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CHEMISTRY EXAM

Chapter 6

Thermochemistry

Section 6.1 The Nature of 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.
The total energy content of the universe is constant.
Energy

Section 6.1 The Nature of Energy

Potential energy – energy due to position or composition.
Kinetic energy – energy due to motion of the object and depends on the mass of the object and its velocity.
Energy

Section 6.1 The Nature of Energy

In the initial position, ball A has a higher potential energy than ball B.
Initial Position

Section 6.1 The Nature of Energy

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) and to the increase in the potential energy of B.
Final Position

Section 6.1 The Nature of Energy

Heat involves the transfer of energy between two objects due to a temperature difference.
Work – force acting over a distance.
Energy is a state function; work and heat are not
State Function – property that does not depend in any way on the system’s past or future (only depends on present state).
Energy

Section 6.1 The Nature of Energy

System – part of the universe on which we wish to focus attention.
Surroundings – include everything else in the universe.
Chemical Energy

Section 6.1 The Nature of Energy

Endothermic Reaction:
Heat flow is into a system.
Absorb energy from the surroundings.
Exothermic Reaction:
Energy flows out of the system.
Energy gained by the surroundings must be equal to the energy lost by the system.
Chemical Energy

Section 6.1 The Nature of Energy

Is the freezing of water an endothermic or exothermic process? Explain.

CONCEPT CHECK!

Exothermic process because you must remove energy in order to slow the molecules down to form a solid.

Section 6.1 The Nature of Energy

Classify each process as exothermic or endothermic. Explain. The system is underlined in each example.

Your hand gets cold when you touch ice.

The ice gets warmer when you touch it.

Water boils in a kettle being heated on a stove.

Water vapor condenses on a cold pipe.

Ice cream melts.

Exo

Endo

Exo

Endo

CONCEPT CHECK!

Exothermic (heat energy leaves your hand and moves to the ice)
Endothermic (heat energy flows into the ice)
Endothermic (heat energy flows into the water to boil it)
Exothermic (heat energy leaves to condense the water from a gas to a liquid)
Endothermic (heat energy flows into the ice cream to melt it)
Section 6.1 The Nature of Energy

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

Methane is burning in a Bunsen burner in a laboratory.

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

CONCEPT CHECK!

System – methane and oxygen to produce carbon dioxide and water; Surroundings – everything else around it; Direction of energy transfer – energy transfers from the system to the surroundings (exothermic)
System – water drops; Surroundings – everything else around it; Direction of energy transfer – energy transfers from the surroundings (your body) to the system (water drops) (endothermic)
Section 6.1 The Nature of Energy

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

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

Water is lower in energy because a lot of energy was released in the process when hydrogen and oxygen gases reacted.

Section 6.1 The Nature of Energy

Thermodynamics

The study of energy and its interconversions is called thermodynamics.
Law of conservation of energy is often called the first law of thermodynamics.
Copyright © Cengage Learning. All rights reserved

Section 6.1 The Nature of Energy

Internal energy E of a system is the sum of the kinetic and potential energies of all the “particles” in the system.

To change the internal energy of a system: ΔE = q + w
q represents heat

w represents work

Internal Energy

Section 6.1 The Nature of Energy

Work vs Energy Flow

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Section 6.1 The Nature of Energy

Thermodynamic quantities consist of two parts:
Number gives the magnitude of the change.
Sign indicates the direction of the flow.
Internal Energy

Section 6.1 The Nature of Energy

Internal Energy

Sign reflects the system’s point of view.
Endothermic Process:
q is positive
Exothermic Process:
q is negative
Copyright © Cengage Learning. All rights reserved

Section 6.1 The Nature of Energy

Sign reflects the system’s point of view.
System does work on surroundings:
w is negative
Surroundings do work on the system:
w is positive
Internal Energy

Section 6.1 The Nature of Energy

Work = P × A × Δh = PΔV
P is pressure.
A is area.
Δh is the piston moving a distance.
ΔV is the change in volume.
Work

Section 6.1 The Nature of Energy

For an expanding gas, ΔV is a positive quantity because the volume is increasing. Thus ΔV and w must have opposite signs:
w = –PΔV

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

Section 6.1 The Nature of Energy

Which of the following performs more work?

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

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

They perform the same amount of work.

EXERCISE!

They both perform the same amount of work. w = -PΔV

a) w = -(2 atm)(4.0-1.0) = -6 L·atm

b) w = -(3 atm)(3.0-1.0) = -6 L·atm

Section 6.1 The Nature of Energy

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

a) An endothermic process that performs work.

Δ E = positive

Δ E = negative

CONCEPT CHECK!

a) q is positive for endothermic processes and w is negative when system does work on surroundings; first condition – ΔE is negative; second condition – ΔE is positive

b) q is negative for exothermic processes and w is positive when surroundings does work on system; first condition – ΔE is positive; second condition – ΔE is negative

Section 6.2 Enthalpy and Calorimetry

Change in Enthalpy

State function
ΔH = q at constant pressure
ΔH = Hproducts – Hreactants
Copyright © Cengage Learning. All rights reserved

Section 6.2 Enthalpy and Calorimetry

Consider the combustion of propane:

C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(l)

ΔH = –2221 kJ

Assume that all of the heat comes from the combustion of propane. Calculate ΔH in which 5.00 g of propane is burned in excess oxygen at constant pressure.

–252 kJ

EXERCISE!

(5.00 g C3H8)(1 mol / 44.094 g C3H8)(-2221 kJ / mol C3H8)

ΔH = -252 kJ

Section 6.2 Enthalpy and Calorimetry

Calorimetry

Science of measuring heat
Specific heat capacity:
The energy required to raise the temperature of one gram of a substance by one degree Celsius.
Molar heat capacity:
The energy required to raise the temperature of one mole of substance by one degree Celsius.
Copyright © Cengage Learning. All rights reserved

Section 6.2 Enthalpy and Calorimetry

Calorimetry

If two reactants at the same temperature are mixed and the resulting solution gets warmer, this means the reaction taking place is exothermic.
An endothermic reaction cools the solution.
Copyright © Cengage Learning. All rights reserved

Section 6.2 Enthalpy and Calorimetry

A Coffee–Cup Calorimeter Made of Two Styrofoam Cups

Section 6.2 Enthalpy and Calorimetry

Calorimetry

Energy released (heat) = s × m × ΔT
s = specific heat capacity (J/°C·g)

m = mass of solution (g)

ΔT = change in temperature (°C)

Section 6.2 Enthalpy and Calorimetry

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

The final temperature of the water is:

a) Between 50°C and 90°C

b) 50°C

c) Between 10°C and 50°C

CONCEPT CHECK!

The correct answer is b).

Section 6.2 Enthalpy and Calorimetry

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

The final temperature of the water is:

a) Between 50°C and 90°C

b) 50°C

c) Between 10°C and 50°C

Calculate the final temperature of the water.

23°C

CONCEPT CHECK!

The correct answer is c). The final temperature of the water is 23°C.

- (100.0 g)(4.18 J/°C·g)(Tf – 90.) = (500.0 g)(4.18 J/°C·g)(Tf – 10.)

Tf = 23°C

Section 6.2 Enthalpy and Calorimetry

You have a Styrofoam cup with 50.0 g of water at 10.°C. You add a 50.0 g iron ball at 90. °C to the water. (sH2O = 4.18 J/°C·g and sFe = 0.45 J/°C·g)

The final temperature of the water is:

a) Between 50°C and 90°C

b) 50°C

c) Between 10°C and 50°C

Calculate the final temperature of the water.

18°C

CONCEPT CHECK!

The correct answer is c). The final temperature of the water is 18°C.

- (50.0 g)(0.45 J/°C·g)(Tf – 90.) = (50.0 g)(4.18 J/°C·g)(Tf – 10.)

Tf = 18°C

Section 6.3 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.
Copyright © Cengage Learning. All rights reserved

Section 6.3 Hess’s Law

This reaction also can be carried out in two distinct steps, with enthalpy changes designated by ΔH2 and ΔH3.
N2(g) + O2(g) → 2NO(g) ΔH2 = 180 kJ

2NO(g) + O2(g) → 2NO2(g) ΔH3 = – 112 kJ

N2(g) + 2O2(g) → 2NO2(g) ΔH2 + ΔH3 = 68 kJ

ΔH1 = ΔH2 + ΔH3 = 68 kJ

N2(g) + 2O2(g) → 2NO2(g) ΔH1 = 68 kJ

Section 6.3 Hess’s Law

The Principle of Hess’s Law

Section 6.3 Hess’s Law

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Section 6.3 Hess’s Law

Characteristics of Enthalpy Changes

If a reaction is reversed, the sign of ΔH is also reversed.
The 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.
Copyright © Cengage Learning. All rights reserved

Section 6.3 Hess’s Law

Example

Consider the following data:
Calculate ΔH for the reaction
Copyright © Cengage Learning. All rights reserved

Section 6.3 Hess’s Law

Problem-Solving Strategy

Work backward from the required reaction, using 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.
Copyright © Cengage Learning. All rights reserved

Section 6.3 Hess’s Law

Example

Reverse the two reactions:
Desired reaction:
Copyright © Cengage Learning. All rights reserved

Section 6.3 Hess’s Law

Example

Multiply reactions to give the correct numbers of reactants and products:
4( ) 4( )

3( ) 3( )

Desired reaction:
*

Section 6.3 Hess’s Law

Example

Final reactions:
Desired reaction:
ΔH = +1268 kJ

Section 6.4 Standard Enthalpies of Formation

Standard Enthalpy of Formation (ΔHf°)

Change in enthalpy that accompanies the formation of one mole of a compound from its elements with all substances in their standard states.
Copyright © Cengage Learning. All rights reserved

Section 6.4 Standard Enthalpies of Formation

Conventional Definitions of Standard States

For a Compound
For a gas, pressure is exactly 1 atm.
For a solution, concentration is exactly 1 M.
Pure substance (liquid or solid)
For an Element
The form [N2(g), K(s)] in which it exists at 1 atm and 25°C.
Copyright © Cengage Learning. All rights reserved

Section 6.4 Standard Enthalpies of Formation

A Schematic Diagram of the Energy Changes for the Reaction
CH4(g) + 2O2(g) CO2(g) + 2H2O(l)

ΔH°reaction = –(–75 kJ) + 0 + (–394 kJ) + (–572 kJ) = –891 kJ

Section 6.4 Standard Enthalpies of Formation

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

Problem-Solving Strategy: Enthalpy Calculations

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

ΔH°rxn = ΣnpΔHf°(products) - ΣnrΔHf°(reactants)

4. Elements in their standard states are not included in the ΔHreaction calculations because ΔHf° for an element in its standard state is zero.

Section 6.4 Standard Enthalpies of Formation

Calculate ΔH° for the following reaction:

2Na(s) + 2H2O(l) → 2NaOH(aq) + H2(g)

Given the following information:

ΔHf° (kJ/mol)

Na(s) 0

H2O(l) –286

NaOH(aq) –470

H2(g) 0

ΔH° = –368 kJ

EXERCISE!

[2(–470) + 0] – [0 + 2(–286)] = –368 kJ

ΔH = –368 kJ

Section 6.5 Present Sources of Energy

Energy Sources Used in the United States

Section 6.5 Present Sources of Energy

The Earth’s Atmosphere

Transparent to visible light from the sun.
Visible light strikes the Earth, and part of it is changed to infrared radiation.
Infrared radiation from Earth’s surface is strongly absorbed by CO2, H2O, and other molecules present in smaller amounts in atmosphere.
Atmosphere traps some of the energy and keeps the Earth warmer than it would otherwise be.
Copyright © Cengage Learning. All rights reserved

Section 6.5 Present Sources of Energy

The Earth’s Atmosphere

Section 6.6 New Energy Sources

Exothermic process because you must remove energy in order to slow the molecules down to form a solid.

Water is lower in energy because a lot of energy was released in the process when hydrogen and oxygen gases reacted.

They both perform the same amount of work. w = -PΔV

a) w = -(2 atm)(4.0-1.0) = -6 L·atm

b) w = -(3 atm)(3.0-1.0) = -6 L·atm

a) q is positive for endothermic processes and w is negative when system does work on surroundings; first condition – ΔE is negative; second condition – ΔE is positive

b) q is negative for exothermic processes and w is positive when surroundings does work on system; first condition – ΔE is positive; second condition – ΔE is negative

(5.00 g C3H8)(1 mol / 44.094 g C3H8)(-2221 kJ / mol C3H8)

ΔH = -252 kJ

The correct answer is b).

The correct answer is c). The final temperature of the water is 23°C.

- (100.0 g)(4.18 J/°C·g)(Tf – 90.) = (500.0 g)(4.18 J/°C·g)(Tf – 10.)

Tf = 23°C

The correct answer is c). The final temperature of the water is 18°C.

- (50.0 g)(0.45 J/°C·g)(Tf – 90.) = (50.0 g)(4.18 J/°C·g)(Tf – 10.)

Tf = 18°C

[2(–470) + 0] – [0 + 2(–286)] = –368 kJ

ΔH = –368 kJ

322

222

NH()N()H() H = 46 kJ

2 H()O()2 HO() H = 484 kJ

¾¾®+D

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ggg

2223

2 N()6 HO()3 O()4 NH()

+¾¾®+

gggg

223

222

N()H()NH() H = 46 kJ

2 HO()2 H()O() H = +484 kJ

+¾¾®D-

¾¾®+D

ggg

223

222

2 N()6 H()4 NH() H = 184 kJ

6 HO()6 H()3 O() H = +1452 kJ

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