b) Now calculate the pre-exponential factor for the collision rate constant unde
ID: 873174 • Letter: B
Question
b) Now calculate the pre-exponential factor for the collision rate constant under these same conditions.
c) The experimental pre-exponential factor for this reaction is 2.06x10-21cm3s-1molecule-1 and the exponential rate constand is 5.80x10-47cm3s-1molecule-1. Determine the Arrhenius activiation energy for this reaction in kJ/mol.
d) Now use the value of Ea obtained in part c) to obtain a value of E*, the collision theory activation energy in kJ/mol.
e) Use this value of E* and the collision theory pre-exponential factor you calculated in part b) to obtain a value of kcol in units of cm3s-1molecule-1. Compare this rate constant to experiment and obtain the ration kcol/kexp.
Explanation / Answer
(a) Collision frequency ZAB = nAnB (dav)2 [(MA + MB) 8Pi RT/MAMB]1/2
A = H2 B = C2H4
(dav)2 = (dA +dB)/2 = (2.7 x 10-8 + 4.3x 10-8 )/2 cm = 3.5 x 10-8 cm
MA = 2 MB = 2x12 + 4x1= 28 (Molecular masses)
Pi = 3.14
R = 8.314 x 107 JK-1mol-1
T = 250C = 298K
nA = number of molecules of A (H2)
nB = number of molecules of B(C2H4)
Number of molecules of A (H2) present in 22400 cm3 at 273 K and 1bar = 6.023 x1023 (Avogadro number)
Number of molecules of A (H2) present in 1 cm3 at 273 K and 1bar = 6.023 x1023/22400 = 2.6 x 1019
Number of molecules of A (H2) present in 1 cm3 at 2298 K and 1bar approximately equal to 1019
ZAB = (1019)2 (3.5 x 10-8)2[(2 + 28) 8 x 3.14 x 8.314 x 107x 298 /2x28]1/2
(1038) x 1.22 x 10-15 x 1.8 x 106 = 4 x 10 29 collisions s-1 cm-1
(b) Pre exponential factor A = 4x1029 x p(dav)2[T(MA+MB)/MAMB]1/2 (the steric factor p is taken as 1)
=4x1029 x 1x (3.5 x 10-8)2[298(2+28)/2x28]1/2 = 6 x 1015dm3 mol-1s-1
(c) k = A e-Ea/RT
e-Ea/RT = k/A
-Ea/RT = ln k/A
Ea = - RT ln k/A
=- 8.314 x 298 ln 5.80 x10-47/2.06x10-21
= 145777 Jmol-1
= 145.777 kJmol-1
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