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THERMO - PART 3
BOND ENERGIES IN COVALENT MOLECULES
NOTES:
(Refer to section 9.11 in text)
Covalent Bonds – bonds resulting from the sharing of electrons
between atoms.
e.g., Cl – Cl, H–C C–H, H–(C=O)–H
The distance between the atoms is the bond length (depends on the
atoms). Not all covalent bonds are equal – they involve different nuclei
and different electron distributions, (e.g, C-C, C=C, C C, etc.). Hence the
energy to break bonds can differ.
Bond Enthalpy (or Energy) (BE)
or Bond Dissociation Enthalpy (or Energy) (BDE)
- the energy required to __________ a mole of bonds in the gas phase only
(does not hold for liquids or solids). Note: The products are neutral species
NOT ions!
- is always ___________________________, since enthalpy change is
associated with breaking a covalent bond (energy must be supplied). BE
values are average values.
- is a measure of the _____________________ of a covalent bond.
BE can be defined as the enthalpy change ( HBE) for the dissociation of the
bond into its constituent atoms or free radicals in the gas phase.
e.g., H–H(g)
2H•(g); HBE = +436 kJ
i.e., to disrupt, break, pull apart, 1 mole of H–H bonds we must provide 436 kJ.
We can then say that the bond enthalpy of H2 is 436 kJ.
Note:
HBE = BE(H–H) = +436 kJ
2 Hf [H(g)]
Hf [H(g)]
Also known as ____________________________________
(Generally, BE are not as exact as Hf values.)
What about F2(g), Cl2(g), Br2(g), N2(g), O2(g)?
No Problem!
Cl2(g)
2 Cl(g), HBE = 242 kJ/mol
O2(g)
2 O(g), HBE = 496 kJ/mol
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NOTES:
What about an O–H bond?
H2O(g)
2H(g) + O(g); HRxn = 926 kJ
ONE mole of O–H bonds would requires ________________________________
What about C–H? (no stable C–H molecule)
CH4(g)
C(g) + 4H(g), HRxn = +1656 kJ
i.e., to break FOUR moles of C–H bonds cost 1656 kJ, therefore bond enthalpy of
C–H =
Using BE’s ( HBE) and their Relationship to Hf
EXAMPLE: Estimate the Hf for NH3 given HBE (N2) is 945 kJ,
HBE (H2) is
435 kJ, and HBE (NH) is 390 kJ. (Aside: Table 9–5 provides several values.)
Strategy: What equation do we want?
What we need to do is break the N–N & H–H bonds and reform 3(N–H) bonds.
? =
Hf (NH3) =
H1 + H2
? = [½BE(N–N) + 3/2BE(H–H)] + [– 3BE(N–H)]
? =
? = – 45 kJ/mol (compare Hf (NH3) exact = – 46.11 kJ)
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GENERALISATION:
HRxn = (energy added to break bonds) + (energy released (–) when bonds formed)
OR
HRxn = BEReactant Bonds – BEProduct Bonds
_____________________________________________________________
NOTES:
ASIDE: Same question but with different data:
Estimate the Hf for NH3 given:
Rxns for values given:
Hf = 472 kJ/mol; Rxn:
Hf = 218 kJ/mol; Rxn:
NH3(g) N(g) + 3H(g);
N(g) Hf = 472 kJ/mol
H(g) Hf = 218 kJ/mol
BE(N–H) = HBE = 390 kJ/mol
HRxn =
Hf (NH3) = – 44 kJ/mol
___________________________________________________________________________________________________
What about C–C? (there are no C–C molecules!)
Given:
H f {C2H6(g)} = –85 kJ,
BE of C–H = +415 kJ,
BE for H2(g) = +435 kJ,
H Atomisation {C(s)} = +720 kJ
Calculate a value for the C–C bond enthalpy.
(Ans: BE (C–C )
340 kJ/mol)
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NOTE: Bond Dissociation Enthalpy calculations are for GASES ONLY
(therefore all reactants AND products must be in the gaseous state).
NOTES:
We can build up a table of bond enthalpies to use (refer to Table 9–5, page 361).
Typical Average Bond Energies (BE, HBE)
C–H
C–C
C–N
C–O
C–F
C–I
411
346
305
358
485
213
O–H
H–H
etc.
459
432
C=C
C C
C=N
C=O
N=N
N N
602
835
615
745
418
942
For benzene, resonance causes the CC bond enthalpy to be 518 (~ 1 bond + ½).
Note also that 1 C C ≠ 3 C–C. Bond energies are useful in determining weak
points in a molecule.
e.g., When CFC–11 is exposed to UV radiation in the stratosphere, the C–Cl bond
(BE ~ 338 kJ/mol) breaks not the C–F bond (BE ~ 484 kJ/mol).
CFCl3(g)
CFCl2(g) + Cl(g)
then, Cl(g) + O3(g) OCl(g) + O2(g)
Bond Lengths and Bond Enthalpies:
Bond
Bond length
Bond enthalpy
C–N
1.47Å or 147 pm
305 kJ
C=N
1.38Å or 138 pm
615 kJ
C N
1.16Å or 116 pm
887 kJ
In comparing single, double and triple bonds, we see that the bond length grows
___________________ while the bond enthalpy gets larger.
Problems:
1. Given the following, determine a value for the N–N bond enthalpy in
hydrazine, N2H4(g) (a rocket fuel which is discussed on p. 251 (p. 253) in the text).
Heat of formation of liquid hydrazine is +51 kJ mol– 1
Heat of vaporisation of liquid hydrazine is +48 kJ mol–1
Bond Energies: N N is +942 kJ
H–H is +432 kJ
N–H is +386 kJ
Answer: Refer to Table 9–5, pg 361.
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2. Given the following (kJ) data, estimate a value for the CS bond enthalpy NOTES:
in CS2(g). [CS2 is isostructural with CO2].
H(combustion) CS2(l) –1108, C(s) –393, S(s) –297;
H(atomisation) C(s) +711, S(s) +280 kJ
H(vaporisation) CS2(l) +27 kJ
Ans: 562 kJ
IONIC BONDING and LATTICE ENERGIES
Refer to section 9.1 (Energy Involved in Ionic Bonding – p. 331) in text
Ionic Bonds
– these result from a complete transfer of an electron
– formation of ionic solids from elements is an exothermic process.
e.g., Na + Cl Na+ + Cl–
NaCl(s)
Let’s say we want to make NaCl(s) starting from its constituent elements.
5 STEP PROCESS:
1. Vaporise Na metal:
2. Ionise Na(g):
Na+(g) + e–;
Na(g)
H2 = 496 kJ (
1)
3. Dissociate Cl2(g) into atoms:
4. Add electrons to Cl(g):
Cl(g) + e–
Cl– (g);
H4 = –349 kJ (EGE = EA)
5. Bring Na+(g) & Cl–(g) together:
(Energy is released in this process.)
Step 5 is the important step (energy wise) in forming the ionic solid and is known as
the REVERSE of the LATTICE ENERGY for NaCl ( H5 = – HLE )
“LATTICE ENERGY” is defined (in Ebbing & Gammon) as the change in
energy that occurs when an _______________________ is separated into
isolated gaseous _________________________.
e.g., NaCl(s)
Na+(g) + Cl– (g); HLE = +781 kJ
(Ebbing & Gammon uses the symbol “U” to represent the lattice enthalpy.)
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__________________________________________________________________
ASIDE: Some texts define lattice enthalpy as the energy RELEASED when separate ions
NOTES:
come together to form one mole of an ionic crystalline solid (values are negative).
“ELECTRON AFFINITY” is defined (in Ebbing & Gammon) as the energy change for
the process of adding an electron to a neutral atom in the gaseous state to form a
negative ion. This definition is also known as the “electron-gain enthalpy”.
Refer to Table 8.4 in text.
NOTE: Some texts define electron affinity as the energy RELEASED when an electron is
added (therefore the values are negative of the values from other texts).
____________________________________________________________________________________________________
So, let’s add up our 5 steps and cancel out any intermediates:
Total of steps 1
5
1. Sublimation of Na:
Na(s)
Na(g);
+
2. First Ionization Energy of Na(g):
Na(g)
Na+(g) + e–;
+
3. One-half the Bond Enthalpy for Cl2(g): ½Cl2(g)
+
4. Electron-gain Enthalpy for Cl(g):
+ 5. Reverse Lattice Enthalpy:
(Energy is released in this process.)
–
Cl(g) + e
Na+(g) + Cl–(g)
H1 = 108 kJ
Cl(g);
–
Cl (g);
H2 = 496 kJ
H3 = 120 kJ
H4 = –349 kJ
NaCl(s); H5 = –781 kJ
Therefore,
H1 + H2 + H3 + H4 + H5
( Hsubl) + (IE1) + (½ HBE) +(
1
) + (– HLE)
= – 406 kJ/mol ( Hf {NaCl(s)}
This is a specific example of the use of Hess’ Law known as a
BORN-HABER CYCLE (refer to see Figure 9-2).
Magnitude of Lattice Enthalpies
- depends on the ________________________________ of the ions and their
______________________
- Lattice Enthalpy is dependent upon Coulomb’s law (i.e., electrostatic
interaction)
V = {k × Q1 × Q2 }/ r
 k is a physical constant, equal to 8.99×109 Jm/C2
 Q's are the charges on the ions, and
 r is the _____________________ between the ions.
If the charges on the ions are of opposite sign, energy is released
(exothermic) when the two ions are brought together to form an ionic
compound. If the charges on the ions have the same sign, then energy is
absorbed (endothermic). The smaller the value of r, the bigger is V.
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Lattice Enthalpies in kJ/mol
LiF 1046 ....
NaF 929 ....
KF
826 ....
LiI
NaI
KI
NOTES:
759
700
645
MgF2 2961 .... MgI2 2327
MgO 3850 .... BaO 3114
MgS 3406 .... BaS 2832
Due to the electrostatic interactions, lattice enthalpies are
___________________ for short M+X– separations and larger ionic charges.
EXAMPLE:
Determine the lattice enthalpy of magnesium oxide from the following information:
Enthalpy of formation for Mg(g):
+ 147 kJ/mol
st
1 ionization energy for Mg(g):
+ 738 kJ/mol
nd
2 ionization energy for Mg(g):
+1451 kJ/mol
enthalpy of formation for O(g):
+ 249 kJ/mol
electron-gain enthalpy for O(g):
– 141 kJ/mol
–
electron-gain enthalpy for O (g):
+ 917 kJ/mol
enthalpy of formation for MgO(s): – 601 kJ/mol
PROBLEMS:
1. Compare CaO and BaO. Which one has the more positive lattice enthalpy?
2. Calculate the lattice enthalpy of calcium sulphide from the following information:
enthalpy of formation for Ca(g):
+ 178 kJ/mol
first ionization energy for Ca(g):
+ 590 kJ/mol
second ionization energy for Ca(g):
+1145 kJ/mol
enthalpy of formation for S(g):
+ 279 kJ/mol
electron-gain enthalpy for S(g):
– 200 kJ/mol
electron-gain enthalpy for S–(g):
+ 532 kJ/mol
enthalpy of formation for CaS(s):
– 482 kJ/mol
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3. Calculate the lattice enthalpy of calcium bromide. Here is some information to
help you complete this calculation:
enthalpy of formation of CaBr2:
– 683 kJ/mole
bond enthalpy of Br2(g):
+ 112 kJ/mole
electron-gain enthalpy for Br(g):
– 325 kJ/mol
NOTES:
4. Place the following compounds in order of increasing Lattice Enthalpies?
A) LiCl, NaCl, BeCl2
B) MgO, MgF2, MgBr2
C) MgCl2, BeO, BeCl2
Answers:
2.
3.
4.
1. CaO since Ca has a smaller radius than Ba.
3006 kJ/mol
2089 kJ/mol
NaCl < LiCl < BeCl2; MgBr2 < MgF2 < MgO;
MgCl2 < BeCl2 < BeO
ENTROPY
(Refer to Chapter 18/chpt 19 in 8th edition )
Is there a thermodynamic function that will indicate whether or not a
reaction will occur spontaneously?
By “spontaneously“ we mean that the reaction will occur on its own
with no external intervention.
Systems spontaneously change in such a manner as to ______________
their capacity for change, i.e., approach a state of equilibrium,
e.g., gases mixing, solutes dissolving, ice melting and cards shuffled.
The REVERSE of these processes is NOT spontaneous,
i.e., they are non–spontaneous or irreversible. Non–spontaneous
processes require external work to make the process occur.
e.g., electrolysis of water: H2O(l)
H2(g) + ½O2(g)
None of these above processes are forbidden by the 1 st Law – energy
is still conserved. In some spontaneous processes H is:
Positive, e.g., H2O(s)
NaCl(s)
H2O(l);
Na+(aq) + Cl–(aq);
Negative, e.g., 4Fe(s) + 3O2(g)
2H2(g) + O2(g)
H
H
= + 6 kJ
Rxn = +3.9 kJ
Rxn
2Fe2O3(s); H Rxn = – 1648 kJ
2H2O(l);
H Rxn = –570 kJ
Zero, e.g., mixing of 2 gases or expansion of a gas into a vacuum
Consideration of H alone is not enough. The spontaneous processes are
accompanied by an increase in disorder or randomness in the system, and
this leads to the idea of some other property.
Let’s look at heat flow: HOT
COLD
Heat is the product of two terms:
1) an _________________ factor, i.e., T
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2) a quantity (or capacity) factor i.e., an extensive factor
NOTES:
The second term is known as ________________________
Heat (q) = T S OR
S = qRev/T, where
S has units of J K–1 OR J mol– 1 K– 1
qRev – refers to heat supplied ________________________.
(Reversible implies that the system can be restored to the original
state by exactly reversing the change, e.g., melting & freezing)
T – refers to the temperature at which the transfer takes place.
Entropy (S) – a STATE function (independent of the path).
a quantitative measure of __________________ or disorder of a system.
measures _________________________ or number of equivalent
arrangements of the same energy.
is dependent upon the size of the system, i.e., extensive
DISORDER can be of two types:
1) Positional – distribution of species in ________________
e.g., mixing of two gases or the expansion of a gas into a vacuum
2) Thermal – distribution of _________________ among species, or
distribution of species over energy levels.
e.g., heat flow from hot to cold objects
Second Law of Thermodynamics:
In any spontaneous process,
the TOTAL ENTROPY of the UNIVERSE MUST INCREASE
STOTAL = Suniverse = Ssystem + Ssurroundings > 0
i.e. the Universe tends towards a more RANDOM state (e.g., your room!).
e.g., 2Mg(s) + O2(g)
2MgO(s); H
Rxn
= –1203 kJ (highly exothermic!)
This reaction goes from solid and gas to only solid, i.e., an increase in order.
However, the surroundings become highly disorganized due to the heat
released.
So we can have THREE scenerios:
1) Spontaneous processes, Suniv ____________
2) Processes at equilibrium, Suniv ___________
3) Non–spontaneous processes, Suniv ________
(spontaneous in the reverse direction)
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NOTES:
ENTROPY TRENDS:
For any substance entropy rises with temperature. This is due to:
– more atomic/molecular motions and,
– more disorder
( H does not vary much with temperature, but S does!)
However,
When a perfect crystalline substance is at
absolute zero (T=0 K), S = 0.
This is the 3rd Law of Thermodynamics.
Since we can define a situation where S = 0, then absolute values of S can be
determined (see Appendix C), unlike H and U where only changes can be
measured. S values are ALWAYS _____________________ and for stable forms
of pure elements at 298K and 1 atm are __________________________ (unlike
H f values of elements).
Phase
Solid
Solid
Liquid
Liquid
Gas (1 atm)
Gas (1 atm)
Gas (1 atm)
S (J)
0
59.0
67.2
94.2
169.8
192.9
205.1
T
0
54.4
54.4
90.1
90.1
200.0
298.2
We can define a STANDARD MOLAR ENTROPY, Sm
It is important that you realize what the means in terms such as Sm . It means
that for gases we have a pressure of 1 atm and for solutes we have 1 M
concentrations. Liquids and solids are in their "pure" or "normal" state. NO
mention of temperature – the convention is 25 C
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NOTES:
Entropies, unlike enthalpies, are strongly dependent on pressure
(gases) and concentration (solutes).
Typical Sm values:
e.g. H2(g) 130.57; F2(g) 202.67; Br2(l) 152.23; I2(s) 116.14 J K–1 mol–1
Factors affecting Sm
1. Sm ____________________ with melting and vaporization
Sm (s) < Sm (l) < Sm (g)
e.g., for water 47.5 < 69.9 < 188.7 J K–1 mol–1
Note: Sm (g) decreases with increasing pressure.
2. Sm ____________________ when a solid or liquid is dissolved in a solvent.
e.g., NaCl(s) vs. NaCl(aq)
72.4 < 115.4 J K–1 mol–1
3. Sm ____________________ when a gas is dissolved in a solvent.
e.g., CO2(g) vs. CO2(aq)
214 > 121 J K–1 mol–1
For similar substances in the same physical state,
the following factors are useful to know
4. Sm _____________________ with increasing mass (energy levels are more
closely spaced as mass increases).
He(g) vs. Ne(g) vs. Ar(g)
126 < 146 < 155 J K–1 mol–1
5. Sm _______________________ with increasing number of atoms in the
molecule (increasing complexity, numbers of conformations, etc.)
CH4(g) vs. C2H6(g) vs. C3H8(g) vs. n–C4H10(g)
186
<
230
< 270
< 310 J K–1 mol–1
6. Sm _______________________ with increasing molecular freedom or
flexibility.
cyclobutane C4H8(g) vs. n–C4H10(g)
265
<
310 J K–1 mol–1
36
QUESTION:
NOTES:
Which substance in each pair has the higher molar entropy?
(a) CH4(g) versus C3H8(g)
(b) KCl (aq) versus KCl (s)
(c) O2(g, 2 atm, 300 K) versus O2(g, 1 atm, 300 K)
(d) N2(g, 1 atm, 273 K) versus N2(g, 1 atm, 450 K)
Examples: Predict the SIGN of S
1. 2H2(g) + O2(g)
2H2O(l)
Answer: S is NEGATIVE – why?
2. 3H2(g) + N2(g)
2NH3(g)
Answer:
3. H2O(l)
H2O(g)
4. CaCO3(s)
5. O2(g)
Answer:
CaO(s) + CO2(g) Answer:
O2(aq)
Answer:
USING TABULATED Sm VALUES TO
CALCULATE ENTROPY CHANGES
For any reaction we can write
SRxn =
n Sm (Products) –
n Sm (Reactants)
and we can use data from tables of Sm to calculate S
Refer to data in Appendix C.
rxn.
Remember: Sm values for elements in their most stable forms at 298 K, 1 atm are
NOT ZERO (unlike Hf ) and are ALWAYS __________________________.
BUT note that Sm
(aq)) = 0, by definition
37
___________________________________________________________________
NOTES:
Problems:
1. Hydrazine and hydrogen peroxide are used as rocket fuel. Using tabulated data,
calculate S for the reaction:
N2H4(l) + 2H2O2(l)
N2(g) + 4H2O(g), H = –642 kJ mol–1
S =
=
= +607 J K–1 mol–1
{large and positive as you would have predicted qualitatively}
2. Calculate the entropy changes associated with
(a) the vaporization of 1 mole of liquid water.
(b) the combustion of 1 mole methane.
(a) Chemical equation is:
S =
=
= 118.81 J K–1 (+ve, as expected)
(b) Chemical equation is:
S =
=
= –242.76 J K–1 (–ve, as expected)
EXAMPLE TO CONTEMPLATE:
For, 2H2O(l)
2H2(g) + O2(g) we predict S to be positive
(0 moles gas 3 moles gas)
SRxn =
=
=
38
What about the Reverse Reaction?
2H2(g) + O2(g)
NOTES:
2H2O(g), SRxn = –88.8 J/K
… but we know the system goes
Entropy-wise this reaction is not favourable for the system.
What about the surroundings?
We need to know SSurroundings (b/c SUniverse =
SSurroundings + SSystem > 0)
we could get this from SSurroundings = qSurroundings/T
we can figure out the heat given off in the reaction:
Rxn
= 2 Hf H2O, g) – [2 Hf H2, g) + Hf
2,
g)]
=
= –483.6 kJ
Since
= –
then
Surroundings = + 483.6 kJ
SSurroundings = +485.6 kJ/298K = 1620 J/K
SUniverse = SSurroundings + SSystem = –88.8 + 1620 = 1531.2 J/K
So, although the system entropy is BAD, SUniverse is GOOD.
System
Surroundings,
____________________________________________________________________________________________________
ENTROPY CHANGES FOR PHASE CHANGES
(e.g., melting and boiling)
Since S = qRev/T, then ______________________
(i.e., S fus = H fus /Tmelt & S vap = H vap /TBoil )
(At the melting and boiling points the system is at EQUILIBRIUM.)
can therefore calculate S and T from H values for these
processes, e.g., Tmelt = H fus(J/mol)/ S fus(J/K mol)
this means the mpt’s increase with increasing H fus and with
decreasing S fus, i.e., the greater the intermolecular attractions
and the smaller the increase in disorder, the higher the mpt.
Example:
Pentanes
n-pentane
CH3CH2CH2CH2CH3
neo-pentane
CH3
H3C–C–CH3
CH3
H
fus
S
fus
Tmelt
(kJ/mol)
(J/K mol)
8.4
58.7
143 K
3.3
12.7
256 K
39
EXAMPLE:
With access to data tables ( Hf & S values estimate
a) the boiling point of liquid bromine at 1 atm and
b) the melting point of ice at 1 atm.
(a) when bromine boils we have the equilibrium: Br2(l)
Br2(g)
Hvaporization =
= _______ – ______ = 30.9 kJ
Svaporization =
= _______ – ___________ = 93.2 J K–1 = 9.32 10–2 kJ K–1
TBoil = qRev/ S = Hvap/ Svap
=
= __________ = 59 C [actual value is 58.78 C]
(b) for melting ice: H2O(s)
H2O(l)
TMelt = Hfusion / Sfusion
=
= ________ K = ______ C [cool!]
To write enthalpy of formation thermochemical equations, one must know the reference forms
of the elements. The reference forms, or the most stable forms (physical state and allotrope) of
the elements under standard thermodynamic conditions, are:
H2(
He(
g)
Li(s
g)
Be(s
)
)
Na(
Mg(
s)
s)
K(s
Ca(s
Sc(
Ti(
)
)
s)
s)
Rb(
Sr(s
Y(s
Zr(
Mn(
Fe(
Co(
Ni(
Cu(
Zn(
B(s)
C(s)
N2(
O2(
F2(g
Ne(
g)
g)
)
g)
Al(
Si(s
P4(s
S8(s
Cl2(
Ar(
s)
)
)
)
g)
g)
Ga(
Ge(
As(
Se(
Br2(
Kr(
V(s)
Cr(s
)
s)
s)
s)
s)
s)
s)
s)
s)
s)
s)
l)
g)
Nb(
Mo(
Tc(s
Ru(
Rh(
Pd(
Ag(
Cd(
In(s
Sn(
Sb(
Te(
I2(s)
Xe(
s)
)
)
s)
s)
s)
)
s)
s)
s)
s)
s)
)
s)
s)
s)
Cs(
Ba(s
La(
Hf(
Ta(
W(s
Re(s
Os(
Ir(s)
Pt(s
Au(
Hg(
Tl(s
Pb(
Bi(s
Po(
At2(
Rn(
g)
s)
)
s)
s)
s)
)
)
s)
)
s)
l)
)
s)
)
s)
s)
g)
Note the following reference forms: Carbon - C(s, graphite); Phosphorus - P4(s, white); Sulphur - S8(s, rhombic)
Clicker Questions
Qu#1: Which of the following rxns has H Rxn equal to the bond enthalpy of the bond
that is broken?
A. HCl (g)
H +(aq) + Cl –(aq)
B. HCl (g)
H +(g) + Cl –(g)
C. HCl (g)
H (g) + Cl (g)
40
Qu#2: Given the following average bond energies: 432 (H–H); 154 (F–F); 565 (H–F) kJ/mol
Estimate H for: H2(g) + F2(g)
2HF(g).
A) −21 kJ
B) 21 kJ
C) 544 kJ
D) –544 kJ
E) 1151 kJ
Qu#3: Which molecule has the lowest nitrogen-nitrogen bond energy ?
A) N2H4
B) N2F2
C) N2
D) All of the bonds have the same bond energy since they all involve two nitrogens.
Qu#4: Consider the reaction: 2HCl(g) + 185 kJ
H2(g) + Cl2(g)
Which of the following is true?
A) The chemical bonds in the products are weaker than those in the reactants.
B) If thermally isolated, the reaction mixture will get cooler as the reaction proceeds.
C) The HCl(g) has a negative heat of formation.
D) The reaction is endothermic.
E) All of these are true.
Qu#5: Which of the following ionic compounds has the largest lattice energy?
A) CsI
B) NaCl
C) LiF
D) CsF
E) MgO
Qu#6: Which of the following processes would be expected to have the SMALLEST increase
in S°rxn per mole of benzene?
A) C6H6(s)
C6H6(l)
D) C6H6(l) + O2(g)
6CO(g) + 3H2O(g)
B) C6H6(s)
C6H6(g)
E) C6H6(l) + O2(g)
6CO2(g) + 3H2O(g)
C) C6H6(l) + Br2(l)
C6H5Br(l) + HBr(g)
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