CLEAR EXPLANATION PLEASE! b) Now calculate the pre-exponential factor for the co
ID: 873416 • Letter: C
Question
CLEAR EXPLANATION PLEASE!
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) We will use the relation for collison frequency Zab,
Zab = Na.Nb.(ra + rb)^2 . sq.rt.(8pi(kB)T/?ab)
with
Reduced mass = ? AB = m A . m B / m A +m B
= 2.0158 x 28.05/2.0158+28.05
= 1.9
Substituting the values we get,
Zab = 1 x 1 (2.7 + 4.3)^2 (sq.rt. (8 x 3.14 x 1.38 x 10^-23 x 298 / 1.9)
= 1.14 x 10^-8
b) Pre-exponential factor for the collison rate found in a) will be = 1.14 x 10^-8
c) We will use the equation for activation energy Ea,
ln k = ln A - Ea/RT
Substuting the given A and k values at 25 C we get,
ln (5.8 x 10^-47) = ln (2.06 x 10^-21) - Ea / 0.0831 x 298
solving for Ea = 1458 kJ/mol
d) the collison theory activation energy for this reaction is thus, 1458 kJ/mol
e) ln k = ln(1.14 x 10^-8) - 1458/0.0831 x 298)
solving for kcol = 3.04 x 10^-34
kcol/kexp = 3.04 x 10^-34/1.14 x 10^-8
= 2.7 x 10^-26