CHEMISTRY EXAM-Thermochemistry

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

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Section 6.1
The Nature of Energy

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  • 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

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Section 6.1
The Nature of Energy

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

Initial Position

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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

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Section 6.1
The Nature of Energy

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  • 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

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Section 6.1
The Nature of Energy

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

Chemical Energy

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Section 6.1
The Nature of Energy

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  • 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

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Section 6.1
The Nature of Energy

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

CONCEPT CHECK!

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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

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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

Endo

Exo

Endo

CONCEPT CHECK!

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  • 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

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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!

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  • 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

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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!

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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.

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Section 6.1
The Nature of Energy

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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

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Section 6.1
The Nature of Energy

Work vs Energy Flow

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Section 6.1
The Nature of Energy

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

Internal Energy

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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

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Section 6.1
The Nature of Energy

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  • 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

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Section 6.1
The Nature of Energy

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  • 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

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Section 6.1
The Nature of Energy

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  • 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

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Section 6.1
The Nature of Energy

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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!

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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

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

a) An endothermic process that performs work.

  • |work| > |heat|
  • |work| < |heat|

b) Work is done on a gas and the process is exothermic.

  • |work| > |heat|
  • |work| < |heat|

Δ E = negative

Δ E = positive

Δ E = positive

Δ E = negative

CONCEPT CHECK!

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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

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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

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EXERCISE!

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(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.

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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.

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Section 6.2
Enthalpy and Calorimetry

A Coffee–Cup Calorimeter Made of Two Styrofoam Cups

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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)

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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

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CONCEPT CHECK!

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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

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CONCEPT CHECK!

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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

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CONCEPT CHECK!

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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.

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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

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Section 6.3
Hess’s Law

The Principle of Hess’s Law

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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.

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Section 6.3
Hess’s Law

Example

    • Consider the following data:
  • Calculate ΔH for the reaction

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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.

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Section 6.3
Hess’s Law

Example

    • Reverse the two reactions:
  • Desired reaction:

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Section 6.3
Hess’s Law

Example

  • Multiply reactions to give the correct numbers of reactants and products:

4( ) 4( )

3( ) 3( )

  • Desired reaction:

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Section 6.3
Hess’s Law

Example

    • Final reactions:
  • Desired reaction:

ΔH = +1268 kJ

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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.

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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.

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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

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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.

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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.

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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

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EXERCISE!

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

ΔH = –368 kJ

Section 6.5
Present Sources of Energy

  • Fossil Fuels
  • Petroleum, Natural Gas, and Coal
  • Wood
  • Hydro
  • Nuclear

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Section 6.5
Present Sources of Energy

Energy Sources Used in the United States

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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.

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Section 6.5
Present Sources of Energy

The Earth’s Atmosphere

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Section 6.6
New Energy Sources

  • Coal Conversion
  • Hydrogen as a Fuel
  • Other Energy Alternatives
  • Oil shale
  • Ethanol
  • Methanol
  • Seed oil

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

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  • 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)

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  • 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)

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

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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

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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

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

ΔH = -252 kJ

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The correct answer is b).

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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

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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

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

ΔH = –368 kJ

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322

222

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

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

¾¾®+D

+¾¾®D-

ggg

ggg

2223

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

+¾¾®+

gggg

223

222

13

22

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

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

+¾¾®D-

¾¾®+D

ggg

ggg

223

222

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

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

+¾¾®D-

¾¾®+D

ggg

ggg

Chapter 6 Thermochemistry Study Guide

1. A gas absorbs 0.0 J of heat and then performs 30.7 J of work. The change in internal energy of the gas is

A)61.4 J
B)30.7 J
C)–61.4 J
D)–30.7 J
E)none of these

ANS: D

2. What is the kinetic energy of a 1.56-kg object moving at 94.0 km/hr?

A)5.32  102 kJ
B)6.89  103 kJ
C)5.32  10–4 kJ
D)1.06  103 kJ
E)2.04  101 kJ

ANS: A

3. Which of the following statements correctly describes the signs of q and w for the following exothermic process at P = 1 atm and T = 370 K?

H2O(g)  H2O(l)

A)q and w are negative.
B)q is positive, w is negative.
C)q is negative, w is positive.
D)q and w are both positive.
E)q and w are both zero.

ANS: C

4. Which of the following statements is correct?

A)The internal energy of a system increases when more work is done by the system than heat was flowing into the system.
B)The internal energy of a system decreases when work is done on the system and heat is flowing into the system.
C)The system does work on the surroundings when an ideal gas expands against a constant external pressure.
D)All statements are true.
E)All statements are false.

ANS: C

5. For a particular process q = –17 kJ and w = 21 kJ. Which of the following statements is false?

A)Heat flows from the system to the surroundings.
B)The system does work on the surroundings.
C)E = +4 kJ
D)The process is exothermic.
E)None of the above is false.

ANS: B

6. One mole of an ideal gas is expanded from a volume of 1.00 liter to a volume of 8.93 liters against a constant external pressure of 1.00 atm. How much work (in joules) is performed on the surroundings? Ignore significant figures for this problem. (T = 300 K; 1 L·atm = 101.3 J)

A)402 J
B)803 J
C)2.41  103 J
D)905 J
E)none of these

ANS: B

7. A fuel-air mixture is placed in a cylinder fitted with a piston. The original volume is 0.310-L. When the mixture is ignited, gases are produced and 935 J of energy is released. To what volume will the gases expand against a constant pressure of 635 mmHg, if all the energy released is converted to work to push the piston?

A)10.7 L
B)8.02 L
C)11.4 L
D)11.0 L
E)1.78 L

ANS: C

8. Calculate the work associated with the compression of a gas from 121.0 L to 80.0 L at a constant pressure of 13.1 atm.

A)–537 L atm
B)537 L atm
C)3.13 L atm
D)–3.13 L atm
E)101 L atm

ANS: B

9. A gas absorbs 825 J of heat and then has 841 J of work done upon it. The change in internal energy of the gas is

A)1666 J
B)16 J
C)–16 J
D)–1666 J
E)none of these

ANS: A

10. Of energy, work, enthalpy, and heat, how many are state functions?

A)0
B)1
C)2
D)3
E)4

ANS: C

11. Which of the following properties is (are) intensive properties?

I. mass

II. temperature

III. volume

IV. concentration

V. energy

A)I, III, and V
B)II only
C)II and IV
D)III and IV
E)I and V

ANS: C

12. For the reaction H2O(l)  H2O(g) at 298 K and 1.0 atm, H is more positive than E by 2.5 kJ/mol. This quantity of energy can be considered to be

A)the heat flow required to maintain a constant temperature
B)the work done in pushing back the atmosphere
C)the difference in the H–O bond energy in H2O(l) compared to H2O(g)
D)the value of H itself
E)none of these

ANS: B

13. Consider the reaction:

C2H5OH(l) + 3O2(g)  2CO2(g) + 3H2O(l); H = –1.37  103 kJ

Consider the following propositions:

I. The reaction is endothermic

II. The reaction is exothermic.

III. The enthalpy term would be different if the water formed was gaseous.

Which of these propositions is (are) true?

A)I
B)II
C)III
D)I, II
E)II, III

ANS: E

14. How much heat is required to raise the temperature of a 5.75-g sample of iron (specific heat = 0.450 J/g°C) from 25.0°C to 79.8°C?

A)2.54 J
B)315 J
C)700 J
D)848 J
E)142 J

ANS: E

15. A 32.5 g piece of aluminum (which has a molar heat capacity of 24.03 J/°C·mol) is heated to 82.4°C and dropped into a calorimeter containing water (specific heat capacity of water is 4.18 J/g°C) initially at 22.3°C. The final temperature of the water is 24.2°C. Ignoring significant figures, calculate the mass of water in the calorimeter.

A)212 g
B)5.72 kg
C)6.42 g
D)1.68 kg
E)none of these

ANS: A

16. A 45.9 g sample of a metal is heated to 95.2°C and then placed in a calorimeter containing 120.0 g of water (c = 4.18 J/g°C) at 21.6°C. The final temperature of the water is 24.5°C. Which metal was used?

A)Aluminum (c = 0.89 J/g°C)
B)Iron (c = 0.45 J/g°C)
C)Copper (c = 0.20 J/g°C)
D)Lead (c = 0.14 J/g°C)
E)none of these

ANS: B

17. You take 295.5 g of a solid at 30.0°C and let it melt in 425 g of water. The water temperature decreases from 85.1°C to 30.0°C. Calculate the heat of fusion of this solid.

A)160 J/g
B)166 J/g
C)331 J/g
D)721 J/g
E)cannot solve without the heat capacity of the solid

ANS: C

18. 30.0 mL of pure water at 282 K is mixed with 50.0 mL of pure water at 306 K. What is the final temperature of the mixture?

A)294 K
B)297 K
C)342 K
D)588 K
E)24 K

ANS: B

19. Consider the reaction

H2(g) + O2(g)  H2O(l) H° = –286 kJ

Which of the following is true?

A)The reaction is exothermic.
B)The reaction is endothermic.
C)The enthalpy of the products is less than that of the reactants.
D)Heat is absorbed by the system.
E)Both A and C are true.

ANS: E

20. What is the specific heat capacity of a metal if it requires 178.1 J to change the temperature of 15.0 g of the metal from 25.00°C to 32.00°C?

A)0.590 J/g°C
B)11.9 J/g°C
C)25.4 J/g°C
D)1.70 J/g°C
E)283 J/g°C

ANS: D

21. A 140.0-g sample of water at 25.0°C is mixed with 111.7 g of a certain metal at 100.0°C. After thermal equilibrium is established, the (final) temperature of the mixture is 29.6°C. What is the specific heat capacity of the metal, assuming it is constant over the temperature range concerned?

A)0.34 J/g°C
B)0.68 J/g°C
C)0.22 J/g°C
D)2.9 J/g°C
E)none of these

ANS: A

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

A)boiling
B)completely vaporized
C)frozen solid
D)decomposed
E)still a liquid

ANS: E

23. A chunk of lead at 91.6°C was added to 200.0 g of water at 15.5°C. The specific heat of lead is 0.129 J/g°C, and the specific heat of water is 4.18 J/g°C. When the temperature stabilized, the temperature of the mixture was 17.9°C. Assuming no heat was lost to the surroundings, what was the mass of lead added?

A)1.57 kg
B)170 g
C)204 g
D)211 g
E)none of these

ANS: D

24. What is the specific heat capacity of silver if it requires 86.3 J to raise the temperature of 15 grams of silver by 25°C?

A)4.3 J/g°C
B)0.23 J/g°C
C)0.14 J/g°C
D)0.60 J/g°C
E)none of these

ANS: B

25. If a student performs an endothermic reaction in a calorimeter, how does the calculated value of H differ from the actual value if the heat exchanged with the calorimeter is not taken into account?

A)Hcalc would be more negative because the calorimeter always absorbs heat from the reaction.
B)Hcalc would be less negative because the calorimeter would absorb heat from the reaction.
C)Hcalc would be more positive because the reaction absorbs heat from the calorimeter.
D)Hcalc would be less positive because the reaction absorbs heat from the calorimeter.
E)Hcalc would equal the actual value because the calorimeter does not absorb heat.

ANS: D

26. Consider the reaction:

When a 21.1-g sample of ethyl alcohol (molar mass = 46.07 g/mol) is burned, how much energy is released as heat?

A)0.458 kJ
B)0.627 kJ
C)6.27  102 kJ
D)2.89  104 kJ
E)2.18 kJ

ANS: C

27. The total volume of hydrogen gas needed to fill the Hindenburg was 2.11  108 L at 1.00 atm and 24.7°C. How much energy was evolved when it burned?

A)8.64  106 kJ
B)2.98  1010 kJ
C)3.02  104 kJ
D)2.47  109 kJ
E)4.94  109 kJ

ANS: D

28. What is the enthalpy change when 49.4 mL of 0.430 M sulfuric acid reacts with 23.3 mL of 0.309 M potassium hydroxide?

H2SO4(aq) + 2KOH(aq)  K2SO4(aq) + 2H2O(l)H° = –111.6 kJ/mol
A)–0.402 kJ
B)–3.17 kJ
C)–2.37 kJ
D)–0.803 kJ
E)–112 kJ

ANS: A

29. How much heat is liberated at constant pressure when 2.35 g of potassium metal reacts with 5.68 mL of liquid iodine monochloride (d = 3.24 g/mL)?

2K(s) + ICl(l)  KCl(s) + KI(s)H° = –740.71 kJ/mol
A)2.22  103 kJ
B)8.40  101 kJ
C)1.28  102 kJ
D)2.23  101 kJ
E)7.41  102 kJ

ANS: D

30. Which of the following statements is/are true?

I. q (heat) is a state function because H is a state function and q = H.
II. When 50.0 g of aluminum at 20.0°C is placed in 50.0 mL of water at 30.0°C, the H2O will undergo a smaller temperature change than the aluminum. (The density of H2O = 1.0 g/mL, specific heat capacity of H2O = 4.18 J/g°C, specific heat capacity of aluminum = 0.89 J/g°C)
III. When a gas is compressed, the work is negative since the surroundings are doing work on the system and energy flows out of the system.
IV. For the reaction (at constant pressure) 2N2(g) + 5O2(g)  2N2O5(g), the change in enthalpy is the same whether the reaction takes place in one step or in a series of steps.
A)I, II, IV
B)II, III
C)II, III, IV
D)II, IV
E)All of the above statements are true.

ANS: D

31. Consider the following processes:

2A  (1/2)B + CH1 = 5 kJ/mol
(3/2)B + 4C  2A + C + 3DH2 = –15 kJ/mol
E + 4A  CH3 = 10 kJ/mol

Calculate H for: C  E + 3D

A)0 kJ/mol
B)10 kJ/mol
C)–10 kJ/mol
D)–20 kJ/mol
E)20 kJ/mol

ANS: C

32. At 25°C, the following heats of reaction are known:

H (kJ/mol)
2ClF + O2  Cl2O + F2O167.4
2ClF3 + 2O2  Cl2O + 3F2O341.4
2F2 + O2  2F2O–43.4

At the same temperature, calculate H for the reaction: ClF + F2  ClF3

A)–217.5 kJ/mol
B)–130.2 kJ/mol
C)+217.5 kJ/mol
D)–108.7 kJ/mol
E)none of these

ANS: D

33. Given the heats of the following reactions:

H°(kJ)
I.P4(s) + 6Cl2(g)  4PCl3(g)–1225.6
II.P4(s) + 5O2(g)  P4O10(s)–2967.3
III.PCl3(g) + Cl2(g)  PCl5(g)–84.2
IV.PCl3(g) + O2(g)  Cl3PO(g)–285.7

Calculate the value of H° for the reaction below:

P4O10(s) + 6PCl5(g)  10Cl3PO(g)

A)–110.5 kJ
B)–610.1 kJ
C)–2682.2 kJ
D)–7555.0 kJ
E)None of these is within 5% of the correct answer.

ANS: B

34. One of the main advantages of hydrogen as a fuel is that:

A)The only product of hydrogen combustion is water.
B)It exists as a free gas.
C)It can be economically supplied by the world’s oceans.
D)Plants can economically produce the hydrogen needed.
E)It contains a large amount of energy per unit volume of hydrogen gas.

ANS: A

35. Consider the following standard heats of formation:

P4O10(s) = –3110 kJ/mol

H2O(l) = –286 kJ/mol

H3PO4(s) = –1279 kJ/mol

Calculate the change in enthalpy for the following process:

P4O10(s) + 6H2O(l)  4H3PO4(s)

ANS:

–290 kJ

36. For the complete combustion of 1.000 mole of ethane gas at 298 K and 1 atm pressure, H° = -1560 kJ/mol. What will be the heat released when 4.42 g of ethane is combusted under these conditions?

A)–230 kJ
B)230 kJ
C)10588 kJ
D)–10588 kJ
E)none of these

ANS: B

37. For the complete combustion of 1.000 mole of propane gas at 298 K and 1 atm pressure, H° = -2220 kJ/mol. What will be the heat released when 4.13 g of propane is combusted under these conditions?

A)–208 kJ
B)208 kJ
C)23651 kJ
D)–23651 kJ
E)none of these

ANS: B

38. A 36.2 g piece of metal is heated to 81°C and dropped into a calorimeter containing 50.0 g of water (specific heat capacity of water is 4.18 J/g°C) initially at 21.7°C. The empty calorimeter has a heat capacity of 125 J/K. The final temperature of the water is 29.7°C. Ignoring significant figures, calculate the specific heat of the metal..

A)1.439 J/gK
B)0.900 J/gK
C)0.360 J/gK
D)0.968 J/gK
E)none of these

ANS: A

1

Practical examples[edit]

One joule in everyday life represents approximately:

· the energy required to lift a small apple (with a mass of approximately 100 g) vertically through one metre of air.

· the energy released when that same apple falls one metre to the ground.

· the energy required to accelerate a 1 kg mass at 1 m·s−2 through a 1 m distance in space.

· the heat required to raise the temperature of 1 g of water by 0.24 K.[6]

· the typical energy released as heat by a person at rest every 1/60 second (approximately 17 ms).[7]

· the kinetic energy of a 50 kg human moving very slowly (0.2 m/s or 0.72 km/h).

· the kinetic energy of a 56 g tennis ball moving at 6 m/s (22 km/h).[8]

· the kinetic energy of an object with mass 1 kg moving at √2 ≈ 1.4 m/s.

· The amount of electricity required to light a 1 watt LED for 1 s.

Since the joule is also a watt-second and the common unit for electricity sales to homes is the kW·h (kilowatt-hour), a kW·h is thus 1000 (kilo) watt × 3600 seconds = 3.6 MJ (megajoules).

One joule can also be defined as:

· The work required to move an electric charge of one coulomb through an electrical potential difference of one volt, or one ‘”coulomb volt” (C·V). This relationship can be used to define the volt.

· The work required to produce one watt of power for one second, or one “watt second” (W·s) (compare kilowatt hour – 3.6 megajoules). This relationship can be used to define the watt.

Conversions[edit]

Main article: Conversion of units of energy

1 joule is equal to:

· 1×107 erg (exactly)

· 6.24150974×1018 eV

· 0.2390 cal (gram calories)

· 2.390×10−4 kcal (food calories)

· 9.4782×10−4 BTU

· 0.7376 ft·lb

· 23.7 ft·pdl (foot-poundal)

· 2.7778×10−7 kilowatt-hour

· 2.7778×10−4 watt-hour

· 9.8692×10−3 l·atm (litre-atmosphere)

· 11.1265×10−15 gram (by way of mass-energy equivalence)

· 1×10−44 foe (exactly)

Units defined exactly in terms of the joule include:

· 1 thermochemical calorie = 4.184 J[14]

· 1 International Table calorie = 4.1868 J[15]

· 1 watt hour = 3600 J (or 3.6 kJ)

· 1 kilowatt hour = 3.6×106 J (or 3.6 MJ)

· 1 watt second = 1 J

· 1 ton TNT = 4.184 GJ

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