Unformatted text preview: HW #3: Problem 1
Wanted: For the gaseous reaction A + B → P, the reactive crosssection obtained from the experimental value of the
preexponential factor is 8.7*1022 m2. The collision crosssections of A and B estimated from the transport
properties are 0.88 and 0.40 nm2, respectively. Calculate the steric factor for the reaction. Solution: A ...... C * The steric factor is equal to the reactive crosssection divided by the total reaction crosssection B = 0.882 = 2 = 0.42 = 2 = ∴ = 0.529
∴ = 0.357 *Circle C represents the total reaction crosssection between molecules A and B
∴ =
2 2 + 2 = 2 = 0.6172 = 6.1710−19 2 8.710−22 2
=
= 6.1710−19 2 ∴ = 0.0014 HW #3: Problem 2
Wanted: Use simple collision theory to calculate the secondorder rate constant for the elementary bimolecular
reaction D2(g) + Br2(g) → 2DBr(g) at 450K. Use 3.930mu for the reduced mass, 200 kJ mol1 for the activation energy,
and 0.075 nm and 0.234 nm for the molecular radii of D2 and Br2, respectively.
Given: = 200 2 = 0.234 2 = 0.075 ( ) = 3.93 = 3.93 1.66110−27 = 6.5310−27 Solution: D
2
....... Br2 2 () + 2 () −9 ( ) = 110 ∴ = 0.309 2 ∴ = 2.9910−19 2 1556.52 8 [2]
= 2 [2 ] 2() ∗ ℎ : = = 2 + 2 = 0.309 = 450 −∗
2 8 1.380610−23 2
450
2 6.5310−27 −53.45 6.0221023 −200
1
1.380610−23 6.0221023
450 ∴ = 1.70910−15 3 HW #3: Problem 3
Wanted: For the reaction A + B ⇌ (AB)‡ → C
a) Calculate the number of translational, vibrational, and rotational degrees of freedom for A, B, and (AB)‡.
b) Use the DOF from part a) to derive equations for qt, qr, qv, and qe.
Solution
a) * The table below is derived from the discussion in chapter 3, where the number of transitional DoF is a 3 for
molecules (single and multiple), and rotational and vibrational only apply to multi molecule complexes, where N is the
number of atoms in the molecule
Molecule Trans.
Atoms
Linear
Non linear Rotation Vibration 3
3
3 0
2
3 0
3N5
3N6 * By applying this table to our reaction, we get the following results
Molecule
A
B
AB Trans. Rotation Vibration 3
3
3 0
0
3 0
0
3(2)5 = 1 * AB is assumed to be a linear molecule b) *Equations for qt, qr, qv, and qe can be taken from Table 3.1
‡ = 3
2 2
= ℎ3 () = 8
ℎ2 = 1 = ℎ 1 − − Translational *Note, the volume compenet is ignored in the following form
2 ℎ3 = 2 ℎ3 3
2 3
2 2 ℎ3 ∴ = 3
2 = − ℎ 1−
1∗1 1 − − Electronic
− = σ −
σ −
σ 1 ∴ = 3
2 ℎ3 2 3
2 Rotational * Values for A and B are 1
8 8 ℎ2 ∴ = =
ℎ2 1∗1 Vibrational * Values for A and B are 1
1 ℎ −
HW #3: Problem 3.7
Wanted: Given: 1 −
∗ ′ ≥ + (1 − cos())
Solution: , = 2
2 ′ − ∗ = න , 2 ∗ , , ℎ 1 − = න 2
2 ′ − ∗
2 = 2
න
′ 1− ∗
2
− 2 ∗ ℎ ℎ ℎ . ℎ ′ ℎ , = 0 = 1 − 2
2 2
∴ ′(2) = 1− ≥ ∗ + ′ (1 − cos( )) ∴ = 2
2 ′ 1− ′(2)
1 ∗
= + ′ 1 − 1
2 ∗ ∗ 0 ′ ′
2 = ′ න 0 − ′ 1− 2 ∗
3
2
− 2 = ′ ∗ ′(2) ′(4)
− 2 1
2 = 2
2 ′ 1− 1− ∗ ∗ 2
4 ′
− 2 2 4 0 ∗
2 1− 1 − ∗
4 1− 2
2 2 −1 HW #3: Problem 3.7 Continuted ∴ = ′ 1− 2 2 ∗
2 1− ∗ 2 2 =
′ − 2 ∗
1− 2 1− 1
2 ∗ 2 ∴ = 2 1−
′
2 ∗ , ℎ , ℎ 3.4
∴ = න 1
wℎ = 2 ∴ =
2
2 2 ∗ =න
1
−
2 ′ 2 =
2 ′ 2 2 1
2 2
2 2 1 =
2 ′ 3
2 1
wℎ = 2 2
1
∗ 1
2 3
2 න 1 −
1
2 න − ∗ 2 3
2 ∗ − 2
ℎ ℎ − 2 − − − ∗ ℎ − ∗ ℎ ℎ ∗ ℎ 2 2
∴ = ′ 1
2 −∗ න − ∗ −∗ 2 −( − ∗ ) ∗ ℎ ℎ 3 ℎ ℎ . − ∗ = ℎ ℎ . ∗ ∞, ℎℎ ℎ ℎ 2 ∴ =
′ = − ∗ 3 1
2 2 − ∗ න − ∗
= 2 2 2
∴ =
′ 1
2 ∴ =
− ∗ ∞ න − ∗ 2 0 0 2
4 ∗ 2 ∴ = ′ 1
2 2 − 2 − ∗ −( − ∗ ) = − ∗ ∞
∞ 0
∗ − ∞ න 2 ∗ wℎ = 8 ∗ − 2
∴ = ′ HW #3: Problem 3.10
Wanted: Given: − ∗
2 = = 2 (ln ) 8 = Solution:
a) ∗ ℎ . ∴ = 2 − ∗
2 8
2
= 2 ∗ − = 2 ℎ 1
ln 2 = ln + ln 2 ℎ , + −∗ − ∗ ln( ) log() log() 1
log =
, ℎ ℎ
= 1
− ∗ − ∗ . ℎ 1 = −1 ℎ 2
∴ = 2 1
− ∗
ln + ln + 2 = 2 1
∗ 1
+
2 2 = 2 + 2 ∗
2 2 % = ( ) − () − ℎ
= ( ) b) ∴ = + ∗
2 ∗
∗
2 + − % =
= 2 + ∗
2 Reaction
#1
#2
#3 Table 2.2
Ea/R (K)
Range (K)
6250
3002,250
14520
2983,000
500
200300 ℎ ∗ ∗ = −
1 % = 2 Mean T (K)
1400
1649
250 Reaction
#1
#2
#3 % Error
11.2%
5.7%
25.0% 2 HW #3: Problem 3.13
Wanted: *Crosssectional area of molecule pairs Solution:
− ∗
− ∗
2 a) = = 2 = = ∴ 2 = ∴ = 2
∴ = 8 = 298 2
19(
)
2 =
=
= 1.81 2 + 2
+ 19(
) = 1.81
= 3.010−27 1000 6.0221023 ...... ...... ...... H2 = F2
2.710−19 2
= 2.9310−10 1.810−19 2
= 2.3910−10 8 = 5.3210−10 = 1.66210−15 2 3 6.0221023 1000 3 8 = 2 + = 5.3210−10 1.380610−23 2
298
3
2 −15
=
1.66210 3.010−27 ∴ = 1.001012 ∗ ℎ ℎ ℎ ℎ ℎ . ℎ ℎ ℎ ℎ ℎ , ℎℎ HW #3: Problem 3.13 Continued = −∗ ∴ = 6.971010 b) −6600
1
1.380610−23 6.0221023 298 = 1.001012 = 1.001012 0.0697 ∗ ‡
− ℎ 2 = ‡ = ∴ = ‡ ℎ 3
2 = 6.0711039 8 ℎ2 82 ∗1
ℎ2 2 3
2 1 4.15610−30 2
2 = 1 = 1−
1
1− − − ℎ ℎ2 = ∗1 1− 1− 1− 1− − 1− − = ℎ 3
2 ℎ3
3
2 2 3
2 6.62610−34 ∗ 3 2 1.380610−23
298 3
2 3 = 4.2510−34 3 3
2 6.0221023
3
= 2.5610−7 7.4310−47 2 2
= 32.3
0.4610−47 2 1
∗ where ℎ 2 ℎ ℎ2 ℎ
− ∗ where ℎ 3 2 ℎ − = 3
2 6.62610−34 ∗ = 2 = 21
1000 6.0221023 2
19 = − ∗ ℎ ′ 3
2 1 −∗ 2 ℎ3 2 8 =
ℎ2 ‡ 2 ℎ3
22 ℎ3 () ‡ = = 3
2 2 = ℎ3
‡ = ∗ ℎ 3 1
6.62610−34 ∗ 13.191013 1.380610−23 298 1
6.62610−34 ∗ 12.021013 − 1.380610−23 298 1− 1
6.62610−34 ∗ 1.191013 − 1.380610−23 298 2 = 0.999
0.999 0.853 2 = 1.374 HW #3: Problem 3.13 Continued = −
σ −
σ 2 2 −
σ = 2 = 4
=1
1∗4 ∗ where ℎ ℎ ℎ ∴ = −∗ 3
2.5610−7
ℎ 32.3 1.374 1 1
3
1000 = 6.2091012 1.13610−5
0.0697
= 4.91109
3 e
∴ = 4.91109 HW #3: Problem 3.14
Wanted: Solution: ∗ 3, ℎ = 3
2 ℎ3
2 ∝ 3
2 1 = 3
2 ℎ = = ℎ ℎ ℎ ℎ 1 − − 1 − − 1 − − ℎ = −
σ − σ −
σ ∝1 8 ℎ2 8 8 ℎ2 ℎ2 ∗ 2 1 ∝ 3−5 (3−6) ∝ 3−5 (3−6) 3−5 (3−6) ∗ ℎ = ‡ 2 3 ∗ (−) ∝ 2 σ
σ σ HW #3: Problem 3.15
1 ∴ ∝ a) ⬚ 3
2 3−5 3−6 1 ∗ ℎ ℎ ℎ ⬚ + 2 ՞ ՜ + , ℎ ∝ 1 2 3 2
⬚ 3(3)−5 1 3(2)−5 2 = 3
3 3 ∴ = 2 3 e ∝ 2 2 ⬚ b) + ՞ ՜ + , ℎ − 2
2
3 ∝ 3 2 c) 2 (−) 2 1 ⬚ 3(4)−6 3(2)−5 1 3(2)−5 2 = 6
= 2
4 ∴ e ∝ 2 ⬚ 3 + 3 ՞ ՜ 2 6 , ℎ − 3 ∝ 3 2 3(8)−6 2 (−) 1
3 3 2 (−) 2 (−) 3(4)−6 1 3(4)−6 2 ∴ = 18 15 e ∝ 3 ...
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