Energy and Chemical Reactions

CHAPTER 6 Energy and Chemical Reactions Objectives You will be able to: 1. 2. 3. 4. 5. 6. Write or identify a description of the Law of Conservation...

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CHAPTER 6 Energy and Chemical Reactions

Objectives You will be able to: 1. 2. 3. 4. 5. 6.

Write or identify a description of the Law of Conservation of Energy. Describe the relationship between stability and potential energy. Explain why energy must be absorbed to break a chemical bond. Explain why energy is released when a chemical bond is formed. Identify the joule as the accepted international unit for energy. Convert between the names and abbreviations for joule, J, calorie, cal, and dietary calorie (Calorie), Cal. 7. Write or identify the relative sizes of the joule, calorie, and dietary calorie (Calorie). 8. Write or identify what is meant in terms of average internal kinetic energy when we say that one object has a higher temperature than another object. 9. Describe the changes that take place during heat transfer between objects at a diﬀerent temperature. 10. Write or identify the relationship between the wavelength of radiant energy and the energy of its photons. 11. Write or identify the relative energies and wavelengths of the following forms of radiant energy: gamma rays, X rays, ultraviolet (UV), visible, infrared (IR), microwaves, and radio waves. 12. Write or identify the relative energies and wavelengths of the following colors of visible light: violet, blue, green, yellow, orange, and red. 13. Explain why some chemical reactions release heat to their surroundings. 14. Explain why some chemical reactions absorb heat from their surroundings. 15. Explain why all chemical reactions either absorb or release energy. 16. Convert between the sign of the heat for a reaction and whether the reaction is exergonic or endergonic. 17. Write or identify the factors that aﬀect heats of chemical reaction. 18. Write or identify the reason why ∆H is used to describe heats of chemical reactions not ∆E. 19. Write or identify the temperature and pressure for standard changes in enthalpies. 20. Given a balanced equation and its ∆H, convert between any amount of reactant or product and the amount of heat involved in the chemical reaction of that amount of reactant or product. 21. Given a balanced chemical equation and its ∆H, determine the ∆H for the same reaction with diﬀerent coeﬃcients in the balanced equation. 22. Given a balanced chemical equation and its ∆H, determine the ∆H for the reverse reaction. 23. Explain why the heat absorbed by or evolved from a chemical reaction might be diﬀerent at constant pressure and at constant volume.

105

106

Chapter 6

Energy and Chemical Reactions

24. Write or identify the signs for heat absorbed by the system, heat evolved by the system, work done on the system, and work done by the system on the surroundings. 25. Explain why the heat at constant pressure is greater than the heat at constant volume for the endergonic reaction of potassium chlorate to form potassium chloride and oxygen. 26. Explain why there is more energy released at constant pressure than at constant volume for the reaction that forms ammonia gas from nitrogen gas and hydrogen gas. 27. Explain why the heat at constant pressure is equal to the heat at constant volume for a reaction that forms the same amount of gaseous products as it had gaseous reactants. 28. Given enough information to determine a reaction’s balanced equation, including states, and enough information to get one of the following, calculate the one not given: (1) the ∆E° for the equation and (2) the ∆H° for the equation. 29. Write or identify general descriptions of bomb and open calorimeters. 30. Explain why heats of reactions are usually determined using bomb calorimeters instead of open calorimeters. 31. Given the following for a calorimeter experiment, calculate the heat capacity of the calorimeter (Ccal): (1) enough information to get a balanced chemical equation for a reaction including states, (2) the amount of reactant in terms of mass or volume of solution of a known molarity or volume of gas at a given pressure and temperature, (3) the molar heat of reaction (∆H°), (4) the mass of water in the calorimeter (mw), (5) the temperature before the reaction starts (T1), and (6) the temperature after the reaction is complete (T2). 32. Given the following for a calorimeter experiment, calculate the molar heat of reaction (∆H°): (1) enough information to get a balanced chemical equation for a reaction including states, (2) ) the amount of reactant in terms of mass or volume of solution of a known molarity or volume of gas at a given pressure and temperature, (3) the mass of water in the calorimeter (mw), (4) the temperature before the reaction starts (T1), (5) the temperature after the reaction is complete (T2) and (6) the heat capacity of the calorimeter (Ccal). 33. Given the ∆H°s and chemical equations of reactions that can be combined to yield an overall reaction, calculate the ∆H° for the overall reaction. (Be able to do common Law of Hess problems.) 34. Write or identify the formulas and states for the standard states of all the elements on the periodic table. 35. Given a formula and state for a compound, write the chemical equation that describes its heat of formation reaction. 36. Given a balanced equation and all but one of the following, calculate the one not given; the heat of reaction for the equation and the heats of formation of each compound involved in the reaction. 37. Convert between the deﬁnition and the term for the following words or phrases. Skip Sections 6.11 and 6.12 in the text.

107

Chapter 6 Glossary Energy The capacity to do work. Kinetic energy The capacity to do work due to the motion of an object. Law of Conservation of Energy Energy can neither be created nor destroyed, but it can be transferred from one system to another and changed from one form to another. Potential energy A retrievable, stored form of energy an object possesses by virtue of its position or state. Endergonic (endogonic) change Change that absorbs energy. Exergonic (exogonic) change Change that releases energy. Exothermic change Change that leads to heat energy being released from the system to the surroundings. Endothermic change Change that leads the system to absorb heat energy from the surroundings. Joule (J) The accepted international unit for energy. calorie (with a lowercase c) A common energy unit. There are 4.184 joules per calorie (abbreviated cal). Calorie (with an upper case C) The dietary calorie (abbreviated Cal). In fact, it is a kilocalorie, the equivalent of 4184 joules. Thermal energy The energy associated with the random motion of particles. Temperature A measure of the average internal kinetic energy of an object. Heat The thermal energy transferred from a region of higher temperature to a region of lower temperature due to collisions between particles. Radiant energy or electromagnetic radiation Energy that can be described in terms of either oscillating electric and magnetic ﬁelds or in terms of a stream of tiny packets of energy with no mass. Photons Tiny packets or particles of radiant energy. Wavelength (λ) The distance in space over which a wave completes one cycle of its repeated form. Frequency(ν) The number of cycles of a wave per amount of time (e.g. cycles per second). Change in enthalpy (∆H) heat evolved or absorbed for a change run at constant pressure. Total internal energy The sum of all the kinetic and potential energy. Change in total internal energy (∆E) Heat evolved or absorbed for a change run at constant volume. Standard change in enthalpy (∆H°) Heat evolved or absorbed for a change run at a constant pressure of 1 atm and a constant temperature of 298.15 K. Standard change in total internal energy (∆E°) Heat evolved or absorbed for a change run at constant volume and a constant temperature of 298.15 K. Bomb calorimeter An instrument used to determine heats of reaction at constant volume. Open calorimeter An instrument used to determine heats of reaction at constant pressure. Heat capacity The heat necessary to increase the temperature of an object by one kelvin (or one degree Celsius).

Chapter 6

Energy and Chemical Reactions

Specific heat capacity The heat necessary to increase the temperature of one gram of pure substance by one kelvin (or one degree Celsius). Heat of formation (∆Hf°) The heat involved in the formation of one mole of substance from its elements in their standard states at a constant pressure of 1 atm and a constant temperature of 298.15 K. Standard states of elements The most stable (or most common) form of an element at normal temperatures and pressures. It includes a formula and state, e.g. Cl2(g) for chlorine.

Figure 6.1 Energy is the Capacity to do work.

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Figure 6.2 Conservation of Energy

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Figure 6.3 Relationship Between Attractions, Stability, and Potential Energy

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Figure 6.4 Endergonic Reactions

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Figure 6.5 Exergonic Reactions

110

Chapter 6

Energy and Chemical Reactions

EXERCISE 6.1 - Kinetic and Potential Energy For each of the following situations, you are asked which pair has the higher energy. Explain your answer with reference to the capacity of each to do work and say whether the energy that distinguishes them is kinetic energy or potential energy. a. Nitric acid molecules, HNO3, in the upper atmosphere decompose to form HO molecules and NO2 molecules by breaking a bond between the nitrogen atom and one of the oxygen atoms. Which has higher energy, (1) a nitric acid molecule or (2) the HO molecule and NO2 molecule the come from its decomposition? HNO3(g) → HO(g) + NO2(g) b. Nitrogen oxides, NO(g) and NO2(g), are released into the atmosphere in the exhaust of our cars. Which has higher energy, (1) a NO2 molecule moving at 439 m/s or (2) the same NO2 molecule moving at 399 m/s. (These are the average velocities of NO2 molecules at 80 °C and 20 °C.) c. Which has higher energy, (1) a nitrogen monoxide molecule, NO, moving out your car’s tailpipe at 450 m/s or (2) a nitrogen dioxide molecule, NO2, moving at the same velocity? d. Liquid nitrogen is used for a number of purposes, including the freezing of warts. Which has higher energy, (1) liquid nitrogen or (2) gaseous nitrogen? (Disregard the likely diﬀerence in temperature, and assume that the two systems are at the same temperature.) e. Halons, such as Halon−1301 (CF3Br) and halon−1211 (CF2ClBr), which have been used as ﬁre extinguishing agents, are a threat to our protective ozone layer. When released into the atmosphere, they can migrate into the upper atmosphere where bromine atoms are stripped from the molecules. These bromine atoms react with ozone molecules to form BrO molecules, which can react with NO2 molecules to form BrONO2. Which has higher energy, (1) separate BrO and NO2 molecules or (2) the BrONO2 that they form? f. Alpha particles, which are released in alpha decay of large radioactive elements, such as uranium, are helium nuclei that contain two protons and two neutrons. Which has higher energy, (1) alpha particles that are close together or (2) alpha particles that are farther apart? g. Which has higher energy, (1) an uncharged helium atom or (2) an alpha particle and two separate electrons?

111

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Figure 6.7 A Radiant Energy Wave’s Electric and Magnetic Fields

112

Chapter 6

Energy and Chemical Reactions

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Figure 6.8 Radiant energy Spectrum

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weaker bonds → stronger bonds + energy higher PE → lower PE + energy 2H2(g) + O2(g) → 2H2O(g) + energy stronger bonds + energy → weaker bonds lower PE + energy → higher PE CaCO3(s) + energy → CaO(s) + CO2(g)

113 ��

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�� Figure 6.9 Energy and Chemical Reactions

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114

Chapter 6

Energy and Chemical Reactions

Figure 6.10 Celsius and Fahrenheit Thermometers ��

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Figure 6.11 Comparing Temerature Scales ��

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115 The amount of heat associated with a chemical reaction depends on the following: •

the nature of the reaction, Diﬀerent reactions ↓ Diﬀerent changes in bonds ↓ Diﬀerent changes in strengths of bonds ↓ Diﬀerent changes in potential energy ↓ Diﬀerent amounts of heat absorbed or evolved

•

the amount of reactants, Increased amount of reactant → Increased amount of heat

•

the conditions of the reaction.

A complete, balanced equation can be used to describe the amount of substance involved and the states of each reactant and product. The following equation describes the combustion of ethane gas. 2C2H6(g) + 7O2(g) → 4CO2(g) + 6H2O(l)

∆H° = −3.08 x 103 kJ

The ∆H describes the amount of energy that is associated with the reaction of the number of moles equal to the coeﬃcients in the balanced equation. Thus there are 3.08 x 103 kJ of heat evolved when two moles of ethane react with seven moles of oxygen to yield four moles of carbon dioxide and six moles of water. The ∆H leads to conversion factors that convert between amount of any reactant or product and the amount of heat associated with the reaction. ��

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EXERCISE 6.2 - ∆H as a Conversion Factor When 1.245 x 104 kJ of heat are evolved from the combustion of ethane, what mass of water is formed?

116

Chapter 6

Energy and Chemical Reactions

EXERCISE 6.3 - ∆H and Changing Coefﬁcients What is the ∆H° for the following equation? 4C2H6(g) + 14O2(g) → 8CO2(g) + 12H2O(l )

EXERCISE 6.4 - ∆H and Reverse Equations What is the ∆H° for the following reaction? 4CO2(g) + 6H2O(l ) → 2C2H6(g) + 7O2(g)

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�� Figure 6.12 Heat at Constant Pressure and Heat at Constant Volume for Endothermic Reaction and Increased Moles of Gas

117 ��

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�� Figure 6.13 Heat at Constant Pressure and Heat at Constant Volume for Exothermic Reaction and Decreased Moles of Gas

118

Chapter 6

Energy and Chemical Reactions

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�� Figure 6.14 Heat at Constant Pressure and Heat at Constant Volume for Reaction with No Change in Moles of Gas

119

EXERCISE 6.5 - ∆H and ∆E Nitrogen dioxide gas reacts with liquid water to yield liquid nitric acid and nitrogen monoxide gas. 23.84 kJ of heat are evolved when one mole of NO2 reacts at constant volume and with standard conditions. What is the ∆H° for this reaction?

EXERCISE 6.6 - ∆H° and ∆E° When 3.000 g of ethyl alcohol, C2H5OH(l ), are burned in a bomb calorimeter at 25.00 °C, 88.90 kJ of heat are evolved. Calculate the ∆E° and ∆H° for this reaction. (The reactions in a bomb calorimeter are run at constant volume.)

Ways to get ∆H° • From tables • From bomb calorimeter data • From open calorimeter data • From the Law of Hess • From heats of formation

120

Chapter 6

Energy and Chemical Reactions

Figure 6.15 Bomb Calorimeter

Thermometer Ignition wire

Stirrer

Water Insulated walls

Bomb (Constant volume reaction vessel)

Sample dish

Sample Study Sheet 6.1:

TIP−OFF − You are given mass of substance, the mass of water in a bomb calorimeter

Bomb Calorimeter Problems Calculating ∆H°

and either the heat capacity of the calorimeter (Ccal) or the ∆H° for the reaction. You

(mw), the temperature before the reaction (T1), the temperature after the reaction (T2), are asked to calculate either the heat capacity of the calorimeter (Ccal) or the ∆H° for the reaction. GENERAL STEPS − Calculating ∆H° •

Calculate qv from the following equation. qv = –[Ccal +

•

Calculate ∆E°.

E° = •

0.00418 kJ mw] T g·°C

? kJ mol substance

=

(qv) kJ

(molar mass) g substance

(given) g substance

1 mol substance

Calculate ∆H°. ∆H° = ∆E° + (∆n)RT

121

EXERCISE 6.7 - Calculating ∆H from Bomb Calorimeter Data 3.200 g of ethyl alcohol, C2H5OH(l ), are burned in a bomb calorimeter. It contains 504.5 g of water. The heat capacity of the calorimeter is 0.415 kJ/°C. (The heat capacity is determined in an experiment described in Exercise 6.8. )The temperature rises from 20.15 °C to 56.49 °C. What is the heat of combustion of ethyl alcohol?

Ignition wire

Heat capacity of calorimeter is 0.415 kJ/°C

Stirrer

Temperature increases from 20.15 °C to 56.49 °C.

504.5 g H2O

Excess O2

3.200 g C2H5OH

122

Chapter 6

Energy and Chemical Reactions

Sample Study Sheet 6.2 : Bomb Calorimeter Problems Calculating Ccal

GENERAL STEPS − Calculating Ccal. • Calculate ∆E° from ∆H°. ∆E° = ∆H° − (∆n)RT • Calculate qv from ∆E°. qv = ? kJ = (given) g substance •

1 mol substance

(E°) kJ

(molar mass) g substance

1 mol substance

Calculate Ccal. Ccal = –

qv T

–

0.00418 kJ g·°C

mw

T = T2 – T1 Watch signs!

EXERCISE 6.8 - Calculating Heat Capacity from Bomb Calorimeter Data The heat of combustion of benzene, C6H6(l ), is −2066 kJ/mole. When 0.200 g of benzene are burned in the bomb calorimeter mentioned in Exercise 6.7, which now contains 526.0 g of water, the temperature rises from 22.56 °C to 24.58 °C. What is the heat capacity of the calorimeter?

123

TIP−OFF − You are given mass of substance, the mass of water in an open calorimeter (mw), the temperature before the reaction (T1), the temperature after the reaction (T2), and either the heat capacity of the calorimeter (Ccal) or the ∆H° for the reaction. You are asked to calculate either the heat capacity of the calorimeter (Ccal) or the ∆H° for the reaction. GENERAL STEPS − Calculating ∆H° •

Calculate qp from the following equation. If the heat capacity of the

calorimeter is not given, assume it is zero. qp = –[Ccal + •

0.00418 kJ mw] T g·°C

Calculate ∆H°.

H° =

? kJ mol substance

=

(qp) kJ

(molar mass) g substance

(given) g substance

1 mol substance

GENERAL STEPS − Calculating Ccal. •

Calculate qp from ∆H°.

qp = ? kJ = (given) g substance •

1 mol substance

(H°) kJ

(molar mass) g substance

1 mol substance

Calculate Ccal. Ccal = –

qp T

–

0.00418 kJ g·°C

mw

T = T2 - T1 Watch signs!

We can calculate the ∆H° of one reaction from H°s of other reactions by applying the Law of Hess. The Law of Hess states that if a reaction can be shown to be the sum of two or more reactions, the ∆H° for the overall reaction is the sum of the ∆H°s of the reactions added. This is true because the overall heat of a reaction is only related to the overall diﬀerence in the potential energy of the reactants and products. The heat of reaction is independent of the pathway between reactants and products. It is an example of a state function, a value that is dependent on the initial and ﬁnal states for a change and independent of the pathway between those states.

Sample Study Sheet 6.3: Open Calorimeter Problems

124

Chapter 6

Energy and Chemical Reactions

LAW OF HESS EXAMPLE The reaction of solid tin and gaseous chlorine could proceed by two diﬀerent paths. They could react in a single step reaction that goes directly to products. The ∆H° for this reaction is −545 kJ. Sn(s) + 2Cl2(g) → SnCl4(l ) They could react in a two step reaction. The sum of the ∆H°’s for these steps is equal to −545 kJ. Sn(s) + Cl2(g) → SnCl2(s)

∆H°1 = −350 kJ

SnCl2(s) + Cl2(g) → SnCl4(l )

∆H°2 = −195 kJ

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125

The following describes the steps for Law of Hess problems. STEP 1 You may not be given the complete equations for the reactions. If not, you will need to determine what they are and write them with their ∆H°s. STEP 2 Rearrange the intermediate equations so their sum equals the net equation. •

There are two general changes you might make to the intermediate equations. •

You might reverse them to get the desired formula on the correct side of the equation.

•

You might multiply each coeﬃcient by the number that gives the correct coeﬃcient for the desired formula. This could be an integer or a fraction.

•

Start with the ﬁrst formula in the net equation. •

If it is mentioned in only one intermediate equation, rearrange that equation to put the ﬁrst formula on the correct side of the equation with the correct coeﬃcient.

•

If the formula is mentioned in more than one equation, skip it until later and go on to the next formula.

•

Do the same for each formula in the net equation.

•

When you have done the above for each reactant and product in the net equation, add the intermediate equations to be sure they add up to the net equation.

•

If the intermediate equations do not add up to yield the desired net equation, you may need to use another intermediate equation to eliminate formulas not mentioned in the net equation.

STEP 3 Whatever you did for each equation, do the same to the ∆H° for the equation. •

If you had to reverse the direction of the equation, you will reverse the sign of the ∆H°.

•

If you multiplied the coeﬃcients by a whole number or fraction, you will do the same to the ∆H°.

STEP 4 Add the new ∆H°s for the intermediate equations to get the ∆H° for the net reaction.

Sample Study Sheet 6.4: Law of Hess Problems

126

Chapter 6

Energy and Chemical Reactions

EXERCISE 6.9 - Law of Hess Given the following data, ∆H°combustion C2H2(g) = −1301 kJ/mole ∆H°ombustion C2H6(g) = −1562 kJ/mole H2O(l) → H2(g) + ½O2(g)

∆H° = 286 kJ

calculate the ∆H° for the following reaction. ½C2H2(g) + H2(g) → ½C2H6(g)

127 A special application of the Law of Hess involves the use of heats of formation to calculate heats of reaction. The heat of formation, ∆Hf°, is the ∆H° for the reaction when one mole of a substance is formed from its elements in their standard states. (It is sometimes called the enthalpy of formation or the change in enthalpy of formation.) The standard state of a substance is its state at 1 atm of pressure and 298.15 K. •

The standard states of the metals are described with a single symbol. They are all solids except mercury, which is liquid.

•

The nonmetals must be memorized separately. •

Some nonmetals are diatomic: H2(g), N2(g), O2(g), F2(g), Cl2(g), Br2(l ) and I2(s).

•

The Noble gases are described with single symbols and are gaseous.

•

Other non−metals have the following formulas and states: S8(s), Se8(s), P4(s), As4(s), Sb4(s).

•

Carbon is C(graphite).

The heat of formation of an element in its standard state is zero. The overall heat of reaction is equal to the sum of the ∆H°f’s of the products minus sum of the ∆H°f’s of the reactants. ∆H°rxn = Σ ∆H°f (products) − Σ ∆H°f (reactants)

EXERCISE 6.10 - Heats of Formation Calculate the ∆H° for the following reaction. 2ZnS(s) + 3O2(g) → 2ZnO(s) + 2SO2(g)

128

Chapter 6

Energy and Chemical Reactions

Table 6.1 Heats of Formation of Some Common Inorganic Substances

Substance

State

∆Hf° kJ/mol

Substance

State

∆Hf° kJ/mol

Ag

s

0

H2O2

l

−187.6

AgCl

s

−127.04

Hg

l

0

Al

s

0

I2

s

0

AlCl3

s

−696.6

HI

g

25.94

Al2O3

s

−1669.79

Mg

s

0

Br2

l

0

MgO

s

−601.8

HBr

g

−36.23

MgCO3

s

−1112.9

C(graphite)

s

0

N2

g

0

C(diamond)

s

1.90

NH3

g

−46.25

CO

g

−110.5

NH4ClO4

s

−290.4

CO2

g

−393.5

NO

g

90.37

Ca

s

0

NO2

g

33.85

CaO

s

−635.55

N2O4

g

9.66

CaCO3

s

−1206.88

N2O

g

81.56

Cl2

g

0

O

g

249.4

HCl

g

−92.3

O2

g

0

Cu

s

0

O3

g

142.2

CuO

s

−155.23

S(rhombic)

s

0

F2

g

0

S(monoclinic)

s

0.30

HF

g

−268.61

SO2

g

−296.06

H

g

−218.2

SO3

g

−395.18

H2

g

0

H2S

g

−20.15

H2O

g

−241.83

ZnO

s

−347.98

H2O

l

−285.8

ZnS

s

−202.9

(calcite)

129 Table 6.2 Heats OF Formation for Some Common Organic Substances

Substance

Formula

State

∆Hf° kJ/mol

Acetic acid

HC2H3O2

l

−484.21

Acetaldehyde

CH3CHO

g

−246.81

Acetone

CH3COCH3

l

−246.81

Acetylene

C2H2

g

226.6

Benzene

C6H6

l

49.04

Ethanol

C2H5OH

l

−276.98

Ethane

C2H6

g

−84.68

Ethylene

C2H4

g

52.3

Formic acid

HCO2H

l

−409.20

Glucose

C6H12O6

s

−1274.45

Methane

CH4

g

−74.85

Methanol

CH3OH

l

−238.66

Sucrose

C12H22O11

s

−2221.70

130

Chapter 6

Energy Transfer and Chemical Reactions