Question

Answer the following based on the reaction. At 313 K, the rate constant for this reaction...

Answer the following based on the reaction. At 313 K, the rate constant for this reaction is 1.09×102 /s and at 564 K the rate constant is 6.62×106 /s.

cyclopentane → 1-pentene

1. Determine the activation energy (EA) (in kJ/mol) for this reaction.

2. Determine the pre-exponential factor, A (in /s) for this reaction.

3. Determine the rate constant (in /s) for this reaction at 1218 K.

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Homework Answers

Answer #1

Determine the activation energy (EA) (in kJ/mol) for this reaction.

ln (k2/k1) = (-Ea/R) [(1-T2) - (1/T1)]

ln ( 6.62×106 / 1.09×102) = -Ea / 8.314 [1/564-1/313]

Ea 1.710 e-4 = 11.014

Ea (activation energy) = 6.441 x 104

2. Determine the pre-exponential factor, A (in /s) for this reaction.

k=Ae−Ea/RT

ln k = ln A - Ea/RT

ln 6.62×106 = ln A -   6.441 x 104 / 8.314 x 564

ln A = 15.70 + 13.7363 = 29.4419

A (pre-exponential factor) = 6.08354 x 1012 S-1

3. Determine the rate constant (in /s) for this reaction at 1218 K.

rate constant (in /s) for this reaction at 1218 K

ln (k2/k1) = (-Ea/R) [(1-T2) - (1/T1)]

ln (K2 /6.62×106 ) = -6.441 x 104 / 8.314 [1/1218 - 1/564]

ln K2 = 23.08

K2 = 1.0524812872 x 1010 /S

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