Question

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.

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