Question

LUI
4.23
The rate constant for the decomposition of hydrocarbons is 2.418 x 10 %
1 at 546 K. If the energy of activation is 179.9 kJ/mol, what will be the
value of pre-exponential factor.
Consider a certain
ith k= 2.0 x 10-2g-1. Calculate
4.24​

Answer 1

Answer:

By using arrhenius equation,

⟹logk=

2.303 RT

−E

a

+logA

We get

⟹logA=log(2.418×10

−5

)+

2.303×8.314×546

179900

⟹(log2.418)−5+15.7082)

⟹0.3834−5+17.209=12.6

∴logA=(12.6)=3.9×10

12