat are used to split water into its components

Section 6.6

New Energy Sources

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Hydrogen as a Fuel - Problems

Storage and transportation

Hydrogen decomposes to atoms on metal surfaces, which may pose a potential leakage threat

Atoms may migrate into the metal and cause structural changes that make the metal brittle
Could cause harm to the atmosphere by raising atmospheric hydrogen levels

Relatively small amount of energy is available per unit volume of hydrogen

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 for and 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)

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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Other Energy Alternatives

Oil shale deposits

Consist of a complex carbon-based material called kerogen
Difficult to extract

Ethanol and methanol

Produced through fermentation
Alternative fuel for car engines

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Section 6.6

New Energy Sources

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Other Energy Alternatives (Continued)

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

138

Section 6.6

New Energy Sources