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Transcript
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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109
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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 6.6
Heat Transfer
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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