Chemistry 152B
Name
Final Examination
March 15, 2006
___________________
Section (TA) ________________
The last page contains potentially relevant equations that may be of use to you.
You are welcome to tear that page off and keep it.
There are 13 pages, be sure you have all pages before you begin.
Useful Information
Constant
Avagdro's #
Value
N A = 6.022 ⋅ 1023 mol −1
R = 0.082A − AtmK −1mol −1
g = 9.8m - s -2
RT = 2.48kJ / mol
RT
= 25.7mV / mol
F
ln e 10 = 2.303
−11
Bohr Radius
Boltzmann
e Charge
Faraday
ao = 5.3 ⋅ 10 m
k B = 1.38 ⋅ 10−23 J / K
e − = 1.6 ⋅ 10−19 C
F = 96, 485Cmol −1
Gas Cnst
mass elec
mass prot
R = 8.314 JK −1mol −1
me = 9.1 ⋅ 10−31 kg
m p = 1.67 ⋅ 10−27 kg
Planck
light speed
h = 6.626 ⋅ 10−34 J − s
c = 2.99 ⋅ 108 ms −1
Rydberg
R y = 3.29 ⋅ 1015 Hz
Pages
Points
Gas Cnst
Gravity
1
Cnsts.
2
30
3
26
4
28
1 cal = 4.18 J
5
24
6
34
7
41
8
32
9
35
10
47
11
24
12
13
Eqns.
P.T.
Total Points: 321 (which will be rescaled to 240).
Suggestions:
1) Make sure you set up all calculation problems for the entire test, and then go back and
finish them. You will get more points covering the full test demonstrating how the
problem may be solved. (Hold off using your calculator.)
2) Some additional algebra, using variables rather than numbers, will save you time
calculating unneeded intermediate quantities.
3) If you see a problem you do not understand, skip it and come back to it at the end and
ask a question.
Q1) (9.13) (8 pts.) When is ΔH = 5 / 2 RT ? For an ideal monoatomic gas.
When is ΔE = 5 / 2 RT ? For some (ideal) polyatomic gasses.
When is ΔE = 3 / 2 RT ? For an ideal monoatomic gas.
When is ΔH = ΔE ? In terms of heat capacities, never, in general only when
Δ ( PV ) = 0
What does this say about ΔE and ΔH being state function?
Definitely distinct functions.
Q2) (9.25) (10 pts.) Consider the following reaction:
2 H 2 ( g ) + O2 ( g ) → 2 H 2 O(l ) ΔΗ = −572kJ
A) How much heat is evolved for the production of 1.00 mol of Η 2 O(l )?
572
ΔΗ = −
= −286kJ
2
B) How much heat is evolved when 166 g of oxygen is reacted with excess
hydrogen?
166
5720
ΔΗ = −572 ⋅
=−
= −2860kJ
32
2
C) The total volume of hydrogen gas needed to fill the Hindenburg was 2.0 × 10 8 L at
1.0 atm and 25°C. How much heat was evolved when the Hindenburg exploded,
assuming all of the hydrogen reacted?
PV
2.0 ⋅ 108 ⋅ 1
n=
=
= 8.2 ⋅ 106 moles
A HUGH number.
RT 298 ⋅ 0.082
9
8.2⋅106
ΔΗ = −572 ⋅ 2 = 2.3 ⋅ 10 kJ
Q3) (9.48) (6 pts.) Given the following data:
Fe 2 O3 (s ) + 3CO ( g ) → 2 Fe(s ) + 3CO2 ( g ) ΔΗ = −23kJ
3Fe2 O3 (s ) + CO (g ) → 2 Fe3 O4 (s ) + CO2 ( g ) ΔΗ = −39kJ
ΔΗ = −18kJ
Fe3 O4 (s ) + CO( g ) → 3FeO(s ) + CO2 ( g )
calculate ΔΗ for the reaction FeO(s ) + CO( g ) → Fe(s ) + CO2 ( g )
FeO ( s ) + 13 CO2 ( g ) → 13 Fe3O4 ( s ) + 13 CO ( g )
ΔΗ = 13 ⋅ 18kJ
1 Fe O s + 3 CO g → Fe s + 3 CO g
( )
( ) 2 2 ( ) ΔΗ = − 12 ⋅ 23kJ
2 3( )
2
2
1 2 Fe O s + CO g → 3Fe O s + CO g
( )} ΔΗ = 16 ⋅ 39kJ
3 4( )
2( )
2 3( )
6{
ΔΗ = 13 ⋅ 18 − 12 ⋅ 23 + 16 ⋅ 39 = 1kJ
Q4) (10.2) (6 pts.) For a liquid, which would you expect to be larger:
ΔS fusion or ΔS vaporization ? Explain why?
ΔS vaporization is larger because the change to the gas state opens up so many more possible
positions for the molecules to occupy, the change in density is around 100 fold going
from condensed to gas phase. From solid to liquid on the other hand the density changes
are modest.
2
Q5) (9.86a) (10 pts) Consider the following reaction at 1.0 atm:
CH 3Cl ( g ) + H 2 ( g ) → CH 4 ( g ) + HCl ( g ) For this reaction
Species
the enthalpy change at 25°C is −81.7kJ / mole . At constant
pressure the molar heat capacities ( CP ) for the compounds
CH 3Cl ( g )
are listed. Assume that the CP values are independent of
temperature. By how much would the enthalpy change if the
reaction were run at 325°C ?
CH 4 ( g )
HCl ( g )
H2 ( g )
CP
J
mol − K
48.5
28.9
41.3
29.1
ΔH Δ T = Δ C P Δ T
ΔT = 300 K
ΔH ΔT = {41.3 + 29.1 − 28.9 − 48.5}( 300 ) J / mol = −2.1kJ
Q6) (9.72) (6 pts) A piece of chocolate cake contains about 300 Calories. A nutritional
Calorie is 1000 calories (thermodynamic calories) and 1 cal is 4.18 Joules. How many
times must you lift (or bench-press) 50 pounds to expend the 300 Cal from the piece of
cake? Assume your arms are 2 feet long so that each lift is 0.7 meters, and that 50 pounds
is 22 kilograms.
q = w = mgh
300 ⋅ 4.18 ⋅ 103 J = N ⋅ 22 ⋅ 9.8 ⋅ 0.7 J
N=
Upper body workouts are not very aerobic
300 ⋅ 4.18 ⋅ 10
= 8 ⋅ 103 reps
22 ⋅ 9.8 ⋅ 0.7
3
Q7) (10.3) (10 pts.) Gas Α 2 reacts with gas Β 2 to form gas AB at constant temperature.
The bond energy of AB is much greater than that of either reactant. What can be said
about the
a) sign of ΔΗ ? Exothermic ΔΗ < 0
b) sign of ΔS surr ? ΔS surr = −
ΔΗ
>0
T
c) sign of ΔS ? ΔS > 0
d) Explain how potential energy changes for this process.
Because it is exothermic and DS is positive, the reaction goes in the described direction,
so the potential drops because the AB is much more stable.
e) Explain how random kinetic energy changes during the process.
No change at all because it is at fixed temperature.
3
Q8) (10.8) (6 pts.) Which is larger: ΔS at constant pressure or ΔS at constant volume?
a) Cite a supporting equation: For I.G. at constant P ΔS ∝ CP and at constant V
ΔS ∝ CV
b) Provide a conceptual rationale. So ΔS is larger at constant P because the volume
increases for processes at constant P when T goes up and this gives the molecules more
ways to arrange themselves.
Q9) (10.25) (10 pts.) One mole of an ideal gas is contained in a cylinder with a movable
piston. The temperature is constant at 77°C. The initial pressure is 4.0 Atm.
a) Weights are removed suddenly from the piston to give the following sequence of
pressures: b. P2 = 2.24 atm and c. P3 = 1.00 atm (final state)
A) What is the total work (in joules) in going from the initial to the final state by way of
the preceding two steps?
w = − Pext ΔV
PV = nRT = PV
1 1 = PV
2 2 = PV
3 3 = 1 ⋅ 8.314 ⋅ 350 = 2.9 kJ
− w = P2 (V2 − V1 ) + P3 (V3 − V2 )
⎧ P
P⎫
2.24
1. ⎫
⎧
− w = nRT ⋅ ⎨1 − 2 + 1 − 3 ⎬ = 2.9 ⎨ 2 −
−
⎬ = 2.9kJ
P2 ⎭
4
2.24 ⎭
⎩
⎩ P1
B) What would be the total work if the process were carried out reversibly?
wrev = −nRT ln
Vf
Vi
= nRT ln
Pf
Pi
= −2.9 ⋅ ln 4 = −4kJ
Q10) (10.61) (12 pts.) Consider the reaction: 2NO2 ( g ) R N 2O4 ( g ) .
For each of the following mixtures of reactants and products at 25°C, predict the
direction in which the reaction will shift to reach equilibrium.
o
o
ΔH rxn
= −58kJ / mol ; ΔS rxn
= −176 J / K − mol
a. PNO2 = PN 2O 4 = 1.0 atm.
Q
1
= 2.48 ⋅ ln
= −2.48 ⋅ 2.2kJ < 0 Spontaneous in the direction written,
K
9.4
shifts to the right (makes products).
ΔGrxn = RT ln
o
o
b) (10.71) Calculate the value of K at 25.0°C. Assuming ΔH rxn
are
and ΔS rxn
temperature-independent,
o
o
o
ΔGrxn
ΔH rxn
-T ΔS rxn
58 − 298 ⋅ 0.176
ln K = −
=−
=
; ln K = 2.24 ⇒ K = 9.4
RT
RT
2.48
c) Estimate the value of K at 100.0°C.
o
o
o
ΔGrxn
ΔH rxn
-T ΔS rxn
58 − 373 ⋅ 0.176
ln K = −
=−
=
; ln K = −2.5 ⇒ K = 0.85
RT
RT
3.1
4
Q11) (10.72) (8 pts.) Consider the van't Hoff relation:
o
o
−ΔH rxn
⎛ 1 ⎞ ΔS rxn
⋅⎜ ⎟ +
ln ( K ) =
. The equilibrium constant for some
R
R
⎝T ⎠
hypothetical process was determined as a function of temperature
(in Kelvin) with the results plotted here.
o
o
A) From the plot determine the values of ΔH rxn
and ΔS rxn
o
ΔH rxn
=−
Δ ln ( K )
0 − 40
R=−
R = 110kJ
3 ⋅ 10−3
⎛1⎞
Δ⎜ ⎟
⎝T ⎠
o
ΔS rxn
= 40 R = 325 J / K
B) What would be the major difference in the plot for an endothermic process as
compared to an exothermic process?
The line would slope the other way, but the intercept stays the same, positive slope going
up from the same intercept point.
Q12) ( 11.39) ( 6 pts.) A) Combine the equations
o
o
o
o
to derive an expression for E° as a function
ΔGrxn
= − nFE o and ΔGrxn
= ΔH rxn
− T ΔS rxn
of temperature.
o
o
o
ΔH rxn
− T ΔSrxn
ΔS o
ΔH rxn
Eo =
= rxn T −
− nF
nF
nF
B) Describe how one can graphically determine ΔΗ ° and ΔS ° from measurements of E°
at different temperatures. [Assume that ΔΗ ° and ΔS ° do not depend on temperatures.]
Plot E vs. T, and the slope gives the entropy and the intercept gives the enthalpy.
Q13) (11.50) (10 pts.) Consider the following galvanic cell at 25°C:
Pt | Cr 2 + (0.30 M), Cr 3+ (2.0 M) || Co 2+ (0.20 M) | Co
The overall reaction is 2Cr 2+ (aq ) + Co 2+ (aq) → 2Cr 3+ (aq ) + Co( s ) , and equilibrium
constant value is K = 2.79 ⋅107
A) Calculate the ΔG for the cell reaction at these conditions
o
ΔGrxn
= − RT ln K = −2.48 ⋅ ln ( 2.8 ⋅ 107 ) = −42.5kJ
B) and the cell potential, E, for the galvanic cell.
o
ΔGrxn
42.5 ⋅ 103
Eo = −
=
= 0.22V
nF
2 ⋅ 96480
5
Q14) (11.55) (8 pts.) Consider the concentration
cell shown. Calculate the cell potential at 25°C
when the concentration of Ni 2+ in compartment on
the left is 1.0M, when the compartment on the
right has the following values.
a) 1.0 M
Both sides are the same, so the battery is dead,
Ecell=0.0V
b) 2.0 M [ For part b identify the cathode, the anode, and the direction in which
electrons flow on the diagram.]
RT
25.7
ln Q = 0 −
ln 2 = 9mV Conc on right is too high, so rhs is Ni+2+2eÆNi,
E = Eo −
2
nF
consumes electrons (reduction), electrons flow from lhs to rhs. So rhs is cathode, and lhs
is anode.
Q15) (12.29) (6 pts.) It takes 208.4 kJ of energy to remove one mol of electrons from the
atoms on the surface of rubidium, metal. If rubidium metal is irradiated with 254-nm
light, what is the maximum kinetic energy the released electrons can have?
hν = 12 mv 2 + Φ
hc
λ
= K . E . + 208.4
K .E. =
hc
λ
−
208.4 ⋅ 103 6.626 ⋅ 10−34 ⋅ 2.99 ⋅ 108
208.4
=
−
J
−9
254 ⋅ 10
6.022 ⋅ 1023
NA
KE = 4.32 ⋅ 10−19 J / electron
Q16) (12.72) (8 pts.) Write the expected ground-state electrons configuration for the
following.
a. the element with one unpaired 5p electron that forms a covalent compound with
fluorine. In, or I, prefer I to make a covalent F compound.
EC ( I ) = [ Kr ] 5s 2 4d 10 5 p5
b. the (as yet undiscovered) alkaline earth metal after radium.
EC ( Ra *) = [ Rn ] 7 s 2 5 f 14 6d 10 7 p 6 8s 2
c. the noble gas with electrons occupying 4f orbitals.
EC ( Rn ) = [ Rn ] fills out the 6p
d. the first-row transition metal with the most unpaired electrons.
EC ( ? ) = [ Ar ] 4 s1 5d 5 this is Cr, with 6 unpaired electrons.
6
Q17) (12.78) (12 pts.) Which of the following electron configurations correspond to an
excited state? Identify the atoms and write the ground-state electron configuration where
appropriate.
a. 1s 2 2 s 2 3 p1 Exc EC (GS ) = 1s 2 2 s 2 2 p1
c. 1s 2 2 s 2 2 p 4 3s1 Exc EC (GS ) = 1s 2 2 s 2 2 p 5
b. 1s 2 2s 2 3 p 6 Exc EC (GS ) = 1s 2 2 s 2 2 p 6
d. [ Ar ] 4s 2 3d 5 4 p1 Exc EC (GS ) = [ Ar ] 4 s 2 3d 6
Q18) (12.90) (6 pts.) Order each of the following sets from the least exothermic electron
affinity to the most.
a. F. Cl. Br, I
EA ( F ) < EA ( Cl ) < EA ( Br ) < EA ( I )
b. N,O,F Already in Order
Already in order
EA ( N ) > 0 & EA ( O ) < EA ( F )
EA ( N ) , EA ( O ) , EA ( F )
Q19) (12.94) (6 pts.) The electron affinities of the elements from aluminum to chlorine
are -44, -120, -74, -200.4, and -348.7 kJ/mol, respectively. Rationalize the trend in these
values. As the element become closer to noble gas the EA go down, Cl is most able to
hold an electron (so has the lowest, i.e. largest negative EA). The one exception is P,
which has to add an electron to an orbital with an existing electron. Therefore there is an
up tick in the trend, but P can and will take an electron (unlike N).
Q20) (12.98) (9 pts.) For each of the following pairs of elements,
(C and N)
(Ar and Br)
(Mg and K)
Pick the one with
a) the more favorable (exothermic) electron affinity
C
Br
K
b) the higher ionization energy
N
Ar
Mg
c) the larger size
As an atom C
Br
K
(Q21) (12.101) (12 pts.) Complete and balance the equations for the following reactions.
a. 4 Li ( s ) + O2 ( g ) → 2 Li2O ( s )
c. Cs ( s ) + H 2O (l ) → CsOH + 12 H 2
b. 2K ( s ) + S ( s ) → K 2 S ( s )
d. 2 Na ( s ) + Cl2 ( g ) → 2 NaCl ( s )
Q22) (12.114) (8 pts.) A certain microwave oven delivers 750. Watts (J/s) of power to a
coffee cup containing 50.0 g of water at 25.0°C. If the wavelength of microwaves in the
oven is 9.75 cm, how long does it take, and how many photons must be absorbed, to
make the water boil? The specific heat capacity of water is 4.18 J / D C ⋅ g . Assume only
the water absorbs the energy of the microwaves.
ΔE = mC ΔT = 50 ⋅ 4.18 ⋅ 75 = 16kJ
= W Δt = 750ΔtJ ⇒ Δt = 20sec
2 ⋅ 10−25
= Nhν = N
=N
= N 2 ⋅ 10−24 J ⇒ N = 8 ⋅ 1027 photons
0.0975
λ
hc
7
Q23) (12.122) ( 10 pts.) In ground state of Strontium (Sr),
a. How many electrons have A = 2 as one of their quantum numbers?
d orbitals are A = 2 , so 10 electrons in the 3d.
b. How many electrons have n = 4 as one of their quantum numbers?
4s and 4p contribute, so that is 10 electrons.
c. How many electrons have me = −1 as one of their quantum numbers?
All p and d each have 2 electrons with –1 me QN. 2p, 3p, 4p, 3d, so 4*2=8 electrons
total.
Q24) (12.132) (8 pts.) An atom of a particular element is traveling at 1% of the speed of
light. The de Broglie wavelength is found to be 3.31x10 −3 pm. Which element is this?
Prove it. It must be Ca (or maybe K) because it has a mass of 40 nucleons (assume
protons and neutrons are the same mass).
v = 0.01c
p = mv =
h
λ
=
6.6 ⋅ 10−34
= 2 ⋅ 10−19
3.3 ⋅ 10−15
p 2 ⋅ 10−19
m= =
= 6 23 ⋅ 10−26
6
v
3 ⋅ 10
m 66.66 ⋅ 10−27
=
= 40amu or m ⋅ N A = 40 g / mole
m p 1.67 ⋅ 10−27
Q25) (12.60) (6 pts) For the hydrogen atom, the wave function for the state
3
1 ⎛ 1 ⎞2
r
−σ
2
3
n = 3, A = 0, mA = 0 is: Ψ 300 =
⎜ ⎟ ( 27 − 21σ + 2σ ) e , where σ = , and
ao
81 3π ⎝ ao ⎠
ao is the Bohr radius. Calculate the position of the nodes for this wave function.
0 = Ψ 300 = 27 − 21σ + 2σ 2 = ( 9 − σ )( 3 − 2σ )
9
4.5
2
3
4
σ = 3⇒r=
Å
Q26) (13.3) (8 pts.) Using the periodic table,
a) predict the most stable ion form of each element: Na, Mg, Al, S, Cl, K, Ca, and Ga.
Na
Mg
Al
S
Cl
K
Ca
Ga
+
3+
2−
1−
2+
2+
1+
Na
Al
S
Cl
Ca
Ga 3+
Mg
K
b) Arrange the elements, S, Cl, K, Ca, and Ga., in their stable ion form, from largest to
smallest radius and explain why the radius varies as it does.
R ( S 2 − ) > R ( Cl 1− ) > R ( K + ) > R ( Ca 2+ ) > R ( Ga 3+ )
Q27) (13.4) (6 pts.) The bond energy for the C―H bond is about 413 kJ/mol in CH 4 but
380 kJ/mol in CHBr3 . Although these values are relatively close in magnitude, they are
8
different. A) Explain why they are different. Br is a much larger atom with larger
electronegativity than H, so it crowds the space of the CH bond and pulls electrons out of
the bond, weakening the bond and making the CH distance longer.
B) Suggest why the C―H bond energy is lower in CHBr3 than it is in CH 4 . The energy is
lower for two reasons, H is croweded and has a weaker hold on C and electons that make
the bond are withdrawn some.
Q28) (13.14) (9pts.) Predict which bond in each of the following groups is the most
polar, and give a short reason for you choice:
a. C―F, Si―F, Ge―F Ge is most electropositive and gives up e' more easily so
this bond is most polar. (Trend going down a column)
b. S―F, S―Cl, S―Br S-F is most polar because (S is held constant) and F is
most electronegative of the halogens.
c. Al―Br, Ga―Br, In―Br, Tl―Br Br is held constant so Tl-Br is most polar
because going down a column elements let electrons go more easily. (Same as a).
Q29) (13.18) (10 pts.) Write electron configurations for each of the following.
Sn 2+ ,
Tl + ,
As 3+
a. the cations: Mg 2+ ,
EC ( Mg 2 + ) = [ Ne ] , EC ( Sn 2+ ) = [ Kr ] 5s 2 4d 10 , EC (Tl + ) = [ Xe ] 6 s 2 5d 10 , EC ( As 3+ ) = [ Ar ] 4 s 2 3d 10
b. the anions: EC ( N 3− ) = [ Ne ] , EC ( F − ) = [ Ne ] ,
c. the most stable ion formed by:
Ba ⇒ Ba 2 + EC ( Ba 2+ ) = [ Xe ] ,
EC ( T e 2 − ) = [ Xe ] ,
Se ⇒ Se 2− EC ( Se 2− ) = [ Kr ] ,
I ⇒ I − EC ( I − ) = [ Xe ]
Q30) (13.24) (10pts.) How does the concept of ionic bonding account for the following
properties of ionic compounds?
a) Low electrical conductivity as solids, and high conductivity in solution or when
molten. As a solid the ions cannot move, but when melted the ions can move and so
conduct electricity.
b) Relatively high melting and boiling points. The strong bonding to all the neighbors
gives very large lattice energy stabilities, so it takes high temp to break all the bonds.
c) Brittleness. The solid can be broken up by cleaving along a plane where neither side
ends up with any net charge.
9
Q31) (13.48) (12 pts) Draw a proper Lewis structure (satisfying the octet rule) for each of
the following molecule and ion. In each case the first atom listed is the central atom.
O
|
A) POCl3 Cl −
P
O
O
&
&
−Cl B) XeO4
O=
3−
Xe = O C) PO4
3−
O=
P =O
|
&
&
Cl
O
O
Q32) (13.59) (35 points 2/3/2) Write Lewis structures for the following 5 molecules,
which have central atoms that do not obey the octet rule: PF5 , BeH 2 , BH 3 , Br3− , SF4 .
PF5
BeH 2
BH 3
Br3 −
(13.71) Predict the molecular structure and the bond angles for each.
PF5
BeH 2
BH 3
Br3 −
Linear
Trigonal
bipyramidal
Trigonal
(planar)
Bent
SF4
SF4
SeeSaw
(13.78) Based on the predictions above, state whether the molecule has a net dipole
moment or if by symmetry the dipole moment must be zero.
PF5
BeH 2
BH 3
SF4
Br3 −
NO
Symmetry
forbidden
NO
NO
Yes
Could argue No
because Br-Br
bond is
covalent
YES
Not symmetry
forbidden and
SF bond is
polar
Q33) 13.112 (10 pts.) Think of forming an ionic compound as three steps (this is a
simplification, as with all models): (1) removing an electron from the metal; (2) adding
10
an electron to the nonmetal; and (3) allowing the metal cation and nonmetal anion to
come together.
a. What is the sign of the energy change for each of these three processes?
(1) Remove e from metal
(2) Add e to nonmetal
(3) Ions come together
Negative (Usually)
Very Negative
Positive
b. In general, what is the sign of the sum of the first two processes?
Positive, so it won't happen (endothermic)
c. What must be the sign of the sum of the three processes?
Negative, exothermic, only the exotermic nature will make the bond because it is
entropically unlikely.
d. In summary (of part c) why do ionic bonds occur?
The coulomb attraction of the ions overwhelms the cost of transferring the electron from
the metal to the non-metal.
Q34) )(Guest Lecture) (14 points) Some people have said that the answer to the energy
crisis in this country is to go to a hydrogen based energy source. Comment on this
suggestion in the light of what was discussed by our guest lecturer, Dr. Dudis, from the
Air Force Research Labs. Discuss the merits and drawbacks of this proposal. [If you
want more room write on the back of this page.]
The first point, there is no one answer to the energy problem; there are many
answers for different needs. The second is: Where does the energy come from
to make hydrogen? The third is that the energy density of hydrogen is not a
solved problem. If it is solved it has been solved by hydrocarbons which carry
hydrogen in very high concentrations. So you can't beat hydrocarbons for
storage and energy density of hydrogen.
The main issue is the water to hydrogen cycle. Energy must be put in (to water)
to make water go to oxygen and hydrogen. The source of that energy is critical
to the problem. If the energy were from hydrocarbons then we really are still on a
petroleum economy (or diet). If it is the sun that is used to make hydrogen then
we must take a close look at using the sun's energy to make alternative fuels,
particularly storing electrical energy. In short hydrogen is not a viable energy
source for a variety of reasons. The ultimate and only viable energy source is
the sun. How we control the energy of the sun is the critical factor, and there are
many ways to do that (the least of which is make hydrogen).
11
Relevant Equations
ΔE = q + w
w = − Pext ΔV
ΔH = ΔH rxn ΔX
ΔG = ΔH − T ΔS
K .E. = 12 mv 2
P.E. = mgh
PV = nRT
ΔGrxn = ΔG
ΔH = nCP ΔT
o
ΔGrxn
ln K = −
RT
o
ΔGrxn = ΔGrxn
+ RT ln Q
wrev = − nRT ln
ΔS = nR ln
Vf
Vi
Vf
Vi
ΔS = nCV ln
Tf
Ti
c = λν
c
d
C ] [ D]
[
K=
[ A]a [ B ]b
o
rxn
qrev
T
E = hν Photon
aA + bB R cC + dD
ΔE = nCV ΔT
ΔS =
P.E.Coul
o
−ΔH rxn
ln K =
R
elect
wrev
hν = 12 mv 2 + Φ
⎛ 1 ⎞ ΔS
⋅ ⎜ ⎟ + rxn
R
⎝T ⎠
= ΔG
RT
ln Q
nF
RT
ln K
Eo =
nF
E = Eo −
⎛ 1 1⎞
− 2⎟
2
n
⎝ 1 n2 ⎠
h
h
λ=
p=
λ
p
=
ΔxΔp ≥
2
h2 n2
En = +
8m L2
2
−18 ⎛ Z ⎞
En = −2.178 ⋅ 10 ⎜ 2 ⎟ J
⎝n ⎠
ν = Ry ⎜
+ RT ln Q
ΔGrxn = −nFE
Ze 2
=−
r
o
Δ = 125 ( EN ( A) − EN ( B ) )
μ = QR
μ2
IC % =
3 + μ2
Standard Reduction potentials for Half-Reactions in
Volts:
−
−
2+
−
o
F2 + 2e → 2 F
2.87
Cu + 2e → Cu
0.34
+
−
o
+
−
Ag + e → Ag
1.99
2 H + 2e → H 2
0.00
−
2+
−
o
Cl2 + 2e → 2Cl
1.36
Pb + 2e → Pb −0.13
+
−
O2 + 4 H + 4e → 2 H 2O 1.23
Cr +3 + e − → Cr +2
−.50
−
−
3+
−
o
Br2 + 2e → 2 Br
1.09
Al + 3e → Al
−1.66
+1
−
+
−
o
Ag + 1e → Ag
0.80
Li + e → Li
−3.05
12
2
13