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University of Washington
Department of Chemistry
Chemistry 355
Spring Quarter 2003
Homework Assignment 2: Due no later than 500 pm on 4/14
Text Problems: 2.4, 2.8, 2.10, 2.11, 2.17
2.4)
Carbohydrate: Select sucrose…
C12 H 22 O11 + 12O2 → 12CO2 + 11H 2 O .
Now take the heat of combustion for sucrose from Table 2-1. This is the heat released per
mole of sucrose. The problem wants the heat released per mole of oxygen. For each mole
of sucrose that reacts, 12 moles of oxygen react.
Heat released =5647 kJ/12moles=470kJ/mole oxygen.
Fatty Acid: Select palmitic acid…
C16 H 32 O2 + 23O2 → 16CO2 + 16 H 2 O .
Use the same method as withsucrose…
From the data in Table 2-1…
Heat released=10,035kJ/23 moles oxygen=436kJ/mole oxygen.
Use a peptide with formula C4.3H6.6NO. This “peptide” would have a formula weight
88.2g.
C4.3 H 6.6 NO + 5.45O2 → 4.3CO2 + 3.3H 2 O + 0.5 N 2
22 kJ / g 88.2 g / mole
Heat Released=
= 360kJ / mole oxygen
5.45
Note we could multiply the chemical equation through by 100 to eliminate the fractional
stoichiometric
coefficients… C430 H 660 N 100 O100 + 545O2 → 430CO2 + 330 H 2 O + 50 N 2 …and the
molecular weight would be 8,820g/mole. Then the heat released per mole oxygen is
22 kJ / g 8,820 g / mole
Heat Released=
= 360kJ / mole oxygen
545
2.8)
b
gb
b
gb
g
g
a)
C6 H12 O6 s → 2C2 H5 OH
bg
b g + 2CO b gg
2
b gh
c
∆H cC H OH b gh = −276.98kJ / mole
∆H cCO b g gh = −39351
. kJ / mole
∆H = 2 ∆H cCO b g gh + 2 ∆H cC H OH b gh − ∆H cC H O b sgh
= b2molesgb −276.98kJ / moleg + b2molesgb −39351
. kJ / moleg − b1molegb−1274.45kJ / moleg
∆H C6 H12 O6 s = −1274.45kJ / mole
0
f
0
f
2
5
0
f
2
0
f
2
0
f
2
0
f
5
6
12
6
= −1340.98kJ + 1274.55kJ = −66.53kJ
b) Glucose:
C6 H12 O6 s + 6O2 g → 6CO2 g + 6 H 2 O
bg
bg
bg
bg
∆H cC H O b sgh = −1274.45kJ / mole
∆H cO b g gh = 0
∆H cCO b g gh = −39351
. kJ / mole
∆H c H Ob gh = −28584
. kJ / mole
∆H = 6∆H cCO b g gh + 6∆H c H Ob gh − ∆H cC H O b sgh
= b6molesgb −39351
. kJ / moleg + b6molesgb−28584
. kJ / moleg − b1molegb −1274.45kJ / moleg
0
f
6
0
f
2
0
f
0
f
12
6
2
2
0
f
2
0
f
0
f
2
6
12
6
= −236106
. kJ − 1715.04 kJ + 1274.45kJ = −280165
. kJ
Ethanol:
C2 H5 OH
b g + 3O b gg → 2CO b gg + 3H Ob g
∆H cC H OH b gh = −276.98kJ / mole
∆H cO b g gh = 0
∆H cCO b g gh = −39351
. kJ / mole
∆H c H Ob gh = −28584
. kJ / mole
∆H = 2 ∆H cCO b g gh + 3∆H c H Ob gh − ∆H cC H OH b gh
= b2molesgb −39351
. kJ / moleg + b3molesgb−28584
. kJ / moleg − b −276.98kJ / moleg
0
f
2
0
f
2
0
f
0
f
2
2
2
5
2
2
0
f
2
0
f
2
0
f
2
5
= −1371kJ
The text calculates the enthalpy change for converting two moles of ethanol to carbon
dioxide and water. The answer in the text is thus twice the answer above. Either answer is
acceptable.
2.10)
Heat Capacity = (4.184 J / g ⋅ K ) × (1000 g / kg ) = 4184 J / kg ⋅ K
b
g
∆T in deg rees Celsius or K = 105F − 98.6 F )(1 deg reeCelsius / 18
. deg reeFahrenheit ) = 356
. Celsius(orK )
b
g
b
gb
Heat Input = Heat Capacity × Temp. Rise ⇒ 5kJ / kg ⋅ hr × time = 4.184 kJ / kg ⋅ K 356
. K
Solving... time ≈ 3hr .
2.11)
3/ 2
2
5/ 2
a) ∆H = 386
. 1 + 1991
.
1 − 3.038 1 + 1019
.
1 = 3832
. kJ / mole
d ∆H
3
5
m1/ 2 − 2 3.038 m + 1019
m 3/ 2
= 386
. + 1991
.
.
dm
2
2
b)
d ∆H
3
5
1/ 2
3/ 2
at m = 0...
= 386
. + 1991
.
0 − 2 3.038 0 + 1019
.
0
= 386
. kJ / mole
dm
2
2
d ∆H
3
5
1/ 2
3/ 2
c) m = 1...
= 386
. + 1991
.
1 − 2 3.038 1 + 1019
.
1
= 3.32 kJ / mole
dm
2
2
bg
bg
bg
bg
b g
b g
b g b g
b g
b gb g b gb g b gb g
b g
b gb g b gb g b gb g
2.17 a) Use any reaction...Here is one example:
Nitric acid is used to make many products, including fertilizers, dyes, and explosives.
The first step in the industrial production is the oxidation of ammonia:
4NH3(g) + 5O2(g) → 4NO(g) + 6H2O(g)
Sub-equation 1:
2N2(g) + 6H2(g) → 4NH3(g); ∆H0f = 4(-46.19 kJ/mol)
Sub-equation 2:
2N2(g) + 2O2(g) → 4NO(g); ∆H0f = 4(90.37 kJ/mol)
Sub-equation 3:
6H2(g) + 3O2(g) → 6H2O(g); ∆H0f = 6(-241.83 kJ/mol)
Now, reverse equation 1:
4NH3(g) → 2N2(g) + 6H2(g); ∆H0f = -4(-46.19 kJ/mol)
2N2(g) + 2O2(g) → 4NO(g); ∆H0f = 4(90.37 kJ/mol)
6H2(g) + 3O2(g) → 6H2O(g); ∆H0f = 6(-241.83 kJ/mol)
Addition of these reactions gives the balanced reaction above.
g
∆H0f (reaction) = ∆H0f (products) – ∆H0f (reactants)
[4(90.37 kJ/mol) + 6(-241.83 kJ/mol)]-[-4(-46.19 kJ/mol)] = -1274 kJ/mol
b) C0p (H2O) = 33.58 J/K*mol, C0p (NO) = 28.96 J/K*mol, C0p (O2) = 29.36 J/K*mol
C0p (NH3) = 35.66 J/K*mol
∆C0p = C0p (products) - C0p (reactants)
∆C0p = [4(33.58) + 6(28.96)] – [4(35.66) + 5(29.36)] = 24.04 J/K*mol = 0.024 kJ/K*mol
∆H0373K = ∆H0298K + ∆C0p(∆T) = 1274 kJ/mol+ 0.024 kJ/K*mol(373 K – 298 K) = 1275.8
kJ/mol
Supplementary Problems:
1) In lecture we discussed how the energy ladder is a good model for atomic
vibrational energies. We also discussed how energy level spacing is large
compared to translational or rotational motions. Assume the energy of the nth
vibrational level is En = nhν where h is Planck’s constant and ν is the frequency
of the bond vibration. Assume ν=1013 sec-1.
a) Calculate the energy level spacing typical of an atomic vibrational motion.
That is calculate ∆E = En +1 − En .
Solution.
∆E = En +1 − En = hν ( n + 1) − hν ( n ) = hν
Assume ν ≈ 1013 sec −1 …
hν = ( 6.62 × 10−34 J ⋅ s )(1013 sec −1 ) = 6.62 × 10−21 J
k BT
at T=10K. Is it
2
greater or smaller? Now calculate the probabilities P0 and P1. Are they
close in magnitude?
Solution:
k BT
= ( 0.5 ) (1.38 × 10−23 J / K ) (10 K ) = 0.69 × 10−22 J = 6.9 × 10−23 J
2
kT
Therefore B
∆E .
2
b) Compare the vibrational energy level spacing to
⎧ 6.62 × 10−21 J ⎫
−95.94
e −ε / kBT = exp ⎨ −
≈ 0.0
⎬=e
−23
6.9
10
J
×
⎩
⎭
1
q=
≈ 1 ⇒ P0 ≈ 1 ⇒ P1 ≈ 0
1 − e − ε / k BT
c) Repeat the calculations in part b for T=1000K.
⎧ 6.62 × 10−21 J ⎫
= e −0.48 ≈ 0.619
e −ε / kBT = exp ⎨ −
−21 ⎬
⎩ 2 × 6.9 × 10 J ⎭
1
1
=
= 2.62
− ε / k BT
1− e
1 − 0.619
1
0.619
∴ P0 ≈
= 0.382 and P1 ≈
= 0.236
2.62
2.62
q=
d) Suppose you have a crystal composed of atoms arranged within a regular
lattice. Each atom can vibrate in three dimensions. According to the
equipartition principle, what is the average vibrational energy of each
atom at T=10K? What is the vibrational energy at T=1000K?
Solution: As mentioned in lecture the energy of a harmonic oscillator is
E = K + U = 12 mv 2 + 12 κ x 2 and so there are two degrees of energetic freedom
per dimension or six degrees of freedom total. So from the equipartition
⎛k T ⎞
theorem the average energy is ( 6 ) ⎜ B ⎟ = 3k BT .
⎝ 2 ⎠
∴ At T = 10 K , 3k BT = ( 3) (1.38 × 10−23 J / K ) (10 K ) = 4.14 × 10−22 J
At T = 1000 K , 3k BT = ( 3) (1.38 × 10−23 J / K ) (1000 K ) = 4.14 × 10−20 J
e) The heat capacity of a substance is the change in internal energy resulting
⎛ ∂U ⎞
from a unit change in temperature at constant volume, i.e. CV = ⎜
⎟ .
⎝ ∂T ⎠V
What is the heat capacity for the crystal in part d?
⎛ ∂U ⎞
−23
Solution: CV = ⎜
⎟ = 3k B = 3.39 × 10 J / K or per moles 3R=24.93J/K-mole.
⎝ ∂T ⎠V
f) Do you expect your answer in part e to be greater or lesser than the
experimental heat capacity for a crystal at T=10K. Explain. If the
measured value is different at T=10K from the value predicted in part e,
will the difference get greater or smaller as temperature is increased?
Explain
Answer: Less. The equipartition heat capacity from part e assumes all levels are
accessible statistically. This is not the case at T=10K so the heat capacity will be less than
predicted by equipartition.
2) Use the formula for the partition function of a ladder of energies, the expression for the
⎛ ∂U ⎞
2 ∂ ln q
,
heat capacity CV = ⎜
⎟ and the expression for the average energy E = k BT
∂T
⎝ ∂T ⎠V
calculate the heat capacity of a system at T=100K and using the energy level spacing as
ε = 10−20 J .
Solution:
3)A good yield of photosynthesis for agricultural crops in bright sunlight is 20 kg of
carbohydrate (e.g. sucrose) per hectare per hour. (1 hectare=104 m2). The net reaction
for sucrose formation in photosynthesis is:
12CO2 ( g ) + 11H 2 O( ) ⎯light
⎯→ C12 H 22 O11 ( s) + 12O2 ( g )
(a) Use standard enthalpies of formation (See Appendix 2 )to calculate ∆H0 for the
production of one mole of sucrose at 298K by the reaction above.
Solution:
∆H = ∆H 0f C12 H 22 O11 s − 11∆H 0f H 2 O
c b gh − 12∆H cCO b ggh
= b1molegb −222170
. kJ / moleg − b11molesgb−28584
. kJ / moleg − b12molesgb−39351
. kJ / moleg
c
b gh
0
f
2
= −222170
. kJ + 3144.24 kJ + 4722.12 kJ = 5644.66kJ
(b) Calculate the rate at which energy is stored in carbohyrates (e.g. sucrose) per hectare
as a result of photosynthesis. Note: 1 Watt=1 Joule/s
20kg of sucrose is 20000g/342g/mole=58.48moles. So the total amount of energy stored
is (58.48moles)(5644.66kJ/mole)=330,100kJ per hectare per hour=91.69kJ per hectare
per second=91.69kWatts/hectare.
(c) Bright sunlight corresponds to radiation flux at the surface of the earth of about
1kWatt/m2. What percentage of this energy can be stored in the form of
carbohydrates as a result of photosynthesis?
4 2
2
Energy storage per square meter is 91.69kWatts/10 m =9.169Watts/m .
Percentage energy stored =9.169/1000=0.009…less than 1%.