MSE 618 HW3.pdf - HW#3 Problem 1 Wanted For the gaseous...

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Unformatted text preview: HW #3: Problem 1 Wanted: For the gaseous reaction A + B → P, the reactive cross-section obtained from the experimental value of the pre-exponential factor is 8.7*10-22 m2. The collision cross-sections 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 cross-section B = 0.882 = 2 = 0.42 = 2 = ∴ = 0.529 ∴ = 0.357 *Circle C represents the total reaction cross-section 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 second-order rate constant for the elementary bimolecular reaction D2(g) + Br2(g) → 2DBr(g) at 450K. Use 3.930mu for the reduced mass, 200 kJ mol-1 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 3N-5 3N-6 * 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 300-2,250 14520 298-3,000 500 200-300 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: *Cross-sectional 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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